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2025-03-21T14:48:30.274548
| 2020-04-10T18:28:13 |
357088
|
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|
Stack Exchange
|
Reference request: Differential graded structures in mixed characteristic
I am looking for references/papers on differential graded structures and their applications in mixed characteristic. The following I have discuss differential graded algebras in the general, not in the mixed characteristic case:
https://stacks.math.columbia.edu/download/dga.pdf
https://arxiv.org/abs/1307.0369
https://www.maths.ed.ac.uk/~mbooth/hodgeproject.pdf
https://link.springer.com/chapter/10.1007%2F978-3-658-25338-7_3
In particular, if there are any interesting applications, papers, examples of dg algebras in mixed characteristic, then I would gladly like to know! There doesn’t seem to be an in depth book (from my search so far) about this.
|
2025-03-21T14:48:30.274614
| 2020-04-10T18:52:37 |
357089
|
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"url": "https://mathoverflow.net/questions/357089"
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|
Stack Exchange
|
Comparison of concentrations of different $L^p$-norms of (sub) Gaussian distributions
It's well-known that the Euclidean $2$-norm of subgaussian random vectors concentrates in high dimensions, e.g. when $X \sim \mathcal{N}(0,I_n),$ (or in general $X$ is subgaussian with independent co-ordinates), $||X||_2$ concentrates near $\sqrt{n}.$ (See Theorem 3.1.1. of this book).
My question is about possible results/papers on concentration on $||X||_p, p \ne 2.$ when $X$ is subgaussian or satisfy some kind of tail decay property like $P[||X||_2 >t]\le e^{-\beta(t)}, \beta(t) > 0,$ for example. More particularly, can we say that the larger the $p,$ the stronger/tighter is the concentration of $||X||_p$ ariund $\mathbb{E}||X||_p?$ So to be precise: can we say $P[| ||X||_p - \mathbb{E}||X||_p | > t] \le P[| ||X||_q - \mathbb{E}||X||_q |> t]$ when $p \ge q$? You can assume first that $X \sim \mathcal{N}(0,I_n),$ or at least $X$ has independent co-ordinates which are sub Gaussian.
The reason is I ask this question is motivated by my study of this paper on machine learning, where they show that for high dimensions, the $L^p$ distance has more discriminative power between nearest and farthest points from the origin, if $p$ is less, and this discriminative power decreases when $p$ increases. See Lemma 1, Corollary 2. They in fact advocate the use of $L^p$ pseudo-metrics for $0<p<1$, for in this case, the discriminative power is high. This led me to guess that for Gaussian or subgaussian random vectors, the concentration of $p$-norm is more tight if $p$ is large, and vice versa. Thanks in advance!
|
2025-03-21T14:48:30.274764
| 2020-04-10T19:41:54 |
357092
|
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"Anton Petrunin",
"Ben McKay",
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"Gael Meigniez",
"Ryan Budney",
"Tobias Fritz",
"Willie Wong",
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|
Stack Exchange
|
Name for a standard trick to construct a diffeomorphism
The following construction is standard, and it deserves a name.
Suppose we need to construct a diffeomorphism from a manifold $M$ to itself with some additional properties.
Observe that the flow $\phi^t$ for any reasonable vector field $v(t)$ on $M$ defines a diffeomorphism for any $t$.
So it remains to find a vector field $v(t)$ such that the flow $\phi^1$ satisfies your properties.
Do you know a name for it?
If not, what would be a good name?
Physicists call this type of thing "infinitesimal generator".
Quick question: is the vector field $v$ intended to be really time dependent (written as $v(t)$ in the question)?
@WillieWong, usually not, but sometimes it is useful.
I only ask because after I upvoted @TobiasFritz's comment, I begin to doubt whether the usual use of the "infinitesimal generator" allows for $v$ to be time dependent.
You see people give these flows various names. I use this construction in the time-dependent context. My vector fields are usually defined by an isotopy of a submanifold, you then extend the velocity vector field of the submanifold to the ambient manifold. So I just call it isotopy extension. But I see other people call them "point pushing" or "circle pushing" maps depending on what the submanifold might be. Your context is a little more general than my own, though.
Do you want a name for the construction or for the vector field?
@DeaneYang and or or.
I would say "Moser trick", since it reminds me of the trick explained below for volume forms. There is still not too much named after Moser, so I don't think it would confuse anyone. "Euler trick" or "Arnold trick" or "Newton trick" might confuse people.
This is often called "path method". In particular one has classically the Moser path method.
Let us prove for example (after Moser) that given a closed oriented $n$-manifold $M$ and two volume forms $\omega_0$, $\omega_1$ on $M$ with the same integral, there exists a diffeomorphism $f$ of $M$ such that $$f^*(\omega_1)=\omega_0$$ To this end, the path method considers $$\omega_t=(1-t)\omega_0+t\omega_1$$ and looks for a time-dependant vector field $X_t$ whose flow
$\phi_t$ satisfies the condition $$\phi_t^*(\omega_t)=\omega_0$$ (Then, for $t=1$, $f=\phi^1$ will work)
The condition holds trivially for $t=0$. Deriving this condition with respect to $t$, one finds $$\phi_t^*(L_{X_t}\omega_t+\omega_1-\omega_0)=0$$ where $L$ is the Lie derivative. This amounts to $$L_{X_t}\omega_t=\omega_0-\omega_1$$ i.e. by Cartan's formula
$$d\iota_{X_t}\omega_t=\omega_0-\omega_1$$
Since $\omega_0$ and $\omega_1$ have the same integral, they are cohomologous: one has a $(n-1)$-form $\alpha$ on $M$ such that $$\omega_0-\omega_1=d\alpha$$ Hence it is enough that
$$\iota_{X_t}\omega_t=\alpha$$
The end of the argument is purely (multi)linear algebra: this last equation admits for every time $t$ and at every point $x$ a unique solution $X_t(x)$ since $\omega_t(x)$ is a nonzero $n$-form on $T_xM$.
An analogous method applies to symplectic forms, nonsingular closed $1$-forms and contact forms.
The path method also proves the $MJ^2$ Lemma (Lemma 3.2 in F. Laudenbach, A proof of Reidemeister-Singer's theorem by Cerf's methods. Ann. Fac. Sci. Toulouse Math. (6) 23 (2014), 1, 197–221, Arxiv 1202.1130); and in particular the Morse Lemma.
What is the $MJ^2$ Lemma?
Thank you, but maybe "Moser trick" is better than "path method" (?)
As you like it, but to me, this deserves the name of "method", being more interesting, more general, and involving more the nature of things, than a mere "trick".
why "path" --- say "flow method" would be better.
This is a hugely studied field in the area of medical image processing. Goes by name of (large) diffeomorphic registration or diffeomorphic mapping
https://en.m.wikipedia.org/wiki/Large_deformation_diffeomorphic_metric_mapping
|
2025-03-21T14:48:30.275086
| 2020-04-10T19:57:09 |
357093
|
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|
Stack Exchange
|
Space of representations of surface group into Lie groups
In the context of Goldman's paper The symplectic nature of fundamental groups of surfaces:
Consider a closed oriented surface $S$ with fundamental group $\pi$, and let $G$ be a connected Lie group. The space $\operatorname{Hom}(\pi,G)$ consisting of representations $\pi\to G$ is a real analytic variety a real analytic space (possibly singular). There is a canonical $G$-action on $\operatorname{Hom}(\pi,G)$ obtained by composing representations with inner automorphisms of $G$. When is the representation space $\operatorname{Hom}(\pi,G)/G$ a smooth manifold? I think that if the representation is irreducible, the space is smooth manifold nearby.
Did you ask this previously? It reminds me of some similar question.
I would expect it's only a manifold when $G$ is trivial. But asking for a manifold structure is fairly peculiar on its own. Don't you want it to be related to the real analytic structure on $Hom(\pi, G)$?
There are examples where Hom$(\pi,G)/G$ happens to be a topological manifold, somewhat by accident. For instance, if $S$ is the torus, then $\pi = \mathbb{Z}\times \mathbb{Z}$ and Hom$(\mathbb{Z}\times \mathbb{Z},SU(2))/SU(2)$ is homeomorphic to the 2-sphere $S^2$. This is discussed in various places in the literature, the oldest place I know of being Morgan, John W.; Shalen, Peter B. Valuations, trees, and degenerations of hyperbolic structures. I. Ann. of Math. (2) 120 (1984), no. 3, 401-476.
Oddly, if you replace $SU(2)$ by $PSU(2)$ in the above example, the representation space Hom$(\mathbb{Z}^2, PSU(2))/PSU(2)$ is still homeomorphic to $S^2$, giving another example.
I think the local condition to be a smooth manifold is that the representation is irreducible (i.e. not contained in any proper parabolic) and the centralizer of the image of pi is the center of G (second condition redundant for G=GL_n but not in general).
The question of the structure of $Hom(\pi, G)$ and $Hom(\pi, G)/G$ is very complicated for a general connected Lie group $G$, though less so for a surface group than for a general finitely generated group. To see just how wild things can get, you should look at this paper of Kapovich and Millson http://www.math.umd.edu/~millson/papers/char19.pdf.
For a surface $\Sigma$, if $G$ is Abelian, then
$Hom(\pi, G)\simeq Hom(\pi, G)/G\simeq H^{1}(\Sigma, G)\simeq G^{2\sigma}$
where $\sigma$ is the genus of $\Sigma.$ In particular, in this case, it is naturally a smooth, even real analytic, manifold in a canonical way.
Meanwhile, as mentioned in the above comments, there are situations when $Hom(\pi, G)$ and/or $Hom(\pi, G)/G$ is homeomorphic to a manifold, although this is something of an accident.
Probably the cleanest statement available that lives in some level of generality, is that if $G$ is a connected reductive complex affine algebraic group (like $GL_{n}(\mathbb{C})$), then
$Hom(\pi, G)$ is a complex affine variety with at most quadratic singularities, this is due to Goldman, Millson and Simpson (https://projecteuclid.org/euclid.bams/1183554530), and remarkably this result remains true as long as $\pi$ is the fundamental group of a compact Kahler manifold. But, even in this case, the quotient $Hom(\pi, G)/G$ is a total mess, in particular it is Hausdorff only when $G$ is Abelian.
This can be remedied (to some extent) in this case by studying the GIT quotient $Hom(\pi, G)//G,$ but since you ask about real Lie groups in general, GIT theory again is a bit of a mess since $\mathbb{R}$ is not algebraically closed.
As mentioned in the comments, if you take $G$ a connected semi-simple real Lie group, then the subset $Hom^{\star}(\pi, G)$ consisting of irreducible representations whose centralizer is equal to the center of $G,$ then this is a smooth (even real analytic) manifold upon which $G$ acts with constant stabilizer equal to the center, and properly, and therefore $Hom^{\star}(\pi, G)/G$ is a (real analytic) manifold.
Just as a final remark, there are many very good reasons why one might consider the space of conjugacy classes of all representations $\pi\rightarrow G,$ but as explained above, this is basically hopeless in any reasonable category of classical spaces such as: smooth manifolds, algebraic varieties, real analytic spaces...
One remedy to this hopelessness is to pass to a suitable category of stacks, but of course it heavily depends on what you want to do.
I'll expand a bit on @WillSawin's comment. Sikora (Corollary 50 of the article Character Varieties in Trans. AMS, 2012) proved that if $G$ is a complex linearly reductive algebraic group and $\pi$ is the fundamental group of a closed surface of genus greater than 1, then irreducible representations $\pi\rightarrow G$ whose centralizer is the center of $G$ do indeed lie in the smooth locus of the GIT quotient Hom$(\pi, G)//G$. (As Will said, irreducible here means the image is not contained in a proper parabolic subgroup.) I'm not sure if this always describes the entire smooth locus. (It would be nice to know!)
I don't know that there is any good understanding of these kinds of things for non-reductive Lie groups.
It seems natural to look at, even with $G$ non-reductive.
For instance, let $G=G_m$ be a $2m+1$-dimensional Heisenberg group: I write $G=V\times K$, where $(V,\phi)$ is a $2m$-dimensional symplectic space over $K$ (not of characteristic $2$) and law $(X,T)\cdot (Y,U)=(X+Y,T+U+\phi(X,Y))$. In particular (with suitable sign conventions), the group commutator $[(X,T),(Y,U)]$ is $(0,2\phi(X,Y))$.
Consider an $2n$-tuple in $G^n$, written as $((x,t),(y,u))$ with $(x,t),(y,u)\in G^n$ (so $x,y\in V^n$ and $t,u\in K^n$), $x=(x_i)_{1\le i\le n}$, $y=(y_i)_{1\le i\le n}$, with $x_i,y_i\in V$, and $T=(t_i)_{1\le i\le n}$, $U=(u_i)_{1\le i\le n}$, with $t_i,u_i\in K$.
The condition $\prod_{i=1}^n[(x_i,t_i),(y_i,u_i)]=e$ can be written as $\sum_i\phi(x_i,y_i)=0$. That is, $\Phi(x,y)=0$, where $\Phi(x,y)$ is defined as $\sum_i\phi(x_i,y_i)$, which makes $(V^n,\Phi)$ a $2mn$-dimensional symplectic space. Note that if we ignore the $t_i,u_i$, we obtain the hypersurface $$H_{mn}=\{(x,y)\in V^{2n}:\Phi(x,y)=0\}$$ in the $4mn$-dimensional space $V^{2n}$, which is non-singular outside zero (and the $G$-action by conjugation is trivial).
So if we consider the space $\mathrm{Hom}(\Gamma_n,G_m)$ (where $\Gamma_n$ is the genus $n\ge 1$ surface group), we have its description as $H_{mn}\times K^{2n}$, and it's non-singular outside $\{0\}\times K^{2n}$.
The conjugation action (after renormalization by $2$ and factoring the action through $V$) consists in letting $V$ act on this product as
$$z\cdot (x,y,t,u)=(x,y,t+\psi(z,x),u+\psi(z,y)).$$
Here $\psi(z,x)$, for $x\in V^n$ and $z\in V$, is defined as $(\phi(z,x_i))_{1\le i\le n}$.
Since for $(x,y)\neq (0,0)$, the linear map $z\mapsto\psi(z,x),u+\psi(z,y)$ is injective, it's not hard to prove that the quotient remains smooth outside $\bar{p}^{-1}(\{(0,0)\})$, where $p$ is the projection $\mathrm{Hom}(\Gamma_n,G_m)\to H_{mn}$ and factors through a projection $\bar{p}:\mathrm{Hom}(\Gamma_n,G_m)/G_m\to H_{mn}$.
Note that fibers of $p$ are "translates" of $K^{2n}$; each fiber of $\bar{p}$ naturally carries a structure of affine space of dimension $2n-1$, except the fiber of $(0,0)$ which is still naturally identified to $K^{2n}$.
On generalizations:
Let $G$ be an arbitrary $s$-step nilpotent simply connected Lie group (that is, $G^{s+1}=\{1\}$ where $(G^i)_{i\ge 1}$ is the lower central series), and $\Gamma$ is an arbitrary finitely generated group. Let $\Gamma(s)$ be the real Malcev completion of the nilpotent quotient $\Gamma/\Gamma^{s+1}$ (this is a a simply connected $s$-step nilpotent Lie group in which $\Gamma/\Gamma^{s+1}$ modulo its finite torsion subgroup, sits as a lattice. Then we can identify $\mathrm{Hom}(\Gamma,G)$ to $\mathrm{Hom}_{\mathbf{TopGrp}}(\Gamma(s),G)$, and this identification commutes with the $G$-action. Furthermore, letting $\Gamma[s]$ be the Lie algebra of $\Gamma(s)$ and $\mathfrak{g}$ that of $G$, this can be identified to $\mathrm{Hom}_{\mathbf{R}\text{-}\mathbf{LieAlg}}(\Gamma[s],\mathfrak{g})$.
Hence all the study with nilpotent target reduces to that of spaces of homomorphisms between Lie algebras.
Thanks for this answer: I've thought more than once that's there's probably a lot of rich geometry for non-reductive groups, but it's largely unexplored territory.
|
2025-03-21T14:48:30.275726
| 2020-04-10T20:44:09 |
357097
|
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|
Stack Exchange
|
Conformal isomorphism uniquely determined by boundary identification?
Let $\Gamma$ be a smooth Jordan arc, and let $\Phi \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \Gamma$ be a conformal isomorphism that fixes the point at $\infty$.
Then $\Phi$ extends continuously onto $\partial \mathbb D$ because $\Gamma$ is smooth (Caratheodory), and for every $z \in \Gamma$ except the two endpoints, $\Phi^{-1}(z)$ consists of two points on $S^1$.
Now if I have another smooth Jordan arc $\Gamma_0$, and a conformal isomorphism $\Phi_0 \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \Gamma_0$ that fixes $\infty$ such that $\Phi_0(\Phi^{-1}(z))$ is a singleton for every $z \in \Gamma$, can I necessarily conclude that $\Gamma$ and $\Gamma_0$ are the same curve (modulo affine transformation)?
For smooth curves, the answer is yes. Proof: $\Phi_0\circ\Phi_1^{-1}$ is a conformal map
of $C\backslash\Gamma_0$ onto $C\backslash\Gamma$, and your condition implies that
this conformal map extends continuously to the boundary. So we have a continuous
map (in fact a homeomorphism) of the Riemann sphere, which is conformal in the complement of a smooth
curve. By a theorem of Painleve, such a map is conformal on the whole sphere,
so it is linear-fractional, and since it fixes $\infty$ it is affine.
Painleve's theorem holds for any rectifiable curve, and the proof is elementary, and uses nothing but Cauchy integral formula.
Rectifiability condition is too strong, and there was much research in the end of 20th century with the aim of understanding to which closed sets Painleve's theorem
really extends.
|
2025-03-21T14:48:30.275854
| 2020-04-10T20:46:48 |
357098
|
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|
Stack Exchange
|
Upper bound for an exponential sum involving characters of a finite field
Let $q = p^n $ be a prime power, $\alpha\in\mathbb{F}_{q} $
a primitive element of the finite field $\mathbb{F}_q$ and denote by $\chi $ a non-trivial additive character of $\mathbb{F}_{q} $. Set
$\omega = \exp\lbrace { i\frac{2\pi} {q - 1} } \rbrace $,
I am looking for an upperbound on the following sum
\begin{equation}
\left\vert \sum _{k=0} ^{q - 2} \omega ^{k^2} \chi (\alpha ^k ) \right\vert.
\end{equation}
If we denote by $\psi _c $ the multiplicative character of
$\mathbb{F}^* _{q} = \mathbb{F}_{q}\setminus\lbrace 0\rbrace =\lbrace \alpha ^0 ,\cdots,\alpha^{q-2}\rbrace$ corresponding to $c\in\mathbb{F}_q ^* $
then the sum can be written as
\begin{equation}
\left\vert \sum _{c\in\mathbb{F}_q^*} \psi _c (c) \chi (c) \right\vert.
\end{equation}
The second sum looks much like a Gaussian sum over finite fields, however in this one the multiplicative character changes as well.
ps: The multiplicative character is given by
$\psi _{\alpha ^l} (\alpha ^k ) = \omega ^{lk} = \exp\lbrace i\frac{2\pi}{q-1} lk \rbrace $ .
The additive character corresponding to an element $a\in\mathbb{F}_{p^n} $ is given by
$\chi _a (b) = \exp\lbrace i\frac{2\pi}{p} tr(ab) \rbrace $ for all $b\in\mathbb{F}_{p^n} $,
where the trace $tr : \mathbb{F}_{p^n} \rightarrow \mathbb{F}_p $ is defined by
\begin{equation}
tr(a) = a+a^p + \cdots + a^{p^{n-1}}.
\end{equation}
I am going to assume that by an additive character you mean
an irreducible representation $\chi_\alpha : \mathbb{F}^n_q \longrightarrow \mathbb{C}$, i.e. a group homomorphism from the additive group $(\mathbb{F}^n_q ,+)$ to the multiplicative group $(\mathbb{C},*)$
which we can prove must all take the form
\begin{equation}\chi_\alpha : \beta \mapsto \exp\left( {\frac{2\pi i \left\langle \alpha ,\beta \right\rangle }{p }} \right)\end{equation} where $ \left\langle \alpha ,\beta \right\rangle = \sum_i \alpha_i \beta_i $, see chapter 4 of Tao for a proof of some of these statements and see ch.2 of Serre or ch.2 of Fulton & Harris for a general (non-abelian) overview of the representation theory perspective on characters. The point is the following
If we let \begin{equation} f(x) = \begin{cases} q \psi_x(x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \\ \end{cases} \end{equation} then the sum you are considering is equal to the Fourier transform of $f$ i.e. \begin{equation} \hat{f}(\alpha) = \frac{1}{q} \sum_{c \in \mathbb{F} _q } f(c) \chi_\alpha(c) = \sum_{c \in \mathbb{F} _q^* } \psi_c (c) \chi_\alpha(c) \end{equation} see definition 4.6 in Tao.
We apply the Hausdorff-Young inequality theorem 4.8 in Tao to get that
\begin{equation}
\left(\sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{p'} \right)^{\frac{1}{p'}} \leq \left(\sum_{\alpha \in \mathbb{F} _q } |f(\alpha)|^p\right)^{\frac{1}{p}} = q\left( \sum_{c \in \mathbb{F} _q^* } |\psi_c (c) |^p\right)^{\frac{1}{p}}
\end{equation}
where the LHS is the $l^{q}$-norm, the RHS is the $l^p$-norm, and $p$ satisfies the following $p^{-1} +q^{-1} = 1 \land 1 \leq p\leq 2$. Plugging in $p = 2$ we get that
\begin{equation}
\sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{2} \leq q\sum_{c \in \mathbb{F} _q^* } |\psi_c (c) |^2
\end{equation}
which is equivalent to saying that
\begin{equation}
\mathbb{Var}[\hat f] = \frac{1}{q}\sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{2} \leq \sum_{c \in \mathbb{F} _q^* } |\psi_c (c) |^2\leq q-1.
\end{equation}
Finally, if you can prove that at least $n$ many $\alpha$ give a value $ | \hat f(a)| \geq \sqrt b$ then you get that
\begin{equation}
nb +\sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right|^2 \leq \sum_{\alpha \in S}\left| \hat f(\alpha)\right|^{2} + \sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right|^2 \leq \sum_{\alpha \in \mathbb{F} _q }\left| \hat f(\alpha)\right|^{2} \leq q(q-1)
\end{equation}
which gives you that the maximum value is at most
\begin{equation}
\sup_{\alpha \in \mathbb{F} _q }\left|\sum_{c \in \mathbb{F} _q^* } \psi_c (c) \chi_\alpha(c) \right| = \sup_{\alpha \in \mathbb{F} _q }\left| \hat f(a)\right| \leq \sqrt{q(q-1)-nb}
\end{equation}
Essentially we reduced the problem of finding an upper bound to that of finding a lower bound.
|
2025-03-21T14:48:30.276084
| 2020-04-10T21:40:08 |
357101
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357101"
}
|
Stack Exchange
|
Minimal subshift with some $x \in X$ such that $x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$?
There exists a minimal subshift $X$ with a point $x \in X$ such that $x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$?
We can produce such a subshift by a standard hierarchical construction.
Let $w_{0,0} = 01$ and $w_{0,1} = 011$.
For each $k \geq 0$, define $w_{k+1,0} = w_{k,0} w_{k,0} w_{k,1}$ and $w_{k+1,1} = w_{k,0} w_{k,1} w_{k,1}$.
Define $X$ by forbidding each word that doesn't occur in any $w_{k, b}$.
Since each $w_{k+1,b}$ contains both $w_{k,0}$ and $w_{k,1}$, it's easy to show that $X$ is minimal.
For $b \in \{0,1\}$, define $x^b \in X$ as the "limit" of $(w_{k,b})_{k \geq 0}$ such that the central $01$ or $011$ is at the origin.
Because of the way we defined $w_{k,0}$ and $w_{k,1}$, the only difference between the words, and hence the limiting configurations, is that single extra $1$.
Concretely, $x^0$ and $x^1$ look like
$$
\cdots 0101011\;0101011\;01011011\;0101011\;010.1011\;01011011\;0101011\;01011011\;01011011 \cdots
$$
and
$$
\cdots 0101011\;0101011\;01011011\;0101011\;010.11011\;01011011\;0101011\;01011011\;01011011 \cdots
$$
(Spaces added for clarity.)
It is well-known that the Chacon substitution $\tau$ defined by $\tau(0) = 0010$, $\tau(1) = 1$ produces a minimal subshift, when you take the legal words to be the words that appear in some $\tau^n(0)$. The two-sided fixed point from $0.0$ is $x.y = {...0010001010010.0010001010010...}$ and the one from $0. 10$ is $x.1y$. So they are both in $X$.
Now apply the additional substitution $\alpha(0) = 01$, $\alpha(1) = 1$ and you still get a minimal subshift $Y$ as image (this is a flow equivalence onto its image). We have $\alpha(x).\alpha(y) \in Y$ and $\alpha(x) . 1 \alpha(y) \in Y$, and since $\alpha(x)$ ends with $1$ this is a pair of the kind you are asking for.
|
2025-03-21T14:48:30.276214
| 2020-04-10T23:23:40 |
357103
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357103"
}
|
Stack Exchange
|
Universal bimodule for homotopy biderivations
Recall the commutative story: for a commutative algebra $A$, its module of differentials $\Omega (A)$ is characterized by the universal property that any derivation $\delta \colon A \to M$ is in a unique way a composition $A \to \Omega(A) \to M$, where $d \colon A \to \Omega (A)$ is the canonical derivation.
(In what follows, I will be careless with signs and shifts.) Now let $A$ be a noncommutative DG-algebra over a field $k$, $M$ a DG-bimodule over $A$. Consider the following version of Hochschild cochain complex:
$$C^n(A,M) = \bigoplus_{i+j=n}\operatorname{Hom}_k^i(A^{\otimes_k j},M).$$
The differential $d_H$ consists of two parts, where for $f \in \operatorname{Hom}_k^i(A^{\otimes_k j},M)$,
$$d_1(f)(a_1\otimes \ldots \otimes a_j) = d_mf(a_1 \otimes \ldots \otimes a_j) + \sum \pm f(d_{A^{\otimes_k j}} (a_1 \otimes \ldots \otimes a_j));$$
$$d_2(f)(a_0 \otimes \ldots \otimes a_n) = a_0f(a_1 \otimes \ldots \otimes a_n)+\sum_i \pm f(a_0 \otimes a_i a_{i+1} \otimes \ldots a_n) +\pm f(a_0 \otimes \ldots \otimes a_{n-1})a_n.$$
One can see that for $f \in Hom_k(A,M)$, the equation $d_H(f)=0$ means that $f$ commutes with differentials and $f(ab) = af(b)+f(a)b$. That is, f is a biderivation.
More general closed elements of this complex can interpreted as "homotopy biderivations". Consider $f \in C(A,M)$, where $f = f_0 + f_1 + f_2+ \ldots + f_n$, with $f_i \in Hom^{-i}(A^{\otimes i},M)$. Then $d_H(f) = 0$ says that: 1) $f_0$ is closed, 2) $f_1$ doesn't respect the differentials, but the commutator is the commutator with $f_0$, 3) $f_1$ is a biderivation up to homotopy given by $f_2$, etc. Let's call such $f$ a homotopy biderivation.
Is there a version of universal DG-bimodule for homotopy biderivations, along the lines of $\Omega(A)$ in the commutative story? I would be grateful for any relevant links.
|
2025-03-21T14:48:30.276345
| 2020-04-11T01:39:51 |
357111
|
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"Dieter Kadelka",
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"url": "https://mathoverflow.net/questions/357111"
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|
Stack Exchange
|
$ \{X_n\mathbb{1}_{X_n\in[-n,n]}\}$ is uniformly integrable
Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space.
Suppose $\{X_n\}$ is a sequence of random variables satisfying :
$$
\sup_{n}{\mathbb{E}(|X_n|)} <\infty
$$
Suppose that
$$
\dfrac{M_j}{2}<\int_{j-1<|X_{n}|\leq j}{|X_{n}(t)|d\mathbb{P}(t)}\leq M_j+\dfrac{1}{j^2} \qquad\forall n\geq 1 \text{ and }1\leq j\leq n^2
$$
with $M_j>0$ such that $\sum_{j=1}^{\infty}{M_j}<\infty$.
Show that:
$$
\{X_n\mathbb{1}_{X_n\in[-n,n]}\}\text{ is uniformly integrable}
$$
Any ideas, please?
This is sequel to https://mathoverflow.net/questions/357070/lemma-of-x-n-is-a-sequence-of-random-variables-satisfying-sup-n-m The introduction of $\frac{1}{j^2}$ is completely superfluous. Is this homework?
Note that for any $K, n\in\mathbb{N}$ with $K<n$ we have that
$$\begin{align*}
\mathbb{E}[|X_n|1_{X_n\in[-n, n]}1_{|X_n|>K}] &= \mathbb{E}[|X_n|1_{n \geq |X_n|>K}]\\
&=\int_{n\geq |X_n|>K} |X_n|d\mathbb{P}\\
&= \sum_{j=K+1}^n \int_{j\geq |X_n|>j-1} |X_n|d\mathbb{P}\\
&\leq \sum_{j=K+1}^n M_j+1/j^2\\
&\leq \left(\sum_{j=K+1}^\infty M_j\right)+ \left(\sum_{j=K+1}^\infty 1/j^2\right)
\end{align*}
$$
Note that the final bound does not depend on $n$, so $\sup_n \mathbb{E}[|X_n|1_{X_n\in[-n, n]}1_{|X_n|>K}]\leq \left(\sum_{j=K+1}^\infty M_j\right)+ \left(\sum_{j=K+1}^\infty 1/j^2\right)\overset{K\to\infty}{\longrightarrow} 0$. Hence, we have uniform integrability.
|
2025-03-21T14:48:30.276457
| 2020-04-11T01:41:07 |
357112
|
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"Andrei Sipoș",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357112"
}
|
Stack Exchange
|
How does one prove that the density of unusual numbers is $\ln 2$?
The Wikipedia page for unusual number states that the density of that set is $\ln 2$, and that this was proven by Schroeppel in 1972. The only source that I found for that is the HAKMEM document, and there is no proof given there, just the statement. Does anyone know a reference for a proof? Thanks.
If $n$ is "unusual" then its large prime factor $p$ is unique.
To count "unusual" $n \leq x$, sum over prime $p \leq x$
the number of "unusual" $n$ that are multiples of $p$.
The count is $p-1$ if $p \leq \sqrt x$, and $\lfloor x/p \rfloor$ if
$\sqrt x < p \leq x$. By the Prime Number Theorem (or even Chebyshev),
$\sum_{p \leq \sqrt x} (p-1) \ll x/\log x$. This leaves essentially
$x \sum_{\sqrt x < p \leq x} 1/p + O(x/\log x)$. Now use
$\sum_{p \leq y} 1/p = \log\log y + o(1)$ for $y=\sqrt x$ and $y=x$ to get $x (\log\log x - \log\log x^{1/2} + o(1)) = x \log 2 + o(x)$, QED.
Thank you for the answer!
@NoamD.Elkies : That should really be an answer, not a comment!
Thanks. If the OP [original proposer] is willing to accept it as an answer
I can post it as such. The question as stated asks for a reference,
not a proof, and I didn't give a reference.
@NoamD.Elkies: $\sum_{p \leq y} 1/p$ equals $\log\log y + M + o(1)$, where $M$ is the Meissel–Mertens constant. Of course the presence of $M$ does not matter for your argument.
Finch gives a reference to Greene & Knuth, Mathematics for the Analysis of Algorithms, 3rd ed., pages 95-98. It appears those pages are freely accessible at https://link.springer.com/content/pdf/bbm%3A978-0-8176-4729-2%2F1.pdf
A reference is Greene and Knuth, Mathematics for the Analysis of Algorithms, 3rd ed., pages 95-98. The section of the book containing those pages may be found at https://link.springer.com/content/pdf/bbm%3A978-0-8176-4729-2%2F1.pdf
|
2025-03-21T14:48:30.276625
| 2020-04-11T02:37:17 |
357113
|
{
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"authors": [
"Max Alekseyev",
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|
Stack Exchange
|
Obtaining generating function for multivariate recurrence with non-constant coefficients
Consider a second order recurrence of the form below for some fixed $n$
\begin{align}
(n+s-1) A_{n}(k,s) &= (n-k+1) A_{n}(k-1,s-1) + (k+s-3) A_{n}(k,s-1) \\
A_n(k,s) &= 0 \tag{when $s < k$ or $k > n$} \\
A_n(1,s) &= \boldsymbol{1}[s = 1]
\end{align}
I would like to obtain a closed form expression for a generating function of the form
$$G_n(x,y) = \sum_{k=1}^n \sum_{s=k}^\infty A_n(k,s) x^k y^s$$
I first observed for non-negative integers $p_x, p_y$ that
\begin{align}
x^{p_x} y^{p_y} G_n(x,y) &= \sum_{k=1}^{n} \sum_{s=k}^\infty A_n(k, s) x^{k+p_x} y^{s+p_y} \\
&= \sum_{k=1+p_x}^{n+p_x} \sum_{s=k+p_y}^\infty A_n(k-p_x, s-p_y) x^k y^s
\end{align}
In the same spirit, I noticed the below result , which holds similarly when considering partial derivatives with respect to $y$.
\begin{align}
x \frac{\partial}{\partial x}\left\lbrace x^{p_x} y^{p_y} G_n(x,y)\right\rbrace &= \sum_{k=1}^{n} \sum_{s=k}^\infty x \frac{\partial}{\partial x}\left\lbrace A_n(k, s) x^{k+p_x} y^{s+p_y} \right\rbrace \\
&= \sum_{k=1}^{n} \sum_{s=k}^\infty (k+p_x) A_n(k, s) x^{k+p_x} y^{s+p_y} \\
&= \sum_{k=1+p_x}^{n+p_x} \sum_{s=k+p_y}^\infty k A_n(k-p_x, s-p_y) x^{k} y^{s}
\end{align}
Using these ideas, I tried to formulate a relationship between the generating function and the given recurrence. In particular, notice the following useful relationships
\begin{align}
(n-1)G_n(x,y) + y \frac{\partial}{\partial y}\left\lbrace G_n(x,y)\right\rbrace &= \sum_{k=1}^n \sum_{s=k}^\infty (n+s-1)A_n(k,s) x^k y^s \tag{1} \\
(n+1) xyG_n(x,y) - x \frac{\partial}{\partial x}\left\lbrace x y G_n(x,y)\right\rbrace &= \sum_{k=2}^{n+1} \sum_{s=k+1}^\infty (n-k+1) A_n(k-1,s-1) x^k y^s \tag{2} \\
-3 y G_n(x,y) + x \frac{\partial}{\partial x}\left\lbrace y G_n(x,y)\right\rbrace + y \frac{\partial}{\partial y}\left\lbrace y G_n(x,y)\right\rbrace &= \sum_{k=1}^n \sum_{s=k+1}^\infty (k+s-3) A_n(k,s-1) x^k y^s \tag{3}
\end{align}
If we first simplify the left-hand side when we do $(1) - (2) - (3)$, we can find that
\begin{align}
\text{LHS}\lbrace (1) - (2) - (3) \rbrace &= \left((n - 1) + 3y - (n+1) x y \right) G_n(x,y) + x y \frac{\partial}{\partial x}\left\lbrace (x - 1) G_n(x,y)\right\rbrace + y \frac{\partial}{\partial y}\left\lbrace (1-y) G_n(x,y)\right\rbrace \\
&= \left((n - 1) + 2y - n x y \right)G_n(x,y) + x y (x-1)\frac{\partial G_n}{\partial x}(x,y) + y (1-y)\frac{\partial G_n}{\partial y}(x,y) \tag{4}
\end{align}
We can do similarly for the right-hand side of $(1) - (2) - (3)$ and find that
\begin{align}
\text{RHS}\lbrace (1) - (2) - (3) \rbrace &= x \left\lbrace\sum_{s=1}^\infty (n+s-1) \underbrace{A_n(1,s)}_{1 \iff s=1} y^s + \sum_{s=2}^\infty (s-2) \underbrace{A_n(1,s)}_{0} y^s\right\rbrace+ (n+1)A_n(2,2)x^2 y^2 \\
& + \sum_{k=2}^{n} \sum_{s=k+1}^\infty \underbrace{\left((n+s-1) A_{n}(k,s) - (n-k+1) A_{n}(k-1,s-1) - (k+s-3) A_{n}(k,s-1)\right)}_{0} \\
&= n x y + (n+1) A_n(2,2) x^2 y^2 \tag{5}
\end{align}
where we can use the recurrence to find that $A_n(2,2) = \frac{(n-1)}{(n+1)}$. Putting $(4)$ and $(5)$ together, we find that the generating function must satisfy the following first order linear partial differential equation:
$$\left((n - 1) + 2y - n x y \right) G_n(x,y) + x y (x-1)\frac{\partial G_n}{\partial x}(x,y) + y (1-y)\frac{\partial G_n}{\partial y}(x,y) = n x y + (n-1) x^2 y^2$$
From here, I am not certain about a few things. I am not well enough versed in using generating functions to solve multivariate recurrences with non-constant coefficients. A big question is, what boundary conditions should I be using for this partial differential equation? Given I have some boundary conditions, I am aware that the above equation can be approached using the method of characteristics, but I feel this is a bit complicated (assuming no errors).
Does anyone have any advice on whether this is the right way to work out the generating function $G_n(x,y)$? Is there a better alternative?
Typically boundary conditions from the recurrence initial conditions. E.g., in your case we have $\frac{\partial G_n}{\partial x}(0,y) = y$.
|
2025-03-21T14:48:30.276827
| 2020-04-11T02:38:48 |
357114
|
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"authors": [
"Gerry Myerson",
"Mak Nazečić-Andrlon",
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|
Stack Exchange
|
Characterizing $x y \equiv r \pmod{a^n}$ when $a^{n-1} \leq x, y < a^n$
Let $x, y \in \mathbb{Z}$ such that $a^{n-1} \leq x, y < a^n$ for some fixed positive integers $a$ and $n$. Is there a nice characterization of the remainders $r$ satisfying $x y \equiv r \pmod{a^n}$? Further, given $r$, under what conditions does $x y \bmod a^n < r$ hold?
Number theory isn't my specialty, so any pointers would be appreciated. I would ideally be looking for an answer that doesn't rely on factoring, if possible.
Since the number of $x,y$ pairs greatly exceeds $a^n$, I would expect most values of $r$ to appear as remainders. Have you tried any experiments? Say, $32\le x,y<64$?
@GerryMyerson Indeed, almost all appear. These are the exceptions I've found, written as $f(a, n)$: $f(2, 2) = {3}$, $f(2, 3) = {5, 7}$, $f(2, 4) = {11, 13}$, $f(2, 5) = {19, 29}$. They're primes, which makes sense that some should appear. But for higher $a$ or $n$, it seems to always come up with every possible $r$.
|
2025-03-21T14:48:30.276916
| 2020-04-11T03:20:35 |
357116
|
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"Ali Taghavi",
"Thomas Rot",
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"url": "https://mathoverflow.net/questions/357116"
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|
Stack Exchange
|
Fredholm transversality
Let $M,N$ be two Banach or Hilbert manifolds. Assume that $P\subset N$ is a smooth submanifold of $N$.
We say that a smooth map $f:M \to N$ is Fredholm transversal to $P$ if for every $x\in M$ with $f(x)\in P$ the pair$(Df_x(T_x M), T_{f(x)} P)$ would be a Fredholm pair.
(Recall that a Fredholm pair in a Banach or Hilbert space H is a pair of closed subspaces $(V,W)$ such that $V\cap W$ is a finite dimensional space and $V+W$ is a closed subspace of $H$ with finite codimension.
Is this terminology introduced in some reference?
Let $H$ be a Hilbert space. Is there a submanifold $P\subset H$ for which there is no a surjective map $f:H \to H$ such that $f$ is Fredholm transverse to $P$? In particular does the unit sphere $P$ of $H$ admit a surjective Fredholm transversal map?
I think there is a typo in the definition of the pair.
@ThomasRot I think it is identical to what the libked paper originally wrote.
Quote from the paper:"We recall that a Fredholm pair is a pair of closed subspaces (V,W) such
that V +W is closed and finite codimensional, V ∩W is finite dimensional"
@ThomasRot Why do you think that there is a typo?
Shouldn't $(Df_x(T_x M), T_{f(x)} N)$ be $(Df_x(T_x M), T_{f(x)} P)$?
Otherwise the condition is that Df has finite rank
@ThomasRot Ah yes yes. I am sorry. I revise it. Thank you very much for your comment!
You might be interested in this paper: https://arxiv.org/pdf/math/0309020.pdf
@ThomasRot I realy thank you very much for this very instructive and interesting paper.
|
2025-03-21T14:48:30.277074
| 2020-04-11T03:33:11 |
357117
|
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"authors": [
"JP McCarthy",
"Peter Krauss",
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"Wojowu",
"Wolfgang",
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|
Stack Exchange
|
Domino tiling obtained from space-filling curves, is possible to predict basic properties?
Periodic and aperiodic domino tiling systems can be obtained by the following construction rules:
Draw a regular square grid n×n of n2 cells.
Select a space-filling curve that is consistent with squares: that do a recursive four-partition of the quadrilateral unit grid (refinement ratio 4 in the OGC jargon).
Draw a space-filling curve path through the center of the square cells, using the curve also as index i for each cell.
Merge neighboring cells to obtain a domino with index j=floor(i/2).It results in a domino tiling of n2/2 dominoes.
Example:
In domino tiling problems the domino's relative orientation to its neighboring is important... Is there a way to predict the basic properties of the domino tiling?That is, for each space-filling curve:
predict the type of tiling system, periodic or aperiodic;
predic the fraction of dominoes that will be different orientation.
Notes
The merge rule transforms the original curve into another fractal, so we cannot use directly the properties of the original fractal. The degenerated curve is like a "dual", we can deduce some general properties of the transformation that will be useful for predictions. For example original Morton curve have periodic Z-shape, and degenerated curve preserves the periodicity. Hilbert curve is aperiodic (the U-shape is rotated), and the degenerated curve preserves this aperiodic orientation.
Illustrating with more dominoes, from the same degenerated curves.
By empirical induction we can suppose that, for any n:
Morton curve generates periodic domino tiling, all dominoes in the same orientation.
Hilbert curve generates aperiodic domino tiling, 50% horizontal and 50% vertical.
PS: another interesting question is about fractal classification, can we use this construction rules to group isolated fractals in complementary ones? The "И-shape curve" is the complement (degenerated form) of the Z-shape curve; the Munkres's fractal (defined at "Theorem 44.1" in this book) is the complement of the Hilbert curve.
Is it clear that the dominoes are always well defined?
"Draw a space-filling curve path through the center of the square cells, using the curve also as index i for each cell." None of the curves you have drawn are actually space-filling. They are just some approximations in some constructions of such curves. It's not clear every space-filling curve can be approximated this way, and then (as Wolfgang highlights) it's not clear that you can always pair up points into dominoes (in fact I expect it's not always the case)
Thanks @Wolfgang and Wojowu, sorry, I omitted the class of curves. Edited. It is sufficient to express a constraint in this terms? "space-filling curve that can be used in a recursive four-partition of the quadrilateral unit grid".
Hi @Wojowu, you say that "None of the curves you have drawn are actually space-filling". I am using the finite approximation... Is it the problem? It is usual in ordinary applications and tradition of the description of "Natural phenomena with fractal features".
So if you restrict the space filling curves to those definable by a recursive construction (which makes sense), I guess it is possible to alternate both kinds mentioned in your post. What would those domino patterns look like?
Hello @Wolfgang, I don't understand your definition of "alternate both kinds" and your last question... Don't the illustrations show the "domino patterns look like"?
Jigsaw HAVE become very popular during the pandemic.
Well I wondered whether there is a way to alternate one Iteration step Morton style and one Hilbert style, but probably it won't be possible as Morton has diagonally opposed endpoints while Hilbert doesn't, which causes a different parity.
Just FYI, the word "domino" on the Wikipedia page for "aperiodic tiling" actually does not refer to this type of dominoes. It refers to subshifts of finite type. (Every substitutive subshift is an SFT up to adding some markings, i.e. sofic, but this is highly non-trivial.)
Hi @VilleSalo, I was looking for a proof that “table”-tiling is infinite (can be expanded recursively). Do you have a reference with a proof that it is not, it is a "finite type" only? About Wikipedia, there are a specific Domino tiling article (and also Wang tile).
A proof that it can be expanded recursively? I'm sorry but I have no idea what that means. Solomyak's paper has the definition of the substitution if that's what you're after. Maybe you don't know what "finite type" means. It means "the set of valid tilings with some finite set of Wang tiles, up to isomorphism of topological dynamical systems". The Wikipedia article about domino tiling is about the kinds of domino you talk about here, I was just pointing out that "domino problem" is NOT about that.
The topological isomorphism part is unimportant, just that "SFT" is shorter than "the subshift of valid tilings by a set of Wang tiles". But I could've just said the domino problem is about Wang tiles.
Thanks @VilleSalo, the correct classification is important. I am using here the general concept of aperiodic, used e.g. in Quasicrystals... The aperiodic "degenerated Hilbert curve" domino tiling is a structure that is ordered but not periodic, means that "shifting the domino tiling by any finite distance, without rotation, cannot produce the same tiling". I need to learn about "uniform distribution theory" and your article about SFT... I will edit the question after bounty expires, adding your point about Wang tiles/"domino problem".
I know what aperiodic means, and I have answered your question in an answer below. In the comment I only made a technical point about your choice of WP page to link and terminology.
Also, I have many papers about SFT, but the paper you link is actually about general subshifts: the SFT case of that problem was previously known.
From "bounty expires", I deduce that my answer did not answer your question? Could you explain why?
@VilleSalo I'm sorry for the misunderstanding. You was quick in your good answer (!), we have to wait for the "bounty period" (plus 5 days) for any other possible answers, and in this period it is not polite to edit the question. After period I can use some more days to study all answers... And time to study for my is necessary, I have poor knowledge in this field, I will use the next weekend to it.
I see, indeed it's presumably a good idea to wait until the end of the bounty time. I read "expires" as the bounty not being awarded at all, and got the impression you already perused my answer and figured it's not worth the bounty, in which case I could've tried to improve on it.
I'm not sure I understood the question and answer given by the asker, but I gather that they are interested in decidability questions about some substitutive subshifts.
As far as I can tell, the questions about which dominoes occur more often are in general solved by basic linear algebra; when you have a substitution giving the tiling you just compute some eigenvectors for its abelianization, not really on-topic for this site. In the case of the two example substitutions the solutions are already given by the asker; I don't know what "empirical induction" is, but the facts that horizontal and vertical dominoes appear with the same frequency in Hilbert, and all are horizontal in the other one, can be proved by induction. That said, let me concentrate on aperiodicity, which is more interesting.
I'll give some basic info about that: for any substitutive curve, the aperiodicity question about its dominofied version is decidable (if the class being implied is anything like I think). I didn't actually get what the degenerated stuff was about, but seems to be about some projections or factor images as well, so should be covered by the general theory. I also give a more classical style symbolic dynamics proof that the dominofication of the Hilbert curve is a minimal aperiodic subshift.
Aperiodicity is decidable: an automata theory proof
First, some definitions. Let $m, n \in \mathbb{N}$, $A$ a finite alphabet and $\tau : A \to A^{m \times n}$ a function. We interpret $\tau$ as replacing $a \in A$ by an $m \times n$ matrix, and we call such $\tau$ a ($m$-by-$n$)-substitution. We can apply $\tau$ to $P \in A^{k \times \ell}$ to obtain $\tau(P) \in A^{km \times \ell n}$, in the obvious way by replacing the individual elements of $A$ with $m \times n$ matrices.
If $\tau$ is a substitution, we say a (one-sided) infinite configuration $x \in A^{\mathbb{N}^2}$ is a $\tau$-periodic point if $\tau^k(x) = x$ for some $k > 0$, where we apply a substitution to an infinite configuration $x$ in an obvious way (the origin stays put, everything else blows up in the positive direction). We can find natural $\tau$-periodic points as follows: Start with a symbol $a \in A$, and keep applying $\tau$. The symbol $\tau^n(a)|_{(0,0)}$ evolves eventually periodically, say with some eventual period $p$. Then $\tau^{p\ell}(a)$ actually tends to a limit in an obvious sense (cellwise), and this limit is $\tau$-periodic as an infinite configuration (with period $p$).
Say $x \in A^{\mathbb{N}^2}$ is $n$-automatic if for all $a \in A$, the set $\{v \in \mathbb{N}^2 \;|\; x_v = a\}$ is $n$-automatic. A set of pairs of numbers $N \subset \mathbb{N}^2$ is $n$-automatic if the language $L_n$, of words $w \in (\{0,1,...,n-1\}^2)^*$ which evaluate to a pair of numbers in $N$ when you separately read off the $n$-ary numbers on the two tracks, is a regular language. Regular languages are very robust so I won't give the precise formulas, you can't really guess them wrong. We similarly define $n$-automatic configurations in $A^{\mathbb{N}^d}$ and subsets of $\mathbb{N}^d$.
The following is relatively easy to show, you can find it (or at least its one-dimensional version) in many references and books that discuss automata and substitutions.
Theorem. Let $\tau : A \to A^{n \times n}$ be a substitution and let $x$ be any $\tau$-periodic point. Then $x$ is $n$-automatic.
The proof is not very hard, the idea is the automaton just keeps track of the current symbol and when it reads a digit, that tells it where it moves in the substitutive image (and it looks up $\tau$ to see which symbol is there).
The following is obvious by basic closure properties of regular languages.
Lemma. Let $x \in A^{\mathbb{N}^2}$ be $n$-automatic, and $\pi : A \to B$ a function. Then $\pi(x) \in B^{\mathbb{N}^2}$, defined by $\pi(x)_v = \pi(x_v)$, is also $n$-automatic.
The following is classical, perhaps first invented by Büchi. There is an implementation called Walnut where you can directly input such statements. I get the impression the asker is into computers, so I'll leave it as an exercise to try this out (sometimes Walnut cracks very difficult problems, sometimes it gets stuck on very trivial things, it's all about whether the intermediate DFAs happen to have huge numbers of states, which is hard to predict). The decidability proof is not so hard, the idea is to do quantifier elimination by using the subset construction.
Theorem. Let $x \in A^{\mathbb{N}^2}$ be $n$-automatic, and let $\phi$ be any first-order formula (with constants and free variables) where quantifiers range over vectors in $\mathbb{N}^2$, and you have function sumbols for vector addition, and have a unary predicate for "$x_v = a$" for each $a \in A$, with the obvious interpretations. Then the set of solutions to $\phi$ (possible values for the free variables) are an $n$-automatic subset of $\mathbb{N}^d$, which can be computed effectively; if there are no free variables, whether $\phi$ is a true statement is decidable.
Now, the question of aperiodicity amounts to programming in first-order logic. We say $x \in A^{\mathbb{N}^2}$ is aperiodic if its stabilizer $\{v \in \mathbb{N}^2 \;|\; v+x = x\}$ is trivial, where $v+x$ denotes translation $(v+x)_u = x_{v+u}$. This is not to be confused with $\tau$-periodicity.
Lemma. A configuration $x \in A^{\mathbb{N}^2}$ is aperiodic if and only if it satisfies the following first-order statement (of the type in the previous theorem):
$$ \forall v \neq 0: \exists u: x_{u+v} \neq x_u. $$
Hopefully no need for a proof, since basically I just wrote the definition. This lemma and formula are about periodicity of a full $\tau$-periodic quarterplane, but you can modify the formula to talk about arbitrarily large periodic areas, or periodicity up to skipping some initial rows and columns, or many other things, it's just first-order formula programming. Now, I believe the following theorem solves a relatively general version of your question:
Theorem. For any substitution $\tau : A \to A^{n \times n}$, any $\tau$-periodic $x \in A^{\mathbb{N}^2}$, and any map $\pi : A \to B$, it is decidable whether the configuration $\pi(x)$ is aperiodic.
Proof: We have that $x$ is $n$-automatic, thus $\pi(x)$ is $n$-automatic, and it is decidable whether the formula of the previous lemma is true, thus it is decidable whether $\pi(x)$ is aperiodic. Square.
Now, we can solve your Hilbert curve question as follows:
Take as alphabet the symbols for the cardinal directions $\{ N, E, W, S \}$, which represent the different directions where a basic "$U$-shape/$U$-curve" of the Hilbert curve (by which I mean one of the length-$4$ segments from which the curve consists) may open. Take the Cartesian product with $\{-1, 0, 1\}$ as there are three different ways the curve may continue from the first and last point of the curve (either it continues straight or turns, and it never goes straight from both ends). You can then work out a substitution on $12$ symbols, which carries all the relevant information. (It's a reasonable coding, since it's easy to see that you can tell from a dominofied configuration which are the $2$-by-$2$ blocks coming from a basic $U$-shape.) The map $\pi$ takes $\{N, E\}$-type symbols to $V$ and $\{W, E\}$-type symbols to $H$ (ignoring the $\{0, 1, 2\}$ component).
For example, if $-1$ means "turn on the left", $1$ means "turn on the right" and $0$ means "turn on both ends", then
$$ \tau(N, 0) = \begin{pmatrix} (W, -1) & (E, 1) \\ (N, 1) & (N, -1) \end{pmatrix}. $$
This is the example you give, but I carry the additional information of how the curve would continue, and I do not yet do the final substitutions where you actually write the basic $U$-curves, and where you then substitute them for a pair of dominoes.
So there's an algorithm that solves whether the Hilbert curve domino thingie is aperiodic. I did not run it, and do not claim it's very efficient. Instead...
Aperiodicity of the Hilbert curve: a symbolic dynamics proof
I'll give a manual proof (I did check the combinatorics in Python, but it's my native language so that's much less work than Walnut, and I think it shouldn't be hard to work out on pencil-and-paper).
Now, let me describe another way, which is what people usually do in practice (because it's more fun to do this way when working manually), and then I can give a quick manual proof that the Hilbert curve gives you an aperiodic domino tiling. We are going to replace computation with more conceptual ideas, so we need more definitions.
A subshift is a subset $X \subset A^{\mathbb{Z}^2}$ which is topologically closed in the profinite (Cantor) topology, and is shift-invariant meaning $\forall x \in X: \forall v \in \mathbb{Z}^2: v+x \in X$. (I move to $\mathbb{Z}^2$-configurations form $\mathbb{N}^2$-configurations for convenience, but actually after the topologization everything will turn into questions about finite objects, so this will not matter much.)
If $x \in A^{\mathbb{Z}^2}$ and $P \in A^{k \times \ell}$, write $P \sqsubset x$ for $P$ appearing somewhere in $x$, i.e. $\exists v \in \mathbb{Z^2}: (v+x)|_{k \times \ell} = P$, where $v+x$ denotes translation $(v+x)_u = x_{v+u}$. Similarly we write $P \sqsubset Q$ for $P$ appearing somewhere in $Q$ when $P \in A^{m \times n}$ and $Q \in A^{k \times \ell}$, with the obvious meaning. The subshift generated by $\tau$ from $a \in A$ is the set of all $x \in A^{\mathbb{Z}^2}$ such that for any $k, \ell$ and any such that $P \sqsubset x$, there exists $n$ such that $P \sqsubset \tau^n(a)$.
We also reformulate $\tau$-periodicity for $\mathbb{Z}^2$-configurations. We say $x \in A^{\mathbb{Z}^2}$ is a good $\tau$-periodic point if there exists $R \in A^{2 \times 2}$ such that $R$ appears in $\tau^n(a)$ for some $a \in A$, $n \in \mathbb{N}$, and $x$ is the limit of $R$ obtained by taking the limits $\tau^p(b)$ of the four symbols in $R$ separately, for some $p \in \mathbb{N}$, i.e. each of them separately expands into its own direction. Such $\tau$-periodic configurations again exist by the pigeonhole principle since the set of $2$-by-$2$ patterns is finite.
If $\tau$ is a substitution, write $M_\tau$ for the $|A|$-by-$|A|$ matrix where $(M_\tau)_{a,b} = |\{k \;|\; \tau(a)_k = b\}|$, i.e. rows tell you how many of each symbol appears in each $\tau$-image. We say a matrix $M$ is primitive if there exists $n$ such that $M^n$ has only positive entries. We have that $M_\tau$ is primitive if and only if $b \sqsubset \tau^n(a)$ for any choice $a, b \in A$, and we then also say $\tau$ is primitive.
The following lemma can be found in any reference that discusses substitutions (at least its one-dimensional version, but it's exactly the same in two dimensions since our substitution is rectangle-shaped).
Let $\tau$ be a substitution such that $M_\tau$ is primitive. Then the subshift $X$ generated from $a$ does not depend on the choice of $a$, and the orbit-closure of every good $\tau$-periodic point is $X$. The subshift $X$ is minimal, i.e. the orbit-closure of every point in $X$ is $X$.
By this lemma, and the easily proved fact that the Hilbert curve substitution is primitive, we don't really have to worry about whether things are one-sided or two-sided: minimality means that either every configuration $x \in X$ satisfies $v+x = x$ for some $v \in \mathbb{Z}^2 \setminus \{(0,0)\}$ independent of $x$, or for all $v \in \mathbb{Z}^2 \setminus \{(0,0)\}$ there exists $m$ such that the $v$-periodic is locally broken in all patterns of size $m$-by-$m$ that appear when you iterate this substitution.
A morphism from subshift $X$ to subshift $Y$ is a continuous function $\pi : X \to Y$ which commutes with the shift maps, i.e. $\pi(v+x) = v + \pi(x)$ for all $x \in X$, $v \in \mathbb{Z}^2$. It is easy to see that being aperiodic is preserved under a morphism.
Now, we proceed as follows: Let $X$ be the subshift generated by the Hilbert curve substitution $\tau$.
Define the map $\pi : \{N,E,W,S\} \times \{-1,0,1\} \to \{N,E,W,S\}$, which drops the information about how the curve continues, let the image be $Y$, so we have a morphism $\pi : X \to Y$.
Show that $Y$ is aperiodic.
Show that the map $\pi' : \{N,E,W,S\} \to \{H,V\}$, which extracts the dominoes, is an isomorphism onto its image, denote it as $\pi' : Y \to Z$.
Now, $Z$ has to also be aperiodic, because the inverse $(\pi')^{-1} : Z \to Y$ would preserve any period.
We first show that $Y$ is aperiodic. To see this, we observe that actually we could have in the first place defined the Hilbert curve substitution $\tau$ without using the curve information (I just included it to keep it as close to the original description as possible; I didn't check whether $\pi$ is an isomorphism): $Y$ is the subshift given by the substitution $\tau'$ obtained from $\tau$ that ignores the curve information completely (again $\tau'$ is primitive so $Y$ is minimal).
I proved the aperiodicity as follows: the pattern
$\begin{pmatrix} N & W & E & N \\ S & W & E & S \end{pmatrix}$
can only appear in only even positions of substituted images of patterns (both coordinates even), as you can see by analyzing $(\tau')^n(R)$ for small $n$ and $2$-by-$2$ patterns $R$ ($n = 3$ is enough). Since $\tau'$ is injective, you can "use a local rule to detect whether a given $2$-by-$2$ block you see is actually a substituted image of a symbol, or whether it appears between two such images, in a unique way" (one says the subshift recognizable in the sense of Mossé).
Let me not define that because I don't know a particularly neat way to do that, but you can find this in any reference that discusses recognizability and substitutions. Instead, let me intuitively explain what we do with this local rule: once you can figure out which $2$-by-$2$ blocks come from symbols, in fact by the substitutive nature of the subshift the preimages of these blocks form a configuration of the same subshift $Y$. So you can iterate the local rule, and what you get once you forget all but the "phases" is a continuous shift-invariant map $\phi : Y \to I^2$, where $I$ the $2$-adic integers, and where $\mathbb{Z}^2$ acts on $I^2 = I \times I$ by translation ($\mathbb{Z} \leq I$ is a dense subgroup; the dynamical system $I$ is usually called the $2$-adic odometer). Since $I^2$ is a torsion-free group, $I^2$ under the translation action of $\mathbb{Z}^2$ has only aperiodic points, and thus so must $Y$ (morphisms between general dynamical systems also preserve periods).
Finally, we need to show that map $\pi' : Y \to Z$, is an isomorphism. For this, we argue similarly as above: A short case analysis shows that the pattern
$$ \pi'(\tau'(\begin{pmatrix} W & E \\ N & N \end{pmatrix})) = \begin{pmatrix} V & H & H & V \\ V & H & H & V \\ H & H & H & H \\ V & V & V & V \end{pmatrix} $$
only appears in even positions (both coordinates even), among the patterns $\pi'(\tau'^n(a)$ for any $n$ and $a \in A$. This again gives us a unique $2$-by-$2$ phase. We observe that $\pi'$ is injective on the $\tau'(A)$ so we can deduce the preimage configuration completely, proving $\pi' : Y \to Z$ is actually an isomorphism.
So as discussed earlier, the subshift $Z$ which is the dominofication of the Hilbert curve subshift, must be aperiodic (i.e. every configuration is aperiodic in it). It is also minimal as it is a morphic image of $Y$.
Hi Ville Salo, thanks for the detailed and complete answer (!). As I commented before, I will use next weekend to read and discuss it. What I can say before are just my doubts, to get a better interpretation of your statements, and explain something of my question... 1. "interested in decidability questions", yes; 2. "questions about which dominoes occur more often", you answerd the subquestion, by confirming is a proof by induction, and thanks to the clues showing how to formalize. ...3... .... I will continue later.
3. "I didn't actually get what the degenerated stuff was about", it is only my jargon to refer to the "transformated Hilbert curve", that maybe have another name like "Munkres curve". I liked your jargon, "dominofication of the Hilbert curve"
Oh, the degenerated curve is the curve along the dominoes of course. I don't know how I missed that. As dynamical systems, the original curve and the degenerated one are presumably the same, i.e. you can locally deduce one from the other.
On my cellphone app I keep seeing a comment here by the asker (one starting with an inverted exclamation mark), and then it disappears again, very strange. It's gone again.
(sorry deleted bad English and write again - not my language) Only checking my interpretatios... At the end, at the last proof, the $\pi'$ map transforms "Hilbert level 1" in to "Hilbert level 3". It is not a "dominofication", but we can suppose that the union of adjacent cells will be U-shape-horizontal or U-shape-vertial in a dominofication. In this case the dominoes of the "Degenerated Hilbert level 2.5" where each U-shape was transformated into 2 dominoes... And your matrix predicts correctly the vertial and horizontal dominoes.
I'm sorry there was an unfortunate typo in the types of $\pi$ and $\pi'$, the latter $\times$ should be $\to$. The map $\pi'$ is meant to be from ${N,E,W,S}$ to ${H,V}$, and it turns for example $N$ into $V$, i.e. it turns a length-4 $U$-shaped curve piece (so Hilbert level 1 if I understand correctly) into two vertical dominoes side-by-side.
(So two dominoes side-by-side is represented by just a single symbol $V$.)
Looking at your link, I'd say in $X$ symbols represent Hilbert level $1$, in $\pi(X) = Y$ they still do but you forget how the curve connects these pieces, and in $\pi'(Y) = Z$ you map to from level 1 to level 0.5.
This is not an answer...
This is a solution sketch in a Wiki, please edit here (!) to enhance. Informal clues and perhaps some starting point for good answers.
All space-filling curves that satisfies the constraints
The step2 (second construction rule of the question) is a big constraint for "all" space-filling curves:
it is to be used in a space partitioning: divides a space into non-overlapping regions, each receiving a region label;
only the "split into 4 regions" partition is valid.
So, seems that no other curve exists (!), only the following 3 types, but one is not valid for domino generation:
Labeling suggestion
This section is also for reviewing the description of the problem, not by mathematical theory, but by naming cells.
The step4 (merge rule) was expressed by indexes $j = \lceil{i/2} \rceil$
The merge rule transforms the original curve into another fractal, so we cannot use directly all properties of the original fractal.
Geometrically it is a recursive 4-partition of the cells, them, the natural label is like Natural Number expressed by base 4... For each curve of hierarchical level L we have a grid of $4^L$ square cells labeled by a numeric code of L digits.
But it is not a number because it need to preserve hierarchy: "0" and "00" are distinct labels. Notice also that the dominoes arise at the intermediate levels, L=½, L=1½, L=2½ etc. The dominoes of "half levels" with $L>½$ and index $j = \lceil{i/2} \rceil$ from $i=0...4^{L+½}$ can reuse the labels of level $L-½$ concatenated with a letter:
L=0; no grid, the square region to partitioned.
L=0.5; 4/2=2 dominoes; labels: G H.
L=1; grid of n=41=4; labels: 0 1 2 3.
L=1.5; 16/2=8 dominoes; labels: 0G 0H 1G 1H 2G 2H 3G 3H.
L=2; grid of n=42=16; labels: 00 01 02 03 10 11 12 13 20 21 ... 33.
L=2.5; 64/2=32 dominoes; labels: 00G 00H 01G 01H 02G 02H 10G 10H ... 33G 33H.
L=3; grid of n=43=64; labels: 000 001 002 ... 333.
Any set of labels can be ordered by the lexicographical order of its alphabet: G,H,0,1,2,3.
Note: the illustrations above were obtained with Sfc4q, where you can play with more variants.
Some label-region relationships
The labels above are codes (not numbers but strings of characters), but its syntax is subject of some mathematical analysis. We can express cell label and cell spatial relationships by some algebra...
Suppose that the union of labels has the semantic of geometrical union, "∪", of grid cells. The union of two squares is a domino. The "merging rule", that is the transformation $j = \lceil{i/2} \rceil$, from integer level L to "half level" L+½, can be translated into a label transformation:
Level 1 to level 0.5, L1→L½ : 0 ∪ 1 = G; 2 ∪ 3 = H
Level 2 to level 1.5, L2→L1½ : 00 ∪ 01 = 0G; 02 ∪ 03 = 0H; etc.
Level 3 to level 2.5, L3→L2½ : 000 ∪ 001 = 00G; 002 ∪ 003 = 00H; etc.
Summarizing: the $j = \lceil{i/2} \rceil$ transformation always can be expressed by simple syntactic rule, "preserve prefix and concatenate a letter"that preserves parent cell label's prefix and preserves geometrical region of the parent cell.
For example, as we can see at illustrations, the code 10 at L2 and its position are preserved at levels L2½ (as 10G and 10H) and L3 (as 100, 101, 102 and 103). Same for code 21, preserved at L2½ (21G and 21H) and L3 (210, 211, 212 and 213).
Of course, it not solves the problem, only perhaps simplify demonstrations.
Some other properties of the label transformation, valid for any fractal, Morton or Hilbert:
In any transformation L → L-½ the union of any distinct cells, $x \cup y$, has labels with "even last digit" in $x$ (0, 2, 00, ...) and "odd last digit" in $y$ (1, 3, 01, ...).
The transformation L1→L½ is not representative for induction (we must to start induction proofs by the transform L2→L1½)
Looking for references
There are some mathematical scholar article about the fractals and domino tiling? Seems not easy to find... For example this illustration is a cut from Goodman-Strauss (2016)
There are no citation of Hilbert curve relationship, but maybe we can find some proof that it is a "infinite aperiodic".
It is interesting also to check references proofing that
in the Morton curve, the "Z"-shape is periodic, for any level L.
...
Clues for Morton curve
Seems easy to proof (if no reference to cite) that
the "Z"-shape is periodic, for any level L.
Proofs about the complementary fractal can use induction:
L2 generates 100% of (periodic) horizontal dominoes.
... Any level L generates 100% of horizontal dominoes. All pairs even-odd labels are horizontal, in any level.
Clues for Hilbert curve
Seems easy to proof (if no reference to cite) that suppose the properties P1 and P2 bellow are based on true assertions:
P1. Hilbert curve have infinite aperiodic "U"-shape distribution.
P2. Hilbert curve have constant (rotated) "U"-shape distribution, for any integer level L>1: 50% are "⊐"-shape or "⊏"-shape, 50% are "⊓"-shape or "U"-shape.
After the "merge rule" transformation the property P2 results in a regular distribution of dominoes of different orientation:
"⊐" and "⊏" generates 2 horizontal dominoes;
"⊓" and "⊔" generates 2 vertical dominoes.
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2025-03-21T14:48:30.279204
| 2020-04-11T06:07:48 |
357122
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|
Stack Exchange
|
Decoding Fock spaces
Technically, the Fock space is (the Hilbert space completion of) the direct sum of the symmetric or antisymmetric tensors in the tensor powers of a single-particle Hilbert space H.(Wikipedia)
Question 1.How to write a Hilbert space if it is a Fock space as (the Hilbert space completion of) the direct sum of the symmetric or antisymmetric tensors in the tensor powers of a single-particle Hilbert space H. How to search for H, in computional-complexity sense?.Is there a general known method to search for H for any Fock space?.
Question 2.If there is not such method, could anyone give suggestions or ideas (not necessarily verified) how possibly to do that?.
I am perhaps still not quite understanding the issue: a Fock space is specified by a vacuum state $|0\rangle$ and a set of creation operators $a^\dagger_p$; the single-particle Hilbert space has basis states $a^\dagger_p|0\rangle$.
How is your already-give Fock space supposed to be specified?
I assume the Fock space is formed out of all linear combinations of states $a_{p_1}^\dagger a_{p_2}^\dagger\cdots a_{p_k}^\dagger|0\rangle$, $k=1,2,\ldots$.
I am asking how to write a Hilbert space if it is a Fock space as (the Hilbert space completion of) the direct sum of the symmetric or antisymmetric tensors in the tensor powers of a single-particle Hilbert space H. How to search for H?.
It seems to me that, if you are given a Fock space with its vacuum state and creation operators, then the first comment by @CarloBeenakker answers the question. If you're given only a Fock space, as an abstract Hilbert space, without a specified vacuum state and creation operators, then this is not enough information to recover the one-particle space $H$ as a subspace of Fock space. Even some non-isomorphic $H$'s can produce isomorphic Fock spaces. Example: any infinite-dimensional $H$ and any nontrivial finite-dimensional boson $H$ (i.e., use symmetric powers).
@AndreasBlass,I guess,since we are searching for H in computional-complexity context, H is a finite Hilbert space. Second, the question does not require uniqueness of H. I am not expert in any of all this.
@AndreasBlass, just for curiosity, what extra information we need to add to make H unique; if you know.
From a computational perspective that needs finite inputs and outputs, there's a problem because Fock spaces are often infinite-dimensional. Even if $H$ is one-dimensional (a spin-$0$ boson), the Fock space will be infinite dimensional because of the infinitely many possible occupation numbers. The only way the Fock space can be finite-dimensional is if you're dealing with fermions with a finite-dimensional $H$.
I don't see any plausible way to "find" $H$ within its Fock space by giving "simpler" information then just pointing out $H$. @CarloBeenakker has shown that the vacuum state and creation operators are enough information, but they describe a lot more than just $H$.
Sorry, in my questions and my comments; probably I am writing one thing and thinking completely about different thing. I am writing about Fock space as an object under computation, but in reality I am thinking about Fock space as a tool to do computation. So my question is wrongly phrased, instead I should write as the following:- " Is there a general known method(scheme),to decode a quantum device;if we know about its Fock space,(we have information about its spectrum topology)".I am not sure if this good question,but this summaries what I am thinking.
|
2025-03-21T14:48:30.279567
| 2020-04-11T07:24:50 |
357124
|
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|
Stack Exchange
|
On comparing planar convex regions of equal perimeter and area
Definitions:
The Hausdorff distance between two point sets is the greatest of all the distances from a point in one set to the closest point in the other set.
Given two planar convex regions $C_1$ and $C_2$ both with unit perimeter, we define the difference between $C_1$ and $C_2$ as the least value of Hausdorff distance between $C_1$ and $C_2$ can have when the regions are placed above one another and transformed with isometries (rotation, translation, reflection) to minimise the Hausdorff distance between them.
Questions:
What are the specific pair of unit perimeter regions $\{C_1, C_2\}$ with some equal specified area such that the difference between $C_1$ and $C_2$ is maximum?
What are the specific pair of unit perimeter regions $\{C_1, C_2\}$ with equal specified area and equal specified diameter (diameter of a region is the greatest distance between any two points in the region). such that the difference between $C_1$ and $C_2$ is maximum?
(further versions of the question with $C_1$ and $C_2$ sharing equal values of more global quantities – and also higher dimensional versions – are natural)
My assumption would be that vertically aligning the respective centers of gravity and the axes of inertia will give at least a local optimum; as there are two rotations that align the axes of least inertia, both orientations should be checked.
We suggest an answer to question 1 above, without proof of optimality:
First we note two facts.
Please see https://nandacumar.blogspot.com/2012/11/maximizing-and-minimizing-diameter-ii.html for some more details.
For specified values of area and perimeter, the convex region with maximum diameter is formed with two equal circular arcs (how one draws a convex lens).
For specified area and perimeter, diameter is minimized by a convex closed curve of constant width.
So choosing these 2 figures as the pair, we get at least a difference of (max diameter - min diameter)/2 when the two figures are superimposed.
Note: When the area specified (with perimeter fixed at 1) is less than that of the Reuleux triangle of unit perimeter, we don't have a convex figure of constant width.
|
2025-03-21T14:48:30.279740
| 2020-04-11T09:10:23 |
357132
|
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"Mark Grant",
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|
Stack Exchange
|
Fibre preserving maps of Borel constructions
Let $G$ be a discrete group with universal principal bundle $EG\to BG$, and let $X$ and $Y$ be left $G$-spaces. An equivariant map $\overline{f}:X\to Y$ induces a fibre-preserving map $f:EG\times_G X\to EG\times_G Y$ between Borel constructions, as in the following diagram.
$\require{AMScd}$
\begin{CD}
X @>\overline{f}>> Y\\
@V V V @VV V\\
EG\times_G X @>>f> EG\times_G Y \\
@V V V @VV V\\
BG @>=>> BG
\end{CD}
I believe some sort of converse to be true. That is, if $f:EG\times_G X\to EG\times_G Y$ is a fibre-preserving map between Borel constructions, it induces a $G$-equivariant map $\overline{f}:X\to Y$. Here I am identifying $G=\pi_1(BG,\ast)$, which acts (up to homotopy) on the fibres.
Does anyone know of a reference to a precise statement along these lines?
Edit (in response to user51223's comment): I'm mainly interested in the existence question: Is it true that there exists an equivariant map $\overline{f}:X\to Y$ if and only if there exists a map $f:EG\times_G X\to EG\times_G Y$ over $BG$? (So $\overline{f}$ doesn't have to be the induced map on fibres.)
I don't have an answer ready, but I believe it is a bit more complicated. Take $X$ and $Y$ to be vector spaces for simplicity, then $f$ is just a linear map between (locally trivial) vector bundles. Even if $\operatorname{Hom}_G(X,Y)$ is trivial, I don't see why $f$ should be zero everywhere.
@SebastianGoette: I'm trying to better understand your comment. Let's say $G=\mathbb{Z}/2$ with $X$ the trivial (1-dimensional) representation and $Y$ the sign representation. Then $\operatorname{Hom}_G(X,Y)=0$. Are there any non-zero bundle maps from the trivial line bundle to the non-trivial line bundle over $\mathbb{R}P^\infty$?
I think that the map from X to Y is in general only coherently homotopy equivariant. Have a look at Dror,Dwyer and Kan, Equivariant maps which are self homotopy equivalences, Propositions 2.2 and 2.3 (this paper is available on Bill Dwyer's webpage).
Dear Mark, you can certainly cook up a map that is not everywhere $0$ using a partition of unity, assuming the base field contains $\mathbb R$. But then on the other hand, the space of linear maps is typically contractible. I guess Gustavo's comment is more to the point.
That $f$ always induces $\overline{f}$ is a fact right? And, your question is that if this map is $G$-equivariant? Or are you looking for any $G$-equivariant map that possible induced by $f$?
@GustavoGranja: This is really helpful, thanks. After reading their 2.3 I realised that $f$ induces a $G$-map $EG\times X\to EG\times Y$ (which is obvious from covering space theory). Is this what is meant by "coherently homotopy equivariant"?
@user51223: I'm interested in both versions of the question (hence the "along these lines"), but only really need the weaker existence question. I've edited to make this clear.
@MarkGrant You are very welcome. By coherently equivariant homotopic I meant extending to a homotopy coherent map of diagrams indexed by the category $G$ with one object (and $G$ as the monoid of endomorphisms). I think this is equivalent to the map extending to $EG \times X \to Y$ as $EG\times X \to X$ is a cofibrant replacement for the diagram $X$. Note that since $EG$ is a terminal object in the homotopy category of free $G$-spaces giving a $G$-equivariant map from $EG\times X \to EG \times Y$ is really the same as giving a $G$-equivariant map form $EG \times X \to Y$.
The answer to the question in the edit is no. Take $G=\mathbb Z$, $X$ to be a point and $Y=\mathbb R$ with the action $n\cdot x = x+n$. Then there are no equivariant maps from $X$ to $Y$ but there are many maps from $EG \times_G X = BG = S^1$ to the cylinder $EG \times_G Y$ over $BG$.
In this special case when $X$ is a point, an equivariant map $X \to Y$ is a fixed point, while a map $EG \times_G X \to EG \times_G Y$ is a homotopy fixed point. The fixed points include in the homotopy fixed points, but in general I don't think there is much one can say about the inclusion.
Similar to Granja's answer: let $X$ be a point, and $Y = EG$, and let $f$ be induced by the diagonal on $EG$.
|
2025-03-21T14:48:30.280039
| 2020-04-11T09:13:09 |
357133
|
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|
Stack Exchange
|
A generalization of Feit–Thompson conjecture, for square-free integers
I asked the following question with my account that I have for these sites Mathematics Stack Exchange and MathOverflow. The bounty that I offered in MSE expired without answers. The post that I refer is this MSE 3591614 (March, 23th 2020). See the useful comments that were added in MSE.
Few weeks ago I wondered about if the following conjecture is in the literature or well if it is possible to find a counterexample. I evoke a generalization of a well-known conjecture, I mean the Feit–Thompson conjecture from Wikipedia.
An integer $n>1$ is square-free or squarefree if has no repeated prime factors. For example $n=15$ is squarefree, while than $12$ has repeated prime factors.
Conjecture. There are no distinct square-free integers $p$ and $q$ (both greater than $1$) with $\gcd(p,q)=1$ such that $\frac{p^q-1}{p-1}$ divides $\frac{q^p-1}{q-1}$.
Question. What work can be done about the veracity of this conjecture? Since Feit–Thompson conjecture seems a particular case of this, then I'm asking if previous conjecture is in the literature, or well if you've some approach or heuristic about the Conjecture or well if it is possible to find a counterexample. Many thanks.
I wrote a Pari/GP program and I've tested the Conjecture, let's say for the segments of integers $2\leq p,q\leq 2000$. I hope that there were not mistakes and this post is interesting.
References:
[1] Feit, W. and Thompson, J. G., A Solvability Criterion for Finite Groups and Some Consequences, Proc. Nat. Acad. Sci. USA 48, 968-970, (1962).
To the moderator team of MO/MSE, as soon there is feedback about if this question is interesting for MathOverflow, feel free to handle the MSE 3591614 question as you consider it is right (I say this since I never had a cross-posted question).
Maybe this could be reduced to the original Feit-Thompson conjecture considering that, writing the set of prime factors of $p$ as $p_i$ and the set of prime factors of $q$ as $q_j$ one has $(p_i,q_j)=1$. The only idea I have so far is to consider automorphisms of the rings of $p$-adic and $q$-adic integers.
More precisely, maybe we can try to prove that FTC is equivalent to $p\neq q\Longleftrightarrow Aut(Z_{p})\cap Aut(Z_{q})={e}$.
Many thanks for your ideas and attention. My abstract algebra isn't the best @SylvainJULIEN
@SylvainJULIEN There are (at least) two issues with this. First of all, how do you want to take an intersection of automorphism groups of distinct rings? Second, $\mathbb Z_p$ always has a trivial automorphism group, so your equivalence doesn't hold since the intersection is also trivial for $p=q$.
(1/2) Since few months ago my belief is that there exists a "nice" interpretation of the sum of divisors function $\sigma(n)$ that "moves rectangles to rectangles in a special way" (I can not explain better my belief). I consider rectangles $a\times b$ with $a\geq 1$ and for integers $a<b$ formed by identical unitary squares $1\times 1$. See also the recent post MathOverflow 357376 with title Another kind of primality related to tessellations by polygons, which inspires in my opinion that $\sigma(\text{rectangle})=\text{ a convex}$ with the exception of the sequence A023194 from the
(2/2) OEIS that is precisely the integers $n\geq 2$ such that $\sigma(n)=\text{ a prime}$. Finally see the comment in previous A023194 by Thomas Ordowski (Nov 18 2017). I edit these comments with the intention if you @Wojowu or the user in commments want to think about it (my belief is that maybe it is potentially related to compositions of the sum of divisors function with, in a way that I don't know, the Dedekind psi function or the Euler's totient function, and related to constructible polygons with compass and straightedge, Fermat primes and Mersenne primes). I hope don't disturb.
I write this comment for your attention @SylvainJULIEN in this and previous comments if you want to think about the previous thread of comments. Additionally I believe that the following conjecture is true: If an integer $n\geq 2$ satisfies $\sigma(\sigma(n))=1+\frac{n\operatorname{rad}(n)-1}{\operatorname{rad}(n)-1}$, then its sum of divisors $\sigma(n)$ is a prime number (here $\operatorname{rad}(n)$ is the product of distinct primes dividing $n$). With the help of Ordowski's claim it is easy to check that each term of the sequence A023194 satisfies previous equation. Many thanks.
I don't quite get your point right now but I'll try to think of it and let you know.
just reaching out to you when I read you are losing your internet access from home; we will miss you as a frequent contributor!
Thank you very much, I appreciate this community of users very much @CarloBeenakker , I hope your colleagues (professors) and users of this website are in good health.
@Wojowu please I've edited a (my last on MathOverflow) comment in the post with identificator 407529 , can you think about it? Many thanks.
@SylvainJULIEN please I've edited a (my last on MathOverflow) comment in the post related with an attempt to get an interpretation of the sum of divisors function, prime numbers and chirality. It is the post with identificator 407529 , can you think about it? Many thanks.
|
2025-03-21T14:48:30.280440
| 2020-04-11T10:00:44 |
357135
|
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"Bill Johnson",
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|
Stack Exchange
|
Cartesian product of Banach spaces: all norms such that the inclusion is an isometry are equivalent?
Let $\mathcal{A}$ be an arbitrary (typically infinite-dimensional) Banach space with norm $\|\cdot\|_{\mathcal{A}}$ and let $\mathcal{A}^{n}$ be its Cartesian product. I came across the following statement and wonder whether it is true or not:
All norms $\|\cdot\|_{\mathcal{A}^{n}}$ on $\mathcal{A}^{n}$ such that $\|(0,...,0,\cdot,0,...,0)\|_{\mathcal{A}^{n}}=\|\cdot\|_{\mathcal{A}}$ for all $i=1,...,n$ (i.e. the inclusion is an isometry) are equivalent.
It is clear that for every $x=(x_{1},...,x_{n})\in\mathcal{A}^{n}$ it holds:
\begin{equation} \|x\|_{\mathcal{A}^{n}}\leq\sum_{i=1}^{n}\|x_{i}\|_{\mathcal{A}}=:\|x\|_{1}
\end{equation}
Therefore, $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{\mathcal{A}^{n}}$.
What what about the other direction? Is it true?
Thanks for your help.
It is actually not true in general that such a norm $\Vert \cdot \Vert_{\mathcal{A}^n}$ must be complete, despite the fact that the contrary is presented as fact in reputable sources in the literature (see, e.g., Section B.4.11 of Albrecht Pietsch's book Operator Ideals [the version published in 1980 by North-Holland] for the case $n=2$]).
A reference for the fact that $\Vert \cdot \Vert_{\mathcal{A}^n}$ need not be complete is the paper of Eve Oja and Peeter Oja, On the completeness of Cartesian products of Banach spaces (in Russian). Acta et commentationes Universitatis Tartuensis, 661 (1984), 33−35. The English summary of the Oja-Oja paper can be read here.
I think the Oja-Oja paper is hard to come by, so when I learnt of its existence many years ago I believe I just derived my own example to satisfy myself. The Oja-Oja paper also notes that equivalence (hence also completeness) is assured when an additional condition is assumed. In particular it follows easily from the triangle inequality that if there exists a constant $c>0$ such that $$ \Vert (-x_1,\ldots,-x_{i-1},x_i,-x_{i+1},\ldots,-x_n)\Vert_{\mathcal{A}^n}\leq c \Vert (x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n)\Vert_{\mathcal{A}^n}$$ for all $x_1,\ldots, x_n\in\mathcal{A}$ then $\Vert\cdot\Vert_{\mathcal{A}^n}$ is $(1+c)$-equivalent to the norm $\Vert\cdot\Vert_1$ as defined in your question.
(If I have time I might come back to this later and give some details of how to construct a counterexample, but I recall that it's not particularly difficult).
Edit: Bill Johnson's comment below notes a nice way of obtaining a counterexample.
Wow! I didn't know this!
Take $H$, $K$ to be quasi-complementary, non complementary subspaces of a Hilbert space $J$ and norm $H \oplus K$ with $|(x,y)| := |x+y|_J$.
@BillJohnson: thanks, Bill! That's a nice, simple way of stating a counterexample; I'd had a specific such counterexample in mind (i.e., of quasi-complemented, non-complemented subspaces of $\ell_2$), but hadn't thought about it in such general terms.
@NikWeaver: I seem to recall that I discovered it by accident, when trawling through Eve Oja's papers on Zentralblatt Math as a graduate student. I made a mental note of it at the time ("the norm on a direct sum need not be 'like' a direct sum of norms"), but haven't used it since - until now!
|
2025-03-21T14:48:30.280679
| 2020-04-11T11:14:39 |
357138
|
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|
Stack Exchange
|
Solvability of a system of polynomial equations
What can be said about the solutions of, resp. solving, the following system polynomial equations, in which the $x_i$ and $y_j$ are the variables and $c_{i,j},\,d_i\in\mathbb{R}$ are constants:
$$\begin{matrix}
c_{0,1}x_1+\,\cdots+\,c_{0,n}x_n&=&d_0\\
c_{1,1}x_1y_1+\,\cdots+\,c_{1,n}x_ny_n&=&d_1\\
c_{2,1}x_1y_1^2+\,\cdots+\,c_{2,n}x_ny_n^2&=&d_2\\
\vdots \\
c_{i,1}x_1y_1^i+\,\cdots+\,c_{i,n}x_ny_n^i&=&d_i\\
\vdots\\
c_{m,1}x_1y_1^{m}+\,\dots+\,c_{m,n}x_ny_n^{m}&=&d_{m}
\end{matrix} $$
knowing the conditions and algorithms for efficient numeric or symbolic calculation of the solutions would be the precondtion for an algorithm for non-polynomial splines as requested in this MO question Non-polynomial splines, a non-linear problem
For the special case where all $c_{i,j}$'s are equal to 1 and $m=2n-1$, take a look at Ramanujam's paper: http://ramanujan.sirinudi.org/Volumes/published/ram03.pdf. Needless to say, it is an ingenious method. The steps are the following:
(1) The key idea is to recognize that the coefficients (w.r.t $\theta$, upto the $2n-1$ order) of $\sum_{k=1}^n\frac{x_k}{1-\theta y_k}$ in the series expansion would be the LHS of the nonlinear equations.
(2) Now, $\sum_{k=1}^n\frac{x_k}{1-\theta y_k} = d_1 + d_2\theta + \cdots + d_{2n}\theta^{2n-1} + \cdots = \frac{A_1 + A_2\theta + \cdots + A_{n-1}\theta^{n-1}}{B_1 + B_2\theta + \cdots + B_{n-1}\theta^{n-1}}$.
(3) Multiplying by $B_1 + B_2\theta + \cdots + B_{n-1}\theta^{n-1}$ on either side and comparing coefficients, gives linear equations in $A_i$'s and $B_i$'s.
(4) If a solution exists, one can then determine partial fractions (in $\theta$), and the coefficients would be the answers.
For your case, you might want to think on the same lines with Pade approximants.
All this is only a suggestion for the most general case. Hope it helps.
the linked paper seems to be the door-opener and appetizer for tackling the more general problem; thank you very much for unearthing the reference!
The link in this answer seems to be broken now...
this https://coding.yonsei.ac.kr/ramanujan_paper.pdf is an alternative link to Ramanujan's paper
|
2025-03-21T14:48:30.280846
| 2020-04-11T11:39:48 |
357139
|
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|
Stack Exchange
|
The product of non-primitive matrices with zero positions in common
Def.[1]:
A non-negative $n \times n$ matrix $A$ is called a non-primitive if there is no an integer $k$ such that all entries of $A^k$ are positive.[1]
Def.[2]:
Let ${\bf A}=(a_{i,j})$ and ${\bf B}=(b_{i,j})$ be two $n \times n$ non-negative matrix over $\mathbb{R}$. Let the positions of zero entries of ${\bf A}$ and ${\bf B}$ be the same ( if $a_{i,j}=0$ then $b_{i,j}=0$ and vice versa ). Then, we say all zero positions of ${\bf A}$ and ${\bf B}$ are in common.
Let ${\bf A}_1,{\bf A}_2, \cdots, {\bf A}_k$ be $n \times n$ non-primitive matrices such that all zero positions of ${\bf A}_i$'s are in common. Let ${\bf B}=\prod_{i=1}^k{\bf A}_i.$
My question:
How to show that the matrix B is a non-primitive matrix.
To decide whether a matrix with nonnegative entries is primitive or not, all you have to know is where the positive entries are – their exact values are irrelevant.
If $C$ and $D$ have positive entries in the same places, then $CD$ and $C^2$ will have positive entries in the same places. More generally, if $A_1,A_2,\dots,A_k$ have positive entries in the same places, and $B=A_1A_2\cdots A_k$, then $B$ has positive entries in the same places as $A_1^k$. If $A_1$ is primitive, then so is $A_1^k$, so $B$ is primitive. And if $A_1$ is not primitive, then neither is $A_1^k$, so $B$ is not primitive.
Thanks for your simple and nice answer.
|
2025-03-21T14:48:30.280964
| 2020-04-11T12:55:02 |
357146
|
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|
Stack Exchange
|
Inequality for $AB + BA$ when $A,B\geq0$, reference request
Let $A,B\geq0$ be positive semidefinite matrices of arbitrary size $n\times n$. Denote by $\alpha$ and $\beta$ their largest eigenvalues.
It is well-known that the eigenvalues of the expression $AB + BA$ are bounded by [F. Zhang, "Matrix Theory", Sec. 7.2]
$$
-\frac{1}{4}\alpha\beta I \,\,\leq\,\, AB + BA \,\,\leq\,\, 2\alpha\beta I.
$$
I can show that
$$
- \rm{tr}(AB)I - \rm{tr}(A)B - \rm{tr}(B)A \,\leq\, AB + BA \,\leq\, \rm{tr}(AB)I + \rm{tr}(B)A + \rm{tr}(A)B \quad (\dagger)
$$
Now normalize to $\rm{tr}(A) = \rm{tr}(B) = 1$. When $\rm{tr}(AB) = 0$, that is $A\perp B$ in terms of the Hilbert-Schmidt inner product, the expression reduces to
$$
-(A+B) \,\,\leq\, AB + BA \,\leq\,\, (A + B) \quad (\ddagger)
$$
These inequalities $(\dagger)$ and $(\ddagger)$ look rather simple and are for $n\geq 3$ even tight. However I failed to find them in the literature, including in Bernstein's book on "Matrix Facts" or in the books by Bhatia. Am I missing something, are these known or can they straightforwardly be derived from other known expressions?
edit: they are weaker than the simple sum-of-squares, $(A-B)^2 \geq 0$; see answer below.
It seems both bounds are much weaker than what can be obtained from a simple sum-of-squares: namely, for all Hermitian matrices $A$ and $B$, one has
$(A-B)^2 \geq 0$. Expand
$$ AB + BA \leq A^2 + B^2 $$
Replacing $A\rightarrow -A$ yields also the lower bound
$$ -(A^2 + B^2) \leq AB + BA $$
Because of $\rm{tr}(A) = \rm{tr}(B)=1$, these are stronger than both $(\dagger)$ and $(\ddagger)$ in the question.
|
2025-03-21T14:48:30.281108
| 2020-04-11T13:59:14 |
357152
|
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|
Stack Exchange
|
Provably undecidable number inequality?
Here is a question that popped into my head right as I fell asleep last night.
I was thinking about constructions of irrational numbers, like pi. I was wondering if there are two constructions (any constructions, the pair doesn't have to include pi itself) such that:
We have good reason to believe they are different numbers.
We have a demonstration that, if they are in fact different, then their difference must be small enough that there is no way we will ever find this difference. By this I mean, the number of calculations it would take to find the difference would take more computational resources than the energy of the Universe would allow.
Heads up, I'm an amateur computer science enthusiast, and am neither a mathematician nor a physicist, but this question involves a mix of all three, so some of the phrasing might be awkward, and the whole question could be conceptually confused. If you think it is, please let me know!
The inspiration for this question comes from reading about hash functions in software version control. It seems like one of the reasons that numbers are "useful" from a computational standpoint, is that they are easy to compare for equality. This probably has something to do with the ease with which they are placed in order. So the puzzle I was pondering was, if we know that numbers can be ordered by definition, do we also have numbers that we can't, in physical reality, put into an order?
Is this question reducible to some incompleteness argument? Has it been asked before? Apologies if this is too open-ended a question for Math Overflow. Any and all suggestions would be greatly appreciated.
What exactly do you mean by "find the difference"?
Since you can code proofs with numbers, you say that one number is, e.g. $\pi$ while the other is $\pi + \sum_i a_i/10^i$ where $a_i =1$ if the $i$ corresponds to a proof that $0=1$ (in ZFC say) and $a_i =0$ otherwise. Then you probably cannot prove (in ZFC) that these two numbers are distinct.
@user76284 you make a good point, that language is misleading/ambiguous. What I meant to say was, "we'd never find the first decimal at which point the two numbers were not equal". So, it would always look to anyone who was calculating the numbers as if they were equal up to that point. From your answer down below, I realized how the question I originally posed could also be interpreted as asking "will we ever find the exact value of their difference", which is an equally interesting question! I'll respond in more detail to your answer when I get a moment to follow up today. Thanks!
@ARG But do we have good reason to believe they are different numbers (i.e. that ZFC is inconsistent)? Also, to be precise, if ZFC is consistent it can’t prove they’re the same, but if it is inconsistent it can prove they’re distinct (since it can prove anything).
@MaxWall Does 0 and the reciprocal of googolplex satisfy your criteria?
@user76284 it does not matter too much. There are tons of conjectures and statements about numbers with statement $n$ being true if $a_n=1$ and false if $a_n =0$. For each mathematician (or group of mathematician) [i'm a bit joking here, but hopefully you get the point], you can enumerate the statements which he believes in [or does not] but cannot yet prove. And you got a nice number for each mathematician or group of them. By starting with the most obvious statement, you can make the number as close to each other as you want. There are also conjectures with absurdly big counterexamples...
then pick $a_n=1$ if this integer (it could be the dimension, the numbers of cell in some CW-complex, the number of vertices in a graph, etc...) is a counterexample to the conjecture. Since you know the conjecture is false you know the numbers are distinct, but since the counterexamples are huge, you'll never know which digit is distinct.
I think standard examples are like this ... Enumerate the cases of Goldbach's conjecture. Let $a_n = 0$ if the $n$th case is true, and $a_n = 1$ if false. Consider the number $A = \sum_{k=1}^\infty a_k 2^{-k}$ with the binary expansion $(a_n)$. We may compute $A$ as accurately as we like. But $A=0$ is equivalent to Goldbach's conjecture.
Repeat this with many other undecidable propositions instead of Goldbach.
But this example can die if/when Goldbach's conjecture is ever settled. Maybe better to take $a_n=1$ if the $nth$ case of Goodstein's Theorem is true. Then we know first that $A=1$, and second that with the tools of ZFC, we will never be able to prove it. And we can be pretty confident that for we will never have the computational resources to distinguish $A$ from, $1-\epsilon$ where $\epsilon$ is, say, 1/10^10^10.
Gerald's Goldbach example absolutely captures the intuition I was trying to express, but I have one small concern. While Goldbach is unproven, isn't the intuition that it's likely true? So wouldn't condition 1 of my question be unsatisfied? That we have no reason to think A is not zero? Maybe condition 1 is too vague (what is a good reason?)... I love the Goodstein idea too (new to me), and I think it captures more of the flavor of my second condition, but doesn't it suffer from the same problem? These are both great, by the way! Thanks!
@StevenLandsburg ZFC proves Goodstein's theorem - it's only undecidable in PA. Of course, using Gödel's theorem you can do a similar thing for ZFC or any other theory $\mathcal{T}$ to which Gödel's theorem applies, by encoding "n is not the Gödel number of a proof of 0=1 in $\mathcal{T}$".
@RoberFurber: Gah. Yes, I meant to say PA. Thanks.
I think your question needs some clarification (in particular, what you mean by "find the difference"). Let $x$ be Chaitin's constant for some prefix-free universal computable function.
We have good reason to believe $x$ and 0 are different numbers (because they are, in fact, different numbers).
Their difference $x - 0$ is uncomputable, so in a sense we will never "find the difference". (A real number is computable iff its Dedekind cut is at level $\Delta_1$ of the arithmetical hierarchy.)
Another example: Let $x$ be a super-omega. Then $x - 0$ is not even limit-computable (i.e., in $\Delta_2$).
Another example: Fix an enumeration of first-order formulas that define arithmetical sets. Consider the real number $x \in [0, 1]$ whose $n$th binary digit after the decimal point is 1 iff $n$ is not in the set defined by the $n$th formula. Then $x - 0$ is not even arithmetical. (See here.)
Alternatively, we could use a construction analogous to Rayo's number: Fix a theory T. Let $\varepsilon$ be a positive real number smaller than any positive real number that can be named by an expression in the language of T with at most googolplex symbols. Then, by definition, there does not exist a positive real number $\delta$ that can be named by an expression with at most googolplex symbols which lower-bounds $\varepsilon$:
$$ 0 < \delta < \varepsilon$$
|
2025-03-21T14:48:30.281691
| 2020-04-11T15:27:41 |
357159
|
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|
Stack Exchange
|
Set of primes $p_{1}\equiv 3 \bmod p_{2}$ such that $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod p_1$ with $p_1,p_2\equiv 3\bmod 4$?
let $p_1$ and $p_2$ be positive primes such that $p_1,p_2 \equiv 3\bmod 4$
and $\phi$ is the Euler totiont function , I want to find the Set of primes $p_{1}\equiv 3 \bmod p_{2}$ such that $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod p_1$ ? I can't find such pairs satisfying the titled claim , I have used the fact that : $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod \frac{{p_1}-3}{p_2}$ but this dosn't give any thing to determine all pairs $(p_1,p_2)$ for which the titled claim w'd be satisfied .
Edit I have Edited the question to avoid any complication and working with ordinary prime which are $3$ modulo $4$ in the same time gives the definition of Gaussian primes
Note:The motivation of this question is to know more about behavior of Euler totiont function with Gaussian primes
If $\frac{|z_1|-1}{|z_2|}$ is an integer, then $|z_{1}|\equiv 1 \bmod|z_{2}|$.
primes of which form?
@FedorPetrov primes of the form $z= b i$ (Gaussian primes) , I have montioned that
Why not just ordinary primes congruent to 3 mod 4, but multiplied by $i$?
Multiplied by i beacuse I meant Gaussian prime and it works also as you claimed
@FedorPetrov, Ok since it works as all positive primes which are 3 mod 4 are Gaussian I think I can edit the question just I take primes such that conguent to 3 mod 4
|
2025-03-21T14:48:30.281811
| 2020-04-11T15:40:15 |
357160
|
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|
Stack Exchange
|
What are some efficient ways to keep a note of results when reading a research article in mathematics?
I learn and produce mathematics. In that process, I had to read quite a number of research articles.
Question :
What are some efficient ways to keep a note of results when reading a research article in mathematics?
I keep a note of definitions (in detail) and results (with out proofs) for each paper I read.
Are there any other efficient methods to keep track of what results does a paper contain?
Reading introduction or abstract of a paper does not count as they are written in cryptic manner.
I asked the same question in Academia but it was not sufficient.
I find your question quite confusing. What is "keep a note of results of a research article"? The article precisely keeps a note of your research. If you mean keeping note of side results, auxiliary proofs, which are not included in the article, please be more explicit. I have no idea what's your point on introductions and abstracts and why you find them "cryptic", if of any relevance.
@YCor "your research".. Not my research.. This question is asking about reading some one else's research article... I have specified in the edit .. Is it less confusing now?
Introductions really ought not to be written in a cryptic manner - for many articles, finding the introduction impenetrable is a good sign that one is not yet ready to read the article (although there are also plenty of papers with horribly written introductions).
I know a colleague who has boxes filled with small, A6 sized, cards on which he writes the generic information about the article (e.g. title, author, year, etc) and then a few notes about the results which he found useful in there. He keeps them sorted in there in various ways -- I really like this offline approach myself, and have tried setting one up myself. The key point about it is that the info he puts in is what he finds interesting or relevant to his own research, not just a summary of the articles.
@NoahSchweber I am not sure if I could have written more clearly.. This feeling of introduction and abstract being "cryptic" comes after I read the full article.. I feel may be the author could have said one/two lines extra about some thing in the introduction itself... But I see this is the procedure followed in most of the papers :D
@Carl-FredrikNybergBrodda If your colleague do not mind, can you post a photo of how it is written as an answer...
@PraphullaKoushik Sure -- but due to the current situation it'll probably be quite a while before I can see him to ask!
@Carl-FredrikNybergBrodda It is ok. When you can, ask him :)
In my own notes, my general approach for summarizing a paper is the following:
I copy the abstract verbatim, aside from correcting blatant typoes. (Every so often there's an exception here for papers with ridiculously long abstracts - which sort of defeats the point of an abstract, after all!)
I generally don't copy definitions exactly; instead I just record the term(s) introduced, the type(s) of object(s), and the intended role. (That said, particularly snappy/amazing definitions do also get stated in full.)
I copy the main results verbatim, again aside from correcting blatant typoes. I use my own judgment as to what constitutes a "main theorem" here, and that doesn't always coincide with the authors' although they usually do.
I never copy proofs in their entirety, but I don't fully omit them either: for each theorem I mention I include a summary in 1-10 sentences of its proof. This summary may only be limited to a description of the major obstacles; I err on the side of brevity. This is often quite hard, but is a super useful exercise.
(In each case above I also record the page number(s) of course.)
Finally, I also record some information about the background of the article. I write a summary of the bibliography (or at least part of it), recording how the various sources are relevant. I'll also record the general gist of each paper, even if it goes beyond how it's used at the moment: if Paper 1 just needs one lemma from Paper 2, I'll still give a very brief summary of Paper 2 itself as well as mentioning the lemma. This summary is again on the order of a few sentences, usually taken from the abstract or introduction (I don't need to read Paper 2 itself for this). Finally, I'll make a list of a small number of very relevant sources from the bibliography. This list could change over time as I understand the topic better, and often is pretty meaningless initially.
This is quite a bit of work in general, but note that the only step which requires deeply engaging with the article is the proof-summary bit; the rest can be done fairly mechanically or instinctively. The result is usually a 1-3 page document. A huge amount of information is generally lost by doing this, of course, but I find the whole process extremely useful and the result is still quite helpful in recalling the paper much later.
Thanks... This is definitely a good approach... Do you mind to share an example? I can send an email if you do not want to put it here,,,
@PraphullaKoushik I prefer not to - they're fairly messy, and I don't like sharing unpolished stuff in general. (And I don't have the time at the moment to polish one up.)
That is understandable.. Never mind :) :) Thanks for the answer..
|
2025-03-21T14:48:30.282200
| 2020-04-11T15:55:53 |
357161
|
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|
Stack Exchange
|
Every element of $A$ and $B$ differ in at least $k$ positions
Let $m,n$ be positive integers, $m,n>1$ and $X = \{(x_1,x_2, ..., x_m) \in \mathbb{Z}^m :1 \le x_i \le n, \forall 1 \le i \le m\}$.
$A$ and $B$ are two disjoint subsets of $X$, such that if $a \in A$ and $b \in B$ then $a$ and $b$ differ in at least $k$ positions (i.e. $a=(a_1,a_2, ..., a_m)$, $b=(b_1,b_2, ..., b_m)$ and there are at least $k$ values $i$ such that $a_i \ne b_i$).
What conditions should $|A|$ and $|B|$ satisfy?
Given that there are positive integers $r, s$ satisfying $|A| \ge r$ and $|B| \ge s$, what is the miminimum value $n$ could have as a function of $k,m,r,s$?
For example, let $m=2$, $X = \{(x_1,x_2) \in \mathbb{Z}^2 :1 \le x_i \le n, \forall 1 \le i \le 2\}$.
If $k=1$ then the relation $|A|+|B| \le n^2$ is sharp.
If $k=2$ it's not hard to see that the following relation holds and it's sharp: $\sqrt{|A|} + \sqrt{|B|} \le n$ (note that if $A_{1}$ is the set of the possible values for $a_1$, and defining $A_2$, $B_1$, $B_2$ in the same way, one has $|A_1|+|B_1| \le n$ and $|A_2|+|B_2| \le n$, where $|A| \le|A_1||A_2|$ and $|B| \le|B_1||B_2|$).
However the problem becomes significantly harder even for $m=3$.
Does anyone know any reference where this problem has been studied or know how to proceed in the general case?
I'm interested on any known results, even for small values of $m$, such $m=3$ or $m=4$.
Here is a construction. Let $m=m_a+m_b,$ where you can choose $m_a\geq 1$ and $m_b\geq 1$ as you wish and consider $\mathbb{Z}^m=\mathbb{Z}^{m_a} \times \mathbb{Z}^{m_b}.$
First work over $\mathbb{Z}_2.$ Choose $k=k_a+k_b$ as the number of coordinates that you want the length $m$ codewords to be different. Define two binary codes as
$$
C_a=\{(v|| \mathbb{0} ): w_H(v)\geq k_a\}
$$
where $||$ denotes concatenation, and note that you have $2^{m_a}-V_{k_a-1}(m_a)$ codewords in $C_a$ where $V_r(\ell)$ denotes the volume of the binary Hamming sphere of radius $r$ in $\ell$ dimensions. This volume can be approximated by the entropy function depending on the ratio $k_a/m_a,$ if you wish, asymptotically. For example, it is easy to see that if $k_a\approx m_a/2$
you can have $|C_a|\gg 2^{m_a}.$
A similar argument can be used for $C_b$ defined below:
$$
C_b=\{( \mathbb{0}||u ): w_H(u)\geq k_b\}
$$
So the concatenated binary code
$C_2 \subset \mathbb{Z}_2^m$ given by $C_a \times C_b$ has $\gg 2^{m},$ codewords asymptotically.
Now even if you confine yourself to bounded integers, $\mathbb{Z} \bigcap [0,2v],$ you can use the above code and its translates by 2, say
$$
C_{\mathbb{Z}}=\bigcup_{j=1}^v (C_2+2j)
$$
to get an even larger code. Here we use the notation $C_2+2j=\{c+(2j,2j,\ldots,2j): c \in C_2\}.$
It seems to me choosing $m_a=\lfloor m/2 \rfloor,$ may be optimal in the sense of getting the largest code, for a given $v.$
Old Answer for Reference:
Note that there may be a better way of doing this, but your question can be addressed within coding theory as below.
I will use $d$ instead of $k$ (coding theory notation). Take any $[n,k,d]$ linear code $C$ over $GF(q).$ So you will be restricted to prime powers for your $m=q,$ and thus $|C|=q^k.$
You can then use the known bounds on these parameters, Hamming, Singleton, Gilbert Varshamov etc.
Decompose $C=A \cup B,$ where $A \cap B=\emptyset.$ Then try to optimize whatever measure $\alpha |A|^a+\beta |B|^b,$ or similar by selecting $|A|,$ since $|B|=q^k-|A|.$
In fact, since the code is linear, all its translates $C_u=C+u$ are distinct (let $C_0=C,$ and each will satisfy the same distance properties as a set. If $d$ is much smaller than $n,$ then translating by a vector $u$ of Hamming weight greater than $d$ will enable you to use the above approach and decompose $C_u \cup C_0,$ improving your parameters with respect to these bounds.
For the binary alphabet, you can try to use set systems with controlled symmetric differences from design theory, taking your vectors to be the characteristic functions of those sets.
|
2025-03-21T14:48:30.282475
| 2020-04-11T16:05:01 |
357163
|
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|
Stack Exchange
|
Invariant subspaces of Markov operators
I am currently working on some kind of graph theoretic problem and the following question came up:
Suppose you have a Markov operator $T$ on $\ell^\infty$, that is a positive, bounded operator such that $T \mathbf 1 = \mathbf 1$, where $\mathbf 1$ denotes the constant $1$-sequence.
Now suppose that the subspace $c_0$ of sequences coverging to $0$ is $T$-invariant, i.e, $T c_0 \subseteq c_0$. How can this invariance be characterized?
In this setting, one can clearly think of $T$ as a positive infinite matrix such that the sum of each row is equal to $1$. I would suppose that one would have this property if one imposes some kind of decay on the coefficients of the matrix which is uniform for all rows but:
I am not sure if that would be a natural condition for the system I am interested in.
Such a assertion would be maybe sufficient but is it necessary?
So is their a known characterization for Markov operators on $\ell^\infty$ which leave $c_0$ invariant? I would assume that is a question that turns up naturally in ergodic theory and/or probability theory.
Moreover, one could generalize this question to the more general setting of AM-spaces with unit. Suppose that $T$ is a Markov operator on some AM-space $E$ with unit, that is a positive, bounded operator on $E$ such that $T \mathbf 1 = \mathbf 1$, where $\mathbf 1$ denotes the unit of $E$. Suppose that $F$ is a closed sublattice such that $TF \subseteq F$.
Is there a useful characterization of the set of Markov operators that leave $F$ invariant?
Good question! But I have to contradict your claim that "one can clearly think of $T$ as a positive infinite matrix": Let $\varphi \in (\ell^\infty)^*$ denote you favourite Banach limit and define $Tf = \langle \varphi, f \rangle \mathbf 1$ for each $f \in \ell^\infty$. Then $T$ satisfies all your properties, but it cannot be represented as a matrix since it maps all canonical unit vector to $0$.
Another comment. Even if your operator is represented by a matrix, a decay condition isn’t really going to work for you. For example if the (I,j) entry of the matrix is $2^{-j}$, then the rows decay nicely, but it doesn’t map $c_0$ into $c_0$.
$T$ preserves $c_0$ if and only if $T e_n \in c_0$ for all n, where $e_n$ is the canonial basis. In case of an infinite matrix, this means that each column tends to 0.
|
2025-03-21T14:48:30.282659
| 2020-04-11T16:55:48 |
357165
|
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|
Stack Exchange
|
Claim on divisibility of a power sum
Let $x,y,z$ are integer and $x,y>0$
Define $S(x,y)=1^y+2^y+3^y+...+x^y$
Can it be shown that
If given $z\ne0$ then $S(x,y)\equiv z\pmod{x}$ have finitely many solution of $x$ with respect to $y$.
Example
Let $z=\pm1$ and $y\equiv 1\pmod2$ then $x=1,2$
"...have finitely many solution of $x$ with respect to $y$." What does this mean?
@Wojowu $y$ and $z$ given then there are finitely many $x$ satisfy $S(x,y)\equiv z\pmod{x}$.
This is Faulhaber’s Formula which include Bernouli numbers such that are congruent ot Z and fermat little theorem w'd be work here as well
From Faulhaber's formula we can see that $S(x,y)*(y+1)*lcm _{2i\leq y} den (B_{2i})$ is divisible by $x$, so for $x>z*(y+1)*lcm _{2i\leq y} den (B_{2i})$ our expression $S(x,y)$ gives remainder more than $z$ modulo $x$ provided $z\not =0$.
|
2025-03-21T14:48:30.282751
| 2020-04-11T17:22:55 |
357166
|
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|
Stack Exchange
|
The stabilizers of the canonical boundary action of hyperbolic groups
My question is that
Is every stabilizer of the canonical boundary action of a hyperbolic group on its Gromov boundary a finitely generated group?
I guess every stabilizer is a (finitely generated) virtually cyclic group, but I do not have a proof nor a reference.
More generally, let G be a countable group that is relatively hyperbolic to subgroups $P_1,\ldots, P_n$. Under which conditions we can conclude that
every stabilizer of the canonical boundary action on its boundary is a (finitely generated) virtually cyclic group or finitely generated virtually abelian group?
Please see section 2 in https://arxiv.org/pdf/1502.04834.pdf for relative hyperbolic groups and their boundaries.
Thank you very much!
Yes (to your first question), it's virtually cyclic. Indeed, it has a loxodromic element (every infinite subgroup of a hyperbolic group has a loxodromic, = infinite order, element). So, if infinite, its action at infinity is axial or focal. In both case, it's quasi-convex. Focal is not possible in this context (it would be if you were considering locally compact compactly generated groups). I'll leave to others to write this in a more standard language (although my wording is more faithful to Gromov's original approach).
Hyperbolic dynamics is unrelated to hyperbolic groups (at least not in an obvious way), I removed the tag.
Thank you Yves. Is it possible to provide me a reference?
@m07kl If I understand your questions correctly, an answer follows from the classification of acylindrical actions on hyperbolic spaces; see Theorem 1.1 in Osin's paper arxiv:1304.1246.
@AGenevois: Thank you and I will take a look into it
@AGenevois And you have to add that the stabilizers of parabolic limit points are exactly the peripheral subgroups, which follows directly from Bowditch definition of relative hyperbolicity in terms of geometrically finite actions.
@m07kl Why would the stabilizers be virtually abelian ? Maybe you are assuming that the peripheral subgroups are virtually abelian, but you didn't say so. Using the acylindrical action on the coned-off graph and AGenevois' comment + the fact that stabilizers of parabolic limit points are exactly the peripheral subgroups, you can deduce that the stabilizers are exactly virtually cyclic or are conjugates of the groups $P_1,...,P_n$.
@M.Dus: Right, I assume that P_1..P_n are virtually abelian or virtually cyclic.
Do you mean subconjugate to one of P_n?
@m07kl What do you mean by subconjugate ? I mean that for every parabolic point $\xi$ in the boundary, there exists $g\in G$ and there exists $1\leq k \leq n$ such that the stabilizer of $\xi$ is $gP_kg^{-1}$. BTW, I'm assuming that the boundary you are considering is the Bowditch boundary, right ?
@M.Dus: Thank you. "What do you mean by subconjugate ?" I mean the stabilizer can be a subgroup of $gP_kg^{-1}$?
How about the stabilizer of another points which are not parabolic in the boundary (yes, we talk about Bowditch boundary)?
@m07kl Oh okay. No the stabilizer of parabolic limit points are exactly the $gP_kg^{-1}$, not subgroups of them. This is getting to long for a comment, I'm writing an answer.
This is following my comment, which was getting too long.
Using the notations of the paper you are citing, there are two kinds of points in the boundary : elements of the Gromov boundary $\partial \Gamma$ of the fine graph $\Gamma$ on which $G$ acts and elements in $V_\infty$, which are vertices of $\Gamma$ of infinite valence. The former are called conical limit points and the later are called parabolic limit points.
This boundary equivariantly agrees with the Gromov boundary $\partial X$ of any proper Gromov hyperbolic space on which $G$ acts via a geometrically finite and minimal action (if you want to remove the word minimal, you need to take the limit set $\Lambda G$ of $G$ instead of $\partial X$). This is Proposition 9.1 combined with Theorem 9.4 in Bowditch's paper relatively hyperbolic groups. In particular, you can choose any fine graph $\Gamma$ on which $G$ acts which satisfies Bowdtich's Definition 2 (which is the definition in the paper you're referring to).
To simplify the following, choose $\Gamma$ to be the coned-off graph with respect to the parabolic subgroups $P_1,...,P_n$, or if you prefer Osin's formulation, the Cayley graph $\mathrm{Cay}(G,S\cup P_1\cup ... \cup P_n)$, where $S$ is any finite generating set, which is quasi-isometric to the coned-off graph. Then, the action of $G$ on this graph $\Gamma$ is acylindrical (this is Proposition 5.2 in Osin's paper that AGenevois indicated in their comment).
Now take any point $\xi$ in the Gromov boundary of $\Gamma$ and let $H$ be its stabilizer. Then, the action of $H$ on $\Gamma$ also is acylindrical and $H$ cannot contain infinitely many independent loxodromic elements, so it is virtually cyclic, by Theorem 1.1 of Osin's paper. This settles conical limit points. On the other hand, let $\xi$ be a parabolic limit point. By point (3) of Bowditch's Definition 2, the stabilizer of $\xi$ is exactly one of the peripheral subgroups, that is, with our notations, is one of the conjugates of the $P_k$.
Thank you very much for your answer. So the answer to my question is that
1)When P_i are all virtually cyclic, so are all stabilizer groups.
") When P_i are all finitely generated virtually abelian, so are all stabilizer groups :)
@m07kl You're welcome and yes you're right ! There is one small thing missing in my argument which I had implicitely in mind. Theorem 1.1 of Osin's paper tells you that the stabilizer of a conical limit point either is virtually cyclic or has bounded orbits on the coned-off graph. To conclude that the group is finite in the later case you need to say that if an element acts with bounded orbits on the coned-off graph, then for the action on the space $X$, it is either eliptic (so has finite order) or is parabolic (which cannot happen here since it already fixes a conical limit point).
@m07kl By the way, if you're interested in groups that are hyperbolic relative to infinitely generated subgroups, you might take a look at Gerasimov and Potyagailo's paper https://arxiv.org/abs/1008.3470. I'm not used to this framework, but I'm pretty sure that the result in my answer still holds : stabilizers are either virtually cyclic or are a conjugate of one of the $P_k$.
|
2025-03-21T14:48:30.283156
| 2020-04-11T18:18:38 |
357167
|
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|
Stack Exchange
|
What is the definition of homotopy flat connections?
What is a definition of a homotopy flat connection - in the context of differential forms with values in a dg algebra
Could you give a reference where such an object is referred to, or elaborate a bit on what you might expect from such a definition?
I think, it should be "homotopy flat" connection.
One possible definition (which yields an ∞-Riemann–Hilbert correspondence) is in Block and Smith's paper A Riemann--Hilbert correspondence for infinity local systems
Andy, It's precisely because I'd like to cite it if it exists,
One place where this notion appears (without a formal definition) is in a remark after Corollary 4 in
Cohen, Ralph L.; Stacey, Andrew, Fourier decompositions of Loop bundles, Goerss, Paul (ed.) et al., Homotopy theory: relations with algebraic geometry, group cohomology, and algebraic (K)-theory. Papers from the international conference on algebraic topology, Northwestern University, Evanston, IL, USA, March 24–28, 2002. Providence, RI: American Mathematical Society (AMS). Contemporary Mathematics 346, 85-95 (2004). ZBL1068.55016.
Remark. The homotopy type of the map of based loop spaces, $\Omega f_\zeta: \Omega M \to U(n)$ can be obtained by taking the holonomy map of a connection on $\zeta$. Therefore this corollary can be interpreted as saying that in order for $L\zeta$ to have a Fourier decomposition, $\zeta$ must admit
a “homotopy flat” connection.
Deciphering what they are saying, my guess is that a collection $\nabla$ on a (say) principal $G$-bundle $P\to B$ with simply-connected base $B$ is homotopy flat if the holonomy map
$$
\Omega(M)\to G
$$
is null-homotopic. (In the case of a flat bundles this map is, of course, constant.) Maybe one of the authors, Andrew Stacey (@user:45) can confirm (or deny) that this is what they meant.
I always thought that in the flat case, the holonomy map $\Omega M \to G$ should factor as $\Omega M \to \pi_1(M) \to G$, where the first map is obtained by taking $\pi_0$ and the second is a homomorphism. Equivalently, it is locally constant.
@JohnKlein: In the answer I was explicitly assuming (as the authors of the quoted paper) that the base is simply-connected. Otherwise, the right thing to do is to restrict to null-homotopic loops or work in the universal covering space.
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2025-03-21T14:48:30.283359
| 2020-04-11T18:56:28 |
357168
|
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|
Stack Exchange
|
Subring of quaternion algebra
I am following the book Introduction to Quadratic Forms over Fields by T. Y. Lam. In section VI.2, the author proves that, over an arbitrary local field $F$, there is a unique quaternion division algebra, namely $D=\left(\frac{\pi,u}{F}\right)$ where $\pi$ is a uniformizer and $u$ is such that $F(\sqrt{u})$ is the unique unramified quadratic extension of $F$. (This has been proved before for nondyadic local fields; the point here is to extend the statement to the dyadic case.) The proof starts as follows:
Step 1: Define a homomorphism $w':E\setminus\{0\}\to\mathbb{Z}$ by $w'(x)=v(\mathrm{N}(x))$, where $\mathrm{N}$ denotes the (anisotropic) norm form of $E$. Let $d$ be the unique positive integer such that $w'(E\setminus\{0\})=d\mathbb{Z}$. Since $w'(\pi)=v(\mathrm{N}(\pi))=v(\pi^2)=2$, $d$ must be either $2$ or $1$. We may "normalize" $w'$ by setting $w(x)=\frac{w'(x)}{d}$ for $x\in E\setminus\{0\}$, and (by convention) $w(0)=\infty$. Let $B=\{x\in E:w(x)\geq 0\}$, which is a subring of $E$ (...).
I don't immediately see that the part on bold is true (namely, the fact that $E$ is closed under taking sums). Is there any obvious reason why this is so? (Non-obvious reasons are also welcome.)
"I don't immediately see" -- as long as you saw it you can congratulate yourself. (The "immediately" notion belongs to trivia).
@WlodAA Please read and understand something like this.
@RobertFurber, what does your link have to do with @WlodAA's comment, which seemed to be purely to the effect that it's OK if the amount of work required to understand a claim is not proportional to the length of the claim?
@RobertFurber, I don't see how you have contributed to "Quality, Quantity, Relation, Manner"? Was your comment to the OP post helpful to user50139 (or to anybody at all? -- that was a rhetoric question since the answer is clear, the answer is "NO"). #### Sorry, everybody, for getting distracted by RF.
That's because $w$ is a valuation (the main point being the nonarchimedean triangle inequality), which can be proved by reducing to the commutative case; see for instance Lemma 13.3.2 in https://math.dartmouth.edu/~jvoight/quat.html.
For the sake of completeness I will give the argument here:
$w$ is a valuation on every commutative subfield of $E$.
$w$ is multiplicative by definition.
Let $a,b\in E$ with $b\neq 0$. Then $w(a+b) = w((ab^{-1}+1)b) = w(ab^{-1}+1)+w(b)$ by multiplicativity. Then $w(ab^{-1}+1) \ge \min(w(ab^{-1}),0)$ since $w$ is a valuation on $F(ab^{-1})$, and finally $\min(w(ab^{-1}),0)+w(b) = \min(w(a),w(b))$ again by multiplicativity. Put together we get $w(a+b)\ge\min(w(a),w(b))$ as desired.
|
2025-03-21T14:48:30.283940
| 2020-04-11T19:56:58 |
357171
|
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|
Stack Exchange
|
How to find the elliptical arc that corresponds to the cubic bezier curve
Let's assume I have a cubic bezier curve that is provided with A, B, C, D points, where
A is the start of the curve
B is the first control point
C is the second control point
D is the end of the curve.
The curve's parametric equation is given like this:
$x(t) = A_x + 3(B_x - A_x)t + 3(A_x - 2B_x + C_x)t ^ 2 + (3(B_x - C_x) + D_x - A_x) t ^ 3, 0\leqslant t\leqslant 1$
$y(t) = A_y + 3(B_y - A_y)t + 3(A_y - 2B_y + C_y)t ^ 2 + (3(B_y - C_y) + D_y - A_y) t ^ 3, 0\leqslant t\leqslant 1$
Let's assume this curve is (UPD: almost) identical to the arc of the ellipse with the center in the point $O$.
Question:
What is the best strategy to find the center $O$, radii and rotation of (UPD: the approximated) ellipse that represents the corresponding arc?
Unfortunately the hypothesis that the curve is
"identical to the arc of the ellipse" is impossible.
Bezier curves can come close enough to circular or elliptic arcs
to be visually indistniguishable from them; but a bezier curve
cannot exactly coincide with an arc of an ellipse.
If it did, for some ellipse $E$ given by an equation
$Q(x,y)=0$, then $x(t)$ and $y(t)$ would satisfy $Q(x(t),y(t))=0$
for all $t$ with $0 \leq t \leq 1$, and thus for all real $t$.
But then $E$ would contain points whose coordinates
$x(t)$ and $y(t)$ are arbitrarily large (by taking $t \to \infty$),
which is impossible $-$ unless both $x$ and $y$ are constants,
in which case the "curve" is just a single point.
0≤≤1 corresponds to the limits of $x$ and $y$ for the elliptic arc, so this puts the limits onto $x$ and $y$ that can be supplied into the ellipse equation (,)=0. Are you stating that even within those limited ranges x(0)≤x≤x(1) and y(0)≤y≤y(1) for the ellipse equation (,)=0 they are still not identical?
The point is that $Q(x(t),y(t))$ is a polynomial in $t$, so if it's
exactly zero for all $t$ in an interval such as $0 \leq t \leq 1$
then it must be identically zero $-$ whence $Q(x(t),y(t)) = 0$
holds for all real $t$, whether inside or outside the interval.
Thanks, I'll note that. Still, I am looking for the best approximation into the elliptic arc. For example, in this paper, the method for the inverse problem is described. The error of approximation of elliptic arc to bezier curve is negligible. I am looking for the same kind of method, but inverse to the one described in that paper.
I'm also interested in this. There seems to be many ways (posts and websites) for approximating a Bezier curve from an ellipse. So why not the other way?
|
2025-03-21T14:48:30.284162
| 2020-04-11T20:58:56 |
357174
|
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|
Stack Exchange
|
Cremona transformation $\sigma: \mathbb{P}^2 \dashrightarrow \mathbb{P}^2 $ and pushforward of divisor
Let consider the birational Cremona transformation $\sigma: \mathbb{P}^2 \dashrightarrow \mathbb{P}^2 $ defined by $$(X:Y:Z) \mapsto (X^{-1}:X^{-1}: Z^{-1})=(YZ:XZ:XY)$$.
In language of birational geometry the sections $YZ,XZ,XY \in H^0(\mathbb{P}^2, O_{\mathbb{P}^2}(2))$ induce it as
$$\sigma: \mathbb{P}^2 \backslash B = \mathbb{P}^2_{YZ} \cup \mathbb{P}^2_{XZ} \cup \mathbb{P}^2_{XY} \to \mathbb{P}^2$$.
with the base locus $B= \operatorname{Supp}V(YZ) \cap \operatorname{Supp}V(XZ) \cap \operatorname{Supp}V(XY) $ where $\sigma$ is not defined and $\mathbb{P}^2_{YZ}:= \{p \in \mathbb{P}^2 \ \vert \ (YZ)_p \neq 0 \}$ the non vanishing locus. Similar for $XZ$ and $XY$.
By construction $\mathbb{P}^2_{YZ}= \sigma^{-1}(D_+ (X)), \mathbb{P}^2_{XZ}= \sigma^{-1}(D_+ (Y)), \mathbb{P}^2_{XY}= \sigma^{-1}(D_+ (Z))$. Obviously $\sigma$ is not defined at points $(0:0:1), (0:1:0),(1:0:0) \in B$.
Up to now we introduced some notations; now the essential part of my question:
Let $V(F)=C \subset \mathbb{P}^2 $ be a curve defined by homogeoneous polynomial $F(X,Y,Z)= \sum_{i+j+k=n}a_{ijk}X^iY^jZ^k$ with $deg(F)=n$.
We assume that $C$ has at point $(0:0:1)$ multiplicity $d$, $(0:1:0)$ multiplicity $e$ and $(1:0:0)$ multiplicity $f$.
Recall, how multiplicity is defined: $(0:0:1) \in D_+(Z)$ and when we dehomogenize $F$ with respect $Z$ we get a new polynomial $f^Z$ in variables $x=X/Z,y=Y/Z$ that has structure
$$f^Z(x,y)= \sum_{i+j+k=n}a_{ijk}x^iy^j= \\ \sum_{i+j+k=n, i+j=d}a_{ijk}x^iy^j + \sum_{i+j+k=n, i+j >d}a_{ijk}x^iy^j =: f^Z_d +f^Z_{>d} $$
Recall again that by definition of multiplicity $d$ is minimal natural number with this property: for all $a_{ijk} \neq 0$ we have $i+j \ge d$. This is equivalent to say that $d$ is maximal with the property that the $d$-th power $(x,y)^d$ of the maxiaml ideal $(x,y) \subset k[x,y]$ contains $f^Z(x,y)$.
Analogously for multiplities $e$ and $f$ with respect to $(0:1:0)$ and $(1:0:0)$.
QUESTION: Why the pushforward of $C$ by $\sigma$ is the vanishing set
$$ \sigma_*C = (X^{-f}Y^{-e}Z^{-d}F(YZ,XZ, XY) =0) \ \ \ ?$$
By symmetry reasons it is clear that it suffice to show locally that this holds for restriction to affine open subscheme $D_+(Z)= \operatorname{Spec} \ k[X/Z,Y/Z] \subset \mathbb{P}^2$.
We know from construction that
$$\sigma_*C(D_+(Z))=C(\sigma^{-1}D_+(Z))= C(\mathbb{P}^2_{XY})= \\ C(D_+(XY))= C \vert _{D_+(X)}(D(y)) = k[y,z]_y/f^X(y,z).$$
Recall, $k[y,z]_y= k[y^{\pm 1},z]$ is the localization
by $y$. If we now denote
$H:= X^{-f}Y^{-e}Z^{-d}F(YZ,XZ, XY)$ and $S:= V(H)$, then $H$ localized
at $Z$ is $h^Z(x,y):=x^{-f}y^{-e} F(y,x, xy)$ and thus
$$S(D_+(Z))= k[x,y]/h^Z(x,y).$$
And we have to verify $k[y,z]_y/f^X(y,z)=k[x,y]/h^Z(x,y)$.
Although up to now I only used definitions (in correct way?) the equation I have to verify seems not to make any sense. I think that somewhere I have already done a hard error but I can't find it. Can anybody help? Is there an easier way to verify the claim in the question?
Thanks for your help.
I think that you need to assume that $F$ is not divisible by $X$, $Y$ or $Z$. This being done, the equation of $\sigma_*(F)$ on the open subset where $XYZ\not=0$ is given by $F(YZ,XZ,XY)$. Hence, $\sigma_*(F)$ is equal to
$$F(YZ,XZ,XY)X^aY^bZ^c$$
for some integers $a,b,c\in \mathbb{Z}$. As you want to remove $X=0$, $Y=0$ and $Z=0$ from the polynomial $F(YZ,XZ,XY)$, you can choose $a,b,c$ to be $\le 0$ and then define them as $-\alpha,-\beta,-\gamma$ where $\alpha,\beta,\gamma$ are the multiplicities of $F(YZ,XZ,XY)$ along $X=0,Y=0,Z=0$ respectively.
Let us do the case of $X$, the rest being symmetrical. Write $$F=\sum_{i\ge f} X^{n-i} a_i(Y,Z)$$ where each $a_i$ is homogeneous of degree $i$ and $a_f\not=0$ (as $f$ is for you the multiplicity of $F$ at $[1:0:0]$). Then $$F(YZ,XZ,XY)=\sum_{i\ge f} (YZ)^{n-i} a_i(XZ,XY)=X^f(\sum_{i\ge f} (YZ)^{n-i} X^{i-f}a_i(Z,Y))$$
which proves that $\alpha=f$. Similarly, you find $\beta=e$ and $\gamma=d$, giving the desired result.
Hi, thank you for your answer. Two aspects I not completely understand. Firstly, you say that the equation of $\sigma_(V(F)) \vert {D+(XYZ)}$ is given by $F(YZ,XZ,XY)$. Could you explain why?
Indeed, $\sigma^{-1}(D_+(XYZ))=\sigma^{-1}(D_+(X) \cap D_+(Y) \cap D_+(Z))= D_+(YZ) \cap D_+(XZ) \cap D_+(XY)$, right? But the way how you conclude from this that the restriction of $\sigma_(F)$ to $D_+(XYZ)$ coinsides with vanishing set of $F(YZ,XZ,XY)$ I not see yet.
And secondly, when we know that $F(YZ,XZ,XY)$ describes $\sigma_(V(F)) \vert {D+(XYZ)}$, then cleary $\sigma_(F)$ is the vanishing set of $F(YZ,XZ,XY)X^aY^bZ^c$ with a priori unknown integers $a,b,c$. You wrote: "As you want to remove $X=0$, $Y=0$ and $Z=0$ from the polynomial $F(YZ,XZ,XY)$, you can choose $a,b,c$ to be $\le 0$ and then define them as $-\alpha,-\beta,-\gamma$ where $\alpha,\beta,\gamma$ are the multiplicities of $F(YZ,XZ,XY)$ along $X=0,Y=0,Z=0$ respectively." I not understand why we have to require that $F(YZ,XZ,XY)X^aY^bZ^c$ must be not divisible by $X, Y$ and $Z$?
$\sigma$ restricts to an automorphism of the open subset of $\mathbb{P}^2$ given by $XYZ\not=0$. I think that it is $D_+(XYZ)$ in your notation.
Then, $\sigma_(F)$ is the closure of the image of $F$ by $\sigma$ on this open set. As $\sigma$ is given by $[X:Y:Z]\mapsto [YZ:XZ:XY]$ on $D_+(XYZ)$, the equation of $\sigma_(F)$ on any point of $D_+(XYZ)$ is given by $F(YZ,XZ,XY)$. The closure on $\mathbb{P}^2$ does not contain $X=0$ or $Y=0$ or $Z=0$, that is why we remove them and that is why $F(YZ,XZ,XY)X^aY^bZ^c$ is not divisible by $X$, $Y$ or $Z$.
About the claim that $\sigma_*(F)$ on $D_+(XYZ)$ is given by $F(YZ,XZ,XY)$ I'm a bit confused. You argue that it's because $\sigma$ is given by $[X:Y:Z]\mapsto [YZ:XZ:XY]$. Let me point out where I still see a subtile problem. In following we consider implicitly only the the restrictions of $\sigma$ and $C=V(F))$ to $D_+(XYZ)$ without make mention of it by $... \vert {D+(XYZ)}$ notation. Also, everywhere where I write in following $F$ I mean $F(X,Y,Z)$ in contrast to $F(YZ,XZ,XY)$.
Probably, we using different notations: what is your concrete definition of $\sigma_(F)$? Your argument that $\sigma_(F)$ is given by $F(YZ,XZ,XY)$ follows from $[X:Y:Z]\mapsto [YZ:XZ:XY]$ leads me to suspicion that you maybe use $\sigma_*(F)= \sigma^{-1}V(F)$ as definition.
The reason that leads me to this
suspicion was the basic fact that $\sigma^{-1}V(F)=V(F(YZ,XZ,XY))$.
Sorry, if I misunderstood your argument.
Maybe exactly here might lie my understanding problem, since with this pushforward notation $\sigma_*C=\sigma_*V(F)$ I mean the the birational transform. And in this case I not see why $[X:Y:Z]\mapsto [YZ:XZ:XY]$ should
imply that $\sigma_*V(F) =V(F(YZ,XZ,XY))$. I tried to approach it as follows,
but this lead me to a wrong result as explaned below.
By general definition of pushforward if we want to determine completly $\sigma_*V(F)$ we need an affine cover of $\bigcup_i U_i$ of $D_+(XYZ)$, then the local sections $\sigma_*V(F)(U_i)=V(F)(\sigma^{-1}(U_i))$ completly determine $\sigma_*V(F)$. But we are lucky here and since $D_+(XYZ)$ is affine and $\sigma^{-1}(D_+(XYZ))= D_+(XYZ)$ the $\sigma_*V(F)$ is completly determined by global sections $$\sigma_*V(F)(D_+(XYZ))=V(F)(\sigma^{-1}(D_+(XYZ)))=V(F)(D_+(XYZ))$$
But this lead me to wrong "conclusion" that $\sigma_*V(F(X,Y,Z))=V(F(X,Y,Z))$. Do you see where I making the weird thinking error?
$D_+(XYZ)$ is an affine open subset and the restriction of $\sigma$ on it is an involution. So $\sigma_$ and $\sigma^$ are the same in restriction to $D_+(XYZ)$. I hope it helps.
So involution here is cruical and if we have in general a rational map $f: \mathbb{A}^n \dashrightarrow \mathbb{A}^n, (X_1,...,X_n) \mapsto (f_1(X_1,...,X_n),..., f_n(X_1,...,X_n))$ and a divisor $V:=V(F) \subset \mathbb{A}^n$ defined by a polynomial $F$, then $f_*V=V(F(f_1,...,f_n))$ is wrong if we not impose some additional conditions on $f$ like beeing involution as in our case above? Or does there exist a more widespreaded class of rational morphisms $f: \mathbb{A}^n \dashrightarrow \mathbb{A}^n$ which satisfy $f_*V=V(F(f_1,...,f_n))$ for all divisors defined by a sigle polynomial?
In general, f_*(V)=V(F(g_1,...,g_n)) on the open subset where you have an isomorphism, where $(X_1,\ldots,X_n)\mapsto (g_1(X_1,...,X_n),...,g_n(X_1,...,X_n))$ is the inverse of $f$.
I think I got it now. Thank you a lot for explanations!
|
2025-03-21T14:48:30.284631
| 2020-04-11T21:06:56 |
357175
|
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|
Stack Exchange
|
Is this an error in Loomis and Sternberg?
In Loomis and Sternberg's Advanced Calculus section 3.3 Continuity, they make this comment (just before Theorem 3.2, [pp. 128-129 in my copy):
A linear map $T : V \rightarrow W$ is bounded below by $b$ if $\|T(\xi)\| \geq b \|\xi\|$ for all $\xi$ in $V$ [...] If $V$ is finite-dimensional, then it is true, conversely, that if $T$ is bounded below, then it is invertible (why?), but in general this does not follow.
The statement that is followed by "(why?)" seems to me it must be false. For example, the inclusion $\mathbb{R}^2 \rightarrow \mathbb{R}^3$ is a linear map bounded below (with $b = 1$), but it is not invertible.
Perhaps they mean $T$ is invertible considered as a map $V \rightarrow \mathrm{im}\ T$? But in this case, the result holds for infinite-dimensional vector spaces too. (A linear map bounded below must have trivial kernel, and so be injective.)
Is this a mistake in the book, or have I misunderstood something?
You are correct about all of this. But if you change $W$ to $V$ then it becomes sensible, true in finite dimensions, and false in infinite dimensions, so I think that must be what they meant.
|
2025-03-21T14:48:30.284748
| 2020-04-11T21:08:39 |
357176
|
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|
Stack Exchange
|
Chains and homotopy type
Let $$C^{\ast}:\mathbf{sSet}\rightarrow E_{\infty}\text-\mathbf{dgAlg}$$
be the cochain contravariant functor from the category of simplicial sets to the category of $E_{\infty}$-dg-algebras (over $\mathbb{Z}$).
Theorem (Mandell): Suppose that $X$ is of finite type and $Y$ is nilpotent space of finite type. Then $C^{\ast}:[X,Y]\rightarrow [C^{\ast}(Y),C^{\ast}(X)]$ is
split injective. Moreover, there exists a natural map $\epsilon: [C^{\ast}(Y),C^{\ast}(X)]\rightarrow [X,Y]$ such that: $\epsilon\circ C^{\ast}=id$ and if $ f,g \in [C^{\ast}(Y),C^{\ast}(X)]$ with $\epsilon(f)=\epsilon(g)$ then $H^{\ast}(f)=H^{\ast}(g)$.
Here $[X,Y]$ is the set of homomorphisms in the homotopy category of simplicial sets and $[C^{\ast}(Y),C^{\ast}(X)]$ is the set of homomorphisms in the homotopy category of $E_{\infty}$-algebras (over $\mathbb{Z}$).
question: Is the Mandell theorem still valid if $Y=K(G,1)$ where $K(G,1)$ is the Eilenberg-MacLane space and $G$ is a finite group.
You might hope to relax the finite type restriction a little bit, but not all the way down to just connectedness. Whenever $X$ is an acyclic space, then $C_{\ast}(X) \simeq \mathbb{Z}$ and the injectivity will fail for $X = Y$.
@PiotrPstrągowski you are right, I will put some restrictions...
For any question of this general type, the first thing to do is to ask what counterexamples you can get from the Kan-Thurston theorem.
|
2025-03-21T14:48:30.284882
| 2020-04-11T21:55:23 |
357181
|
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|
Stack Exchange
|
Counterexample for absolute summability of autocovariances of strictly stationary strongly mixing sequence
Suppose $(X_i)_{i\in\mathbb{Z}}$ is a strictly stationary, strongly (i.e. $\alpha-$)mixing sequence of real random variables. If we have $\mathbb{E}[|X_1|^{2+\epsilon}]<\infty$ for some $\epsilon>0$ and if we have for the mixing coefficients that there exists $C>0, \gamma\geq \frac{(2+\epsilon)(1+\epsilon)}{\epsilon}$ such that for large enough $k:$
$$\alpha(k) := \alpha\left(\sigma((X_i)_{i\leq 0}), \sigma((X_i)_{i\geq k})\right) \leq Ck^{-\gamma},$$
then Davydov's inequality (see for example Corollary A2 in Hall and Heyde (1980)) implies that the autocovariances are absolutely summable:
$$\sum_{k=1}^\infty\left\lvert \operatorname{Cov}\left(X_0, X_k\right)\right\rvert<\infty.$$
I'm currently reading a paper where we only assume $\mathbb{E}[X_1]=0, \mathbb{E}[X_1^2] = 1,$ but higher moments might be infinite. It seems to me that this is not enough to guarantee absolutely summable autocovariances, but I find it hard to give a counterexample. In particular, I'm interested in an example of a strictly stationary and strongly mixing sequence with, $\mathbb{E}X_1^2<\infty$ and (for $k$ large enough) $\alpha(k)\leq Ck^{-\gamma}$ for some $C>0, \gamma\geq 3+2\sqrt{2}$ (the minimum of $\epsilon\mapsto\frac{(2+\epsilon)(1+\epsilon)}{\epsilon}$), but with autocovariances that are not absolutely summable. I find it hard to construct an example and the literature I found seems to either assume a higher moment exists without giving a counterexample to show its necessity, or assume some other condition to guarantee absolute summability. An example or a reference to one would be much appreciated.
What is the title of the paper that you are reading?
We can construct a strictly stationary sequence $\left(X_k\right)_{k\in\mathbb Z}$ having the following properties:
$X_0$ has finite moments of any order.
$\beta(k)\leqslant Ck^{-1+\delta}$ for some $\delta\in (0,1)$.
The series $\sum_{k\in\mathbb Z}\lvert\operatorname{cov}\left(X_0,X_k\right)\rvert$ diverges.
Note that it is not exactly the wanted counterexample since the mixing rates are not as fast as the opening poster wants, but it is almost the best that we can get with finite moment of any order.
Let $\left(e_{k,i},k\geqslant 1,i\in\mathbb Z\right)$ be an independent family of random variables such that for all $k$ and $i$, $e_{k,i}$ takes the values $\pm 1$ with probability $1/(2n_k^2)$ and $0$ with probability $1-1/n_k^2$, where $(n_k)$ is an increasing sequence of integers such that $\sum_{k\geqslant 1}n_k^{-2}$ converges. Let
$$
X_j=\sum_{k\geqslant 1}\sum_{i=1}^{n_k}e_{k,j-i}, \quad j\in\mathbb Z.
$$
Item 1. can be seen by an application of Rosenthal's inequality.
For a fixed $j$, let us compute $\operatorname{cov}\left(X_0,X_j\right)$ for $j>0$. By definition, since $e_{k,i}$ is centered for all $k$ and $i$,
$$
\operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\sum_{i=1}^{n_k}
\sum_{k'\geqslant 1}\sum_{i'=1}^{n_k}\mathbb E\left[e_{k,-i}e_{k',j-i'}\right].
$$
Observe that if $k\neq k'$, then $e_{k,-i}$ is independent of $e_{k',j-i'}$ hence
$$
\operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\sum_{i,i'=1}^{n_k}
\mathbb E\left[e_{k,-i}e_{k ,j-i'}\right].
$$
If $j-i'=-i$, then $ \mathbb E\left[e_{k,-i}e_{k ,j-i'}\right]=n_k^{-2}$ and
if not, then $ \mathbb E\left[e_{k,-i}e_{k ,j-i'}\right]=0$ hence
$$
\operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\frac 1{n_k^2}\sum_{i,i'=1}^{n_k}
\mathbf{1}\{j-i'=-i\}.
$$
Moreover, $\sum_{i =1}^{n_k}
\mathbf{1}\{j-i'=-i\}=\mathbf{1}\{1\leqslant i'-j\leqslant n_k \}$. If $j>n_k$, then $\sum_{i'=1}^{n_k}\mathbf{1}\{1\leqslant i'-j\leqslant n_k \}=0$ and if $j\leqslant n_k$, this sum is $n_k-j$. We thus got
$$
\operatorname{cov}\left(X_0,X_j\right)=\sum_{k\geqslant 1}\frac{n_k-j}{n_k^2}
\mathbf 1\{j\leqslant n_k\}.
$$
Summing this over $j$ gives a divergent series.
For the mixing rates, we can use the steps in the papers
Giraudo, Davide, and Dalibor Volný. “A Strictly Stationary β-Mixing Process Satisfying the Central Limit Theorem but Not the Weak Invariance Principle.” Stochastic Processes and their Applications 124.11 (2014): 3769–3781.
Giraudo, Davide. An improvement of the mixing rates in a counter-example to the weak invariance principle. C. R. Math. Acad. Sci. Paris 353 (2015), no. 10, 953--958
We have to choose $n_k$ such that $n_{k+1}\geqslant n_k^{1+\gamma}$ for some positive $\gamma$.
I have found a paper that proves the existence of a counterexample (Theorem 2 of Bradley 1983 "Information regularity and the central limit question"). The construction seems quite complicated, but it does provide exactly what I'm looking for.
|
2025-03-21T14:48:30.285157
| 2020-04-11T22:15:23 |
357184
|
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|
Stack Exchange
|
Are "étalé spaces" a thing for probability spaces?
Let $PX$ be a $\sigma$-algebra on the set $X$, and let $j : PX \to {\sf Set}_{/X}$ be the tautological functor that sends an event $E\subseteq X$ to itself, regarded as a function with codomain $X$. Now, the category ${\sf Set}_{/X}$ is cocomplete, thus $j$ has a unique cocontinuous extension to a pair of functors
$$
J : [PX^\text{op},{\sf Set}] \leftrightarrows {\sf Set}_{/X} : N
$$ here, $J$ is the left Kan extension of $j$ along the Yoneda embedding and $N$ is its right adjoint.
Most of you will have noticed that I am copying the exact procedure that yields the equivalence between sheaves on a topological space $X$ (so, a subcategory of $[OX^\text{op},{\sf Set}]$: open subsets instead of events, but the idea is the same) and étalé spaces (those $(E,p : E\to X) \in {\sf Top}_{/X}$ that are local homeomorphisms).
I am interested in the properties of the adjunction $(J,N)$.
I have no exact request apart a little bit of help wrapping my head around this construction, with particular attention to what is different from the topological case.
For the moment, let me just add something about the functor $J = \text{Lan}_yj$: the Kan extension can be written acting on the presheaf $F$ as
$$
\int^{E\in PX} FE \otimes jE
$$
(here $\otimes$ is a tensor in ${\sf Set}_{/X}$, so –I think– a coproduct of as many copies of $E \hookrightarrow X$ as there are elements in $FE$; now you perform a suitable quotient on the coproduct of all these $FE \otimes jE$; the colimit is done in Set and then the universal property there yields a unique structure of object in Set/X for the colimit, because –if I remember well that a category is connected if and only if its twisted arrow category is– the colimit defining the coend is over a connected category).
I know this might appear a naive question, but I have always found this construction very specific to sheaf theory (to the point that the very name "sheaf" comes from a pictorial representation of how the functor $[OX^\text{op},{\sf Set}] \to {\sf Top}_{/X}$ acts: all in all, the colimit breaks into the coproduct $\coprod_{x\in X} \text{colim}_{U\ni x} FU$ of all the "stalks" of $F$, that are fibers "stemming" from the "root" $x$... The stalks are then tied together by a certain topology on the disjoint union).
Is there a similar visual intuition for how $J$ acts on a... well, how would you call it? a pre—?
I beg to disagree with the edit. Etalé space is the correct spelling (https://ncatlab.org/nlab/show/étalé+space); or rather, étalé: it's the past participle of the verb "s'étaler" (something like "to sprawl"?); étale, and in partic. étale space, is a different thing, afaicr. Cf. the French "faire étalage"; "la mer s'étale sous nos yeux".
Fine: the spelling "etalé" from the original post couldn't be correct anyway.
Etale Spaces can be used to analyze Giry-algebras ($\mathcal{G}$-algebras), and hence (for a fixed object $X$) probability spaces on $X$ as follows. First note that your functor $j$ above should read $j: \Sigma_X \rightarrow \mathbf{Meas}/X$, which is analogous to the topological case (requiring continuous functions rather than just set functions). Here, $\mathbf{Meas}$ is the category of separated measurable spaces - meaning $(2, Discrete)$ is a coseparator of the elements of $X$. Now suppose $\pi_X:\mathcal{G}(X) \rightarrow X$ is a Giry-algebra. (The reason we require separated measurable spaces is that because if $X$ is not separated, then there are no $\mathcal{G}$-algebras on $X$.)
Now the slice category $\mathbf{Meas}/X$ is cocomplete, and one has the same construction as you have noted above, which is just Thm. 2, pp41-42 of Sheaves in Geom. & Logic (SGL), so we have the cited adjunction between the left-Kan extension, $J$, and the functor $N$. Now fix the object $\pi_X$ in $\mathbf{Meas}/X$, and using the adjunct pair $J \dashv N$, look at the universal arrow from $J$ to the object $\pi_X$, i.e., the counit of the adjunction at $\pi_X$. $N(\pi_X)$ is the ''sections functor'', i.e., $N(\pi_X)(U) = \{s: U \rightarrow \mathcal{G}(X) \, | \, \pi_X \circ s = id_U\}$, and $J(N(\pi_X)) = \pi_X$. (Provided I haven't done something foolish, this is just applying the argument in equation 8-10 of the text SGL, p 42, taking $E=\pi_X$ and $P$=sections functor.)
OK. This is all ''standard fare'', and I haven't said anything that answers your question regarding how you interpret presheaves, etc. - and I've yet to work it out. Sheaf spaces are constructed with a horizontal and vertical slice interpretation. Toward that same end note the following two points:
(1) every $\mathcal{G}$-algebra, such as $\pi_X$, specifies a super convex space structure on the underlying set of $X$, via $\sum_{i=1}^{\infty} \alpha_i x := \pi_X( \sum_{i=1}^{\infty} \alpha_i \delta_{x_i})$. More specifically, there is a functor, $\mathbf{Meas}^{\mathcal{G}} \rightarrow \operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}$, from $\mathcal{G}$-algebras to $\mathbb{R}_{\infty}$-coseparated super convex spaces. ($\mathbb{R}_{\infty}$ is the one-point extension of the real line $\mathbb{R}$ by a point ''$\infty$'', and that set has the obvious super convex space structure, i.e., $(1-r) u + r \infty = \infty$ for all $r \in (0,1]$.) (The object $\mathbb{R}_{\infty}$ ''arises'' as follows. Every convex space is either a geometric convex space (meaning it embeds into a real vector space), a discrete convex space, or a mixture of the two (which is most common). A geometric space is coseparated by the unit interval $[0,1]$. A discrete space is coseparated by $\mathbf{2}$. In $\mathbf{SCvx}$ there is a map $\mathbf{2} \rightarrow \mathbb{R}_{\infty}$, taking $0 \mapsto \infty$ and $1 \mapsto 0$. The space $\mathbb{R}_{\infty}$ can therefore coseparate any super convex space. (Borger & Kemp showed $\mathbb{R}_{\infty}$ is a coseparator for $\mathbf{Cvx}$, and by restricting to $\operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}$ it is a coseparator for that category also.))
(2) The object $\pi_X: \mathcal{G}(X) \rightarrow X$ is a (weak) terminal object in $\mathbf{Meas}/X$ because if $f: Y \rightarrow X$ is an object in $\mathbf{Meas}/X$ then the composite $\eta_X \circ f: Y \rightarrow \mathcal{G}(X)$ is an arrow to $\pi_X$. We know how the $\sigma$-algebra structure of $\mathcal{G}(X)$ is constructed - via the evaluation maps $ev_U: \mathcal{G}(X) \rightarrow \mathbb{R}$.
Now to the main point. The idea is that the fibers over $x \in X$, which are the ''vertical slices'', specify a (coseparated) super convex space, while the horizontal slices specify the measurable structure. Taking $X = \mathbb{R}_{\infty}$, the $\mathcal{G}$-algebra is the expectation operator,
$\mathbb{E}: \mathcal{G}(\mathbb{R}_{\infty}) \rightarrow \mathbb{R}_{\infty}$, sending $P \mapsto \int_{x \in \mathbb{R}_{\infty}} x dP$. (Taking $P$ to be the half-Cauchy distribution, it is clear why we need $\infty$.)
Now suppose $X$ is an arbitrary (separated) measurable space with the $\mathcal{G}$-algebra $\pi_X$. Then a commutative square, corresponding to a $\mathcal{G}$-algebra morphism $\hat{f}: \pi_X \rightarrow \mathbb{E}$ is specified by a measurable function $f: X \rightarrow \mathbb{R}_{\infty}$, which under the induced super convex space structures on $X$ and $\mathbb{R}_{\infty}$ is also a countably affine map (which is easy enough to verify directly).
This gives the basic idea of how you interpret (part of) the construction you are referring to.
Let me add some context. Your coend formulation is correct - but you can view it from a slightly different point of view. (The coend formulation is Prob. 5 on Page 223, CWM, MacLane.) Let me use MacLanes's notation. Let S be any presheaf, $S: \Sigma_X^{op} \rightarrow \mathbf{Set}$, and take $T: \Sigma_X \rightarrow \operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}/\mathcal{G}(X)$ to be given by $U \mapsto (\mathcal{G}(U) \hookrightarrow \mathcal{G}(X))$. Since $\operatorname{\mathbb{R}_{\infty}-\mathbf{SCvx}}$ is cocomplete, so is the slice over $\mathcal{G}(X)$, and an element of that slice category is any ''kernel map'' $k: A \rightarrow \mathcal{G}(X)$. Then the tensor product of $S$ and $T$, which is the coend, is valued in $\operatorname{ \mathbb{R}_{\infty}-\mathbf{SCvx}} / \mathcal{G}(X)$, i.e., the tensor product is a kernel map. (Note that the functor $T$ is just the composite of $j$ and the functor $\hat{\mathcal{P}}: \mathbf{Meas}/X \rightarrow \operatorname{ \mathbb{R}_{\infty}-\mathbf{SCvx}} / \mathcal{G}(X)$ which is induced by the functor $\mathcal{P}: \mathbf{Meas} \rightarrow \operatorname{ \mathbb{R}_{\infty}-\mathbf{SCvx}}$ which is just the Giry monad viewed as a functor into the category of super convex spaces.
I will need time to parse your entire answer. But this is precious and extremely time-saving!
|
2025-03-21T14:48:30.285725
| 2020-04-11T22:55:25 |
357186
|
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|
Stack Exchange
|
Find a complex matrix on a unit sub-spheres
I am new to optimization theory. I have a following question. For a given $X = [x_1 x_2 \ldots x_N] \in \mathbb{C}^{N \times N}$, where $x_i \in \mathbb{C}^{N\times 1}$ for $i \in \{1,\ldots,N\}$, $U = [u_1 u_2 \ldots u_N]\in \mathbb{C}^{N \times N}$, is there any possibility of obtaining a closed form solution of
$$\min_{U \in \mathbb{C}^{N\times N}}\{\|U-X\|_F^2: \|u_i\|_2^2=1\}$$
where $\|\cdot\|_2$ and $\|\cdot\|_F$ denotes the Euclidean and Frobenius norm respectively. I know how to find the solution for $\|U\|_2^2=1$, however, I do not see how to solve it when we have $\|u_i\|^2=1$.
I may be overthinking. I can just take the $X$ and normalize each column. I am not sure that would be the right approach.
We have $|U-X|_F^2=\sum_i |u_i-x_i|_2^2$ and your approach would be correct.
Great! Thanks so much!
|
2025-03-21T14:48:30.285937
| 2020-04-11T23:06:35 |
357187
|
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"Yi Wang",
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|
Stack Exchange
|
What do conjugacy classes of involutions like in finite simple group $E_7(q)$?
Are there any refences for conjugacy classes of involutions in finite simple group $E_7(q)$?
For $q$ odd see:
D. Gorenstein, R. Lyons and R. Solomon,The classification of the finite simple groups, Number 3,Mathematical Surveys and Monographs, vol. 40, Amer. Math. Soc., 1998 MR1490581
For $q$ even:
Aschbacher, Michael M., and Gary M. Seitz. "Involutions in Chevalley Groups over Fields of Even Order." Nagoya Mathematical Journal 63 (1976): 1-91. Web. MR0422401
Thank you very much!
|
2025-03-21T14:48:30.286007
| 2020-04-12T01:23:12 |
357196
|
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|
Stack Exchange
|
Stably deforming vector bundles
Let $X$ be a smooth projective variety. $V_1$ and $V_2$ are two vector bundles on $X\times \mathbb{A}^1$ such that $V_1|_{X\times \{0\}}\cong V_2|_{X\times \{0\}}$ and $V_1|_{X\times \{1\}}\cong V_2|_{X\times \{1\}}$. Is it possible to deform $V_1\oplus K$ to $V_2\oplus K$ for some vector bundle $K$? More precisely is there a vector bundle $V$ on $X\times \mathbb{A}^1\times\mathbb{A}^1$ such that $V|_{X\times \mathbb{A}^1\times\{0\}}\cong V_1\oplus K$ and $V|_{X\times \mathbb{A}^1\times\{1\}}\cong V_2\oplus K$? If so is it possible to find a deformation that it stays constant on the lines $X\times \{0\} \times \mathbb{A}^1$ and $X\times \{1\} \times \mathbb{A}^1$? ($X$ is a variety over a finite field if that matters.)
Edit 1: The answer to the first question is yes but still I don't know the answer to the second one. The conditions of the problem implies that $V_1$ and $V_2$ are equal to each other in $K_0(X\times \mathbb{A}^1)$. A trick due to Heller implies that whenever two elements in $K_0$ are equal you can find $K$ such that $V_1\oplus K$ and $V_2\oplus K$ are extensions of same two vector bundles. Now it is well known that you can deform any two extensions to each other.
Edit 2: I had another similar question, I decided to add it here instead of asking a new question. Let $V_1$, $V_2$ and $V_3$ be vector bundles on $X$. Assume there are two different non-isomorphic short exact sequences $SE1$ and $SE2$ of the form of $0\rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$. Is there a deformation (or a sequence of deformations) $SE1_t$ and $SE2_t$ both of the form $0\rightarrow V_{1t}\rightarrow V_{2t} \rightarrow V_{3t} \rightarrow 0$, such that it gives us $SE1$ and $SE2$ at $t=0$. At $t=1$ it gives us two isomorphic short exact sequences.
I'll appreciate if you have any ideas how to approach the problem.
Deforming both to the split extension gives a family over $\mathbf A^1 \amalg_{\operatorname{pt}} \mathbf A^1$, not over $\mathbf A^1$, right? But you can also do it directly because the space of extensions is an affine space.
You are correct. Even if you do both of them to the split one. You can also direct sum the split extension into the $K$. Thus you can get the desired deformation with the modified $K$.
|
2025-03-21T14:48:30.286345
| 2020-04-12T01:50:36 |
357197
|
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Stack Exchange
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Conway's lesser-known results
John Horton Conway is known for many achievements:
Life, the three sporadic groups in the "Conway constellation," surreal numbers, his "Look-and-Say" sequence analysis, the Conway-Schneeberger $15$-theorem, the Free-Will theorem—the list goes on and on.
But he was so prolific that I bet he established many less-celebrated
results not so widely known. Here is one:
a surprising closed billiard-ball trajectory in a regular tetrahedron:
Image from Izidor Hafner.
Q. What are other of Conway's lesser-known results?
Edit: Professor Conway passed away April 11, 2020 from complications of covid-19:
https://www.princeton.edu/news/2020/04/14/mathematician-john-horton-conway-magical-genius-known-inventing-game-life-dies-age
The latest issue of the Monthly has a paper by Conway, Paterson and Moscow. This lovely paper was written more than 40 years back, and published in a difficult-to-find festschrift for Lenstra. On this sad day, this article brings out yet again the sheer joy of Conway.
https://mobile.twitter.com/DavidSpergel/status/1249119023234789377
:(
@ToddTrimble News is circulating that he passed away because of coronavirus.
Oh no... sad day indeed.
Coxeter and Conway constructed the beautiful "frieze patterns" in the 70s from polygonal triangulations, seemingly out of nowhere; their work was only "properly" understood many years later in the context of cluster algebras. (Probably this counts as a "well-known" result, though.)
How well known is Conway's ZIP proof?
@MarkSapir You know, some and perhaps many of his results may be buried in newsgroup posts, etc., or made known to others by means other than papers.
This is a good question, I think his Wikipedia doesn't really say a lot about his achievements.
I also found this, which might be the spiral generated by Conway's billiard path: https://en.wikipedia.org/wiki/Boerdijk%E2%80%93Coxeter_helix
@IanAgol: The architect Arata Isozaki designed a building of stacked, spiraling tetrahedra. Image at this MESE link.
Just found out about his death. This video is so much more special now.
I fear we are missing out on some interesting answers here because we have no precise definition of “lesser-known”, and it’s often hard to think of something one knows oneself as being lesser-known.
Maybe a criteria for lesser know results could be what doesn't appear on "know for" on wikipedia bio, also I wanted to share for those like me who doesn't understand anything Conway work on knots an idea can be found on this ams feature column
xkcd in his style has put an homage to Conway which is worth spending a (almost literally) second to see.
Link to the billiards result?
@Meekohi: See the link in the figure caption above. That link will take you to a Mathematica Demo which cites Wells' book as the source.
"The latest issue of the Monthly has a paper by Conway, Paterson and Moscow." @Lucia, that's how the authors are listed, but in fact Moscow is not a person but the location where Conway and Paterson did the work (or perhaps they were on their way to Moscow when they did the work, my memory is not clear about this).
Conway discovered that the right triangle with sides $(1, 2, \sqrt{5})$ can be subdivided into five congruent triangles similar to the original one:
Performing this subdivision repeatedly leads to the non-periodic “pinwheel tiling” of the plane by such triangles, in which the triangle appears in infinitely many different orientations:
This tessellation is occasionally incorrectly credited to Radin¹, although Radin’s paper itself clearly attributes it to unpublished work of Conway.
Radin, Charles. “The Pinwheel Tilings of the Plane.” Annals of Mathematics, vol. 139, no. 3, 1994, pp. 661–702.
I wonder, does it appear in every orientation?
@BlueRaja There are only countably infinitely many triangles in this tiling (it's fairly straightforward to construct a counting scheme), so the set of orientations must have measure zero.
I would also wager that, say the cosine of the orientation angles are all algebraic. Maybe there's a nice way to describe them.
Well, it’s easy to see from the relative orientation of the large triangle to the five constituents that all the angles belong to the abelian group generated by $\pi/2$ and $\arctan 2$.
@BlueRaja The set of orientations is countable, but it does distribute uniformly on the unit circle.
As a side note, if one replace the diagonal in the rectangle with the other diagonal, the tiling becomes periodic.
@probably_someone Since each triangle has nonempty interior and therefore contains a point with rational coordinates, one doesn't need to construct a counting scheme; counting schemes for the points with rational coordinates are available "off the shelf".
Just for completeness, and for those looking to read further on this tiling substitution, this is Conway and Radin's 'pinwheel tiling', for which there is a huge literature and still many open questions.
I regret learning about this. This is another tiling to add to my maze project at https://github.com/mdchaney/jsmaze
I am curious what is special about the $1$-$2$-$\sqrt{5}$ right triangle? Any right triangle yields a decomposition into $5$ congruent pieces in this way, giving a tiling of the plane. I suppose for some special ones such as isosceles right triangle it will be a periodic tiling, but presumably for "most" right triangles it will be aperiodic. Has this been studied?
@ZachTeitler If you actually draw it, you will quickly find out that if you start with a right triangle not similar to the 1–2–$\sqrt5$ one, you’ll end up with 4 congruent triangles, and a fifth noncongruent triangle.
@EmilJeřábek Thank you. All five of the triangles are similar, but one of them is a different size than the other four. Yes, it is obvious. I feel very stupid; I should have noticed that myself.
Conway's office at Cambridge was notoriously messy. One day, he got tired of how hard he had to struggle to find a paper in there, and shut himself away for a few hours to come up with a solution to the problem. He proudly showed a sketch of his solution to Richard Guy, who said, "Congratulations, Conway – you've invented the filing cabinet."
https://twitter.com/robinhouston/status/1249302645556289537
Richard Guy had a surprisingly organized office. (At least compared to the other math professors at Calgary. And for being a centenarian.) Sadly, Dr Guy also passed away earlier this year. What an amazing person he was! I watched him give a talk at a conference in Alberta when he was 99. It was complete with a song and dance.
Wow, Dick guy really shut him down, he sounds like such a ...
@ham Guy & Conway were friends and collaborators, and I'm sure Conway appreciated the wit in Guy's remark.
Although it is well known that Conway was able to quickly calculate the day of the week of any given date, it is less well known that one part of the algorithm is easy to remember and useful in practice: In any given year, the following dates all fall on the same day of the week: 4/4, 6/6, 8/8, 10/10, 12/12, 5/9, 9/5, 7/11, 11/7, and the last day of February. For example, in 2020, all these dates fall on a Saturday. Conway, in his characteristically colorful way, would say that the Doomsday of 2020 is Saturday. Knowing this fact allows you to calculate fairly quickly in your head, with no special training, the day of the week for any date in 2020.
The full algorithm tells you how to calculate the Doomsday of any given year, but in everyday life, one is mostly interested in the current year, so you can just remember this year's Doomsday, and update that fact once a year.
there's also a mnemonic for the last four: "I work a nine to five job at the Seven-Eleven", although its memorability may be less in some regions of the world.
Are your dates in month/day or day/month format? :=)
He also had a much less-known mneumonic for converting between Hebrew and Gregorian dates. I wonder if I should post about that.
I guess the reason for omitting pi day (3/14) is that it only works in day/month format?
@bof : You're free to add other days if you want (Boxing Day, July 4, etc.) but what is desired is a maximally easy-to-remember set that covers as much of the year as possible.
@WillSawin in some other regions it might be even easier to remember. In USSR May 9 was Victory Day and Nov 7 was October Revolution Day. In some xUSSR countries these holidays are still celebrated.
It has always bothered me that January is awkward. The "official" (Winning Ways) suggestion for January is January 31 except in leap years, when it's January 30. Recently, it occurred to me that in the U.S. at least, a more convenient solution for January is that 1/23 falls on the Doomsday of the previous year. 123 is easy to remember. It may seem inelegant to revert to the previous year's Doomsday, but if you're using Doomsday in "real time" (i.e., for the current year), it just means waiting until February to update Doomsday. Anyway, I've found that it works for me!
In leap years it's surely January 32 (that is, February 1), which is particularly inelegant.
@MichaelLugo Whoops...you're right of course! I've been using 1/23 for some years now and haven't found anything better.
A convoluted set of discussions on the newsgroup geometry.puzzles in October and December 2001 seems to be due to Conway (the various threads were a mess), with the conclusion that the lines which bisect the area of a triangle do not all cross the centroid but instead form an envelope making up a deltoid whose area is $\frac{3}{4} \log_e(2) - \frac{1}{2} \approx 0.01986$ times the area of the original triangle, and affine transformations show this a constant for all triangles
As an illustration:
This is not difficult to show, so counts as minor and lesser known. I once asked here if there was any direct relationship between the deltoid and $$\sum_{n=1}^{\infty}\frac{1}{(4n-1)(4n)(4n+1)} = \frac{3}{4} \log_e(2) - \frac{1}{2}$$ apart from giving the same value
A recent related MO question: On 'fair bisectors' of planar convex regions.
The Conway base 13 function is a function $f : \mathbb{R} \rightarrow \mathbb{R}$ that takes on every real value in every interval. It is thus discontinuous at every point.
Oh, I love that one. It's absolutely magnificent!
How about the Conway-Gordon theorems? Any embedding of a six-clique in $\mathbb{R}^3$ contains a nontrivial link; any embedding of a seven-clique in $\mathbb{R}^3$ contains a nontrivial knot. My very first published paper was based on this!
I love the works of Prof Conway and I am so sad because of this happening. This is not an answer, but a nice things about him that shows his attractive personality. This was told by Prof Peter Cameron in his blog:
This happened at a conference somewhere in North America. I was chairing the session at which he was to speak. When I got up to introduce him, his title had not yet been announced, and the stage had a blackboard on an easel. I said something like "The next speaker is John Conway, and no doubt he is going to tell us what he will talk about." John came onto the stage, went over to the easel, picked up the blackboard, and turned it over. On the other side were revealed five titles of talks. He said, "I am going to give one of these talks. I will count down to zero; you are to shout as loudly as you can the number of the talk you want to hear, and the chairman will judge which number is most popular."
So he did, and so I got to hear the talk I wanted to hear.
RIP John, the world is a poorer place without you.
A very nice memorial about Prof Conway by Princeton university:
https://www.princeton.edu/news/2020/04/14/mathematician-john-horton-conway-magical-genius-known-inventing-game-life-dies-age
I had a similar experience at Northwestern University in 1985-1986. He offered three possible talks and asked for a show of hands.
At Mathcamp, where he was a regular visitor for several years, he had an even more extreme version: campers would suggest and vote on a series of topics, and he would only see the list of suggestions and votes at the beginning of his talk. It was, to put it mildly, extremely impressive.
Conway and Peter Doyle found a lovely proof of Morley’s trisector theorem, using only elementary geometry. Morley’s theorem says that if you take any triangle, trisect its angles, and extend the trisectors so that adjacent trisectors meet pairwise in three points, those points always form an equilateral triangle:
Conway-Doyle’s proof begins with the equilateral triangle in the centre, and shows how to construct an arbitrary triangle around it. The details are given in Conway’s lecture The Power of Mathematics.
It’s unusual for mathematicians to publicly discuss the provenance of joint work; but Conway was an unusual mathematician, and in a talk at MOVES 2015 he explained his view of the matter:
Peter Doyle had a rather worse proof – a distinctly worse proof – and I took his worse proof and tidied it up and made this proof. So I got this, so to speak, out of Doyle’s rather worse proof. I deliberately use language reminiscent of the language used in horse racing and so on: this proof is by Conway, out of Doyle. I’ve never dared to say that in print.
thanks for sharing these fantastic notes!
I don't know if it's lesser known, but it is certainly not on par with some of the other results on this page.
Theorem. (Doyle–Conway) Assume $\sf ZF$. If there is a bijection between $3\times A$ and $3\times B$, then there is a bijection between $A$ and $B$.
This is nontrivial. There's no reason to a priori believe that this is true without the axiom of choice. But it is. You can find the paper on arXiv.
The proof in the paper is Doyle and Conway’s, but the theorem is due to Lindenbaum and Tarski (unless you believe Bernstein).
@Todd Tarski talks about it. Several of their results were written decades after the fact, because the war got in the way of publication.
@Andrés: And in the way of Lindenbaum's life, which probably hindered publication even further.
@ToddTrimble While the original Lindenbaum’s proof claimed in the Lindenbaum and Tarski 1926 paper is only known from hearsay, Tarski undeniably published a proof of the result in 1949.
The Lindenbaum–Tarski cancellation theorem was for any positive integer $n$, not just $n=3$.
@bof: If my memory serves me right, the Doyle–Conway paper proves just that. The case $n=3$ is the first difficult one, though.
Asaf, Conway's paper Effective implications between "finite" choice axioms might also be worth discussing. Although I've cited this a couple of times in recent years (here and here), you're certainly more qualified than I am to comment on the significance and interest of this paper.
Regarding my previous comment, just now I came across Andrés E. Caicedo's answer to Axiom of binary choice vs Axiom of finite choice.
@EmilJeřábek, now I am curious: what did Bernstein say on the matter?
@MarianoSuárez-Álvarez Bernstein claimed to have proved it for all $n$ (two decades before Lindenbaum and Tarski’s work), but apparently no one can make sense of his arguments. See the first paragraph on p. 5 of https://arxiv.org/abs/math/0605779, or p. 20 in https://arxiv.org/abs/1504.01402 .
@EmilJeřábek, Tarski's «Bernstein gave only a rough outline of a proof, the understanding of which presents some difficulties» is a tour de force in elegance!
Conway had an analysis of the notorious Steiner-Lehmus theorem, arguing that no "equality-chasing proof" is possible. MO user Timothy Chow initiated a discussion about Conway's analysis on the FOM list some years back; see here (where Conway's argument is quoted).
For what it's worth, Wikipedia mentions a recent (2018) article that argues a direct proof of this theorem must exist (without giving the proof however!).
See also John Conway and Alex Ryba, The Steiner-Lehmus Angle Bisector Theorem, Math. Gazette 98 (2014), 193-203 (also reprinted in The Best Writing on Mathematics 2015). In this article, they argue that asking whether there is a "direct proof" is actually the wrong question to ask.
Penney's game is a non-transitive competitive two-player coin tossing game, and a method known as Conway's algorithm provides a method for calculating the probabilities of each player winning; a description is given in Plus magazine and elsewhere. But this is not something for which I can or particularly want to remember the details.
Where I do remember the details (but seems to be less widely mentioned) is the simpler question of the expected number of tosses of a fair coin until a particular pattern appears; you might naively guess it is simply the reciprocal of the probability the pattern appears immediately and for the pattern HHHHHT this is correct, being an expected $\frac1 {2^{-6}}=64$ tosses. But for the same length pattern HHHHHH it is almost twice as high at $126$.
Here Conway's algorithm for calculating the expectation is easier to remember: you see whether the length $n$ string on the left of the pattern matches the length $n$ string on the right; if so then add $2^n$ to the result (clearly this happens at least when $n$ is the full length since the string matches itself).
So for example
HHHHHH has $2^1+2^2+2^3+2^4+2^5+2^6=126$ expected tosses because everything matches
HHHHHT has $2^6=64$ expected tosses because only the full length matches
HHTHHH has $2^1+2^2+2^6=70$ expected tosses (the matches are H, HH and HHTHHH)
HHTHHT has $2^3+2^6=72$ expected tosses (the matches are HHT and HHTHHT).
For me the nice part of this is that it does not have to involve coins. Dice work too by changing $2^n$ to $6^n$. So throwing the pattern $1\, 1\, 5\, 1\, 1\, 5$ has an expected number of $6^3+6^6=46872$ throws until the pattern appears. Neat and easy.
Nice. I teach this in my class. And the intuitive reason that if a pattern matches one of its shifts it has a higher propensity to occur in clusters, delaying the first occurrence, in expectation.
And if the letters are not equiprobable, add the reciprocal of the product of probabilities of letters in the $n$ string...
Conway Circle
Extending the sides of a triangle as shown, the six points lie on the same circle, with center at the incenter (center of the inscribed circle). If Conway was the Euler figure of modern times, this could be likened to the discovery of the Euler line, for it could have been known to the ancients.
Just one representative of his work in "classical" geometry of the triangle. He and Steve Sigur had been writing a "definitive" book on the triangle, titled The Triangle Book, but perhaps no one was holding their breath after the untimely death of Steve (a high school math teacher who would visit Princeton each summer to collaborate on the book) in 2008. I remember seeing a few sample pages on Steve's school webpage but it's gone now.
I wish I could say more. (This particular result is from a quick Google search, so may not necessarily represent his best work in this "elementary" area.)
Thanks for reminding me of this nice circle! Steve Sigur's website, or parts of it, can still be found at https://web.archive.org/web/20081230033236/http://www.paideiaschool.org/Teacherpages/Steve_Sigur/interesting2.htm .
@darijgrinberg thank you for digging this up (and I see that you contributed to many MathWorld entries)! Feel free to edit this answer.
@darijgrinberg: Unfortunately, I never got the people at MathWorld to edit the denominator of the expression inside the radical symbol in equation (2) of their entry about the Conway Circle: https://mathworld.wolfram.com/ConwayCircle.html
@JoséHdz.Stgo.: Oh yes, something is clearly wrong with that formula. I have no hot wire to MathWorld these days; my last communication with Weisstein was ca. 2004. (I never had anything like edit rights -- "contributor" means someone exchanging emails with Weisstein.)
@José and darij: the last time I talked with Eric, it seems maintenance has not been a high priority as of late. Nevertheless, I don't think it'd hurt to send another correction.
Here is a generalization of Conway circle https://mathoverflow.net/questions/363139/generalization-of-tucker-circle-conway-circle-and-van-lamoen-circle
Apparently back in the late 90's, Conway convinced Princeton to put in a $1.2M bid at the auction for the Archimedes Palimpsest! See, here for example. He was worried that the manuscript would be hidden again, in an inaccessible vault and unavailable to researchers, as I understand.
Of course we all know the palimpsest sold for \$2M to a then-anonymous bidder, who has since allowed research and restoration of the manuscript. So both the Princeton bid and the worries of Conway may have been moot. Further this may be sideways from other answers, in that it's more an anecdote than a mathematical result, but still, a nice footnote on the interest and advocacy of the man.
How about the shortest paper ever written with Alexander Soifer which proved that for small enough $\epsilon>0$, in order to cover an equilateral triangle of side length $n+\epsilon$, $n^2+2$ unit equilateral triangles suffice.
I'd say, side length $n+\epsilon$, for some $\epsilon>0$.
Thanks for your comment Gerry but isn't that precisely equivalent to $>n$ but requiring more characters? Do you have a specific reason for preferring the longer version?
The way you have it, I would take it to mean for all side lengths greater than $n$, which would be nonsense.
@GerryMyerson Great, thanks for the clarification - I'll update my answer.
That paper is reproduced in https://mathoverflow.net/a/29459/5340
Conway has some well-known work around the monster simple group (which he named), such as his proposal of the Monstrous moonshine conjecture with Norton, and his simplified construction of the monster, which is sketched in a chapter near the end of SPLAG. However, the following construction is not so well-known and kind of miraculous, with additional hints of more miracles in Allcock's A monstrous proposal.
Conway conjectured the $Y_{555}$-presentation of the bimonster, namely the 2-fold wreath product $\mathbb{M} \wr 2 = (\mathbb{M} \times \mathbb{M}) \rtimes (\mathbb{Z}/2\mathbb{Z})$ of the monster. This was later proved independently by S. P. Norton and A. A. Ivanov. Here, $Y_{555}$, which he later called $\mathbb{M}_{666}$, is a connected graph with a central vertex of degree 3 attached to three chains of length 5. The corresponding infinite Coxeter group, generated by the 16 reflections, surjects to the bimonster, with kernel generated by the "spider" relation: $$(ab_1c_1ab_2c_2ab_3c_3)^{10}.$$ Here, $a$ is the reflection attached to the central vertex, and $b_i, c_i$ are reflections attached to the nearby vertices in the spokes.
Conway also noted that $Y_{555}$ embeds into the 26-vertex incidence graph of $\mathbb{P}^2(\mathbb{F}_3)$, and that the corresponding Coxeter group has a homomorphism to the bimonster that extends the $Y_{555}$-map. The kernel is given by "deflating" all free 12-gons to generate copies of $S_{12}$, instead of the affine Weyl group $\mathbb{Z}^{11} \rtimes S_{12}$. Furthermore, the symmetries of the projective plane, including the duality between points and lines, extend to automorphisms of the bimonster.
SPLAG = Sphere Packings, Lattices and Groups
Conway studied the following recurrence relation (OEIS), purportedly studied originally by Hofstadter (of G.E.B. fame):
$$a(k)=a(a(k-1))+a(k-(a(k-1)))$$
with initial conditions $a(1)=a(2)=1$.
(image from MathWorld)
Conway was able to show that
$$\lim_{k\to\infty}\frac{a(k)}{k}=\frac12$$
He offered a \$10,000 prize to anyone who could discover a value of $k$ such that
$$\left|\frac{a(j)}{j}-\frac12\right|<\frac1{20},\quad j > k$$
Collin Mallows from Bell Labs found $k=3173375556$, 34 days after Conway's initial talk on the sequence, and the prize was awarded by Conway after "adjusting" it to the "intended" value of \$1,000.
(See also this and this.)
There's no "k" in the inequality?
@Quant, read the second condition as "for all $j$ greater than $k$".
OK, thanks, but isn't math a precise language? =D
@Quan Indeed, so I'm not certain why you missed the $k$.
The FRACTRAN esoteric programming language.
Although it is related with computer languages, it is not a traditional one, because it is based more on mathematical properties than on typical programming structures
Well, it's basically a register machine.
Easy to write a program (in usual computer language) https://observablehq.com/@liuyao12/fractran
There's one I originally learned about in this excellent answer here at Math Overflow.
Complemented modular lattices satisfying a finiteness condition are exactly the lattice of subspaces of projective spaces. This raises the question of whether we can reverse the process, and associate a geometry with every modular lattice satisfying the same finiteness condition. There are several versions of this idea, but one particularly simple one is found in Benson and Conway, Diagrams for Modular Lattices.
All of the versions share two basic ideas. We already have one clue for what a geometry should look like for a distributive lattice by considering Birkhoff's representation theorem -- join-irreducible elements are points, and these points have a natural partial order on them. What's new in the modular case is that we also have lines, which are when you have three or more join-irreducible elements such that any two of them have the same join. A complete version of this idea was already found in Faigle and Hermann, but Benson and Conway is essentially a rediscovery, but the paper itself explains the idea very clearly.
Since Conway was more famous for his work on the other kind of lattice, I was curious how many of them were about this kind of lattice. Based on a quick search of paper titles it looks like the answer is: one.
@ShahroozJanbaz People who know him are reporting it on Twitter: https://twitter.com/CardColm/status/1249038195880341505
It's not a joke, he has passed away in Princeton.
The Angel problem is an interesting contribution to the pursuit-evasion branch of game theory, one of those where Conway laid out initial results, and playfully managed to spark further interest resulting in stronger bounds.
A related new MO question: The Angel and Devil problem with a random angel.
Conway's Soldiers. And an interesting special case Reaching row 5 in Solitaire Army.
This is amazing. I didn't know about the version with infinite pegs
The moving sofa problem and the Conway car
Conway worked on the moving sofa problem (find the shape of the largest sofa that can turn a right-angle corner in a corridor).
In Another Fine Math You've Got Me Into, Stewart writes:
« You’re in trouble,” said Wormstein. “You’ve landed yourself with an old chestnut and it’s a tough nut to crack. Nobody even knows where the question came from. Certainly John Horton Conway asked it in the ‘60s, but it’s probably a lot older. At that time the object being moved was a piano, but in view of the obvious piano-sofa isomorphism I think we can conclude that the optimal piano must have the same shape as the optimal sofa. The first published reference that I know is by Leo Moser in 1966. The shape you found [Figure 116] was published soon after by J. M. Hammersley, as part of a tirade against ‘Modern Mathematics,’ and he conjectured that it is optimal. But at a meeting on convexity theory in Copenhagen (some say Ann Arbor) a group of seven mathematicians, including Conway, G. C. Shephard, and possibly Moser, did some informal work on the problem. In fact they worked on seven different variations—one each!” Two are shown in Figure 117; you might like to think about them for yourselves. “And they quickly proved that Hammersley’s answer is not optimal, much as you did.”
In his proposed optimal solution (Geometriae Dedicata volume 42, pages 267–283 (1992)), Gerver cites his private correspondence with Conway.
The variation alluded to by Stewart and considered by Conway is the following: what is the optimal shape of a car that can turn around at a T-junction. The exact solution is, I think, unknown, but the solution is named the Conway car. See Stewart (loc. cit.) and Gibbs : A Computational Study of Sofas and Cars.
(I learned all this from my daughter's project on the topic.)
Great stories! Related MO links: Gerver's sofa. Sofa in a snaky 3D corridor.
The Conway car problem is closely related to the ambidextrous sofa problem mentioned in the Wikipedia link
John Conway and Neil Sloane collaborated often (at least 55 times by MathSciNet's count). One observation they made together answered a previously unanswered question in lattice theory, namely whether there are lattices which are generated by their minimal vectors which have the additional property that the minimal vectors do not contain a basis for the lattice.
They showed that such lattices appear in dimensions as small as $d=11$ by an explicit construction. Later Jacques Martinet and Achill Schürmann discovered a new example in dimension $d=10$ and proved that phenomenon cannot happen for $d\leq 9$ settling the question of for which dimensions lattices of the above type may exist.
Conway had a more intuitive, although informal, proof of classification of compact surfaces, called the "ZIP proof", where ZIP stands for "Zero Irrelevancy Proof".
https://web.archive.org/web/20100612090500/http://new.math.uiuc.edu/zipproof/zipproof.pdf
Francis, George K., and Jeffrey R. Weeks. "Conway's ZIP proof." American Mathematical Monthly 106, no. 5 (1999): 393-399.
This is a puzzle rather than a theorem, but I think it fits in this wonderful list:
Conway’s Wizards, as discussed here by Tanya Khovanova.
Last night I sat behind two wizards on a bus, and overheard the
following:
A: “I have a positive integral number of children, whose ages are
positive integers, the sum of which is the number of this bus, while
the product is my own age.”
B: “How interesting! Perhaps if you told me your age and the number of
your children, I could work out their individual ages?”
A: “No.”
B: “Aha! AT LAST I know how old you are!”
Now what was the number of the bus?
The paper by John H. Conway and Joseph Shipman on "extreme" proofs of irrationality of $\sqrt{2}$,
"We shouldn’t speak of ‘‘the best’’ proof, because different people will value proofs in different ways. [...] It is enjoyable and instructive to find proofs that are optimal with respect to one or more such value functions [...] Indeed, because at any given time there are only finitely many known proofs, we may think of them as lying in a polyhedron [...] and the value functions as linear functionals, as in optimization theory, so that any value function must be maximized at some vertex. We shall call the proofs at the vertices of this polygon the extreme proofs.
Terence Tao mentions this paper here, and describes his interaction with some of Conway's contributions to mathematics and with Conway himself. He closes his post with
Conway was arguably an extreme point in the convex hull of all mathematicians. He will very much be missed.
ADDENDUM:
CONWAY published an interesting paper with R.H. Hardin, and N.J.A. Sloane
regarding Packings in Grassmannian Space and it were adressed this question how should N n-dimensional subspaces of m-dimensional Euclidean space be arranged so that they are as far apart as possible? , He gives a way to describe $n$-dimensional subspaces of $m$-space as points on a sphere in dimension $(m-1)(m+2)/2$
That blog post by Tao just seemed like he was trying to turn the conversation back to himself, he said very little about Conway's concrete achievements.
I am sorry but you said (I quote) 'Tao said and described here all Conway contributions to mathematics [sic]'. In that post he says almost nothing about Conway's contributions and admits lack of familiarity with most of his work, so what you said is not accurate.
Tom, I'm sorry, but that sounds like a pretty uncharitable reading of Tao's post. I read it more as a brief personal snapshot of a genius. I think many people here may have reminiscences of the man in that spirit.
@ToddTrimble, Really Tao is a humbel man and am sorry that Tom missed that Tao is the first man who is probably constructed MO and he has a huge contributions here and out of this space like his blog
Ccnway-Shipman is hidden behind a paywall. Can you give some idea of what they proved, please?
Made some substantial edits to the post to ease its reading and clarify the point of the paper being mentioned.
Art Benjamin shared Conway's smart methods for finding by hand small prime factors of 3- and 4-digit numbers in Factoring Numbers with Conway's 150 Method, College Mathematics Journal 49 (2018) 122-125. In the acknowledgment, he thanks Conway "for being such a large prime factor in the mathematics community."
John Conway considered himself a classical geometer, so it seems good to mention that in 1965, he and Michael Guy discovered an anomalous uniform 4-polytope called the Grand Antiprism (pictured here with Jenn3D).
It is a beautiful object with two dual rings of 10 pentagonal antiprisms, connected to each other with 300 tetrahedra. One way to construct it is by diminishing the regular 600-cell.
In the theory of formal languages, Conway's problem asks, if the greatest solution $X$ of $LX = XL$, for some finite language $L$, is regular. Now, we know that this does not have to be the case, but it was an open problem for many years.
It goes back to his book Regular algebra and finite machines, which grew out of the work of one of his PhD students. In the book, he gave a proof of Parikh's theorem that is quite short and elegant. His student published the proof. The original proof is very long and technical.
I studied mathematics, and did some group theory classes. So surely I knew about John Conway. As I started my PhD in theoretical computer science, it was a little bit surprising to find out that he had done some work in formal language theory. The book has a somewhat unconventional take on it. As far as I remember, in it he introduced biregular relations, which seemed to be quite similar to what was later introduced as an algebraic treatment of transductions. Also, he introduced the factor matrix of some regular language, which is also called the universal automaton.
As a graduate student, Conway proved that any integer is the sum of at most
$37$ integer $5$-th powers.
I think I read this in Siobhan Roberts'
Genius at Play,
which I cannot access now.
Otherwise, I have not been able to find a citation for this result.
I would appreciate any confirmation of this $5$-th powers theorem.
This story is definitely in Genius at Play, which I have in front of me. Previously it was mentioned in Charles Seife’s 1994 article Impressions of Conway. There may well be earlier sources too.
This is known as Waring's problem: https://en.wikipedia.org/wiki/Waring%27s_problem#The_number_g(k) . The result g(5) = 37 is usually attributed to Chen Jingrun, who published before Conway wrote up his proof. https://en.wikipedia.org/wiki/John_Horton_Conway#Number_theory The relevant discussion appears on page 42 of "Genius at Play".
The link to the Seife article no longer works, but see here - seems to be the same article under a different title. Proper citation: Seife, C. (1994) 'Mathemagician: presenting the legendary John Horton Conway; trickster, group theorist, inventor of the game of life', The Sciences, 34(3), 12+, available: https://link.gale.com/apps/doc/A15361712/AONE?u=loyoland_main&sid=googleScholar&xid=6eb11292 [accessed 12 Aug 2024].
It has been some time since I read about it and I find it difficult to find a reference for it but I recall Conway asked, if we call a triangular array of hexagons with $n$ hexagons along its sides $T_n$, which $T_n$ can be tiled by copies of $T_2$, which he named "tribones"?
The answer was some condition on $n$ modulo 12, proved via "Tiling Groups", but he remarkably showed that this result couldn't obtained through any colouring argument!
EDIT: It appears to be mentioned in this overview of tiling results and they provide a helpful illustration:
The condition is $T_n$ can be tiled iff $n=0,2,9,11$ modulo 12.
Many years ago, Conway told me that during his high school years he kept a notebook of his discoveries in triangle geometry. Much later he introduced "Conway triangle notation"--see MathWorld for the standard version and Wikipedia for extensions.
Conway once intended to publish a triangle-shaped triangle book, as recalled by Richard Guy (https://arxiv.org/pdf/1910.03379.pdf): "This might have been titled The Triangle Book, except that John Conway already has a project in hand for such a book. Indeed, Conway’s book might well have been completed but for the tragically early death of Steve Sigur. It might also have been finished, had I been in closer proximity to John."
In addition to the Conway circle (https://mathworld.wolfram.com/ConwayCircle.html), there are also several Conway triangles and a Conway point: see X(384) in the Encyclopedia of Triangle Centers (https://faculty.evansville.edu/ck6/encyclopedia/ETC.html). The Conway point, among the named points on the Euler line of a triangle, has remarkably simple barycentric coordinates:
$$a^4+b^2c^2: b^4+c^2a^2 : c^4+a^2 b^2$$
I'll mention one more of Conway's contributions to triangle geometry: extraversion. Conway wrote, "There's a pun, of course, since I invented the term." Extraversion involves "extraverting" a triangle or turning it inside out, but it also produces "extra versions" of various entities. (from Katherine Merow's "Let's Bring Back That Gee-om-met-tree! (https://www.maa.org/let-s-bring-back-that-gee-om-met-tree).
Here's Conway himself talking about extraversion in 2015: https://youtu.be/O1GhzHmjpDQ
I believe that the theorem to which he referred to as the murder weapon has not been mentioned so far.
The murder weapon is the main theorem in a paper he coauthored with H. S. M. Coxeter and G. C. Shephard in the 1970's and whose title is "The centre of a finitely generated subgroup" (Commemoration volumes for Prof. Dr. Akitsugu Kawaguchi's seventieth birthday, Vol. II.
Tensor (N.S.) 25 (1972), 405-418; erratum, ibid. (N.S.) 26 (1972), 477.).
I first read about this peculiar contribution of his in the interview of I. Hargittai with him that appeared in the March 2001 issue of The Mathematical Intelligencer (pp. 7-14). This is what Conway told Hargittai about the theorem under discussion:
Coxeter came to Cambridge and he gave a lecture, then he had this problem for which he gave proofs for selected examples, and he asked for a unified proof. I left the lecture room thinking. As I was walking through Cambridge, suddenly the idea hit me, but it hit me while I was in the middle of the road. When the idea hit me, I stopped and a large truck ran into me and bruised me considerably, and the man considerably swore at me. So I pretended that Coxeter had calculated the difficulty of this problem so precisely that he knew I would get the solution just in the middle of the road. In fact I limped back afer the accident to the meeting. Coxeter was still there, and I said, "You nearly killed me." Then I told him the solution. It eventually became a joint paper. Ever since, I've called that theorem "the murder weapon" One consequence of it is that in a group if $a^{2} = b^{3} = c^{5} = (abc)^{-1}$, then $c^{610}=1$.
You can take a look at this previous discussion in MO if you feel like hearing of some additional exploits of his...
So did Randall Munroe get this joke from Conway?
I don't know the answer to that question... Anyway, thanks for sharing that link to xkcd with me.
In addition to his well-known Doomsday Algorithm for calculating, he had thoughts on various other calendrical systems and dates, some of which are described here. In particular, he once told me (in person) about an algorithm that allowed him to convert between Hebrew and Gregorian dates in his head (which took him about 1.5 times as long as his Doomsday Algorithm).
I don't remember all the details, and I would love it if someone can please fill them in. I remember that there were three constants called "he", "she", and "it" that you have to memorize to do the computation.
Some brief mentions to this are found here and here, as well as p.388 of his biography.
Little did I know, less than a month after I posted the above answer, this came out: https://slusky.ku.edu/wp-content/uploads/2020/08/CONWAY-AGUS-SLUSKY-PDF.pdf
The Conway immobilizer problem
The set-up:
There are three positions (left, middle, right) on the table and there are three cards labelled $A,B,C$.
Each card is face-up and occupies one of the three positions.
A position might contain zero, one, two. or all three cards.
Only the top card in each position is visible. If exactly two cards are visible, it is not known which of them does conceals the hidden third card.
The game:
The goal is to have all three cards in the middle position with $A$ above $B$ above $C$. (As soon as the goal is reached, a bell rings.)
A move consists in moving one card at a time from the top of one position to the top of another position.
The crux is that the player has no memory of what he has done in the past. The player must decide his moves based only on what is currently visible.
This puzzle has been popularized by Peter Winkler, and it has been discussed in one of Winkler's books.
A full solution to the puzzle can be found in:
John Horton Conway, Ben Heuer:
All solutions to the immobilizer problem.
Mathematical Intelligencer 36 (2014), no. 4, 78-86.
In "THE SEQUENCE SPACES $l(p_\nu)$ AND $m(p_\nu)$" by S. Simons there is the following theorem about sequence spaces of $l_p$-type with varying exponents which is attributed (without a more precise reference) to H.T. Croft and Conway:
For a sequence $p_\nu$ of positive numbers, $l(p_\nu)$ denotes the space of sequences $(a_n)$ such that $\sum_\nu |a_\nu|^{p_\nu}$ is finite and $l_1$ is the usual space of absolutely summable sequences.
Theorem: We suppose that $0 < p_\nu \leq 1$ for all $\nu$, and write $\pi_\nu$ for the conjugate index of $p_\nu$, i.e. $(l/p_\nu) +
> (1/\pi_\nu) = 1$, giving $\pi_\nu$ the value $-\infty$ when $p_\nu =
> 1$. Then the following are equivalent:
$l(p_\nu) = l_1$
$\sum_\nu N^{\pi_\nu} < \infty$ for some integer $N > 1$.
Conway polynomials are irreducible polynomials that provide a basis for finite fields of order $p^n$.
While there is a unique finite field of order $p^n$, there are many ways of representing its elements as polynomials, all resulting in the same field (up to isomorphism).
Conway polynomials provide a standardize choice of basis.
They are nice because they satisfy a certain compatibility condition with respect to subfields (i.e., fields of order $p^m$ with $m$ dividing $n$).
Formally, the Conway polynomial for $\mathbb{F}_{p^n}$ is defined as the lexicographically minimal monic primitive polynomial of degree $n$ over $\mathbb{F}_p$ that is compatible with the Conway polynomials of all its subfields (see Wikipedia for more details).
Of course, imposing the lexicographical ordering is a convention and is necessary to make them unique.
Conway polynomials are very useful when performing computations using computer algebra systems.
They also provide portability among different systems.
Computing them in general is hard, however for many small cases they have been tabulated
(e.g., see the extensive tables computed by Frank Lübeck).
These tables are available, for example, in Sage.
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2025-03-21T14:48:30.289993
| 2020-04-12T02:31:42 |
357199
|
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|
Stack Exchange
|
Seeking a Lyapunov function for a SIR model with immunity loss
We add the immunity loss to the SIR model and obtain the following autonomous system.
$$
\begin{align}
s' &= -is+\alpha r \\
i' &= i s - \gamma i\\
r' &= \gamma i-\alpha r
\end{align}
\tag1
$$
with $$(s+i+r)\big|_{t=0}=1,\ s(0)\ge0,\ i(0)\ge0,\ r(0)\ge0,$$
where prime denotes derivative w.r.t. time, $s,i,r$ represent the proportion of “susceptible”, “infected” and “recovered” individuals, $\beta$ is the kinetic constant of infectiousness, $\gamma$ that of recovery, and $\alpha$ the speed of the immunity loss. Suppose all the coefficients are positive. It is easy to supply the candidates for the long term ($t\to\infty$) asymptotic steady solution, which is to simply set the derivatives on the left hand side of the differential equations to zero, and obtain two solutions.
$$s_\infty=1,\ i_\infty=0,\ r_\infty=0;\tag2$$
or
$$s_\infty=\min(\gamma,1),\ i_\infty=\frac{(1-\gamma)_+}{1+\frac\gamma\alpha},\ r_\infty=\frac{(1-\gamma)_+}{1+\frac\alpha\gamma}.\tag3$$
I conjecture that Solution (2) is achieved either when the initial condition is exactly that, and that Solution (3) is achieved under all other conditions.
The function
${\scr L}(s,i)=s-s_\infty\ln s+w(i-i_\infty \ln i)$ for some positive $w$ seems to fail to play the role of a Lyapunov function. What is an appropriate Lyapunov function for this autonomous ODE?
What do the eigenvalues of linearized system look like at the two fixed points?
@PiyushGrover: I looked at it first but was deterred by the apparent complexity for the second fixed point. I will examine it in detail more. Do the eigenvalue help in finding an appropriate Lyapunov function?
It will help clarify your conjecture (it could be true if all eigenvalues for 2 are non-negative and for 3 are non-positive). In any case, you cannot have a global lyapunov function simply because 2 is a fixed point and hence 3 cannot be globally stable fixed point.
@PiyushGrover: I have put in the local stability analysis. Please review. Are you saying if there are more than one fixed points, there can not exist a global Lyapunov even if one is not locally stable and the second one is locally stable?
@PiyushGrover: Even if there are two locally stable fixed points, we should be able to construct a Lyapunov function just for a specified (compact) domain containing a unique locally stable fixed point, right?
Yes, you can have local lyapunov function, but not global even if the other fixed point is unstable.
For global stability of a fixed point (say 3 here), you need derivative of Lyapunov function to be negative definite everywhere. However, any function you pick will have derivative zero at the "other" fixed point (2).
@PiyushGrover: Yes. I just came back to say the same. Now, since Fixed Point $(2)$ is right on the boundary of the domain of definition $s+i\le1,\ s\ge0,\ i\ge0$, could there be a global Lyapunov function in the interior?
I believe if (2) has all eigenvalues with positive real parts, then it is possible. The trace is not enough to judge that.
@PiyushGrover: You are right. I was lazy and trying to be clever looking only at the trace. One has to examine the eigenvalues themselves. I have edited my local stability analysis. Please review. Fixed Point (2) is unstable for $\gamma<1$. So there is no global stability for Fixed Point (3) in the interior of the domain of definition. Is there a maximal open domain with containing Fixed Point (3) such that Fixed Point (3) is globally asympstotically stable in that open domain?
I don't see how the first two equations form the largest set of independent equations because $s'$ depends on $r$ so you need the third equation as well?
@SampleTime: $r=1-s-i$.
@Hans But only at $t = 0$, at least that is what you wrote in the question. Is this equation supposed to hold for all $t$?
If so, the characteristic polynomial is $\alpha(1 - \gamma)s^2 + \frac{\alpha (\alpha + 1)}{\alpha + \gamma} s + 1$ so this confirms your result $0 <\gamma < 1$ because otherwise some coefficients would be negative.
@SampleTime: Adding all three equations in $(1)$, you get $(s+i+r)'=0$. Does this answer your question?
Thanks for the clarification I didn't notice that. Then you can just use the characteristic polynomial since for a second order system that is sufficient.
We examine the local stability of this system. Since the first two equations of System $(1)$ form the largest set of independent equations, the Jacobian of this system is
$$J(s,i) := \begin{bmatrix}
-i-\alpha & -s-\alpha \\
i & s-\gamma
\end{bmatrix}.
$$
At Fixed Point $(3)$, the eigenvalues are
$$x=-\frac{1+\alpha}{2(1+\frac\gamma\alpha)}(1\pm\sqrt{1-4\delta}),\quad \delta:=\frac{1-\gamma}\alpha\Big(\frac{\alpha+\gamma}{1+\alpha}\Big)^2.$$
$\gamma<1\iff\delta>0\implies \mathbf{Re}(1\pm\sqrt{1-4\delta})>0$. So the system is locally stable there.
At Fixed Point $(2)$, the eigenvalues are
$$x_1:=-\alpha,\ x_2:=1-\gamma.$$
For $\gamma<1$, the fixed point is locally unstable.
Therefore, as Piyush Grover suggests in his comment above, the system is not globally stable for the domain $\big\{(s,i)\,\big|\,s+i\le1,s\ge0, i\ge0, (s,i)\ne(1,0)\big\}$ when $\gamma<1$.
|
2025-03-21T14:48:30.290387
| 2020-04-12T02:38:03 |
357200
|
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|
Stack Exchange
|
Is a gluing of homeomorphic Mazur manifolds diffeomorphic to $S^4$?
A recent paper proves the existence of homeomorphic but not diffeomorphic Mazur manifolds (see also examples of exotic pairs of contractible Stein manifolds).
Let's call them $M_1$ and $M_2$. If we glue $W= M_1 \cup_{\partial M_1=\partial M_2} M_2$, then we get a manifold homeomorphic to $S^4$ by Freedman's theorem. Doubling $M_1$ or $M_2$ yields the standard smooth $S^4$ by a theorem of Mazur. But I'm wondering if $W$ is diffeomorphic to $S^4$?
Actually, I don't know if twisting the double of an Akbulut cork can yield an exotic $S^4$? Any exotic pair of manifolds are related by twisting along a cork, so I guess I'm asking whether anything more is known when the complementary contractible manifolds are homeomorphic?
It can be worth saying that the double is guaranteed to be S^4 only if the Mazur manifold M you start with is made of 0-, 1-, and 2-handles along an Andrews-Curtis trivial presentation. This is not guaranteed for any contractible M, not even if made only of 0-, 1-, and 2-handles.
@BrunoMartelli I see, yes I just had in mind the classic sort of example with one 0,1 & 2-handle like the examples in the paper.
It is diffeomorphic to $S^4$. I have drawn the Kirby picture, please let me know if it is not clear.
(Also I'm sorry that I don't know how to draw Kirby picture in computer so I do old fashioned drawing on notebook)
The Key ideas of the proof are,
1)When we upside down a compact 4 manifold $M$ with 0,1 and 2 handle, then 0 handle becomes 4 handle and 1 handle become 3 handle. Now when we upside down the 2 handle, the co-core become the new attaching circle, which is an unknotted meridian to the orginial attaching cirle of the 2 handle. And the new attaching framing is trivial,i.e, 0. [This is what we use to draw the kirby picture of $M_1\cup M_2$, using the fact that they have identical boundary]
2) 0 framed meridian of a knot always helps to resolve all the crossings of a knot (handle slides).[ This is what we use to simplify the Kirby picture]
3) If a 1 handle and a 2 handle (in the picture a dotted 1 handle and the attaching circle of the 2 handle) geometrically intersect at one point, then they cancel each other. Similarly a 0 framed unknotted 2 handle get canceled by a 3 handle. [this is what we used in last two steps to cancel all but only one 0 handle and one 4 handle and thus we get $S^4$].
Okay, thanks, I think I get the idea.
|
2025-03-21T14:48:30.290584
| 2020-04-12T03:47:29 |
357204
|
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|
Stack Exchange
|
Limit of Hankel function for large complex order, fixed real argument
Consider the Hankel function $H_\nu(z)$ where $\nu=re^{i\theta}$ (real $z>0$, $r>0$, $0\leq\theta<\pi$) as $r\rightarrow\infty$. I am aware that the Bessel function $J_\nu(z)$ has the following dependence, obtained from its series definition:
$$ J_\nu(z)\rightarrow \frac{(z/2)^\nu}{\Gamma(\nu+1)}(1+\mathcal{O}(\nu^{-1})) $$
and that the Hankel function can be obtained from
$$ H_\nu(z)=\frac{J_{-\nu}(z)-e^{-i\pi\nu}J_\nu(z)}{i\sin(\pi\nu)} \qquad \theta\neq 0.$$
But the limit of $J_\nu(z)$ depends on $\theta$ :
for $r\rightarrow\infty, \theta\rightarrow0$, we have $J_{-\nu}>>J_\nu$,
for $r\rightarrow\infty, \theta\rightarrow \pi/2$, we have $J_\nu>>J_{-\nu}$.
This paper
[HIGH FREQUENCY SCATTERING BY AN IMPENETRABLE SPHERE, W. E. I. Sha and W. C. Chew, 2009] mentions an anti-Stokes line at $\theta=\pi/3$, which I believe means there is a drastic shift in the limit for $\theta$ either side of these lines. But the paper does not cite where this comes from.
Numerically it seems for $\pi/3<\theta<\pi/2$, that
$$H_\nu(z)\rightarrow 2J_\nu(z)\rightarrow \frac{2(z/2)^\nu}{\Gamma(\nu+1)}.$$
Is this true analytically?
Did you look at the DLMF? The asymptotic is given explicitly, and the limit $\theta=\pi/3$ is discussed.
|
2025-03-21T14:48:30.290690
| 2020-04-12T05:28:30 |
357209
|
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|
Stack Exchange
|
About the role of total variation measure on boundary reflected stochastic processes
I am reading this paper about stochastic differential equations with reflecting boundary conditions. In page 165, an example equation with an explicit solution is presented. However, I can't see that condition 2.3 in the same page, namely:
The set $\\{ t\in \mathbb{R}^+ : \xi(t)\in D\\}$ has $\mathrm{d}|\varphi|$-measure zero.
holds for every instance of such example. Taking $\xi$ to be any continuous path such that $\xi[0,t_0] \subseteq \partial D$, $\xi(t_0, +\infty) \subseteq D$, the total variation $|\varphi|$ would be constant and positive in $[t_0, +\infty)$, so
$$ \int_{t_0}^{+\infty} \mathrm{d}|\varphi|(s) = \int_{t_0}^{+\infty} |\varphi|(s) \mathrm{d}s > 0$$
and this would be a counterexample. What am I reading incorrectly?
|
2025-03-21T14:48:30.290889
| 2020-04-12T05:47:34 |
357210
|
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|
Stack Exchange
|
Continuous dependence on initial parameters of an ODE for non-Lipschitz functions?
For ODEs, the standard theorem of continuous dependence of initial parameters deals only with functions that are Lipschitz. Do there exist more general results holding for non-Lipschitz functions? If so, could someone direct me to some resources on the topic?
Welcome to MathOverflow! Hmm, if the vector field is not Lipschitz, the solution will not be unique, in general. So it seems to me that it is not even clear what one means by "continuous dependance on the initial parameter", since there could be many solutions for each initial value. (+1 anyway, since I find the question to be an interesting thought.)
For non-Lipschitz differentil equations, there is no uniqueness theorem. So we cannot speak of "dependence" on initial conditions at all. There are some generalizations of uniqueness theorem to more general functions, but it is something very close to the Lipschitz condition.
https://mathoverflow.net/questions/234183/solution-set-of-non-unique-solutions-to-first-order-odes/234198#234198
When we lose uniqueness, maybe we can do with an "existence" formulation of continuous dependence? Something along the lines of "for every bounded open $\Omega\subset\mathbb{R}^N$ there exists a time interval $(-T,T)$ and a continuous function $f: \Omega\times(-T,T) \to \mathbb{R}^N$ such that $\frac{d}{dt} f(\alpha,t) = F(f(\alpha,t))$ with initial data $f(\alpha,0) = \alpha$"?
In regards to my previous comment: I think in at least the case of the Peano theorem (where the vf is only continuous), the usual proof by looking at the integral formulation and using compactness should go through.
The continuous dependence on initial conditions and parameters (and even on the right-hand side in the compact-open topology) is a consequence of the uniqueness. See Theorem 3.2 of Chapter II in Hartman's "Ordinary differential equations" (I call this statement the Kamke lemma). Note that without the uniqueness, the question of continuous dependence (in the classical sense) is incorrect. So when you have this correctness, you automatically have the continuous dependence.
|
2025-03-21T14:48:30.291066
| 2020-04-12T08:02:05 |
357218
|
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|
Stack Exchange
|
Efficient computation of $\sum_{i=1}^{\sqrt{n}} i^2\cdot\left\lfloor{\frac n{i^2}}\right\rfloor$
I need to compute efficiently the sum
$$
\sum_{i=1}^{\sqrt{n}} i^2\cdot\left\lfloor{\frac n{i^2}}\right\rfloor.
$$
We can do this in $O({\sqrt{n}})$ but I need a faster algorithm: for example, it would be fine an algorithm of complexity $O(\sqrt[3]{n})$ (cube root in time) or $O(\log n)$ whatever, but however less than the square root in time.
edit1:
So far what i have got
$$
\sum_{i=1}^{\sqrt{n}} i^2\cdot\left\lfloor{\frac n{i^2}}\right\rfloor= n *\left \lfloor {\sqrt{n}} \right \rfloor - \sum_{i=1}^{i=\left \lfloor {\sqrt{n}} \right \rfloor} n \mod i^{2}
$$
Now how can we efficiently compute
$$
\sum_{i=1}^{i=\left \lfloor {\sqrt{n}} \right \rfloor} n \mod i^{2}
$$
edit2:
We can look at it by taking
$\left\lfloor\frac{N}{i^2}\right\rfloor=1$ whenever $1\leq\frac{N}{i^2}<2$. So whenever $\sqrt{N}\geq i>\sqrt{\frac{N}{2}}$. There are $\left\lfloor\sqrt{N}\right\rfloor-\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor$ such values of $i$.Now how can $i^2$ can be multiplied to the above terms.
This seems to be an active question on codechef competition. It has been flooding math.stackexchange.
Following by comment of Alexey Kulikov we could split our sum in the next way:
$$
\begin{split}
\sum_{i=1}^{[\sqrt{n}]} i^2\left [\frac{n}{i^2}\right ] &=
\sum_{[n/i^2]>[\sqrt[3]{n}]} i^2\left [\frac{n}{i^2}\right ]+\sum_{[n/i^2]\leq [\sqrt[3]{n}]} i^2\left [\frac{n}{i^2}\right ]\\
& =\sum_{i=1}^{\Big [\sqrt{\frac{n}{[\sqrt[3]{n}]+1}}\Big ]} i^2\left [\frac{n}{i^2}\right ]+\sum_{j=1}^{[\sqrt[3]{n}]} j \sum_{i=\big[\sqrt{n/(j+1)}\big]+1}^{\big[\sqrt{n/j}\big]} i^2,
\end{split}$$
while last sum can be computed effectively:
$$
\begin{split}
\sum_{i=\big[\sqrt{n/(j+1)}\big]+1}^{\big[\sqrt{n/j}\big]} i^2 &=\sum_{i=1}^{\big[\sqrt{n/j}\big]} i^2-\sum_{i=1}^{\big[\sqrt{n/(j+1)}\big]} i^2\\
& =\frac{1}{6}\left \{\left[ \sqrt{\frac{n}{j}}\right ]\cdot\left (\left[ \sqrt{\frac{n}{j}} \right ] +1\right ) \cdot\left (2\left[\sqrt{\frac{n}{j}}\right ] +1\right ) \right.\\
& \quad\quad- \left[ \sqrt{\frac{n}{j+1}}\right ]\cdot \left.\left (\left[ \sqrt{\frac{n}{j+1}}\right ] +1\right ) \cdot \left (2\left[ \sqrt{\frac{n}{j+1}}\right ] +1\right ) \right \}.
\end{split}
$$
|
2025-03-21T14:48:30.291197
| 2020-04-12T08:03:45 |
357219
|
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|
Stack Exchange
|
Small linear relations between primitive Pythagorean triples $\mathsf I$
Say $a^2+b^2=c^2$ is a primitive Pythagorean triple. Then consider the Linear Diophantine Equation
$$ua^2+vb^2+xab+ybc+zca=0$$ where $(u,v,x, y, z)\in\mathbb Z^4$ are variables. If $(u,v,x, y, z)\neq(0,0,0,0,0)$ then can we say anything about $\|(u,v,x, y, z)\|_\infty$ or the probability distribution of $\|(u,v,x, y, z)\|_\infty$?
By this I mean can $\|(u,v,x, y, z)\|_\infty$ be much smaller than $\sqrt{\max(a^2,b^2)}$?
Sorry, how do you want to get any small solution if $a|y,b|z$ and $c|x$ provided $(a,b)=1$?
$ybc=-a(xb+zc)$, so $a|ybc$. If $a,b,c$ are pairwise coprime then $a|y$ is nesessary.
@PavelKozlov I updated and for this I am pretty sure we can do better that $\sqrt{\max(a^2,b^2)}$.
Yes, when $m>n>0$ and
$$ a = m^2 - n^2 $$
$$ b = 2mn $$
$$ c = m^2 + n^2 $$
then
$$ -n a^2 +(m-n)b^2 - n ab +(m-n)bc - n ca = 0 $$
or quintuple
$$ -n, m-n, -n, m-n, -n $$
There is a second pattern that gives the same optimum when $n$ is small, quintuple
$$ -2n, m-n, m-n, m-n, -2n $$
Do we always such $m,n$?
@VS. of course. It's well-known that every primitive Pythagorean triple is given by those formula (with $\gcd(m,n)=1$, and $m,n$ not both odd).
@WillJagy So $\sqrt{\max(|a|,|b|)}$ is the correct scale up to constants?
@WillJagy Now I am really curious. What about $ua+vb+zc=0$ with $(u,v,z)\neq(0,0,0)$? Is $|(u,v,z)|_\infty$ also at least $\sqrt{\max(|a|,|b|)}$ or should the scale (disregarding constants) be smaller (perhaps $\sqrt[3]{\max(|a|,|b|)}$)?
@WillJagy https://mathoverflow.net/questions/357444/small-linear-relations-between-primitive-pythagorean-triples-mathsfii
|
2025-03-21T14:48:30.291325
| 2020-04-12T11:13:13 |
357227
|
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"Avi Steiner",
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|
Stack Exchange
|
On Grothendieck's abstract definition of differential operators
I have heard that there is the following abstract definition due to Grothendieck of differential operators on a module $M$ over a commutative associative unital algebra $A$ over a field of characteristic 0.
The definition is inductive in the order of differential operator.
Degree zero operators are precisely endomorphisms of $M$ as $A$-module.
Given $f\in A$, let us denote by $\hat f\colon M\to M$ the operator of multiplication by $f$.
A linear map $D\colon M\to M$ is called differential operator of degree at most $d$ if for any $f\in A$ the commutator
$[D,\hat f]$ is differential operator of degree $d-1$.
(If $A$ is the algebra of functions on a smooth affine variety, and $M$ is a space of sections of an algebraic vector bundle then the Grothendieck's definition is equivalent to the standard one.)
I am wondering whether this notion is considered to be useful when $A$ might have nilpotent elements. Are there any interesting examples/ applications of such a situation?
You might consider taking a look at https://arxiv.org/abs/1711.03960v2 by Jack Jeffries a long with the references there
Indeed, the case when $A$ has nilpotent elements is essential to the theory! Check out Grothendieck's article "Crystals and the de Rham cohomology of schemes" in the collection "Dix exposes...": http://scholar.google.com/scholar_url?url=http://cm2vivi2002.free.fr/AG-biblio/AG-43.pdf&hl=en&sa=X&scisig=AAGBfm0Ci64BjfUOe_NAgqgTuIpm2M-5QQ&nossl=1&oi=scholarr Essentially, degree $d$ differential operators on $A$ are certain endomorphisms of $A \otimes A/I^d$, where $I$ is the augmentation ideal.
|
2025-03-21T14:48:30.291460
| 2020-04-12T11:25:11 |
357229
|
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|
Stack Exchange
|
Explicit locally free resolution of a perfect complex $E\oplus F\to (E\oplus F)\otimes \mathcal{O}_X(D)\to (E\otimes \mathcal{O}_X(D))|_D$
Let $X$ be a smooth projective variety over $\mathbb{C}$. Let $E,F\to X$ be 2 holomorphic vector bundles and $D\hookrightarrow X$ be a smooth divisor. Denote by $\mathcal{O}_X(D)$ the line bundle associated to the divisor and let $s: \mathcal{O}_X\to \mathcal{O}_X(D)$ be a section, such that $s^{-1}(0) = D$. Assume also that we have an isomorphism $E|_D\cong F|_D$ of the restrictions to $D$.
Consider the following perfect complex $\mathcal{E}^\bullet$ on $X$:
\begin{equation*}
E\oplus F\xrightarrow{
\begin{pmatrix}
\text{id}\otimes s& 0\\
0&\text{id}\otimes s
\end{pmatrix}}
(E\oplus F)\otimes \mathcal{O}_X(D)\xrightarrow{
\begin{pmatrix}
\rho_E&\rho_F
\end{pmatrix}
} (E\otimes \mathcal{O}_X(D))|_D\,,
\end{equation*}
where $\rho_{(-)}$ corresponds to restriction of sections to $D$, and we use the isomorphism $F|_D\cong E|_D$.
Now it is well known that there exists a locally free resolution of this complex on $X$. However, is it possible to write it down explicitly from what we know?
Edit: By a locally free resolution I mean a complex of vector bundles $L^\bullet$ with a quasi-isomorphism $L^\bullet\to \mathcal{E}^\bullet$. It is important to me that I have such a map, as I would like to use it to construct a differential operator.
Projections to the second summands define a morphism from that complex to the complex
$$
F \stackrel{s}\to F(D)\tag{*}
$$
of locally free sheaves. The cone of this morphism is the complex
$$
0 \to E \stackrel{s}\to E(D) \stackrel{\rho_E}\to E(D)\vert_D \to 0
$$
which is acyclic. Therefore, $(*)$ is a locally free resolution of the original complex.
EDIT. Alternatively, let
$$
K = \mathrm{Ker}\Big((\rho_E, \rho_F) \colon E(D) \oplus F(D) \to E(D)\vert_D\Big).
$$
Then $K$ is locally free, the morphism $(s, s) \colon E \oplus F \to E(D) \oplus F(D)$ factors through $K$, and the complex
$$
E \oplus F \to K
$$
is quasiisomorphic to the original one (via the morphism, which is identical on $E \oplus F$ and is the natural embedding on $K$).
For me a resolutions is a complex of vector bundle $L^\bullet$ with a map $L^\bullet \to \mathcal{E}^\bullet$ which is a quasi-isomorphism. The map goes the wrong way in your case.
For me a resolution is a quasiisomorphism, no matter which direction it goes.
I have changed the question to be more explicit now. It is important to me that it goes the way that I have mentioned, otherwise it is trivial as your answer shows.
Thank you for the edit Sasha. Could you please add some argument explaining why $K$ is locally free? I am struggling to see why that is true.
Because the projective dimension of a locally free sheaf on a Cartier divisor is 1.
|
2025-03-21T14:48:30.291658
| 2020-04-12T11:40:42 |
357231
|
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"George S Cowan",
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|
Stack Exchange
|
History of well founded relations
I have changed the title in the hope of attracting the attention of someone who knows about the history of set theory as well as intuitionistic logic:
Who was the first to state the definition of well-foundedness intuitionistically as the induction scheme?
$$ \forall \phi.\quad\frac{\forall x.(\forall y. y\prec x\Rightarrow \phi y)\Rightarrow \phi x}{\forall x.\phi x} $$
While I'm here, are the following historically correct?
Euclid's Elements Book VII Proposition 31 (the Euclidean algorithm for prime factorisation) says that an infinite descending sequence is impossible amongst the natural numbers.
What other forms of induction and recursion were stated before the 17th century?
Fermat, Pascal and Wallace in the 1650s stated induction in the form of the base case and induction step.
Cantor 1897 (earlier?) proved that, for any two well ordered sets, one is uniquely equivalent to an initial segment of the other.
Mirimanoff 1917 was the first to recognise the importance of the absence of infinite descending sequences in the consistency of set theory. I have made an English translation of this paper.
von Neumann 1925 proposed the first version of the axiom of foundation, that the system of set theory is the minimal one.
Zermelo 1930 first asserted the axiom of foundation as the lack of infinite descending sequence of elements.
Zermelo 1935 was the first to study well-foundedness in the abstract, as a tool for proof theory.
von Neumann 1928 was the first to prove the recursion theorem for ordinals.
I am currently working on the categorical reformulation of von Neumann's recursion theorem, for well founded coalgebras:
http://www.paultaylor.eu/ordinals
This includes the full bibliographical details of the above references.
I can read French and Italian reasonably fluently, but unfortunately not German.
Even some guesses would be appreciated!
Postscript: There is a historical introduction in my papers Well Founded Coalgebras and Recursion, which is now with referees and can be found on the webpage mentioned above, along with follow-up work on Ordinals as Coalgebras and infrastructural work on Pataraia's fixed point theorem.
As for the first person to state well-foundedness as opposed to well-ordering, Mirimanoff 1917 uses the idea without naming it, but otherwise the earliest I could find was:
Ernst Zermelo, Grundlagen einer allgemeinen Theorie der mathematischen Satzsysteme, Fundamenta Mathematicae 25 (1935) 135--146, with an English translation in volume I of his Collected Works, edited by Heinz-Dieter Ebbinghaus et al.
What about Dedekind arithmetic (1888), which states induction for the natural numbers with successor, and Peano's (1889) subsequent beautiful and popular use of it to develop the basics of number theory. Also, Cantor made essential use of transfinite recursion in the Cantor-Bendixson theorem (1872), and indeed, these ideas are what led him to the ordinals.
@JoelDavidHamkins. Yes of course. But I was looking specifically for the abstract notion well founded relation, as classically stated. There's another question about who first stated it intuitionisticallly, ie just as the induction scheme. However, that is much more difficult because of the habit of intuitionists to state classical definitions and then argue at length about why they're wrong, instead of just giving the correct constructive definition.
I was thinking it gives a somewhat fuller historical picture, since not everything on your list is about that abstract well-founded concept. In particular, Dedekind's 1888 categoricity result for arithmetic (basically, any two inductive successor operations are isomorphic) amounts to a precursor and special case of the comparability result you mention of Cantor. And Cantor's earlier use of transfinite recursion in the Cantor-Bendixson theorem is the first nontrivial use of transfinite recursion, which I think of as absolutely central and indeed seminal in the history of well-foundedness.
It would be better to tell the story backwards. You would be sure to finish at some point.
As the Postcript says, there is a historical introduction in my paper "Well founded coalgebras and recursion" that is perhaps already too long. The purpose of this Question was to fill in one specific gap in that history. Of course it would be nice to have a comprehensive history of induction, but I am not a qualified historian. Writing an up-to-date such history backwards would, I think, be even more difficult than writing it forwards!
@PseudoNeo I saw what you did there.
Emmy Noether must fit in there somewhere. Computer Scientists always mention her when talking about the foundations for making sure that iterative and recursive algorithms terminate. Unfortunately, I don't know of any translations of her work from the German. Bibliography at https://enacademic.com/dic.nsf/enwiki/9878553 where the only relevant work that I see has the note "By applying ascending and descending chain conditions to finite extensions of a ring, Noether shows that the algebraic invariants of a finite group are finitely generated even in positive characteristic.".
Oh. I see you're way ahead of me on that.
Welcome to MO. My other question attracted a lot of discussion about Noether, but the relationship is still a historical blur to me. In this question I was looking for the reference for a very specific formulation. It might have been one of her German contemporaries, such as Goedel or Gentzen.
But isn't the legitimacy of recursive definition due to Dedekind in 1888? He showed on the basis of axioms of successor that recursive definitions succeed, and this was how he defined addition and multiplication etc. from successor and it was the basis for his categoricity theorem for the second-order theory of the successor operation.
|
2025-03-21T14:48:30.292052
| 2020-04-12T12:09:34 |
357232
|
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|
Stack Exchange
|
Degree of morphisms and isogenies
$\renewcommand{\J}{\mathrm{Jac}} \renewcommand{\F}{\mathbb{F}}$
I am reading this paper by B. Gross, and there is something I don't understand on p. 945. Here is the context: fix a prime $p \equiv 3 \pmod 4$, and define the (hyper)elliptic curves over $\F_p$ given by the (affine) equations
$$X_1 : y^2 = x^p - x,\quad X_2 : y^2 = x^{p+1}-1,\quad
E_1 : y^2 = x^3-x,\quad E_2 : y^2=x^4-1.$$
I checked (using Tate isogeny theorem) that there is a non-zero isogeny $\alpha : \J(X_1) \to \J(X_2)$ between the jacobian varieties (actually both are isogenous to $E_1^{(p-1)/2}$ over $\F_p$), and there is a non-zero isogeny $\beta : E_1 \to E_2$.
Question: There is a morphism $f_2 : X_2 \to E_2, (x,y) \mapsto (x^{(p+1)/4}, y)$ which has degree $(p+1)/4$. Then it is claimed that we therefore get a morphism $f_1 : X_1 \to E_1$ of degree $(p+1)/4$, but I don't see why/how.
Thoughts:
I know that $f_1$ induces a morphism $\phi_2 : \J(X_2) \to E_2$, we get a morphism $\beta \circ \phi_2 \circ \alpha^{\vee} : \J(X_1) \to E_1$, hence a morphism $f_1 : X_1 \to E_1$, but I believe that it has degree at least the degree of $f_2$.
Maybe there is a clever way to compose $\phi_2$ with other isogenies to get the equality of degrees?
In general, given a non-constant morphism $f_2 : X_2 \to E_2$, it might not be possible to get a morphism $f_1 : X_1 \to E_1$ of same degree as $f_2$: just take $X_2 = E_2 = X_1, f_2 = \mathrm{id}$ and $E_1$ an elliptic curve isogenous but not isomorphic to $E_2$.
I am probably missing something easy, but I prefer asking for clarifications.
It's easier if we forget about isogenies: $E_1$ and $E_2$ are isomorphic,
and $X_1$ and $X_2$ are isomorphic, so the cover $X_2\to E_2$
induces a cover $X_1 \to E_1$ of the same degree.
To make this more explicit:
let $d = (p+1)/4$,
and rewrite the curve equations in separate coordinate systems:
\begin{align*}
X_1: Y^2 & = X^p - X \,,
&
X_2: V^2 & = U^{4d} - 1 \,,
\\
E_1: y^2 & = x^3 - x \,,
&
E_2: v^2 & = u^4 - 1 \,.
\end{align*}
Let $i$ be a square root of $-1$ in $\mathbb{F}_{p^2}$.
Then there is an isomorphism $\psi: X_1 \to X_2$
defined by
$$
\psi: (X,Y)
\longmapsto
(U,V) = \left(
\frac{X + i}{X - i}
,
\frac{(i+1)Y}{(X -i)^{2d}}
\right)
$$
(here we need $i^p = -i$),
the degree-$d$ cover $f_2: X_2 \to E_2$ defined by
$$
f_2: (U,V) \longmapsto (u,v) = (U^d, V)
\,,
$$
and an isomorphism $\phi: E_2 \to E_1$
defined by
$$
\phi: (u,v)
\longmapsto
(x,y) = \left(
-i\cdot\frac{u + i}{u - i}
,
\frac{(i+1)v}{(u-i)^2}
\right)
\,.
$$
Composing,
we get a degree-$d$ cover $f_1 = \phi\circ f_2\circ\psi: X_1 \to E_1$,
which is what we wanted... Well, almost what we wanted, because we would probably want $f_1$ to be defined over $\mathbb{F}_p$.
But expanding, we see that $f_1$ is defined by
$$
f_1: (X,Y)
\longmapsto
(x,y)
=
\left(
-i\cdot\frac{
(X + i)^d + i(X-i)^d
}{
(X + i)^d - i(X-i)^d
}
,
\frac{ 2i Y }{
((X + i)^d - i(X-i)^d)^2
}
\right)
\,,
$$
Both of the rational functions
are symmetric with respect to $i \leftrightarrow -i$,
so they are defined over $\mathbb{F}_p$,
and therefore so is $f_1$.
All four curves have plenty of automorphisms (some over $\mathbb{F}_p$, some over $\mathbb{F}_{p^2}$) which you can compose with these morphisms to produce more solutions.
Dear Ben Smith, thank you so much for your helpful answer.
But why is $f_1$ defined over $\Bbb F_p$ ? According to my computation, the second component is not invariant under swapping $i$ and $-i$ (set e.g. $X=0, Y=1, p=7$, so $d=2$).
I think that taking $\dfrac{X + i}{X - i}$ in the first coordinate of $\psi$, and by taking $$\phi(u,v) = \left(
-i \cdot\frac{u + i}{u - i},
i \cdot\frac{(i-1)v}{(u-i)^2}
\right),$$ does give a morphism $f_1$ which is defined over $\Bbb F_p$.
@Watson you're right, I compounded a few typos in my working. I think you might have an $i$ too many on the $y$-coordinate of $\phi$, too. Will update my answer accordingly.
|
2025-03-21T14:48:30.292308
| 2020-04-12T13:00:39 |
357238
|
{
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"authors": [
"Asvin",
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|
Stack Exchange
|
Classification of finitely generated modules over non-commutative rings
Let $\Lambda$ be a commutative integral ring with an automorphism $\sigma$ (I have in mind $\mathbb Z_p[[t]]$ and $\sigma(t) = (1+t)^\alpha - 1$ with $\alpha \in \Lambda^\times$) and $R = \Lambda\{F\}$ with $F\lambda = \sigma(\lambda)F$ for $\lambda \in \Lambda$.
Is there a classification of finitely generated modules over $R$ that are free and finite as modules over over $\Lambda$? I allow faithfully flat base changes of $\Lambda$ so that we can assume it's fraction field is algebraically closed (among other things).
Ultimately, I am only interested in the eigenvalues of F, if that makes sense.
When $\Lambda$ is a field, there is a classification similar to the standard one over PID's in chapter three of "The theory of rings" by Nathan Jacobson.
What about the general case or at least my specific example? Or even when $\Lambda$ is a PID? Ideally, I would want any finitely generated module to be isomorphic to a direct sum of modules generated by one element, perhaps up to finite kernel and cokernel.
Do you mean that $R$ is a skew-polynomial ring in the sense of https://en.wikipedia.org/wiki/Polynomial_ring#Differential_and_skew-polynomial_rings, so that typical elements are of the form $\sum_{i=0}^n \lambda_i F^i$ with $\lambda_i\in \Lambda$? I think you must but wanted to check.
Yes, that's exactly right. I didn't know the name for it.
I don't know about your specific example, but for PIDs in general: for $\Lambda=\mathbb{Z}$ and $\sigma=\text{id}$, so you're looking at $\mathbb{Z}[t]$- modules, I believe it's known that the classification of these (even if you restrict to those that are finitely generated and free over $\mathbb{Z}$) is a wild problem.
Really? At least for similar examples to that, I thought I had an atgument where you tensor with Q, use the classification over PIDs and finally use the freeness over Z to classify such modules. I must have made a mistake somewhere. Do you have a reference for modules over Z[t] that are free as Z modules?
@Jeremy Rickard I am not sure why it is wild. If you restrict to finitely-generated free it seems to be covered by Latimer-MacDuffee: https://en.wikipedia.org/wiki/Latimer%E2%80%93MacDuffee_theorem
I'm not sure right now of a reference for wildness, but there's a pair of papers of Heller and Reiner from the early 60s in Annals Representations of cyclic groups in rings of integers, I and II where they at least show that there are finitely many $\mathbb{Z}$-free indecomposable $\mathbb{Z}C_{p^k}$-modules iff $k<3$ (and such modules are the same as those $\mathbb{Z}$-free $\mathbb{Z}[t]$ where $t$ acts invertibly with order dividing $p^k$.
@BugsBunny I think that is just reducing it to another wild classification problem.
No. Take a matrix by which $t$ acts, compute its characteristic polynomial. Then Latimer-MacDuffee tells you that there are finitely many possibilities with this characteristic polynomial. It does not look wild to me (and it looks like the wild-tame-finite trichotomy is not applicable)
Yes, I see that now. That's a wonderful theorem
@BugsBunny But you have to classify the finitely many possibilities for all the infinitely many possible characteristic polynomials simultaneously to get a classification.
@Asvin If you are happy to assume, assume. Otherwise, you cannot classify finitely-generated $\Lambda$-modules for an arbitrary $\Lambda$ already.
On the other hand, I am okay with extending scalars and in that case, it does seem like it would decompose into a sum of cyclic modules.
@Jeremy Rickard Yes, but this does not make it wild. No relation to pairs of matrices. It sound cotame, hence, the trichotomy may not be applicable.
Okay, I modified the problem to make it (hopefully, much) simpler in response to the comments.
@BugsBunny There's a reference in my answer to this question to a paper of Nagornyı where he claims to prove that a classification of $4n\times4n$ matrices over $\mathbb{Z}/p^2\mathbb{Z}$ up to conjugacy would imply a classification of pairs of $n\times n$ matrices over $\mathbb{Z}/p\mathbb{Z}$ up to simultaneous conjugacy. So even classifying similarity classes of integer matrices mod $p^2$ seems to be a wild problem.
I haven't thought about base changes, but the original problem for $\alpha=1$ (so $\sigma$ is the identity map and $R=\mathbb{Z}_p[[t]][F]$ is just a polynomial ring over $\mathbb{Z}_p[[t]]$, and classifying $R$-modules that are finitely generated and free over $\mathbb{Z}_p[[t]]$ up to isomorphism is equivalent to classifying square matrices over $\mathbb{Z}_p[[t]]$ up to conjugacy) is a "wild" problem (i.e., if you could classify these then you could classify pairs of matrices over some field up to simultaneous conjugacy), and so is probably intractable.
In fact, Theorem 2 of
Gudivok, P. M.; Oros, V. M.; Rojter, A. V., Representations of finite $p$-groups over the ring of formal power series with integral $p$-adic coefficients, Ukr. Math. J. 44, No. 6, 678-689 (1992); translation from Ukr. Mat. Zh. 44, No. 6, 753-765 (1992). ZBL0787.20006.
shows that the classification of representations of the cyclic group $C_{p^2}$ over $\mathbb{Z}_p[[t]]$ is a wild problem, and this is the subproblem of classifying those matrices whose $p^2$th power is the identity.
For $\alpha\neq1$, I think it should still be a wild problem, as the problem of classifying $R$-modules that are finitely generated and free over $\mathbb{Z}_p[[t]]$ should be at least as hard as classifying representations of finite cyclic groups over $\mathbb{Z}_p$, and this is a wild problem for $G=C_{p^3}$ ($p$ odd) and $C_{16}$ ($p=2$) (see the main theorem of
Dieterich, Ernst, Group rings of wild representation type, Math. Ann. 266, 1-22 (1983). ZBL0506.16021.)
|
2025-03-21T14:48:30.292781
| 2020-04-12T13:38:30 |
357240
|
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|
Stack Exchange
|
Pointwise functional derivative as partial derivative
Suppose $x_{1},...,x_{n} \in \mathbb{R}^{d}$ are fixed and $f: \mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$ is given by:
$$ f(\phi) = e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}$$
with $\alpha_{1},...,\alpha_{n} \in \mathbb{C}$. I'd like to know if there is any kind of derivative whose derivative of $f$ would be:
$$ \sum_{j=1}^{n}\alpha_{j}e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}$$
In other words, it would be as if we're considering partial derivatives $\partial/\partial \phi(x_{j})$, $j=1,...,n$, such that
$$\sum_{j=1}^{n}\frac{\partial}{\partial \phi(x_{j})} e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})} = \sum_{j=1}^{n}\alpha_{j}e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}.$$
Is there a way to define such objects in a precise/rigorous way?
A functional derivative is what you need:
$$
\begin{split}
\frac{\partial f}{\partial \phi} &\triangleq \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon} f(\phi+\varepsilon\varphi)\right|_{\varepsilon=0} = \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}e^{\sum_{j=1}^{n}\alpha_{j}[\phi(x_{j})+\varepsilon\varphi(x_i)]}\right|_{\varepsilon=0}\\
& =\sum_{j=1}^{n}\alpha_{j}\varphi(x_j)e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}
\end{split}
$$
and choosing $\varphi\in\mathcal{S}(\Bbb R^d)$ such that $\varphi(x_j)=1$ for all $j=1,\ldots,n$ does the Job.
Notes
The choice of the variation $\varphi$. Since $\{x_j: j=1,\ldots,n\}\subsetneq\Bbb R^d$, the choice of a suitable $\varphi(x)$ is always possible and moreover, you could extend the above formula even to the case $j\in\Bbb N_+=\Bbb N\setminus \{0\}$, provided the indexed (numerable) family of points $\{x_j\}_{j\in\Bbb N_+}$ is contained in a suitable bounded set $M\subsetneq\Bbb R^d$: in all these cases, you can choose a sufficiently large compact set $K\Supset M$ and mollify its characteristic function by the heat kernel (if you do not want a compactly supported function), i.e.
$$
\varphi(x)=\varphi(x,t)=\int\limits_K \mathscr{K}(x-y, t)\,\mathrm{d}y
$$
where
$$
\mathscr{K}(x, t)=\frac{1}{\sqrt{(4\pi t)^d}}e^{-\dfrac{\langle x,x\rangle}{4t}}.
$$
On the domain of definition of $f$(and thus again on $\varphi$). As user131781 noted in the comments, $f$ is defined on a much wider space than the "sole" $\mathcal{S}(\Bbb R^d)$. Its domain is at least the whole $\mathcal{E}(\Bbb R^d)\equiv C^\infty(\Bbb R^d)$, thus also the functional derivative $\frac{\partial{f}}{\partial{\phi}}$ is defined and meaningful for every $\varphi\in C^\infty(\Bbb R^d)$ and the choice $\varphi(x) \equiv 1$ is perfectly licit. However, the choices of $\varphi$ described above show that the very same solution holds in more stricter spaces where, perhaps by unstated context consideration, the functional $\varphi$ is bounded to act.
Does $\varphi = 1 \in \mathcal{S}(\mathbb{R}^{d})$?
No, however you can use a function $\varphi(x)$ such that $\varphi(x_j)=1$ for all $j=1,\ldots, n$. I'll clarify this point in the answer.
I think I got it! Just one more thing. Once $x_{1},...,x_{n}$ are fixed, another approach would be to find functions $\varphi_{j} \in \mathcal{S}(\mathbb{R}^{d})$ such that $\varphi_{j}(x_{j}) = 1$ and $\varphi_{j}(x_{k}) = 0$ for $k\neq j$ right? Then I would have $\sum_{j=1}^{n}\frac{\partial f(\phi)}{\partial\varphi_{j}(x)} = \sum_{j=1}^{n}\alpha_{j}f(\phi)$.
The problem is how to find these functions
@MathMath no, it is not a problem: choose the standard mollifier $$\varphi(x) = \begin{cases} e^{-1/(1-|x|^2)}/I_n& \text{ if } |x| < 1\ 0& \text{ if } |x|\geq 1\end{cases}$$ where $I_n$ is the standard normalizing factor. Then putting $$\epsilon=\frac{1}{10}\sup_\overset{\displaystyle{i,j =1,\ldots,n}}{\displaystyle{ i\neq j}}|x_j-x_i|,$$ you can create the sought for function as a sum of the functions $$\varphi^\epsilon_j(x)=\varphi_j(x,\epsilon) =\epsilon^{-d}\varphi\left(\frac{x-x_j}{\epsilon}\right), \quad j=1,\ldots, n.$$
Great! You helped me so much! Thanks!
I think I got another example of such functions. Take your $\varphi$ with $I_{n} = e$, so that $\varphi(0) = 1$. Then $\mbox{supp} \varphi \subset \bar{B}{1}(0)$ (the closed ball centered at zero and radius 1). Take $\epsilon{j} := \frac{1}{10}\min_{j} |x_{i}-x_{j}|$ and define $\psi_{j}(x) := \varphi\bigg{(}\frac{x-x_{j}}{\epsilon_{j}}\bigg{)}$. Then $\psi_{j}(x_{j}) = \varphi(0) = 1$, $\psi(x_{i}) = 0$ if $i \neq j$ because $\mbox{supp} \psi_{j} \subset \bar{B}{\epsilon{j}}(x_{j})$ and $\varphi$ is $C^{\infty}$.
@MathMath nice example. By the way, I note that you correctly deleted the $varepsilon^{-d}$ term: I should have done the same in my post above.
Again, thank you so much for the help and assistance!
I hope it is not rude to say that this answer, while correct, is much too complicated. The situation is that you have a continuous linear functional on a lcs (a linear combination of $\delta$-distributions) composed with the one dimensional smooth exponential function. Both of these are clearly differentiable and hence so is their composition whose derivative can then be calculated by the chain rule. The use of mollifiers is spurious since this functional is continuous on much larger spaces (a more rigorous condition) which contain $1$, for example, the continuous functions.
Of course, once you know that it is differentiable in this much stronger sense, it is perfectly legitimate to compute it as a directional derivative as in the original answer (without the further considerations).
@user131781 I understand your point: it is simple to see that $f(\varphi)$ is defined on the whole $\mathcal{E}(\Bbb R^d)\equiv C^\infty(\Bbb R^d)$, thus I can readily define the functional derivative in a wider, stronger sense on such spaces (where $1$ is an admissible variation). I'll add a note on your comment to the answer. However, I bounded myself to $\mathcal{S}(\Bbb R^d)$ since we may be right in saying that the domain of definition of $f$ is wider, but MathMath may be bounded to consider this more restrict space by unstated context considerations
|
2025-03-21T14:48:30.293150
| 2020-04-12T13:58:03 |
357242
|
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|
Stack Exchange
|
Two measures of noncompactness of operators
Let $X,Y$ be Banach spaces. We denote by $\mathcal{K}(X,Y)$ the space of all compact operators from $X$ into $Y$. For an operator $T:X\rightarrow Y$, we let $$\|T\|_{e}:=\inf\{\|T-K\|:K\in \mathcal{K}(X,Y)\},$$ and $$\|T\|_{m}:=\inf\{\|T|_{M}\|:codim M<\infty\},$$ where $M$ represents the finite co-dimensional subspace of $X$. It is known that $T$ is compact if and only if $\|T\|_{m}=0$.
Let $X$ be a Banach space. We define $J:X\rightarrow l_{\infty}(B_{X^{*}})$ by $\langle Jx,x^{*}\rangle=\langle x^{*},x\rangle, x\in X, x^{*}\in B_{X^{*}}.$ Let $T:Z\rightarrow X$ be an operator. My question is:
Question. Does $\|JT\|_{e}=\|T\|_{m}$ hold?
Could you, maybe, add a reference for the fact that $T$ is compact iff $|T|_m = 0$?
My go-to books for things like this are Pietsch's History of Banach Spaces and Linear Operators and Spectral Theory and Differential Operators by Edmunds and Evans. Both discuss s-numbers at length. If I remember right $\Vert T \Vert_m$ looks like it might be connected with the so-called `Gelfand numbers' of $T$
The answer is yes. Indeed, let us denote $Y_h=\ell_\infty(B_{Y^*})$. For every $L\in\mathcal{K}(X,Y_h)$,
$\|T\|_m=\|JT\|_m=\|JT-L\|_m\leq \|JT-L\|$. Hence $\|T\|_m\leq \|JT\|_e$.
Conversely, since $Y_h$ is $1$-injective, for every finite codimensional closed subspace $M$ of $X$, then there exists an extension $\hat T:X\to Y_h$ of $JT|_M$ with $\|\hat T\|=\|T|_M\|$. Since $JT-\hat T$ is compact, $\|JT\|_e\leq \|T\|_m$.
|
2025-03-21T14:48:30.293280
| 2020-04-12T14:38:04 |
357245
|
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"authors": [
"Ben McKay",
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"zeraoulia rafik"
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|
Stack Exchange
|
Is this $a(p)=\lim_{r\to \infty} \frac{VolS(p,r)}{e^{h r}}$ exists and applied for manifolds with positive curvature?
In $1969$, Margulis proved, for suitable constant $h>0$and $r$ is a positive constant that :
$a(p)=\lim_{r\to \infty} \frac{VolS(p,r)}{e^{h r}}$ with ($(S(p,r)$ is geodesic spheres), exists at each point $p$ in manifolds of negative curvature which it is the main result implies purely exponential growth of volume of geodesic spheres.Really I'am curiouse to know if this limit does exist for Manifolds with positive curvature ,we may look to an example for application this in unit tangent bundle of $S^4$ and probably cohomology $CP^3$ which they admit Riemannian metrics with positive sectional curvature almost everywhere ? and what about continuity of $a(p)$ in this case ?
What is $S(p,r)$? What is $\operatorname{exp}(hr)$?
I meant Volume of spheres
I think volume of balls is governed by Ricci, so should be no problem. Try Gromov, Sign and geometric meaning of curvature.
@BenMcKay Surely you mean the Bishop-Gromov comparison theorem? The denominator usually is the analogous volume in a comparison space. So I believe that the denominator $e^{hr}$ will produce less significant results ($0$ whenever $\operatorname{Ric}\ge 0$ if I am not mistaken). And am I right that $r$ is not meant to be constant?
When $M$ has positive curvature, the limit should always be 0. Indeed if $M$ has non-negative Ricci curvature, Bishop-Gromov tells you that $\frac{Vol S(p, r)}{n\omega_nr^{n-1}}\leq 1$, so the volumes grow at most polynomially.
Bishop's inequality: in any complete Riemannian manifold of Ricci nonnegative curvature, if balls around a point have radii $r$ and $r'=\lambda r$, then their volumes have $\lambda^n V \ge V'$. This is for balls, not for spheres, but I think it is what you are looking for.
|
2025-03-21T14:48:30.293432
| 2020-04-12T15:01:21 |
357247
|
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|
Stack Exchange
|
When two isomorphic subgroups are conjugate?
I would like where could I find a reference in which this question is answered.
Let us consider $H,H'\leq G$ two isomorphic subgroups of a Lie group $G\in \{\mathbf{SO}(n),\mathbf{SU}(n),\mathbf{Sp}(n)\}$. When can we ensure that $H$ is conjugate to $H'$ in $G$?
I know that if these subgroups are maximal among connected proper subgroups and $G\in \{\mathbf{SU}(n),\mathbf{Sp}(n)\}$, they must be conjugate. Some more particular questions related to this are the following:
(1) When can we ensure that two isomorphic subgroups which are maximal among connected proper subgroups, are conjugate in $\mathbf{SO}(n)$?
(2) For instance, is there some way to see how many subgroups of $\mathbf{SO}(2n)$ are isomorphic to $\mathbf{U}(n)$ up to conjugation?
Could you be more precise by "maximal"? Do you mean (maximal among proper subgroups and connected) or "maximal among connected proper subgroups"?
I was thinking about the second: maximal among connected proper subgroups. Would it be a big difference with respect to the first one?
Yes it makes a big difference, just think of subgroups of $\mathrm{SO}(3)$.
But in this case, I don't see the difference. If you have a subgroup of $\mathbf{SO}(3)$ maximal among proper subgroups and connected must be isomorphic to $\mathbf{SO}(2)$ and if it is maximal among connected proper subgroups must be isomorphic to $\mathbf{SO}(2)$, either.
The difference is that $\mathbf{SO}(2)$ is (maximal proper connected), but is not maximal proper (it's contained as subgroup of index 2 in its normalizer).
Ok now I understand, I will edit the question.
For (2), classifying $2n$-dimensional reps of $\mathfrak{sl}_n$ should make some cleaning. For $n\ge 6$, there's only combinations of the standard rep $V$, its dual, and copies of the trivial 1-dim rep, and requiring the existence of a non-degenerate invariant quadratic form narrows to $V\oplus V^*$ and the trivial $2n$-dim rep. This should imply that the only possible embedding $\mathfrak{su}(n)\to\mathfrak{so}(2n)$ is the standard one. I haven't looked at $3\le n\le 5$ but the list of maximal proper subalgebras of $\mathfrak{so}(2n)$ up to conjugation should be known.
What's a reference for the statement for the case $G\in {SU(n), Sp(n)}$?
@YCor: If you don't mind, I'm still struggling with your proof sketch that my answer was wrong. Is the result that the normalizer of $U(n)$ is the normalizer of $\langle J\rangle$ some kind of general property of normalizers-of-centralizers, or something unique to this $U(n)$? Also, it's still not clear to me why this would force the index to be at most 2. (I do understand the result of the argument. Sorry to continue to be a bother, and thanks for your time!)
@JasonDeVito call J-element an element $J$ satisfying $J^2=-1$, and let $J_0$ be the multiplication by $i$. Then the only J-elements $J$ in $\mathrm{GL}_{2n}(\mathbf{R})$ preserved by $U(n)$ are $J_0^{\pm 1}$ (this is because first $J$ commutes with $J_0$, hence belongs to $\mathrm{GL}_n(\mathbf{C})$; since $U(n)$ acts irreducibly its centralizer therein is reduced to scalars, i.e., $\mathbf{C}^*$, in which these are the only elements squaring to $-1$). The normalizer of $U(n)$ preserves the set of J-elements preserved by $U(n)$ (this is general), and hence this ${J_0^{\pm 1}}$ pair.
@YCor: It doesn't seem to be true that a J-element in $\mathrm{GL}{2n}(\mathbf{R})$ necessarily commutes with $J_0$ (though certainly all J-elements in $\mathrm{GL}{n}(\mathbf{C})$ do!). Consider $J_0 = \begin{bmatrix} 0 & 1 \ -1 & 0\end{bmatrix}$ and $J = \begin{bmatrix} 0 & 2\ -1/2 & 0\end{bmatrix}.$ Anyway, I'll continue to think about this tomorrow. (If you'd like to continue via email, I'd be happy to. I'm also happy to continue in the comments, but the OP may be getting sick of all the pings.)
@JasonDeVito I didn't say this. I said that a J-element $J$ preserved by $U(n)$ has to commute with $J_0$. That $J$ is preserved by $U(n)$ means that $J$ centralizes $U(n)$. Since $U(n)$ contains the "scalar" matrix $J_0$, $J$ has to commute with $J_0$. (Better here than email for me – if the discussion becomes longer the system will propose the chat mode)
@YCor: Your last message was all I needed - sorry for misunderstanding your previous claim. And thanks for your patience. (And for when I read this in the future: if $A,B\in N_{O(2n)}(\langle J_0\rangle)\setminus \langle J_0\rangle$, then it's easy to compute that $AB$ commutes with $J_0$, hence $AB\in U(n)$, hence $U(n)\subseteq N_{O(2n)}(\langle J_0\rangle)$ has index at most $2$. Since $U(n)\subseteq N_{O(2n)}(U(n))\subseteq N_{O(2n)}(\langle J_0\rangle)$, the index $2$ claim now follows.)
|
2025-03-21T14:48:30.293735
| 2020-04-12T15:20:50 |
357250
|
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|
Stack Exchange
|
Equivalence classes of a circle of n bits upon flipping 3 consecutive 0s to 1s or vice versa
Consider a circle of n-bits and define the equivalence relation as follow:
Two configurations A and B of the n-bits circle are equivalent if they can be transformed into each other by performing a sequence of "3-bit flip" operations.
Here the "3-bit flip" can only flip 3 consecutive 1s to 0s or vice-versa, namely 111<->000.
The question is how to compute the equivalence class of the n-bits string on a circle? Note that the only operation allowed is the "3-bit flip", any other operations such as overall rotation are not allowed in the equivalence relation. I guess that there will be a lot of invariants that characterize the equivalent classes, but I can only construct a small number of them (such as the numbers of 0s module 3).
Any help is appreciated, thanks in advance.
The sequence begins (with n=3): 7, 8, 13, 27, 32, 51, 98, 130, 210, 374, 542, 872, 1505, 2268, 3663, 6197, 9552, 15429, 25880, 40298, 65146, 108588, 170266, 275296, 457119, 719864, 1164165, 1927235, 3044980. Code to calculate this: https://github.com/Pazzaz/equivalence-classes . (I reposted this comment as I noticed I had made a mistake)
Hi Pazzaz, thank you for the answer! I used BFS to classify all binary strings and got the same answer!
It's unclear if you consider $n$-bit circular strings up to rotations, and this would affect the answer. Still, in either case a rough idea is as follows.
Transform $111$ to $000$ until no $111$ remains in the string. Then the string is formed by runs of $0$'s separated by $1$ or $11$. Notice that we can move $000$ from one run to a neighboring run with two $3$-bit flips, and so we can assume that there is at most one run of $0$'s of length $\geq 3$. Hence, there are two major types of equivalence classes:
There is no run of $0$'s of length $\geq 3$. Such classes are classified by a sequence of run lengths $(u_1,z_1,\dots,u_k,z_k)$ for some $k\geq 1$, where $u_i\in\{1,2\}$ is the $i$-th run length of $1$'s and $z_i\in\{1,2\}$ is the $i$-th run length of $0$'s, with $u_1+z_1+\dots+u_k+z_k=n$.
There exists a run of $0$'s of length $\geq 3$. Such classes are classified by a sequence of run lengths $(u_1,z_1,\dots,u_k)$ for some $k\geq 0$, where again $u_i,z_i\in\{1,2\}$, and $u_1+z_1+\dots+u_k\leq n-3$.
Thank you for your answer! Previously I thought the circular strings that only differ by rotation are in the same class, then I realized that it may not be the case. Thank you for pointing it out! I have edited the post to clarify this.
Hi Max, while I agree with the classification of your first-type classes ("frozen class"), I think some classes in your second-type classes may be equivalent to each other. For example, for n=8 the string 00000101 and 00001001 are in fact in the same class, because they can be transformed into each other by moving around 000: 00000101->00101000->00001001. I think this is because when we try to eliminate all 111 in a string, there may be multiple ways to do this and may result in multiple "classes" that are in fact equivalent.
As I said, it's a rough idea. To make the second case more accurate one needs to consider two subcases: (i) all runs of $0$'s (including the longest) have length $\equiv 1\pmod{3}$; and (ii) the longest run has length $\equiv 2\pmod{3}$.
|
2025-03-21T14:48:30.293977
| 2020-04-12T15:21:18 |
357251
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daniel Santiago",
"Pete L. Clark",
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/1149",
"https://mathoverflow.net/users/149325",
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"url": "https://mathoverflow.net/questions/357251"
}
|
Stack Exchange
|
Do there exist general conditions for cyclicity of unit groups of quotient rings (generalizations of the primitive root theorem)?
Let $R$ by a commutative ring with $1$, and $I \subset R$ a non-zero integral ideal in $R$. When $R$ has finite quotients, and $I = P$ is prime in $R$, the group of units $(R/P)^{\times}$ of the finite ring $R/P$ is cyclic as $R/P$ is a finite field. Do there exist known sufficient and necessary conditions on $R$ and $I$ in general or for certain classes of unital rings for cyclicity of $(R/I)^{\times}$ ? In particular, do there exist more general analogues of the primitive root theorem, which answers this question for $R = \mathbb{Z}$ in terms of number-theoretic criteria on the positive generators of the principal ideals $I = (n)$?
For anyone else who reads this post, this is also discussed in an old SE post: https://math.stackexchange.com/questions/73498/is-the-group-of-units-of-a-finite-ring-cyclic
Throughout, let $R$ be a Noetherian ring and $I \subseteq R$ an ideal such that $R/I$ is finite. Then $R/I$ is Artinian, so we may write $I = I_1 \cdots I_r$ with $I_i = \mathfrak m_i^{n_i}$ where $\mathfrak m_1, \ldots, \mathfrak m_r \subseteq R$ are pairwise distinct prime ideals and $\mathfrak m_i^{n_i} \subsetneq \mathfrak m_i^{n_i-1}$. Let $k_i = R/\mathfrak m_i$ and $p_i = \operatorname{char} k_i$, and write $q_i$ for the order of $k_i$ and $a_i$ for the order of $(R/I_i)^\times$.
Theorem. Let $R, I, I_i, \mathfrak m_i, n_i, k_i, p_i, q_i, a_i$ as above. Then $(R/I)^\times$ is cyclic if and only if all of the following hold:
The $a_i$ are pairwise coprime;
If $n_i > 1$, then $k_i$ is the prime field $\mathbf F_{p_i}$ and $\mathfrak m_i/\mathfrak m_i^2$ has dimension $1$ (over $\mathbf F_{p_i} = k_i$);
When $p_i \neq 2$, either $n_i \leq 2$ or $R/I_i \cong \mathbf Z/p^{n_i}$;
When $p_i = 2$, we have $n_i \leq 3$ and $R/I_i \not\cong \mathbf Z/8$.
If these hold, then $a_i = q_i - 1$ if $n_i = 1$ and $a_i = \phi(p_i^{n_i}) = p_i^{n_i-1}(p_i - 1)$ if $n_i > 1$, and the order of $(R/I)^\times$ is the product of the $a_i$.
For $I = (m) \subseteq \mathbf Z$, the criteria are satisfied if and only if $m \mid 4$ or there exists an odd prime $p$ such that $m = p^n$ or $m = 2p^n$.
Indeed, criterion 2 is automatically satisfied. Criteria 3 and 4 give $n_i \leq 2$ if $p_i = 2$. We have $a_i = \phi(p_i^{n_i})$, which is even as soon as $p_i > 2$ or $n_i \geq 2$. This gives the result for $R = \mathbf Z$.
Notation. Given an Artinian local ring $(R,\mathfrak m)$ with $\mathfrak m^n = 0$ and $\mathfrak m^{n-1} \neq 0$, write $v \colon R \to \{0,\ldots,n\}$ for the function such that $(r) = \mathfrak m^{v(r)}$ for all $r \in R$. If $p = \operatorname{char}(R/\mathfrak m)$, write $e = v(p)$ (if $R = \mathcal O_K/\mathfrak m^n$ for a finite extension $\mathbf Z_p \to \mathcal O_K$, then $e$ is the minimum of $n$ and the ramification index of $\mathbf Z_p \to \mathcal O_K$).
Proof of Theorem. The Chinese remainder theorem gives
$$R/I \cong \prod_{i=1}^r R/I_i,$$
so we get the same statement for the unit groups. Since a product $\prod_i A_i$ of finite abelian groups is cyclic if and only the $A_i$ are cyclic of pairwise coprime degrees (again by the Chinese remainder theorem!), we get criterion 1 and reduce to the case $r = 1$. We will drop all subscripts $i$ and write $I = \mathfrak m^n$ with residue field $k$ of characteristic $p > 0$. Replacing $R$ by $R/I$ we may assume that $\mathfrak m^n = 0$ and $\mathfrak m^{n-1} \neq 0$, and we use the notation of Notation above. For $0 \leq i \leq j \leq n$ we get a short exact sequence
$$1 \to \frac{1+\mathfrak m^i}{1+\mathfrak m^j} \to \big(R/\mathfrak m^j\big)^\times \to \big(R/\mathfrak m^i\big)^\times \to 1.\tag{1}\label{1}$$
Moreover, for $1 \leq i \leq j \leq 2i$ we have an isomorphism
\begin{align*}
\psi \colon \frac{\mathfrak m^i}{\mathfrak m^j} &\stackrel\sim\to \frac{1+\mathfrak m^i}{1+\mathfrak m^j}\tag{2}\label{2}\\
x &\mapsto 1+x.
\end{align*}
Indeed, it is clearly a bijection, and the formula
$$(1+x)(1+y) = 1 + x + y + xy \equiv 1 + x + y \pmod{\mathfrak m^{2i}}$$
shows that $\psi$ is a homomorphism.
Criteria 2, 3, 4 are necessary
We will first show that criteria 2, 3, and 4 are necessary. We will use repeatedly that subquotients of cyclic groups are cyclic. For criterion 2, if $n > 1$ then the sequence (\ref{1}) and the isomorphism (\ref{2}) for $(i,j) = (1,2)$ show that
$$\mathfrak m/\mathfrak m^2 \hookrightarrow \big(R/\mathfrak m^2\big)^\times.$$
Thus if $R^\times$ is cyclic, so are $(R/\mathfrak m^2)^\times$ and hence $\mathfrak m/\mathfrak m^2$, so
$$\dim_{\mathbf F_p} \mathfrak m/\mathfrak m^2 = 1.$$
This also forces $k = \mathbf F_p$ since $\mathfrak m/\mathfrak m^2$ is actually a $k$-vector space, proving criterion 2. This also implies that $\mathfrak m^i/\mathfrak m^{i+1} \cong \mathbf F_p = \mathbf Z/p$ for $i < n$, which together with the sequence (\ref{1}) and the isomorphism (\ref{2}) proves the formula
$$a = \begin{cases}q-1, & n = 1, \\ p^{n-1}(p-1), & n > 1. \end{cases}$$
For criterion 3, assume $p>2$. If $e = 1$, then $(p) = \mathfrak m$, so the unique map $\mathbf Z/p^n \to R$ is surjective (see e.g. Tag 00DV(11)), hence an isomorphism by length considerations. Thus it suffices to show that if $e > 1$ and $n \geq 3$, then $(R/\mathfrak m^3)^\times$ is not cyclic. The truncated exponential
\begin{align*}
\exp \colon \mathfrak m/\mathfrak m^3 &\to \big(R/\mathfrak m^3\big)^\times\\
x &\mapsto 1 + x + \tfrac{x^2}{2}
\end{align*}
is an injective group homomorphism (here we use $p > 2$). Since $e \geq 2$, every element in $\mathfrak m/\mathfrak m^3$ is killed by $p$, so we conclude that $(R/\mathfrak m^3)^\times$ contains $\mathfrak m/\mathfrak m^3 \cong \mathbf Z/p \oplus \mathbf Z/p$, hence cannot be cyclic. This shows criterion 3.
For criterion 4, it is clear that $(\mathbf Z/8)^\times$ is not cyclic. Similar to the above, we see that $e = 1$ iff $R = \mathbf Z/2^n$, so it suffices to show that $(R/\mathfrak m^4)^\times$ is not cyclic if $n \geq 4$ and $e \geq 2$. For $x \in \mathfrak m^2$, we get
$$(1+x)^2 = 1 + 2x + x^2 \in 1 + \mathfrak m^4$$
since $2 \in \mathfrak m^2$. Thus, all $4$ elements of $(1+\mathfrak m^2)/(1+\mathfrak m^4)$ have order $2$, so $(R/\mathfrak m^4)^\times$ is not cyclic. This shows criterion 4.
Criteria 2, 3, 4 are sufficient.
Conversely, given a finite Artinian local ring $(R,\mathfrak m)$ satisfying criteria 2, 3, and 4 (where $n = v(0)$ is the smallest integer such that $\mathfrak m^n = 0$), we have to show that $R^\times$ is cyclic. Clearly the case $n = 1$ is good, since $\mathbf F_q^\times$ is cyclic of order $q-1$. The case $n = 2$ is also good, by the sequence (\ref{1}) and the isomorphism (\ref{2}): by assumption $\mathfrak m/\mathfrak m^2$ is cyclic of order $p$, and $k^\times$ is cyclic of order $p-1$. Then the sequence (\ref{1}) for $(i,j) = (1,2)$ splits and the middle term is cyclic by the Chinese remainder theorem.
For $p > 2$ we have to show that $(\mathbf Z/p^n)^\times$ is cyclic. This follows since the $p$-adic exponential
\begin{align*}
\exp \colon p\mathbf Z_p &\to 1 + p\mathbf Z_p\\
x &\to \sum_{i=0}^\infty \frac{x^i}{i!}
\end{align*}
converges (in general, it converges when $v(x) > \tfrac{e}{p-1}$, so we're using that $e = 1$ and $p > 2$) and defines isomorphisms $p^i\mathbf Z_p \cong 1 + p^i\mathbf Z_p$ for all $i \geq 1$, hence an isomorphism
$$\frac{1 + p\mathbf Z_p}{1+p^n\mathbf Z_p} \cong \frac{p\mathbf Z_p}{p^n\mathbf Z_p} \cong \mathbf Z/p^{n-1}.$$
Then the sequence (\ref{1}) again splits (this time with $(i,j) = (1,n)$), and the Chinese remainder theorem shows that $(\mathbf Z/p^n)^\times$ is cyclic. (In fact the $p$-adic exponential gives $\mathbf Z_p^\times \cong \mu_{p-1} \times p\mathbf Z_p$, where $\mu_{p-1}$ are the $(p-1)^{\text{st}}$ roots of unity, so $\mathbf Z_p^\times$ is procyclic with generator $(\zeta_{p-1},p) = \zeta_{p-1}\exp(p)$ for a primitive $(p-1)^{\text{st}}$ root of unity $\zeta_{p-1}$.)
For $p = 2$ we have to show that $R^\times$ is cyclic if $n = 3$ and $e > 1$. We claim that $R^\times$ is generated by $1+\pi$ for any $\pi \in \mathfrak m \setminus \mathfrak m^2$. Indeed, $(1+\pi)^2 = 1 + 2\pi + \pi^2 \equiv 1 + \pi^2 \pmod{\mathfrak m^3}$ since $2 \in \mathfrak m^2$. Thus, $(1+\pi)^2 \neq 1$, so $1 + \pi$ has order $4$, hence generates.
This shows that the criteria are sufficient. We already saw that they are necessary and that the final statement holds. $\square$
This is amazing! Thank you so much! Out of curiosity for the infinite case, does there exist a ring $R$ with non-zero integral ideal $I$ such that $(R/I)^{\times} \cong \mathbb{Z}$? I think that such an example could come from group rings but I'm not certain
A commutative ring $R$ with $R^\times = \mathbf Z$ necessarily has characteristic $2$, since $-1$ has order $2$ unless $-1 = 1$. It is necessarily reduced since $1+x$ has order $2$ if $x \neq 0$ but $x^2 = 0$. An example is $R = \mathbf F_2[x,x^{-1}]$, which is the group ring $\mathbf F_2[\mathbf Z]$. Any such $R$ contains $\mathbf F_2[x,x^{-1}]$. Whenever $R$ is an example, so is the polynomial ring $R[x_i\ |\ i \in I]$ for any set $I$. Classifying such rings is almost certainly impossible (as is for example writing down all $\mathbf C$-algebras with $R^\times = \mathbf C^\times$).
$\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$
[Throughout this answer all rings will be commutative (and unital!).]
It seems that van Dobben de Bruyn has essentially rediscovered a theorem of Gilmer:
Gilmer, Robert W., Jr. Finite rings having a cyclic multiplicative group of units. Amer. J. Math. 85 (1963), 447-452.
A couple of preliminary comments: (i) In van Dobben de Bruyn's answer, we may as well take $I = (0)$: that is, he is giving necessary and sufficient conditions on a finite commutative ring to have cyclic unit group. (ii) A finite ring $R$ is indeed Artinian, hence a finite product $\prod_{i=1}^r R_i$ of local rings $R_i$, each of which must have prime power order. As seen in his answer, we quickly find that $R^{\times}$
is cyclic iff each $R_i^{\times}$ is cyclic and $\# R_1^{\times},\ldots, \# R_r^{\times}$
are pairwise coprime. Thus the critical case is the classification of finite local rings with cyclic unit group. Here is Gilmer's result:
Theorem Let $R$ be a finite local ring. Then $R^{\times}$ is cyclic iff $R$ is isomorphic to one of the following rings:
(A) A finite field $\F$.
(B) $\Z/p^a \Z$ for an odd prime number $p$ and $a \in \Z^+$.
(C) $\Z/4\Z$.
(D) $\Z/p\Z[t]/(t^2)$ for a prime number $p$.
(E) $\Z/2\Z[t]/(t^3)$.
(F) $\Z[t]/\langle 2t,t^2-2 \rangle$, a $\Z/4\Z$-algebra of order $8$.
To compare Gilmer's classification to van Dobben de Bruyn's it is helpful to observe that the local rings of order $p^2$ are $\F_{p^2}$, $\Z/p^2\Z$ and $\Z/p\Z[t]/(t^2)$ and to know the six local rings of order $8$.
By the way, Gilmer's Theorem appears as Theorem 5.14 in this expository note of mine, where it used to derive Theorem 5.15, a 2013 result of Hirano-Matsuoka that explicitly determines the product over all elements of the unit group of a finite ring. (Thus it is a generalization of Wilson's Theorem that $(p-1)! \equiv -1 \pmod{p}$. It seems weird that it is so recent.) I wanted to include the proof of Gilmer's Theorem in the note, but it is rather long and computational. The proof of van Dobben de Bruyn looks a bit shorter!
A final comment leading to a question: It turns out that all the rings in Gilmer's classification are principal, i.e., every ideal is principal. (This is obvious except for (F), in which case you can see my paper if you don't want to do the calculation yourself.) In other words, for a finite ring $R$ the property that the unit group be cyclic forces every $R$-submodule of $R$ to be cyclic. Is that just a coincidence, or can it be proved directly?
Wow thank you so much!! I had been told the classification results that I asked for along the lines of Gilmer's work exist but I had trouble finding them. It is amazing to see old results be rediscovered like this.
For your final question: as in my proof, it is necessary that $\mathfrak m/\mathfrak m^2$ has dimension $0$ or $1$ over $R/\mathfrak m$. Such rings are always principal; see e.g. Prop. 8.8 in Atiyah–MacDonald. (In fact, using the Cohen structure theorem plus a little computation, any finite local Artinian ring with $\dim \mathfrak m/\mathfrak m^2 = 1$ can be written as a quotient of a finite extension of $\mathbf Z_p$ ― this was my first approach until I realised it's really a question about Artinian rings.)
Thanks, that's helpful.
|
2025-03-21T14:48:30.295183
| 2020-04-12T16:21:41 |
357256
|
{
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"Aleksei Kulikov",
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"Giorgio Metafune",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357256"
}
|
Stack Exchange
|
Poincaré inequality under weighted average condition
Let $\Omega=[0,1]^2$ be the unit square and $a>0$.
1) I would like to know one estimate of the constant $C(a)$ such that
$$
\forall u\in W^{1,2}(\Omega)\quad \int_\Omega u^2\le C(a)\int_\Omega |\nabla u|^2,
$$
under the weighted average and periodicity conditions
$$
\int_\Omega e^{-ay}u(x,y)\, dx dy=0,\qquad u(0,y)=u(1,y).
$$
Or, at least, a proof of existence.
2) How is the Euler-Lagrange equation of the associated variational problem? Is
$$
\Delta u+\lambda u+\mu e^{-az}=0,\quad \int_\Omega u^2=1,\quad \int_\Omega e^{-ay}u=0
$$
correct?
The first should be true (even without the periodicity condition) by expanding everything into the sines and cosines (aka eigenvalues of the laplacian). But I don't understand the second question.
The second question is about the calculation of the optimal constant $C$ and the function $u$ realizing the infimum of $ |\nabla u|^2/ |u|^2$
Can you give an (even very rough) estimate of $C$ by means of the method of expansion you mentioned?
The first follows by contradiction (without the periodicity conditions) as for $a=0$. If false, you get a sequence $(u_n)$ of norm one functions and gradients tending to 0 in $L^2$. By compactness a subsequence converges in $L^2$ to a norm 1 function whose gradient is zero, hence to a constant. However, the mean zero condition (which is inherited by the limit) forces the constant to be 0 and contradicts norm 1.
Thank you for the explanation
My back of the envelope computation suggests that $C(a) = C\sqrt{a+1}$ works.
Actually, a more precise estimates suggest that $C(a)$ is uniformly bounded!
Interesting. When you are comfortable you can post your answer. Thanks
For $u\in W^{1, 2}(\Omega)$ set $L(u) = \int_\Omega e^{-ay}u(x, y)dxdy$. We want to prove that if $L(u) = 0$ then $||u||_{L^2}\le C||\nabla u||_{L^2}$ for some universal constant $C$. We will first deal with the more interesting case $a \ge 1$ for simplicity.
Since $W^{1, 2}(\Omega)$ is a Hilbert space and $L$ is a continuous linear functional there exists $f\in W^{1, 2}(\Omega)$ such that $L(u) = \langle u, f\rangle_{W^{1, 2}}$. First of all let us prove that $||f||_{W^{1, 2}} \le \frac{C}{a}$. We have $||f||_{W^{1, 2}} = \sup_{||u||_{W^{1, 2}} = 1} |\langle u, f\rangle|_{W^{1, 2}}$.
$$\langle u, f\rangle_{W^{1, 2}} = \int_\Omega e^{-ay}u(x, y)dxdy = \int_\Omega \frac{\partial u}{\partial y}(x, y)\frac{1}{a}e^{-ay}dxdy + \int_0^1\frac{1}{a}u(x, 0)dx - \int_0^1\frac{e^{-a}}{a}u(x, 1)dx.$$
From this (and the boundedness of the trace map) we can easily see that $|\langle u, f\rangle_{W^{1, 2}}| \le \frac{C}{a}$ if $||u||_{W^{1, 2}} = 1$.
On the other hand we have $\langle 1, f\rangle_{W^{1, 2}} = \int_\Omega e^{-ay}dxdy = \frac{1-e^{-a}}{a} \ge \frac{1-e^{-1}}{a}$.
Let's expand everything into the eigenvectors of the Laplacian $v_0 = 1, v_1, v_2, \ldots $ with eigenvalues $0 = \lambda_0 < \lambda_1 \le \lambda_2 \le \ldots$(in our case they are sines and cosines but we do not need this). We normalize them so that $||v_n||_{L^2} = 1$.
Let $u = \sum_{n = 0}^\infty a_nv_n$, $f = \sum_{n = 0}^\infty b_nv_n$. We have $0 = \langle u, f\rangle_{W^{1, 2}} = \sum_{n = 0}^\infty (\lambda_n + 1)a_n \overline{b_n}$. Therefore
$$|a_0| = |-\frac{1}{\overline{b_0}} \sum_{n = 1}^\infty (\lambda_n + 1)a_n\overline{b_n}| \le \frac{1}{|b_0|} \sqrt{\sum_{n = 1}^\infty (\lambda_n + 1)|a_n|^2}\sqrt{\sum_{n = 1}^\infty (\lambda_n + 1)|b_n|^2} \le C\frac{1}{|b_0|}||\nabla u||_{L^2} *||f||_{W^{1, 2}},$$
where in the first inequality we used Cauchy-Schwarz and in the second we used the standard Poincare inequality for the zero-mean function $u - a_0$ (or the inequality $\frac{1+\lambda_n}{\lambda_n} \le \frac{1+\lambda_1}{\lambda_1} < \infty$).
Recall that $|b_0| = |\langle 1, f\rangle_{W^{1, 2}}| \ge \frac{c}{a}$ and $||f||_{W^{1, 2}} \le \frac{C}{a}$. Therefore we get (denoting all constants and their product by $C$) that
$$|a_0| \le C||\nabla u||_{L^2}.$$
Therefore we have
$$||u||_{L^2} = \sum_{n = 0}^\infty |a_n|^2 = |a_0|^2 + \sum_{n = 1}^\infty |a_n|^2 \le C^2||\nabla u||^2_{L^2} + C'||\nabla u||^2_{L^2} = C_1||\nabla u||^2_{L^2},$$
where in the inequality step we used the above bound for $|a_0|$ and once again standard Poincare inequality for $u-a_0$. This is exactly what we want.
In the case $0 < a < 1$ we can see directly from the definition (without integration by parts) that $|\langle u, f\rangle_{W^{1, 2}}| \le ||u||_{L^2} \le ||u||_{W^{1, 2}}$, therefore $||f||_{W^{1, 2}}\le 1$. On the other hand $\langle 1, f\rangle_{W^{1, 2}} = \frac{1-e^{-a}}{a} \ge c$ for some universal $c > 0$. The rest of the proof is exactly the same.
Ok. One more question: when you write $ \langle u, f\rangle_{W^{1, 2}} = \sum_{n = 0}^\infty (\lambda_n + 1)a_n \overline{b_n}$ I guess there has been an integration by parts on $\int\nabla u\cdot\nabla f$. What about the surface terms?
@AlekseiKulikov I like the proof but I wonder what happens for $p \neq 2$ (also for similar situations). If $p>2$ one can use Sobolev embedding to deduce $L^p$ estimates from $L^2$ (hence also here $C$ does not depend on $a$) but I do not know for $p<2$. Do you see how to do?
@DiegoG7 I don't think it is integration by parts... Like, we have $\langle v_n, v_m\rangle = \delta_{n, m} (\lambda_n + 1)$ by the definition of the eigenvalues and eigenvectors, I believe. I guess the right words here are Neumann eigenvalues
@DiegoG7 Aleksei is using the Laplacian with Neumann boundary conditions.
This means that $v_n$ (hence $u$) have homogeneous Neumann conditions (they would be fine for me)?
@GiorgioMetafune Actually, there is a proof completely avoiding any eigenvalues and working for all $1\le p\le \infty$. Here is a sketch: we can decompose $u = c + u_0$ with $u_0$ zero-mean. Then we have $L(u) = L(c) + L(u_0) = cL(1) + L(u_0)$. Therefore $|c| = |-\frac{L(u_0)}{L(1)}| \le C||u_0||$ for some universal $C$ in our case (by the bounds for $||L||$ and $L(1)$ from the post). Thus $||u|| \le c + ||u_0|| \le (C+1)||u_0|| \le (C+1)C'||\nabla u_0|| = (C+1)C' ||\nabla u||$, where in the second step we used Poincare.
@AlekseiKulikov This proof is very clear. Just a remark: the first norm of $u_0$ (when you estimate $c$) should be that of $W^{1,2}$ where $L$ has a better bound, or just the norm of its gradient.
|
2025-03-21T14:48:30.295574
| 2020-04-12T16:52:12 |
357259
|
{
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"authors": [
"Mikhail Borovoi",
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"kneidell"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357259"
}
|
Stack Exchange
|
Is the connected centralizer of a semisimple element in a connected reductive group also a centralizer?
Let $G$ be a connected reductive algebraic group defined over an algebraically closed field and let $g\in G$ be semisimple. Write $C=\mathrm{C}_G(g)$ and $C^\circ=\mathrm{C}_G(g)^\circ$ for the centralizer of $g$ and for its identity component, respectively.
Question Does there exist $h\in G$ such that $\mathrm{C}_G(h)=C^\circ$?
The statement is true in the case where $C^\circ$ is the Levi factor of a parabolic subgroup of $G$. The proof I know is by computing dimensions.
Fix $T$ to be a maximal torus containing $g$, and let $\Sigma$ (resp $\Phi$) denote the root system of $C^\circ$ (resp of $G$) with respect to $T$. Then, for $h\in T$, it holds that $\mathrm{C}_G(h)=C^\circ$ if and only if $\{\alpha
\in \Phi:\alpha(h)=1\}=\Sigma$ and $\mathrm{Stab}_W(h)$ (the stabilizer of $h$ in the Weyl group of $G$) is generated by the reflections $s_\alpha$ for $\alpha\in \Sigma$.
Given a root systems $\Sigma\subseteq \Psi\subseteq\Phi$ and subgroups $\langle{s_\alpha:\alpha\in \Sigma}\rangle\subseteq S\subseteq W$, put
$$T_\Psi^S=\{h\in T: \{\alpha\in \Phi:\alpha(h)=1\}=\Psi\text{ and }\mathrm{Stab}_W(h)\supseteq S\}.$$
Then, using the assumption that $\Sigma$ is the root system of a Levi subgroup in several key steps, one can show that $\dim T_\Psi^S$ attains a maximum if and only if $\Psi=\Sigma$ and $S=\langle{s_\alpha:\alpha\in \Sigma}\rangle$. Therefore $T_\Sigma^{\langle s_\alpha:\alpha\in \Sigma\rangle}\setminus (\bigcup T_\Psi^S)$ is non-empty, and contains the $h$ we are seeking.
However, this proof falls apart completely if we take $\Sigma$ to be an arbitrary (closed) subsystem of $\Phi$.
I would very much appreciate if anyone could either suggest a different proof for this statement, which hopefully extends to general centralizers, or otherwise provide a counterexample for my question. Thank you.
Note that $\Sigma$ is not an arbitrary closed subsystem, but is the root system of a pseudo-Levi. As such it corresponds to a subset of the simple affine roots. I think this should lead to a similar proof in general - I will try to think this through carefully later if there are no other answers.
You write: $\langle{s_\alpha:\alpha\in \Phi}\rangle\subseteq S\subseteq W$. However, it seems that $\langle{s_\alpha:\alpha\in \Phi}\rangle= W$. Is there a typo in the formula?
@Sam Gunningham Thank you, I was unaware of this terminology, but I think I do know what you mean (I called these Deriziotis root systems, but i guess that's uncommon). In any case, one thing that comes into play in my proof is that the root system of a Levi (as opposed to a pseudo Levi) is not contained in any subsystem of the same rank.. I would be very interested if you could supplement this.
@Mikhail Borovoi: sorry, you're correct. I meant $\alpha\in\Sigma$..
So after thinking about this for a few extra days, I think I found a counterexample. I'm recording it here, as community wiki, in case it would be of interest for anyone in the future.
Let $k$ be an algebraically closed field of characteristic not $2$ and $G=\mathrm{PGSp}_{2n}(k)=\mathrm{GSp}_{2n}(k)/k^\times$, where $\mathrm{GSp}_{2n}(k)$ is the group of similitudes of the standard symplectic form, i.e.
$$\mathrm{GSp}_{2n}(k)=\{x\in\mathrm{GL}_{2n}(k):x^tJx=\lambda J\text{ for some }\lambda\in k^\times\}\text{ where }J=\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix},$$
(I don't know if this notation is common).
$G$ is simple of adjoint type with maximal torus $T=\lbrace d(t_1,t_2):=\left[\begin{smallmatrix}t_1\\&t_2\\&&t_1^{-1}\\&&&t_2^{-1}\end{smallmatrix}\right]:t_1,t_2\in k^\times\rbrace$, (here $[\cdot]$ denotes the class mod $k^\times$ of a matrix; note that $d(\lambda t_1,\lambda t_2)=\lambda d(t_1,t_2)$ implies $\lambda=\pm 1$) and root system with simple roots:
$$\alpha(d(t_1,t_2))=t_1/t_2\text{ and } \beta(d(t_1,t_2))=t_2^2$$
(the other positive roots are $\alpha+\beta$ and $2\alpha+\beta$).
Consider the subsystem $\Sigma=\lbrace\pm \beta,\pm(2\alpha+\beta)\rbrace$ (viz. the long roots). Then $\Sigma$ is the root system of a pseudo-Levi subgroup of $G$ which is isomorphic to $(\mathrm{GL}_n(k)\times\mathrm{GL}_n(k))/k^\times$, and one can easily verify that
$$(\star)\quad \beta(d(t_1,t_2))=(2\alpha+\beta)(d(t_1,t_2))=1\:\iff\: d(t_1,t_2)\in\lbrace\left[\begin{smallmatrix}1\\&1\\&&1\\&&&1\end{smallmatrix}\right],\left[\begin{smallmatrix}1\\&-1\\&&1\\&&&-1\end{smallmatrix}\right]\rbrace.$$
Let $g$ be the non-central element in this set. Then, a standard computation, taking into account that $g=-g$ in $G$, shows that $C_G(g)$ is disconnected (the non-identity connected component is generated by the coset of the Weyl group element permuting $t_1$ and $t_2$). On the other hand, $Z(C_G(g))$ consists of precisely the two elements on the RHS of $(\star)$, so there exists no $g\ne h\in Z(C_G(g))$ such that $C_G(h)\subseteq C_G(g)$, and, in particular, the question above has a negative answer in this case.
I just realized this counterexample too! Glad you figured that out, and sorry if I sent you down the wrong path.
|
2025-03-21T14:48:30.295907
| 2020-04-12T17:53:44 |
357264
|
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"Greg Martin",
"Matt Groff",
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}
|
Stack Exchange
|
Average size of iterated sumset modulo $p-1$,
Given a prime $p$, what is the average size of the iterated sumset, $|kA|$, modulo $p-1$, with $p$ a prime, and $k$ given, with $A$ chosen at random?
You can pick any type of prime you like for $p$, but we must be able to get arbitrarily large $p$'s of this type. Note that the elements are chosen independently from the uniform distribution over $\mathbb{Z}_p$, and we assume that the size of $A$ is fixed.
How, precisely, is $A$ chosen at random? Is there any reason to think that the modulus being one less than a prime is important?
@GregMartin: I'm really trying to determine the iterated product set modulo a prime, but this is equivalent, AFAIK. You can assume that the values in $A$ are evenly distributed. In essence, this is the question I came up with when trying to answer https://math.stackexchange.com/questions/3619521/a-multiplicative-order-bound
"the values in $A$ are evenly distributed" is not specific enough. Is the number of elements of $A$ fixed, or is the probability that each $n\in A$ fixed? Is $A$ chosen with respect to the prime $p$, or as a set of integers?
@GregMartin: Oh, sorry. The number of elements of $A$ is fixed. $A$ is chosen with respect to the prime $p$. Every element is chosen independently of all other elements.
|
2025-03-21T14:48:30.296019
| 2020-04-12T18:24:02 |
357266
|
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"authors": [
"David White",
"Evgeny Shinder",
"Patrick Elliott",
"R. van Dobben de Bruyn",
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|
Stack Exchange
|
To what extent is the derived category of coherent sheaves on a scheme a "homotopy type" of the scheme?
It is well known that the derived category of coherent sheaves (unbounded, bounded, and all cousins) on a scheme $X$ contain most - if not all (depending on specifics) - of the cohomological information about $X$.
I know there is no formal statement to this effect, but how inaccurate is to to think of the derived categories living over X as a kind of homotopy type of $X$?
Is there any hope of a "decomposition" of $X$ into a system $(X_{i})$ (say by iterated blowups) in such a way that the corresponding system of derived categories is an analogue of the Postnikov tower?
I apologise for how vague these questions are, I am just trying to centre myself in the theory.
Small comment: if $X$ is a variety over $\mathbf C$, we can also view it as topological space $X(\mathbf C)$. The quasicoherent derived categories (or even the isomorphism type of $X$ as an abstract scheme!) do not know everything about the classical homotopy type. For example, Serre showed that for a smooth projective variety $X$ over $\mathbf C$ and an abstract field automorphism $\sigma$ of $\mathbf C$, the spaces $X(\mathbf C)$ and $^\sigma!X(\mathbf C)$ can have different fundamental groups.
Maybe you're not asking about the comparison with the classical homotopy type, but if not then I'm not sure what you mean by 'a kind of homotopy type of $X$'.
True I have not been very specific. I guess what I mean by "homotopy type" ought to be a minimal collection of data encoding the information accessible by cohomological methods. Another essential piece is that this data be amenable to some sort of truncation or variation operation in the way the spaces have Postnikov decompositions.
I suppose I am picturing a world in which increasingly complex singularities are analogous to higher homtopical data. In this way the chain of blowups in a resolution of singularities (in characteristic 0) would be something like a Postnikov decomposition in reverse.
Perhaps I should ask about the singularity/homotopy analogy somewhere else, because after cursury thought it's not that ridiculous. After all, both higher homotopical features and singularities provid obstructions and complexity to algebraic avatars of the space. On the other hand this might be nonsense.
Informally, both derived categories and (stable) homotopy types provide a functor from a nonlinear category of spaces to a more linear category. Both formalisms relate to Grothendieck's idea of motives which is supposed to be a universal such functor. Indeed there are conjectural links between motives and derived categories, as well as adjoint functors between triangulated category of motives and A^1-stable homotopy category. Finally, one analogy for Postnikov towers in derived category is the concept of a semiorthognal decomposition.
Does this paper of Ayoub help at all? Just after (3.65) is a tower of blowups, the bottom of page 132 is about the limit of such a tower, and Lemma 2.4.5 is about the analogous Postnikov towers of sheaves. https://arxiv.org/pdf/2010.15004.pdf To me, as a model category theorist, I'd relate the two by saying both are right Bousfield localizations. To get at questions of convergence of the corresponding sequence of functors, I'd think about Goodwillie calculus. But I feel this is more homotopical than you wanted.
There are also several papers that try to recover $X$ from the (derived) category of coherent sheaves over $X$, e.g., this recent PhD thesis: https://hal.science/tel-04018741/document. On page 9 it also references the non-derived version of that question (Gabriel-Rosenberg reconstruction). Sequences of blow-ups, and the corresponding functors between model categories are also treated by Orlov: Smooth and proper noncommutative schemes and gluing of DG categories
|
2025-03-21T14:48:30.296292
| 2020-04-12T20:37:20 |
357275
|
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"sort": "votes",
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|
Stack Exchange
|
Upper bounds on the genus of the surface produced by Seifert's algorithm
Let $K$ be a knot with genus $g$. Seifert's algorithm produces a surface of genus $k$ whose boundary is $K$. In general $k$ may be larger than $g$, but are there any bounds on how much larger it can be? That is, does there exist a function $\phi \colon \mathbb{N} \rightarrow \mathbb{N}$ such that $k \leq \phi(g)$?
Just a note about terminology: the minimal genus of a Seifert surface arising from Seifert's algorithm to a diagram of $K$ is called the canonical genus of $K$.
This paper by Brittenham and Jensen proves that the Whitehead double of an alternating pretzel knot $K$ has canonical genus equal to the crossing number of $K$.
Since Whitehead doubles have genus 1 and alternating pretzel knots have arbitrarily large crossing number, this shows that you cannot even define $\phi$ for $g=1$.
(In fact, there was an earlier paper of Tripp, The canonical genus of Whitehead doubles of a family torus knots, proving the same statement for the torus knots $T(2,2n+1)$.)
About the paper: the upper bound is given by applying Seifert's algorithm for a specific diagram, while the lower bound is given by Morton's inequality giving a lower bound on the canonical genus from the HOMFLY polynomial.
|
2025-03-21T14:48:30.296399
| 2020-04-12T21:33:22 |
357280
|
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"url": "https://mathoverflow.net/questions/357280"
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|
Stack Exchange
|
Space contained in the Interpolation of $L^\infty$ and the Wiener Algebra $\mathcal{F}(L^1)$
Let $\ell^p$ be the space of sequences with power $p$ summable to $\ell^\infty$, $L^p = L^p(\mathbb{R^d})$ be the Lebesgue spaces and $\mathcal{F}$ be the Fourier $d$-dimensional Fourier transform.
My problem is the following,
I have a linear operator $T$ continuous from $L^\infty$ to $\ell^2$ and from $\mathcal{F}(L^1)$ to $\ell^\infty$, and I am looking for the sharpest space $X_p$ such that $T : X_p \to \ell^p$ for $p\in(2,\infty)$ (i.e. I would like the space $X_p$ to have the less possible regularity constaints).
What I have been able to get is telling that $B^0_{\infty,1}⊂L^∞$ and $B^{d/2}_{2,1}⊂ \mathcal{F}(L^1)$, so that by complex interpolation
$$
B^{d/r}_{r,1} ⊂ X_p \qquad \text{ with } r=\frac{2\,p}{p-2}.
$$
Is it possible to get a sharper estimation? For example decrease the $d/r$ exponent, get a bigger second index than $1$ or get a Bessel-Sobolev space $H^{s,r}$ with $s\leq d/r$?
|
2025-03-21T14:48:30.296500
| 2020-04-14T10:17:11 |
357434
|
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"authors": [
"Brendan McKay",
"https://mathoverflow.net/users/9025"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357434"
}
|
Stack Exchange
|
Minimal and maximal degrees in self-complementary graph
If $\ X\ $ is a set, we let $\ \binom X2\,=\, \big\{\{a,b\}: a \neq b \in X \big\}.\ $ Given a simple, undirected graph $\ G=(V,E),\ $ we let $\ \delta(G)\ $ be its minimum degree, and $\ \Delta(G)\ $ its maximum degree. We say that $\ G\ $ is self-complementary if $\ G \cong \bar{G}\ $ where $\ \bar{G} = \left(V, \binom V2\setminus E\right)$.
Given $\ N\in\mathbb{N},\ $ is there a self-complementary graph $\ G\ $ with $\ \Delta(G) \geq N\cdot \delta(G)\,$?
For $n=4,5,8,9,12,13,16$ there is a self-complementary graph with $\delta=1$ and (therefore) $\Delta=n-2$. It shouldn't be hard to identify the general construction. Tomorrow, if nobody gets in first.
Self-complementary graphs exist if $n\equiv0,1\pmod 4$.
Take a self-complementary graph $G$ with $n$ vertices. Append to it a path $u{-}v{-}w{-}x$ of 4 vertices and join $v$ and $w$ to all of $G$. Now you have a self-complementary graph with $n+4$ vertices, $\delta=1$ and $\Delta=n+2$. This gives all sizes from 4 onwards.
This is really just a comment on Brendan McKay's answer, but I'd like to point out that there are also infinite self-complementary graphs with minimum degree $\delta=1$. Here are two different ways to see that.
I. From the existence of arbitrarily large finite self-complementary graphs with $\delta=1$ it follows, by the compactness theorem of first-order logic, that there is an infinite self-complementary graph with $\delta=1$. This is because the class of structures $(V,E,f)$, where $(V,E)$ is a graph with $\delta=1$ and $f:V\to V$ is an isomorphism from that graph to its complement, can be characterized by first-order sentences.
II. Start with any self-complementary infinite graph $G$. (For example the random infinite graph or Rado graph is self-complementary; of course every vertex of Rado's graph has infinite degree.) Now you can construct a self-complementary infinite graph with $\delta=1$ by appending to $G$ a $4$-point path, as in Brendan McKay's answer.
|
2025-03-21T14:48:30.296649
| 2020-04-14T10:20:12 |
357436
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357436"
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|
Stack Exchange
|
Conditions such that norm of matrix vector can be written as the derivative of the norm of the vector for some convex fonction
Problem statement:
Let $A$ be a matrix $\mathbb{R}^{d \times d}$, I want to find some conditions on $A$ such that there exists a differentiable convex function $f: \mathbb{R_{+}} \rightarrow \mathbb{R}$ which verifies:
\begin{equation}
\forall x \in \mathbb{R}^{d}\ \|Ax\|_{2}^{2}=f'(\|x\|_{2}^{2})
\end{equation}
Some examples:
If $A$ is a orthogonal matrix coming from some isometry then we can take $f(t)=t^{2}/2$, if $A=cI$ for some $c\geq0$ we can take $f(t)=c t^{2}/2$. Are these the only cases ?
I have two strategies so far :
First approach I want to restrict the space of research for $f$. Suppose that $Ker(A)\neq {0}$, then it implies that $f$ must be constant on $[0,\sup_{x\in Ker(A)}(\|x\|_{2}^{2})]$ otherwise it would break the convexity of $f$. Appart from this segment we know that the derivative is bounded: $f'(t)\leq \|A\|_{2}t$. Can we go further on this ?
Second approach, maybe it is useful to decompose $A$ with $A=U\Sigma V^{T}$ where $\Sigma=diag(\lambda_{1},\cdots,\lambda_{d})$, with the change of variable $v=V^{T}x$, the problem is equivalent as finding $f$ satisfying $f'(\|v\|_{2}^{2})=\sum_{i} \lambda_{i}^{2}\|v_{i}\|_{2}^{2}$.
Here is an answer: $\|Ax\|_{2}^{2}=x^{T}A^{T}Ax$, we can write $A^{T}A=P D P^{T}$ where $D$ diagonal and $P$ orthogonal, then it is equivalent to find a $g=f'$ such that:
(E) $||D^{1/2}v ||_{2}^2=\sum_{i} |\lambda_{i} v_{i}|^{2} = g(||v||_{2}^2)$
(with $\lambda_{i}\geq 0$ and we note $v=P^Tx$)
If there is one $\lambda_{i}=0$ (the first one without loss of generality) then for all $t\geq 0$ we can take $v=(\sqrt{t},0,..,0)$ then $g(||v||_{2}^2)=g(t)=0$ which is not possible (unless $A=0$).
In this way all $\lambda_{i}>0$. We will show that they are all equal to some $c$. Indeed let $e_{i}$ be one eigenvector of $A^{T}A$ then $\|Ae_{i}\|_{2}^{2}=e_{i}^{T}A^{T}A e_{i}=\lambda_{i}=g(\|e_{i}\|_{2}^{2})=g(1)$. So $\lambda_{i}=g(1)=c$ for all $i$.
So $A^TA=cI$. Overall it implies that $A$ can be written as $A=c_{0} O$ with $O$ an orthogonal matrix.
|
2025-03-21T14:48:30.296815
| 2020-04-14T12:15:02 |
357444
|
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"authors": [
"Noam D. Elkies",
"VS.",
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|
Stack Exchange
|
Small linear relations between primitive Pythagorean triples $\mathsf{II}$
WillJagy answered a linear relation question on Pythagorean Triples in Small linear relations between primitive Pythagorean triples $\mathsf I$.
Now let $a^2+b^2=c^2$ be a primitive Pythagorean triple and then consider the Linear Diophantine Equation $ua+vb+zc=0$ where $(u,v,z)\in\mathbb Z^3$ are variables. If $(u,v,z)\neq(0,0,0)$ then:
Is $\|(u,v,z)\|_\infty$ at least $\sqrt{\max(|a|,|b|)}$ up to constant factors or should the scale (disregarding constants) be smaller (perhaps $\sqrt[3]{\max(|a|,|b|)}$)?
What is the distribution of $\|(u,v,z)\|_\infty$?
Note if it were $ua^2+vb^2+zc^2=0$ then the answer is $O(1)$ since $(u,v,z)=(1,1,-1)$ suffices.
This is what I have
$$ a = m^2 - n^2 $$
$$ b = 2mn $$
$$ c = m^2 + n^2 $$
then
$$ n(m^2 - n^2 ) +(-m)(2mn) + n(m^2 + n^2) = 0 $$
or triple
$$(u,v,z)=(n,-m,n)$$ works and this gives morally $\sqrt{\max(|a|,|b|)}$ ($(m,n,-m)$ also works to give morally $\sqrt{\max(|a|,|b|)}$). Could there be something smaller?
Yes, the minimal $\|(u,v,z)\|_\infty$ is within a constant factor of
$\sqrt{|c|}$ (equivalently, of $\sqrt{\max(|a|,|b|)}$.
The orthogonal complement of $(a,b,c) = (m^2-n^2, 2mn, m^2+n^2)$
contains the independent integer vectors
$v_1 := (n,-m,n)$ (which you found) and $v_2 := (m,n,-m)$.
Their $\bf Z$-span is the full integral complement of $(a,b,c)$,
for example because that span has discriminant
$$
\|v_1\|_2^2 \|v_2\|_2^2 - \langle v_1, v_2 \rangle^2
= 2(m^2+n^2)^2
$$
which equals $a^2+b^2+c^2 = \|(a,b,c)\|_2^2$.
Moreover, $(v_1,v_2)$ is a reduced basis, because
$|\langle v_1, v_2 \rangle| = mn$ is less than $(m^2+n^2)/2$,
and thus less than both $ \frac12 \|v_1\|_2^2$ and $\frac12 \|v_2\|_2^2$.
Therefore the minimal $\|(u,v,z)\|_2$ exceeds
$\sqrt{m^2+n^2} = \sqrt{|c|}$, whence the same is true
of $\|(u,v,z)\|_\infty$ within a constant factor. QED
@NoamElkies Fastest accept. However I am wondering if we take unbalanced diophantine equations $au+bv+c^2z=0$ or $a^2u+b^2v+cz=0$ or any $a^tu+b^{t'}v+c^{t''}z=0$ where $t,t',t''\in{1,2}$ and $1<tt't''<8$ also will we get similar $\sqrt{\max(|a|,|b|)}$ bounds? At lower limit $tt't''=1$ we get this bound and at upper limit $tt't''=8$ we get $O(1)$. Perhaps there is some interpolating rule? Or perhaps $\sqrt{\max(|a|,|b|)}$ until at $8$? Should I post a new post?
That's a different question (and I'm can't guess the motivation for
those particular "unbalanced diophantines"). Once we have a reduced
basis for the lattice, you can use it to start analyzing such equations
where you require one or more of the coordinates to be square.
I see the point of analyzing that way however posted https://mathoverflow.net/questions/357507/small-linear-relations-between-primitive-pythagorean-triples-mathsfiii. Motivation is to see if it might give a smaller relations that cannot be ordinarily seen with just probabilistic analysis. I think only at $8$ it breaks down and at every value from ${1,2,4}$ it remains $\Omega(\sqrt{\max(|a|,|b|)})$ but I do not see a reason why it should fall only at $8$.
Denote $u_1,\dots,u_{n-1}$ to be basis of orthogonal complement of a vector $v\in\mathbb Z^n$. 1.) Then is there a definition of discriminant wrt the basis that is defined? 2. Also discriminant same as $|v|_2^2$ implies $\mathbb Z$-span of basis spans the full integral complement? 3. Also is there a general relation between reduced basis and discriminant? Any references for these that will help?
Discriminant of ${\bf Z}$-span of any vectors $v_1,\ldots,v_m$ =
determinant of Gram matrix $(\langle v_i, v_j\rangle)$.
This is the formula for any quadratic form. 2) It is known that
if $v \in {\bf Z}^n$ is primitive (not $cv'$ for some $c>1$ and
integer vector $v'$) then its orthogonal complement has discriminant $|v|_2^2$.
So if you've found $n-1$ vectors with the same discriminant
you've got everything. (A superlattice of index $d>1$ would have discriminant
$d^2$ times smaller.) 3) Not sure what exactly the first question means.
For the second, [cont'd]
these days I'm even less able to dig up references than usual, but
all this can surely be found in "SPLAG" = Conway and Sloane's
Sphere Packings, Lattices, and Groups, not far from the beginning.
$(v_1,v_2)$ is reduced basis since inner product magnitude is less than $\frac{\min_i|v_i|2^2}2$. For $n$ dimensions is it pairwise inner product $|\langle v_i,v_j\rangle|<\min{k\in{1,\dots,}}\frac{|v_k|_2^2}2$ or is it just $|\langle v_i,v_j\rangle|<\min(\frac{|v_1|_2^2}2,\frac{|v_2|_2^2}2)$?
Related: https://mathoverflow.net/questions/365032/how-small-do-linear-diophantine-solutions-shrink-when-solutions-are-small-linear
|
2025-03-21T14:48:30.297212
| 2020-04-14T12:47:00 |
357445
|
{
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"authors": [
"ABIM",
"Francois Ziegler",
"LSpice",
"Martin Sleziak",
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|
Stack Exchange
|
Which topological spaces admit embeddings into Euclidean spaces
I'm interested in the dual question to:
continuous images of open intervals, about surjections onto open intervals.
Namely, if $X$ is a topological space, when can we guarantee that there exists a topological embedding of $X$ into some Euclidean space?
There is ambiguity between the title question and body question. (“Embedding” usually means that $f$ is a homeomorphism $X\to f(X)$ with subspace topology on the latter.)
Another thing which should be clarified is whether $d$ represents finite number or whether products of infinitely many copies are allowed, too.
@MartinSleziak should be finite.
It's fine I'll accept the answer but would be open to other posts (purely for scientific interest)
@FrancoisZiegler, the only change in my edit was to delete one of the two 'which's in the title "Which topological spaces which admit continuous injections into Euclidean spaces". That seemed like a clear typo, but, if it changed meaning, I have no problem with its being rolled back. (@New_Topologist_On_The_Block had made an edit before me, which may be what you saw.)
I know, but the result is so nice that I have to accept it :0
@New_Topologist_On_The_Block Then I suggest changing the question & title question back to “embedding” so that it matches the answer you accepted.
There's an old theorem of Deák that gives an interesting characterization.
Given a topological space $X$, define a relation on subsets of $X$ as follows: we write $U \sqsubseteq V$ if and only if $\overline{U} \subseteq V$.
Theorem (Deák): A separable metrizable space is homeomorphic to a subset of $\mathbb R^n$ if and only if its topology has a subbasis generated by $\leq n+1$ collections of open sets, each totally ordered by $\sqsubseteq$.
For example, to generate the topology of $\mathbb R^2$ with $3$ collections of this kind, think of how $3$ families of open half-planes can be used to form a small open triangle around every point of the plane.
This theorem appears in
J. Deák, "A new characterization of the class of subspaces of a Euclidean space," Studia Sci. Math. Hungar. 11 (1980), pp. 253-258.
I don't have a link to this paper, but the result is discussed in section 2 of this paper, which contains a few other related things as well.
My favorite part of this theorem is that by changing it a little, one arrives at an interesting (to me) conjecture:
Conjecture: A separable metrizable space is homeomorphic to a subset of $\mathbb R^n$ if and only if its topology has a subbasis generated by $\leq n+1$ collections of open sets, each totally ordered by $\subseteq$.
The conjecture is true for $n=1$, but I don't think it's known for larger $n$. (Does the Klein bottle have a subbasis generated by $4$ nested collections of open sets? Even this special case does not seem trivial to me.)
Note that the question was originally about the existence of a continuous injection, which is a priori weaker (for non-separable metrizable spaces it's weaker, just considering an uncountable discrete space of at most continuum cardinal). (But the question might be amended retroactively...)
|
2025-03-21T14:48:30.297449
| 2020-04-14T12:50:45 |
357446
|
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"Adrien",
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|
Stack Exchange
|
Abelian category from the category of Hopf algebras
The kernel of a Hopf algebra map $\phi:H_1 \to H_2$ is in general not a Hopf
sub-algebra of $H_1$. Is there some replacement or alteration of the notion
of a kernel in the Hopf algebra setting. Same question for cokernels. So can we construct an abelian category from the category of Hopf algebras (over a fixed field $k$).
Kernels are not the issue (not the only one at least) the same objection would otherwise apply to the category of commutative and cocommutative Hopf algebras which s abelian, see e.g. https://arxiv.org/abs/1502.04001v2 and references therein.
@Adrien's link: Vespa and Wambst - On some properties of the category of cocommutative Hopf algebras.
$\DeclareMathOperator\Hker{Hker}\DeclareMathOperator\Hcoker{Hcoker}\DeclareMathOperator\Im{Im}\DeclareMathOperator\coIm{coIm}\DeclareMathOperator\Id{Id}$The category $\mathcal{H}$ of finite dimensional, commutative, cocommutative hopf algebras is an abelian category.
The set $\mathcal{H}(F,G)$ of all hopf algebra maps $F\to G$ is an abelian group with sum given by the convolution product. (The convolution inverse for $f\in F$ is: $f\!\circ\! S_F=S_G\!\circ\! f$ and the neutral element is $\eta_G\!\circ\!\varepsilon_F$).
We have also product of maps, given by composition (and composition distributes over convolution).
This is a classic result. I have been told it was first shown by Grothendieck. I know you can find details in Sweedler's book, ch. XVI, p. 314 (basically the whole chapter, p.303–315 presents a detailed proof of the above).
Remark 1: You are right to mention that in the general case the definitions of the kernels and the cokernels seem to pose a problem when one tries to figure out which hopf algebras would be good to make an abelian category from.
For $F$, $G$, f.d., commut, cocommut, hopf algebras and $\omega:F\to G$ a hopf algebra map, then in $\mathcal{H}$, kernels are defined through: $$\Hker\omega=\{f\in F|(I\otimes\omega)\Delta (f)=f\otimes 1\}$$
which -due to cocommutativity- is the same with $\{f\in F|(\omega\otimes I)\Delta (f)=1\otimes f\}$,
and cokernels through:
$$\Hcoker\omega=G\big/\big(\omega (F^+)G\big)$$
where $F^+$ is the augmentation ideal and $\omega (F^+)G$ denotes the right ideal (which by commutativity is a two-sided ideal), of $G$ generated by $\omega(F^+)$.
Under the above defs, it can be shown that $\Hker\omega$ is a sub-hopf algebra of $F$, $\Hcoker\omega$ is a quotient hopf algebra of $G$ (i.e. $\omega (F^+)G$ is a hopf ideal) and any hopf algebra map $\omega\in\mathcal{H}(F,G)$ "factorizes" as:
$$
\Hker\omega \overset{i_{\omega}}{\hookrightarrow} A
\overset{\pi_{i_{\omega}}}{\twoheadrightarrow} \coIm\omega \cong \Im\omega
\overset{i_{\pi_{\omega}}}{\hookrightarrow} B
\overset{\pi_{\omega}}{\twoheadrightarrow} \Hcoker\omega
$$
where:
$\Hker\omega \overset{i_{\omega}}{\hookrightarrow} A$ is the kernel of $A\overset{\omega}{\to}B$,
$B\overset{\pi_{\omega}}{\twoheadrightarrow} \Hcoker\omega$ is the cokernel of $A\overset{\omega}{\to}B$,
$A\overset{\pi_{i_{\omega}}}{\twoheadrightarrow} \coIm\omega$ is the cokernel of $\Hker\omega \overset{i_{\omega}}{\hookrightarrow} A$ and
$\Im\omega\overset{i_{\pi_{\omega}}}{\hookrightarrow} B$ is the kernel of $B
\overset{\pi_{\omega}}{\twoheadrightarrow} \Hcoker\omega$.
Remark 2: There are even more detailed things to be said about the category $\mathcal{H}$ of finite dimensional, commutative, cocommutative hopf algebras over a field: We can construct a functor $\mathcal{G} : \mathcal{H} \Rrightarrow \mathcal{Ab}_{\text{fin}}$ from $\mathcal{H}$ to the category $\mathcal{Ab}_{\text{fin}}$ of finite, abelian groups (assigning to each hopf algebra of $\mathcal{H}$ the group of its grouplikes) and another functor $\mathcal{F} : \mathcal{Ab}_{\text{fin}} \Rrightarrow \mathcal{H}$ (assigning to each fin, abelian group its group hopf algebra). It is not difficult to show that these functors satisfy
$$
\begin{array}{cccc}
\mathcal{G} \mathcal{F} = \Id_{\mathcal{Ab}_{\text{fin}}} & & & \mathcal{F} \mathcal{G} \cong
\Id_{\mathcal{H}}\\
\end{array}
$$
constituting thus an equivalence of the categories $\mathcal{H}$, $\mathcal{Ab}_{\text{fin}}$.
Is remark 2 definitely correct? If these categories are equivalent then they have the same isomorphism classes of objects, but there is only one finite abelian group of order $p$ and typically more than one isomorphism class of fin.dim. com. cocom. hopf algebras of order $p$ (for example take the group algebra and its dual over characteristic $p$)? Forgive me if I have missed something basic.
@James, although i am not sure i have fully understood what you mean with "hopf algebras of order $p$... " (maybe you mean hopf algebra of dim $p$?), i think you have a point in the following sense: remark 2 is meanignful for hopf algebras over an algebraically closed field of char $0$. Not over arbitrary fields. One can further relax these conditions (to include finite fields), but i think they guarantee that things will work ok (under such circumstances, the group algebra and its dual are isomorphic).
This is a delicate matter, and there are in principle many possible answers depending on exactly what you wish to apply things to.
Frequently, one of two variations of the Hopf subalgebra mentioned by Konstantinos is used: Given $\pi\colon H\to K$ a morphism of Hopf algebras, we have the left coinvariants and right coinvariants (of $\pi$) defined respectively by:
$$ {}^{\text{co}\,\pi}H = \{ h\in H \ | \ (\pi\otimes\operatorname{id})\Delta(h) = 1\otimes h\},\\
H^{\text{co}\,\pi} = \{ h \in H \ | \ (\operatorname{id}\otimes\pi)(\Delta(h))= h\otimes 1\}.$$
These need not be the same subobjects of $H$, however.
One could then define a short exact sequence of Hopf algebras (over $k$)
$$ k\longrightarrow K \overset{i}{\longrightarrow} H \overset{\pi}{\longrightarrow} L\longrightarrow k$$
to be a sequence of morphisms of Hopf algebras such that
$i$ is injective and $\pi$ is surjective;
$\ker(\pi) = H\, i(K)^+$;
$i(K) = {}^{\text{co}\,\pi}H$.
Note that the more classical notion of kernel is still involved: it will generally not be enough to conduct any of the usual arguments you'd like to do with a "short exact sequence" to assume only the first and third (or first and second) conditions. The first and third are enough when $H$ is faithfully coflat, and the first and second are enough when $H$ is faithfully flat.
This is a good definition for short exact sequences and results that normally rely on them, such as looking for Jordan-Holder analogues. As such, coinvariants-as-kernels will be prominent when working in a category of Hopf algebras. But as I said it's not the only answer, and in other categorical contexts (generalizing/lifting to functors between representation categories, namely) it is the wrong answer. The categorical kernels and cokernels of $\pi\colon H\to K$ are
$$ \text{Hker}(\pi) = \{ h\in H \ | \ h_{(1)}\otimes \pi(h_{(2)})\otimes h_{(3)} = h_{(1)}\otimes 1 \otimes h_{(2)}\},\\
\text{Hcoker}(\pi) = K/(K \pi(H^+) K).$$
And this kernel (and cokernel) is not necessarily the same thing as the left or right coinvariants (and associated cokernel). By applying counits to the left/right of the defining relation, we see that $$\text{Hker}(\pi)\subseteq {}^{\text{co}\,\pi}H\cap H^{\text{co}\,\pi},$$ at least. Though equality (of all three subobjects) can happen, and assuring it does is usually the key to making sure attempts at defining exact sequences of tensor categories works out well.
Sonia Natale recently posted a review of these notions, and the related issues with tensor categories, that should be helpful.
Two questions: First, in the commutative, cocommutative case, the left and right coinvariants are the same: ${}^{\text{co},\pi}H = H^{\text{co},\pi}$. Are they also equal to your second definition of the kernel: $\text{Hker}(\pi)$? Are there other interesting situations in which the second definition is identified with the one posted in my answer?
And second, you claim that "categorically speaking (the first def) is the wrong answer". I guess you refer to the non-(co)commutative case. what is the advantage of the second definition for $\text{Hker}(\pi)$ in the general case? Does it enable one, to enlarge somehow the class of hopf algebras which make an abelian category?
@KonstantinosKanakoglou Such Hopf algebras would be faithfully flat and coflat, so everything would coincide nicely. Proposition 4.2 and subsequent remark in the linked paper should answer the remaining matters.
If I am correct $S({}^{co\pi}H)=H^{co\pi}$ and I wonder whether $\mathrm{Hker}(\pi)$ can be characterized thorugh some $S$-closedness condition. After all if you really want to be Hopf the antipode should play a role here...
@NicolaCiccoli Well, when one moves from Hopf algebras to representations thereof to more general categorical settings like fusion categories, the antipode sort of disappears as a core feature. What it normally gives us is the existence of duals in the representation category, which as a categorical property is called "rigid", but it doesn't do much else for the category. And in the categorical setting the rigid property is typically assumed, rather than derived, and you don't necessarily need functors to "respect" underlying antipodes.
|
2025-03-21T14:48:30.298004
| 2020-04-14T13:34:08 |
357450
|
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|
Stack Exchange
|
Subharmonic in any holomorphic coordinates = Plurisubharmonic?
An upper semi-continuous function $u : \Omega \to \mathbb{R}$, $\Omega \subseteq \mathbb{C}^n$ is said to be subharmonic if it satisfies the submean inequality $u(a) \leq \mu_S(u;a,r)$, where $\mu_S(u;a,r)$ is the mean value of $u$ on the sphere $S(a,r)$ with center $a$ and radius $r$ (for any $a$,$r$ such that the closed ball $\overline{B(a,r)}$ is contained in $\Omega$).
The function is said to be plurisubharmonic if it is subharmonic restricted to any complex line. It is straight-forward to see that plurisubharmonic implies subharmonic, see for example Demailly, "Complex Differential and Algebraic Geometry", §I.5.A. It is also not so difficult to see that plurisubharmonicity is invariant under holomorphic changes of coordinates, see for example Demailly, Theorem I.5.11.
The entry Encyclopedia of Mathematics: Pluripotential theory states that "Plurisubharmonic functions are precisely the subharmonic functions invariant under a holomorphic change of coordinates". One inclusion is clear from above, but does anyone have an idea of how to prove the other inclusion, i.e.: if $u$ is subharmonic with respect to any local holomorphic coordinates, then $u$ is plurisubharmonic?
It is actually enough to assume that your function $u$ defined on a neighborhood of $0\in \mathbb C^n$ remains subharmonic after composing with any linear transformation.
Indeed, let $0\neq \lambda \in \mathbb C^n$ and set, for $\sigma \in \mathbb C^*$,
$$A(\sigma) = \begin{pmatrix}
\lambda_1 & \sigma a_{12} & \cdots & \sigma a_{1n}\\\
\vdots & \vdots & & \vdots \\\
\vdots & \vdots & &\vdots\\\
\lambda_n & \sigma a_{n2} & \cdots & \sigma a_{nn}
\end{pmatrix}$$
with $a_{ij} $ chosen such that the matrix is invertible. Then one finds, at the origin
$$\begin{eqnarray}
\Delta (u \circ A(\sigma))&=&\sum_{k=1}^n \frac{\partial^2(u \circ A(\sigma))}{\partial z_k \partial \bar{z_k}}(0)\\\
&=&\sum_{k=1}^n \sum_{1\leq i,j\leq n}A(\sigma)_{ik} \overline{A(\sigma)_{jk}} \frac{\partial^2 u}{\partial z_i \partial \bar {z_j}}(0)\\\
&=&Hu_{0}(\lambda) + |\sigma|^2 \sum_{\ell=2}^n Hu_{0}((a_{j \ell})_{j})
\end{eqnarray}$$
where $H$ denotes the complex hessian. This quantity is non-negative by assumption, and one gets the result by making $\sigma$ approach $0$.
Are you assuming that the function is smooth? That argument can be modified to work with non-smooth functions.
Yes, I was assuming that $u$ was $\mathcal C^2$. If $u$ is not $\mathcal C^2$, take a convolution $u_\varepsilon=u\ast \rho_{\varepsilon}$ of $u$ with a smoothing kernel $\rho_{\varepsilon}$. Then, $u_\varepsilon$ remains subharmonic after composing with any affine transformation and the argument above shows that $u_\varepsilon$ is psh, hence so is $u$.
Hi, nice computation but I guess you want the $\lambda$'s to be on the first column of $A$, not the first row.
@user111 You are perfectly right; I edited the answer accordingly. Thanks.
|
2025-03-21T14:48:30.298205
| 2020-04-14T13:48:31 |
357452
|
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|
Stack Exchange
|
Absolute value of a diffusion
Suppose $B_t$ is a standard Brownian motion on a filtered probability space $\langle \Omega, \mathcal F, \{\mathcal F_t\}_t, \mathbb P\rangle$. Consider two SDEs below.
Suppose, $X_0 = Y_0 = 0$
\begin{align*}
dX_t =& sign(X_t) dt + d B_t\\
dY_t =& \alpha_t dt + dB_t
\end{align*}
where $\alpha_t$ is some $[-1,1]$-valued, $\mathcal F_t$ measurable process.
I am wondering if the following is true. And if yes, how might one approach this thing?
$$\mathbb E[\vert X_T\vert] \ge \mathbb E[\vert Y_T \vert]$$ for all $T \ge 0$.
Intuitively, the reason for this conjecture is the following: Both the processes start from $0$. Say, at some time they are equal to some $x > 0$. Now, $X_t$ has an upward drift of $1$ while the upward drift of $Y_t$ is weakly less than $1$. So, the moment the two processes have the same sign, it seems that $X_t$ runs farther away compared to $Y_t$. But, I have no idea how to make this precise, even if true.
Thanks!
What happens in the case $\alpha_t =1$ for all $t$?
Do you conjecture that with $\alpha_t = 1$ for all $t$, $\mathbb E \vert X_T \vert \le \mathbb E \vert Y_T \vert$? I would be surprised. Heuristically, if we get a negative shock at $t=0$ of size $\sqrt{d t}$, then, $\alpha_t = - 1$ there makes it $-d t - \sqrt{ dt}$, while $\alpha_t = +1$ makes it $dt - \sqrt{ dt}$. So, on the negative side, negative push via drift should be better, and on the positive a positive shock is a vague reasoning I am employing.
The comparison theorem of N. Ikeda & S. Watanabe (https://projecteuclid.org/download/pdf_1/euclid.ojm/1200770674) tells us that $X_t \le Y_t$ for all $t$, a.s., when $\alpha_t =1$.
Yes, square the processes then $$ dX_t^2 = 2X_t dX_t + (dX_t)^2 = 2|X_t| dt + 1 + 2X_t dB_t$$ and similarly for the other. Then you see the the drift on this process must be bigger than that of the $Y_t$ process at the same level. Then you can finish it off with a comparison theorem that says $X_t^2$ will be stochastically larger than $Y_t^2$ and the same is therefore true about the absolute value as well. There is a chapter on the comparison theorems in Ikeda and Watanabe , I don't know where else they can be found. If this is true, the I acknowledge that I learned this trick from Ionnis Karatzis about 1986, but if I'm muddled, i's just me.
A very neat trick! Actually, the comparison theorem yields that here $|X_t|\ge |Y_t|$ pathwise, not just stochastically.
Pathwise? Is this under an additional Markovian assumption?
@e.lipnowski, the comparison theorem is also valid for random (adapted) coefficients (however, I do not know any classic reference for that, so I had to prove it for our paper, see Theorem 8).
Thanks @mike This is great, and very useful!
|
2025-03-21T14:48:30.298421
| 2020-04-14T13:48:46 |
357453
|
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|
Stack Exchange
|
Finite groups such that non-central, commuting elements have the same stabilizer
Let me say that a finite, non-abelian group $G$ has the property $(P)$ if
for any two elements $x, \, y \in G - Z(G)$ such that $y \in C_G(x)$, one has $C_G(y)=C_G(x)$.
For instance, it is easy to see that both the dihedral group $D_8$ and the quaternion group $Q_8$ have the property $(P)$.
Instead, the symmetric group $S_n$, with $n \geq 5$, does not have it: indeed, setting $x=(1\, 2)$, $y=(3 \,4)$, we have $y \in C_G(x)$, but $C_G(y) \neq C_G(x)$, since $(2 \, 5) \in C_G(y)$ but $(2 \, 5) \notin C_G(x)$.
I am not an expert in Group Theory (actually, I was led to consider these groups by studying a problem in Algebraic Geometry) and I suspect that this property is already known. So, let me ask the following
Question. Was property $(P)$ already studied in the literature? Is there a classification of groups having the property $(P)$, at least for small order? If this is the case, what are relevant references?
When the center is trivial it's known as "commutative-transitive". Your property, satisfied by $\mathrm{GL}_2(K)$ for any field, passes to subgroups. In particular, for $K$ field of characteristic $2$, $\mathrm{SL}_2(K)$ is commutative-transitive.
Also $S_4$ doesn't satisfy the property: $(1234)$ commutes to its square $(13)(24)$, which commutes to $(12)(34)$ but $(1234)$ and $(12)(34)$ don't commute; similarly $A_6$ is not commutative-transitive. Still $A_4$ and $A_5$ are commutative-transitive.
@YCor: If it passes to subgroups, it should remain true for $\mathrm{SL}_2(K)$ for any field. Am I missing something?
Yes, your property is true for $\mathrm{SL}_2(K)$. What characteristic 2 ensures is that the center is trivial.
Ah, ok. Commutative-transitive means that the commutativity relation is transitive on $G-{1}$, and if $Z(G)$ is non-trivial this implies that $G$ is abelian.
Well one often assumes commutative-transitive to imply center-free... which makes it only pass to center-free subgroups (= subgroups that are not non-trivial abelian) too. In any case you provided examples showing that your property can be interesting for non-abelian groups with non-trivial center.
There's a large quantity of 2-step nilpotent groups with your property. For instance the free 2-step nilpotent group of exponent $p$ (odd prime) on $2$ generators has this property.
@YCor: many thanks. As I said in the question, I am not an expert, in fact I do not know this terminology ("free 2-step nilpotent group"). Could you please provide some introductory reference?
I have no idea of a reference, it's just familiarity playing with these groups. (Still you reach literature if you search "commutative-transitive group" or so.)
When $Z(G) = 1$ and $G$ is finite, his means that $G$ is a $CA$-group (group in which each non-identity element has Abelian centralizer). Simple $CA$-groups were studied and classified by M. Suzuki in the 1950s and 60s. For such groups which are not simple, I am not sure about the extent of the literature, but others might be. However, if $Z(G)$ is non-trivial, the problem becomes rather different.
I’d be curious to hear the algebro-geometric motivation behind this question
@SantanaAfton: groups with property (P) cannot be finite quotients (+ a little extra condition) of some surface braid groups. I'm studying these quotients in order to construct some fibred algebraic surface, in the spirit of this paper: https://arxiv.org/abs/2002.01363
It's always interesting when distinct communities use distinct terminology for the same notion (CA-group vs commutative-transitive). Btw CA-group is largely used by people studying (infinite) locally finite groups. Note: both CA-group and ZCA-group (OP's definition) are obviously (finitely axiomatizable) first-order properties, and in particular they pass to ultraproducts, as well as their negations.
|
2025-03-21T14:48:30.298681
| 2020-04-14T13:52:38 |
357454
|
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|
Stack Exchange
|
Showing an "obviously-optimal" control is optimal (without smoothness assumptions)
Let $\mathcal{A}\subseteq\mathbb R$ be a compact interval, $T\in\mathbb R_+$ be a finite horizon, and $g:\mathbb R\to\mathbb R_+$ be a continuous function with $g\leq 1+|\cdot|$. Consider an optimal control problem where, observing a standard Brownian motion $\{B_t\}_t$, I choose an $\mathcal A$-valued control $\{a_t\}_t$ to control the drift. I want to maximize $\mathbb E g(X_T)$, where $X_t:=B_t + \int_0^t a_s \text{ d}s$.
I have proven the following:
1) The optimal value function $v:[0,T]\times\mathbb R\to\mathbb R_+$ (taking a time $t$ and current level of $X_t$ to an optimal continuation value) is finite-valued.
2) There is a continuous cutoff function $\kappa:[0,T]\to\mathbb R$ such that, for every $t\in[0,T)$, the function $v(t,\cdot)$ is strictly increasing on $[\kappa(t),\infty)$ and strictly decreasing on $(-\infty,\kappa(t)]$.
I want to show that there is an optimal control where I choose $\max \mathcal A$ whenever $X_t > \kappa(t)$ and $\min \mathcal A$ whenever $X_t < \kappa(t)$. Ideally, I would show that any other optimal control almost surely agrees with this one at almost every time.
This seems, intuitively, like it must be true, but I don't see an easy proof (or know a reference), especially given that I've made no differentiability assumptions.
"The optimal value function $v$ ... is finite-valued" you certainly need some assumptions about the growth of $g$.
Yes, sorry for being unclear. I meant that I had an instance of this problem in which I had already proven it was finite-valued. In my instance, $g\leq 1+|\cdot|$, so that finite-valuedness is easy. I'll edit to add this feature to the problem.
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2025-03-21T14:48:30.298932
| 2020-04-14T13:53:43 |
357455
|
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|
Stack Exchange
|
Reference / Survey for finite field analog number theory
It is folklore that many number theoretic results on prime numbers have a simpler-to-prove finite field analog. For example, on the one hand, the proof of the Prime Number Theorem
$$\#\{\text{prime numbers}\leq x\} \sim \frac{x}{\log x}, \;\;\,\; x \to +\infty,$$
requires complex analysis and the study of the Riemann zeta function (or the elementary, albeit complicated, machinery of Selberg), while, on the other hand, the proof of
$$\#\{\text{irreducible monic polynomials in $\mathbb{F}_q[x]$ of degree} \leq n\} \sim \frac{q^n}{\log_q q^n}$$
requires only some combinatorial reasonings (which also gives an error term of the size of the one expected for the error term of the prime numbers counting function under the Riemann's hypothesis).
Do you know where I can find a survey or some other kind of collection of these finite field analog? Thanks
In general, we say "function field" analogue. I don't know a good survey, but maybe concerning objects this could be a good start. See also this question: https://mathoverflow.net/questions/1367/global-fields-what-exactly-is-the-analogy-between-number-fields-and-function-fi?noredirect=1&lq=1, especially this reference.
See also the non-analogies: https://mathoverflow.net/questions/177234
Arguably counting prime polynomials of degree $n$ is not really the right analogue of the prime number theorem. A better one might be prime polynomials with the leading $k$ coefficients fixed. This requires nonvanishing on $s=1$ for a Dirichlet $L$-function, and can be proven by the same kind of complex-analytic argument. (Or by one of the four different proofs of the Riemann hypothesis, of course - but none of them are as trivial as the "some combinatorial reasonings" you mention, which give the wrong impression IMO.)
One of the best references I am aware of, is the book by Rosen ‘Number theory in function fields’ https://www.springer.com/gp/book/9780387953359.
A good survey, which also concerns the important connection with random matrix theory, is the following paper by Rudnick, although you won’t find proofs there: http://www.math.tau.ac.il/~rudnick/papers/ICM-Proceedings-Rudnickv5.pdf
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2025-03-21T14:48:30.299106
| 2020-04-14T14:40:38 |
357459
|
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|
Stack Exchange
|
Extend (Lie) group action from the boundary to the entire manifold
Let $M$ and $W$ be smooth manifolds such that $\partial W=M$. Let $G$ be a group acting on $M$.
Can one generally extend the action of $G$ to $W$? If not, under which conditions on $W$ and/or $G$ does it work? How do you construct such an extension?
EDIT: suppose in addition that $G$ is a connected Lie group acting smoothly and that the extension should be smooth as well.
EDIT: suppose furthermore that the Lie group is acting by isometries if a Riemannian metric is given on $M$ and the extension should again be acting isometrically.
If you don't care about any properties of the action, then let $G$ act trivially on $W \setminus M$. Together with the existing action on $M$, this will give you an action on all of $W$. But if, say, $G$ is a Lie group acting smoothly on $M$ and you want to preserve this smoothness, then this construction of course fails horribly. Perhaps edit the question to clarify what properties you want the extension to satisfy?
Indeed certainly you want a continuous extension, but the conclusion might differ according to the degree of differentiability. In the case of the circle boundary of the disc, the topological extension is obvious but the differentiability issue seems non-trivial. The purely topologic question is interesting too. Also restricting to $G$ cyclic is very rich. Maybe give more context and focus a bit...
Thank you @SophieMacDonald for your comment. I edited my question to add the more specific question of Lie group actions, but I would also be interested in a more general answer.
@YCor I would actually also like to know if there is some reference discussing as many cases as possible.
@Kafka91 To elaborate on Sophie MacDonald and Ycor's comments, you can also ask the action on $W$ to be by isometries if $W$ is given a Riemannian metric structure. The case of an action on the circle extending as an action on the Poincaré disk is already very interesting.
About the circle-disc case: the action of $\mathrm{PSL}_2(\mathbf{R})$ extends smoothly (since we have the action on the Poincaré disc). The action of $\mathbf{R}$ on the circle = $\mathbf{R}^2-{0}$ is already unclear to me: the naive extension for a matrix $A$ and $x$ in the disc, is then $A\cdot x=\frac{|x|}{|Ax|}Ax$, but this is continuous, and not smooth, at $0$ (even for $A$ fixed, say diagonal $\neq\pm 1$). Maybe some trick works anyway (compactitifying suitably the action on $\mathbf{R}^2$?). But then what about the faithful action, say of a 3-fold covering of $\mathrm{PSL}_2$?
@Kafka91 I see that you incorporated the isometry extension of your question. Thank you for that. One more comment about that : it's not likely that the action of $M$ is by isometries. In what I suggested, only $W$ was given a Riemannian structure. If you take the example of $PSL_2(\mathbb{R})$ acting on the Poincaré disk, the action on the circle is by Mobius homeomorphisms.
Edit: I might have misinterpreted the question, which asks whether the action on $M$ extends to a specific bounding manifold $W$. The answer below concerns whether the action extends to some bounding manifold.
Your question can be rephrased in the language of algebraic topology as: If $M$ is a null-bordant manifold with a $G$-action, is $M$ equivariantly null-bordant? Equivariant (co)bordism was a very active subject in the 1960s and 70s, dating back at least as far as the monograph by Conner and Floyd:
Conner, P. E.; Floyd, E. E., Differentiable periodic maps, Ergebnisse der Mathematik und ihrer Grenzgebiete. Neue Folge. 33, Reihe: Moderne Topologie. Berlin-Göttingen-Heidelberg: Springer-Verlag. VII, 148 p. (1964). ZBL0125.40103.
Part I gives an introduction to bordism theory, and Part II studies $G$-equivariant bordism with $G=\mathbb{Z}/p$. Some of the results apply to general Lie groups $G$. For example: if $G$ acts freely on a closed manifold $M$ of dimension $m$, the bordism class of the pair $(M/G,f)$ in the bordism group $\Omega_m(BG)$, where $f:M/G\to BG$ classifies the principal $G$-bundle $M\to M/G$, provides an obstruction to extending the action to a free action on an $(m+1)$-dimensional manifold with boundary. This can be studied using characteristic numbers.
Thank you very much for your answer! However I was indeed asking under what conditions on $W$ the action can be extended on it. I will edit my question, but your answer is very instructive anyway.
Here is an example of a connected Lie group acting on $M$, such that the action does not extend to $W$ but extends to some $W'$ with $\partial W'=M$. All the existing actions are analytic isometric, and all obstructions are continuous, so the problem is not regularity.
Take $G=M=\mathbb T^1$ the 1-torus, acting on itself in the obvious fashion, by translations. Clearly, if $M$ is seen as the boundary of the disc $W'=\mathbb D^2$, then the action on $M$ extends to an action on $W'$, again by rotations. (In the following I prefer to think of $\mathbb T^1$ as $\mathbb R/\mathbb Z$.)
However, let $W$ be a 2-torus with a disc removed, and identify $M$ with the boundary of $W$. Assume by contradiction that the induced action on $\partial W$ is the restriction of an action on $W$.
Let $Z$ be the manifold constructed as the gluing of $[0,1]\times M$ with $W$ along $\lbrace1\rbrace\times M$ and $\partial W$. Define an action of $\mathbb R$ on $Z$ in such a way that the boundary of $Z$ (which corresponds to $\lbrace0\rbrace\times M$) is fixed, and the restriction to $W$ is the action induced by that of $\mathbb T^1$ on $W$. On $[0,1]\times M$, we have to continuously Dehn-twist the collar to match the rotation of $\lbrace1\rbrace\times M$.
Now the action $f$ at time $1$ is homotopic to the identity of $Z$ relative to its boundary (in the strong sense), but $f$ is just a Dehn twist of $[0,1]\times M$, since $f$ is the identity in $W$ (1 has image 0 in $\mathbb T^1$). It is probably classical that this Dehn twist is not homotopic to the identity relatively to the boundary, personally I convinced myself on a picture that the induced map $f_*$ on $\pi_1(M,x)$ for $x$ on the boundary is not the identity; in any case, this is a contradiction.
Let me try a (very) wild guess: the action $\rho$ of a connected Lie group $G$ on a manifold $M$ extends to every $W$ such that $\partial W\simeq M$ if and only if there is an action on the cylinder $M\times[0,1]$ such that the action restricted to $M\times\lbrace0\rbrace$ (resp.$M\times\lbrace1\rbrace$) is $\rho$ (resp. the trivial).
|
2025-03-21T14:48:30.299569
| 2020-04-14T15:46:48 |
357465
|
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|
Stack Exchange
|
non-archimedean valuations on graded rings
Let $R$ be a commutative (non-trivially) graded ring. By a non-archimedean valuation I mean a map $v: R \to \Gamma \cup {0}$ such that for all $x,y \in R$, we have $v(x+y) \leq \max\{v(x),v(y)\}$, $v(xy)=v(x)v(y)$ and $v(0)=0,v(1)=1$. Here $\Gamma$ is a totally ordered commputative group and $0 \leq g$ for all $g \in \Gamma$.
Now, one says two valuations $v,w$ are equivalent for all $x,y \in R$ $$v(x) \geq v(y) \iff w(x) \geq w(y).$$ Now, my question is, can I check two valuations are equivalent by just looking at homogeneous elements, in other words if for all $a,b$ homogeneous elements in $R$ we have $$v(a) \geq v(b) \iff w(a) \geq w(b)$$ can we deduce that $v,w$ are equivalent?
It wouldn't be suprised if the answer is no, but I don't know much about this. If the answer is indeed no, can one add some extra assumtions to $R$ to make this the case, e.g free or something.
Thanks
Let $K$ be a field, and set $R=K[x]$ with the usual grading by degree. For each irreducible polynomial $p(x)\in R$, we get a valuation $v_p$ on $R$ given by
$$
v_p(f)=2^{-\operatorname{ord}_p(f)},
$$
where
$\operatorname{ord}_p(f)\in\mathbb{Z}_{\geq 0}\cup\{\infty\}$ is the power to which $p(x)$ occurs in the prime factorization of $f$. Suppose $p,q\in R$ are non-associate irreducibles, neither of which is associate to $x$. Then $v_{p}$ and $v_{q}$ are not equivalent, since $v_{p}(q)\geq v_{p}(1)$ but $v_{q}(q)<v_{q}(1)$. However, for every non-zero homogeneous element $f\in R$, we have $v_{p}(f)=v_{q}(f)=1$.
|
2025-03-21T14:48:30.299682
| 2020-04-14T18:06:39 |
357475
|
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|
Stack Exchange
|
For pseudo-Cauchy sequence, does there always exist a pseudo-Cauchy subsequence of $(x_n)$ having distinct terms?
Given a metric space $(X,d),$ let us call a sequence $(x_n)$ in $X$ to be pseudo-Cauchy if $\lim\limits_{n\to\infty}d(x_n,x_{n+1})=0.$
For example, the sequence $\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)_n$ is pseudo-Cauchy in $\mathbb R$ without being Cauchy.
Now I would like to ask the following question:
Given a pseudo-Cauchy sequence $(x_n)$ in a metric space $(X,d)$ having no constant subsequence, does there always exist a pseudo-Cauchy subsequence of $(x_n)$ having distinct terms?
Intuitively it is appearing to me that there should not always exist such a subsequence; however I failed to construct a counterexample or give a proof.
Please help me!
The answer to the question is yes. I struggled for a while trying to find the right growth rate so that terms in sequences are sufficiently spread but also not too much, until I realized that the following basic trick is more suited to the problem:
The hypothesis of having no constant subsequence means that each value of the sequence occurs a last time in the sequence. Start with the value $x_0$ and set $\varphi(0)$ as the largest index with $x_0=x_{\varphi(0)}$. Having defined $\varphi(n)$, define $\varphi(n+1)$ as the largest index with $x_{\varphi(n)+1}=x_{\varphi(n+1)}$. This insures at the same time that $x_{\varphi(n+1)}$ is distinct from all $x_{\varphi(k)},k\leq n$, and that two consecutive terms of the subsequence $(x_{\varphi(k)})_{k \in \mathbb{N}}$ are consecutive terms of $(x_k)_{k \in \mathbb{N}}$ in an orderly way. Hence $\varphi$ is a solution.
The best kind of answer: here's what I tried that didn't work, and here's how it guided me to the solution that did work.
|
2025-03-21T14:48:30.299820
| 2020-04-14T18:11:05 |
357477
|
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|
Stack Exchange
|
CW structure on infinite-dimensional manifolds
It is well-known (due to this work of Palais, I believe) that Banach manifolds are dominated by countable CW complexes. It then follows (due Whitehead, as indicated by Milnor in this work) that they have the homotopy type of CW complexes.
Are there conditions ensuring that a Banach or Fréchet manifold admits a genuine CW complex structure?
What are the main examples of that?
Actually, I'm working with Banach fiber bundles, which are Serre fibrations. I would like to regard them as Hurewicz fibrations to use HLP (homotopy lifting property). Due to this answer of Peter May (and the references cited by him), I know that it is enough to get a CW structure.
Any help is welcome.
In your application, what are you doing homotopy lifting against? Is it really an arbitrary space, or is there some smoothness requirement?
I'm trying to use global analysis to get general results ensuring existence of emergence phenomena between parameterized Lagrangian field theories. Those structures appear basically as the space of parameters of the theories. Thus, a priori, they are arbitrary (if we think in finding C^0 emergence phenomena).
There is a triangulation theorem for Hilbert manifolds (=separable metric spaces locally homeomorphic to $\ell^2$). Namely, any Hilbert manifold is homeomorphic to the product of $\ell^2$ and some locally finite simplicial complex. No smoothness is needed.
Thank you. But it seems to have some contradiction between the triangulation theorem and Thomas Rot's answer. Am I forgetting something?
@math-phys-cat: I don't think so. Note that $l^2$ is not a CW complex.
I see. Thanks. So, in your answer (together withe the comments after it) you are suggesting that a Hilbert manifold $M$ cannot admit a CW structure in which there is a point $x\in M$ meeting only finite-dimensional cells. Right?
No I’m claiming it does not admit a cw structure at all.
Let $M$ be a Hilbert manifold. Then there exists a Riemannian metric $g$ on $M$ such that the induced metric $d$ is complete (See https://arxiv.org/pdf/1610.01527.pdf by Biliotti and Mercuri). Thus $M$ has the topology of a complete metric space. By the Baire category theorem $(M,d)$ is a Baire space: It cannot be written as a countable union of closed sets with empty interior.
If $M$ had the structure of a CW complex, it does. It must clearly be an infinite dimensional $CW$-complex, but such a thing is the countable union of its $n$-skeleta, which have empty interior.
With this argument you can treat also other infinite dimensional manifolds: using the fact that seperable Frechet spaces are homeomorphic to seperable Hilbert space, one can show that a Frechet manifold is homeomorphic to a Hilbert manifold.
maybe I am a bit nitpicking right now, but the n-skeleta of infinite dimensional CW-complexes can in general have nonempty interior. It depends on whether one attaches higher cells to each $n$-cell. This should be the case here, since we look at Banach manifolds.
@HenrikRüping: Yes, I should argue a bit more about that. I think that if $x$ only meets cells up to a finite dimension, then the local homology $H_*(M,M\setminus {x})$ shouldn't vanish. But the local homology of Hilbert manifolds does vanish.
Thank you for the answer and for the comments.
Thomas, could you look at my comment below the comment of Igor Belegradek? Thank you.
This final step (about embedding Fréchet manifolds into Hilbert spaces) is well-known on the separable case. Have you some idea of how to extend this to the general case or if there are obstructions to do that?
No I don’t know the details about that.
Ok. Thank you for all the help. I will mark your answer as solving my question.
@ThomasRot this seems to imply that any (T2) separable Banach manifold is metrizable and hence (by Palais) paracompact. Is my understanding correct?
@Overflowian: This seems correct to me.
|
2025-03-21T14:48:30.300125
| 2020-04-14T19:12:20 |
357484
|
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|
Stack Exchange
|
Asymptotically tight concentration of norms of subgaussian random vectors with independent coordinates, as the dimension $n \to \infty?$
Let $X=(X_1 \dots X_n)\in \mathbb{R}^n,$ be a subgaussian random vector so that $X_i$'s are independent, $\mathbb{E}X_i = 0, \mathbb{E}X_i^2=1.$ Before we pose our question, let's state the following:
Definition notation and background: A random variable $V \in \mathbb{R}$ is called subgaussian, if $P[|V|\ge t]\le 2 e^{-c(V)t^2}, c(V)>0$ depends on $V$ only. One can show that: $\mathbb{E}\left[exp({\frac{V^2}{t^2}})\right ]\le 2$ for some $t>0,$ which makes us define the Orlicz norm: $||V||_{\psi_2}:= inf_{t > 0}\mathbb{E}\left[exp({\frac{V^2}{t^2}})\right ]\le 2.$ Note that: then we have: $P[|V|\ge t]\le 2 e^{-ct^2/||V||_{\psi_2}^2}, c>0$ is an absolute constant.
A random vector $X\in \mathbb{R}^n$ is called subgaussian if $<X,x>_{\mathbb{R}^n}$ is subgaussian random variable for every constant $x \in \mathbb{R}^n.$
With the above, let's state the question, $c, C$ below are absolute:
We know the following about the concentration of $ ||X||_2,(\mathbb{E}X_i=0, \mathbb{E}X_i^2=1$)
$$P[ |\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}| \ge t] \le 2e^{-ct^2/K^4}, K=max_{1 \le i \le n}||X_i||_{\psi_2}.$$
Or equivalently:
$$||\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}||_{\psi_2}\le CK^2 > 0, K=max_{1 \le i \le n}||X_i||_{\psi_2}$$
where ${\psi_2}$ denotes the subgaussian norm. But we note that the right side of the inequality is dimension-free, i.e. there's no function of $n$ on the right that goes to zero as $n \to \infty.$ So my question is: do we necessarily have tighter concentration of the norm around $\sqrt{n}$ as $n \to \infty?$ That is: do we have:
$$(1a) \hspace{1mm} lim_{n \to \infty} |\mathbb{E}||X||_2 - \sqrt{n}|=0? $$
$$(1b) \hspace{1mm} lim_{n \to \infty} | \frac{\mathbb{E}||X||_2}{\sqrt{n}} - 1|=0? $$
$$ (2a) \hspace{1mm} lim_{n \to \infty}||\hspace{1mm}{||X||_2 - \sqrt{n}} \hspace{1mm}||_{\psi_2}=0?$$
$$ (2b) \hspace{1mm} lim_{n \to \infty}||\hspace{1mm}{\frac{||X||_2}{\sqrt{n}} - 1 } \hspace{1mm}||_{\psi_2}=0?$$
Next, if we don't assume that the co-ordinates are independent, are (1a,b) and (2a,b) still true?
What is your definition of a subgaussian random vector?
@IosifPinelis Thanks for your comment! A random variable is subgaussian if it satisfies: $P[|X| > t] \le 2 e^{-ct^2}, t>0.$ A random vector $X \in \mathbb{R}^n$
is called sub-gaussian if the one-dimensional marginals $<X,x>$ are sub-gaussian
random variables for all deterministic $x \in \mathbb{R}^n.$
Consider first the case when the $X_i$'s are independent.
In view of your definition of a sub-gaussian random vector, we appear to have the condition
$$s^2:=\sup_i Var(X_i^2)<\infty.$$
Let $N:=\|X\|_2$. We have this key identity:
$$N-\sqrt n=\frac{N^2-n}{2\sqrt n}-R_n,\tag{1}$$
where
$$R_n:=\frac{(N^2-n)^2}{2\sqrt n(N+\sqrt n)^2}.$$
Moreover,
$$0\le R_n\le\frac{(N^2-n)^2}{n^{3/2}},$$
whence
$$|ER_n|\le\frac{E(N^2-n)^2}{n^{3/2}}=\frac{Var(N^2)}{n^{3/2}}\le\frac{s^2}{n^{1/2}}\to0$$
(as $n\to\infty$). So, by (1),
$$EN-\sqrt n=-ER_n\to0,$$
so that your condition (1a) holds, which also obviously implies (1b).
Your condition (2b) immediately follows from your second displayed inequality, $\|N-\sqrt n\|_{\psi_2}\le C$, which implies $\|\frac N{\sqrt n}-1\|_{\psi_2}\le C/n$.
Your condition (2a) does not hold even when the $X_i$'s are iid standard normal -- because then, by (1) and the central limit theorem (say), the distribution of $N-\sqrt n$ converges to $N(0,1/2)$.
Thus, in the "independent" case, your conditions (1a), (1b), and (2b) hold, whereas (2a) does not hold in general.
Consider now the "dependent" case, when the $X_i$'s are not necessarily independent. Let e.g. $X_i=X_1$ for all $i$, where $X_1$ is any zero-mean unit-variance random variable such that $a:=E|X_1|$ is strictly less than $1$. Then $N=\sqrt n\,|X_1|$. So, $\frac{EN}{\sqrt n}=a-1\not\to0$, so that (1b) fails to hold, and hence (1a) fails to hold.
Also, here $\|\frac N{\sqrt n}-1\|_{\psi_2}=\||X_1|-1\|_{\psi_2}\not\to0$, so that (2b) fails to hold, and hence (2a) fails to hold.
Thus, in the "dependent" case, none of your conditions (1a), (1b), (2a), (2b) holds in general.
I appreciate your answer, it's helpful! I've a few following remarks/questions though. First, apologies, as I made a subtle mistake in the constant in the first and second concentration inequality (the ones that I already gathered and not askign to prove) I stated before: the constant I stated before wasn't absolute, but rather depending on the subgaussian vector itself; to clarify this, I modified it correctly now: $P[ |\hspace{1mm}{||X||2 - \sqrt{n}} \hspace{1mm}| \ge t] \le 2e^{-ct^2/K^4},K=max{1 \le i \le n}||X_i||_{\psi_2}.$ [contd.]
With this modification where the constant $CK^2$ in the second assumed inequality has a part depending on $n$, namely, $K^2 \equiv K(n)^2= (max_{1 \le i \le n}||X_i||_{\psi_2})^2,$ and this can go to $\infty$ as $n \to \infty$ (correct me if I'm wrong!). So, in this case:we've $ || N - \sqrt{n}|| \le CK^2, || \frac{N}{\sqrt{n}} - 1 || \le C \frac{K(n)^2}{\sqrt{n}}$ may not go to $0,$, but it will, if $X_i$'s are iid, not assumed in the question. Correct me if I'm wrong!
[contd.] Also, regarding the part where you wrote: "whence
$|ER_n|\le\frac{E(N^2-n)^2}{n^{3/2}}=\frac{Var(N^2)}{n^{3/2}}\le\frac{s^2}{n^{1/2}}\to 0$," I'm not sure I see why: $s^2= max_{1 \le i \le n} Var[X_i^2]= \mathbb{E}{X_i}^4 - 1$ is finite for each $n$ for sure, but why does it stay bounded as $n \to \infty,$ i.e. why do we assume $lim sup s^2(n)$ is not $\infty ?$ In this case, $\frac{s^2}{n^{1/2}}$ may not go to $0,$ rigth?
Also, I didn't understand the Central Limit Theorem argument: I see one can apply it on $N^2,$ but how does it helps us see $N - \sqrt{n} \to N(0,1/2)?$
@Learningmath : If you don't have a boundedness condition on the moments of the $X_i$'s of a high enough order (at least on an average), then of course the results won't hold. (My condition was $s^2<\infty$.) Moreover, all your questions will obviously lose any meaning if you don't specify the rate of growth of the moments of the $X_i$'s that you want to allow. Therefore, if you can specify such conditions, then I suggest you state them in another, separately posted question.
@Learningmath : As for the central limit theorem (CLT) argument, you do first apply it to $N^2$, and then just use (1) and the fact that $R_n\to0$ in probability/distribution to get the CLT for $N$. (That $R_n\to0$ in probability and hence in distribution follows by Markov's inequality, because $ER_n\to0$.)
|
2025-03-21T14:48:30.300762
| 2020-04-14T20:15:22 |
357487
|
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|
Stack Exchange
|
Optimization with parametric constraints: solution maps
For constrained optimization problems
$$ \begin{array}{ll} \min\limits_{x \in \mathbb R^n} & f(p, x) \\
\text{s.t.} & x \in C \end{array} $$
where $p \in \mathbb R$ is a parameter, we can deduce, under convexity of $f, C$, existence and Lipschitzness (in $p$) of solution maps $x^*(p) \in \arg\min\limits_{x \in C} f(p, x)$, say, by applying Theorem 2F.10 from this Rockafellar's book. Here, I attached an excerpt from there:
Question: under what conditions Lipschitz-dependence of the solution map can be derived when $C$ depends on $p$? Say, if it's described by $g(p, x) \le 0$ with whatever properties required, like continuity in $p, x$, (strong) convexity in $x$ for each $p$ etc.
The said book does not seem to contain a result in such a generic form.
One particular problem I see there is that the subgradients be pivoted to the static constraint. The argument collapses when the constraint changes
I'm almost sure that if the cost function is convex and the constraint doesn't change arbitrarily fast, the optimum must be continuous with a bounded rate of change
This is a repost of my MSE question that, unfortunately, didn't lead anywhere, so I decided to try my luck here.
Not sure if I understand correctly. Lets stay one-dimensional and take $f(x)=-x^2$ and $C(p)=[p-1,p+1]$ then $argmin_{x \in C(p)} f(x)=p+sign(p)$. So even though $f$ and $C$ are smooth in $x$ and $p$ the solution map is not. Doesn't this show that it is impossible allow $C$ to depend on $p$ while still have smoothness of the solution map?
@user35593 I specified the question better. It should be possible to achieve to continuity and even Lipschitzness when $f,g$ are convex and $g$ depends on $p$ continuously
I came up with the following counterexample to show that continuity or convexity in x for each p is not enough. Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ resp. $g: \mathbb{R}^2\times \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=x_1^2$ resp. $g(x,p)=(x_1-px_2)^2-1$. Then the minimizer is $x^*=(0,1/p)$ for $p\neq 0$. Hence it is not continuous at zero.
@user35593 But $g$ is not convex in $x$ for every $p$
it is not strictly convex but it is convex. I actually meant $g(x,p)=(x_1-2-px_2)^2-1$ so that $C$ does not contain the global minimum of $f$.
|
2025-03-21T14:48:30.300940
| 2020-04-14T20:16:42 |
357488
|
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"Francis",
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"url": "https://mathoverflow.net/questions/357488"
}
|
Stack Exchange
|
Given a finite set of points, does there exist a linear function pass through a point and strictly below the other points for all the points?
I guess my question is a follow up question of this one: usul, Existence of a strictly convex function interpolating given gradients and values, version: 2019-04-13.
In usul's question, the answer proves the existence of such a convex function. I think the assumption the question has - "For each point, we are given a linear function through it and strictly below the others" is very interesting. It's not difficult for us to construct such a convex function if the assumption is true. However, I am more interested in what conditions could make the assumption to be true?
In other words, I think my question is: suppose we are given a finite set $C$ with cardinality $T$ of pairs $(x,y) \in C$, with $x \in \mathbb{R}^d$ and $y \in \mathbb{R}$, under what conditions of $X = \{x_1, \dotsc, x_T\}$ and $Y = \{y_1, \dotsc, y_T\}$, can we construct $T$ linear functions $L =\{l_1, \dotsc, l_T\}$ such that each linear function $l_i$ passes through $(x_i, y_i)$ and is strictly below other points?
The mathematical formula for the above question may be written as the following: let each linear function $l_i = k_i(x - x_i) + y_i$, then the above question is to decide whether there exist a sequence of $K = \{k_1, \dotsc, k_T\}$ such that $l_i(x_i) > l_j(x_i)$, i.e.,
$$
y_i > k_j(x_i - x_j) + y_j \,\, \forall i\neq j.
$$
Then I think the question is to decide the existence of such a sequence of $K$ which I am not sure about the answer. Also, I don't know if the above mathematical representation is the best way to solve this problem. Feel free to come up with your own method if my initial thinking doesn't work.
Updated: an image from usul's question so this question could be more intuitive:
Maybe I don't understand the question ... isn't the condition just that each point should lie below the line joining the two points to either side of it?
MathJax supports links, so I combined the separate text-and-URLs into a single link. Since you were concerned with pointing to a specific revision, I linked that, too. Also, the problem has two answers; I picked the one that seemed more likely and linked that. I hope that all this is agreeable.
Nik's understanding is correct, but my question is what kind of sequences of $X$ and $Y$ can make this condition (i.e. each point should lie below the line joining the two points to either side of it) be satisfied? For example, give some points ${(0,1), (1,0), (2,1)}$ (i.e. $f(x) = (x-1)^2$), we know we can construct 3 linear functions as desired. However, if we are given a set of ${(0,-1), (1, 0), (2, -1)}$ then we can't construct such linear functions. Therefore, my question is: if we are given a set of points, can we construct such linear functions?
To be more specific, my question is like the following: given variable sequence ${ x_t }{t \in [T]} $ and function value sequence ${ y_t }{t \in [T]} $, when would it be possible to find a concave function $u$ such that $u(x_t) = y_t$.
I think another problem here is that my $x$ is in $d$ dimension. Therefore, we don't have such kind of "neighboring relationship from Nik's comment" for the points, i.e. we don't have an ordering for the ${x_t}_{t\in T}$ so we don't have neighboring points for each point.
|
2025-03-21T14:48:30.301275
| 2020-04-14T20:22:05 |
357489
|
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"Benjamin Steinberg",
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|
Stack Exchange
|
Are most semigroups nilpotent of degree 3?
A semigroup is nilpotent of degree 3 if every product of 3 elements gives the same result. In 2012, Andreas Distler and James D. Mitchell wrote that:
It is part of the folklore of semigroup theory that almost all finite semigroups are nilpotent of degree 3.
Here's a way to make this precise:
Conjecture. If $S(n)$ is the number of isomorphism classes of semigroups with $n$ elements, and $S_3(n)$ is the number of isomorphism classes of semigroups with $n$ elements that are nilpotent of degree 3, then
$$ \lim_{n \to \infty} \frac{S_3(n)}{S(n)} = 1. $$
Has anyone proved this, or a similar result?
Semigroup theorists seem to enjoy counting semigroups up to equivalence, meaning up to isomorphism or anti-isomorphism, so maybe someone has proved a similar result but counting semigroups up to equivalence rather than up to isomomorphism.
In 2012, Andreas Distler, Chris Jefferson, Tom Kelsey, and Lars Kottho counted the semigroups with 10 elements and found
$$ 12,418,001,077,381,302,684 $$
of them. Of these, all but
$$ 718,981,858,383,872 $$
were nilpotent of degree 3. So, about 99.994% were nilpotent of degree 3.
By the way, this calculation still took a lot of work.
I believe if you count labeled multiplication tables rather than isomorphism then the result is known. See https://www.ams.org/journals/proc/1976-055-01/S0002-9939-1976-0414380-0/S0002-9939-1976-0414380-0.pdf for labeled semigroups.
Since 3-nilpotent is anti-isomorphism-invariant and passing from isomorphism classes to equivalence classes is $\le 2$-to-1, counting up to isomorphism or equivalence doesn't matter.
Sorry. I deleted too many old comments. I'll rewrite. Nobody has ever proved this folklore conjecture. It is like the conjecture that almost all finite groups of order less than n are 2-groups. Everybody believes it but no ideas how to prove it.
|
2025-03-21T14:48:30.301419
| 2020-04-14T21:16:22 |
357491
|
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"Aleksei Kulikov",
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|
Stack Exchange
|
Is there a simple proof that proves $C^1[0, 1]$ is $\Sigma^1_1$ in $C[0, 1]$?
In his book, "Descriptive Set Theory", Moschovakis states $C^1[0, 1]$ is $\boldsymbol{\Sigma}^1_1$ in $C[0, 1]$ in the exercise 1E.8.
Here, $C[0, 1]$ is the space (metrized by the sup norm) of continuous functions from $[0, 1]$ to $\mathbb{R}$ and $C^1[0, 1]$ is the set of continuously differentiable functions from $[0, 1]$ to $\mathbb{R}$.
Anush Tserunyan says in her lecture note actually that $C^1[0, 1]$ is $\boldsymbol{\Pi}^0_3$ in $C[0, 1]$, that is stronger claim than Moschovakis'.
There is a proof of $\boldsymbol{\Pi}^0_3$-ness in Anush's note (Example 11.6).
But I think there may be a shorter proof of $\boldsymbol{\Sigma}^1_1$-ness than the proof of $\boldsymbol{\Pi}^0_3$-ness.
I suspect the following claim is true: for all $f \in C[0, 1]$,
$$
f \in C^1[0, 1] \iff (\exists g \in C[0, 1], \forall x \in [0, 1] \cap \mathbb{Q}, f'(x) = g(x)).
$$
But I have not be able to prove this claim.
Is the claim true? Or is there another proof of $\boldsymbol{\Sigma}^1_1$-ness?
I know nothing about descriptive set theory so can not comment on the second question, but your claim is false, see this question
Thank you. I did not know such an interesting function.
You can even make it everywhere differentiable. See Pompeiu's derivative. https://en.wikipedia.org/wiki/Pompeiu_derivative
Your claim is false, as pointed out in the comments. And clearly a function is $C^1$ iff $f-f(0)$ is in the image of the continuous operator $g \mapsto (x \mapsto \int_0^x g(t) dt) $, which gives an analytic definition. I don't think your question is suitable for this forum.
Thank you. This proof of $\Sigma^1_1$-ness is very simple and smart, which I have looked for.
|
2025-03-21T14:48:30.301569
| 2020-04-14T21:17:40 |
357492
|
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"zeraoulia rafik"
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"url": "https://mathoverflow.net/questions/357492"
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|
Stack Exchange
|
Inverse of the incomplete elliptic integral of the second kind
The incomplete elliptic integral of the second kind $E(\varphi \, | \,k)$ is defined as follows:
$$E(\varphi \, | \,k) = \int_0^\varphi \sqrt{1-k^2\sin^2\theta} \, \mathrm{d}\theta $$
Where $0<k^2 < 1$.
Five years ago, this MSE post was made asking about an inverse to this function (with respect to $\varphi$.) Wolfram is (or, perhaps was, I am not sure) also looking for such a function.
I am posting this to ask if there has been any progress as to defining a function $E^{-1}$ so that
$$E(\varphi \, | \,k) = v \implies \varphi = E^{-1}(v \, | \,k)$$
Or approximating $E^{-1}(v \, | \,k)$.
Edit: This article provides a numerial method to obtain a high-accuracy estimate of $E^{-1}(v \, | \,k)$.
Have you looked to write your integral as power series or formal power series ? then you can use inverse function theorem to get your coeffecients
For a numerical representation of the inverse in terms of the angle $\varphi$
where $E(\varphi \, | \,k) = \int_0^\varphi \sqrt{1-k^2\sin^2\theta} \, \mathrm{d}\theta $ is the
elliptic integral for the second kind, one could expand $E(\varphi |k)$ in a power
series around $\varphi=0$,
$ E(k |\varphi) := \varphi
-1/6 k^2\varphi^3
+1/5(1/6k^2-1/8k^2)\varphi^5
+1/7(-1/45 k^2+1/12k^4-1/16k^6)\varphi^7
+1/9(1/630k^2-1/40k^4+1/16k^6-5/128k^8)\varphi^9 \cdots$
let $k^2=m$,The expansion coefficients in front of the order $\varphi^{n+1} ,(n=2,4,6,...)$ are
$\sum_{r=2}^{n}U(r,n)F(r,m)/r!]/(n+1)!$, $r$ is even integer and
$U(r,n) = (-1)^{(r+n)/2}/2^{r-n} \sum_{l=0}^{r}(-1)^l (l-r/2)^n \binom{r}{l}]$ and $F(r,m) = -m^{n/2}[(r-1)!!]^2/(r-1)$ ,
with $(r-1)!! = 1\times 3\times 5\times 7\times \cdots\times(r-1) $
Then invert this as outlined in chapt 3.6.25 of the book edited by M Abramowitz
and I Stegun you get finally your inverse :
$\varphi := E(\varphi,m)
+1/6mE(\varphi,m)^3
+1/120m(13m-4)E(\varphi,m)^5
+1/5040m(493m^2-284m+16)E(phi,m)^7
+1/362880m(37369m^3-31224m^2+4944m-64)E(\varphi,m)^9 ...$
you may use Mathematica code up to $\varphi=12$ you may try the following code for numerical verification:
In[2]:= InverseSeries[Series[EllipticE[z, m], {z, 0, 12}]]
Out[2]= z + (m*z^3)/6 + (1/120)*(-4*m + 13*m^2)*z^5 + ((16*m - 284*m^2 +
493*m^3)*z^7)/5040 + ((-64*m + 4944*m^2 - 31224*m^3 +
37369*m^4)*z^9)/362880 + ((256*m - 81088*m^2 + 1406832*m^3 - 5165224*m^4 +
4732249*m^5)*z^11)/39916800 + O[z]^13
|
2025-03-21T14:48:30.301719
| 2020-04-14T22:11:03 |
357497
|
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|
Stack Exchange
|
Signature of a non-compact manifold
Let $v_0,\dots,v_n\in\mathbb{Z}^2$ be integer vectors which satisfy the condition $\det\begin{pmatrix}v_{k-1}&v_k\end{pmatrix}=(-1)^k$, whose relevance will become apparent in a moment. We may then construct a $4$-manifold $M$ as follows.
For $2\leq k\leq n-1$, let $\tilde{M}_k$ be the set of points $(z,w)\in\mathbb{C}^2$ for which $-2\leq|w|^2-|z|^2\leq 2$. For $k=1$, do the same thing except only requiring $|w|^2-|z|^2\leq 2$, and similarly, for $k=n$, only require $-2\leq |w|^2-|z|^2$.
Making use of the determinant condition stated at the beginning, we may write $$\begin{pmatrix}v_{k-1}&v_k\end{pmatrix}^{-1}\begin{pmatrix}v_k&v_{k+1}\end{pmatrix}=\begin{pmatrix}0&1\\1&m_k\end{pmatrix}$$ for some $m_k\in\mathbb{Z}$. Let $M$ be the disjoint union of the sets $\tilde{M}_k$, modulo identifying $(z,w)\in\tilde{M}_{k+1}$ with $((\frac{z}{|z|})^{m_k}w,z)\in\tilde{M}_k$, whenever $|w|^2-|z|^2=-2$.
One way to view $M$ is as an "$\mathbb{R}^4$ with $n$ origins", since it is possible to give a torus action on $M$ whose fixed points are exactly the origins in the sets $\tilde{M}_k$. The manifold $M$ can quite easily be given a natural smooth structure and a natural orientation.
Anyway, I am trying to calculate the topological signature of $M$, and part of this is of course to make sure I am working with the "right" definition of signature. My motivation for doing this is to assume the existence of a Ricci-flat metric on $M$ with appropriate asymptotic structure, in order to apply (a version of) the Hitchin-Thorpe inequality (see here, Section 6.2). The definition of signature given on Wikipedia seems to only discuss the compact case. I highly suspect that the definition of signature in terms of compactly supported cohomology should be the correct one in order to work with the Hitchin-Thorpe inequality I linked (does this seem correct?). In other words, we define the signature $\tau(M)$ as the signature of the bilinear form $H_c^2(M,\mathbb{R})\times H_c^2(M,\mathbb{R})\to H_c^4(M,\mathbb{R})\cong\mathbb{R}$ given by the cup product.
Assuming this is the case, I am a bit stuck on how to calculate this signature, since I'm not an algebraic topologist. Using the Mayer-Vietoris sequence with compact supports, I proved that the dimension of the second compactly supported cohomology of $M$ is $n-1$, which therefore reduces the problem to finding $n-1$ independent generators (for instance, compactly supported closed $2$-forms, if we're working with de Rham cohomology), in order to compute the cup product on those. But I'm having a hard time constructing such generators. Is there any simple approach for this signature calculation that I am missing?
One thing that I noticed is that for each $k\in\{1,\dots,n-1\}$, the set of points $(0,w)\in\tilde{M}_k$ together with the set of points $(z,0)\in\tilde{M}_{k+1}$ forms a $2$-sphere $R_k\subseteq M$, and in addition, these $2$-spheres form a basis for the second homology with real coefficients. I got this by simply constructing (non-compactly supported) closed $2$-forms $\omega_k$, such that the integral of $\omega_k$ over $R_l$ is $\delta_{kl}$. Does this help us in any way?
To summarize, I have two main questions:
Am I working with the right notion of signature?
How should I approach the calculation of this signature?
|
2025-03-21T14:48:30.301993
| 2020-04-14T22:20:14 |
357498
|
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"authors": [
"Akiva Weinberger",
"Alex Nelson",
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"Emil Jeřábek",
"John Baez",
"Timothy Chow",
"Todd Trimble",
"YCor",
"benrg",
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"https://mathoverflow.net/users/50073",
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"https://mathoverflow.net/users/88172",
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|
Stack Exchange
|
Bourbaki's definition of the number 1
According to a polemical article by Adrian Mathias, Robert Solovay showed that Bourbaki's definition of the number 1, written out using the formalism in the 1970 edition of Théorie des Ensembles, requires
2,409,875,496,393,137,472,149,767,527,877,436,912,979,508,338,752,092,897 $\approx$ 2.4 $\cdot$ 1054
symbols and
871,880,233,733,949,069,946,182,804,910,912,227,472,430,953,034,182,177 $\approx$ 8.7 $\cdot$ 1053
connective links used in their treatment of bound variables. Mathias notes that at 80 symbols per line, 50 lines per page, 1,000 pages per book, this definition would fill up 6 $\cdot$ 1047 books. (If each book weighed a kilogram, these books would be about 200,000 times the mass of the Milky Way.)
My question: can anyone verify Solovay's calculation?
Solovay originally did this calculation using a program in Lisp. I asked him if he still had it, but it seems he does not. He has asked Mathias, and if it turns up I'll let people know.
(I conjecture that Bourbaki's proof of 1+1=2, written on paper, would not fit inside the observable Universe.)
I'm going to move comments to chat. Honestly, I'm finding some of the language used in comments pretty abusive (such as saying that the OP is "ranting" when in fact he's quoting someone, or saying the style of the question is "awful"). Let's tone it down, please.
I was able to confirm Matthias' calculations that the ordered pair $(x, y)$ has a length of 4545 symbols. I can continue trying to confirm the estimates for the length of the Cartesian product $X\times Y$, and then...whatever else.
The Cartesian product is a term represented by 3,184,591,216,053,048,167 symbols(!), confirming the number reported by Matthias.
Has anyone tried this exercise with Principia Mathematica's definition of 1?
These calculations have been carried out by José Grimm; see [1] as well as [2].
According to one version of the formalism in the original Bourbaki, Grimm gets
$$16420314314806459564661629306079999627642979365493156625 \approx 1.6 \times 10^{55}$$
(see page 517 of [1, version 10]). The discrepancy with Solovay's number is probably due to some subtle difference of interpretation of some detail. Note that the English translation of Bourbaki introduces some "small" changes and Grimm gets a rather different value:
$$5733067044017980337582376403672241161543539419681476659296689 \approx 5.7 \times 10^{60}$$
EDIT: As suggested in the comments, here are the full citations for Grimm's papers.
[1] José Grimm. Implementation of Bourbaki’s Elements of Mathematics in Coq: Part Two; Ordered Sets, Cardinals, Integers. [Research Report] RR-7150, Inria Sophia Antipolis; INRIA. 2018, pp.826. inria-00440786v10. doi:10.6092/issn.1972-5787/4771
[2] Grimm, J. (2010). Implementation of Bourbaki's Elements of Mathematics in Coq: Part One, Theory of Sets. Journal of Formalized Reasoning, 3(1), 79-126. doi:10.6092/issn.1972-5787/1899
Might be good to add the titles of the linked articles in case the links break at some point in the future.
Grimm also rederives Solovay's exact numbers: see part 2, page 514. Solovay defines a singleton set as ${x:x=a}$ and uses Bourbaki's definition of $1$. The $5.7\times 10^{60}$ figure comes from defining a singleton set as ${x:x=a\vee x=a}$ and using an allegedly fixed definition of $1$. I don't understand the assumptions behind the $1.6\times 10^{55}$ figure.
There are some interesting issues here, as I will elaborate. Let me first make a full quote of §4.2 "Bourbaki on formalization" from Thomas Hales' 2014 Bourbaki seminar "Developments in formal proofs" ((1)).
Over the past generation, the mantle for Bourbaki-style mathematics has passed to
the formal proof community, in the way it deliberates carefully on matters of notation
and terminology, finds the appropriate level of generalization of concepts, and situates
different branches of mathematics within a coherent framework.
The opening quote claims that formalized mathematics is absolutely unrealizable. Bourbaki objected that formal proofs are too long ("la moindre démonstration
. . . exigerait déjà des centaines de signes" [translation by YCor: "any proof... would require hundreds of signs")], that it would be a burden to forego
the convenience of abuses of notation, and that they do not leave room for useful
metamathematical arguments and abbreviations ((2)).
Bourbaki is correct in the strict sense that no human artifact is absolutely trustworthy
and that the standards of mathematics evolve in a historical process, according to available technology. Nevertheless, the technological barriers hindering formalization have
fallen one after another. Today, computer verifications that check millions of inferences
are routine. As Gonthier has convincingly shown in the Odd Order theorem project,
many abuses of notation can actually be described by precise rules and implemented
as algorithms, making the term abuse of notation really something of a misnomer, and
allowing mathematicians to work formally with customary notation. Finally, the trend
over the past decades has been to move more and more features out of the metatheory
and into the theory by making use of features of higher-order logic and reflection. In
particular, it is now standard to treat abbreviations and definitions as part of the logic
itself rather than metatheory.
So Bourbaki's point (at that time, namely in the few years before 1970) was that writing formal proofs, although precisely defined, is a hopeless task. For this reason, Bourbaki made no effort of efficiency.
To give an illustrating example, assume that I'm writing an exercise program in some language, computing $n\mapsto \sum_{k=1}^nf(k+a_n)$, where $a(n)$ is the $n$-the decimal of $\pi$, and $f(n)$ say is $\lfloor n^{3/2}\rfloor$, which I previously defined as functions in the same programming language (the only point with this choice is that $f$ is much faster to compute than $a$). If I do it crudely
$$(j:=0; \quad \text{for } k=1..n\;\; j:= j+f(k+a_n);\quad \text{return }j)$$ then I will compute $a(n)$ $n$ times. If instead I write
$$(j:=0;\quad a:=a(n); \quad \text{for } k=1..n\;\; j:= j+f(k+a);\quad \text{return }j)$$
I'll compute $a(n)$ only once and hence this will be far quicker, although at first sight this is the same "algorithm". If I roughly say what should be the principle of the proof, this aspect will not appear.
The point of Bourbaki, once they assume that formalizing proofs is not worth an explicit realization, was therefore not to make this realization any practical. The possible (likely) fact that the size of the resulting formal proof of $1+1=2$ is huge is therefore anecdotical: for instance, if $N$ is the already huge proof of $0+1=1$ (or any related prior result), it's very possible that the formal proof will contain identical copies of this proof $N$ times, resulting in something of size $\ge N^2$. The conclusion is that the Bourbaki's formalization is highly unpractical— this conclusion was already Bourbaki's (yet based on a clear underestimate, which would appear today as efficient). Given that this formalization was written just to exist and without practical concern, this is not a big deal (although the paper advertized by the OP makes a lot of conclusions from this fact).
Most likely, the main ideas of Bourbaki's foundations could be formalized in a more efficient way (more efficient than highly inefficient is not too hard— should we bluster if $10^{54}$ is upgraded to $10^{20}$?), but I have no idea whether they could be formalized in a useful practical way. (There might also be good reasons that Bourbaki's foundations are not prone to efficient formalization; the estimate asked by OP is not a sufficient one, for the reason elaborated in the previous paragraph.) Given that Bourbaki's foundations have an interest which is now essentially reduced to historical (including for the aspects other than this exact formalization), and other foundations have been successfully formalized by others in the last 15 years, I'm not sure if anybody would spend energy on this.
At a historical level, one can wonder whether Bourbaki's 1970 belief that formal proofs are not to be written down, has had any counterproductive effect in the next few decades; this is hard to measure and the paper ((3)) linked by the OP speculates on this with no serious grounds. Let me make a quote from the conclusion of ((3)).
Bourbaki themselves took the first course: as remarked by Corry, they shied away from their own
foundations. I expect that they came to the conclusion that logic is crazy — they had to conclude that
to protect their sanity; but were they aware that the picture of logic they were giving to their disciples
is merely a grotesque distortion and diminution of that subject ? Is it too fanciful to see here, in this
choice of formalism, with its unintuitive treatment of quantifiers, the reason for the phenomenon (which
many mathematicians in various European countries have drawn to my attention whilst beseeching me not
to betray their identity, lest the all-powerful Bourbachistes take revenge by depriving them progressively
of research grants, office facilities and ultimately of employment) that where the influence of Bourbaki is
strong, support for logic is weak ? How does one get the message across, to those who have accepted
the Bourbachiste gospel, that logicians are actually not interested in a formal system of such purposeless
prolixity, still less do they advocate it as the proper intellectual framework for doing mathematics ?
In regard to the fact that Bourbaki invited Hales to talk on formal proofs, I found the "revenge" claim particularly crisp today! ((3)) was written earlier, but whether there was a ban in the 1990s against elaboration and study of formal proofs, I'll leave it to better witnesses of that time and subject.
((1)) Th. Hales. Developments in formal proofs
((2)) N. Bourbaki. Théorie des ensembles [Set theory], 1970.
((3)) A. Mathias. A term of length 4,523,659,424,929 (1999). Link
PS: about the estimate: if we start from $2$ and iterate $n$ times the squaring function, we get $2^{2^n}$, which in 7 steps reaches $2^{128}\sim 10^{39}$ and in 8 steps reaches $2^{256}\sim 10^{77}$. Hence assuming that every step in an intuitive proof roughly squares the size of the formal proof makes such estimates far from surprising.
PPS Here's (on YouTube) Th. Hales' video of the quoted June 2014 Bourbaki seminar, and Thierry Coquand's talk the same day, the latter being in French.
Tangential comment on "definitions as part of the logic itself" (at the end of the quote form Hales): See also the book "A Theory of Sets" by Anthony P. Morse (1965) and the work of Lesniewski in the 1920's and 1930's.
One can eliminate definitions without exponential blowup: https://www.andrew.cmu.edu/user/avigad/Papers/definitions.pdf . (Though I’m not sure if it applies to Bourbaki’s formalization, which I’m told does not use standard first-order logic, but some sort of $\varepsilon$-calculus.)
@EmilJeřábek: How does your last comment square with another answer of yours, since ACA0 is essentially PA plus definitorial expansion, yet has superexponential speedup over PA?
@user21820 I have no idea where you got the idea that ACA0 is essentially PA plus definitional expansion. It’s not.
@EmilJeřábek: Maybe we have different notions of "definitional expansion". ACA0 has the axioms of PA− plus the second-order induction axiom plus arithmetical comprehension axiom. Let's ignore the induction axiom for now, since every instance of the PA induction schema is trivially obtained from it. Every set given by that comprehension axiom is definable over PA, where the defining formula may have set parameters. Is it because of those set parameters that you do not consider ACA0 to be PA plus definitional expansion?
@user21820 See https://en.wikipedia.org/wiki/Extension_by_definitions . Since ACA0 is a two-sorted theory, you’d also need extension by definition of a new sort, which is not treated in the article: this amounts to fixing a formula $\delta(x)$ and defining $\exists X,\phi(X)$ as $\exists x,(\delta(x)\land\phi(x))$ and $\forall X,\phi(X)$ as $\forall x,(\delta(x)\to\phi(x))$. No such thing is possible for ACA0. More generally, you cannot even interpret ACA0 in PA, as ACA0 is finitely axiomatizable and PA is reflexive.
@EmilJeřábek: Ok so your comment was only about that strict kind of extension. In short, extending an FOL theory by adding sorts and comprehension axioms for the first-order definable predicates/functions will not suffer exponential blowup, but a further extension to add comprehension with parameters from the added sorts may lead to exponential speedup. Correct?
You can add individual definable predicates/functions as new symbols, one symbol per predicate. You cannot add a new sort for the collection of all definable predicates, along with a membership predicate. This is a completely different thing. Parameters have nothing to do with it.
I understand that. Perhaps I need to explain why I had in mind a broader kind of definitorial expansion. It is the result of starting with the original theory embedded in a higher-order logic (which as you said needs extra sorts) with no comprehension axioms, and then adding comprehension axioms corresponding to first-order definable predicates/functions (in the higher-order language). As for parameters, if you restrict ACA0's comprehension schema to $∃S\ ∀k\ ( k∈S ⇔ φ(k) )$ for each arithmetical formula $φ$ (with no set parameters), then it really becomes like PA plus definitorial expansion.
@user21820 Whatever you have in mind, this is not what defitional expansion means. The point of extension by definitions is that you DO NOT increase the expressive power of the theory, only adding new names for things that are already definable in the original language. That’s why they are called DEFINITIONS. If you do something else, you need to call it differently. And please, take the discussion elsewhere, as it is completely irrelevant to the original question, YCor’s answer, and my original comment.
@EmilJeřábek: Yes, my initial comment was based on a misunderstanding of what you meant. But I was just asking a question, no need for you to get so riled up as to use caps. Anyway, thanks for your responses.
I asked Robert Solovay if he still had the program he used to compute the length of Bourbaki's definition of "1" (using their 1970 definition of ordered pairs). He didn't, but Adrian Mathias did, and Solovay has allowed me to release it. So, here are three documents. They don't settle every question one might have, but they should be useful to people attempting to check this calculation.
calcusol.pdf. This is a document entitled "The Bourbaki Constant 1". It begins:
This is a private document [for the eyes of RMS and ARDM only] which extends ARDM's computation of the length of the Bourbaki rendition of "The ineffable name of 1" to the case when the Kuratowski ordered pair is employed. My plan is to write programs in Allegro Common Lisp to compute the relevant numbers.
I program using the style of "literate programming" introduced by Knuth. However the Web and Tangle introduced by Knuth [which have been refined to CWEb and CTangle by Levy] are limited to languages closely linked to C or Pascal. So I prefer to use a more flexible literate programming language which permits fairly arbitrary target languages. Currently, I use Nuweb which is available on the TEX archives in the directory /web/nuweb.
One of the nice things about literate programming is that one can write programs in the natural psychological order, but arrange that the output files have the order needed for the target programming language. We will exploit this heavily in what follows.
solfact.txt. This is a short document including the output of the program described in the previous document. Here it is, in its entirety:
From<EMAIL_ADDRESS>Wed Nov 11 13:39:46 1998
Date: Tue, 10 Nov 1998 23:43:04 -0800 (PST)
From: "Robert M. Solovay"
To<EMAIL_ADDRESS>Cc<EMAIL_ADDRESS>Subject: Results
Adrian,
Here is the printout of my calculation of the length of the
Bourbaki term for 1. If we do the original definition, I get approx.
4.524 * 10^{12}
If we use the Kuratowski ordered pair, I get approx.
2.41 * 10^{54}
This is big, but not nearly as big as the 2 * 10^{73} that you
claim. This is certainly related to the smaller estimate that I have
for the size of the Kuratowski ordered pair.
I omitted some trivial lines from this printout where I "gave
the wrong commands to the genie".
USER(1): (setq p 0) ;;;[Doing the original Bourbaki definition where
;;; ordered pair is a basic undefined notion.]
0
USER(3): (load "compute.cl")
; Loading ./compute.cl
T
USER(4): J_length
4523659424929
USER(5): (log J_length 10)
12.65549
USER(6): J_links
1179618517981
USER(7): (log J_links 10)
12.071742
USER(8): (setq p 1) ;;; Now use Kuratowski ordered pair
1
USER(10): (load "compute.cl")
; Loading ./compute.cl
T
USER(11): J_length
2409875496393137472149767527877436912979508338752092897
USER(12): (log J_length 10)
54.381996
USER(13): J_links
871880233733949069946182804910912227472430953034182177
USER(14): (log J_links 10)
53.940456
solpair.pdf. This is a short note on some of Bourbaki's definitions including the definition of ordered pair.
There seems to be a rule of thumb in the logic community: if Solovay says it, it's correct. He seems to have garnered the reputation of being very, very careful. For example, he was the one who alerted Nash to a hole in his (Nash's) embedding theorem.
I was able to reproduce Mathias's results with some Haskell code with some specific details about how many symbols are needed in each term. (As a sanity check, I verified I recovered the same results reported by Mathias term-by-term when the ordered product was primitive.)
When the ordered pair is not primitive, the sizes of the various terms are:
Size of 1 = 2,409,875,496,393,137,472,149,767,527,877,436,912,979,508,338,752,092,897
Size of term A = 15,756,227
Size of term B = 10,006,221,599,868,316,846
Size of term C = 59,308,566,315
Size of term D = 364,936,653,508,895,574,881
Size of term E = 101,217,516,631
One thing worth noting is that, well, this seems dishonest. I mean, there are a lot of double negations which are not simplified, which bloats the size quite a bit (an additional $1.863\times 10^{53}$ symbols or so). I wouldn't be surprised if there were other simplifications which would cut down the bloat further...not that we'd get anything less than $10^{50}$ or so.
If you'd like to check the number of links, I can do that too.
Addendum. The relation "1+1=2" can be computed, and found to have a length of 22,411,322,875,029,037,193,545,441,224,646,148,573,589,725,893,763,139,344,694,162,029,240,084,343,041 (or approximately $2.24113228750290371 \times 10^{76}$). This is using the definitions in Bourbaki of cardinal addition $\mathfrak{a}+\mathfrak{b}$ using the disjoint sum of the indexed family $f\colon\mathrm{Card}(2)\to\{\mathfrak{a},\mathfrak{b}\}$ considered as a graph. It's really convoluted, but the details can be found in Bourbaki's Theory of Sets Chapter II sections 3.4, 4.1, and 4.8 as well as Proposition 5 (in chapter III, section 3.3); this all works with the Kuratowski ordered pair, not a primitive $\bullet A B$ ordered pair.
For what it's worth, computing the size of 1 was nearly instantaneous, whereas computing the size of "1+1=2" took about 7 minutes and 30 seconds.
Hmm...actually, if we simplify it further with substitutions, we can cut the representation of 1 down to $6.011873743380472\times 10^{37}$ symbols.
Do you mean Mathias?
Damn it, man, I'm a mathematician, not a spelling bee champion! Err, I mean, thanks, I'll fix that :p
Nice! By the way, this particular result 2,409,875,496,393,137,472,149,767,527,877,436,912,979,508,338,752,092,897 was calculated by Robert Solovay, not Adrian Mathias. See my own answer for Solovay's software.
And for the sake of completeness, using a primitive ordered pair, the length of "1+1=2" is a modest 19,516,572,617,436,743,593 $\approx 1.9\times 10^{19}$ symbols.
@AlexNelson : If you have the time, I'd be curious to hear your take on Grimm's papers. Do you agree with me that he gets a slightly different number? If so, I have not read carefully enough to understand why.
@TimothyChow Oh, I agree with you. because there are syntactic variants for the same concept -- for example -- Mathias took a shortcut in using $\tau_{z}(\forall y)(y\in z\iff y=x)$ for the singleton ${x}$ whereas Bourbaki uses ${x}:={x,x}$ which would be defined as $\tau_{z}(\forall y)(y\in z\iff (y=x\lor y=x))$ which is semantically identical but syntactically bigger, and expands to a larger number of symbols (albeit redundancies). They're semantically equivalent. There are many small syntactic variations which could be made, which explains the variation in sizes reported.
@TimothyChow This sounds unexciting (after all, what's the difference between 4 symbols among friends?), but when it's repeated 1000 times in the full definition, it starts to add up. Particularly because the Cartesian product is used frequently in the definition of Equipotence, and the unordered pair is used in the Cartesian product. The definitions cascade instances of these simpler definitions, which have several variations in their implementations. I could go on and on, if you like :)
@TimothyChow I have been procrastinating on writing a commentary about Bourbaki's foundations (more for "my future self" than anyone else), and I forgot about Grimm's paper. But reading his first paper, he glosses over a lot, and doesn't catch a number of errors in Bourbaki's first chapter, which makes me think he just skipped ahead to chapter 2. I don't understand why Grimm didn't use Isabelle to implement Bourbaki's system as an object logic ("Isabelle/Bourbaki"), I might have more to say later.
@AlexNelson I skimmed through your draft and it's very cool!
Mathias, Grimm, and some people who've contributed to this Q&A seem to take these numbers as evidence of the impracticality of working formally in Bourbaki's theory. YCor explicitly called it unpractical, and said that Bourbaki shared that belief, which may be true. In light of that, I think it's worth writing a bit about the premise that seems to underlie the question.
If you define $F_1=F_2=1,\; F_{n+2}=F_n+F_{n+1}$, and eliminate $F$ from the expression $F_{250}$ by substitution, you get a tree of about $1.6\times 10^{52}$ nodes, or a string of that many symbols in Bourbaki/Polish notation.
I don't think it follows that the definition is a bad one. It's hard to see how you could avoid the blowup without introducing complications that would interfere with the definition's intended use.
You could object that this isn't a comparable situation because $250>1$, but that scarcely matters. Bourbaki happened to put a bunch of structure underneath $1$, but even if $1$ is primitive in your proof system, as soon as you do anything remotely interesting with it, you will have the same problem. The mere statement of your upper bound in Ramsey theory, never mind the proof, will have over Graham's-number symbols in it.
There is a simple way to get a better measure of the complexity of $F_{250}$ without changing the definition: merge common subtrees. The result is a dag of 249 nodes, of which 248 are additions: one for $F_3$, one for $F_4$, $\ldots$
Here's what happens when you merge common subtrees of the expressions from Grimm page 514. (The one called M is Solovay's, and the one called SS is mentioned in Timothy Chow's answer.)
expression
tree size
tree links
dag size
dag links
SS
57330670440×1050
21634097377×1050
13153
876
SM
315628276×1050
114233082×1050
13015
876
S
171713×1050
64721×1050
8061
544
M
24098×1050
8718×1050
7971
544
I calculated these with a Python program that constructs the complete dags in memory and then collects statistics with and without duplication counts. The construction consumes a whopping several megabytes of RAM due to CPython's inefficient object representation, and takes a noticeable fraction of a second due to CPython being slow. If my code were published in a series of books, and each book weighed 1 kilogram, the total mass of all the books would be a few grams.
The dags are still much larger than the code that made them. The remaining bloat can be blamed on the definition of $\exists$, which uses its body twice in a way that precludes sharing, and on complicated constructions that can't be shared because they have children, particularly ordered pairs. These problems can be solved by introducing parametrized subtrees. One approach is to express the dag as a tree without duplication by introducing a "structural let" that assigns names to subtrees, and then permit the named subtrees to have holes whose values are specified at each use point. (Or you can keep the dag and add $\phi$ nodes – see Appel.) With this extra flexibility you can write any of these expressions in just a few hundred nodes.
I want to stress that these concise expressions are fully formally equivalent to the original strings. Not a single double negative has been eliminated. Given the concise formula and an index, you can compute the symbol at that index in the original string.
In this framework – which is just a small subset of what's available in a proof system like Coq – it hardly matters what's primitive. If Bourbaki's pairing operator $\supset$ isn't primitive, you can define it once at the top of the expression; the space cost is constant regardless of how many times it's used. If the expression is a proposition whose proof will be formally checked, the axioms of $\supset$ need to be proven and checked as theorems only once each, adding a constant startup time regardless of how many times they're used. These costs aren't fundamentally different from the costs of implementing $\supset$ as primitive. Only if you convert to the normal form will you see a difference, and there is no reason ever to do that. It's an abstraction violation.
To summarize:
These gigantic normal forms show up in all formal systems, not just Bourbaki's.
They aren't a good measure of practicality or anything else.
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2025-03-21T14:48:30.303828
| 2020-04-15T00:02:48 |
357500
|
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|
Stack Exchange
|
Weyl symbol of product
Are there explicit formulas for the Weyl symbol of $-f(x)D_x^2 $ where $D_x:=-i\partial_x $ and $\partial_x$ is the derivative and $f$ some sufficiently smooth function?
In the standard quantization the symbol would of course just be $w(x,\xi)=f(x) \xi^2$ but in Weyl quantization this seems to be no longer true.
By quantization, one is usually more trying to find the operator corresponding to a given symbol than the symbol corresponding to an operator. In your case, it seems the Weyl symbol you are talking about is the Wigner transform (the inverse of the Weyl trasnform). For an operator $A$ of kernel $a(x,y)$ it is defined by
$$
w_A(x,\xi) = ∫_{\mathbb{R}} e^{-i\,y\,\xi}\,a(x+y/2,x-y/2)\,\mathrm{d}y.
$$
If your operator is $A = -f(x)\,D^2_x = f(x)\,\partial_x^2$, then its kernel is $a(x,y) = f(x)\,\delta_0''(x-y)$, and so since $(x+y/2)-(x-y/2) = y$, we find (replacing the integral by a duality bracket)
$$
\begin{align*}
w_A(x,\xi) &= \left\langle \delta_0'', e^{-i\,\xi\,\cdot}f(x+\cdot/2)\right\rangle
\\
&= \left.\partial^2_y\left(e^{-i\,y\,\xi}f(x+y/2)\right)\right|_{y=0}
\\
&= -\xi^2 f(x) - i\,\xi\, f'(x) + \frac{1}{4}\, f''(x).
\end{align*}
$$
Yes I did the correction thanks (without the $1/2$ since it is counted twice)
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2025-03-21T14:48:30.303955
| 2020-04-15T01:54:22 |
357508
|
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|
Stack Exchange
|
Chord of fixed length traveling around a Jordan curve
Let $C$ be a Jordan curve with nice enough properties whenever necessary (e.g. smooth, or just rectifiable, perhaps). I am interested in knowing how long can a chord be that "traverses" the Jordan curve once around, with both ends attached to the curve. By this I mean we can move around the chord such that both endpoints pass over every point on $C$ continuously. The chord can intersect the curve. See this article on Holditch's theorem for some nice pictures to give some intuition.
Edit:
A better way to define traversing the curve is as follows. Let $M$ be the space of unordered pairs of points of $C$. It is known that $M$ is a Möbius strip, with boundary corresponding to pairs of identical points. Let $\phi:M \to [0,\infty)$ be the square of the distance between a pair of points. If $C$ is smooth, so is $\phi$. For some regular value $l \geq 0$ of $\phi$, its pre-image consists of embedded circles in $M$.
Now we say that a chord of length $l$ can traverse $C$ if $\phi^{-1}(l^2)$ contains a circle component homotopy equivalent to the boundary.
End edit.
There is a simple upper bound to the length of such a chord for arbitrary Jordan curves. Let $d_\theta$ be the largest distance among the set of pairs of points on the curve that form a line at an angle $\theta$, and let $d$ be the minimum $d_\theta$ over all angles. Then a chord longer than $d$ cannot traverse the whole of $C$. Indeed, let $\theta_0$ be an angle such that $d_\theta = d$. Then if the chord traverses the whole of the curve, it will have to rotate a full $2\pi$ and have at some point an angle $\theta_0$, but $d$ exceeds the maximum length at that angle.
I'm guessing that for convex curves, this bound is attained. I was hoping this would also hold for smooth Jordan curves, but I could find non convex counterexamples. I do not know if there exists some arbitrary Jordan curve where there are no chords of any length (besides the trivial one of length $0$) that can traverse the curve, which perhaps could be the case for curves of infinite length, or other highly pathological examples.
So my main question is:
For a nice enough Jordan curve, what is the longest chord that can traverse it, as defined above?
Why does the chord make a full $\pi$ rotation? Why does it even move continuously in the sense that we can choose the second endpoint continuously depending on the first one?
Sorry, typo, it should be $2\pi$. But a rotation by $\pi$ is enough to ensure it covers all angles. As for continuity, I am assuming that as well, and I should have included that. I'll edit this in, thanks.
But still, I don't understand how to actually prove that the angle makes a full rotation, and not going back at some point. Also, I think the question of continuity is quite interesting.. perhaps it shouldn't be swept under the rug?
If the chord is small enough to traverse the whole curve, it is possible that it "goes back at some point", i.e. one of the points (not both) moves in retrograde. I believe this is discussed in some of the examples in the linked article. If you mean that both points travel backwards at some point, this is part of the problem when the chord is too long to traverse the curve. In this case we do not necessarily have a full rotation.
But perhaps I'm misunderstanding your point. Do you mean that, even when the chord is capable of traversing the whole curve, it might not make a full rotation? To be honest I also assumed this because it seemed to me that it is trivial, but thinking about it, it might require something more. I'll give it a try and see if I can come up with something simple.
Yes, that is what i meant. However it seems i have a proof: we have $\gamma,\delta:S^1\to R^2$ - curves with the same image, but we never have $\gamma(t)=\delta(t)$, since we assume that the chord is non-trivial.
If the vector $\gamma(t)-\delta(t)$ does not make a full rotation WLOG $\gamma(t)-\delta(t)$ is never horizontal. If $\gamma=(\gamma_x,\gamma_y)$, and same for $\delta$, this means that $\gamma_y$ is never equal $\delta_y$.
Since these are continuous functions on a connected set $S^1$, we can assume $\gamma_y>\delta_y$. Since $S^1$ is also compact, both functions attain maximum, and maximum of $\gamma_y$ is strictly greater than that of $\delta_y$. This however contradicts the fact that $\gamma$ and $\delta$ have the same image.
By the way i am not using explicitly that $\delta=\gamma\circ\alpha$ for some continuous $\alpha:S^1\to S^1$. Perhaps it follows automatically?
Nice. I'm assuming you mean that $\gamma$ is the first endpoint, $\delta$ is the second? Anyway, for continuity, I think it is necessary to assume it as part of the problem statement.
Yeah, $\gamma$ is the first endpoint, and $\delta$ is the second. So here is what i am wondering about now: we have that both $\gamma$ and $\delta$ are continuous have the same image and $|\gamma-\delta|$ is constant. Does it follow that $\delta=\gamma\circ\alpha$ for continuous $\alpha$?
I've given it some thought but I couldn't come up with anything.
Your question is related to this paper (and those that followed it): Goodman, Jacob E., János Pach, and Chee K. Yap. "Mountain climbing, ladder moving, and the ring-width of a polygon." The American Mathematical Monthly 96, no. 6 (1989): 494-510. PDF download.
Thank you for this. May I ask where I can find the papers that followed this one?
It seems ring-width of a polygon is similar to your question. If you Google Scholar the paper, you will find 34 later papers cite it, including "The mountain climbers' problem."
|
2025-03-21T14:48:30.304384
| 2020-04-15T02:06:39 |
357509
|
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|
Stack Exchange
|
What are the consequence of Snevily's conjecture to analytic number theory if really there is a connection between them?
Snevily's conjecture it is an open conjecture in Group theory for non cyclic Group and it were proved for abelian groups of prime order using a fairly standard application of the Alon-Tarsi polynomial technique .It states that:
Snevily's conjecture:
let $G$ be an abelian Group of odd order and let $A,B \subseteq G$ satisfy $|A|=|B|=k$ .Then the elements of $A$ and $B$ may be ordered $A=\{a_1\cdots a_k\}$ and $B=\{b_1\cdots b_k\}$ so that the sums $a_1+b_1,a_2+b_2,\cdots a_k+b_k$ are pairwise distinct .
its seems that conjecture has connection to analytic number theory such that , This conjecture were proved for $\mathbb{Z_n}$ a subgroup of the multiplicative group of the field of order $2^{\phi(n)}$ with $\phi(n)$ is the Euler totiont function , Now my question here is : What are the consequence of Snevily's conjecture to analytic number theory if really there is a connection between them?
Addedendum :I have added this paper entitled "Divisors of the number of Latin rectangles" just to show the connection between latine square which it is the source of the titled conjecture and number divisors (number theory) .This means probably we will have some consequence of the titled conjecture to analytic number theory in the futur.
Can you make this more precise? Interesting isn't a well-defined term, and this may be too opinion based for it to attract good answers.
Is $k$ related to $G$ in some way?
Snevily's conjecture was proved in 2009 by Bodan Arsovski. He was a high-school student at the time.
I don't know of any consequences in analytic number theory. On the other hand, there are several papers before and after Arsovski's proof that connect Snevily's conjecture to other (combinatorial) group theoretic questions or discuss variations.
However my question would be close soon but I up vote to this answer because it will encourage me to do more research about connection between that conjecture to analytic number theory ,When I checked the bounds of number of latice square which it is very hard to be computable I think we can write that bound using arithmitic functions
|
2025-03-21T14:48:30.304577
| 2020-04-15T02:12:31 |
357510
|
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}
|
Stack Exchange
|
Functoriality of Hochschild cohomology for Drinfeld quotients
Let $C$ be a dg category and $C \to D$ a Drinfeld localization. Is there an induced pushforward map on $\operatorname{HH}^*(C) \to \operatorname{HH}^*(D)$, where $\operatorname{HH}^*$ denotes the Hochschild cohomology?
I think so, by combining Keller's limited functoriality for functors which are fully faithful on the level of derived categories (\S4 of https://webusers.imj-prg.fr/~bernhard.keller/publ/dih.pdf), and Drinfeld's result that the derived category of $D$ sits in the derived category of $C$ (proposition 4.6(ii) from his dg quotients paper). Is this what you are looking for? I was a bit surprised to see it covariantly, but this is what you get apparently, because one uses the restriction functor to get the fully faithful functor on the derived level.
|
2025-03-21T14:48:30.304672
| 2020-04-15T04:18:24 |
357515
|
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"url": "https://mathoverflow.net/questions/357515"
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|
Stack Exchange
|
Is a domain biholomorphic to the unit ball a Runge domain?
Let $\Omega \subset \mathbb C^n$ be a bounded domain which is biholomorphic to the unit ball $B^n=\{|z|<1 \mid z\in \mathbb C^n\}$. Can we show $\Omega$ must be a Runge domain? By definition, $\Omega$ is a Runge domain if any analytic function in $\Omega$ can be approximated by polynomials.
Notice that J. Wermer gave a counterexample if $\Omega$ is biholomorphic to a bidisk.
The answer to your question is NO. Every domain in $\mathbb{C}^n$ can be embedded into $\mathbb{C}^n$ such that its image is non-Runge.
See:
Wold, E. F.: A Fatou–Bieberbach domain in $\mathbb{C}^2$ which is not Runge.* Math. Ann. 340 (2008) 775–780
Dear @LuKa Thaler, what you claimed seems to be false when $n=1$. For example, for every embedding of the open unit disk into $\mathbb{C}$, the image is simply connected hence Runge in $\mathbb{C}$.
|
2025-03-21T14:48:30.304754
| 2020-04-15T05:33:19 |
357519
|
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|
Stack Exchange
|
Conformal welding of rectifiable curves
In classical conformal welding theory, we start with a homeomorphism $h$ of the unit circle and try to find a Jordan domain $D$ together with two conformal isomorphisms $f_1 \colon \mathbb D \to D$ and $f_2 \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \overline{D}$.
such that $h|_{S^1} = f_1^{-1} \circ f_2$. The existence of these two maps are guaranteed if $h$ is a quasisymmetric homeomorphism, and the uniqueness follows from uniqueness of complex dilatation.
My question is: Let $h$ be a homeomorphism of the unit circle, and let $J(z) = z+1/z$. Under what conditions can we find a rectifiable curve $\Gamma$ of length $4$ together with a conformal isomorphism $f \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \Gamma$ such that $f|_{S^1} = \gamma \circ J \circ h$, where $\gamma \colon [-2,2] \to \mathbb C$ is an arc-length parameterization of $\Gamma$? What about uniqueness?
I would really appreciate any reference of previous work in this direction, as well as ideas and suggestions.
Some motivations for this question:
Right now, I am trying to classify all smooth curves (and eventually rectifiable curves or Jordan arcs) $\Gamma$ with a certain property that reduces to
$$\gamma'(t) = \Phi_1^{-1}(\gamma(t)) \Phi_2^{-1}(\gamma(t)).$$
To explain the symbols, $\gamma$ is an arc-length parameterization of $\Gamma$, and $\Phi_1^{-1}(z), \Phi_2^{-1}(z)$ represent the two pre-images of $z \in \Gamma$ under the conformal isomorphism $\Phi \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \Gamma$. These pre-images are distinct if $z$ is not one of the endpoints of $\Gamma$.
I know curves of constant curvature satisfy this property, and I have some reasons to believe that they are the only ones. I would hope the "uniqueness" of conformal welding of curves, in some appropriate sense, would help me with the classification.
You stated the welding theorem incorrectly. The conformal isomorphisms are from the disk to SOME Jordan region and from the complement of the disk to the complement of this Jordan region.(Condormal isomorphisms of a disk onto itself are trivial: they are fractional-linear).
@AlexandreEremenko That was a careless mistake on me. Already corrected!
Could you add some motivation as to where your question comes from? Just curious...
@MalikYounsi Happy to!
|
2025-03-21T14:48:30.304938
| 2020-04-15T08:49:04 |
357523
|
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|
Stack Exchange
|
Existence of a quasi-isometric residually finite group?
It's, by now, more or less well known that residual finiteness is not a quasi-isometry invariant for finitely generated groups (see here for an example). Thus the following question makes sense:
Question: given a finitely generated group $G$, is there a finitely generated group $H$ such that $H$ is residually finite and $G$ and $H$ are quasi-isometric? That is, does the class of groups quasi-isometric to $G$ always contain residually finite groups?
Acknowledging this question is fairly open-ended, I would be glad with any reference in this direction. Thanks in advance. The following tries to motivate the above via some C*-algebraic questions.
Motivation (and guess): Should the above be known true, then one would be able to manually build so-called quasi-diagonalizing projections for the reduced group C*-algebra of $G$, that is, finite rank projections $p_n \in \mathcal{B}(\ell^2 G)$, with $p_n \rightarrow 1$ in the strong operator topology and
$$ ||p_n \lambda_g - \lambda_g p_n|| \rightarrow 0 $$
for every $g \in G$, where $\lambda$ is the left regular representation of $G$. Indeed, note that if $H$ is QI to $G$ and res. finite and amenable, then, by work of Orfanos there are projections $q_n$ for $H$ as above, and those can be pullback-ed to projections $p_n$ for $G$. Thus, since this C*-question is still open, my guess is that the original question is open as well.
The answer is certainly no, and the question is whether a counterexample (a f.g. group not QI to any RF f.g. group) is already known. The limit is that QI rigidity is known for quite few groups. Possibly Kac-Moody groups (these are simple finitely presented CAT(0) groups) or non-RF Baumslag-Solitar groups would be workable candidates. Also for most non-RF groups I can imagine, there's no natural candidate for a RF group QI to it.
@MarkSapir I googled on White's paper before posting my first comment. Could you provide a reference? I'm aware of an unpublished paper of Whyte, which is known to contain mistakes in its claim on QI-classification (I don't know whether it affects your assertion).
PS it's Whyte, not White, we both misspelled... it's Kevin Wkyte, Coarse bundles arXiv link. Nevertheless I can find in this paper a statement describing arbitrary groups quasi-isometric to $\mathrm{BS}(2,3)$, so I'd be happy to hear about a reference, if any.
@YCor I think you're looking for [Whyte, K.
The large scale geometry of the higher Baumslag-Solitar groups.
Geom. Funct. Anal. 11 (2001), no. 6, 1327–1343.]. Theorem 5.1 in there is the following: a f.g. group $\Gamma$ is qi to $BS(2,3)$ iff it is a graph of virtual $\mathbb{Z}$s which is neither commensurable to $F_n \times \mathbb{Z}$ nor virtually solvable.
Take any finitely-presented group $G$ with undecidable word problem. Then $G$ is not quasi-isometric to any finitely generated group with decidable word problem, in particular, to any residually-finite group. (Note that finite presentability and decidability of the WP are quasi-isometry invariant. The latter is because quasi-isometries preserve the equivalence class of the Dehn function and WP is for a finitely-presented group decidable iff the Dehn function is recursive.)
Since it's implicit, you're using that finitely presented residually finite groups have solvable word problem, and that among finitely presented groups, to have solvable word problem is QI-invariant. Nice answer!
|
2025-03-21T14:48:30.305185
| 2020-04-15T09:20:23 |
357526
|
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|
Stack Exchange
|
Extend fibre bundle
Let $F\rightarrow E\rightarrow B$ be a smooth fibre bundle. Suppose $W$ is a smooth manifold such that $F=\partial W$.
When is it possible to extend the bundle to a bundle over $B$ with fibre $W$?
Do you mean the extension to be a bundle over $B$, or can it be over some manifold containing $B$?
@MarkGrant thanks for the comment, I edited my question.
Perhaps again I am saying things you know. In the universal case, you are asking for the existence of a section of $B\text{Diff}(W) \to B\text{Diff}(F)$. This will not have an answer in full generality, so let me specialize to $W = D^n$ so something can be said. When $n \leq 3$, a section exists (and the space thereof is contractible) by the Smale conjecture. For $n$ large and even, the calculations here by Waldhausen and Farrell-Hsiang give...
that $\pi_4 B\text{Diff}(D^n, \partial) \to \pi_4 B\text{Diff}(D^n)$ cannot be injective, which contradicts the existence of a section. At a glance $n \geq 16$ should suffice, but I was not careful in comparing to the FH result. (I would be surprised to see a section of the above map for any $D^n, n \geq 4$ or $5$ or so.)
@MikeMiller Thanks a lot, I was not aware of that! I'll have a look at it and refine my question if needed.
|
2025-03-21T14:48:30.305422
| 2020-04-15T11:03:58 |
357530
|
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|
Stack Exchange
|
Action on cohomology by automorphisms of ihs manifolds
For all known deformation types of irreducible holomorphic symplectic manifolds (which I call K3, K3$^n$, Kum$^n$, OG$^3$, OG$^5$, the exponent being half the complex dimension), it is known that the map ${\rm Aut}(X) \rightarrow {\rm Aut}(H^*(X,\mathbb Z))$ is injective.
For K3, K3$^n$ and OG$^5$ this follows from the fact that the map ${\rm Aut}(X) \rightarrow {\rm Aut}(H^2(X,\mathbb Z))$ is injective (Beauville, Mongardi--Rapagnetta). For Kum$^n$, this is a theorem by Oguiso. For OG$^3$, this is known to experts.
Is this known in general?
|
2025-03-21T14:48:30.305497
| 2020-04-15T11:24:26 |
357532
|
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|
Stack Exchange
|
Stone–Weierstrass theorem for stronger topologies
The Stone–Weierstrass theorem gives an easy to check criterion on a (algebra) set of functions $D\subseteq C(X)$ which ensures that $D$ is dense.
Are any similar results for density in $C_b(X)$ when it is equipped with the topology of uniform convergence? Here, I'm now assuming that $X$ is a Euclidean space.
Edit: Due to the comments this type of result seems unlikely. However, I'm tempted to ask what are such criteria on $C^1(X)$ or $C^{\infty}(X)$; assuming that $X$ is a compact sub-manifold with boundary of $\mathbb{R}^K$, for some finite $K$. Here we take the completion of the metric topology
$$
d(f,g) :=
\sum_{n =0}^k \frac1{2^{n+A(n)}} \sum_{|\alpha|\leq n}\frac{
\sup_{x \in X} \frac{\partial}{\partial x^{\alpha}} \|f(x)-g(x)\|
}{
1+
\sup_{x \in X} \frac{\partial}{\partial x^{\alpha}} \|f(x)-g(x)\|
}
$$
as described by this nice post.
Do you mean $C_b(X)$ if $X$ is not compact?
Many naive generalizations fail because $C_b(\mathbf{R}^n)$ for $n\ge 1$ is not separable.
Yes, I've encountered this problem while trying. But I immagine there must be a body of literature on the subject
I'm no specialist but I doubt if there are easily practicable criteria. Let $\beta X$ be the Stone-Cech compactification of $X$. Then $C_b(X)$ and $C(\beta X)$ are almost indistinguishable. If you want to apply Weierstrass-Stone you have to work with $H \subset C(\beta X)$, which separates points of $\beta X$ (not of $X$ alone). And here is the problem. Are there "natural" candidates for $H$? We have similar problems even with $\ell_\infty$.
If we instead look in $C^{\infty}$ or $C^1$ with their usual topologies.... I guess we weouldnt have this kind of issue...
Sorry, I know Frechet topology in the context of locally convex spaces. But what is "usual Frechet topology" on a manifold?
There is a natural lc topology on $C^b(X)$ for which the Stone-Weierstraß theorem holds—the so-called strict topology. This was introduced by R.C. Buch in the 50’s for the special case of locally compact spaces and has several natural definitions. For example, it is the finest lc topology which coincides with that of compact convergence on the unit ball for the supremum norm. Another natural property is that the dual space is naturally identifiable with the space of bounded, radon measures. It can also be defined using weighted seminorms (as was done by Buck).
This is the inductive limit topology no (the one you get by looking at $C(K_i)$ with the compact-convergence topology and $K_i \subset K_{i+1}$ and taking the inductive limit in LCSs) no?
Potentially helpful: http://people.math.harvard.edu/~kupers/teaching/272x/book.pdf (Chapter 3)
@AnnieTheKatsu If you are referring to my comment, the answer is indeed “no”. The dual of your space consists of all radon measures, i.e., also unbounded ones.
|
2025-03-21T14:48:30.305720
| 2020-04-15T12:26:34 |
357536
|
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|
Stack Exchange
|
"sparsifying" a binary (over the field F2) matrix
Assume I have a matrix $A \in GF(2)$, i.e., $A_{i,j} \in \{0, 1\}$ and the sum is modulo 2. Is there any known algorithms/methods to sparsify (reduce the number of non-zero entries) $A$ while keeping its row space the same? A simple way, for example, will be to add every two rows and if the resulting sum is of lower weight than one of the two rows we replace them.
Is there any known good/better method?
Thanks
Isn't this what row reduction, aka Gaussian elimination, is for?
It won't necessarily result in a sparse matrix. For example the matrix A = [[1 0 1 1], [0 0 1]] is in reduced form, but the matrix B = [[1 0 1 0],[0 0 1]] spans the same space and is more sparse
I think we got cut off. But is there a guarantee that there is a way to sparsify every matrix?
That's a good question. I haven't found anything concrete online. Playing a bit with some examples it seems like in most cases there is a sparse matrix
Your example, three comments up, I don't get – a matrix with four entries in the first row, and three in the second row?
sorry, A = [[1 0 1 1], [0 0 0 1]] and B = [[1 0 1 0],[0 0 0 1]]
But $A$ isn't in reduced row-echelon form.
why not? a matrix is in row echelon form if
all nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix), and
the leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it. Am I missing something? every [] is a row in $A$
That's row echelon form. Reduced row echelon form has one more requirement; the pivot is the only nonzero entry in its column.
Ok, it is still easy to find counter examples: A = [[1 0 1 1], [0 1 1 1]] B = [[1 1 0 0] ,[0 1 1 1]]
This is equivalent to finding a minimum basis of a matroid. Hence you could indeed use a greedy algorithm, like you describe, and I'd expect you cannot do much better.
We're not far from discussing error-correcting codes here. E.g., the generating matrix for the simplest Hamming code, $$\pmatrix{1&0&0&1&1&0&1\cr0&1&0&1&0&1&1\cr0&0&1&0&1&1&1\cr}$$ All the nonzero vectors in the row space have weight $4$, so no further sparsification can occur.
|
2025-03-21T14:48:30.306166
| 2020-04-15T12:43:01 |
357539
|
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|
Stack Exchange
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Unique representation and sumsets
Let $A$ be a finite, nonempty subset of an abelian group, and let $2A:=\{a+b\colon a,b\in A\}$ and $A-A:=\{a-b\colon a,b\in A\}$ denote the sumset and the difference set of $A$, respectively.
If every non-zero element of $A-A$ has a unique representation as $a-b$ with $a,b\in A$, then all sums $a+b$ are pairwise distinct; as a result, $A$ is a Sidon set and $|2A|=\frac12|A|(|A|+1)$. Suppose now that only, say, $k$ elements of $A-A$ are known to be uniquely representable; how large must $|2A|$ be in this case? I am specifically interested in the situation where $k=|A|+1$.
Another way to cast the problem is as follows. If there is a group element with a unique representation in $A-A$, then $|2A|\ge 2|A|-1$. How large must $|2A|$ be given that $A-A$ has at least $|A|+1$ uniquely representable elements?
Ah, a much better example: take $A = {1 , \ldots , n} e_1 \cup e_2 \cup e_3 \subset \mathbb{Z}^3$. This shows $|2A|$ can be as small as $4|A| - O(1)$.
@GeorgeShakan: Thanks, good examples - but still fall a little short of what I need: can $|2A|$ be smaller than, say, $2.25|A|$?
In the integers, can you use Freiman's $3k-3$ theorem to rule this out? I didn't check.
@GeorgeShakan: yes, this seems to be true for the integers. Suppose that
${0,l}\subset A\subset[0,l]$, and write $|A|=n$ and $|2A|=Cn$. Ignoring the
$O(1)$-terms, since $|2A|<3|A|$, we have $l<(C-1)n$. If $g<l/2$ is uniquely
representable, then $g\le|[1,2g]\setminus A|\le l-n$. Thus, the interval
$[l-n,l/2]$ does not contain any uniquely representable integers, and
similarly for the interval $[l/2,n]$. Hence, there are at most
$2(l-n)<2(C-2)n$ such integers, and this is less than $n$ provided $C<2.5$.
Ah one can improve the $4|A|$ to $3|A|$ in the comment above by working over $\mathbb{F}_2^n$ and replacing the arithmetic progression with a subgroup. It is easy to show that the hard case is $|A\cap A+(s-s')| \geq (1-.25)|A|$ for all $s,s'\in U$ where $U$ is the set of elements with unique representation.
@GeorgeShakan: I wonder whether the concentration of measure can be used: if $a_i-b_i$ are uniquely representable, then $(a_i+A)\cup(b_i+A)$ are large subsets of $2A$. If there are many such subsets, then there must be two of them with a large intersection showing, basically, that $A$ is close to a periodic set.
One can have $|2A|$ as small as $2|A|$. Take $A = H \cup \{g\}$ where $H$ is a subgroup, $g \notin H$ and$g \neq -g$. Then $|A+A| = 2|A| + O(1)$ while
$g+H$ and $H - g$ all have a unique representative in $A-A$.
On the other hand, I can show if the number of uniquely representable elements of $A-A$ is at least $|A|$ and $|A+A| \leq (7/3)|A|$, then there is a subgroup, $H$, of size at most $(3/2)|A|$ such that the unique representatives lie in a coset of $H$. I'll adopt notation from Tao and Vu (mostly Chapter 2). Let
$$U : = \{x \in G : r_{A-A}(x) =1\}.$$
Let $g, h\in U$. By the Bonferroni inequalities, we have
$$ |A+A| \geq |A| + |A+g| + |A+h| - |A\cap(A+g)| - |A\cap(A+h)| - |(A+g) \cap (A+h)|.$$
As $g,h \in U$, we have $|A\cap(A+g)| = |A \cap (A+h)| = 1$ and so
$$|A+A| \geq 3|A| - 2 - r_{A-A}(g-h).$$
Thus if $r_{A-A}(g-h) \leq (1-\epsilon)|A|$
we have
$$|A+A| \geq (2+\epsilon)|A| - 2.$$
So we suppose for all $g,h\in U$,
$$\tag{1}\label{1} r_{A-A}(g-h) \geq (1-\epsilon)|A|.$$
Note that \eqref{1} implies that
$$U-U \subset {\rm Sym}_{1-\epsilon}(A).$$
Markov implies
$$|{\rm Sym}_{1-\epsilon}(A)| \leq \frac{|A|}{1-\epsilon},$$
and so by assumption
$$|U-U| \leq \frac{|A|}{1-\epsilon} \leq \frac{|U|}{1-\epsilon}.$$
Suppose now that $(1-\epsilon)^{-1} \leq 3/2$ (i.e. $\epsilon \leq 1/3$). Then by baby Freiman (see Theorem 1.5.2), we have that
$$U \subset H + t,$$
for some $H \leq G$, with $|H| \leq (3/2)|A|$ and $t \in G$.
Great, thank you! (And please, fix the reference...)
Upon a second look, I am still a little uncertain. Applying Bonferroni, you seem to assume that $A,A+g$ and $A+f$ are subsets of $2A$ - which, in general, is not the case. It is my understanding that in fact, you find $f$ and $g$ so as to have $f=a-c$ and $g=b-c$ with some $a,b,c\in A$, and then consider $(A+a)\cup(A+b)\cup(A+c)$. However, in this case it is not true that $r(g-h)$ is large for any $f$ and $g$ (but only for $f$ and $g$ which can be represented as above). Could you explain?
hmm, seems you are right
Still, there is an interesting property which seems to follow this way: namely, the unique representation graph is triangle-free. Incidentally, in your example the URG is a star.
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2025-03-21T14:48:30.306475
| 2020-04-15T13:41:31 |
357544
|
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Stack Exchange
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Are polynomial sections for unramified principal series generated by the spherical vector?
Let $F$ be a non-archimedean local field, $G = \operatorname{GL}_n\left(F\right)$, $B$ the standard Borel subgroup, $K = \operatorname{GL}_n\left( \mathcal{O}_F \right)$. We have the Iwasawa decomposition $G = BK$. Let $\pi$ be an unramified principal series $\left( \pi, I\left({\underline{s}}\right) \right) = \operatorname{Ind}_{B}^G\left({\left| \cdot \right|^{s_1}} \boxtimes \dots \boxtimes \left|\cdot\right|^{s_n} \right)$, where $s_1, \dots, s_n \in \mathbb{C}$ and $\underline{s} = \left( s_1, \dots, s_n \right)$. Every element $f \in I\left({\underline{s}}\right)$ is a (right smooth) function $f : G \rightarrow \mathbb{C}$ satisfying $f\left(b g\right) = \delta_B\left(b\right)^{\frac{1}{2}} \chi_{\underline{s}}\left(b\right) f\left( g \right)$, for every $g \in G$ and $b \in B$, where $\chi_{\underline{s}} : B \rightarrow \mathbb{C}^{\times}$ is the character we used for induction.
Thinking of $\underline{s}$ as variables, we have the notion of standard sections (or flat sections): these are functions $f_{\underline{s}} : G \rightarrow \mathbb{C}$ such that $f_{\underline{s}}\left( bk \right) = \delta_B\left(b\right)^{\frac{1}{2}} \chi_{\underline{s}}\left(b\right) h \left( k \right)$, for every $b \in B$, $k \in K$, where $h : K \rightarrow \mathbb{C}$ is a smooth function that does not depend on $\underline{s}$. In particular, we have the spherical standard section $f_{\underline{s}}^{\circ}\left( bk \right) = \delta_B\left(b\right)^{\frac{1}{2}} \chi_{\underline{s}}\left(b\right)$.
We know that if $q^{s_i} \ne q^{s_j + 1}$ for any $i, j$, then $\pi$ is irreducible. Therefore if we fix such $\underline{s}$, we can write every element of $I \left( \underline{s} \right)$ as a $\mathbb{C}$-linear combination of right translations of $f_{\underline{s}}^{\circ}$.
Question: is it possible to write every standard section $f_{\underline{s}}$ as a $\mathbb{C}\left[ q^{\pm s_1}, \dots, q^{\pm s_n} \right]$-linear combination of right translations of the standard spherical section $f_{\underline{s}}^{\circ}$? If so, is this result also true for other classical $p$-adic groups?
This can't be true, because for some choices of $\underline{s}$, the spherical vector does not generate $I\left( \underline{s} \right)$ (see Paul Garrett - Representations with Iwahori-fixed vectors, page 10). However, we can write $f^{\circ}_{\underline{s}}$ as a $\mathbb{C}\left( q^{-s_1}, \dots, q^{-s_n} \right)$ linear combination of right translations of the normalized spherical standard section. The proof I give is inspired by G. Mui´c, A geometric construction of intertwining operators for reductive p–adic groups, Lemma 5-12.
To see this, let $G = \operatorname{GL}_n\left( F \right)$. Let $U \le K$ be an open compact subgroup. Then $I \left( \underline{s} \right)^U$, the subspace of $U$-fixed vectors, is finite dimensional, for every choice of $\underline{s}$. Let $s_0$ be such that $I \left( \underline{s_0} \right)$ is irreducible. Let $h^1, \dots, h^r$ be a basis for $I \left( \underline{s_0} \right)^U$. This is equivalent to saying that the restrictions of $h^1, \dots, h^r$ to $K$ form a basis for ${C}^{\infty}\left( K / U \right)$ (the space of smooth functions on $K$ which are invariant to $U$ right translations).
We have the standard sections ${h'}_\underline{s}^1, \dots, {h'}_\underline{s}^r \in I\left( \underline{s} \right)^U$, defined by $${h'}_\underline{s}^j \left( b k \right) = \chi_{\underline{s}} \left( b \right) h^j \left( k \right), $$ for $b \in B$ and $k \in K$. These form a basis for $I \left( \underline{s} \right)^U$. Moreover, any standard section which is $U$ invariant, is a $\mathbb{C}$ linear combination of ${h'}_\underline{s}^1, \dots, {h'}_\underline{s}^r$ (as its value is determined by its restriction to $K$, and the restriction to $K$ of these functions is a basis independent of $\underline{s}$).
Since $I \left( \underline{s_0} \right)$ is irreducible, there exist $\left( \varphi_j \right)_{j=1}^r \subseteq C^{\infty}_c\left( G \right)$ (i.e. compactly supported smooth functions on $G$), such that
$$ h^j \left( g \right) = \left(\pi_{\underline{s_0}}\left( \varphi_j \right) f^{\circ}_{\underline{s_0}}\right)\left(g\right) = \int_G \varphi_j\left(g_0 \right) f^{\circ}_{\underline{s_0}} \left( g g_0 \right) dg_0.$$
Since $h^j$ is right $U$ invariant, we may assume that $\varphi_j \in C^{\infty}_c\left(
U \backslash G \right)$, i.e., that $\varphi_j$ is left $U$ invariant for every $j$.
Define $h^j_{\underline{s}} \in I\left( \underline{s} \right)$ by
$$ h^j_{\underline{s}} \left( g \right) = \left(\pi_{\underline{s}}\left( \varphi_j \right) f^{\circ}_{\underline{s}}\right)\left(g\right) = \int_G \varphi_j\left(g_0 \right) f^{\circ}_{\underline{s}} \left( g g_0 \right) dg_0.$$
$h_{\underline{s}}^j$ lies in the $\mathbb{C}$ span of right translations of $f^{\circ}_{\underline{s}}$: cover the support of $\varphi_j$ with the open covering $\bigcup_{g_0 \in G} {g_0 K}$. Then, since the support of $\varphi_j$ is compact, it admits a finite subcover $\operatorname{supp} \varphi_j \subseteq \bigcup_{i=1}^N g_i K$, and we can write $$h_{\underline{s}}^j \left( g \right) = \sum_{i=1}^N \varphi_j \left( g_i \right) f_{\underline{s}}^{\circ}\left( g g_i \right).$$ Since $\varphi_j$ is $U$ left invariant, this implies that $h_{\underline{s}}^j$ is $U$ right invariant. Therefore the restriction of $h_{\underline{s}}^j$ to $K$ lies in $C^{\infty}\left( K / U \right)$. Since the restrictions $h^1 \restriction_K, \dots, h^r \restriction_K $ form a basis for $C^{\infty}\left( K / U \right)$, there exist $\left( \kappa_{ij}\left( \underline{s} \right) \right) \subseteq \mathbb{C}$, such that
$$ h^j_{\underline{s}} \restriction_K = \sum_{i=1}^r \kappa_{ij}\left( \underline{s} \right)h^i \restriction_K.$$
By the Iwasawa decomposition, we have that for every fixed $g \in G$, the value $f_{\underline{s}}^{\circ}\left( g \right)$ lies in $\mathbb{C} \left[ q^{\pm s_1}, \dots, q^{\pm s_r} \right]$. Since $h^j_{\underline{s}}$ is a $\mathbb{C}$-linear combination of right translations of $f_{\underline{s}}^{\circ}$, we have that for every $k \in K$, the value $h^j_{\underline{s}} \left( k \right)$ lies in $\mathbb{C} \left[ q^{\pm s_1}, \dots, q^{\pm s_r} \right]$. It now follows by inverting a matrix that $\kappa_{ij}\left( \underline{s} \right) \in \mathbb{C} \left[ q^{\pm s_1}, \dots, q^{\pm s_r} \right]$. We get by inverting the matrix of $\left( \kappa_{ij} \left( \underline{s} \right) \right) $ that the restrictions $h^j \restriction_K$ are in the $\mathbb{C}\left( q^{-s_1}, \dots, q^{-s_n} \right)$ span of $h^1_{\underline{s}} \restriction_K,\dots,h^r_{\underline{s}} \restriction_K$. Note that this matrix is invertible for $\underline{s}$ such that the right translations of $f_{\underline{s}}^{\circ}$ span $I \left( \underline{s} \right)$.
This implies that ${h'}_\underline{s}^1, \dots, {h'}_\underline{s}^r$ are in the $\mathbb{C}\left( q^{-s_1}, \dots, q^{-s_n} \right)$ span of $h_{\underline{s}}^1, \dots, h_{\underline{s}}^r$, which is contained in the $\mathbb{C}\left( q^{-s_1}, \dots, q^{-s_n} \right)$ span of $h_{\underline{s}}^1, \dots, h_{\underline{s}}^r$, which is contained in the $\mathbb{C}\left( q^{-s_1}, \dots, q^{-s_n} \right)$ span of right translations of $f_{\underline{s}}^{\circ}$. Therefore we have that all standard sections that are $U$ invariant lie in the space of right translations of $f_{\underline{s}}^{\circ}$. Since this is true for every compact open $U \le K$, we are done.
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