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2025-03-21T14:48:29.884209
| 2020-02-21T08:11:35 |
353224
|
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"Hua Wang",
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|
Stack Exchange
|
$\exists k$ s.t. $k^m\le 1^m+2^m+...+(k-1)^m <2\cdot k^m$?
Can it be shown that,for all $m\in\mathbb{Z}_+$ there exists at least one $k$ with respect to $m$ such that
$$1\le \frac{\sum_{i=1}^{k-1}i^m}{k^m}<2$$
Example: let $m=1$ then $k=\{3,4\}$
This question has already been asked on MSE but it has not received a solution so that's why I posted here. Given claim solve my first claim from this post "Observation on Erdős–Moser equation".
Post on MSC
Does Faulhaber's formula help? At least, it seems to imply that such $k$ exists (and is between $m+1$ and $2(m+1)$) once $m$ is large enough.
Yes. Note that for large $k$ we have $1^m+\dots+(k-1)^m\geqslant k^m$. Choose minimal $k$ with such property. Then $k>2$ and $$2k^m>2(k-1)^m>1^m+\dots+(k-1)^m$$
by minimality.
|
2025-03-21T14:48:29.884293
| 2020-02-21T10:59:13 |
353230
|
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}
|
Stack Exchange
|
Spectral abscissa of symmetric matrix with skew-symmetric perturbation
I am interested in bounds on the minimal distance between the spectral abscissa $\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ of a matrix $A$ and the eigenvalues of its perturbated version $A+S$. In my case, the matrix $A$ is symmetric and the perturbation $S$ is skew-symmetric. The matrix has eigenvalues with algebraic multiplicity higher than 1 (this is an unnice property that excludes some results). The matrix is a $N$-block matrix with $n\times n$ blocks and the perturbation is block diagonal with $n\times n$ blocks.
A known result in the literature states:
If $A$ and $B$ are normal with $\|A-B\|=\varepsilon$ in a unitarily invariant norm $\|\cdot\|$ (e.g. the Frobenius norm), then for any eigenvalue $\lambda$ of $A$ there is an eigenvalue $\mu$ of $B$ such that $|\lambda-\mu|\leq\varepsilon$.
Let $A$ be symmetric and $\nu=\arg\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ (eigenvalue with spectral abscissa of $A$). Let $B=A+S$ where $S$ is skew-symmetric. By the above result, there exists an eigenvalue $\mu$ of $A+S$ such that
$$
|\mathrm{Re}\nu-\mathrm{Re}\mu|\leq|\nu-\mu|\leq\|S\|.
$$
I wonder if there is some results that utilises that one matrix is symmetric and the other skew-symmetric and not just normality? I've found many results about symmetric (Hermitian) matrices, but not the symmetric and skew-symmetric combination.
My intuition says that the skew-symmetric matrix should contribute to the imaginary part of the eigenvalue at first. Therefore, the minimum distance between the spectral abscissa of $A$ and the spectrum of $A+S$ is reduced compared to the case when only normality holds.
Edit: Another result states that for a matrix $C(z)$, continuously differentiable around $z=0$, with eigenvalue $\xi$ and left, right eigenvectors $u,v$, it holds for $C(z)v(z)=\xi(z)v(z)$ that
$$
\xi^\prime(0)=\frac{u^*C^\prime(0)v}{u^*v}
$$
If we let $C(z)=A+Sz$, then $C^\prime(0)=S$, $u=v$ whereby $\xi^\prime(0)=0$. This supports the idea that the change in the eigenvalues of $C$ is smaller for a symmetric matrix with skew-symmetric perturbation than it is in the general case.
The result for normal matrices is misstated---take diagonal matrices. The best I can see is that for any eigenvalue $\lambda$ of $A$, there exists an eigenvalue $\mu$ of $B$ such that $|\lambda - \mu| \leq \epsilon$. If $A$ and $B$ are symmetric, then the eigenvalues are real and can be ordered, and a stronger result is available.
You are right. Sorry about that, I corrected the formulation.
Welcome to MathOverflow! I don't quite see what kind of result you are looking for. By choosing $A=0$ and by choosing a skew-symmetric matrix $S$ that has only $i|S|$ and $-i|S|$ as eigenvalues, you can see that the result that you mentioned (below the paragraph that is written italic) is optimal.
|
2025-03-21T14:48:29.884489
| 2020-02-21T11:20:52 |
353231
|
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"Fernando Muro",
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"nikola karabatic"
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}
|
Stack Exchange
|
Limit of split short exact sequences
Let $X$ be a module over some ring which splits as $$X\cong M_1\oplus S_1\cong M_1\oplus M_2 \oplus S_2 \cong M_1\oplus M_2 \oplus M_3\oplus S_3\cong \ldots$$
where the isomorphisms come from splittings $S_i\cong M_{i+1}\oplus S_{i+1}$.
Let $S=\bigcap S_i$ and $M=\sum M_i$. There is a canonical injection $M\oplus S\rightarrow X$ which is compatible with the projections to $S_i$ and $M_i$ in all of the above splittings.
Question: Is this an isomorphism, i. e. is $X\cong M\oplus S$?
If the answer is "no", I would additionally be interested in some 'conceptual' way to 'measure' the failure of this being an isomorphism.
I guess you want more conditions, like the "canonical" injection being compatible with the projections onto the $M_i$ etc. With your current conditions it's easy to produce counterexamples.
yes that's what I meant by canonical, I will add this to the question, thanks
No. Take $X $ to be the direct product of nonzero modules $M_i $ indexed by the positive integers, and $S_i $ to be the direct product of all but the first $i $ of them. Then $S=0$ and $M $ is the direct sum of the $M_i $.
|
2025-03-21T14:48:29.884596
| 2020-02-21T11:54:56 |
353235
|
{
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"Ennio Mori cone",
"Piotr Achinger",
"TartagliaTriangle",
"Yosemite Stan",
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|
Stack Exchange
|
Nef divisors on abelian varieties are pullbacks of ample ones
It is well known that for any polarization ( that is, ample line bundle) $L$ on an abelian variety $A$, there is an isogeny $\phi\colon A \to B$ to another abelian variety with a principal polarization $M$ such that $\phi^*M=L$.
If instead I take a nef but not ample divisor $D$ on $A$, is it always true that $D$ is the pullback of some ample divisor on a quotient of $A$?
Thanks for help!
No, take a generic degree 0 divisor on an elliptic curve.
If you work in the complex setting I think the Appel-Humbert theorem should give you essentially what you want (https://en.wikipedia.org/wiki/Appell%E2%80%93Humbert_theorem). The only caveat, is that you might have to work "up to translation." Nefness I think corresponds to positive semidefiniteness of the Hermitian form H in the Appell-Humbert theorem. Then the kernel of this form should define a sublattice corresponding to a subcomplex torus. If you quotient by this subtorus, I think it should be straightforward to check that the Hermitian form "descends" and is ample on that quotient torus.
... A minor annoyance is that one might not be able to descend the $\alpha$ in the Appell-Humbert theorem (e.g. in @wnx's answer - indeed this is true for "most" strictly nef line bundles with a given $c_1$). However, if you only want up to topological equivalence, i.e. to say that there is a quotient $A\rightarrow B$ such that $D$ is the image of some ample divisor under $\mathrm{NS}(B)\rightarrow \mathrm{NS}(A)$ then you should be okay I think.
Thinking more algebraically, it seems that the quotient should be taken by (the connected component of) $T(L) = { x\in A ,:, T_x^* L \cong L}$. Indeed, the restriction of $L$ to $T(L)^\circ$ lies in its $\mathrm{Pic}^0$, so by twisting by an element of $\mathrm{Pic}^0(A)$ one should be able to ensure $L$ trivial on $T(L)^\circ$. Then it descends to the quotient and is ample there (I think finiteness of $T(L)$ will ensure this).
Thanks to YosemiteStan and PiotrAchinger for their useful comments. Can you write them as an answer, so I can accept them?
@PiotrAchinger, I don't understand how the divisor descends to quotient and why it should be ample. Also, why $T(L)$ has to be finite? $L$ is not supposed to be ample.
|
2025-03-21T14:48:29.884938
| 2020-02-21T14:32:22 |
353241
|
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"Jeremy Brazas",
"Moishe Kohan",
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|
Stack Exchange
|
Free product decompositions of the fundamental group of Hawaiian Earrings
This is a spin-off of my question here, separated from the older question following Jeremy's suggestion.
Definition. Call a group $G$ essentially freely indecomposable if in every free product decomposition $G\cong G_1\star G_2$, one of the free factors $G_1, G_2$ is free of finite rank.
Question. Let ${\mathbb H}$ denote the Hawaiian Earrings. Is the fundamental group $\pi_1({\mathbb H})$ essentially freely indecomposable?
The only relevant result I could find in the literature is the theorem (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $G=\pi_1({\mathbb H})$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall.
With reasonable conventions the trivial group is not freely indecomposable (as $1$ is not prime).
My original question (comment to the linked question) was slightly distinct: is $\pi_1(\mathbb{H})$ not decomposable as free product of two uncountable groups. You're asking something stronger (well, with weaker negative) but which sounds plausible to me anyway. I'd ask Samuel Corson if it's not answered here.
I asked Sam about this and he shared the answer with me. If I get time later, I'll share it.
@JeremyBrazas: Would be great.
Funny enough, I emailed Sam about this yesterday before this question was posted. Clearly, he is the right person to ask.
This answer is courtesy of Sam Corson who kindly pointed out the following.
Theorem: The Hawaiian earring group $\pi_1(\mathbb{H})$ is essentially freely indecomposable, i.e. if $\pi_1(\mathbb{H})\cong G_1\ast G_2$, then one of $G_1$ or $G_2$ must be a finitely generated free group.
The key is to apply a special case of Theorem 1.3 in
K. Eda, Atomic property of the fundamental groups of the Hawaiian earring and wild locally path-connected spaces, J Math. Soc. Japan 63 (2011), 769-787.
Let $C_n$ be the $n$-th circle of $\mathbb{H}$. Take $\mathbb{H}_{\geq n}=\bigcup_{k\geq n}C_k$ to be the smaller copies of the Hawaiian earring and $\mathbb{H}_{\leq n}=\bigcup_{k=1}^{n}C_k$ to be the union of the first $n$-circles. One of the defining properties of the Hawaiian earring group is that there is a canonical isomorphism $\pi_1(\mathbb{H})\cong \pi_1(\mathbb{H}_{\geq n+1})\ast \pi_1(\mathbb{H}_{\leq n})$ for all $n\in\mathbb{N}$.
Suppose that $\phi:\pi_1(\mathbb{H})\to G_1\ast G_2$ is an isomorphism. In the case of the Hawaiian earring, Eda's theorem cited above implies that for any homomorphism $\phi:\pi_1(\mathbb{H})\to G_1\ast G_2$, there exists an $n\in\mathbb{N}$, $i\in\{1,2\}$, and $w\in G_1\ast G_2$ such that $\phi(\pi_1(\mathbb{H}_{\geq n+1}))\leq w G_i w^{-1}$. Suppose, without loss of generality, that $i=1$. Now if $\gamma(g)=w^{-1}gw$ is conjugation in $G_1\ast G_2$, then $\gamma\circ \phi:\pi_1(\mathbb{H})\to G_1\ast G_2$ is an isomorphism mapping $\pi_1(\mathbb{H}_{\geq n+1})$ into $G_1$. Let $\psi:G_1\ast G_2\to G_2$ be the projection, which is surjective. Then $\psi\circ\gamma\circ\phi:\pi_1(\mathbb{H})\to G_2$ is a surjection and since $\pi_1(\mathbb{H})\cong \pi_1(\mathbb{H}_{\geq n+1})\ast \pi_1(\mathbb{H}_{\leq n})$ and $\psi\circ\gamma\circ\phi(\pi_1(\mathbb{H}_{\geq n+1}))=1$, we must have that $\psi\circ\gamma\circ\phi$ maps $\pi_1(\mathbb{H}_{\leq n})$ onto $G_2$. Therefore, $G_2$ is finitely generated. Moreover, since $G_2$ is isomorphic to a subgroup of the locally free group $\pi_1(\mathbb{H})$, it follows that $G_2$ is free.
Great! This is even stronger, it shows that every free product decomposition is 'geometric.'
|
2025-03-21T14:48:29.885173
| 2020-02-21T15:09:04 |
353245
|
{
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"Yemon Choi",
"efs",
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|
Stack Exchange
|
Left module which cannot be made into a bimodule?
Let $A$ be a noncommutative unital algebra, defined over $\mathbb{C}$ say. What is an example of a left $A$-module $M$ that does not admit a right $A$-module structure giving $M$ the structure of a bimodule?
Perhaps https://mathoverflow.net/questions/348736/left-right-non-bimodule-examples answers your question.
@EFinat-S No, because that question only asks for the existence of some right $A$-module structure that is incompatible with the given left $A$-module structure. This question is asking for something much stronger
Such examples are a plenty. You are asking about non-existence of an algebra map $A\rightarrow End_AM$. Take $A$ simple, at most countably dimensional, and a simple module ${}_{A}M$. Then $End_AM={\mathbb C}$. Bingo!
|
2025-03-21T14:48:29.885247
| 2020-02-21T15:15:13 |
353246
|
{
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"Sebastian Goette",
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|
Stack Exchange
|
Multilinear Morse functions on the n-torus
Consider the $n$-dimensional Torus $T^n = \prod_{i=1}^n S^1$ as a subset of $\mathbb R^{2n} = \prod_{i=1}^n \mathbb R^2$ in the standard way.
Is it true that a generic $n$-multilinear functional on $\mathbb R^2$, i.e. a tensor in $T_0^n(\mathbb R^2)$, defines a Morse function on $T^n$? If so, does a generic element have the minimal number of critical points $2^n$ (among Morse functions)?
Generic here refers to all but a Haar null set of exceptions (dense would also be fine). The statement is reasonable in my opinion since these multilinear functionals seem to be good "height functions" in this context and the statement is indeed true for $n=1,2$. For $n=1$ it is obviuously true for any nontrivial linear function on $\mathbb R^2$. For $n=2$ we can write any $\omega \in T_0^n(\mathbb R^2)$ as
$$
\omega = a\, e_1^* \otimes e_1^* + b\, e_1^* \otimes e_2^* + c\, e_2^* \otimes e_1^* + d\, e_2^* \otimes e_2^*
$$
with respect to the standard basis of $\mathbb R^2$. Define the function $f : \mathbb R \times \mathbb R \to \mathbb R$ by $f(s,t) = \omega(e^{is},e^{it})$ and solve for critical points. For generic $a,b,c,d$ it follows rather quickly (modulo calculation errors) using the addition formulas for sine and cosine
\begin{align*}
s & = \frac{1}{2}\arctan \frac{c+b}{a-d} - \frac{1}{2}\arctan \frac{b-c}{a+d} + m\pi\ , \\
t & = \frac{1}{2}\arctan \frac{c+b}{a-d} + \frac{1}{2}\arctan \frac{b-c}{a+d} + n\pi \ ,
\end{align*}
for $m,n \in \mathbb Z$. This has exactly four solutions $(s,t) \in (-\pi,\pi]^2$. Similarly one can check that for generic $a,b,c,d$, the Hessian matrix of $f$ at critical points $(s,t)$ is nondegenerate.
It is interesting to note that the statement can't hold for all $\omega \neq 0$ because if the coordinates of $\omega$ are for example given by $b = c = 0$ and $a = d = 1$, then the associated function $f(s,t) = \cos(s-t)$ has a continuum of critical points.
Preferably there is an argument that avoids calculations as above for $n > 2$ as this seems to be quite a laborious task, unless I miss something (which is likely). Maybe this question has been considered before or follows directly from more general principles available in the literature, but I wasn't able to find something. Any references in this direction are welcome.
Assuming your torus has the form $\mathbb R^n/\mathbb Z^n$, then multlinear maps with integer coefficients give well-defined maps from $T^n$ to $\mathbb R$. Are you referring to those? Then what does "generic" mean in this context?
I am refering to multilinear maps $\omega : \mathbb R^2 \times \cdots \times \mathbb R^2 \to \mathbb R$ and its restriction to the "standard embedding" $T^n = S^1 \times \cdots \times S^1 \subset \mathbb R^2 \times \cdots \times \mathbb R^2$. Generic is with respect to the Haar measure on the vector space of such multilinear functions.
Thanks for the clarification - I should have read the question more carefully.
But anyway - could one use the maps I referred to as well?
I think the functions on $\mathbb R^n$ that lead to the same class of functions on $T^n = \mathbb R^n /(2\pi \mathbb Z)$ are finite sums of functions of the form $f(x_1,\dots,x_n) = r\cos(x_1 + a_1)\cdots\cos(x_n + a_n)$ for some $r, a_1,\dots, a_n \in \mathbb R$. Don't know if these functions have something to do with those you are suggesting.
|
2025-03-21T14:48:29.885462
| 2020-02-21T15:15:32 |
353247
|
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"David A. Craven",
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|
Stack Exchange
|
Maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic
For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For $E_7$, for example, it is $27$, and is the radical of the $E_6$ parabolic.
What about in characteristic $p$ for $p>0$? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by $29$ or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic $0$.
I don't need the exact bound, just something close will do. (I have a Lie algebra $L$ with an abelian subalgebra of dimension, say, $43$, and I want to know that $L$ cannot be embedded in $E_7$, and in particular is not $E_7$. But I'm in characteristic $7$, for example, or worse, $2$ or $3$.)
Let $\ell\ge0$ be the characteristic of the (algebraically closed) ground field. Let $G$ be semisimple with Lie algebra $\mathfrak g$.
First, the maximal dimension of an abelian subalgebra of $\mathfrak g$ can increase for small $\ell$. Let, e.g., $\mathfrak g=sl(2)$ and $\ell=2$. Then the Borel subalgebra is abelian, hence the maximal dimension is $2$ instead of $1$. The same holds for $\mathfrak g=pgl(2)$ and $\ell=2$ where $\langle e,f\rangle$ is abelian.
Secondly, this is a phenomenon of very small characteristics. The following statement is not the most general one but covers most cases:
Assume that $G$ is of adjoint type and $\ell>3$. Then the dimension of a maximal abelian subalgebra is the same as in characteristic $0$.
Proof: Key is the following deformation argument: Being an abelian subalgebra is a closed condition in the Grassmannian of $\mathfrak g$. In other words, there is a closed subscheme $A_d\subseteq Gr_d(\mathfrak g)$ classifying abelian subalgebras of dimension $d$. The group $G$ acts on $A_d$ by conjugation. Assume that $A_d\ne\emptyset$. Since $A_d$ is projective, any Borel subgroup $B$ of $G$ has a fixed point. This shows, that if $\mathfrak g$ contains an abelian subalgebra $\mathfrak a$ of dimension $d$ then it will also contain one which is normalized by $B$. Assume from now on that this is the case.
Then, in particular, $\mathfrak a$ will be normalized by a maximal torus $T\subseteq B$. Let $\mathfrak g=\mathfrak t\oplus\bigoplus_\alpha\mathfrak g_\alpha$ be the root space decomposition. Then any $T$-stable subspace, so also $\mathfrak a$, has the form
$$
\mathfrak a=\mathfrak a_0\oplus\bigoplus_{\alpha\in S}\mathfrak g_\alpha
$$
where $S$ is a set of roots.
Since $\mathfrak a$ is ablian, its subset of semisimple elements $\mathfrak a_0$ inside $\mathfrak b$ is also normalized by $B$. Hence $\mathfrak a_0$ consists of fixed points of $B$ and therefore of $G$. So $\mathfrak a_0$ sits in the schematic center of $\mathfrak g$. Because $G$ is of adjoint type we infer $\mathfrak a_0=0$.
Thus everything depends on $S$. I argue that the condition on $S$ making $\mathfrak a=\bigoplus_{\alpha\in S}\mathfrak g_\alpha$ abelian is independent of $\ell$, proving our assertion.
For this let $e_\alpha\in\mathfrak g_\alpha$ be a Chevalley generator. Then $[e_\alpha,e_\beta]$ must be zero for all $\alpha,\beta\in S$. If $\alpha+\beta=0$ then $h_\alpha=[e_\alpha,e_\beta]\ne0$ since $\ell\ne2$. So this case must not occur. If $\gamma=\alpha+\beta$ is a root then a famous formula of Chevalley asserts
$$
[e_\alpha,e_\beta]=\pm N_{\alpha\beta}e_\gamma\quad\text{with }N_{\alpha\beta}\in\{1,2,3\}.
$$
Because $\ell>3$, by assumption, we get $[e_\alpha,e_\beta]\ne0$. So this case must not occur either. In the remaining cases we have $[e_\alpha,e_\beta]=0$. The condition on $S$ is therefore: $\mathfrak a$ is an abelian subalgebra if and only if $\alpha+\beta$ is not zero and not a root for all $\alpha,\beta\in S$. This condition is clearly characteristic free.
As you mention one could make a more general statement, I guess this is clear but I want to write this down. Suppose that $\ell \geq 0$ does not divide any of the $N_{\alpha\beta}$. Then the maximal dimension of an abelian subalgebra is $n + k$, where $n$ is the maximal dimension of an abelian subalgebra in char $0$, and $k$ is the dimension of the center of $\mathfrak{g}$. For ADE types this always holds, since $N_{\alpha\beta} = \pm 1$. For types $B$, $C$, $F_4$ assuming $\ell \neq 2$ suffices. For type $G_2$ we need $\ell \neq 2,3$.
There is also an AMS Memoir by Ross Lawther "Maximal Abelian Sets of Roots" which seems to classify the sets of roots $S \subseteq \Phi$ with the property that for all $\alpha, \beta \in S$, we have $\alpha + \beta \neq 0$ and $\alpha + \beta \not\in \Phi$.
Thank you very much for this, and thanks to spin for the comments. I found two days ago, and have been meaining to update this question with, a reference for the p good case. A paper by Julia Pevtsova and Jim Stark called 'Varieties of Elementary Subalgebras of Maximal Dimension for Modular Lie Algebras' proves this, but with the requirement that the abelian subalgebra have trivial p-restriction. This seems to be needed to make sure that you can conjugate $\mathfrak a$ into $\mathfrak u$....
... You seem to have proved more than they did in their paper, which is interesting considering it is more or less the main result of their paper! In particular, they definitely require $p$ to be good, which for $E_8$ for example is a long way from your condition. I will have to read this proof again, as it's a little far outside my research area (I want to apply this result to work with finite groups of Lie type) but it all seems good to me.
@spin It turns out that I need this more general statement. I have $\ell=2$ and an abelian $29$-space for $E_7$. So it just sneaks over the bound. This has been incredibly useful.
|
2025-03-21T14:48:29.885949
| 2020-02-21T15:34:26 |
353250
|
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|
Stack Exchange
|
A stopping time that gives the metric
Let $\Omega$ a finite metric space with $\forall x,y,z\in \Omega:d(x,y)<d(x,z)+d(z,y)$. Does there exists a continuous-time Markov process $X$ on $\Omega$ such that $$\mathbb{E}_x(T_y)=d(x,y)$$ for all $x,y\in \Omega$ with $T_y:=\inf \{t\geq 0, X_t=y \}$ and $\mathbb{E}_x$ is for the conditon $X_0=x$?
Remark: since one can first go to $z$ and then to $y$ we have by the Markov property $\mathbb{E}_x(T_y)\leq \mathbb{E}_x(T_z)+\mathbb{E}_z(T_y)$ so the triangular inequality. But is it the only constraint?
So far, I tried to express $\mathbb{E}_x(T_y)$ using the resolvant of the stochastic matrix or to look for some connection with the resistance network (see for example the book "Markov Chain and Mixing Times" by Levin, Peres and Wilmer) but I haven't find anything very successful.
Do you mean "continuous time Markov process"? If $X$ is actually continuous, i.e. has continuous paths, then it can only be constant.
Is "continuous Markov process" a "continuous-time Markov process"?
Yes "continuous-time", I edited the question.
The are further restrictions — see my answer below — but I have no idea what they are, the counter-example given in my answer does not provide any insight. An unrelated question: why on earth would one call the state space of a Markov chain $\Omega$?
Suppose we have $n$ states $\{1, 2, \ldots, n\}$, and call transition rates $q_{ij}$. Set, as usual, $q_{ii} = -\sum_{j \ne i} q_{ij}$. Also, write $d_{ij}$ for $d(i,j)$. For $i \ne j$ we have
$$ d_{ij} = \frac{1}{-q_{ii}} \biggl( 1 + \sum_{k \ne i} q_{ik} d_{kj} \biggr) , $$
that is,
$$ 0 = 1 + \sum_k q_{ik} d_{kj} . $$
This system can be solved explicitly, and if I typed it correctly to Mathematica, when $n = 5$ and
$$ d_{ij} = \min\{1 + |i - j|, 3\} , $$
we obtain $q_{24} = -\tfrac{1}{4} < 0$, which is not possible.
Mathematica code:
nn = 5;
d[i_, j_] := If[i == j, 0, Min[Abs[i - j] + 1, 3]];
eqns = Flatten@
Table[If[i == j, 0 == Sum[q[i, k], {k, 1, nn}],
0 == 1 + Sum[q[i, k] d[k, j], {k, 1, nn}]], {i, 1, nn}, {j, 1, nn}];
vars = Flatten@Table[q[i, j], {i, 1, nn}, {j, 1, nn}];
q[2, 4] /. Solve[eqns, vars]
Note that clearly $d(x,x) = \mathbb E_x(T_x) = 0$, which is good.
In general, even for reversible Markov chains, equivalently electrical networks, the Levin--Peres--Wilmer book is good for these, as is the Lyons--Peres Probability on Trees; for lecture notes, see the second half of these, written by Sousi.
There is the so-called hitting/commute-time identities (see Lyons--Peres, Proposition 2.20 and Corollary 2.21).
These imply that
$$ \mathbb E_x(T_y) = \sum \pi(z) v_{x \to y}(z)
\quad
\text{where}
\quad
\text{$v_{x \to y}$ is the voltage wrt to a unit current $x \to y$}
$$
and
$$
\mathbb E_x(T_y) + \mathbb E_y(T_x) = R_\mathrm{eff}(x,y) \sum c(e)
\quad
\text{where}
\quad
\text{$c(e)$ is the conductance of edge $e$ and $R_\mathrm{eff}$ is the "effective resistance"}.
$$
For a more precise description, see one of the above references.
In short, it's pretty difficult to me to see how you are going get this voltage-sum to be the same when swapping $x$ with $y$.
In fact, Question 5 here shows, for SRW, that
$$
\mathbb E_x(T_y)
= \textstyle
\tfrac12
\sum_z
\deg(z) \bigl( R_\mathrm{eff}(x,y) + R_\mathrm{eff}(y,z) - R_\mathrm{eff}(x,z) \bigr).
$$
In the SRW case, $\pi(z) = \deg(z) / \sum_u \deg(u)$ and the conductances are all units. Replacing $\deg$ appropriately, you can get the more general formula for any reversible Markov chain.
Of course, if you define $d$ via the commute time, eg
$$
d(x,y)
:=
\mathbb E_x(T_y) + \mathbb E_y(T_x)
\qquad
\text{(or half this, as you see fit)},
$$
then you're in business.
In the case that $X$ is strong Markov and $\mathbb E_x[T_y] = d(x,y) < \infty$ for all $x,y\in \Omega$ we have the following. Take $x,y,z\in\Omega$.
\begin{align*}
\mathbb E_x[T_y]
& = \mathbb E_x[\mathbb E_x[T_y | \mathscr F_{T_z}]] \\
& = \mathbb E_x[\mathbb E_x[T_y - T_z | \mathscr F_{T_z}] + T_z] \\
& = \mathbb E_x[\mathbb E_z[T_y] + T_z] \\
& = \mathbb E_z[T_y] + \mathbb E_x[T_z].
\end{align*}
That is,
$$
d(x,y) = d(z,y) + d(x,z).
$$
This contradicts your requirement that
$$
d(x,y) < d(z,y) + d(x,z).
$$
This requirement also necessitates the assumption that $\mathbb E_x[T_y] = d(x,y) < \infty$ for all $x,y$. You might have luck with a Markov process which isn't strong Markov, since they can have more exotic behaviour with respect to $\mathscr F_0$-measurable events.
In the displayed equation, the third equality is wrong: we have $\mathbb{E}x[T_y \circ \theta{T_z} | \mathscr{F}_{T_z}] = \mathbb{E}z[T_y]$ a.s., but $T_y \circ \theta{T_z} = T_y - T_z$ only on the event ${T_y > T_z}$. Also, there are extremely few functions that satisfy $d(x,y) = d(x,z) + d(z,y)$ for every triple $x, y, z$...
|
2025-03-21T14:48:29.886277
| 2020-02-21T16:31:56 |
353251
|
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|
Stack Exchange
|
Integral expressions for Bessel-like power series
I'm interested in power series of form $$f(z)=\sum_{k=0}^\infty \frac{z^k}{(k!)^\alpha}.$$ When $\alpha=1$, this becomes $\exp(z)$. For $\alpha=2$ this is a Bessel function and for larger integer $\alpha$ we get a hypergeometric series. These special functions ($\alpha>1$) have integral expressions in form of some integral of an elementary function.
Can anything be said for non-integral values of $\alpha$? If $\alpha>1$ is an arbitrary real number, is there a hope to write an integral expression for the sum? In general, has this form of series been studied anywhere in the literature? Any useful techniques to work with them?
even something simple as $\sum_{k=0}^\infty 1/\sqrt{k!}$ doesn't seem to be a known special function...
It may not have a name but I was wondering if it can be written in some form (such as integral expression) to turn the sum into a continuous integral so that one can work with the function in nontrivial ways. It seems like Barnes integrals can be used to produce such expressions, so that may be a potential approach.
Related:
https://mathoverflow.net/questions/84958/is-sum-limits-n-0-infty-xn-sqrtn-positive
[for $\alpha = 1/2$, but fedja's accepted answer applies to all
$\alpha \in (0,1)$]; https://mathoverflow.net/questions/85013
[about positivity of $\sum_{r=0}^n (-1)^r {n \choose r}^{1/2}$,
again with answers that generalize to
$\sum_{r=0}^n (-1)^r {n \choose r}^\alpha$ with $\alpha < 1$].
F. Olver, in 'Asymptotics and Special Functions,' chapter 8, has shown that
$$ F_\rho(x):=\sum_{j=0}^\infty \Big( \frac{x^j}{j!} \Big)^\rho \sim
\frac{\exp(\rho \, x)}{\sqrt{\rho}(2\,\pi\,x)^{(\rho-1)/2}} \Big(1+O(1/x)\Big) $$
for $ 0<\rho\le 4$ and $x \to \infty.$ The OP might get some hints from that analysis.
This function is really known and has its name: Le Roy function, cf.
https://www.tandfonline.com/doi/abs/10.1080/10652469.2018.1472592?journalCode=gitr20
and references therein.
|
2025-03-21T14:48:29.886425
| 2020-02-21T18:00:03 |
353256
|
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|
Stack Exchange
|
At what aspect ratio does the Ruzsa-Szemeredi Theorem begin?
One of the many equivalent phrasings of the Ruzsa-Szemeredi theorem is as follows. Suppose one has a three-layered $n$-node graph $G = (V=L_1 \cup L_2 \cup L_3, E)$, and one can partition $E$ into two-edge paths going across the layers that are each the unique shortest path between their endpoints. The theorem then states that $|E| = o(n^2)$.
If $G$ can have edge weights, then the corresponding statement is false: one can have $|E| = \Omega(n^2)$. The standard construction is to embed the graph nodes in the integer lattice, in $[3] \times [n/3]$, and put edge weights corresponding to Euclidean distance. Each triple of colinear points across the three layers then gives a unique shortest path, and there are $\Omega(n^2)$ such triples in total.
This construction has aspect ratio (max edge weight / min edge weight) $\alpha = \Omega(n)$, whereas the Ruzsa-Szemeredi theorem basically constrains graphs with $\alpha=1$. My question is whether there are any results making progress on this gap on either side. For example, can there be a construction with $\alpha = O(\sqrt{n})$ but $|E| = \Omega(n^2)$?
|
2025-03-21T14:48:29.886547
| 2020-02-21T18:03:55 |
353258
|
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"authors": [
"Martin Brandenburg",
"Robert Furber",
"Simon Henry",
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"varkor"
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"sort": "votes",
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|
Stack Exchange
|
Characterisation of essentially algebraic theories as monads
The following correspondence between algebraic theories and monads on $\mathbf{Set}$ is well-known (see, for example, Algebraic Theories: A Categorical Introduction to General Algebra).
The category of (finitary) $S$-sorted algebraic theories is equivalent to the category of (finitary) monads on $\mathbf{Set}/S$. The category of models for a fixed algebraic theory $T$ is equivalent to the category of algebras for the corresponding monad.
I would like to know if there is a known analogous result in the setting of essentially algebraic (i.e. finite limit) theories. That is, some statement like the following.
The category of (finitary) $S$-sorted essentially algebraic theories is equivalent to the category of [some class of] monads on [some category]. The category of models for a fixed algebraic theory $T$ is equivalent to the category of algebras for the corresponding monad.
There are many characterisations of categories of models of essentially algebraic theories (i.e. locally presentable categories), but I have not been able to find one in terms of a category of algebras for some monad. Given the many generalisation of Linton's result, for example in Notions of Lawvere theory and Monads with arities and their associated theories, it would seem like this result should be a straightforward application of an existing theorem, but, as far as I can tell, the case of essentially algebraic theories is never explicitly treated. (I originally read Theorem 6.7 of Notions of Lawvere Theory as stating that one-sorted essentially algebraic theories also correspond to finitary monads on $\mathbf{Set}$, but this seems unlikely to be correct.)
If the $S$-sorted case is unknown, I am also interested in the correspondence specifically in the one-sorted setting.
Is "finitary monad on a presheaf category" a satisfying answer ? An alternative is iterated monads (considering finitary monads acting on category of another finitary monads on sets)
You can have two distinct essentially algebraic theories such that the monad defined by the free-forgetful adjunction to $\mathbf{Set}$ is the same. For example, both abelian groups and torsion-free groups are essentially algebraic theories, but they define the same monad on $\mathbf{Set}$, because every free abelian group is torsion-free.
Of course, I meant to say "torsion-free abelian groups" in my previous comment.
To expand on Simon Henry's comment, the category of models $C$ of an essentially algebraic theory form a locally finitely-presentable theory, which is a finitarily reflective subcategory of a presheaf category $Set^{C_0^{op}}$ (in particular, the inclusion is monadic). In turn, the presheaf category $Set^{C_0^{op}}$ is monadic, via the obvious forgetful functor, over $Set^{Ob C_0}$. Thus $C$ admits a functor to a slice of $Set$ which is a composite of monadic functors (but typically not itself monadic). I'm not sure how to get a composite of monadic functors to $Set$ itself though...
Looking back at this question, I think the point I ought to have emphasised more was that I was interested in essentially algebraic theories for a fixed set of sorts. One can recover finite limit theories using the technique of Simon Henry, but it is not possible to fix a set of sorts this way. This is then somehow less insightful than the traditional correspondence between algebraic theories and strongly finitary monads on $\mathbf{Set}$.
@varkor Is it still open?
@MartinBrandenburg: a monad correspondence as I originally envisaged it is still open. For a while, partial algebraic theories seemed a possible route to a correspondence, but there are some difficulties with this approach. However, there is a "monad-like" correspondence described in Essentially equational categories. I believe this result can be phrased along the lines of the usual monad–theory correspondence, but have not yet gotten around to working out all the details.
I'm going to give a partial answer to my question, which addresses a misconception I had and illustrates why many of the existing generalisations of theory–monad correspondence are not sufficient to provide a monadic correspondence with essentially-algebraic theories. As far as I know, this indicates that a correspondence is open.
Let $\mathbb C$ be a small category, and denote by $\widehat{\mathbb C}$, $\mathbf{Rex}(\mathbb C)$, $\mathbf{Ind}(\mathbb C)$ and $\mathbf{Lex}(\mathbb C)$ the free cocompletion, finite cocompletion, filtered cocompletion and finite completion of $\mathbb C$, respectively. We have $\mathbf{Ind}(\mathbf{Rex}(\mathbb C)) \simeq \widehat{\mathbb C}$.
We'll instantiate the results of Bourke & Garner's Monads and theories for $K : \mathbf{Rex}(\mathbb C) \hookrightarrow \widehat{\mathbb C}$. This is an example of the so-called "presheaf context". The results follow directly, so I won't spell everything out. $K$ preserves finite colimits and so a $\mathbf{Rex}(\mathbb C)$-theory is a finite colimit-preserving identity-on-objects functor $J : \mathbf{Rex}(\mathbb C) \to \mathcal T$. A model is a finite-limit preserving functor $F : \mathcal T^{\mathrm{op}} \to \mathbf{Set}$. A $\mathbf{Rex}(\mathbb C)$-nervous monad is a monad on $\mathbf{Rex}(\mathbb C)$ whose underlying endofunctor preserves filtered colimits. The categories of $\mathbf{Rex}(\mathbb C)$-nervous monads and $\mathbf{Rex}(\mathbb C)$-theories are equivalent by Theorem 17 of ibid. (and similarly their categories of algebras and categories of models by Theorem 34 of ibid.)
We can now take opposites appropriately to generalise the classical algebraic theory–finitary monad correspondence: the categories of finite limit-preserving identity-on-objects functors from $\mathbf{Lex}(\mathbb C^{\mathrm{op}})$ (which I'll dub "category-sorted lex theories") are equivalent to finitary monads on $\widehat{\mathbb C}$. When $\mathbb C = S$ is discrete, category-sorted lex theories are equivalent to sorted algebraic theories (this can be seen as a consequence of sifted colimits and filtered colimits coinciding on indexed sets, or of the coincidence of finite completion and finite product completion on sets). When $\mathbb C$ is a non-discrete category, the two constructions differ. This essentially gives a classification of finitary and sifted colimit-preserving monads on presheaf categories (at least on small categories), which was suggested by Simon Henry.
I was previously assuming that category-sorted lex theories (at least for discrete categories) was the right notion of essentially algebraic theory. However, I was mistaken: category-sorted lex theories are less expressive than sorted algebraic theories. In particular, the requirement for theories to be identity-on-objects is too restrictive, as the codomains may no longer be finitely complete. An analogous definition of sorted essentially algebraic theory to sorted algebraic theory will look different (as far as I'm aware, no such characterisation has been given in the literature) and, as such, the existing theory–monad correspondences are insufficient to describe a correspondence for essentially algebraic theories. I'll continue to pursue this question according to this line of thinking, but unless anyone can see that I've missed something obvious, I'm satisfied that such a correspondence is at least not known (or easily derivable from results in the literature).
|
2025-03-21T14:48:29.886999
| 2020-02-21T18:48:02 |
353260
|
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|
Stack Exchange
|
Let $f$ be periodic with a continuous image and $a_n = cn$ for some $c > 0$. When is $\{f(a_n)\}$ dense in the image of $f$?
Let $f:\mathbb{R}\to\mathbb{R}$ be a periodic function with period $T$ and continuous everywhere except perhaps on a countable set, and have an image that's an uncountable subset of $\mathbb{R}$. Let $a_n = cn$ for some $c > 0$. When is the subset $\{f(a_n)\}_{n \in \mathbb{N}}$ dense in the image of $f$?
What is a "continuous image"?
A non-countable subset of $\mathbb{R}$, like an interval. A better word escapes me, other than "non-countable subset of $\mathbb{R}$".
I'd reformulate as follows: let $f:\mathbf{R}/\mathbf{Z}\to\mathbf{R}$ be a function whose domain of continuity has a countable complement. Let $c$ be irrational. Is $f(cn)_{n\ge 0}$ dense in $f(\mathbf{R}/\mathbf{Z})$?
Quite clearly there are counterexamples and you should probably rather ask whether the closure of your subset has countable complement in the closure of $\mathrm{Im}(f)$ (indeed, modify a bounded continuous function on a closed countable subset disjoint from the given arithmetic sequence...)
|
2025-03-21T14:48:29.887101
| 2020-02-21T19:53:48 |
353268
|
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|
Stack Exchange
|
Convex triangulations
Given a set of $n$ points in the Euclidean plane of which no three are collinear, does there always exist a convex triangulation and how can one be found algorithmically?
In this context a convex triangulation shall mean a triangulation in which the union of triangles with a common corner-point inside the convex hull constitute to a convex polygon.
Nice idea but it seems not to always exist:
Point $4$'s star is reflex at $3$,
Point $3$'s star is reflex at $4$.
Here's an argument that those are the only two triangulations.
Each interior point ($3$ and $4$) must be degree-$3$ or
degree-$4$: degree-$3$ to span more than $180^\circ$,
and at most degree-$4$ because there are only four other points.
In the left figure point $3$ has degree-$3$ and point $4$ degree-$4$,
and in the right figure point $3$ has degree-$4$ and point $4$ degree-$3$.
There must be a total of $9$ edges, $6$ of them diagonals.
Then the diagonals in the two figures are forced.
It is an interesting question: When does a convex triangulation exist? And what is the computational complexity to decide if a set of points admits a convex triangulation?
Observation: If all neighbors of an interior vertex $v$ in a convex triangulation are interior vertices, then the degree of $v$ is at least $5$.
@JanKyncl: May I ask how you arrived at this interesting fact?
When looking at the figure, I noticed that around a vertex of degree at most $4$, two of the consecutive angles will sum up to more than 180 degrees. So the neighbor along the edge common to those two angles will have a nonconvex union of incident triangles.
|
2025-03-21T14:48:29.887246
| 2020-02-16T14:48:43 |
352868
|
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|
Stack Exchange
|
Pseudo-intersections, splitting families, and ultrafilters
Suppose $U$ is a non-principal ultrafilter on $\omega$, and let us define $\tau(U)$ to be the minimum cardinality of a family $\mathcal{X}\subseteq U$ such that $\mathcal{X}$ does not have an infinite pseudo-intersection, that is, there is no infinite $A$ such that $A\setminus B$ is finite for all $B\in \mathcal{X}$.
Claim: $\tau(U)\leq\mathfrak{s}$ for any $U$.
The point is that $U$ will contain a splitting family of cardinality $\mathfrak{s}$, and this splitting family has no infinite pseudo-intersection. (To see the first statement, note that if $\mathcal{X}$ is a splitting family and we replace some of the elements of $\mathcal{X}$ by their complements, then the resulting collection is still a splitting family. Since an ultrafilter will contain one of $X$ and $\omega\setminus X$ for each $X\in\mathcal{X}$, we may as well assume $\mathcal{X}\subseteq U$.)
Question: Is there (in ZFC) an ultrafilter $U$ on $\omega$ for which $\tau(U)=\mathfrak{s}$?
I conjecture that the answer is "no", and that this negative answer will be witnessed in the original Blass-Shelah model of NCF from the paper below.
Blass, Andreas; Shelah, Saharon, There may be simple $P_{\aleph _ 1}$- and $P_{\aleph _ 2}$-points and the Rudin-Keisler ordering may be downward directed, Ann. Pure Appl. Logic 33, 213-243 (1987). ZBL0634.03047.
Is it true that $\tau(U)$ is a regular cardinal, for every non-principal ultrafilter $U$ on $\omega$?
I don't know the answer to that one. A priori, I don't see why this would need to occur, but I have not thought deeply about it.
I think another good candidate for a counterexample could be the Mathias model.
I guess it’s really about reflection dealing with questions like “If every small subfamily has a pseudointersection, does the whole family?”
The answer is no -- it is consistent that every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$.
I had an idea for proving this earlier today, using the Mathias model. I couldn't quite make things work, and I ended up talking about the problem with Alan Dow for a good part of the afternoon. (1) We still think the Mathias model could work, but it seems tricky. (2) There's a fix: if you interleave Laver forcings with Mathias forcings in a certain way, then the resulting iteration does work. (I'll sketch this below.) (3) I learned from Alan that your question has been studied already. The consistency of "every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$" is already known (via a different argument than the Mathias-Laver iteration sketched below), and the characteristic $\tau(U)$ has been studied quite a bit.
A rich source of information on $\tau(U)$ is the following paper by Brendle and Shelah:
Jörg Brendle and Saharon Shelah, ``Ultrafilters on $\omega$ -- their ideals and their characteristics,'' Transactions of the AMS 351 (1999), pp. 2643-2674. (available here)
What you call $\tau(U)$ is in this paper called $\pi \mathfrak{p}(U)$. They prove, among other things, that
$\bullet$ If $U$ is not a $P$-point, then $\tau(U) \leq \mathfrak{b}.$
$\bullet$ If $U$ is not a $P$-point, then the cofinality of $\tau(U) \leq \mathfrak{b}$ is uncountable. (This provides a partial answer to Santi's question in the comments.)
$\bullet$ A characterization of $\tau(U)$ is given in terms of an ideal defined from Ramsey-null sets.
$\bullet$ It is consistent that $\tau(U) < \mathfrak{s}$ for all $U \in \omega^*$.
The first two results are in Section 2, the next in Section 3, and the last in Section 7. The last result answers your question, of course, but I should also mention another relevant paper:
Alan Dow and Saharon Shelah, ``Pseudo P-points and splitting number,'' Archive for Mathematical Logic 58 (2019), pp. 1005-10027. (available here)
In the Brendle-Shelah paper, they prove $\sup_{U \in \omega^*}\tau(U) < \mathfrak{s}$ is consistent, but the gap between $\sup_{U \in \omega^*}\tau(U)$ and $\mathfrak{s}$ is only one. In the Dow-Shelah paper, they use a more complicated matrix iteration to make the gap between $\sup_{U \in \omega^*}\tau(U)$ and $\mathfrak{s}$ arbitrarily large.
Finally, let me sketch the idea I mentioned above. The idea is to do a countable support iteration that uses Laver forcing at limit steps of cofinality $\omega_1$, and uses Mathias forcing everywhere else. The iteration is of length $\omega_2$ and CH holds in the ground model. Let $V[G]$ denote the result of such an iteration, and let $U \in \omega^*$ in $V[G]$. By reflection, there is some intermediate model $V[G_\alpha]$ with $\alpha < \omega_2$ where $U \cap V[G_\alpha]$ is an ultrafilter in $V[G_\alpha]$, and where $\alpha$ has cofinality $\omega_1$. At this stage, we force with Laver forcing, and this adds a length-$\omega_1$ tower to $U \cap V[G_\alpha]$. All the subsequent Laver forcings and Mathias forcings preserve the fact that this tower has no pseudo-intersection, and so this tower is a witness to the fact that $\tau(U) = \aleph_1$ in the final extension. (See Theorem 7.11 from this paper of Alan's for more detail on the last two sentences.) Finally, the Mathias forcings enable us to get $\mathfrak{s} = \mathfrak{c}$. This is well-known to be true in the Mathias model (although not in the Laver model), and it's true in this model for essentially the same reasons.
Perfect! And thank you for the references!
This can be modified to push up $\mathfrak{h}$ as well by sprinkling in more Mathias forcing on a stationary/co-stationary set of ordinals of cofinality $\omega_1$. Can the Dow-Shelah construction accomplish the same?
|
2025-03-21T14:48:29.887656
| 2020-02-16T16:28:49 |
352875
|
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|
Stack Exchange
|
Extensions of infinite loop spaces
I've been reading Rognes' paper "Algebraic K-theory of 2-adic integers" and I'm confused about something he's doing regarding infinite loop spaces (see Theorem 8.1 of that paper). Let me try to phrase more generally what confuses me:
Suppose we have a fiber sequence of connective spectra
$X\to Y\to Z$. We can think of $Y$ as an extension of $Z$ by $X$ in some sense and (supposedly) the homotopy type of $Y$ is determined by maps of spectra $\Sigma^{-1}Z\to X$.
Why is this the case? Is there an analog of this in the world of group extensions? I've never seen a result like this before and I haven't been able to figure myself why it should be true. Any help is appreciated!
Because the stable homotopy category is triangulated, and a fiber sequence is in particular a distinguished triangle in that category; with that in mind the result follows from a general result in triangulated categories ($\Sigma^{-1}Z\to X\to Y \to Z$ is a distinguished triangle, so the isomorphism type of $Y$ is determined by the map $\Sigma^{-1}Z\to X$). You can think of it via the long exact sequence you get when applying $\hom(x,-)$
Think about chain complexes instead of group extensions. What does the shifted sequence look like for the exact sequence of complexes $\Bbb Z \to \Bbb Z \to (\Bbb Z/2)$?
Puppe showed that the homotopy fiber of $X \to Y$ is $\Sigma^{-1} Z$, and the homotopy cofiber of $Y \to Z$ is $\Sigma X$, so that $Y$ can be recovered as the homotopy cofiber of $\Sigma^{-1} Z \to X$, or as the homotopy fiber of $Z \to \Sigma X$:
$$\Sigma^{-1} Z \to X \to Y \to Z \to \Sigma X$$
If $X = HA$, $Y = HB$, $Z = HC$ are Eilenberg-Mac Lane spectra, then you have a short exact sequence $0 \to A \to B \to C \to 0$, and the homotopy class of $Z \to \Sigma X$ in $[Z, \Sigma X]{H\mathbb{Z}-mod}$ corresponds to a class in $Ext^1{\mathbb{Z}}(C, A)$ that classifies this extension.
@JohnRognes Ok that clarifies things a lot. It's even easier than I was thinking. Thanks!
|
2025-03-21T14:48:29.887825
| 2020-02-16T16:37:54 |
352878
|
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|
Stack Exchange
|
Calabi-Yau threefold with an automorphism of infinite order
I am looking for a (hopefully simple) example of a Calabi-Yau threefold (projective, simply connected, with trivial canonical bundle) admitting an automorphism of infinite order.
A Schoen manifold $X$ is a generic complete intersection in $\mathbb{P}^2 \times \mathbb{P}^2 \times \mathbb{P}^1 $ of two hyper-surfaces of degree (3,0,1) and (0,3,1) respectively. Alternately, you can describe this Calabi-Yau threefold as the fiber product of two generic rational elliptic surfaces $X = S\times _{\mathbb{P}^1} S'$ where $S$ and $S'$ are hyper-surfaces in $\mathbb{P}^2\times \mathbb{P}^1$ of degree (3,1) and projection to $\mathbb{P}^1$ induces the elliptic fibration. The surface $S$ has an infinite automorphism given by fiberwise addition by a non-zero section $\sigma: \mathbb{P}^1 \to S$ and this induces an infinite automorphism on $X$.
More generally, there are lots of CY3s which admit elliptic fibrations and I would expect there to be lots of examples where the Mordell-Weil group of sections is infinite.
Certainly. My problem was to get an explicit construction — I guess the one you give is reasonably explicit.
For sure there are many examples of CY3s with elliptic fibration of positive MW rank. But typically these translations only give birational automorphisms, not biregular ones.
(For clarity, my comment was in reference to Jim Bryan's comment, not to his answer.)
@abx I think that with a little bit of work, one can get extremely explicit with a Schoen manifold (writing down the equations of $X$ and then writing the automorphism in terms of the variables). I haven't done it, but I would start by writing down the group law for a pencil of cubics. I don't know what you are trying to do, but I would be surprised if having explicit equations is any more useful than the more conception description in terms of translation by a section in the Abelian fibration.
I agree. Actually, what I needed is for instance to compute the action on the Picard group, and that can be done directly from the description as a fiber product.
Another nice example is by Oguiso and Truong, "Explicit Examples of rational and Calabi-Yau threefolds with primitive automorphisms of positive entropy".
Briefly, take $E$ to be an elliptic curve with an order $3$ automorphism $\tau$ and form the abelian variety $A = E \times E \times E$. The quotient of $A$ by the diagonal action of $\tau$ is mildly singular, but it has only isolated singularities and it's easy to check/standard that there's a crepant resolution $X$ that's a CY3.
Then it's easy to get automorphisms of infinite order: $SL(3,\mathbb Z)$ acts on $X$ in an obvious way. In fact, [OT] show that some of these automorphisms are "primitive", meaning they don't preserve any nontrivial fibration (unlike in the elliptically fibered examples). This is done using the theory of dynamical degrees.
(Note that this obviously gives honest automorphisms, not just birational maps; there's nothing subtle to check on singular fibers, unlike what you usually run into looking at elliptically fibered examples, as Bort alludes in the comments above.)
Very useful, thanks!
|
2025-03-21T14:48:29.888175
| 2020-02-16T17:22:25 |
352882
|
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|
Stack Exchange
|
Does this equation have more than one integer solution?
Consider the following diophantine equation
$$n = (3^x - 2^x)/(2^y - 3^x),$$ where $x$ and $y$ are positive integers and $2^y > 3^x$.
Does $n$ have any other integer solutions besides the case when $x=1$ and $y=2$, which give $n=1$?
It is not a solution because of the last inequality, but $x=2$ and $y=3$, so $n=-5$, it is an integer solution of the equation (which I am reluctant to say is Diophantine).
Something that may help: $n + 1 = \frac{2^y - 2^x}{2^y - 3^x}$, so there is such an $n$ iff $2^y - 3^x | 2^y - 2^x$. The denominator is odd, so there is such an $n$ iff $2^y - 3^x | 2^{y - x} - 1$. Notice that this implies that $2^{y - 1} < 3^x < 2^y$, so for each $y$, there's only one $x$ to check.
This follows quickly from the observations of user44191. Check each $1 \leq x \leq 66$ and note that, for $x \geq 67$, we have $y < 1.6x$. Applying lower bounds for linear forms in two complex logarithms (as in, say, Theorem 5.2 of de Weger's thesis), we have that
$$
2^y - 3^x \geq 3^{0.9x},
$$
since $3^x > 10^{15}$. From the fact that $2^y-3^x \mid 2^{y-x}-1$, it follows that
$$
2^{0.6x}-1 \geq 3^{0.9x},
$$
a contradiction.
|
2025-03-21T14:48:29.888294
| 2020-02-16T21:13:53 |
352891
|
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|
Stack Exchange
|
Largest absolute value of a polynomial of degree $n$ on $\{0,1,\ldots,n\}$
Consider a polynomial $P_n(x)\in\mathbb{R}[x]$, of degree $n\geq1$, of the form
$$P_n(x)=c_0+c_1x+c_2x^2+\cdots+c_{n-1}x^{n-1}+x^n.$$
To illustrate the question, take $P_1(x)=c_0+x$ so that $P_1(0)=c_0$ and $P_1(1)=c_0+1$. If $\vert c_0\vert<\frac12$ then $\vert c_0+1\vert\geq1-\vert c_0\vert>1-\frac12=\frac12$. That means, $\pmb{\max}\{\vert P_1(0)\vert,\vert P_1(1)\vert\}\geq\frac12$.
In general,
QUESTION. is this true?
$$\pmb{\max}\{\vert P_n(0)\vert,\vert P_n(1)\vert,\vert P_n(2)\vert,\dots,\vert P_n(n)\vert\}\geq\frac{n!}{2^n}.$$
you mean $P_1(\color{red}{1})=c_0+1$ instead of $P_1(\color{red}x)=c_0+1$, right?
Surely you mean $\max$ and not $\min$? And also probably an absolute value somewhere in there.
Both of you are correct; edited as such too. Thank you!
Just a little remark: the first thing that I thought was trying to link this with the fact that Chebyshev polynomials are those that have the minimum maximum absolute value on [-1,1] and principal coefficient 1.
edited the title, hope you don't mind.
Thank you, kodlu.
You can write your polynomial as
$$ P(x) = \sum_{k = 0}^n P(k) L_{n,k}(x) ,$$
where $L_{n,k}$ are the Lagrange interpolation polynomials with nodes at $0, 1, \ldots, n$. Note that $(-1)^{n - k} L_{n,k}$ has positive coefficient at $x^n$. Thus, with the constraint $$\max\{|P(k)| : k = 0, 1, \ldots, n\} \leqslant 1,$$ the largest possible value of the coefficient of $P$ at $x^n$ is attained by
$$ \bar P(x) = \sum_{k = 0}^n (-1)^{n - k} L_{n,k}(x) , $$
and the coefficient of $\bar P$ at $x^n$ is equal to
$$ \sum_{k = 0}^n \prod_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} \, .$$
In other words, if the coefficient at $x^n$ is to be equal to $1$, the least possible value of $\max\{|P(k)| : k = 0, 1, \ldots, n\}$ is
$$ \biggl(\sum_{k = 0}^n \prod_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} \biggr)^{-1} .$$
It remains to note that
$$ \sum_{k = 0}^n \prod_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} = \sum_{k = 0}^n \frac{1}{k! (n - k)!} = \frac{1}{n!} \sum_{k = 0}^n \binom{n}{k} = \frac{2^n}{n!} \, .$$
Lagrange interpolation suggested by Mateusz Kwaśnicki is perfectly ok, but in this case it is probably easier to use the finite difference formula $$\sum_{i=0}^{n} (-1)^{i}{n\choose i}P(t+n-i)=\Delta^n P=n!$$
for $t=0$, where $\Delta:f(t)\to f(t+1)-f(t)$ is a finite difference operator.
Also this very statement is well known, in case if you need references.
Nice! It would be good to get any reference.
It was proposed by Vietnam to IMO 1977 (for arbitrary integers, not consecutive, Mateusz' proof works for this case too), see longlist of 1977 in any edition of IMO Compendium.
I will look it up, thanks.
|
2025-03-21T14:48:29.888494
| 2020-02-16T21:52:57 |
352894
|
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|
Stack Exchange
|
HNN-extension as a 2-colimit
In the spirit of this question, it would be interesting to give a characterization of HNN extensions as a 2-colimit. If $G$ is a group and $\alpha:H \xrightarrow{\cong} K$ is an isomorphism between two subgroups of $G$, then I think that the HNN extension $G_{\alpha}$ has the following universal property: if we let $i_1:H \hookrightarrow G,i_2:K \hookrightarrow G$ be the canonical inclusions, then the set of group homomorphisms $G_{\alpha} \to T$ are in natural bijection to pairs $(f,t)$ where $f:G \to T$ is a group homomorphism and $t:f \circ i_2 \circ \alpha \Rightarrow f \circ i_1$ is a 2-morphism. Here we're considering $\mathbf{Grp}$ as a full subcategory of $\mathbf{Cat}$ which carries a standard 2-category structure.
I'm just wondering if this universal property can be phrased as some kind of 2-colimit.
I assume you are meaning to think of groups as one-object categories? It's not obvious to people who haven't thought about it this way before.
David, I think this is clear from the last sentence "we're considering Grp as a full subcategory of Cat".
Assuming your universal property is true, it exactly says that the HNN extension is the coinserter of $(i_2 \circ \alpha,i_1) : H \rightrightarrows G$.
|
2025-03-21T14:48:29.888610
| 2020-02-16T22:09:21 |
352895
|
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|
Stack Exchange
|
obstruction cocycle for nonsimple spaces using local coefficients
This question is similar to here but I was hoping for a concrete theorem statement surrounding the obstruction cocycle for non-simple spaces.
I'm hoping for a theorem like the following:
Let $A \subset X$ such that $\pi_1(A)=\pi_1(X)$ and $f:A \to Y$ be a function. Letting $\pi_n(Y)$ be a $\mathbb Z[\pi_1(A)]$ module, where $\pi_1(A)$ acts via $f_*$ and the usual action of $\pi_1(Y)$. Suppose that $H^*(X,A,\pi_n(Y))=0$ for all $n \in \mathbb N$. Then there exists an extension of $f$ to all of $X$. More generally if $\pi_1(X)=\pi_1(A)/N$ and $N \subset \ker f_*$, then if $H^*(X,A,\pi_n(A)/N)=0$, there is an extension.
The slightly awkward generalization is due to my desire to prove the universal property for Quillen's plus construction presented in these notes, prop 1.1.2.
A naive guess on my part would be to apply the usual obstruction theorem to the universal cover of $Y$, and then if we use local coefficients, we can ensure that $f:A \to \tilde{Y}$ is a lift of $f:A \to Y$, but I'm not completely sure.
Note that this is a crosspost from math.SE. If there are any issues with this question, I am happy to edit or remove it.
Given maps $i\colon A\rightarrow X$ and $f\colon A\rightarrow Y$ of pointed connected complexes, once you fix a homomorphism $\phi\colon \pi_1 X\rightarrow \pi_1 Y$ extending $f_\ast$, obstructions to extending $f$ up to homotopy along $i$ to a map $g\colon X\rightarrow Y$ with $g_\ast = \phi$ live in the cohomology groups $H^{n+1}_{\pi_1 X}(X,A;\pi_n Y)$. This comes from the Postnikov tower of $Y$, but I don't know a good reference; see also https://mathoverflow.net/q/328022/ .
See Baues's obstruction theory book, also the algebraic homotopy book, I think you'll find it in both.
Dear Fernando, please feel free to leave that as an answer, it is theorem 4.2.9 in the obstruction theory book. Assuming no errors, it is stated as follows: Let $p:E \to X$ be a fibration with path-connected fiber and let $X_n=X^n \cup L$ and $E^(n)=p^{-1} (X^{(n)})$. Then for $n>1$ and( $n=1$ if $\pi_1(F)$ is abelian), given a section $u:X^{(n)} \to B^{(n)}$ there is an obstruction cocycle $c(u) \in C^{(n+1)}(X,L,\pi_n(F))$ so that the section can be extended to the n+1 skeleton if and only if it vanishes....
Taking the trivial bundle $p:X \times Y \to X$ and any map $f:X \to Y$ we write a section as $(x,f(x)) \in X \times Y$ and the previous theorem applies.
|
2025-03-21T14:48:29.889082
| 2020-02-17T00:14:09 |
352902
|
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|
Stack Exchange
|
Conceptual reason that monadic functors create limits?
Let $U: Alg_T \to C$ be the forgetful functor from the category of algebras of $T: C \to C$ ($T$ could be a monad; I'm happy to think about the simpler case where $T$ is just an endofunctor or pointed endofunctor). Then a very simple diagram chase shows that
$U$ creates limits.
In other words, if I have a diagram in $Alg_T$ whose image in $C$ has a limit, there is an obvious way to lift the limit cone to $Alg_T$, and a straightforward diagram chase verifies that indeed, this is a limit in $Alg_T$.
This is all very easy, but I'm unhappy with the state of affairs for a few reasons:
I'm very lazy and I hate doing diagram chases.
This proof does not obviously generalize to other contexts. For one thing, I have to rehash variations on the same diagram chase for each of the cases where $T$ is a monad, an endofunctor, a pointed endofunctor, etc. For another, even if I stick with just monads, say, I have to rehash the same diagram chase if I want to generalize to other contexts such as enriched or internal category theory. Of course, doing the same work over and over is supposed to mean there's a bigger picture I'm missing.
I find it remarkable that in order for $U$ to create limits, one need not assume any kind of limit-preservation hypotheses about $T$. There's something to be explained, and the proof via diagram chase doesn't accomplish that.
The second point may have real weight -- I haven't checked very diligently, but it seems that it might not actually be known whether this this theorem remains true in full generality in the enriched context, for example! (Although a moment's reflection makes me think I could run the same diagram chase with no fuss if it weren't for point (1) above.) So here's my
Question: What is the conceptual reason for which monadic functors (and other forgetful functors from "categories of algebras") create limits?
One point is that the Eilenberg-Moore object of a monad $T$ on an object of a 2-category $\mathcal K$ is the lax limit of $T$, when viewed as a lax functor from the terminal 2-category. Since representables preserve lax limits, this implies that for any $X$, if $T$'s base is $Y$, then $\mathcal K(X,\mathrm{EM}(T))$ is the EM category of the monad $\mathcal K(X,T)$ on the ordinary category $\mathcal K(X,Y)$. For reasonable examples of $\mathcal K$, limits in $Y$ are hopefully some kind of thing that turns into a limit in a category upon applying a representable functor, no?
The above is more relevant for generalizing the 2-category in which you take a monad, rather than generalizing away from monads. I'm not sure exactly how, but surely there must be a lax limit definition for these other categories of algebras too.
@KevinCarlson Hmm... there's something delicate here, maybe a level-shifting thing. The co-EM object for a comonad is an oplax limit, and its creates colimits rather than limits. I don't think I understood your argument well enough to see where you used that the limit is lax rather than oplax.
I don’t think I did. Rather the schematic argument is that forgetful functors out of EM/coEM categories for monads/comonads/endomorphims should generally create whatever (co)limits they do in $\underline{\mathbf{Cat}}$, since both lax and oplax limits in a 2-category are created by representables, as are at least certain notions of limit in an object in a 2-category. Forone specific example, $x$ has (co)limits indexed by a category $J$ if the canonical morphism $x\to x^J$ into the cotensor by $J$ has a right (left) adjoint. The latter argument certainly needs a more precise formulation.
The heart of the diagram chase is: All the operations and equations involved in an algebra structure on X have target X. (More generally: the target of each operation/equation could be any limit-preserving functor of the algebra-specified-so-far.)
For each of the kinds of “algebra” you describe, the structure is built up by sequentially adding operations and equations.
Each “add an operation” step fits into the template of taking an inserter. You have two categories and two parallel functors $\newcommand{\C}{\mathbf{C}}\newcommand{\D}{\mathbf{D}}F, G : \C \to \D$; then the inserter category $\newcommand{\Ins}{\mathrm{Ins}}\Ins(F,G)$ consists of objects $X$ of $\C$ equipped with a map $x : FX \to GX$, and a map $f:(X,x) \to (Y,y)$ is a map $f:X \to Y$ in $\C$ satisfying $(Gf)x = y(Ff)$.
(If $T$ is a monad on $\newcommand{\E}{\mathcal{E}}\E$, then the first stage of defining monad algebras is equipping objects with a map $TX \to X$, i.e. taking the inserter of $\Ins(T,1_\E)$.)
Similarly, each “add an equation” stage fits into the template of taking an equifier: you have categories and functors $F,G : \C \to \D$ as before, plus natural transformations $\alpha,\beta : F \to G$, and the equifier $\newcommand{\Eqf}{\mathrm{Eqf}}\Eqf(\alpha,\beta)$ is the full subcategory of $\C$ of objects $X$ for which $\alpha_X = \beta_X$.
(Back with our monad $T$ on $\E$: after adding the operation, getting the category $\Ins(T,1_\E)$ with a forgetful functor $U$ to $\E$, we can now impose the associativity axiom by taking the equifier for the functors $T^2U, U : \Ins(T,1_\E) \to \E$, with the natural transformations sending $(X,x)$ to the maps $x(Tx), x\mu_X : T^2 X \to X$.)
So monad algebras, and their forgetful functor, can be built up as $\newcommand{\Alg}{\mathrm{Alg}}\Alg_\E(T) = \E_3 \to \E_2 \to \E_1 \to \E_0 = \E$, where each step $\E_{n+1} \to \E_n$ “adds an operation or equation”, i.e. is the forgetful functor from an inserter or equifier category. And moreover, “the target of each operation/equation on $X$ is just $X$”, i.e. the target functor $G$ of each inserter/equifier is the composite forgetful functor $\E_n \to \E$. (And algebras for endofunctors or pointed endofunctors can be built up in the same way.)
Now for creation of limits:
Proposition.
Given $F,G : \C \to \D$, the forgetful functor $\Ins(F,G) \to \C$ creates all limits that exist in $\C$ and are preserved by $G$. Dually, it creates all colimits that exist in $\C$ and are preserved by $F$.
Given $\alpha, \beta : F \to G : \C \to \D$, the forgetful functor $\Eqf(\alpha,\beta) \to \C$ creates all limits that exist in $\C$ and are preserved by $G$. Dually, it creates all colimits that exist in $\C$ and are preserved by $F$.
The proof of this proposition is that diagram chase you don’t want to do — it has to be done sooner or later. But once it’s done here, it applies to give creation of limits for monad algebras, endofunctors, etc, by their presentation in terms of inserters/equifiers as above. It also (dually) gives creation of colimits for coalgebras over comonads, endofunctors, etc.; and also directly gives creation of limits/colimits for various other structures, e.g. monoids/comonoids in a monoidal category, without needing to show they’re (co)monadic.
So I think it quite satisfactorily answers your questions (1) and (3). It doesn’t answer your question (2) — I’m surprised to learn that this doesn’t generalise straightforwardly to the enriched setting, and haven’t worked it through enough to understand why — but it should help clarify what happens there, since the decomposition of algebras in terms of inserters and equifiers still holds, even if the proposition about creation of limits fails.
+1, this is very instructive.
Thanks! Regarding the generalization to enriched category theory, it occurs to me that what probably happened in the discussion I linked to is that somebody read about this topic in an old reference from way back before the general definition of a weighted limit was given by Borceux and Kelly (after all, a substantial amount of enriched category theory was originally developed using just conical limits and cotensors) and so there simply wasn't the language available to show this. Surely it must have been noted in the literature at some point since then that it (presumably) works out...
I am not able to give the high-tech answer you are clearly hoping for. For me this statement is just a straight forward generalization of the known fact that, say, $\mathsf{Grp} \to \mathsf{Set}$ creates limits, and the same proof can be used. The statement is so basic that you should better watch out if any high-tech answer actually already uses this statement in its proof or in the proofs of the results that are used.
I think the following related example should answer in particular question 3, because it makes visible what is responsible for the limit creation, namely a second functor.
Let $T : \mathcal{A} \to \mathcal{C}$, $S : \mathcal{A} \to \mathcal{C}$ be two functors. Consider the inserter category $\mathrm{Ins}(T,S)$ with object class $\{(A,h) : A \in \mathcal{A}, h : T(A) \to S(A)\}$ and the evident morphisms. We have the forgetful functor $E : \mathrm{Ins}(T,S) \to \mathcal{A}$.
Lemma. If $S$ preserves limits, then $E$ creates limits.
In particular, when $T : \mathcal{C} \to \mathcal{C}$ is an endofunctor, then the forgetful functor $\mathrm{Alg}(T) = \mathrm{Ins}(T,\mathrm{id}_{\mathcal{C}}) \to \mathcal{C}$ creates limits.
When $\mathbf{T}=(T,\eta,\mu)$ is a monad, then the inclusion $\mathrm{Alg}(\mathbf{T}) \hookrightarrow \mathrm{Alg}(T)$ creates limits as well. Hence, $\mathrm{Alg}(\mathbf{T}) \to \mathcal{C}$ creates limits.
From an abstract point of view, the reason is that the monad $T$ always preserves any limits that exist colaxly and colax preservation is what is required.
(This answer is closely related to Peter's answer, but describes some published results on the topic.)
The following is Proposition 4.11 of Limits for Lax morphisms by Steve Lack.
If $T$ is a $2$-monad on a $2$-category $C$ then the forgetful $2$-functor $U:T-Alg_c \to C$ from strict algebras and colax morphisms to the base creates lax limits.
Note the switch between lax limits and colax morphisms. The sense of creation is that the projections from the limit should be strict maps.
The Eilenberg-Moore object of a monad or the category of algebras for a pointed endofunctor are both examples of lax limits.
One can take $T$ the $2$-monad for categories with a class $D$ of limits and the result then applies to your setting.
Another instance would be to take as $T$ the $2$-monad for monoidal categories. Then the result becomes that if you have an opmonoidal monad on a category, it lifts to a monoidal structure on the category of algebras.
Let's see if I've got this -- Let $\mathbb T: Cat \to Cat$ be the 2-monad for categories with $D$-limits. Then $\mathbb T-Alg_c$ is the 2-category of categories with $D$-limits and all functors between them. By Lack's result the forgetful functor $\mathbb T-Alg_c \to Cat$ creates lax limits. In particular, a monad $T: C \to C$ in $\mathbb T-Alg_c$ is the same as a monad in $Cat$ whose underlying category $C$ has $D$-limits. So Lack's result says that the EM object for $T$ in $\mathbb T-Alg_c$ is the usual category of algebras $C^T$ as in $Cat$. In particular, $C^T$ has $D$-limits. Neat!
Of course, this is a bit weaker than than the result I asked about, in a few respects: (1) we get that the existence of certain limits in $C^T$, but not the full strength of "creating a limit" (2) we need to assume that $C$ has all $D$-limits, not just the limit of a particular diagram (3) we don't get the case of algebras for an endofunctor. I think (1) is not a big deal since one can show separately that the forgethful functor preserves limits and is conservative. (2) and (3) are more significant weakenings. Nevertheless, it's really great to get this far at such an abstract level!
Yes, that's right. With regards (3), it does cover the case of an endofunctor too -- that's the lax limit of an endomorphism. I forgot to mention this. I agree it doesn't answer your (2). With regards (1) one can certainly capture creation in the 2-monad sense too though, much as for (2), it won't concern an "individual limit".
An especially cute thing about this is that it's an example of the microcosm principle: we show that categories of algebras for 1-monads (i.e. monads in 2-categories) create limits by showing that 2-categories of algebras for 2-monads (i.e. monads on 2-categories, i.e. monads in 3-categories) create limits.
You can deduce the corresponding result for limits of a particular diagram using the Yoneda embedding, which preserves and reflects all existing limits. A monad $T$ on $C$ induces by left Kan extension a monad $\hat{T}$ on the presheaf category $\hat{C}$, and the latter is complete. Thus the category of $\hat{T}$-algebras has all limits created in $\hat{C}$, and a $\hat{T}$-algebra is a $T$-algebra precisely when its underlying object in $\hat{C}$ is representable. See Prop. 5.6 of https://arxiv.org/abs/1104.2111 for a similar argument.
I can't resist mentioning that another cute application of this theorem is a proof of the Knaster-Tarski fixed point theorem. The proof I have in mind proceeds in two steps, first viewing complete lattices as inf-complete, and then viewing them as sup-complete.
This is just to flesh out the approach using inserters and equifiers discussed in the answers, in a way that doesn't quite go down to the level of diagram chasing. I fear, however, that some things implicit in this argument woule require a diagram chase to carefully check.
Also, there's something I still find mysterious: what is it about inserters and equifiers which makes it so there is a particular leg of the limit cone and particular elements of the diagram such that certain hypotheses on these elements in the diagram ensure the forgetful functor down the special leg creates limits? To put a finer point on it: can we give a better description of the class of restriction functors among limits which (under certain partial limit preservation conditions) create limits? A better description than "whatever can be built up from inserters and equifiers"?
Lemma: Let $F,G: C\rightrightarrows D$ be functors, and let $Ins(F,G)$ be the inserter. Then the forgetful functor $Ins(F,G) \to C$ creates any limits that $G$ preserves.
For the proof, note that in in general, if $(c',\phi'), (c,\phi) \in Ins(F,G)$, then
$$ Hom((c',\phi'), (c,\phi)) = Hom(c',c) \times_{Hom(Fc',Gc)^2} Hom(Fc',Gc)$$
where the pullback is over the diagonal map.
Proof: Consider a diagram $(c_i, F(c_i) \xrightarrow {\phi_i} G(c_i))_{i \in I}$ in $Ins(F,G)$ such that $G(\varprojlim_i c_i) = \varprojlim_i G(c_i)$. Then $(F(\varprojlim_i c_i) \to F(c_i) \xrightarrow{\phi_i} G(c_i))_{i \in I}$ is a cone, and so induces a map $\phi: F(\varprojlim_i c_i) \to \varprojlim_i G(c_i) = G(\varprojlim_i c_i)$. We claim that $(\varprojlim_i c_i , \phi)$ is a limit of our diagram. Indeed,
$$Hom((c',\phi'), (c,\phi)) = Hom(c',c) \times_{Hom(Fc',Gc)^2} Hom(Fc',Gc) \\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \varprojlim_i Hom(c',c_i) \times_{\varprojlim_i Hom(Fc',Gc_i)^2} \varprojlim_i Hom(Fc',Gc_i) \\
\qquad \qquad \qquad \qquad \qquad \qquad = \varprojlim_i (Hom(c',c_i) \times_{Hom(Fc',Gc_i)^2} Hom(Fc',Gc_i)) \\
\qquad \quad = \varprojlim_i Hom((c',\phi'),(c_i,\phi_i))$$
where we have used that limits commute with limits.
Lemma: Let $\phi,\psi: F \rightrightarrows G : C \rightrightarrows D$ be a diagram of categories, and let $Eq(\phi,\psi)$ be its equifier. Then the full subcategory inclusion $Eq(\phi,\psi) \to C$ is closed under any limits preserved by $G$.
Proof: This boils down to the limit of equal morphisms being equal.
Well, PIE limits are well studied and have nice characterizations as in Bourke and Garner’s paper on flexible, semi flexible, and pie. However I’m not sure which legs of PIE limits are supposed to create limits...certainly one needs all the legs for products. More familiar is a result that flips roles around-the forgetful functor out of algebras for a 2-monad, with the pseudo morphisms, itself creates PIE limits!
@KevinCarlson Yeah, this business about assuming certain parts of the diagram preserve the limit is hard to put in a larger context. In the paper of Lack mentioned by john in his answer above, this pattern is repeated: he shows that $U: T-Alg_c \to C$ creates any comma objects where the appropriate leg is strong, etc. Maybe there's something to say about weighted $\mathcal F$-enriched limits... BTW does the forgetful functor from algebras for a 2-monad create even all flexible limits?
Steve and I characterized the limits that lift to $\mathcal{F}$-categories of algebras and pseudo/lax/colax morphisms for a 2-monad in https://arxiv.org/abs/1104.2111 -- which was in fact the original purpose of introducing $\mathcal{F}$-categories.
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2025-03-21T14:48:29.890145
| 2020-02-17T07:01:21 |
352910
|
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|
Stack Exchange
|
Effect of removing an edge on Hadwiger number
If $G=(V,E)$ is a finite, simple, undirected graph, then by $\eta(G)$ we denote the maximum integer $n\in \mathbb{N}$ such that $K_n$ is a minor of $G$. If $e\in E$ we write $G\setminus e$ to denote the graph $(V, E \setminus \{e\})$.
Is there a finite graph $G=(V,E)$ and $e\in E$ such that $\eta(G\setminus e) < \eta(G)-1$?
No, there is no such graph. Suppose $\eta(G)=n$. Let $T_1, \dots, T_n$ be a collection of vertex disjoint trees in $G$ such that for all distinct $i,j \in [n]$, there is an edge $e(ij) \in E(G)$ between $T_i$ and $T_j$. Consider an arbitrary edge $e \in E(G)$. If $e=e(ij)$ for some $i,j$, then removing $T_i$ (or $T_j$) yields a model of $K_{n-1}$ in $G \setminus e$. If $e \in E(T_i)$ for some $i$, then removing $T_i$ yields a model of $K_{n-1}$ in $G \setminus e$. Otherwise, $\eta(G \setminus e) = n$.
Thanks Tony for this great argument!
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2025-03-21T14:48:29.890351
| 2020-02-17T12:07:05 |
352919
|
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|
Stack Exchange
|
Birational model of a log smooth pair
Given a log smooth pair $(X,B)$ with a reduced boundary divisor $B$, consider a birational model $\pi:X' \to X$ and a boundary divisor $B'$ which is given by $K_{X'}+B'=\pi^*(K_X+B)$. Here is my question:
(1) can we construct an example that $(X',B')$ is not sub-dlt?
(2) is there an lc center of $(X',B')$ which is not an irreducible component of an intersection locus of some certain components of $B'^{=1}$ (this means the coefficient one part of $B'$)?
What about $\mathbb{A}^2$ with the two lines, and then you blow-up $(x,y^n)$? So I think the answer to the first one is yes. The second answer should be no, the morphism $X'\rightarrow X$ should be toroidal around the divisors with coefficient one.
The second answer should be no, the morphism $X′\to X$ should be toroidal around the divisors with coefficient one. Thank you! This is exactly what I need. Could you please offer a reference or something? The term "toroidal" is really confusing to me..
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2025-03-21T14:48:29.890454
| 2020-02-17T12:28:29 |
352920
|
{
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"Eduardo Longa",
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"url": "https://mathoverflow.net/questions/352920"
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Stack Exchange
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Minimizing area in relative homology class
A well known result in geometric measure theory asserts that if $(M^{n+1}, g)$ is a closed Riemannian manifold and $\alpha \in H_n(M)$ is a nonzero homology class, then there exists a closed embedded minimal hypersurface $\Sigma$ (smooth outside a singular set of dimension less than or equal to $n-7$) minimizing area in $[\Sigma]=\alpha$.
My question is if this result can be (or has been) generalized for manifolds with nonempty boundary. Precisely, I wonder if the following is true:
Let $(M^{n+1}, g)$ be a compact Riemannian manifold with nonempty boundary. Given $\alpha \in H_n(M, \partial M)$ a nonzero relative homology class, there exists an embedded and free boundary minimal hypersurface $\Sigma$ (smooth outside a singular set of dimension less than or equal to $n-7$) minimizing area in $[\Sigma]=\alpha$.
Yes, it works: double your manifold across the boundary and observe that the classical argument works equivariantly, with respect to the natural involution of the double.
Will the result surface be free boundary?
Doubling a manifold with boundary results in a manifold without boundary. Thus, the double of $\Sigma$ has no boundary.
Yes, I know that. What I meant was: apply the procedure for the double and obtain a hypersurface $\Sigma’ \subset D(M)$. Now let $\Sigma_0 = \Sigma’ \cap M$. Is $\Sigma_0$ minimal free boundary inside $M$?
Yes: Compare it with a competitor $\Sigma_1\subset M$. Then $Area(\Sigma_1)=Area(D\Sigma_1)/2\ge Area(D\Sigma_0)/2=Area(\Sigma_0)$, where $D\sigma_i$ is the double of $\Sigma_i$ inside $DM$, the double of $M$.
Another question: is the minimizer surface stable as a free boundary minimal surface?
Well, since it is an area-minimizer in its homology class....
I realized that the boundary of a minimizer may be empty, so it is not a free boundary hypersurface.
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2025-03-21T14:48:29.890601
| 2020-02-17T12:48:18 |
352921
|
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Stack Exchange
|
Approximating ring maps of finite Tor-dimension
Let $R$ be a commutative ring, and let $S$ be a finitely presented $R$-algebra of finite Tor-dimension over $R$. Can $R \to S$ be realized as the base change, along some ring map $R_0 \to R$, of a finitely presented ring map $R_0 \to S_0$ of finite Tor-dimension with $R_0$ Noetherian? Can we furthermore arrange that $R_0$ is actually of finite type over $\mathbf{Z}$?
Coming directly from the facebook page "Derived Memes for Spectral Schemes", reading that comment under a commutative algebra question was an absolutely fabulous moment.
Thanks, Riza! That's exactly what I was looking for.
I'm sorry for my incompetence but, could somebody translate that Proposition in Jacob Lurie's book to the language of commutative algebra (that is, what says that proposition for the case of commutative rings).
First of all, I strongly suspect that the answer to the question is: no.
Secondly, I want to give an example to show that the answer by Riza Hawkeye is incorrect as it stands (see also the comment by Denis Nardin for why the result is quoted incorrectly from Lurie). Namely, consider the system of rings
$$
A_0 \to A_1 \to A_2 \to A_3 \to \ldots
$$
where $A_i = \mathbf{Z}[x, z_i]/(xz_i)$ and where the maps send $x$ to $x$ and $z_i$ to zero. Then we see that the colimit of the system is $A = \mathbf{Z}[x]$. Now set $B_0 = A_0/xA_0$. Clearly, we see that $B_i = B_0 \otimes_{A_0} A_i$ is equal to $A_i/xA_i$ which has infinite tor dimension over $A_i$ for all $i$. On the other hand, we have $B = A/xA$ and this has tor dimension $1$ as a module over $A = \mathbf{Z}[x]$.
@Thank you, Johan and Denis Nardin. I understood from the comment by Denis that the case n=0 holds. Is there any reference from that case ($B$ flat over $A$)? It is difficult for me to understand Lurie's proof without understanding the language of E$_\infty$-rings, mainly lemma <IP_ADDRESS>.
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2025-03-21T14:48:29.890780
| 2020-02-17T13:28:14 |
352923
|
{
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"Kimball",
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|
Stack Exchange
|
Subquotient of principal series
Let $F$ be a local field of characteristic 0.
I am wondering whether an unramified principal series representation of $\operatorname{GL}_n(F)$ can have 1-dimensional quotient when $n>1$.
In some paper, the author claims that it can’t.
Do you know the reason?
This is true for the $\mathrm{GL}_2(F)$ principal series containing the special representation as a subrepresentation.
@Peter, How about unramified principal series representation of $GL_n(F)$ for $n>1$? Is this still don’t have 1-dimesional reps as a subquotient?
I just answered the case $n=2$. I don’t know the answer off the top of my head for $n>2$.
@Peter, Even in when n=2 case, unramified principal series representation of () can have 1-dimensional representation in some case? It’s so strange!
Look up the Steinberg representation
@PeterHumphries It's the same thing for general $n$---for the analogous induction of powers of absolute values from the Borel, one irreducible subquotient is Steinberg and one is trivial. The trivial one is of course the unique unramified subquotient of the corresponding unramified reducible principal series.
@Kimball, Oh! Thank you very much! Every unramified principal series representation has at most 1-dimensional subquotient and that should the trivial representation. Thank you very much!
@Monty no, it's not necessarily trivial---you can also get unramified twists.
Why [automorphic-forms]?
Think of $Ind_B^G 1$ as smooth functions on $G/B$ and look at the subspace of constant functions.
The constant function forms a 1-dimensional subrepresentation of $Ind_B^G 1$. But how about quotients? Does it can have 1-dimensional quotients?
Look at the contragriedient representation, it will be unramified principal series and ti will have trivial rep as a quotient.
@user152491, perhaps it would be worthwhile to enlarge your answer to include the point about taking contragredients...
@Monty, your question currently asks about subquotients, so it's probably best to edit it if you only want to discuss quotients.
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2025-03-21T14:48:29.890940
| 2020-02-17T13:33:53 |
352924
|
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"Flyingpanda",
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|
Stack Exchange
|
Log canonical centers of toric (and toroidal) varieties
Q1: Let $(X,B)$ be a toric variety. There exists a toric resolution of singularities $f:(Y,E) \to (X,B)$. Here is my question:
Is any lc center of $(X,B)$ an irreducible component of an intersection locus of some certain components of $B$?
Q2: Consider a toroidal variety $(X,B)$ (for definition, see [AK] [Kawamata])
https://arxiv.org/pdf/alg-geom/9707012.pdf
and https://arxiv.org/pdf/1008.1489.pdf
if $(X,B)$ is quasi-smooth (i.e.each cone $\sigma_x$ is simplicial, or each local toric model has only abelian quotient singularities.), then $X$ is $\mathbb{Q}$-factorial klt. Here is my question:
(1) according to [AK, pg.5], $(X,B)$ has a stratification by intersections of components of $B$. Does it imply all lc centers are strata of this structure?
(2) since each component of $B$ is required to be normal, how far is "quasi-smooth" from "dlt"?
For log toric pairs dlt implies log smooth. $\mathbb{Q}$-factorial and dlt just implies finite quotient of a log smooth point. The answer of $(1)$ is yes.
Thank you Joaquin! Where can I find a reference of (1)? So the singularities of a toric variety as a log pair are always like "intersecting not transversally" of boundary divisors?
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2025-03-21T14:48:29.891043
| 2020-02-17T14:02:27 |
352925
|
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"Ben McKay",
"https://mathoverflow.net/users/13268"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352925"
}
|
Stack Exchange
|
Applying analytic coordinate changes to singular function germs
Suppose we are given a function germ
\begin{align}
f = \sum a_{ijk}x^iy^jz^k
\end{align}
such that $f\in \mathfrak{m}^2$, where $\mathfrak{m}$ is the ideal in $\mathbb{C}\{x,y,z\}$ of holomorphic functions vanishing at 0.
I am currently reading an expository text on the du Val surface singularities in which the author sometimes simplifies a germ like this by using analytic coordinate changes. Stuff like: "If the 2-jet of $f$ is $x^2$, an analytic coordinate change can be used to remove any further appearances of $x$ in $f$" or "if $J_2(f)=x^2 + y^2$ then the existence of an $a_{ijk}\neq 0$ with $i + j <2$ implies that at least one term of the form $z^m$ or $xz^m$ or $yz^m$ appears in $f$, and then an analytic coordinate change can be used to make $f = x^2 + y^2 + z^{n+1}$."
I don't really understand which coordinate changes are applied here. Does anyone have a reference explaining techniques like this in further detail?
Analytic change of coordinates in $(\mathbb{C}^3,0)$ must take $0$ to $0$. So they are induced by (local)
$\mathbb{C}$-automorphisms of $\mathcal{O}_{\mathbb{C}^3,0}=\mathbb{C}\{x,y,z\}$ taking $\mathfrak{m}=(x,y,z)$ to itself. It is not difficult to check that those maps are always of type $$x=a_{1,1}x_1+a_{2,1}y_1+a_{3,1}z_1+\ldots$$
$$y=a_{1,2}x_1+a_{2,2}y_1+a_{3,2}z_1+\ldots$$
$$z=a_{1,3}x_1+a_{2,3}y_1+a_{3,3}z_1+\ldots$$ where $A=(a_{i,j})\in GL_3$ and stand $\ldots$ for 'higher order terms'.
Being very practical, in each case you should express a given function $f(x,y,z)$ in the new coordinates $(x_1,y_1,z_1)$ in a way that simplify its form. Most of the times it reduces to a linear algebra question...
Of course this also works for every dimension $n$.
I don't think that your observation helps with any of the particular examples that were presented in the question.
|
2025-03-21T14:48:29.891181
| 2020-02-17T14:46:22 |
352928
|
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|
Stack Exchange
|
Measure invariant under circle maps
Consider continuous bijections (may even assume these are homeomorphisms or diffeomorphisms if it helps) from the circle onto itself given by $x \mapsto x + s_i(x)$ where $i = 1,2$ or $3$. (I'm identifying the circle with $\mathbb{R}/\mathbb{Z}$.)
The basic question would be, is there an measure which is invariant under these bijections? (i.e. under the monoid generated by these maps)
I'm fairly sure the answer is negative most of the time:
take $s_1$ to be an irrational constant and $s_2$ not to be constant (the only invariant measure for $s_1$ is the uniform measure).
But there are also conditions which would ensure the existence of such a measure, e.g.:
if $s_1$ and $s_2$ are both rational constants and $s_3$ is arbitrary [this boils down to the fact that a finite extension of $\mathbb{Z}$ is amenable]
another example would be that $s_2$ and $s_3$ already belong to the monoid generated by $s_1$.
Question: Are there other conditions which ensure the existence of an invariant measure (on the circle)? [Or is the action of these monoids almost never amenable?]
As a side note, the $s_i$ in my problem are such that, for all $x$, $s_1(x) < s_2(x) < s_3(x)$ and $s_i(x) \in [a_i,b_i]$ where $a_i >0$ and $b_i < \tfrac{1}{2}$
A continuous permutation of a compact space is a self-homeomorphism, so this is just a question about subgroups of $\mathrm{Homeo}(R/Z)$ generated by three elements. Subgroups of $\mathrm{Homeo}(R/Z)$ preserving a Borel probability measure (I assume this is what you call measure, otherwise the counting measure is invariant) are well-understood and very restricted (I'd need to track the precise statement).
Some information can be found in Navas' survey: Groups of circle diffeomorphisms. For instance, see Theorem 2.3.2, Exercise 2.3.11, Corollary 3.2.6.
@Ycor Indeed, I meant probability measure. Many thanks for the reference(s). It seems that (up to conjugacy) the options I listed cover already most of the (simple) positive cases.
Probably rather up to semi-conjugacy. Indeed, let $f$ be a homeo with no fixed point, with some open interval $I$ such that the $f^n(I)$ are pairwise disjoint. Then the whole wreath product $G=\mathrm{Homeo}^+(I)\rtimes \langle f\rangle$ acts with an invariant proba (supported by the complement of $\bigcup_{n\in\mathbf{Z}} f^n(I)$) and $G$ is not amenable (it has free subgroups).
|
2025-03-21T14:48:29.891364
| 2020-02-17T14:47:36 |
352929
|
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|
Stack Exchange
|
What's the essential definition of resonance of Schrodinger operator?
Rencently, I am reading some articles about time decay estimates or Strichartz estimates for Schrodinger equations with potential. When considering Strichartz estimates for potential $V$ with decay $|x|^{-2-\varepsilon}$, Rodnianski & Schlag said in Time decay for solutions of Schrödinger equations with rough and time-dependent potentials that the Strichartz estimte (nonendpoint case) holds if we assume the initial data $u_0 \in P_{[0,\infty)} X$, where $X$ is the total space. Moreover, in view of $0$ will contribute negative effection on the esimates, we should assume that $0$ is neither eigenvalue nor the resonance of $H=-\Delta+V$.
I am wondering what's the precise definition of resonance? Because I have seen at least three definitions of it, but what definition is the most essential?
definition (version 1: Rodnianski & Schalg: Time decay for solutions of Schrödinger equations with rough and time-dependent potentials) A resonance is a distribution solution to $H \psi=0$ so that $\psi \notin L^2$ but $(1+|x|^2)^{-\frac{\sigma}{2}} \psi(x) \in L^2$ for any $\sigma>\frac{1}{2}$.
definition (version 2: Artzi & Klainerman: Decay and regularity for the Schrödinger equation ) zero is said to be a resonance of $H$ if for every $\varepsilon >0$ there exists a $0 \not= \psi \in H^{2,-1-\varepsilon}$ such taht $H\psi=0$.
definition (version 3: Nakanish & Schalg: Invariant manifolds and Dipersive Hamiltonian Evolution Equations) in odd dimensions, the reslovent $(H-z^2)^{-1}$ can be expanded by $(H-z^2)^{-1}=B_{-2}z^{-2}+B_{-1}z^{-1}+B_0+B_1z +... B_k z^k +R_k(z)$, where the expansion is in $L^{2,-1/2-\varepsilon} \to L^{2,1/2+\varepsilon}$ sense, we say $0$ is neither eigenvalue nor the resonance of $H$ iff $B_{-2}=B_{-1}=0$.
and so on...
I just know the aim of ruling out the resonance is to ensure the "good" behavior of $(H-z)^{-1}$ near $z=0$.
However, are the three definitions of resonance above equivalent to each other? Especially for definition 2 and definition 1. I think the definition 2 is a little peculiar, but Rodnianski and Schalg： "time decay estimates..." cited the results in Artzi & Klainermann: "decay and regularity..." to support their result. But at least we note that $\text{definition 1} \subset \text{definition 2}$, thus $\text{(definition 2)}^c \subset \text{(definition 1)}^c$. So we can see that if $0$ is not resonance in def 1, the conclusion that $0$ is not resonance in the sense of def 2 will not hold necessarily.
I am struggled about it, if someone has known something about it, please tell me, thank you!
|
2025-03-21T14:48:29.891544
| 2020-02-17T15:24:13 |
352930
|
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|
Stack Exchange
|
Flatness directions of the operator norm
It is known that the standard operator norm $\|\cdot\|_2$ over ${\bf M}_n({\mathbb R})$ is very flat, as is any operator norm (= subordinated norm) actually. The set of extremal points of the unit ball is the very small set ${\bf O}_n({\mathbb R})$ and the unit sphere contains faces (convex subsets) of dimension $(n-1)^2$.
I discovered that the directions of flatness actually are singular matrices: if a segment $[M,N]$ is included in the unit sphere, then $\det(M-N)=0$. I hardly pretend that this is an original result.
Is there any reference in the literature for the above result ?
I should be also interested in any related statement for other operator norms.
Notation. I use to write $\|\cdot\|_p$ for the norm over ${\bf M}_n({\mathbb R})$ induced by the $\ell^p$ norm over ${\mathbb R}^n$.
(You write $\|\cdot\|_2$ but say "operator norm", and I assume that is what you mean.)
I haven't seen this exact statement, but it follows easily from known facts about the facial structure of operator unit balls. The basic reference is Akemann and Pedersen, Facial structure in operator algebra theory, Proc. London Math. Soc. 64 (1992), 418–448.
Akemann and Pedersen show that the weakly closed faces of the unit ball of a von Neumann algebra are precisely the sets of the form $v + (1 - vv^*)B(1 - v^*v)$ where $B$ is the full unit ball and $v$ is a partial isometry. In finite dimensions all faces are weakly closed, of course. The difference of two elements of such a face would have the form $(1 - vv^*)x(1 - v^*v) = pxq$ where $p$ and $q$ are orthogonal projections. That can be nonsingular only if $p = q = 1$, which would imply $v = 0$, making the "face" be all of $B$ (if you consider that to be a face).
Your result is stated for real matrices, but any face of the unit ball of $M_n(\mathbb{R})$ is contained in a face of the unit ball of $M_n(\mathbb{C})$, so nothing more needs to be said.
|
2025-03-21T14:48:29.891700
| 2020-02-17T16:03:24 |
352931
|
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"authors": [
"Brendan McKay",
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|
Stack Exchange
|
About a generalization of complete graphs
Does anyone know what are called (if there is any nomenclature for this class of graphs in the literature) the connected graphs such that each of their edge belongs to some triangle? For example, every complete graph with more than two vertices satisfies this property. Is there any reference to the subject?
triangle-covered
I've found this in the literature: "The k-truss of G is the largest subgraph of G in which every edge is contained in at least (k-2) triangles within the subgraph.". See:
https://arxiv.org/pdf/1205.6693.pdf
http://sariyuce.com/sem/NSATR08.pdf
So your graph needs to be a connected 3-truss.
|
2025-03-21T14:48:29.891778
| 2020-02-17T16:45:20 |
352934
|
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|
Stack Exchange
|
Is this integral finite and how does it decay to zero?
I would like to know if the following is convergent/finite (it represents a bound from a truncated Legendre series approximation)
\begin{equation}
\varepsilon_n \leq \int_{-1}^1 \left(\int_{n\gg 1}^\infty \dfrac{1}{k(1 + k^2(1-x^2))^{1/4}}\mathrm{d}k\right)^4 \mathrm{d}x.
\end{equation}
where I suspect
$$\lim_{n\to\infty} \varepsilon_n = \mathcal{O}(n^{-2})\qquad ?$$
With a change of variables $\tilde{k} = k\sqrt{1-x^2}$ we can re-arrange the inner integrand to be
\begin{equation}
\int_{\tilde{n}}^\infty \dfrac{1}{\tilde{k}(1 + \tilde{k}^2)^{1/4}}\mathrm{d}\tilde{k}.
\end{equation}
This can be integrated analytically to give the summation of an $\arctan$ and $\mathrm{arctanh}$, but Mathematica evaluates this to a nasty function which I can't seem to get anything sensible out of when using AsymptoticIntegrate.
Ultimately I'm hoping the integral is finite, and if so, then how fast does it decay to zero as $n\to \infty$.
That the inner integral is finite - doesn't it follow from the ratio test, say by comparing with $1/klog(k)$?
When $\tilde{k}$ is large, we have $1 + \tilde{k}^2 \sim \tilde{k}^2$ and so your $\tilde{k}$ integral is approximately $\int_{\tilde{n}}^\infty \tilde{k}^{-3/2},d\tilde{k} = 2\tilde{n}^{-1/2}$. Then the fourth power explains how you end up with $n^{-2}$ in the end.
@auniket The inner integral is finite a.s.
@NateEldredge $\tilde{k}$ is not large everywhere, and if I neglect the "$1 + \ldots$" then the outer integration produces a divergent integral. Hence I need to keep the small "$1+\ldots$" term to try and keep everything finite.
|
2025-03-21T14:48:29.891908
| 2020-02-17T17:07:59 |
352936
|
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|
Stack Exchange
|
formal smoothness and McQuillan formal schemes
Let $k$ be an algebraically closed field, $A\rightarrow B$ be a continuous map of weakly admissible topological $k$-local algebras.
We assume that it is formally smooth and topologically of finite type in the sense of https://stacks.math.columbia.edu/tag/0ALL
does it imply that $B\simeq A[[t_1,..,t_r]]$ for some $r$? If $A$ and $B$ are both noetherian, this is well-known, so the issue is about a generalisation in a non-noetherian setting.
|
2025-03-21T14:48:29.891966
| 2020-02-17T17:08:46 |
352937
|
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|
Stack Exchange
|
Is a $G$-bundle over $\mathbb{R}$ a $G$-fibre bundle?
Let $G$ be a Lie group with a smooth (non-transitive) action on a connected manifold $M$ (none of them need to be compact). Let further $f\in C^\infty(M,\mathbb{R})$ be $G$-invariant. Suppose that for every two open intervals $I,J\subset\mathbb{R}$ the $G$-bundles $f^{-1}(I)$ and $f^{-1}(J)$ are isomorphic, i.e., there is a map $\Phi_{I,J}\in C^\infty(f^{-1}(I),f^{-1}(J))$ which is $G$-equivariant and $f\circ\Phi_{I,J}=\phi_{I,J}\circ f$ for a map $\phi_{I,J}\in C^\infty(I,J)$.
Qustion: Is it true that $M\overset{f}\to\mathbb{R}$ is isomorphic to the trivial $G$-bundle $\mathbb{R}\times f^{-1}(0)$? Or does there exist a map $\Phi\in C^\infty(M,f^{-1}(0))$ which is $G$-equivariant and $f\circ\Phi=1_\mathbb{R}$?
Discussion: Since every locally trivial (fibre) bundle over $\mathbb{R}$ is trivial, it is enough to find only a local trivialization.
It is hard to check if the $G$-action on $M$ is proper, so the orbit structure of an invariant leaf $f^{-1}(t)$ may not be very neat. But it is the same for all such leaves, and the question basically boils down to finding a smooth section in the "orbit space bundle", which need not be a smooth bundle? This becomes too subtle, and I hope that there are higher level arguments to show this.
If $M$ admits a smooth cross section to the $G$-action, that is, a submanifold $S\subset M$ such that $\{gS\}_{g\in G}$ is a partition of $M$, and if $S\overset{f}\to\mathbb{R}$ is a trivial bundle then I think that the answer is positive. But I am not sure if this would be too optimistic.
Thank you.
|
2025-03-21T14:48:29.892083
| 2020-02-17T17:49:27 |
352940
|
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|
Stack Exchange
|
Intuition about how Voronoi formulas change lengths of sums
In reading the literature one encounters countless examples of Voronoi formulas, i.e., formulas that take a sum over Fourier coefficients, twisted by some character, and controlled by some suitable test function, and spits out a different sum over the same Fourier coefficients, twisted by some different characters, and this time controlled by some integral transform of the test function.
The reason one wants to do this in practice is that the second sum is somehow better, of course, which in my (admittedly limited) experience tends to boil down to the length of the second sum having changed significantly to the better.
I'll give an example (from Xiaoqing Li's Bounds for GL(3)×GL(2) L-functions and GL(3) L-functions, because it is what I happen to have in front of me).
In this case we have the GL(3) Voronoi formula
$$
\sum_{n > 0} A(m, n) e\Bigl( \frac{n \bar d}{c} \Bigr) \psi(n) \sim \sum_{n_1 \mid c m} \sum_{n_2 > 0} \frac{A(n_2, n_1)}{n_1 n_2} S(m d, n_2; m c n_1^{-1}) \Psi \Bigl( \frac{n_2 n_1^2}{c^3 m} \Bigr),
$$
where $\psi$ is some smooth, compactly supported test function, $\Psi$ as suggested above is some integral transform of it, $A(m, n)$ are Fourier coefficients of (in this case) an SL(3) Maass form, $(d, c) = 1$, and $d \bar d \equiv 1 \pmod{c}$.
(I've omitted lots of details here, but the details, I think, aren't relevant to my question.)
Doing so essentially transforms the $n$-sum into the $n_2$-sum, where, as is evident in the formula, the $n_2$-sum has a very different argument in its test function.
What happens in practice now is that, once we get to a point where applying the Voronoi formula is appropriate, we transform the sum and study the integral transform, chiefly by means of stationary phase analysis in order to find what length of the new $n_2$-sum is.
In the particular example at hand, this, after identifying the stationary phase and playing along, this takes us from an $n$-sum on $N \leq m^2 n \leq 2 N$, i.e., $n \sim \frac{N}{m^2}$, to an $n_2$-sum on
$$
\frac{2}{3} \frac{N^{1/2}}{n_1^2} \leq n_2 \leq 2 \frac{N^{1/2}}{n_1^2},
$$
i.e., $n_2 \sim \frac{N^{1/2}}{n_1^2}$, which then means that the arguments in the test function are now of size $\frac{N^{1/2}}{c^3 m}$.
I can go through the motions of performing this stationary phase analysis and so on, but my question is this: is there any intuition to be had about how and in what way these Voronoi formulas alter the lengths of sums?
First of all, the description of $\psi$ after the first display is confusing (assuming OP meant $\psi$ is supported around $N$, otherwise conclusion form the first display does not make sense). I went to the relevant part (end of p.318) of Li's paper and found that $\psi$ is not just any test function, it has a weight of $n^{-3/4}$, and, more importantly, it has oscillation. A necessary display of the LHS would be (taking $m=d=1$)
$$\sum_{n}A(1,n)e\left(\frac{n}{c}+2\sqrt{n}-\frac{1}{\sqrt{n}}\right)\psi(n/N),$$
where $\psi$ is a test function supported on $[1,2]$.
To have an intuition about what the length of the dual sum would be one can follow the general heuristic formula (HF) below.
$$\text{length of the dual sum} = \frac{\text{total conductor}}{\text{length of the original sum}}.$$
Here total conductor is the conductor of the oscillating object taken all the twisting into account. For example, the total conductor of $L(1/2+it,\pi'\otimes\pi)$ where $\pi$ and $\pi'$ are fixed $\mathrm{GL}(n)$ and $\mathrm{GL}(m)$ automorphic representations, is $t^{nm}$.
Conductor of $e_q(x):=e(x/q)$ is $q$. In this case the denominator in the entry of $e()$ is of size $c\sqrt{N}$. But there is twist by a $\mathrm{GL}(3)$ Hecke eigenvalue. So the total conductor is $c^3N^{3/2}$. Applying the above formula one obtains the length of the dual sum.
One can check that the above HF also occurs in the formula of the approximate functional equation of the central $L$-value. There are two sums in the formula corresponding to the representation and its contragredient. One can check that the product of the length of the sums equals to the conductor of the $L$-function.
(The main reason of this HF to work is the automorphy under some suitable Weyl element of the underlying automorphic form, which is, indeed, the key ingredient to prove the approximate functional equation and Voronoi formula.)
Wonderful! This is precisely the sort of heuristic I need, thank you, and thank you also for correcting and clarifying my hasty and poor presentation of the question.
|
2025-03-21T14:48:29.892493
| 2020-02-17T19:38:09 |
352944
|
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|
Stack Exchange
|
Matrix smoothly parametrized by t has eigenvalues (0, $\lambda$), eigenvector $v$. Is $\lambda v$ smooth?
Let $C(t)$ be a symmetric, two-by-two real matrix whose entries are smooth functions of $t \in \mathbb{R}$. Suppose that $C(t)$ point-wise has eigenvalues $\lambda$ and $0$. Then $\lambda(t)$ is a smooth function too (since $\lambda(t)$ is the trace of $C(t)$). However, in general the unit-length eigenvector $v(t)$ corresponding to $\lambda(t)$ is not smooth. The problematic points are wherever $\lambda(t) = 0$ where $v(t)$ is not even well-defined even up to sign.
Is $\lambda(t) v(t)$ a smooth vector field?
Since $v(t)$ is only ever defined up to sign, I really mean to ask whether there is a smooth vector field of length $\left|\lambda(t)\right|$ that is point-wise an eigenvector of $C(t)$ of eigenvalue $\lambda(t)$.
Dieci and Eirola, 1999 Theorem 3.3 implies that if $\lambda(t)$ never goes to zero to infinite order, then $v(t)$ is in fact smooth. So we are interested in the case when $\lambda(t)$ goes to zero to infinite order and are hoping that $\lambda(t)$ then goes to zero fast enough to kill off any problems occurring in $v(t)$.
$\newcommand{\la}{\lambda}
\newcommand{\R}{\mathbb{R}}$Such a smooth field $(\la(t)v(t))$ does not exist in general.
Indeed, let
$$C=\begin{bmatrix}f&fg\\fg&fg^2\end{bmatrix},$$
where $f$ and $g$ are the (nonnegative) functions defined in this answer.
The eigenvalues of $C$ are $\la:=f+fg^2$ and $0$.
The eigenvectors of $C$ belonging to the eigenvalue $\la$ that are of length $|\la|[=\la]$ are of the form $w:=hf\sqrt{1+g^2}\,(1,-g)$ for some function $h\colon\R\to\{-1,1\}$. As detailed in the mentioned answer, the first coordinate $hf\sqrt{1+g^2}$ of this vector field $w$ cannot be smooth for any choice of a $\pm$-function $h$. Therefore, the vector field $w$ cannot be smooth for any choice of $h$.
Thanks for working on this. I don't believe this answers the question I'm asking. My $C$ is symmetric and so there is no counter-example with $\lambda = 1$ a constant since in that case the Schur decomposition is smooth and the eigenvector itself is smooth (eg: Prop 2.4 in the paper I linked in the question). Also, I do not believe that your $c$ is smooth at $t = 0$ - it seems to have a $sin(1/t)/2$ term. Please correct me if I am mistaken.
@user32157 : Indeed, I missed the fact that $c$ was not continuous at $t=0$. I have now rewritten the answer completely, proving that one can choose a continuous field $(\lambda(t)v(t))$, even without assuming the matrix $C(t)$ to be symmetric. This choice seems unique, up to a global sign change. So, if there is a completely smooth version of your field, it must be this, again up to a global sign. However, to show the complete smoothness, it seems lots of calculations will be needed.
I think this is correct, but continuity of the vector field is not the problem; I'm concerned about smoothness. Even differentiability is doable (though showing that's it's even continuously differentiable has stumped me) by the last remark in my question and considering just the points where $\lambda(t)$ goes to zero to infinite order. A bounded function times a function going to zero to infinite order also goes to zero to infinite order, hence has zero derivative at that point.
Also consider whether $(a|c)$ and $(b|d)$ are the same. They're clearly parallel but could still give opposite sides, for example $a = 1, b = -1, c= -1, d=1$.
@user32157 : Thank you for your further comments. I'll think more about this.
@user32157 : Now your question has been fully answered.
Thanks, this looks good to me. Minor typo: the eigenvector should be a multiple of $(-g,1)$ instead of $(1,g)$. I feel like I'm still missing some intuition, since my instinct is to take $r = 0$ on the example and avoid the infinite sum. However doing that leaves you with a very simple matrix $\begin{pmatrix} x^4 & x^3 \ x^3 & x^2 \end{pmatrix}$ which has no problems.
@user32157 : I have corrected the typo. As for the role of $r$ in fedja's answer (and my detalization of it), it is still a bit mysterious for me too -- you might want to ask fedja about how such a phenomenon could be found.
|
2025-03-21T14:48:29.892785
| 2020-02-17T20:04:54 |
352945
|
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|
Stack Exchange
|
Eigenvalues of structured matrices
Let $A=(a_{i,j})$ be an $n\times n$ matrix with $a_{j,j+1}>0,\; 1\leq j\leq n-1,$ and $a_{j,j-2}>0,\; 3\leq j\leq n$, the rest of the entries are zeros.
Is the following fact known:
All roots of the characteristic polynomial, except possibly $0$, are simple
and their arguments are $2\pi k/3$, where $k\in\{0,1,2\}$?
(Zero occurs as a root when $n$ is not divisible by 3, and it is of multiplicity 2 when $n\equiv 2\; (\mathrm{mod}\, 3)$ and simple when $n\equiv 1\; (\mathrm{mod}\, 3)$.)
If known, what is a reference?
(A proof of the highlighted statement was communicated to me by Vitaly Tarasov on my request, but we don't know whether the fact is new.)
Edit. It seems that a straightforward generalization is true for all "2-diagonal" matrices with positive entries: one strictly positive diagonal immediately above the main, and
other strictly positive somewhere below the main, the rest are zeros.
If $p_k(x)$ is the characteristic polynomial of the upper left $k$-by-$k$ block, then cofactor expansion at the bottom right gives a recurrence for $p_k$ in terms of $p_{k-1}$ and $p_{k-3}$. Given that $p_k(x)=x^jq_k(x^3)$ for some polynomial $q_k$, there could be an argument that $q_k$ has positive real roots by showing $q_{k-1}$ and $q_{k-3}$ have interlacing roots.
And on third reflection, what I thought is relatively obvious: $A\Sigma = \Sigma A = e^{2 \pi i/3} A$, where $\Sigma$ is the clock matrix (diagonal matrix with $e^{2 \pi i k/3}$ on the diagonal), so if $\vec{v}$ is an eigenvector, then so is $\Sigma \vec{v}$, so each eigenvalue has 2 "buddies" (same magnitude, argument changed). Doesn't help with simple, or to stop other arguments from existing, sorry.
@MTyson: yes. How do you prove that all roots of $q_k$ are simple and positive? Tarasov used a Sturmian argument by induction in triples of polynomials.
With respect to your edit: have you checked whether the generalization seems to work for the 4-version (that is, 2 diagonals, one 1 above, one 3 below)?
I think an equivalent formulation without matrices is: let $a_i$ be a sequence of $n - 2$ positive numbers. Let $A_j$ be the set of increasing sequences of indices that are separated by at least $3$ (e.g. if $n = 7$, then $A_2 = {(1, 4), (1, 5), (2, 5)}$). Let $c_j$ be the sum of all products of the $a_i$ indexed by an element of $A_j$ (continuing the example, $c_2 = a_1 a_4 + a_1 a_5 + a_2 a_5$). Then the polynomial $\sum c_j x^{\lfloor \frac{n - 2}{3}\rfloor - j}$ has simple, positive roots.
@user44191: On your second comment: no, I did not check.
|
2025-03-21T14:48:29.892986
| 2020-02-17T20:18:50 |
352946
|
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|
Stack Exchange
|
Moments of a positive random variable
Suppose one is handed a list of $K$ numbers, with a claim that these numbers are the first $K$ moments of a positive random variable $X$ (meaning there is 0 probability that $X<0$).
What is the strongest possible test that one could run on this list to test this claim? (We do not know any additional information about $X$.)
The most obvious thing to check first is that all the moments are positive.
A better test would involve checking that Jensen’s inequalities are satisfied. What is the most powerful test?
In general, there is a convex "allowed region” in the $K$-dimensional space of possible moments of $X$. Is there a good way to characterize this space?
Your question is too unspecific.In which field of research do you need the answers?
What is the region for $K=2,3,4$?
I'm a bit confused -- do you want to test if it is possible for there to exist any $X$ with these moments, or are you also given a particular $X$ (say you are given i.i.d. observations) and you want to test if it has these moments?
You are aware of the k infinite versions of this problem ? EG, https://en.wikipedia.org/wiki/Stieltjes_moment_problem
@usul I am interested in the former, e.g., does there exist any positive real $X$ with these moments. I have updated the question to clarify it slightly.
@mike is the multi-variate existence version of the problem also known for infinite k?
@Seva - my answer gives what seems to be the region for $K=1,2,3,4$
This is known as the truncated Stieltjes moment problem, and there is a necessary and sufficient condition taking the form of a semidefinite program. See Section 5 of the classic paper by Curto and Fialkow.
Seva asked for the region when $K=2,3,4$
Empirically it seems
$$m_1 > 0$$
$$m_2 > m_1^2$$
$$m_3 > \dfrac{m_2^2}{m_1}$$
$$m_4 > \dfrac{m_3^2+m_2^3-2m_1 m_2 m_3}{m_2-m_1^2}$$
and while it is possible to turn one of these inequalities into an equality for a particular moment, that then fixes every higher moment, with
$m_1=0 \implies m_2=0$ and $m_3=0$ and $m_4=0$
$m_2=m_1^2 \implies m_3=m_1^3$ and $m_4=m_1^4$
$m_3=\frac{m_2^2}{m_1} \implies m_4=\frac{m_2^3}{m_1^2}$
It also seems empirically that it is possible for find an example for $X$ with the first $K$ given moments where $X$ can take $\lceil (K+1)/2\rceil$ possible non-negative values with associated probabilities (if $K$ is even then one of the values can be $0$), where this example can then give the boundary for the next higher moment. Finding the example involves solving a set of polynomial simultaneous equations.
|
2025-03-21T14:48:29.893195
| 2020-02-17T20:24:58 |
352950
|
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|
Stack Exchange
|
Approximative extension of the autohomeomorphism of the complement of the trivial knot?
Let $S^1\subset \mathbb{R}^3$ be the unit circle and suppose $h\colon \mathbb{R}^3\setminus S^1\to \mathbb{R}^3\setminus S^1$ is a homeomorphism. Clearly it might be that $h$ cannot be extended to $S^1$. But can it be "extended" in an approximative way? More precisely:
Given an $\varepsilon>0$, does there exist a homeomorphism $\hat h\colon\mathbb{R}^3\to\mathbb{R}^3$ such that $\hat h(x)=h(x)$ for all $x$ with $d(x,S^1)>\varepsilon$.
Below are my incomplete thoughts on this so far.
Here is one way I think this could be proved. Morton Brown has proved the following generalization of the Jordan-Schönflies theorem: If $f\colon S^2\times[0,1]\to S^3$ is an embedding, then the components of $S^3\setminus f[S^2\times\{\frac{1}{2}\}]$ are topological $3$-cells. SUPPOSE this result was proved for $S^2$ replaced by the standard torus, i.e. saying that "if $g\colon T^2\times [0,1]\to S^3$ is an embedding, then the complements of $g[T^2\times\{\frac{1}{2}\}]$ are solid tori". Then we can apply this theorem to the torus $\{x\mid \varepsilon/2<d(x,S^1)<2\varepsilon\}$ and conclude that the "inside" of $h[\{x\mid d(x,S^1)=\varepsilon\}]$ is homeomorphic to a torus.
BUT I haven't found this generalization of this Brown's theorem. Nor any version of Jordan-Schönflies theorem or Jordan-Brouwer Separation Theorem that I have found seems to help (at least I don't see how).
Or perhaps someone can cook-up an argument using Dehn's surgery?
"...then ONE of the complements is a standard torus", I assume you mean. (This is true smoothly, I'm sure it's true in your topologically flat setting as well; see Hatcher's 3-manifold notes for a proof in the smooth case.) One complement could be knotted.
Interesting, thanks. I will look into it. Indeed, the torus can be knotted in general (not in my case I think).
I do not assume flatness explicitly, but I suppose that by the triangulation results by Moise I can assume that the homeomorphism is piecewice linear (or smooth) (but the derivatives explode as we appriach $S^1$)
The answer to my question seems to be Exercise 3 on page 9 of Hatcher's 3-manifold notes (so thanks!) and it says that it is a theorem of Alexander without reference. Do you @MikeMiller , or someone, have reference?
I did not see originally that your circle is the unit circle, I agree both sides are unknotted in your case. For a reference, I get from this Manifold Atlas page: J. W. Alexander, On the subdivision of 3-space by polyhedron, Proc. Nat. Acad. Sci. USA, 10, (1924) 6–8.
|
2025-03-21T14:48:29.893394
| 2020-02-17T21:34:42 |
352956
|
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|
Stack Exchange
|
$S$ subring of $R$ implies $\mathrm{rank}_R M\leq \mathrm{rank}_S M$?
This is a question that I've posted in MSE but I've got no answers.
The statement is as follows:
Let $R$ be a ring with identity, and $S$ be a subring with $1_S=1_R$. Let $M$ be a free unitary $R$-module such that there exists $\mathrm{rank}_R M$ (i.e. every $R$-basis of $M$ has the same cardinality).
Is it true that $\mathrm{rank}_S M$ exists?
And, in the case that $\mathrm{rank}_S M$ exists, is it true (as in vector spaces) that $\mathrm{rank}_R M\leq \mathrm{rank}_S M$?
Any help will be appreciated!
How are you defining rank of a general module? For finitely generated free modules, the condition you describe is known as IBN (invariant basis number), that is, $R^n $ isomorphic to $R^m$ implies $m = n$, and this property is obviously preserved by taking unital subrings.
If any two R-bases for M have the same cardinality, then the rank_R of M is the number of the elements of any R-basis
First question: Probably no, consider $M=R$.
A more natural question would be: Assume that $\mathrm{rank}_R(M)$ and $\mathrm{rank}_S(R)$ exist. Does $\mathrm{rank}_S(M)$ exist and do we have $$\mathrm{rank}_S(M) = \mathrm{rank}_S(R) \cdot \mathrm{rank}_R(M)?$$I think this equation is quite straight forward once we know that $\mathrm{rank}_S(M)$ exists (i.e. all $S$-bases of $M$ have the same cardinality).
|
2025-03-21T14:48:29.893545
| 2020-02-17T22:07:01 |
352957
|
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|
Stack Exchange
|
Examples of plane algebraic curves
There are many interesting sequences of polynomials which contain
polynomials of arbitrarily high degree, for example classical
orthogonal polynomials. Most of them arise as characteristic polynomials
of some sequences of operators, or as polynomial solutions
of some differential equations.
What are some natural specific
sequences of plane (affine or projective) algebraic
curves which contain curves of arbitrarily high degree and genus?
One such example is Fermat's curves $x^n+y^n=1$. Lissajous
(a.k.a. Chebyshev)
curves are of arbitrary degree but they have zero genus. Sequences of hyperelliptic curves occur in the theory of integrable systems. What else?
I looked to the Catalog of Plane curves by D. Lawrence (Dover, 2014)
and to the book of Brieskorn and Knörrer, Plane algebraic curves,
and found only Lissajous curves, epitrochoids and
hypotrochoids (all of genus zero) as examples of arbitrarily high degree.
I understand that many examples can be constructed. But I am asking on some naturally occuring sequences, whatever it can mean. Of some historical significance or appearing in applications.
EDIT. Thanks to all who answered or commented. I am not marking this question as "answered" for a while, hoping for more examples. Of course, classical modular curves belong here, thanks to Felipe Voloch.
Let me mention my motivation for this question. For some time I am studying what can be called "Lamé modular curves" (surprisingly, there is no established name for them). Lamé functions
are solutions of Lamé's differential equation whose squares are polynomials. Existence of
such a solution imposes a polynomial equation connecting the modulus of the
torus $J$ and an "accessory parameter". These polynomials define a family of plane affine algebraic curves
which contains curves of arbitrary degree and genus, and their coefficients are integers.
I’m ashamed to say, I don’t know if the nth caustic by reflection (of rays through a point) at a circle (Holditch 1858; pictures) is algebraic. The 1st is, by St-Laurent (1826, p. 134) or Cayley (1857, pp. 353-355) who also cites a paper of Lagrange I was never able to find. There may be something to Thm 6 of Glaeser-Schröcker (2000).
@vrz: frankly speaking, I don't know.
There will probably be many examples from combinatorics, e. g. Tutte Polynomials (https://en.m.wikipedia.org/wiki/Tutte_polynomial). I do not know/have not tried to prove that there are Tutte Polynomials with arbitrarily high genus, but seems entirely plausible.
@auniket: do zeros of Tutte polynomials have any significance?
@AlexandreEremenko: You got me there - I don't know.
Not sure why modular curves were edited out. The modular curves $X_0(n)$ have a natural plane model. https://en.wikipedia.org/wiki/Classical_modular_curve
And the modular curve $X_1(n)$ for $n\ge$ has an even more elementary description. Let $E_{a,b}:y^2+axy+by=x^3+bx^2$. Write down the condition for the point $P=(0,0)\in E_{a,b}$ to have exact order $n$. This gives a polynomial equation $P_n(a,b)=0$ that is an affine model for $X_1(n)$, albeit with some singular points corresponding to $(a,b)$ values at which $E_{a,b}$ is singular.
Another slightly silly example is to fix a polynomial $f(X)$ having at least two roots, and look at the sequence of curves $C_n: Y^n=f(X)$. Lang called these superelliptic curves. The degree and genus go to $\infty$, although the gonality never exceeds $\deg(f)$, since $(X,Y)\to Y$ is a map to $\mathbb P^1$ of degree at most $\deg(f)$. As a place they might come up, I think the $abc$ conjecture (over number fields) impiies that for all sufficiently large $n$, the set $C_n(\mathbb Q)$ contains only points with $Y\in{0,\pm1}$.
Even though it doesn't really answer the question because not all curves there are algebraic, and there are few families of arbitrarily high degree, I think the web site mathcurve.com (in French) deserves at least a mention here, if only because it certainly matches the title of the question.
How about the affine plane curves $\Phi_n(c,t)=0$ that classify $(c,t)$ such that $t$ is a point of exact period $n$ under iteration of the quadratic map $f_c(X)=X^2+c$? These are often called dynatomic curves and have been much studied in recent years, especially since describing their rational points is related to the dynamnical uniform boundedness conjecture. These curves are irreducible (Bousch) and there is a nice formula for their genus (Morton) showing the genus goes to infinity. There is even some work (Poonen, Doyle, ...) showing that the gonality also grows. For the basic construction, you can see for example Sections 4.1 and 4.2 of my book The Arithmetic of Dynamical Systems. More generally, people study the dynatomic curves for $X^d+c$.
(I've cheated a little, one needs to include a few extra points on the curve where the point $t$ has "formal period $n$," but actual period smaller than $n$. This is Milnor's terminology.)
Thanks. This example I should have known:-)
The caustic by n-th order reflection from the circle was advanced by Francois Ziegler in a comment. It is indeed known to be algebraic. As pointed out, the caustic curve of n-th order reflection from arbitrary point sources (including at infinity) was derived by Holditch "On the nth Caustic, by Reflexion from a Circle", The Quarterly Journal of Pure and Applied Mathematics, vol. 2, London, 1858, pp. 301–322. This paper does include a proof that his class of curves is indeed algebraic (see p. 322 section "The Equation").
Sadly this contribution is somewhat underappreciated/overlooked leading to rediscoveries of partial results later on. For example, the case of parallel light rays (source at infinity) and the point source of the light rays on the circle for arbitrary order of reflections has been rediscovered and shown to be algebraic by Bromwich "The Caustic, by Reflection, of a Circle." American Journal of Mathematics, 1904, Vol 26, 33-44. Specifically pp. 43-44. As Bromwich points out these cases are equivalent to epitrochoids with given radius relations.
As a word of caution regarding the naturality of the Holditch caustic. Rays reflect at different lengths as order increases. This introduces a discrepancy in order between rays in the ray bundle. So the equality of order in Holditch's derivation is not physical if one accounts for traveling distance (say via finite traveling speed). Hence the nth order reflection curve according to Holditch has to be broken down into different order segments to achieve a physical caustic. In short, the Holditch caustics contain all the information needed to recover the physical phenomena, but there is a need for accounting for reflection order discrepancies (see Essl "Computation of wave fronts on a disk I: numerical experiments." Electronic Notes in Theoretical Computer Science 161 (2006): 25-41.)
Given any algebraic curve as reflector, Josse and Pene ("On the degree of caustics by reflection." Communications in Algebra 42.6 (2014): 2442-2475.) give the order of the caustic by reflection being an algebraic curve. This gives a different handle on the order of the algebraic curve. While the order of the Holditch caustic is directly related to the order of reflection, here it enters as the order of the reflector.
Has anyone found the genus of this caustic ? From the abstract of your last reference I conclude that it is a rational curve for the case of circle. So it does not satisfy one of my criteria.
Ah yes, I'm sorry. Holditch already proved his caustic to be rational. This is not an example. There is another reason why I thought this might be applicable (from someone who is interested in separable PDEs). The Holditch caustic relates to separability in polar coordinates (it describes the singularities of the characteristic cone under reflection of a linear hyperbolic PDE and the separated ODE of interest being Bessel). The Lame ODE springs from separation of ellipsoidal coordinates. But this probably is way too loose an association to be of interest.
What you call Fermat curves, are also referred to as Lamé curves, since Lamé proposed these as tools for physics, in particular crystals. I started using them in the 1990s to describe the shapes of plants. Recently we tested this on many botanical samples, see e.g.
Huang, Weiwei; Li, Yueyi; Niklas, Karl J.; Gielis, Johan; Ding, Yongyan; Cao, Li; Shi, Peijian, A Superellipse with Deformation and Its Application in Describing the Cross-Sectional Shapes of a Square Bamboo, Symmetry 12 no. 12 (2020) 2073 https://doi.org/10.3390/sym12122073
Yours,
Johan
I doubt this is what you seek, but the minimal polynomial for a packing of $n$
congruent disks in a square can have arbitrarily high degree:
Szabó, Péter Gábor, Mihály Csaba Markót, and Tibor Csendes. "Global optimization in geometry—Circle packing into the square." In Essays and Surveys in Global Optimization, pp. 233-265. Springer, Boston, MA, 2005. PDF download.
The minimal polynomial for $n=13$. p.17 of Szabó et al.
The minimal polynomial is derived from a series of quadratic equations describing
the circle contacts.
Whether these polynomials are "naturally occurring" is a judgement call.
but where is a curve here? What I see is a polynomial in 1 variable. A curve is $F(x.y)=0$ where $F$ is a polynomial in two variables.
@AlexandreEremenko: I see now that this example does not fit your criteria.
|
2025-03-21T14:48:29.894175
| 2020-02-17T22:09:12 |
352958
|
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|
Stack Exchange
|
What is known about lower etale cohomology of unirational varieties?
Which information is currently known about $H^1_{et}(X,\mathbb{Z}_l)$ and $H^2_{et}(X,\mathbb{Z}_l)$, where $X$ is a smooth unirational variety over an algebraically closed field of finite characterstic $p\neq l$? Actually, I am interested in the case where $X$ is a quotient of a rational variety $Y$ by a free action of a finite group (whose order is prime to $p$). Does the situation differ much from the complex variety case; are there any examples where it does?
Certainly, my question is closely related to (the torsion) of $\operatorname{Pic}(X)$ and to the Brauer group of $X$.
If you want your varieties to behave like in characteristic $0$, it's probably better to say separably unirational. In positive characteristic there are all sorts of 'strange' unirational varieties; for example there exist unirational K3 surfaces. Maybe this doesn't matter for your question, but it seems that you're thinking about the separable case anyway.
Concerning the $H^2$, for $X$ a smooth projective rationally chain connected variety over an algebraically closed field $k$ and $\ell \in k^\ast$, it follows from
Theorem 1.2 in https://arxiv.org/abs/1703.05735 that
$$\mathrm{NS}(X)\otimes \mathbb{Z}_{\ell} = \mathrm{H}^2_{et}(X,\mathbb{Z}_{\ell}(1))$$
|
2025-03-21T14:48:29.894287
| 2020-02-17T22:10:13 |
352959
|
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"url": "https://mathoverflow.net/questions/352959"
}
|
Stack Exchange
|
Transverse knots with knot types of strongly quasi-positive knots
In 2008, Etnyre and Van Horn-Morris proved that if $L$ is a fibered strongly quasi-positive link, there is a unique (up to transverse isotopy) transverse link with the knot type of $L$ in the standard tight contact structure of $S^3$ with maximal self-linking number.
What is known for (not necessarily fibered) strongly quasi-positive links? If $L$ is strongly quasi-positive, is there a classification of transverse links with the knot type of $L$, with maximal self-linking number? Any comments or references would be appreciated.
The twist knot $K_{-6} = m(7_2)$ is strongly quasipositive, not fibered, genus-1, and it's Legendrian non-simple; it has five Legendrian isotopy classes with Thurston-Bennequin number 1, rotation number 0. A good source of material for low-crossing knots is the Legendrian knot atlas, by Chongchitmate and Ng.
Legendrian twist knots have been classified by Etnyre, Ng, and Vértesi (J. Eur. Math. Soc. 15 (2013)). I doubt that there exists a classification for arbitrary strongly quasipositive knots.
|
2025-03-21T14:48:29.894506
| 2020-02-17T22:56:45 |
352961
|
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"url": "https://mathoverflow.net/questions/352961"
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|
Stack Exchange
|
Lifting a complete intersection in $\mathbb{P}^n_{\mathbb{F}_p}$ to $\mathbb{Z}_p$
Suppose that you are given a (not necessarily smooth) projective variety $X \subseteq \mathbb{P}^n_{\mathbb{F}_p}$ of codimension $d$ that is a complete intersection. In other words, it can be defined by exactly $d$ homogeneous polynomials and no set of polynomials of cardinality less than $d$ has $X$ as its zero set. Let $f_1$, ..., $f_d$ be a set of polynomials defining $X$. If I lift these polynomials arbitrarily to $\mathbb{Z}_p$ (call the lifts $F_1$, ..., $F_d$), I will get a projective algebraic set $X' \subseteq \mathbb{P}^n_{\mathbb{Z}_p}$.
My question is this: will this $X'$ be flat over $\mathbb{Z}_p$? Equivalently, is the module $\mathbb{Z}_p[x_0,...x_n]/(F_1, ..., F_d)$ flat over $\mathbb{Z}_p$? I know that $\mathbb{Z}_p$ is a DVR, so flatness is equivalent to being torsion-free but I just cannot see how to prove that it is either. If $X$ is smooth, then $X'$ is definitely flat: $X'$ will also be smooth by the Jacobian criterion and hence flat.
Have you looked at any non-smooth example already?
@MartinBrandenburg To be honest, I have not. I was expecting this to work even in the non-smooth case. Do you think that it fails to be flat for non-smooth complete intersections in general?
This is really a statement in commutative algebra: a local complete intersection ring is Cohen-Macaulay, but if the ideal of torsion elements were nonzero then there would be an embedded prime (which cannot exist for a Cohen-Macaulay ring).
@ulrich Thank you for the reply. I am sorry but I am confused.
Is it absolutely necessary that the coordinate ring of $X'$, call it $A$, will be a Cohen-Macaulay (CM) ring?
Assume that $A$ is CM. That means that it will be a CM module over itself. Does that mean that it will be a CM module over the $p$-adic integers? We require flatness over $\mathbb{Z}_p$.
I don’t see why it is necessary for $A$ to be finite as a module over $\mathbb{Z}_p$. Is finiteness not required to conclude that a CM module has no embedded primes? See [link] (https://stacks.math.columbia.edu/tag/0BUS).
You can get a quick proof by using Hartshorne, Theorem III.9.9, which says that it's sufficient to show that the two fibres $X$ and $X' \times_{\mathbb{Z}_p} \mathbb{Q}_p$ have the same Hilbert polynomial. But this is true, since they're both complete intersections defined by polynomials of the same degree.
This is true, and here is one possible proof. There might be easier ones; however I suspect that it will always involve some algebra (not just geometry).
Write $S = \operatorname{Spec} \mathbf Z_p$ and $\mathbf P = \mathbf P^n_{\mathbf Z_p}$ with structure map $\pi \colon \mathbf P \to S$, and let $X \subseteq \mathbf P$ be a complete intersection with generic fibre $X_\eta \subseteq \mathbf P_\eta$ and special fibre $X_s \subseteq \mathbf P_s$. Assume $X$ is cut out by sections $f_1,\ldots,f_r$ of $\mathcal O_{\mathbf P}(d_1), \ldots, \mathcal O_{\mathbf P}(d_r)$ repesctively, such that the $f_i$ remain a regular sequence modulo $p$ (this is equivalent to the setting you're starting with).
Set $\mathscr E = \bigoplus_i \mathcal O_{\mathbf P}(-d_i)$, and consider the Koszul resolution
$$0 \to \wedge^r \mathscr E \to \wedge^{r-1} \mathscr E \to \ldots \to \mathscr E \to \mathcal O_{\mathbf P} \to \mathcal O_X \to 0,$$
which is exact since $X$ is a complete intersection (Tag 062F). Since the $\wedge^i\mathscr E$ are locally free, the sheaves $\mathcal Tor_i^{\mathcal O_{\mathbf P}}(\mathcal O_X, \mathcal O_{\mathbf P}/p)$ are computed by the cohomology of
$$0 \to \wedge^r \mathscr E \underset{\mathcal O_{\mathbf P}}\otimes \mathcal O_{\mathbf P}/p \to \wedge^{r-1} \mathscr E \underset{\mathcal O_{\mathbf P}}\otimes \mathcal O_{\mathbf P}/p \to \ldots \to \mathscr E \underset{\mathcal O_{\mathbf P}}\otimes \mathcal O_{\mathbf P}/p \to \mathcal O_{\mathbf P}/p \to 0.$$
This sequence is still exact since the $f_i$ modulo $p$ are still a regular sequence, so we conclude that $\mathcal Tor_i^{\mathcal O_{\mathbf P}}(\mathcal O_X, \mathcal O_{\mathbf P}/p) = 0$ for all $i > 0$. $\square$
This is an expanded version of my comment:
The main claim is that the question can be answered from basic facts in (local) commutative algebra. In particular, the same statement holds for a complete intersection in affine space, which I explain below.
We work over any dvr $R$ with residue field $k$. Let $X$ be of codimension $d$ in $\mathbb{A}^n_k$ defined by equations $f_1,f_2,\dots,f_d$. Let $F_1,F_2,\dots,F_d$ be elements of $R[x_1,x_2,\dots,x_n]$ such that $F_i$ is a lift of $f_i$ and let $X'$ be the subscheme of $\mathbb{A}^n_R$ defined by the ideal $(F_1,F_2,\dots,F_d)$. Then $X'$ is a local complete intersection scheme since the uniformizer of $R$ is a nonzero divisor in the polynomial ring and $X$ is a local complete intersection. In particular, $X'$ is Cohen-Macaulay, so it has no embedded points.
Now suppose the coordinate ring $A$ of $X'$ has non-zero $R$-torsion elements. The set of all such elements is a non-zero ideal in $A$ and the support of this ideal (as an $A$-module) is contained in $X'$ (viewed as a subset of $X$). This implies that $A$ has an embedded prime, so we get a contradiction.
|
2025-03-21T14:48:29.895083
| 2020-02-17T23:04:30 |
352962
|
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|
Stack Exchange
|
Greatest common divisor in $\mathbb{F}_p[T]$ with powers of linear polynomials
Let $n>1$ and $p$ be an odd prime with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ in $\mathbb{F}_p[T]$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$.
Question. Is $T^p - T$ actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$ in $\mathbb{F}_p[T]$?
I have verified this with computer algebra software for $n \leq 7000$ (code link). For many $n$ actually $u=0,1$ are sufficient.
I tried to find a proof, but my first idea didn't work. The only thing I know so far is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is clearly false without the $p^k-1$-requirement.
Here is a wild guess. Let $D$ be the difference operator,i.e. $Df(x)=f(x)-f(x-1)$. Then the upper gcd is also the gcd of $f,Df,D^2f,..,D^{p-1}f$ for $f=T^n-T$. Maybe that gcd is already the gcd of $f$ and $D^{p-1}f$. Could you run a computer experiment to check this ? Then when computing $D^{p-1}$ for $p>2$, we have to derive twice,i.e. the last summand in $T^n-T$does not matter. This could be why the behaviour is different for $p=2$.
Thanks Henrik, and nice to hear from you! I will try that.
Unfortunately $T^p - T = \mathrm{gcd}(f,D^{p-1}(f))$ is not true for the following pairs $(n,p)$ for $n \leq 200$: $(33,5),(73,7),(81,5),(109,7),(113,5),(127,7)$.
@MartinBrandenburg Oh sorry, add one to all the numbers that I said.
Can you try examples like $p=3,n=1+(3^{11}−1)/23,1+(3^{23}−1)/47,1+(3^{29}−1)/59$?
Ok. The first example works, but the second is too large for GAP, I'm afraid. There is an error message.
The first example also works with https://www.wolframalpha.com, input is PolynomialGcd(x^(7703)-x,(x+1)^(7703)-(x+1),(x+2)^(7703)-(x+2),Modulus->3). But with the second exponent it says, for some reason, "Wolfram|Alpha doesn't understand your query".
@MartinBrandenburg I think I figured out how to abstractly show the existence of a counterexample in this case - see my answer.
This is false.
Let $p$ be an odd prime, let $\ell$ be another prime, and let $m$ be a small prime divisor of $p^{\ell}-1$, that doesn't divide $p-1$. Let $n= 1 + \frac{ p^{\ell}-1}{m}$.
Then $n-1$ is a multiple of $p-1$, is not a multiple of $p^{\ell}-1$, and is not a multiple of $p^{k}-1$ for any other $k$ because $p^{\ell}-1$ is not a multiple of $p^{k}-1$ for any $1 < k < \ell$.
Then $x \in \mathbb F_{p^\ell}$ is a root of $T^{n } - T$ if and only if $x$ is an $m$'th power in $\mathbb F_{p^\ell}$. So roots of the gcd of $(T+u)^n - (T+u)$ for all $u$ in $\mathbb F_p$ are exactly those $x \in \mathbb F_{p^\ell}$ such that $x = y_0^m, x+1 = y_1^m, \dots, x+p-1 = y_1^{m}$ for some $y_0, \dots, y_{p-1}$ in $\mathbb F_{p^\ell}$.
Thus, to find a counterexample, it suffices to check that the number of $\mathbb F_{p^\ell}$ points of the curve $C_m$ with variables $x,y_0,\dots, y_{p-1}$ and equations $x +i = y_i^m$ is greater than the number $p m^{p-1}$ of solutions with $x \in \mathbb F_p$. Then the $x$ coordinates of the extra points will be roots of the gcd but not roots of $T^p- T$.
By Riemann-Hurwitz, the genus $g$ of $C$ satisfies $$2-2g = 2 m^p - (p+1) m^{p-1} (m-1)$$ since the degree over $\mathbb P^1$ (under the map $x$) is $m^p$, there are $p+1$ branch points $0,1\dots, \infty$, and each branch point has ramification of order $m$ on each point lying above it. There are $m^{p-1}$ missing points at $\infty$.
So by Weil's theorem, the number of $\mathbb F_{p^\ell}$-points of $C$ is at least $$p^\ell - p^{\ell/2} ((p+1) m^{p-1} (m-1) + 2 - 2 m^p) +1 - m^{p-1}$$
with the first term the main term, the second term coming from the Frobenius eigenvalues, and the last term coming from the missing points.
Thus, as long as
$$ p^\ell > p^{\ell/2} ((p+1) m^{p-1} (m-1) + 2 - 2 m^p) + m^{p-1} + p m^{p-1},$$ there is a counterexample.
Plugging into Wolfram Alpha, this inequality fails for $p=3, \ell=11, m=23$ but succeeds for $p=3,\ell=23, m=47$ and $p=3, \ell=29, m=59$.
I found these by looking at the sequence of multiplicative orders of $3$ modulo primes and looking for prime values, which became $\ell$, with the modulus prime becoming $m$.
It seems there are many examples with $\ell$ not much smaller than $m$, which as long as both are much larger than $p$, means this inequality is easily satisfied, since anything of the form $p^\ell$ beats anything of the form $m^p$.
In the comments, François Brunault found an explicit example: The polynomial $$T^{23}-T^{22}-T^{21}-T^{20}-T^{19}+T^{18}-T^{16}+T^{13}+T^{12}+T^{11}-T^{10}+T^8+T^6+T^4-T^2-T-1$$ divides $$\gcd( T^{n} - T, (T-1)^n - (T-1) , (T+1)^n - (T+1))$$ in $\mathbb F_3[T]$ when $n= 1 + \frac{3^{23}-1}{47}$ and provided the following Pari/GP code to check it:
P = Mod(x^23-x^22-x^21-x^20-x^19+x^18-x^16+x^13+x^12+x^11-x^10+x^8+x^6+x^4-x^2-x-1, 3); n = 1 + (3^23-1)/47; t = Mod(x, P); print(t^n == t & (t+1)^n == t+1 & (t-1)^n == t-1);
Thank you, Will! This is so fascinating and quite unexpected for me, I mean using arithmetic geometry to produce a counterexample. Unfortunately I don't have enough background to understand it. I hope that others will say something about your answer. When it's correct, I will accept it of course.
@MartinBrandenburg If no one else looks at it I can also try to fill in some of the details. When this is done it should all be very elementary algebraic geometry + statements taken as a black box.
@MartinBrandenburg Another approach is to check these examples by doing a computer search over $\mathbb F_{3^\ell}$, rather than a gcd calculation, which might be easier as you only have one number in memory at a time, but I don't have the knowledge of computer algebra systems to say for sure.
Does this produce an easily verifiable counterexample that you can write down (or at least check by computer), or is it more complicated?
Unfortunately my GAP program dies during the computation for $\ell=23,m=47$ in $\mathbb{F}_{3^\ell}$.
@R.vanDobbendeBruyn This doesn't give any clue for what the solutions are, only a lower bound for how many there are.
Very nice! Could you explain how you find the points at infinity of $C_m$? (In which space are you working so that the closure is non-singular?) Otherwise, I found an explicit solution in the case $(p,\ell,m)=(3,23,47)$ using Pari/GP: $P=x^{23}-x^{22}-x^{21}-x^{20}-x^{19}+x^{18}-x^{16}+x^{13}+x^{12}+x^{11}-x^{10}+x^8+x^6+x^4-x^2-x-1$.
To check it: P = Mod(x^23-x^22-x^21-x^20-x^19+x^18-x^16+x^13+x^12+x^11-x^10+x^8+x^6+x^4-x^2-x-1, 3);
n = 1 + (3^23-1)/47;
t = Mod(x, P);
print(t^n == t & (t+1)^n == t+1 & (t-1)^n == t-1);
@FrançoisBrunault The point is to calculate the local monodromy at infinity. When we have a fiber product of coverings, the Galois group is (contained in) the product of the Galois groups, and a generator of the local monodromy is the product of generators of the local monodromy of each covering. We have $p$ coverings $y_i^m = x+i$, each with Galois group $\mathbb Z/m$ and generator of the local monodromy at $\infty$ going to a generator of the group. So the local monodromy at $\infty$ is the elements $(1,\dots, 1)$ of $(\mathbb Z/m)^p$, which acts on $m^{p-1}$ orbits of size $m$.
@FrançoisBrunault Alternately, one can embed it into affine space with coordinates $y_0,\dots, y_{p-1}$ and equations $y_i^m + 1 = y_{i+1}^m$, and then take the projective closure, which has $m^{p-1}$ points at infinity, all smooth. WRT your example: Nice! Thanks for de-geometrizing my proof. Did you find it by an exhaustive search over degree 23 irreducible polynomials, or was it something more intricate?
@WillSawin Thanks for the explanations. Yes, it is just exhaustive search (of course just testing $t^n=t \bmod{P}$). But enumerating all the elements of $\mathrm{GF}(3^{23})$ takes very long time, so your lower bound on the number of solutions is essential: it tells us a solution will be found in reasonable time, long before testing all elements. E.g. my computer checked that there is no solution in $\mathrm{GF}(3^\ell)$ with $\ell$ prime $\leq 17$, for $\ell=17$ it took several hours...
|
2025-03-21T14:48:29.895645
| 2020-02-17T23:28:41 |
352965
|
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|
Stack Exchange
|
reference request: unbounded operators on normed spaces
I'm looking for books where the theory (basic properties, adjoints etc.) of unbounded linear operators between locally convex spaces or at least Banach spaces is developed. In Brezis' functional analysis, Sobolev spaces and partial differential equations there are some limited results for the Banach space case and all other related books I found only treat such operators on Hilbert spaces.
Thanks in advance!
The second volume of Köthe’s “Topological vector spaces” has about 50 pages on non-continuous linear mappings between locally convex spaces ($$ 35-38).
@user131781 thank you for this reference!
You can try to use: Goldberg, Seymour Unbounded linear operators. Theory and applications. Reprint of the 1985 corrected edition [MR0810617]. Dover Publications, Inc., Mineola, NY, 2006. viii+199 pp. ISBN: 0-486-45331-6
|
2025-03-21T14:48:29.895748
| 2020-02-18T00:20:54 |
352967
|
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|
Stack Exchange
|
Is this a new strange attractor?
I recently made some experiments in programming strange attractors, and I found this (very simple) equations, which create a nice strange attractor:
xn=x+dt*(z-y)
yn=y+dt*(x/2-1)
zn=z+dt*(-xy/2-z)
You can see it in action on my Youtube channel:
https://youtu.be/Bm_M6mUGjtg
My question: Is this a variation of the Lorenz- or Rössler attractor - or did I stumble upon something new?
EDIT:
Meanwhile I programmed a little 3D View for this attractor:
You can see/move the view on this applet (Java needed):
https://cerumen.de.cool/attractor/index.html
(Here you also find the source code for Processing)
And here the Javascript-Version: https://cerumen.de.cool/attractor/js/index.html (with processing.js... bit slow)
Perhaps my question was also asked too amateurishly.
I was simply surprised by the simplicity of the system of equations I have found.
Therefore I would like to know if this strange attractor is a descendant of one of the well known ones (Lorenz / Rössler).
Edit 2:
I have now brought the system of equations into a more general form:
xn=x+dt*(z-y)
yn=y+dt*(ax-b)
zn=z+dt*(-axy-z)
with a in range [0 to 1], b in range [0.5 to 1]
This makes it more interesting.
Here some sample images for different values for a and b:
Edit 3:
Here a video with the generalized equations and constantly changing parameters a and b: https://youtu.be/gxusM8pmNwU
I think you can see here quite well how the system goes from order through bifurcation into chaos...
I believe it is not that difficult to cook up new fractals - there is plenty of software products out there that does this, Flamethyst for example (although the method is a bit different).
It's certainly very close to the Rossler system: two coordinates with linear differential equations, one 'slightly' nonlinear one, with a quadratic perturbation term. While the specific form of the linear piece is slightly different, I wouldn't be surprised if the two are fundamentally equivalent in some fashion.
Thank you, Steven, I kind of suspected that too. But so far I have not found a way to transform the Rössler equations into my system of equations. I think this is the core of my question.
As Per Alexandersson suggests, strange attractors are actually very common for nonlinear systems in three variables. You can literally make up new ones just by making up a system of nonlinear differential equations and then using some graphing software to check whether it has a strange attractor. There are certain "famous" strange attractors which are famous mostly for historical or scientific reasons, but there are lots of unnamed strange attractors associated with various systems. This is a particularly nice one and you've made some great pictures of it, but it's not really very novel.
Thank you, @Jim, for your answer. It seems that the answer to my question is: This is just another new strange attractor. But I still love the simplicity of the equation system, which really surprised me when I programmed this thing.
In another forum a user drew my attention to the publication of
J. C. Sprott, Some simple chaotic flows, Phys. Rev. E 50, R647-R650 (1994), doi:10.1103/PhysRevE.50.R647, author pdf.
This shows that there are many very simple chaotic systems of equations.
I guess this answers my question...
You can see some of the equations here, and here are the corresponding graphs.
But thank you all for your interest.
I just figured I added some Mathematica code and a picture for the attractor in the question.
With[{dt = 0.001},
iter[{x_, y_, z_}] := {x, y, z} +
dt {(z - y), x/2 - 1, -x y/2 - z}
];
pts = NestList[iter, {0.1, 0.1, 1/2}, 500000];
ListPlot[{#1, #2} & @@@ pts[[1 ;; ;; 5]], PlotRange -> All,
Axes -> False, PlotStyle -> {Opacity[0.9], PointSize[Tiny], Orange},
AspectRatio -> 1, Background -> Gray, ImageSize -> {800, 600}]
I only plotted every fifth of the points, as it is a bit quicker, and the image is a bit more pleasant with this variant.
EDIT:
With some more creative edits,
$$
(x_{n+1}, y_{n+1}, z_{n+1}) =(x_n, y_n, z_n)+dt
(z - y, -1 + x + 6 \sin(\pi/4 + 10 x/ z), -x y/2 - z)
$$
one can produce the following picture.
Adding any non-linear disturbance, and making sure that it does not diverge, or converge to something boring, it is rather easy to cook up exotic variations that give rise to chaotic behavior.
Per, thank you for your detailed explanations. But my question is not how to create chaotic systems with increasing complexity.
What I want to know is whether my really simple system of equations is some sort of derivative of already known attractors.
|
2025-03-21T14:48:29.896054
| 2020-02-18T03:01:33 |
352972
|
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|
Stack Exchange
|
Bound on the number of unlabeled tree on n vertices
By the Cayley's Theorem, the number of labeled tree on n vertices is at most n^{n-2}. On the other hand, what is the bound on the number of unlabeled tree on n vertices?
See https://oeis.org/A000055.
It is at most $n4^n$. One way to do this is to count unlabeled rooted trees; these can be uniquely encoded using balanced parantheses and orderings. If the subtrees at the children of the root have encodings $e_1,e_2,\ldots,e_k$ with $e_1 \leq e_2 \leq \ldots \leq e_k$, then the encoding for the tree is $((e_1)(e_2) \ldots (e_k))$.
|
2025-03-21T14:48:29.896149
| 2020-02-18T03:14:21 |
352974
|
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"Kevin Casto",
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|
Stack Exchange
|
Significance of the length of the Perron eigenvector
Let $A$ be a positive square matrix. Perron-Frobenius theory says that there exist $\lambda,v$ with $Av=\lambda v$ and $\lambda$ equals the spectral radius of $A$, $\lambda$ is simple, and $v$ is positive.
Now consider also the left Perron eigenvector $u^T A=\lambda u^T$. Another result of Perron-Frobenius theory is that
$$\lim_{m\to \infty} \frac{A^m}{\lambda^m} = \frac{v u^T}{u^T v}.$$
Suppose $\|v\|=1$. The above result says that the "correct" normalization for u is $u^T v=1$ rather than the more usual $u^T u=\|u\|^2=1$. This motivates the question: what is the significance of the ratio
$$\frac{u^T v}{u^T u} ?$$
Are there matrices $A$ for which this ratio is arbitrarily large? Arbitrarily small? Does this ratio determine any properties of $A$? Note that if $A$ is symmetric, then $u=v$ and this ratio is always equal to $1$, but that's not the case in general for arbitrary $A$. Could it be the case that this ratio is measuring how far $A$ is from being symmetric?
Note too that this normalization is necessary so that the limit $\frac{v u^T}{u^T v}$ is a projection matrix (i.e. that its only non-zero eigenvalue is one). In this context, I understand why the normalization is necessary, but I'm interested in the amount of normalization necessary with respect to the length of $u$.
Any pointers appreciated. Thanks!
EDIT In the comments, it is argued that the real quantity of interest in this setup is
$$\frac{\left( u^T v \right)^2}{\left(u^T u \right) \left( v^T v \right)}.$$
This quantity is also of interest to me, and an acceptable replacement for my original question.
No significance. It's just a question of scaling. $u^Tv=1$ ensures that $u$ and $v$ are dual eigenvectors. With this scaling, there is still one degree of freedom: you can multiply $u$ by a constant $t$ and divide $v$ by the same constant (notice that this rescaling scales your ratio by a factor of $1/t^2$).
Perhaps the true quantity to understand is $(u^T v)^2/((u^T u)(v^T v))$ then?
@AnthonyQuas the ratio is still uniquely defined as stated (with $v$ having unit length). In any case, Kevin Casto proposes an alternative.
@KevinCasto, yes, studying that alternative would also be of interest!
That quantity $s = \frac{|u^Tv|}{\|u\|\|v\|}$ is the inverse of the eigenvalue condition number. The smaller it is, the more sensitive to perturbation the Perron value is.
More precisely, any perturbed matrix $A+E$ with $\|E\| \leq \varepsilon$ has a Perron value $\tilde{\lambda}$ that satisfies $|\tilde{\lambda}-\lambda| \leq \frac{\varepsilon}{s} + \mathcal{O}(\varepsilon^2)$. See e.g. Section 7.2.2 of Golub and Van Loan's Matrix Computations 4th ed.
In addition, note that if $A$ is normal then $s=1$ (its maximum possible value) and the Perron value is perfectly conditioned; while if $\lambda$ is a defective eigenvalue (e.g. $A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$) then $s=0$. So rather than a "distance from symmetric" I'd say that $1-s$ is a "distance from normal" or $s$ is a "distance from defective".
This answers my question. Thanks!
I'm not sure how helpful, but you can say the following: in general, $u$ (or the 1-dimensional eigenspace spanned by it) is the orthogonal complement to the span of all the (right-) eigenvectors except $v$, call this $W$. The quantity I mentioned is $\cos^2$ of the angle between $v$ and $u$, which is $1 - \cos^2$ of the angle between $v$ and $W$.
So if you know things about just the right eigenvectors, you can say things about this quantity. For example, it's close to 0 if $v$ is close to $W$, and close to 1 if $v$ is close to being perpendicular to $W$.
This is very helpful! I take it the other way around though: knowing only $v$ and $u$ can say something about $v$ and $W$. Now I'm wondering whether $v$ being close to $W$ affects the convergence rate of power methods for computing eigenvalues...
For example, if $$ \eqalign{A &= \pmatrix{1 & t\cr 1 & 1\cr},\
v = \pmatrix{\sqrt{t}\cr 1},\ u =\pmatrix{1\cr \sqrt{t}},\cr
\frac{(u^T v)^2}{(u^T u)(v^T v)} &= \frac{4t}{(1+t)^2} \to 0 \ \text{as}\ t \to \infty} $$
By Cauchy-Schwarz we always have
$$ 0 < \frac{(u^T v)^2}{(u^T u)(v^T v)} \le 1$$
with equality on the right iff $u$ is a scalar multiple of $v$.
Note also that if $A$ is doubly stochastic, $u = v = (1,\ldots,1)^T$. Not all doubly stochastic matrices are symmetric.
Interestingly, the ratio equals $1$ exactly when the matrix is symmetric.
well $u=v$ if matrix is symmetric, so it is not a surprise ?
I didn't say it was a surprise, I said it was interesting in light of my conjecture (see original post) that this ratio measures how far $A$ is from being symmetric.
@RobertIsrael - I just saw your latest edit - perhaps this ratio being equal to $1$ is an indication of normality rather than symmetricity.
Doubly stochastic matrices need not be normal either.
Woops I was thinking of permutation matrices.
|
2025-03-21T14:48:29.896499
| 2020-02-18T03:44:10 |
352975
|
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|
Stack Exchange
|
The converse of a Poincaré's result on regular boundary points
Let $V$ be a bounded open set in $\mathbb{R}^n$ with $n>1$. According to a well known result due to Poincaré, if $x$ is a point in the boundary $\partial V$ and there exists a ball $B$ such that $x\in\partial B$ and $B\cap V=\emptyset$, then $x$ is a regular point for the Dirichlet problem.
Is there a converse for this result? More generally, under which condition(s) can we say that if $x\in\partial V$, there exists a ball $B$ such that $x\in\partial B$ and
$B\cap V=\emptyset$?
The answer is no if $n=2$. Indeed, every Jordan domain in the plane is regular for Dirichlet problem.
The answer is no. Take the unit ball in $R^3$, and remove from it the halfplane
$x_3=0, x_1\geq 0$. This is regular.
|
2025-03-21T14:48:29.896593
| 2020-02-18T04:15:02 |
352976
|
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"Dominic van der Zypen",
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|
Stack Exchange
|
Calculate the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of the poset of nontrivial finitary partitions of $\omega$
Let $(P,\le)$ be a poset. For a point $x\in P$ let
$${\downarrow}x=\{p\in P:p\le x\}\quad\text{and}\quad{\uparrow}x=\{p\in P:x\le p\}$$be the lower and upper sets of the point $x$, and for a subset $S\subset P$, let
$${\downarrow}S=\bigcup_{s\in S}{\downarrow}s\quad\text{and}\quad{\uparrow}S=\bigcup_{s\in S}{\uparrow}s$$be the lower and upper sets of the set $S$ in $P$.
Now consider the following cardinal characteristics of $P$:
$\bullet$ the $\downarrow$-cofinality ${\downarrow}(P)=\min\{|C|:C\subseteq P\;\wedge \;{\downarrow}C=P\}$;
$\bullet$ the $\uparrow$-cofinality ${\uparrow}(P)=\min\{|C|:C\subseteq P\;\wedge\;{\uparrow}C=P\}$;
$\bullet$ the $\uparrow\downarrow$-cofinality ${\uparrow}{\downarrow}(P)=\min\{|C|:C\subseteq P\;\wedge \;{\uparrow\downarrow}C=P\}$;
$\bullet$ the $\downarrow\uparrow$-cofinality ${\downarrow}{\uparrow}(P)=\min\{|C|:C\subseteq P\;\wedge\;{\downarrow\uparrow}C=P\}$.
Proceeding in this fashion, we could define the $\downarrow\uparrow\downarrow$-cofinality ${\downarrow\uparrow\downarrow}(P)$ and $\uparrow\downarrow\uparrow$-cofinality ${\uparrow\downarrow\uparrow}(P)$ and so on.
It is clear that $$\max\{{\uparrow\downarrow}(\mathfrak P),{\downarrow\uparrow}(\mathfrak P)\}\le \min\{{\downarrow}(\mathfrak P),{\uparrow}(\mathfrak P)\}.$$
I would like to know the values of the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of the poset $\mathfrak P$ of nontrivial finitary partitions of $\omega$.
By a partition I understand a cover $\mathcal P$ of $\omega=\{0,1,2,\dots\}$ by pairwise disjoint sets.
A partition $\mathcal P$ is defined to be
$\bullet$ finitary if $\sup_{P\in\mathcal P}|P|$ is finite (i.e., the cardinalities of the cells of the partition are upper bounded by some finite cardinal);
$\bullet$ nontrivial if the subfamily $\{P\in\mathcal P:|P|=1\}$ is finite (i.e., $\mathcal P$ contains infinitely many cells of cardinality $\ge 2$).
The family $\mathfrak P$ of all nontrivial finitary partitions of $\omega$ is endowed with the refinement partial order $\le$ defined by $\mathcal P_1\le\mathcal P_2$ if each cell of the partition $\mathcal P_1$ is contained in some cell of the partition $\mathcal P_2$.
It can be shown that $${\uparrow\downarrow\uparrow}(\mathfrak P)=1={\downarrow\uparrow\downarrow}(\mathfrak P),$$ so only four cofinalities (with at most two arrows) can be infinite.
Using almost disjoint families of cardinality continuum, it can be shown that ${\uparrow}(\mathfrak P)=\mathfrak c$.
Problem 1. Calculate the $\downarrow$-cofinality ${\downarrow}(\mathfrak P)$ of the poset $\mathfrak P$. In particular, is ${\downarrow}(\mathfrak P)=\mathfrak c$? Or ${\downarrow}(\mathfrak P)=\mathfrak d$?
Remark 1. It can be shown that ${\downarrow}(\mathfrak P)\ge\mathfrak d$.
Problem 2. Evaluate the cardinal characteristics ${\downarrow\uparrow}(\mathfrak P)$ and ${\uparrow\downarrow}(\mathfrak P)$ of the poset $\mathfrak P$.
Re: Compatibility: Two partitions ${\cal A}, {\cal B}$ are compatible if there is a directed family ${\frak D}$ of partitions with respect to the refinement ordering and ${\cal A}, {\cal B}\in{\frak D}$ - is that correct? (Just trying to make a link to a previous question of yours that involved directed families of partition orderings.)
Yes, exactly! This is equivalent.
Doesn't "cardinal-characteristics" make "independence-results" redundant? while it's a pity not to tag it "infinite-combinatorics" where this question primarily belongs.
23rd version of the question.
@GerryMyerson Simply I am thinking on this problem myself, so some information appears, which I add to the questions in edits. For example, Problem 3 disappeared as I was able to show that ${\uparrow}{\downarrow}(\mathfrak P)=\mathfrak j_2=\mathfrak j_{2:1}$ thus reducing the problem to another my question about $\mathfrak j_{2:1}$. Since there was no answers, I hope such revisions of the problems are not grave. But anyway, thank you for turning attention to this my problem.
What is $X$ in your definitions supposd to be? Is that supposed to be $P$?
@StevenStadnicki Exactly, those $X$'s should be $P$'s. Now it is corrected. Thank you for your attention and comment.
At the moment we have the following information on the cofinalities of the poset $\mathfrak P$ (see Theorem 7.1 in this preprint).
Theorem.
1) ${\downarrow}\!{\uparrow}\!{\downarrow}(\mathfrak P)={\uparrow}\!{\downarrow}\!{\uparrow}(\mathfrak P)=1$.
2) ${\downarrow}(\mathfrak P)={\uparrow}(\mathfrak P)=\mathfrak c$.
3) ${\downarrow}\!{\uparrow}(\mathfrak P)\ge \mathrm{cov}(\mathcal M)$.
4) $\mathsf \Sigma\le{\uparrow}\!{\downarrow}(\mathfrak P)\le\mathrm{non}(\mathcal M)$.
Here $\mathrm{non}(\mathcal M)$ is the smallest cardinality of a nonmeager set in the real line, and
$\mathsf \Sigma$ is the smallest cardinality of a subset $H$ in the permutation group $S_\omega$ of $\omega$ such that for any infinite sets $A,B\subseteq \omega$ there exists a permutation $h\in H$ such that $h(A)\cap B$ is infinite.
By Theorem 3.2 in this preprint, $$\max\{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal N)\}\le\mathsf\Sigma\le\mathrm{non}(\mathcal M).$$
The cardinal $\mathsf\Sigma$ is equal to the cardinal $\mathfrak j_{2:2}$, discussed in this MO-post.
However I do not know the answer to the following
Problem. Is ${\downarrow\!\uparrow}(\mathfrak P)\le\mathrm{non}(\mathcal N)$?
Here $\mathrm{non}(\mathcal N)$ is the smallest cardinality of a subset of the real line, which is not Lebesgue null.
I’m confused. What is $\downarrow\downarrow$ and $\uparrow\uparrow$? And how does Theorem 6.1 give ${\downarrow}(\mathfrak P)=\mathfrak c$?
@EmilJeřábek3.0 Ups! Thank you for the comment. I have just copy-pasted the theorem from my preprint and there were two more cardinal characteristics ${\uparrow\uparrow}(\mathfrak P)$ and ${\downarrow\downarrow}(\mathfrak P)$, which are analogs of the cellularity. Unfortunately, arXiv is a bit slow and the new version of the paper with all these symbols will appear only tomorrow (I hope). Till that moment the new version of the paper with proper form of Theorem 6.1 can be found at https://www.researchgate.net/publication/339416314_Small_uncountable_cardinals_in_large-scale_topology
Great, thank you.
|
2025-03-21T14:48:29.897095
| 2020-02-18T08:18:39 |
352978
|
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|
Stack Exchange
|
Tightness of a uniformly bounded sequence of functions integrated with respect to a semimartingale
I am reading this paper by Jacod, Jakubowski and Mémin. In the proof of Theorem 1.3 the authors define, for each $n\geq1$ the function $\phi_n$ by
$\quad\phi_n(s)=i+1-ns,\quad\text{if } \frac{i}{n}<s\leq\frac{i+1}{n},\quad\phi_n(0)=0$.
Then, for an arbitrary semimartingale $X$, they claim that the sequence of semimartingales $(\tilde{Y}'^n(X))_{n\geq1}$ is tight (in the Skorokhod J1-topology), where
$\quad\tilde{Y}'^n(X)_t=\int_0^t \phi_n(s)\,dX_s$.
They support this claim with a reference to Theorem VI.5.10 in Jacod & Shiryaev (1987).
Can anyone help me understand how this theorem is applied to get the desired tightness? Condition (i) is clearly satisfied (since $\tilde{Y}'^n_0=0$) but I have trouble verifying the remaining conditions. My only idea so far is to somehow use the fact that $|\phi_n(s)|\leq1$ for all $n\geq1,s\geq0$.
Can you state Theorem VI.5.10 in Jacod & Shiryaev? For those who do not have the book.
I see your point but unfortunately this is not straightforward. The total formulation of the theorem takes up around two pages (there are several conditions of which at least one must be satisfied) and it uses fairly complicated notation introduced earlier in the book.
|
2025-03-21T14:48:29.897243
| 2020-02-18T08:40:45 |
352980
|
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"authors": [
"Giorgio Metafune",
"Hua Wang",
"Iosif Pinelis",
"Ivan Meir",
"Luc Guyot",
"Rick Sternbach",
"Steven Stadnicki",
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|
Stack Exchange
|
Linear combination of sine and cosine
I was explaining to my students the other day why $\cos(2x)$ is not a linear combination of $\sin(x)$ and $\cos(x)$ over $\mathbb{R}$. Besides the canonical method of using special values of sine and cosine, I noticed something interesting. In the following, all vector spaces are over $\mathbb{R}$.
Consider the linear space $C^\infty_b(\mathbb{R})$ of real-valued bounded smooth functions on $\mathbb{R}$, and take any $c > 0$. We say a function $f \in C^\infty_b(\mathbb{R})$ has property $P(c)$, if for all $k \in \mathbb{N}$ (including $0$), we have
$$\sup f^{(k+1)} = c \sup f^{(k)} = -\inf f^{(k+1)} = -c \inf f^{(k)}.$$
Here, the supremum and infimum are of course taken over $\mathbb{R}$, and $f^{(k)}$ is the $k$-th derivative of $f$, with the convention that $f^{(0)}=f$.
Define
$$S(c) = \{f \in C^\infty_b(\mathbb{R}) \,\vert\, f \text{ has property } P(c)\}.$$
Since for fixed $a,b$ and all $x$, we have $a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \sin(x + \theta) $ for some fixed $\theta$, it is clear that all linear combinations of $\sin(cx)$ and $\cos(cx)$ belong to $S(c)$. In particular, linear combinations of $\sin(x)$ and $\cos(x)$ are all in $S(1)$, while $\cos(2x)$ is not.
Question: is it true that $S(c) = \operatorname{Vect}\bigl(\sin(cx), \cos(cx)\bigr)$?
If $f \in S(1)$ and $f$ is periodic, then using Fourier series, I can prove that $f$ is indeed $2 \pi$-periodic and $f \in \operatorname{Vect}\bigl(\sin(x), \cos(x)\bigr)$ with some work. Although I haven't checked this yet, I also believe that the periodic case for $S(c)$ where $c>0$ is arbitrary could be established by a more elaborate Fourier series argument (of course, I could be wrong). So the real interest lies in treating the non-periodic case, i.e., answering the following
Special case: does $f \in S(c)$ imply $f$ is periodic?
At first, I suspect the answer to the above special case is negative. But after some experiment, I am not so sure. Note that the radius of convergence of the Taylor series (say around $0$) for all $f \in S(c)$ is infinite, so the Taylor series of $f$ converges to $f$ itself. In particular, all functions in $S(c)$ are automatically analytic, so one does not have much freedom when trying to construct a (counter-)example.
If the answer turns out to be negative, then can one at least assert that $S(c)$ is a linear subspace? What if we only consider periodic functions for some fixed period in case $S(c)$ is not a linear subspace? Of course, these probably depend on the explicit form of the answer which is not yet known to me, and all of these are just some (perhaps stupid and naive) speculation on an old exercise of a first-year undergraduate. But it seems interesting, and any thought is appreciated.
Edit: I was a bit careless in formulating the question since the questions for all different $c$ are equivalent merely by rescaling, so one can simply assume $c = 1$, in which case we still have much work to do.
Edit 2: Proof of the periodic case can be found here in case anyone is interested.
Could you modulate $\sin(x)$ by some sufficiently smooth bump function $\varphi(x)$ with $\varphi(x)\to 1$ as $x\to\infty$? At first glance that seems like it should work, and since you're in $C^\infty$, I would think that the usual bumps are sufficient for this...
All such functions are analytic, so once it vanishes on some interval, it is zero everywhere. There's unfortunately no analytic bump functions!
@RickSternbach I didn't pay sufficient attention to the notes about analyticity later in the post! That said, it seems like one could still modulate by a Gaussian, something along the lines of $\phi(x)=1-e^{-x^2}$...
@Zero Just consider the remainder of the Taylor expansion to the order $n$, and you will see immediately that due to the restrictions of the higher derivatives, the remainder tends to zero as $n$ goes to infinity, which means exactly that the Taylor series converges pointwise to the function itself and the function is indeed given by a power series (its Taylor series).
@HuaWang : Can you let us know how you dealt with periodic $f$?
@IosifPinelis I will only treat the case $S(1)$, the general case follows by rescaling as remarked in the edit. Basically there are two steps. 1. One shows that such $f$ must be a trigonometric polynomial. More precisely, using integral par parts repeatedly, and by the restrictions on the higher derivatives, one shows that the Fourier coefficients vanishes once the order is high enough. 2. Now that $f$ is a trigonometric polynomial, taking the derivative repeatedly, one shows that all supremum and infimum involved are exactly determined by the coefficients of highest order, the rest follows.
@IosifPinelis Sorry I couldn't be clearer due to the restrictions of characters in the comment. I could elaborate on each of the above step if needed. But I believe with some work, one can finish the periodic case using this rough outline.
@HuaWang : You wrote in your post: "If $f \in S(1)$ and $f$ is periodic, then using Fourier series, I can prove that $f$ is indeed $2 \pi$-periodic and $f \in \operatorname{Vect}\bigl(\sin(x), \cos(x)\bigr)$ with some work." If can indeed prove this, I think it could well be useful for some of us to see that proof, which you could write elsewhere, and here provide a link to that proof.
@IosifPinelis OK. Following your advice, I wrote a detailed version of the periodic case, and put it on my google drive. Here is the link https://drive.google.com/file/d/1Vyp5bcaGVXHvdxuvksNaaXQcrNlG4pFT/view?usp=sharing
@IosifPinelis And as one can see now, the periodic case is really just an exercise of Fourier series (albeit a nice one imho), and I am more interested in treating the non-periodic case...
@HuaWang : Thank you for your write-up for the periodic case. Yes, it is nice (I think there is a typo in the first display on p. 4, though). It seems that you use the periodicity very essentially. I hoped a Paley--Wiener theorem (which corresponds to the first part of your proof, before Lemma 2) together with what you did in the periodic case could be close to sufficient, but now I see you use some pretty delicate arithmetic, seemingly not available in the non-periodic case. I wish someone solves your problem, which could be a good learning experience for interested users.
@IosifPinelis Thanks for pointing out this typo. I will fix it when I get home (I was stupid enough to include only the pdf in my google drive, not the tex file, which is only available in my home computer). Concerning the general case, I was thinking that one might begin with the property that $f^{(4)}$ should be $f$ itself (or $f^{(2)}=-f$) for such functions, but couldn't proceed further.
I have only some comments. First of all, expanding $f(x+iy)$ in power series around $x$ one sees that it is bounded by $e^{|y|}$ hence, using Paley-Wiener, its Fourier transform, as a tempered distribution, has support in the interval $[-1,1]$. To conclude we should exclude the interior but this uses only $|f^{(n)}| \le 1$ for all n. Using Ascoli-Arzela' and the identity $\hat{f^{(k)}}(\xi)=(i\xi)^k\hat{f}(\xi)$ one (probably) can show that $f^{(4k)}$ converges uniformly on compact sets to $g$ and $g^4=g$...but then maybe some body else sees how to go on.
As noted by the OP we can replace $f$ by $af(bx)$ for suitable $a,b\in\mathbb{R}$ so that wlog we can take $c=1$ and ensure that $\sup f=-\inf f=1$.
Firstly we note that $f(z)$ is infinitely differentiable on $\mathbb{R}$ so we can form the taylor series at 0, $f(z)=\sum_{i=0}^{\infty}\frac{f^{(i)}(0)}{i!}z^i$. Since $|f^{(i)}(0)|\leq 1$ for all $i\geq 0$ we have $\sum_{i=0}^{\infty}|\frac{f^{(i)}(0)}{i!}z^i|=\sum_{i=0}^{\infty}\frac{|f^{(i)}(0)|}{i!}|z|^i\leq \sum_{i=0}^{\infty}\frac{|z|^i}{i!}$ which converges for all $z$ to $e^{|z|}$.
Hence $F(z)=\sum_{i=0}^{\infty}\frac{a_i}{i!}z^i$ is an absolutely convergent series defining an entire function on $\mathbb{C}$ agreeing with $f$ on $\mathbb{R}$ s.t. $\sup f^{(k)}=-\inf f^{(k)}=1$ for all $k\in \mathbb{Z}_{\geq0}$ and $|F(z)|\leq e^{|z|}$ for all $z\in \mathbb{C}$.
We now determine the form of $F$ given the preceding conditions.
First note that Bernstein proved the following (see Rahman and Tariq$^1$) as an extension of his related inequality for polynomials:
Theorem Let $g$ be an entire function of exponential type $\tau>0$ such that $|g(x)|\leq M$ on the real axis. Then $$\sup_{-\infty<x<+\infty}|g^{'}(x)|\leq M\tau.$$
Now $|F(z)|\leq e^{|z|}$ for all $z\in \mathbb{C}$ and therefore we know that $F$ is of exponential type 1. In addition $|F(x)|=|f(x)|\leq 1$ on the real axis.
We are therefore interested in the conditions of equality in the above.
Fortunately in their book "Analytic Theory of Polynomials"$^2$ Rahman and Schmeisser prove (Theorem 14.1.7) that equality holds in the above if and only if $g(z)=ae^{i\tau z}+be^{-i\tau z}$ where $|a|+|b|=M$. (You can access the relevant pages 513-514 on google books)
$|F(x)|\leq 1$ for $x\in\mathbb{R}$ and hence in the above theorem we may take $M=1$, $\tau=1$ and $g=F$ since $|F|$ is of exponential type 1. Also $\sup_{x\in\mathbb{R}}|F^{'}(x)|=1$ which means we have the case of equality.
Thus $F(z)=ae^{iz}+be^{-iz}$ where $|a|+|b|=M=1$.
On the real axis $F$ is real valued and agrees with $f$ so we must have $f(x)=ae^{ix}+be^{-ix}$, $x\in \mathbb{R}$ with $a=\bar{b}$. Setting $a=c+id$ we obtain $f(x)=2c\cos x-2d\sin x$ with $|a|=|b|=2\sqrt{c^2+d^2}=1$. Rewriting with $C=2c$, $D=-2d$, we obtain
$$f(x)=C\cos x+D\sin x$$
for some constants $C,D\in \mathbb{R}$, $C^2+D^2=1$.
This proves the OP's conjecture. Note that the condition $C^2+D^2=1$ is due to the bound we imposed on $f$ due to our normalisation which is not part of the definition of $S(c)$.
1 Rahman, Q. I.; Tariq, Q. M., On Bernstein’s inequality for entire functions of exponential type, J. Math. Anal. Appl. 359, No. 1, 168-180 (2009). ZBL1168.30002.
2 Rahman, Q. I.; Schmeisser, G., Analytic theory of polynomials, London Mathematical Society Monographs. New Series 26. Oxford: Oxford University Press (ISBN 0-19-853493-0/hbk). xiv, 742 p. (2002). ZBL1072.30006.
Reading your solution, it looks as if the condition $\inf = -\sup$ is actually superfluous and the result follows only from the conditions $\sup \vert f^{(k)} \vert = 1$ for all $k \ge 0$. So, you did prove a stronger result (unless I have overlooked something).
@LucGuyot I think you can lose all the sup and inf conditions except one of them say $\sup |f^{(i)}|=1$ or $\inf|f^{(i)}|=-1$, $i>0$ as long as you have $|f^{(k)}|\leq1$ for all $k\geq 1$. You need the bounds on the derivatives to get the $|F(z)|\leq e^{|z|}$ so that $F$ has exponential type 1, essential for this solution.
@LucGuyot Note that if you have $\sup |f^{(i)}|=1$ or $\inf|f^{(i)}|=-1$, $i>0$ then you just apply the same method to $f^{(i-1)}$ to get the result.
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2025-03-21T14:48:29.897915
| 2020-02-18T08:44:25 |
352981
|
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|
Stack Exchange
|
Quantum entropy Venn diagrams
We know that in classical information theory the relation between different entropies can be depicted by Venn Diagram as given below.
Can we create such Venn-diagrams for the quantum information theory case, where we consider Von-Neumann Entropy? I feel such Venn diagrams are not possible because in Quantum Information, some entropies can be negative. What can we do in such a scenario. Is there any research being performed in this direction, to create a geometrical picture of quantum information theory?
Even for three discrete random variables, the central region in the Venn diagram may represent a negative entropy. See for instance 8.31 in MacKay's book: https://www.inference.org.uk/itprnn/book.pdf.
@Mark Wildon Thanks, I did not know that the representation of entropies in terms of Venn diagrams is misleading. A lot of books on Quantum Information follow this approach of using Venn Diagrams, also a lot of them do not consider 3 random variables. I had not read this book so I did not know about this.
Conditional entropies can be negative, so the sign is not an issue here. An introduction to quantum Venn diagrams that you may like is given by Chris Adami. Here is an example of a quantum system Q measured by a classical measurement device A that consists of two pieces $A_1$ and $A_2$. The shared information (central region) is always zero, which is what prevents cloning of a quantum state.
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2025-03-21T14:48:29.898062
| 2020-02-18T09:35:44 |
352982
|
{
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"Asaf Shachar",
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|
Stack Exchange
|
Is there a volume-preserving diffeomorphism of the disk with prescribed singular values?
This is a cross-post. While working on a variational problem, I have reached to the following question.
Let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1\sigma_2=1$, and let $D \subseteq \mathbb{R}^2$ be the closed unit disk.
Question: Does there exist a smooth map $f:D \to D$ such that $df$ has everywhere the fixed singular values $\sigma_1,\sigma_2$ and $\det(df)=1$?
Is there such a diffeomorphism of $D$?
The linear map $x \to \begin{pmatrix} \sigma_1 & 0 \\\ 0 & \sigma_2 \end{pmatrix}x$ does not satisfy the requirement; it gets outside of $D$, as $ \sigma_2 > 1$. Exclude a ray from $D$, there is such a map, given by $re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$.
Edit:
Here is a summary of the results from Dmitri Panov's great answer:
For every choice of $\sigma_1 \in (0,1)$ he constructs an example for a diffeomorphism $D\setminus \{0\} \to D \setminus \{0\}$ with the singular values $(\sigma_1, \frac{1}{\sigma_1})$:
$f_c: (r,\theta)\to r(\cos(\theta+c\log(r)), \sin(\theta+c\log (r))),\;\; $
(for every non-zero $c ֿ\in \mathbb R$, $f_c$ is an example).
I still want to know whether there exits a diffeomorphism defined on all of $D$.
He constructs an example for a (non-injective) smooth map $D \to D$ that satisfies the requirements, whenever $\sigma_1 < \frac{1}{2}$.
Here are the details:
Let $D_0$ be the unit disk centered around zero, and let $D_a$ be the unit disk centered around $(a,0)$ where $a>1$. (so $D_a$ does not contain the origin).
Consider the map $f: re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$. $f(D_a)$ is contained in the disk of radius $\sigma_1(1+a)$, centred at $(0,0)$. Thus, if $\sigma_1(1+a)\le 1$, the map $x \to f(x+(a,0))$ sends $D_0$ to $D_0$ and has the desired properties.
Since any $a>1$ will do, and we want to optimize $a$ in order to maximize the range for $\sigma_1$, we can take $a \searrow 1$, and construct an example for any $\sigma_1 < \frac{1}{2}$.
I still want to know whether there exits such maps defined on all of $D$, for all values of $\sigma_1 \ge \frac{1}{2}$.
Panov proves that whenever $\sigma_1 \ge \frac{1}{\sqrt{2}}$, every smooth map $D \to D$ with the singular values $(\sigma_1, \frac{1}{\sigma_1})$ must be a diffeomorphism. (but we still don't know whether such diffeomorphisms exist).
Conclusion from items $(2),(3)$:
For $\sigma_1 < \frac{1}{2}$ there are non-diffeomorphic examples. For $\sigma_1 \ge \frac{1}{\sqrt{2}}$ every potential example is a diffeomorphism. We still don't know what happens when $\sigma_1 \in [\frac{1}{2},\frac{1}{\sqrt{2}})$.
This reminds me of the study of quasi-conformal homeomorphisms.
This answers the first (simple) half of the question, asking just about a smooth map. In fact, you've already given an answer to it, in some sense. Apply the map $f: re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$ to a unit disk that doesn't contain $(0,0)$, say radius $1$ disk $D$, centred at $(2,0)$. Then, the image $f(D)$ is contained in the disk of radius $3\sigma_1$, centred at $(0,0)$. So if $\sigma_1<\frac{1}{3}$, the map $f+(2,0)$ sends $D$ to $D$ and has the desired properties.
PS. Concerning the second part of the question about a diffeomorphism, I can't give an example, but can give something that looks almost like an example (though one can show that this can not be perturbed to an example by a perturbation which is $C^1$-small close to the boundary of the disk). I'll construct a one parameter family of maps $f_c:D\to D$ (two maps $f_c$ for each $\sigma_1\in (0,1)$). These maps are diffeos on the complement to $(0,0)\in D$, and have all required properties on $D\setminus (0,0)$, though they are not differentiable at $(0,0)$. In the radial coordinates the formula is as follows:
$$f_c: (r,\theta)\to r(\cos(\theta+c\log(r)), \sin(\theta+c\log (r))),\;\; c\in \mathbb R.$$
In order to see that these maps satisfy the necessary conditions, it is enough to notice that they have the following two properties:
1) Each circle $x^2+y^2=r^2$ is sent isometrically to itself.
2) Each radial segment $(\cos(\theta)t, \sin(\theta)t)$ ($t\in [0,1]$) is sent to a spiral $e^s(\cos(\psi+cs),\sin(\psi+cs))$ parameterized by $s$ (where $\psi$ is a constant that depends on $\theta$).
All the conditions are satisfied because the family of spirals and the family of circles form the same angle at all points in $D\setminus (0,0)$, and the map is obviously a symplectomorphism and has unit norm on all the circles. For each $\sigma_1<1$ correspond exactly two $c$ (that differ by a sign).
PPS. 22/02/2020 I would like to propose one more statement (of which I am very happy), concerning the maps with $\sigma_1\in (\frac{1}{2},1]$.
Lemma. Suppose that we have a smooth map $f: D\to \mathbb R^2$ from the unit disk to $\mathbb R^2$ with fixed $\sigma_1<1$ and $\sigma_2=\frac{1}{\sigma_1}$. Then $f$ is univalent (i.e. a diffeo on its image) in case $\sigma_1>\frac{1}{\sqrt{2}}$.
The main tool of the proof is the isomperimetric inequality that says that a simple closed curve $\eta$ on $\mathbb R^2$ bounds area nor more than $\frac{l(\eta)^2}{4\pi}$. It is also surprising, that the constant $\frac{1}{\sqrt{2}}$ is exact, i.e. for $\sigma_1<\frac{1}{\sqrt{2}}$ the map doesn't need to be univalent!
Proof. Assume the converse. Denote by $D_r\subset D$ a disk of radius $r\le 1$ centred at $(0,0)$. Clearly, for $r$ small enough the restriction map $f: D_r\to D$ is a diffeomorphism onto its image. Hence there is a minimal $t\in (0,1]$, such that this map is not a diffeo on its image. Let $S_t$ be the boundary of $D_t$ (a circle of radius $t$). Clearly, the curve $f(S_t)$ touches itself at some point. There could be more than one point where it touches itself, but the argument will not change, so we will assume that $f(S_t)$ self-touches once.
Let $x$ and $y$ be the two points in $S_t$ such that $f(x)=f(y)$. Let $(xy)$ and $(yx)$ be the two arcs into which $x$ and $y$ cut $S_t$. Without loss of generality we assume that the arc $(xy)$ is longer than $(yx)$. Denote by $\gamma_{xy}$ and $\gamma_{yx}$ the images $f((x,y))$ and $f((y,x))$. Both these images are simple closed loops. Let's prove first that the loop $\gamma_{xy}$ contains the loop $\gamma_{yx}$ in its interior.
Indeed, assume the converse. Note that by definition, the length $l(\gamma_{yx})$ satisfies
$$l(\gamma_{yx})\le \frac{1}{\sigma_1}l([yx])\le \sqrt{2}\cdot \pi t.$$
One the other hand, the curve $\gamma_{yx}$ encloses the whole image $f(D_t)$ of $D_t$ and the disk bounded by $\gamma_{xy}$, that doesn't belong to $f(D_t)$. So, since $f$ is area preserving, we see that $\gamma_{yx}$ encloses area more than $\pi t^2$. This contradicts the isomperimetric inequality.
We conclude, that $\gamma_{xy}$ encloses $\gamma_{yx}$, and moreover $l((xy))>\pi t$.
Let's go on to get a contradiction. Let $z$ be the midpoint of the chord $[xy]$ in $D_t$ that joins $x$ and $y$ (don't confuse it with the curvy arc $(x,y)$ that lies in $S_t$!). Let $z$ be the midpoint of $[xy]$. Consider the circle $S_z$ centred at $z$ that passes through $x$ and $y$. Then one half of this circle lies inside $D_t$. Denote this half-circle by $\eta$. Obviously, $l(\eta)=\pi\frac{l([xy])}{2}$. Denote further by $D_t'$ the connected component of $D_t\setminus \eta$ that contains the shorter arc $(y,x)$ of $S_t$. Note finally, that
$${\rm area}(D')>\frac{1}{2}\pi\left (\frac{l([xy])}{2}\right)^2,$$
since $D'$ contains a half-disk of radius $\frac{l([x,y])}{2}$ centred at $z$.
Now, to get the contradiction, we apply again the isoperimetric ineqaulity, this time to the disk bounded by the simple curve $f(\eta)$. By construction, the disk bounded by $f(\eta)$ contains in its interior the image $f(D')$, so it has area more than ${\rm area}(D')$. At the same time the length of $f(\eta)$ is less than $\frac{\pi}{\sqrt{2}} l(xy)$. END of proof.
Moral. If there are self-maps with $\sigma_1>\frac{1}{\sqrt{2}}$, they are diffeos... But I still don't know if such diffeos exist:)
PPPS. 25/02. I would like to address Asaf's question, stated in the comments. Namely, is it possible to find any symplectomorphism $f:D\to D$ ($D$ is the unit disk) that has distinct singular values at any points. It seems to me that the infinitesimal version of this question is equivalent to the following funny question:
Question. Is it possible to construct a smooth function $H$ of $D$, vanishing at the boundary of $D$, and such that the system of equations
$$\frac{\partial^2 H}{\partial x^2}-\frac{\partial^2 H}{\partial y^2}=0=\frac{\partial^2 H}{\partial x\partial y}$$
has no solutions in the disk?
Indeed, if we consider the Hamiltonian flow for $H$, satisfying the above conditions, I believe that for small time the flow map will have distinct singular values. What is good about this system, is that it should not be hard to program (for someone who unlike me knows how to do this), to look for examples. Namely, one can fix a degree $d>0$ and consider all polynomials $H=(x^2+y^2-1)P_d$, where $P_d$ is a poly of degree at most $d$. The space of such polynomials (for all $d$) will probably be dense in the space of all smooth functions. So if there is a counterexample to the question, it should be possible to find it among polynomials. Maybe its degree will not be very large (according to Arnold's "Topological Economy Principle in Algebraic Geometry", ). Or, on the contrary, if there is no counterexample among polynomials, there will be no among all functions, this would be quite exiting. It is easy to check that for $d=1$ indeed there is no counterexamples.
Dear Asaf, indeed it is true that one of two curves $\gamma_{xy}$ or $\gamma_{yx}$ should contain the other in the interior. I am using here the following lemma: Suppose $f: D_t\to \mathbb R^2$ is an immersion and it is a one-to-to one map onto the image apart from two points $x,y\in \partial D_t$ that are mapped to the same point. Then the point $f(x)=f(y)$ cuts $f(\partial D_t)$ into two loops one of which contains the other. In particular, the bouquet of two circles can never be the boundary of an immersed disk. Are you happy with the lemma? Does this answer your question?
Yes, thank you. I am happy with the lemma. I reviewed more thoroughly your argument again, and I find it fantastic, really. Last two remarks: (1) Are you sure that the constant $\frac{1}{\sqrt{2}}$ is sharp? As far as I can see your original construction can produce such non-diffeomorphic maps whenever $\sigma_1 < \frac{1}{2}$ (see my edit in the question), while your claim that such maps must be diffeomorphisms holds whenever $\sigma_1 >\frac{1}{\sqrt{2}}$. So, we are still left with the regime $(\frac{1}{2},\frac{1}{\sqrt{2}})$ where we don't know whether such non-diffeomorphic maps exist
Concerning sharpness, there are two ways to state it, for your original question, indeed I don't know how to construct maps $D\to D$ with $\sigma_1>\frac{1}{2}$. But as you see, I have removed the condition from the lemma that we have a map $D\to D$. I prefer to state this lemma just for maps $D\to \mathbb R^2$. In this case, if we take a disk of radius $1$ centred at the point $(1+\varepsilon, 0)$ and apply the standard map with $\sigma_1<\frac{1}{\sqrt{2}}$, then for small $\varepsilon$ the image $f(D)$ will overlap with itself. Though it will not be contained in a disk of radius $1$.
Concerning just constructing a diffeo with distinct singular values, it is an interesting question, I agree that one should first think about it. Probably I would first try to construct an example. If this doesn't work, I would try to connect it to $2$-dimensional symplectic geometry. Anyway, if an example comes to my mind, I'll tell you. One immediate idea that comes to mind, is to try to construct a symplectomorphism $D\to D$ that dilatetes (stretches) a certain non-zero vector field $v$ in $D$. It is not clear why such a symplectomrphism would not exist. But who knows...
I realised, that this question about constructing an infinitesimal symplectomorphism resembles a bit an infinitesimal version of https://en.wikipedia.org/wiki/Carath%C3%A9odory_conjecture . This conjecture studies umbilic points of a convex surface in $\mathbb R^2$. Namely if you consider the graph of the Hamiltonian multiplied by $\varepsilon$, $z=\varepsilon H(x,y)$, then in the limit $\varepsilon\to 0$, the umbilic points on the graph should converge (it seems) to the points with two equal singular values of $H$ in the disk $D$. I don't know whether this is just a superficial connection.
I am curious to know how one could use this question to construct a symplectomorphim with distinct singular values. I can only see how that can give an obstruction, but I am not sure that in the end such a topological obstruction will be enough to prove that such examples don't exist, if this is indeed the case. (Who knows though...)
Thank you for your answer on the topological obstruction. Indeed, I was sloppy in my previous phrasing; naively, the topological information only gives a necessary condition for the existence of such a symplectomorphim. I have elaborated on the nature of the obstruction here: https://math.stackexchange.com/questions/3557359/does-there-exist-a-self-diffeomorphism-of-the-disk-with-no-conformal-points (see the edited question). Perhaps you would have additional insights about this.
To be honest, I doubt that this topological obstruction will work. On the positive side, one can use it to show that the singularity in the example in the PS above can not be smoothened. On the other hand, I guess it should not be hard to construct a symplectomorphism from a neighbourhood of the boundary of the disk for which this obstruction equals zero (I think, if necessary, I can do this, just a bit techincal). But then it is completely unclear why such a symplectomorphism extends to the whole disk... I would first try to use computer, as in the above Question, to look for an example.
@DmitriPanov Regarding the proof the lemma, why is that the disk bounded by $f(\eta)$ contains in its interior the whole image $f(D')$ ? Also, could you please explain the inequality $l(\gamma_{yx})\le \frac{1}{\sigma_1}l([yx])$ ? Thank you !
Did you find out the answer to the original question ?
I came across this* work (pg. 775, conjecture 7.1) where precisely that question is formulated as a conjecture (I ignore whether or not the author managed to prove or refute it (What he calls cps-self homeomorphism is precisely your mapping with constant singular values). 2. and 3. in your summary are very interesting contributions but still, I'd be glad to know if there are examples of diffeomorphism).
*Gevirtz, Julian, Boundary values and the transformation problem for constant principal strain mappings, Int. J. Math. Math. Sci. 2003, No. 12, 739-776 (2003). ZBL1015.35062.
Thank you, that is an interesting reference. No, at the moment I don't know whether there exist a diffeomorphism of the entire disk with constant, distinct singular values. It seems that there is no 'continuous family' of such diffeomorphisms though, see the end of the answer here: https://mathoverflow.net/a/389718/46290
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2025-03-21T14:48:29.899173
| 2020-02-18T10:07:19 |
352984
|
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|
Stack Exchange
|
Is $\mathfrak j_{2:1}=\mathfrak{j}_{2:2}$ in ZFC?
A function $f:\omega\to\omega$ is called
$\bullet$ 2-to-1 if $|f^{-1}(y)|\le 2$ for any $y\in\omega$;
$\bullet$ almost injective if the set $\{y\in \omega:|f^{-1}(y)|>1\}$ is finite.
Let us introduce two critical cardinals, related to $2$-to-$1$ functions:
$\mathfrak{j}_{2:1}$ is the largest cardinal $\kappa\le\mathfrak c$ such that for any family $F\subset \omega^\omega$ of $2$-to-$1$ functions with $|F|<\kappa$ there exists an infinite subset $J\subset\omega$ such that for any $f\in F$, the restriction $f{\restriction}J$ is almost injective;
$\mathfrak{j}_{2:2}$ is the largest cardinal $\kappa\le\mathfrak c$ such that for any family $F\subset \omega^\omega$ of $2$-to-$1$ functions with $|F|<\kappa$ there are two infinite sets $I,J\subset\omega$ such that for any $f\in F$ the intersection $f(I)\cap f(J)$ is finite.
It can be shown that $\max\{\mathfrak s,\mathfrak b\}\le\mathfrak j_{2:1}\le\mathfrak j_{2:2}\le\mathrm{non}(\mathcal M)$.
I would like to have more information on the cardinals $\mathfrak j_{2:1}$ and $\mathfrak j_{2:2}$.
Problem 0. Is $\mathfrak j_{2:1}=\mathfrak j_{2:2}$ in ZFC?
Problem 1. Is $\mathfrak j_{2:2}=\mathrm{non}(\mathcal M)$ in ZFC?
Problem 2. What is the value of the cardinals $\mathfrak j_{2:1}$ and $\mathfrak j_{2:2}$ in the Random Model? (In this model $\mathfrak b=\mathfrak s=\omega_1<\mathfrak c=\mathrm{non}(\mathcal M)$, see $\S$11.4 in this survey paper of Blass).
Remark. It can be shown that the cardinal $\mathfrak j_{2:1}$ (resp. $\mathfrak j_{2:2}$) is equal to the smallest weight of a finitary coarse structure on $\omega$ that contains no infinite discrete subspaces (resp. contains no infinite asymptotically separated sets). In this respect $\mathfrak j_{2:1}$ can be considered as an asymptotic counterpart of the cardinal $\mathfrak z$, defined as the smallest weight of an infinite compact Hausdorff space that contain no nontrivial convergent sequences. The cardinal $\mathfrak z$ was introduced by Damian Sobota and deeply studied by Will Brian and Alan Dow.
The similarity between $\mathfrak j_{2:1}$ and $\mathfrak z$ suggests another
Problem 3. Is $\mathfrak j_{2:1}=\mathfrak z$ in ZFC?
A theorem of Koppelberg states that $\mathrm{cov}(\mathcal M) \leq \mathfrak{z}$. It's consistent that $\mathrm{non}(\mathcal M) < \mathrm{cov}(\mathcal M)$ (this happens in the Cohen model), and you've proved $\mathfrak{j} \leq \mathrm{non}(\mathcal M)$. So it is consistent that $\mathfrak{j} < \mathfrak{z}$. If you feel there's some connection between $\mathfrak{j}$ and $\mathfrak{z}$, maybe Problem 3 should ask whether $\mathfrak{j} \leq \mathfrak{z}$. If you could prove this upper bound, it would also solve Problem 1, since $\mathfrak{z}$ and $\mathrm{non}(\mathcal M)$ are incomparable.
Actually, Alan Dow has proved (though it's unpublished) that $\mathfrak{z} = \aleph_1$ in the Laver model. But we have $\mathfrak{b} = \aleph_2$ in the Laver model and hence $\mathfrak{j} = \aleph_2$ as well. So $\mathfrak{z} < \mathfrak{j}$ is consistent. Combined with my previous comment, this shows there is no ZFC-provable inequality between $\mathfrak{j}$ and $\mathfrak{z}$.
I can answer problems 2 and 3, although I still don't know the answer to problems 0 and 1.
The main point is that
$\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model.
I'll sketch a proof of this below. (It's a bit long, but I've tried to make it readable.) The proof actually shows a little more: it gives you $\mathrm{cov}(\mathcal{N}) \leq \mathfrak{j}_{2:1}$.
This result also answers problem 3, because we know that $\mathfrak{z} = \aleph_1$ in the random model. (This was first proved by Alan Dow and David Fremlin here. It is also a corollary to Theorem 4.2 in this paper by me and Alan.) Therefore $\mathfrak{z} < \mathfrak{j}_{2:2},\mathfrak{j}_{2:1}$ is consistent. On the other hand, Koppelberg proved that $\mathfrak{z} \leq \mathrm{cov}(\mathcal{M})$. (Actually, she proved the dual statement in the category of Boolean algebras here. Stefan Geschke wrote a purely topological proof here.) Because you have proved that $\mathfrak{j}_{2:2},\mathfrak{j}_{2:1} \leq \mathrm{non}(\mathcal{M})$, and because $\mathrm{non}(\mathcal{M}) < \mathrm{cov}(\mathcal{M})$ in the Cohen model, it follows that $\mathfrak{j}_{2:2},\mathfrak{j}_{2:1} < \mathfrak{z}$ is consistent. Thus there is no inequality between $\mathfrak{z}$ and either of $\mathfrak{j}_{2:2}$ of $\mathfrak{j}_{2:1}$ that is provable in $\mathsf{ZFC}$.
(I know I gave a different argument for this in the comments. I don't like that argument as much because it relies on Alan's unpublished -- and mostly unwritten -- argument that $\mathfrak{z} = \aleph_1$ in the Laver model. I'm sure he's right. But I like that the argument here relies on the fact that $\mathfrak{z} = \aleph_1$ in the random model, and you can go read one or two proofs of this if you like.)
Now let's sketch the proof that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model. For the sake of clarity, I'm going to avoid forcing jargon and give a probabilistic argument that (I hope) will give you the right idea.
To show that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model, let's first recall how random real forcing works. Roughly, we imagine ourselves to live in a universe $V$ of sets, containing real numbers, subsets of $\mathbb N$, lots of $2$-to$1$ functions, and whatever else. But we know that our universe is about to get bigger -- this is the forcing -- by the introduction of a "truly random" real number $r$. The new, bigger universe is called $V[r]$.
The first observation I'd like to make is that all continuous measures on uncountable Polish spaces are essentially isomorphic. This means that it doesn't matter whether we view $r$ as a random element of $\mathbb R$, or of $[0,1]$, or of $2^\omega$ with the standard product measure, or whatever. For this problem, we want to view $r$ as an infinite sequence of random selections from larger and larger finite sets $I_n$, where $I_n$ has size $n!$. We select, at random, only a single element from each set. (This can be formalized by saying that we'd like $r$ to be a random element of the Polish space $\prod_{n = 0}^\infty \{1,2,\dots,n!\}$, equipped with the usual product measure. But let's keep it informal.) So our universe is about to get bigger by introducing a truly random sequence of selections from some sets $I_0, I_1, I_2, \dots$ with $|I_n| = n!$.
Within $V$, we can try to anticipate objects that will be constructible from $r$ in $V[r]$. For example, we can anticipate that once we get $r$, we can build a set $J \subseteq \mathbb N$ according to the following recipe: first identify $I_n$ with the interval $[1+1+2+\dots+(n-1)!,1+1+2+\dots+(n-1)!+n!) \subseteq \mathbb N$, and then let the $n^{\mathrm{th}}$ element of $J$ be whatever $r$ randomly selects from this interval.
Now I claim that this set $J$ described above has the following property: if $f$ is any $2$-to-$1$ function in the ground model $V$, then the restriction of $f$ to $J$ is almost-injective. To prove this, it suffices to argue that it's true with probability $1$, given that $r$ makes its selections randomly. This suffices because this is precisely what we mean when we say that $r$ is a "truly random" addition to $V$: if there is a randomness test defined in $V$ (such as one defined from any $f \in V$), then $r$ is random with respect to that test.
So let's argue probabilistically.
Fix a $2$-to-$1$ function $f \in V$.
If $f(a) = f(b)$, we may view this as a "guess" that $f$ is making about our set $J$: the guess is that $a$ and $b$ are both in $J$. In other words, $f$ gets to guess at pairs from $J$ infinitely many times, and it is our job to prove that, with probability $1$, only finitely many of these guesses are correct.
So what is the probability that $f$ correctly guesses a pair of elements from $J$?
If $f$ identifies a member of some $I_m$ with a member of some $I_n$, where $m \neq n$, then there is a probability of exactly $\frac{1}{m!n!}$ that $f$ will have correctly guessed a pair from $J$. When $f$ makes other kinds of guesses (not identifying some member of some $I_m$ with a member of some $I_n$, where $m \neq n$), then the probability is $0$ that $f$ will have correctly guessed a pair from $J$.
If $m < n$, then $f$ gets at most $|I_m| = m!$ chances to guess a pair from $J$ with one member in $I_m$ and the other in $I_n$. By the previous paragraph, the probability of one of these guesses being correct is $\leq\! m!\frac{1}{m!n!} = \frac{1}{n!}$. Summing over all $n > m$, it follows that the probability of $f$ correctly guessing any pair of elements from $J$ with one member in $I_m$ and the other in $I_n$ for some $n > m$ is
$$\leq\! \sum_{n = m+1}^\infty \frac{1}{n!} < \frac{1}{(m+1)!} \sum_{k = 0}^\infty \frac{1}{(m+1)^k} = \frac{1}{(m+1)!}\frac{m+1}{m} < \frac{1}{m!m}.$$
Now fix $k > 0$. Summing over all $m > k$, we see that the probability of $f$ correctly guessing a pair of elements from $J$ with one member in $I_m$ and the other in $I_n$ for some $n > m > k$ is
$$\leq\! \sum_{m = k+1}^\infty \frac{1}{m!m} < \sum_{m = k+1}^\infty \frac{1}{m!} < \frac{1}{k!k}.$$
Therefore the probability of $f$ correctly guessing a pair of elements from $J \setminus (I_0 \cup \dots \cup I_k)$ is at most $\frac{1}{k!k}$. For any fixed $\varepsilon > 0$, we can choose $K$ large enough that $\sum_{k = K}^\infty \frac{1}{k!k} < \varepsilon$. This means that for $K$ large enough, the probability of $f$ correctly guessing more than ${K+1} \choose 2$ pairs of elements of $J$ is less than $\varepsilon$. Therefore the probability of $f$ correctly guessing infinitely many pairs of elements of $J$ is less than $\varepsilon$. As $\varepsilon$ was arbitrary, the probability of $f$ correctly guessing infinitely many pairs of elements of $J$ is $0$.
This shows that our set $J$ in $V[r]$ "should" (probabilistically) have the property that $f \restriction J$ is almost injective for every $f \in V$. But as we said earlier, this means $J$ really does have this property.
Why does this mean $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model? The random model is $V[G]$, where $G$ is a "random" element of the measure algebra $2^{\aleph_2}$. If $\mathcal F$ is any set of $\aleph_1$ $2$-to-$1$ functions in $V[G]$, then a standard "nice names" argument shows that there is some weight-$\aleph_1$ subalgebra $X$ of $2^{\aleph_2}$ such that $\mathcal F$ is already in the intermediate model $V[X \cap G]$. Because $|X| = \aleph_1$, there will be random reals added in moving from the intermediate model $V[X \cap G]$ to the final model $V[G]$ -- random over $V[X \cap G]$, not just over $V$. We've just showed that the addition of these random reals adds some $J$ that "works" for every $f \in \mathcal F$.
Why does this mean $\mathrm{cov}(\mathcal{N}) \leq \mathfrak{j}_{2:1}$? There are a few ways to see this. The easiest is probably just to go through the above argument and convince yourself that what we've really proved is that every $2$-to-$1$ function $f$ is "solved" by a measure-$1$ set of $J$'s in the Polish space $\prod_{n = 0}^\infty \{1,2,\dots,n!\}$. Equivalently, the set of $J$'s that fail to work for a given $2$-to-$1$ function $f$ is a null set $N_f$.
Therefore, if $\mathcal F$ is any size $<\! \mathrm{cov}(\mathcal{N})$ family of $2$-to-$1$ functions, $\bigcup_{f \in \mathcal F}N_f$ does not cover our Polish space, and so there is some $J$ that works for every $f \in \mathcal F$.
Thank you Will, very much, for the detail explanation. But if I understood correctly, everything follows from the lower bound $\mathrm{cov}(\mathcal N)\le j_{2:1}$ which holds in ZFC. And then we can just apply the know information (see the Table 4 in the Blass' survey) that $\mathrm{cov}(\mathcal N)=\mathfrak c$ in the Random Model. Right?
Yes, that's right. We prove the lower bound by considering the construction of $J$ from a random sequence as described above, and then showing that a "sufficiently random" sequence makes $J$ work for a given $2$-to-$1$ function. To be honest, I phrased the whole thing in terms of forcing because that's how I thought of it, and by the time realized the argument can be modified to give us a ZFC inequality, I'd already written most of it down and didn't have the energy to go back and modify it too much!
Ok, but maybe the proof of the lower bound $\mathrm{cov}(\mathcal N)\le \mathfrak j_{2:1}$ actually yields $\mathrm{cov}(\mathcal E)\le\mathfrak j_{2:1}$ where $\mathcal E$ is the $\sigma$-ideal generated by closed null sets?
What is unclear to me is why the measure of resolving sets $J$ for a given 2-to-1 function has measure 1. Just take the 2-to-1 function $\varphi$ such that $\varphi(2n)=\varphi(2n+1)$ for every $n\in\omega$. For a set $J\subset \omega$ to be resolving this $\varphi$ the intersection $J\cap{2n,2n+1}$ should be equal to $\emptyset$, ${2n}$ or ${2n+1}$. This yields probablity $3/4$ to be resolving on the doubleton ${2n,2n+1}$. And for infinitely many such doubleton you will get $(3/4)^\omega=0$. So, the measure of resolving is zero, not 1. How this can be explained?
@TarasBanakh: You would be correct if $J$ were defined by just randomly deciding (via a coin flip, for example) with probability $\frac{1}{2}$ whether any given integer is in $J$. This is what is often meant by "a random real" and it won't work here for the reason you state. Our set $J$ is constructed differently, by selecting just one member of each interval $I_k$, where $|I_k| = k!$. The probability that some particular $n$ is in our set $J$ is equal to $\frac{1}{k!}$, where $k$ is the unique number with $n \in I_k$. So the probability of selecting the doubleton ${2n,2n+1}$ is very small.
@TarasBanakh: I looked back at how I phrased things, and it was a bit confusing. I've added a comment to clarify what I mean in the construction of $J$.
Great! Now I understand. Thank you.
The lower bound $\mathrm{cov}(\mathcal N)\le \mathfrak j_{2:1}$ cannot be improved to $\mathrm{cov}(\mathcal E)\le\mathfrak j_{2:1}$ as it would imply the lower bound $\mathrm{cov}(\mathcal M)\le\mathrm{non}(\mathcal M)$ which does not hold in ZFC.
Now we have the bounds $\max{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal N)}\le\mathfrak j_{2:1}\le\mathfrak j_{2:2}\le\mathrm{non}(\mathcal N)$ and in order to distinguish the cardinals $\mathfrak j_{2:1}$ and $\mathfrak j_{2:2}$ we have to look at models with $\max{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal N)}<\mathrm{non}(\mathcal N)$. I noticed that no model in the Table 4 of the Blass survey has this property. Do you know any model of ZFC satisfying such strict inequality?
A good reference for that sort of question is the book by Bartoszynski and Judah -- Chapter 7 shows you how to separate any two cardinals in Cichon's diagram. There are three ways listed to get $\mathfrak{b},\mathrm{cov}(\mathcal N) < \mathrm{non}(\mathcal M)$, although none of them is particularly simple to describe. I would imagine all of them have $\mathfrak{s} = \mathfrak{b}$, although this isn't stated explicitly in the text. I'm not sure what the $\mathfrak{j}$'s would be in these models, but I think that would be a good place to look.
|
2025-03-21T14:48:29.900196
| 2020-02-18T10:10:32 |
352985
|
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|
Stack Exchange
|
Is the rectifiability of currents independent of the choice of Riemannian metric?
I apologize if this is a trivial question – GMT is not my area of expertise but I'm working my way through a proof that makes extensive use of GMT and I haven't been able to find an answer to my question in the usual books.
Suppose $(M^n, g)$ is a Riemannian manifold, for $j ≥ 0$ denote by $ℋ^j$ the induced $j$-dimensional Hausdorff measure. Let $T ∈ D_j(M)$ denote an (integer-multiplicity) rectifiable $j$-current in $(M, g)$, that is, for all compactly supported $j$-forms $ω ∈ D^j(M)$:
$$
T(ω) = ∫_A ⟨ ω, \vec{T} ⟩ \, θ \,dℋ^j \;\;,
$$
where $A ⊂ M$ is a $j$-rectifiable set (i.e. up to an $ℋ^j$-nullset it is contained in a countable union of $j$-dimensional submanifolds of $M$), $\vec{T}: M → Λ^j(TM)$ is an $ℋ^j$-measurable section (a simple $j$-vector made of orthonormal vectors that orient the approximate tangent space $T_x A$ of $A$ at a.e. $x$) and $θ: A → ℕ$ is locally $ℋ^j$-integrable. (I'm following Leon Simon's definition here.)
Suppose I picked a different Riemannian metric $g'$ on $M$. Is $T$ then still a rectifiable current w.r.t. $(M, g')$ and the induced Hausdorff measure $ℋ'^j$? Heuristically speaking, when the metric changes, both $\vec{T}$ and $dℋ^j$ change and these changes should cancel out. At least for currents given by smooth submanifolds this is definitely the case. Is it true in general, though?
UPDATE: I'm increasingly sure that one can use a parametrization of $A$ by a countable family of Lipschitz functions (living on Borel sets in $\mathbb{R}^j$) together with the area formula to prove this. I'll be looking into this. In the meantime, in case anyone's got a reference or can give a definite answer to my question, I'd still very much appreciate it!
UPDATE 2: The proof indeed goes through by taking local parametrizations of the submanifolds containing the rectifiable set and then considering the pullback of the ambient metric under such parametrizations and using the area formula. I'll post a detailed proof as soon as I find the time.
The way you define it, your class corresponds to the class of "locally integral currents" in Federer's book (see Section 4.1.24 for the definition in Euclidean space). This class is not affected by the change of a metric on a Riemmanian manifold because as metric spaces they are locally bi-Lipschitz equivalent (one can take the identity map). The class of integral currents (assuming they are allowed to have noncompact support) changes in general because a current may have finite mass with respect to one metric but not with respect to another.
@rozu Thanks for your comment! However, I think you meant locally rectifiable currents, right? After all, I didn't say anything about the boundary. And yes, the mass (and its (un-)boundedness in case the current has non-compact support) can obviously change when the metric changes. Finally, thanks for pointing me at the bi-Lipschitz property – I'll look into that!
Yes indeed, I meant locally rectifiable.
|
2025-03-21T14:48:29.900820
| 2020-02-18T12:12:51 |
352988
|
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|
Stack Exchange
|
Example of a non log-canonical pair for an abelian variety with polarization of degree >2
Let $(A,L)$ be a polarized abelian variety of dimension $g$, with an indecomposable polarization of degree $\chi(L)=d$. There is a theorem of Debarre and Hacon about the singularities of pairs:
Theorem: Let $D \in |mL|$ for some $m\ge 2$ be a divisor, and suppose $d=2$. Then the pair $(A,(1/m)D)$ is log-canonical.
In his notes (https://www.math.ens.fr/~debarre/DivVarAb.pdf) Debarre says that, for $d>2$ and $m\ge d-1$, there are pairs $(A,(1/m)D)$ which are not log-canonical anymore, and gives an example:
Let $(A_1,L_1)$ be a general polarized abelian variety of type $(d-1)$ and let $E$ be an elliptic curve. Pick an isomorphism $\psi\colon K(L_1) \to E[d-1]$ and consider the quotient $A$ of $A_1 \times E$ by the subgroup $\{(x,\psi(x))~|~x \in K(L_1)\}$. There is a divisor $\Theta$ on $A$ that defines a principal polarization and $\mathcal{O}_A(\Theta)$ restricts to $L_1$ on $A_1$. The line bundle $L:=\mathcal{O}_A(\Theta+A_1)$ defines an indecomposable polarization of degree $d$ on $A$ that is indecomposable if $d\ge3$.
The linear system $|(d-1)\Theta-A_1|$ is nonempty, hence so is the linear system $|m\Theta-(m'-m)A_1|$ for $\frac{d}{d-1}m\ge m' > m\ge d-1$. If $D'$ is in that linear system, $D=D'+m'A_1$ is in $|mL|$ and the pair $(A,(1/m)D)$ is not log-canonical since it has a component of multiplicity $>1$.
I have some problem understanding this example, in particular:
Why does such a $\Theta$ on $A$ exists?
Why is $L$ an ample line bundle and why is its degree as a (indecomposable?) polarization equal to $d$?
Why is the linear system nonempty?
I don't know if this can be seen as a (partial) answer (or maybe an edit), so I post it as a comment. For the first question, I've found the construction behind $\Theta$ in another article of Debarre (here), construction 9.2. Still, I'm not sure about $L$: is it obvious that is an ample line bundle? Why is the degree $d$ and the polarization indecomposable?
|
2025-03-21T14:48:29.901008
| 2020-02-18T13:26:03 |
352994
|
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|
Stack Exchange
|
Zeroes of linear combination of sines
Let us consider $$f(z):=\sum\limits_{j=1}^{j=n}a_j\sin(\lambda_jz) $$
where all $a_j$ and $\lambda_j$ (of course, $\lambda_j$ are distinct) are real numbers and $n \in \mathbb{Z},\, n \ge 3$. The question is: are all the zeroes of $f(z)$ real?
Obviously not. Any 3 complex numbers are linearly dependent over reals. So, as long as $n\geq 3$ you can choose non-zero $a_j$ to make your expression zero at any given complex number.
I think also for $n=2$ there are counter examples: $\sin z+3\sin(z/2)=0$ for $z=2\pi+2i\log(\tfrac{3}{2}+\tfrac{1}{2}\sqrt{5})$
@Carlo Beenakker: Thank you. How about $n \ge 3$? I refined my question.
@Oltg Eroshkin: Can you kindly elaborate your comment in details?
there are complex roots for any $n$; for example, for $n=3$ try $\sin z+\tfrac{1}{2}\sin 2z+\sin(3z/2)$, which vanishes at $z=4.57903+0.785214 i$.
@Carlo Beenakker: do you mean $- 4.57903271180618+ 0.785214119641865,i$ up to the command of Maple RootFinding:-Analytic(sin(z) + 1/2sin(2z) + sin((3*z)/2), z = -5 - I .. 0.5 + I)? Can you present your answer for the general case?
@Oleg Eroshkin: I am not sure about " non-zero $a_j $ ".
@Carlo Beenakker: All that is a too general direction. Can you kindly give an example for e.g. $n=10$? TIA.
Down votes are made In the best practice of some MO users. What is wrong in my question?
Note that $\sinh$ is a strictly increasing positive unbounded function on $(0,\infty)$ so for say $n \ge 3$, the equation $\sinh 1+ \sinh 2+...\sinh (n-1) =\sinh x$ has a unique positive solution $x_n> n-1$.
Using that $\sin {iy}=i\sinh y$ for real $y$, the above means that $i$ is a root of the equation $\sin z+\sin 2z +..\sin {(n-1)z}-\sin x_nz=0$
For $n=2$ we can adapt this and use that $2\sinh 1=\sinh x_2$ for a unique $x_2>1$ and get that $i$ is a root of $2\sin z-\sin x_2z$
Proof of the existence of a complex root in the general case, expanding Oleg's remark:
Think of the $n$ complex numbers $\sin(\lambda_j z)$ as vectors in the 2D plane; for $n\geq 3$ and fixed $z$ add the first $n-2$
of these vectors with coefficients $a_j\neq 0$
such that the sum is neither a real multiple of $\sin\lambda_n z$ nor of $\sin\lambda_{n-1}z$; then adjust $a_n\neq 0$ and $a_{n-1}\neq 0$ to cancel the full sum.
an example for $n=10$ (requested by the OP):
$$\sum_{n=1}^8 \sin nz +\tfrac{1}{2}\sin 9z+\sin 10z=0$$
for $z=0.6142+0.0627\,i$.
To demonstrate that this zero is not a numerical artefact, here is a plot in the complex plane of contours where the real part of the sum vanishes (orange) and of contours where the imaginary part vanishes (blue); the contours intersect, demonstrating the existence of a complex root.
The Mathematica code (contour plot in the complex plane of the roots of the real and imaginary parts of sum of sines) :
ContourPlot[{Cosh[y] Sin[x] + Cosh[2 y] Sin[2 x] +
Cosh[3 y] Sin[3 x] + Cosh[4 y] Sin[4 x] + Cosh[5 y] Sin[5 x] +
Cosh[6 y] Sin[6 x] + Cosh[7 y] Sin[7 x] + Cosh[8 y] Sin[8 x] +
1/2 Cosh[9 y] Sin[9 x] + Cosh[10 y] Sin[10 x] == 0,
Cos[x] Sinh[y] + Cos[2 x] Sinh[2 y] + Cos[3 x] Sinh[3 y] +
Cos[4 x] Sinh[4 y] + Cos[5 x] Sinh[5 y] + Cos[6 x] Sinh[6 y] +
Cos[7 x] Sinh[7 y] + Cos[8 x] Sinh[8 y] + 1/2 Cos[9 x] Sinh[9 y] +
Cos[10 x] Sinh[10 y] == 0},
{x, 0.614, 0.6145}, {y, 0.06265, 0.06275}, WorkingPrecision -> 30]
the OP also requests an example for arbitrary $n$, let me take $z=\pi +i$ and $\lambda_k=k$, and all coefficients $a_k$ equal to unity except $a_n=a$,
$$S_{n}=\sum_{k=1}^{n-1}\sin k(\pi +i)+a\sin[n(\pi+i)]=$$
$$\qquad\qquad=-\frac{i}{2 \cosh\left(\frac{1}{2}\right)} \left[(-1)^{n} \sinh \left(n-\tfrac{1}{2}\right)+\sinh \left(\tfrac{1}{2}\right)\right]+a(-1)^ni\sinh n,$$
which vanishes for a nonzero real $a$.
Thank you for you interest to the question. Hope you understand (these are nuts and bolts of numerical analysis), that $z=0.614+0.0627,i$ for $\sum_{n=1}^8 \sin nz +\tfrac{1}{2}\sin 9z+\sin 10z=0 $ proves nothing even in view of its numerical verification with Maple: eval(add(sin(jz), j = 1 .. 8) + 1/2sin(9z) + sin(10z), z = 0.614 + 0.0627*I) which performs $- 0.0001542513- 0.0016096971,i $. I will be waiting for a more solid answer. Thank you anyway.
Also the command of Maple Digits := 15;
RootFinding:-Analytic(add(sin(jz), j = 1 .. 8) + 1/2sin(9z) + sin(10z), z = -1 - I .. 1 + I); performs $ 0+0,i$, not confirming your statement.
Hope you understand that a plot without a command is not any argument.
+1 for the idea, Thank you for the added code. I can reproduce it. The matter is interesting: two different CASes produce different results. We still havt no counterexample for $n=10$.
Can you ground your statement " it's just not the only root."? I repeat numerical calculations do not confirm $z=0.6142+0.0627,i$, producing $ - 0.0001542513- 0.0016096971,i$ instead of zero.
The code of Maple Digits := 15;
RootFinding:-Analytic(add(sin(jz), j = 1 .. 8) + 1/2sin(9z) + sin(10z), z = 0.614 + 0.06265*I .. 0.6145 + 0.06275); produces $0.614170494092837 + 0.0627047401490220,i $, confirming your result. The general case of an arbitrary positive integer $n$ remains open.
Sorry, but your latest example is a slightly changed example by @Conrad. The edit was made after his answer. Here https://www.dropbox.com/s/7z8nfnj45y9qbz8/screen18.10.19.docx?dl=0 is a screen with timings.
|
2025-03-21T14:48:29.901475
| 2020-02-18T14:05:52 |
352997
|
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|
Stack Exchange
|
Vertex operator algebras and isomorphism of graded vector spaces
I have two vertex operator algebras and I would like to show that as graded vector spaces, they are isomorphic, rather than as algebras.
The issue is I have not found anything in the literature that has done this for a particular case. One idea I had was since there is a cohomology theory for VOAs, there are tools to compare invariants of them as algebras, and I am hoping there may be something in the literature which uses this (or some other method), to make a statement about them as graded vector spaces.
I would highly appreciate if anyone can point me to a paper that develops any of these ideas, that is, if one can 'descend' a statement of two VOAs to something about them as graded vector spaces.
The universal enveloping affine algebra associated two two different Lie algebras of the same dimension will have the same graded dimension and will be different VOAs.
That just means that the "characters" $\operatorname{tr}(q^{L_0})$ are the same. For example, the VOA's associated with the $E_8\oplus E_8$ and the $D_{16}^+$ lattice have the same characters and thus are isomorphic as graded vector spaces.
@MarcelBischoff So actually I have fermionic generators so you get a $(-1)^F$ that allows for cancellations to happen. I don't think in that case it's enough for characters to match.
This was answered by Marcel Bischoff in a comment. For any vertex operator algebra, there is an invariant called the "character" or "graded dimension", that captures precisely the isomorphism type of the underlying graded vector space. In many cases, this is a modular form, and you may find yourself using tools from the theory of modular forms to identify characters of vertex algebras.
I am a bit unfamiliar with this, but quick question: does this method of finding the characters circumvent the need to know the null vectors/relations for each vector space?
@JamalS Perhaps it would help if you explained why you find yourself confronted with a computation involving null vectors.
I think the reason I haven't found this method in the literature used is because with the VOAs I am dealing with, they have fermionic generators as well as bosonic, and so cancellations may occur in computing the characters, such that they can no longer be used to establish isomorphism on the level of graded vector spaces.
@JamalS I see. In that case, you can resolve the problem by using a vector-valued character. One of the components is the graded trace of the parity operator, and one of the components is total dimension.
Thanks for your quick reply. Do you have a reference to a paper or textbook that provides a straightforward example of this?
@JamalS See work by John Duncan on Conway moonshine.
|
2025-03-21T14:48:29.901721
| 2020-02-18T15:24:57 |
352999
|
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|
Stack Exchange
|
On functional equation $f\circ \exp=\exp \circ Df$ on a Riemannian manifold or a Lie Group
Let $M$ be a Riemannian manifold or a Lie group whose corresponding exp map (in corresponding context) is denoted by "exp" which is a map $\exp:TM\to M$
We search for the set $\mathcal{H}$ of all smooth maps $f:M\to M$ which satisfy $$f\circ \exp=\exp \circ Df$$
Is there a natural finite dimensional manifold structure on $\mathcal{H}$? In the case that $M$ is a Lie group, is there a Lie group structure on $\mathcal{H}$? In either case what can be said about the dimension of $\mathcal{H}$?
Example: For Lie group $M=\mathbb{R}$ we get $\mathcal{H}=\{ax+b\mid a,b \in \mathbb{R}\}$ the additive group of all affine linear maps on $\mathbb{R}$, a $2$ dimensional Abelian lie group isomorphic to $\mathbb{R}^2$.
Yes, it's a Lie group -- the defining equation you mention, that the maps commute with exponentiation this tells you your maps are determined by their behaviour on a basis for one tangent space. So the Lie group is at most $m + m^2$ dimensional. Presumably this is only realized for $\mathbb R^m$. I imagine for a compact Riemann manifold this is the isometry group.
@RyanBudney For a Riemannian manifold every constant map satisfies this equation so the last part of your comment is not true. Let me think about the first part of your comment. Please read my question again in particular the example in the last part of my post.
@RyanBudney note that a possible group structure on $\mathcal{H}$ is not probably defined via composition but it possibly comes from the group structure of $G$, as it is the case in the example of the post.
@RyanBudney appart from constant map you can find other maps for example $z\mapsto z^2$ on circle, right?
My Lie group comment refers to the maps where Df is invertible. I suppose I imagined that was an assumption you were making. It's certainly a manifold by the frame bundle argument. When the derivative is invertible and the manifold has no self-covering then it's a Lie group. If you allow degenerate maps then I suppose the question is more open-ended. But I don't see a reason for there to be a group structure when allowing degenerate maps, unless you are in some very special case.
Presubably the images of the degenerate maps will be something like Riemann submersions to totally geodesic submanifolds. A fairly special thing.
@RyanBudney after your interesting comment I realized that the equation is preserved by composition. In the case that M is a Lie geoup, is the equation preserved by group multiplication? Namely let $f_1,f_2$ satisfy the equation does their (pointwise) product $f_1f_2$ satisfy the eqiation, too. O guess the answer is affirmatove on abelian case.
In fact your emphasis on "isometric group" and "composition" lead me to realize that the equation is preserved by composition. Thanks for your attention to my question. BTE even if Df is invertible on can not say that f is an isometry for example $z\mapsto z^n$ on circle.
It's probably not a group for $M = S^2 \times S^2$.
What do you mean by $\exp:TM\to M$ in the Lie group context? Do you want something like $X_g\mapsto g\exp(\mathrm{L}_{g^{-1}*}X_g)$, where $\mathrm{L}_g$ denotes left-translation by $g$?
@RobinGoodfellow Yes in the Lie group case we use the translation of exp at e to all other elements. In the other word let X_g be a tangent vector at g. There is a unique left invariant ector field on the group whose restriction to g is X_g. Now we start from g by flow of this vectoe field and stop at t=1. The resulting point is exp. I think that is equivalent to what you say
Let us begin with the “Riemannian” case. In truth, it doesn’t have much to do with the Riemannian structure at all: $\mathcal{H}$ is the space of all affine maps on $(M,\nabla)$, with the affine connection $\nabla$ given by the Levi-Civita connection.
As you note, this will never have a natural Lie group structure unless $M$ is 0-dimensional, since it will always contain the constant maps. If we make the additional restriction that $f$ be a diffeomorphism, however, we get the group of affine transformations, which has been extensively studied and is well-known to be a Lie group when $M$ has finitely many connected components. For compact Riemannian manifolds, the identity component of the group of affine transformations coincides with the identity component of the isometry group.
Regardless of whether we use diffeomorphisms, it is a standard exercise in differential geometry to prove that affine maps are determined entirely by their value and derivative at a single point, so the dimension of $\mathcal{H}$ will be at most $\dim(M)+\dim(M)^2$.
It isn’t a very cohesive space most of the time, though: it will always contain both a copy of $M$ (the constant maps) and the identity map, and a path between a constant map and the identity would constitute a homotopy, which can only occur for $M$ contractible. Thus, in the generic case, it’ll probably just be $M$ together with an isolated point corresponding to the identity map.
In the Lie groups case, it’s essentially the same.
Note that if $f$ satisfies the condition, then so does $\mathrm{L}_g\circ f$. Thus, $f$ is of the form $\mathrm{L}_g\circ f_0$, where $f_0$ fixes a point, which we will predictably choose to be $e$. I think it follows that the space you end up with will be something like $G\times\mathrm{Hom}(G,G)$. Again, this won’t naturally be a Lie group outside of the discrete case, but if we restrict to diffeomorphisms, I think we get the Lie group $G\rtimes\mathrm{Aut}(G)$.
Thank you for your answer. Your first paragraph:" $\mathcal{H}$ is the group of affine maps" as you noted in your answer it is not realy the case because of constant maps and also $z\mapsto z^k$ on circle. So O wonder if there are some other kind of pathological maps. Any way the example in the post shows that the identity is not isolated.
What is this dimension for circle torus or spher? BTW what is your definition of affine maps when the map is not at least local diffeomorphism?
For having a Lie structure, the case 0 dimensional is not the only relevant case. Please look at the example
@AliTaghavi I’m not sure what you’re referring to in my first paragraph. Perhaps you misread? Also, I don’t think there is much chance of pathological examples; as far as smooth maps go, affine maps are exceptionally well-behaved. Finally, the example in your post is rather specific in that you have chosen a contractible Lie group, so in that specific case you can find a homotopy from the identity to a constant map. As I said, this will almost never happen. Try, for example, any Lie group that isn’t diffeomorphic to $\mathbb{R}^n$.
So we get the holomorph of group?(your last word). But I wonder what is the reason and detail of your answer? Can you dummarize your answer in a precise manner in one short psragraph? Thanks again for your attention. my underestanding is that the equation implies that f maps the geodesocs to geodesics. But I wonder what is the details of your answer.
In the lie group case there is an alternative candidate for group structure on $\mathcal{H}$, pointeise multiplication. Os it really closed under this multiplication? May I ask you to read the previous comment conversations and the question agaon carefully?
I’m starting to get the impression that you aren’t actually reading the things I’m writing. Again, I would encourage you to compute a nontrivial example (try $\mathrm{SL}_2(\mathbb{R})$ or even $\mathbb{R}^2\rtimes\mathrm{SO}(2)$) for yourself before continuing to ask me these questions. As for distilling everything into a short paragraph, mathematics cannot always be spoon-fed to you. If you are struggling with concepts like affine maps, then you may find more luck on our sister site, where more elementary mathematical inquiries are probably better suited.
I read your answer but it does not contain about pointwise multiplication in the group case. Right?
I did not worked much with affine maps but i now that it is a map which pull back the connection to itself. Regarding the sister site, the participant of MO can judge about a question and they vote either in + or vite to close. The point is that it is better that we refrain from non mathematical comments in MO
So if you realy find this question as a trivial post please vote to close
You have misunderstood me. I’m saying that the numerous computational questions regarding affine maps, like the ones you are asking in the comments here, are probably more suited to our other site. You should recall the elementary fact that all the connected components of a Lie group are necessarily diffeomorphic, so my answer does say that you won’t get any kind of Lie group in most cases. I continue to ask for you to actually compute a nontrivial example because you do not seem willing to listen to me and instead insist that I have not read what you have written carefully enough.
I should admit that most part of your answer and your comment is not clear for me. In particular there is no a proof in your answer.Could you please give a reference investigate this equation.or can we start a chat?
For affine maps, you can probably find what you want to know in Chapter VI (“Transformations”) of Kobayashi and Nomizu’s Foundations of Differential Geometry, Vol 1. For contractibility, you can probably find that in any elementary topology textbook. For the Lie groups version, note that if $f(e)=e$, then $f(\exp(X))=\exp(f_*X)$, so you should probably be able to say $f(g)f(\exp(X))=f(g)\exp(f_*X)=f(g\exp(X))$.
Very helpful comment thank you
|
2025-03-21T14:48:29.902424
| 2020-02-18T15:42:50 |
353000
|
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|
Stack Exchange
|
Near permutation $n\mapsto n+1$ not conjugate to its inverse on the Stone-Čech remainder?
Let $\beta\omega$ be the Stone-Čech compactification of the discrete infinite countable space $\omega$, and $\beta^*\omega=\beta\omega\smallsetminus \omega$ is the Stone-Čech remainder.
The map $j:n\mapsto n+1$ extends to an self-injection of $\beta\omega$, which itself restricts to a self-homeomorphism $\phi$ of $\beta^*\omega$.
In ZFC+CH, is it true that $\phi$ and $\phi^{-1}$ are not conjugate in $\mathrm{Homeo}(\beta^*\omega)$?
Indeed in Shelah's model ("forcing axiom"), in which CH fails, there exists a homomorphism $\mathrm{Homeo}(\beta^*\omega)\to\mathbf{Z}$ mapping $\phi$ to $1$. So the non-conjugacy of $\phi$ with $\phi^{-1}$ is consistent. But under CH, the group $\mathrm{Homeo}(\beta^*\omega)$ is simple (Rubin) so the non-conjugacy couldn't be attested by a homomorphism to $\mathbf{Z}$ as above.
Note: Boolean algebraic translation through Stone duality: consider the endomorphism of the Boolean algebra $2^\omega$ of subsets of $\omega$ given by $A\mapsto \{a\in\omega:a+1\in A\}$. It induces an automorphism $\Phi$ of the quotient Boolean algebra $2^\omega/\mathrm{fin}$, where $\mathrm{fin}$ is the ideal of finite subsets. Is (under ZFC+CH) $\Phi$ non-conjugate to its inverse in $\mathrm{Aut}_{\mathrm{Ring}}(2^\omega)$?
Indeed Stone duality yields (in ZFC) an isomorphism $\mathrm{Homeo}(\beta^*\omega)\to\mathrm{Aut}_{\mathrm{Ring}}(2^\omega)$ mapping $\phi$ to $\Phi$.
Further comments:
A side question is whether it is consistent with ZFC that $\phi$ and $\phi^{-1}$ are conjugate, I don't know either (but I'm primarily interested in the CH case).
Also in ZFC it is easy to check that $\phi$ is not conjugate to $\phi^n$ for any $n\ge 2$.
YCor, have you looked at the analogous C*-algebra question: are the unilateral shift and its adjoint related by an automorphism of the Calkin algebra? Will Brian lists several facts about $\beta\omega\setminus\omega$ whose Calkin algebra analogues aren't familiar to me (but maybe experts would know better).
@NikWeaver Nice question. Farah has results in this direction. For instance (Ann. Math. 2011 link) he proved the consistency of ZFC + all automorphisms are inner. In such a (quite exotic) model, they're non-conjugate since the Fredholm index distinguishes them. I don't know whether CH implies they're conjugate (if true this might be easier than the set-theoretic counterpart).
Yes ... I'm aware of Ilias's paper. It followed a paper by Chris Phillips and me where we showed that CH implies outer automorphisms exist. Our techniques wouldn't be helpful for this problem though.
@NikWeaver But isn't existence of nontrivial outer automorphisms immediate from Rudin? Rudin proved under CH that $|\mathrm{Aut}(S_\omega/\mathrm{fin})|=2^c$. It immediately implies (since $\mathrm{Aut}(S_\omega/\mathrm{fin})$ embeds into $\mathrm{Aut}$(Calkin)) that $\mathrm{Aut}$(Calkin) has the same cardinality (and hence has Out of the same cardinality, since $\mathrm{Inn}$(Calkin) has cardinal $c$).
No, it's a lot harder for the Calkin algebra. Phillips and Weaver, The Calkin algebra has outer automorphisms, Duke Math. J. 139 (2007), 185–202.
(Look again at whether ${\rm Aut}(S_\omega/{\rm fin})$ embeds in ${\rm Aut}(Q(l^2))$.)
Is that why everyone has heard of Ilias's direction and no one has heard of mine? Because they assume my direction was trivial? Yikes.
@NikWeaver oops, sorry, actually $\mathrm{Aut}{\mathrm{grp}}(S\omega/\mathrm{fin})$ has cardinal $c$, not $2^c$ (which is not trivial by the way – Alperin-Covington-MacPherson/Truss). So my "immediate" argument fails in any case. Whether $\mathrm{Aut}_{\mathrm{BA}}(2^\omega/\mathrm{fin})$ embeds into $\mathrm{Aut}$(Calkin), I would have expected it too but without serious grounds. Sorry again!
Ah, I was mistaken as well --- as you seem to have guessed, I read "${\rm Aut}(S_\omega/{\rm fin})$" as "${\rm Aut}(P(\omega)/{\rm fin})$".
Although it's maybe less palatable these days, I think that the usual terminology for what you call the "Stone–Čech remainder" is the corona.
@LSpice "Stone-Cech corona" yields 20 times less Google occurences than "Stone-Cech remainder". One advantage of "Stone-Cech remainder" is that you can guess the meaning assuming that you know what "Stone-Cech compactification" is. I've actually encountered "corona" never in the meaning of this Wikipedia page (to which I'd recommend renaming), but in generalizations such such as the Higson-Roe corona, or binary corona of a metric space.
@YCor Obliged. $ $
Update: The answer is yes -- if $\mathsf{CH}$ is true then $\phi$ and $\phi^{-1}$ are conjugate in the group of self-homeomorphisms of $\omega^*$.
I've written this up in a new paper, which you can find on the arXiv: (link)
For the sake of anyone who's interested in how the proof goes (but not that interested), let me try to summarize some of what goes into it here.
Ultimately, the proof relies on a transfinite back-and-forth argument. This recursive argument needs to deal with $\mathfrak c$ tasks in succession, but the hypotheses of the recursion do not survive more than $\omega_1$ stages. Thus the argument can only succeed if $\mathfrak c = \omega_1$, i.e., if $\mathsf{CH}$ holds. (This is the only point in the entire proof where $\mathsf{CH}$ is needed.)
The limit steps of the recursion are trivial, and all the difficulty lies in the successor steps.
At successor steps, we wish to take a conjugacy between countable substructures of $\langle \mathcal P(\omega) / \mathrm{fin},\phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\phi^{-1} \rangle$ and extend it to a conjugacy between strictly larger substructures.
Furthermore, we must have at least some control over the growth of these substructures as the recursion progresses, so that we can ensure they cover $\mathcal P(\omega) / \mathrm{fin}$ in $\omega_1$ steps.
The standard approach to dealing with this kind of thing is known to model theorists as ``$\aleph_1$-saturation".
Roughly, if the structures $\langle \mathcal P(\omega) / \mathrm{fin},\phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\phi^{-1} \rangle$ were $\aleph_1$-saturated, then a conjugacy between two of their countable substructures could always be extended in the way we want.
Unfortunately, these structures are not $\aleph_1$-saturated (this is proved in Section 5 of the paper).
Our recursion asks us to perform $\omega_1$ tasks of a certain kind, but this lack of saturation means that some tasks of this kind are undoable.
So we cannot just launch into our recursion and deal with any instance of this task as it arises. Instead, we do the recursion in a particular way, carefully avoiding ever running into these undoable tasks.
The idea for avoiding these undoable tasks uses the model-theoretic idea of elementarity: if one of our two countable substructures is an elementary substructure, then the kind of extension described above can always be done.
Showing that this elementarity idea actually works is the hardest part of the proof, and it's really the heart of the whole thing. I won't say much about this piece of the argument here, except that it has less to do with set theory or model theory: it's just really tricky combinatorics, a careful analysis of the finite directed graphs that each capture some finite amount of information about $\langle \mathcal P(\omega) / \mathrm{fin},\phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\phi^{-1} \rangle$.
Let me mention that the final section of the paper includes two other related results, corollaries to the main theorem.
The first is that $\langle \mathcal P(\omega) / \mathrm{fin},\phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\phi^{-1} \rangle$ are elementarily equivalent. Unlike the main theorem, this corollary does not assume $\mathsf{CH}$.
The second states that $\mathsf{CH}$ implies there is an automorphism $\psi$ of $\mathcal P(\omega) / \mathrm{fin}$ that conjugates $\phi$ with itself in a nontrivial way: i.e., $\psi \circ \phi = \phi \circ \psi$, but $\psi \neq \phi^n$ for any $n \in \mathbb Z$.
$$$$
Original post:
This is a great question -- and it's wide open. Here's what I know about it:
$\bullet$ As you mentioned, it is consistent that $\phi$ and $\phi^{-1}$ are not conjugate. This observation was first made by van Douwen, soon after the publication of Shelah's result that you mention in your question. You mentioned forcing axioms, so let me point out that the non-conjugacy of $\phi$ and $\phi^{-1}$ follows from $\mathsf{MA}+\mathsf{OCA}$, which is a weak form of $\mathsf{PFA}$. This is due to Boban Velickovic.
$\bullet$ If it is consistent with $\mathsf{ZFC}$ that $\phi$ and $\phi^{-1}$ are conjugate, then it is consistent with $\mathsf{ZFC}+\mathsf{CH}$. (Proof sketch: If $\phi$ and $\phi^{-1}$ are conjugate in some model, then force with countable conditions to collapse the continuum to $\aleph_1$ and make $\mathsf{CH}$ true. Because this forcing is countably closed, it won't change much about the Boolean algebra $\mathcal P(\omega)/\mathrm{fin}$, and will preserve the fact that $\phi$ and $\phi^{-1}$ are conjugate.)
$\bullet$ Even better, the existence of certain large cardinals implies that if it is possible to force "$\phi$ and $\phi^{-1}$ are conjugate" then this statement is already true in every forcing extension satisfying $\mathsf{CH}$. This follows from a theorem of Woodin concerning what are called $\Sigma^2_1$ statements about the real line (explained further here). The assertion "$\phi$ and $\phi^{-1}$ are conjugate" is an example of such a statement. (Very roughly, this theorem seems to suggest that if this statement is consistent, then it should follow from $\mathsf{CH}$. At any rate, trying to prove it from $\mathsf{CH}$ seems like a reasonable strategy.)
$\bullet$ In fact, Paul Larson has pointed out to me that the statement "$\phi$ and $\phi^{-1}$ are conjugate" is a now very rare example of a $\Sigma^2_1$ statement about the real line whose status we do not know under $\mathsf{ZFC}+\mathsf{CH}$ (plus large cardinal axioms).
$\bullet$ I proved a partial result a few years ago, showing that $\mathsf{CH}$ implies $\phi$ and $\phi^{-1}$ are semi-conjugate:
$\qquad$Theorem: Assuming $\mathsf{CH}$, there is a continuous surjection $Q: \omega^* \rightarrow \omega^*$ such that $$Q \circ \phi = \phi^{-1} \circ Q.$$
The paper is "Abstract $\omega$-limit sets," Journal of Symbolic Logic 83 (2018), pp. 477-495, available here. In the same paper, I show that the forcing axiom $\mathsf{MA}+\mathsf{OCA}$ implies $\phi$ and $\phi^{-1}$ are not semi-conjugate. (Or rather, I show that this is a corollary to a deep structure theorem of Ilijas Farah.)
$\bullet$ Finally, in a more recent paper (to appear in Topology and its Applications, currently available here), I show that there is no Borel set separating the conjugacy class of $\phi$ and the conjugacy class of $\phi^{-1}$ (in the space of self-homeomorphisms of $\omega^*$ endowed with the compact-open topology). Roughly, this shows that if $\phi$ and $\phi^{-1}$ fail to be conjugate, it's not "for any real reason" -- or at least not for any nicely definable topological reason.
Thanks for this detailed answer and for the links! (Side note: I had to open the second link to figure out that "isomorphism class" denotes what I know as "conjugacy class" — I agree that from a certain categorical viewpoint it's also an isomorphism class, namely in the category of spaces endowed with a homeo; actually my colleagues in dynamical systems talk of conjugacy of dynamical systems even when acting on different spaces)
You're welcome! This is a question that I've thought a lot about, so I'm happy to be able to share.
Do you know if $(\mathrm{Clopen}(\beta^\omega),\Phi)$ and $(\mathrm{Clopen}(\beta^\omega),\Phi^{-1})$ are elementary equivalent (as structures encoding a Boolean algebra endowed with an automorphism)? I might post a separate question if it's not immediate or an immediate consequence of your result (it's not clear to me that the set of homeomorphisms $\alpha$ satisfying some sentence $u(\alpha)$ is Borel).
That's a good question. I don't think it's known. My guess is that the answer is yes, and we might be able to prove they are elementarily equivalent by showing that they're potentially isomorphic (https://en.wikipedia.org/wiki/Potential_isomorphism). I don't think that the set of all homeomorphisms satisfying some fixed sentence needs to be Borel -- this isn't true for sets of reals, anyway, because unbounded quantifiers push us out of the Borel sets into the projective hierarchy.
Thanks, I eventually posted the question https://mathoverflow.net/questions/353074/
I take back what I said about potential isomorphisms. (But I don't take back what I said about your question being a good question.)
The question is answered by Will Brian arXiv, Feb. 6 2024.
|
2025-03-21T14:48:29.903245
| 2020-02-18T16:03:54 |
353002
|
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|
Stack Exchange
|
Are $f\sqrt{1+g^2}$ and $fg\sqrt{1+g^2}$ smooth if $f,fg,fg^2$ are smooth?
Suppose that $f$ and $g$ are functions from $\mathbb R$ to $\mathbb R$ such that the functions $f,fg,fg^2$ are smooth, that is, are in $C^\infty(\mathbb R)$. Does it then necessarily follow that the
functions $f\sqrt{1+g^2}$ and $fg\sqrt{1+g^2}$ are smooth?
Of course, the problem here is that the function $g$ does not have to be smooth, or even continuous, at zeroes of the function $f$.
One may also note that the continuity of the
functions $f\sqrt{1+g^2}$ and $fg\sqrt{1+g^2}$ (at the zeroes of $f$ and hence everywhere) follows easily from the inequalities $|f\sqrt{1+g^2}|\le|f|+|fg|$ and $|fg\sqrt{1+g^2}|\le|fg|+|fg^2|$.
No.
Set
$$ f(x) = \exp(-2/|x|^2) \operatorname{sign} x, \qquad g(x) = \exp(1/|x|^2) \sqrt{|x|} \operatorname{sign} x $$
for $x \ne 0$, and, of course, $f(0) = g(0) = 0$. Then clearly
$$ \begin{aligned}
f(x) & = \exp(-2/|x|^2) \operatorname{sign} x , \\
f(x) g(x) & = \exp(-1/|x|^2) \sqrt{|x|} , \\
f(x) (g(x))^2 & = x
\end{aligned} $$
are infinitely smooth, but
$$ f(x) g(x) \sqrt{1 + (g(x))^2} = \sqrt{|x| \exp(-2/|x|^2) + |x|^2} = |x| (1 + o(1)) $$
is not even differentiable at $0$.
Thank you for this counterexample. Just one typo, think: It seems to me that for your $f$ and $g$ we have $f(x)g(x)^2=x$ for all $x$.
There is now a follow-up question at https://mathoverflow.net/questions/353015/are-pm-f-sqrt1g2-and-pm-fg-sqrt1g2-smooth-if-f-fg-fg2-are-smoot
@IosifPinelis: Sure! Corrected, thanks!
|
2025-03-21T14:48:29.903476
| 2020-02-18T16:33:32 |
353006
|
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|
Stack Exchange
|
(1, 2) stability and Hausdorff dimension
Recall that a set $E \subset \mathbb{R}^n$ is called (1,2) stable if it satisfies the following:
$$
W_0^{1,2}(E) = W_0^{1,2}(E^0),
$$
where $E^0$ is the interior of $E$ and $W_0^{1,2}(E)$ is defined by
$$
W_0^{1,2}(E) = \{ u \in W_0^{1,2}(\mathbb{R}^n) : u = 0 \text{ on } \mathbb{R}^n \setminus E \}.
$$
For equivalent definitions, see a book "Function Spaces and Potential Theory" by Adams and Hedberg (Theorem 11.4.1, for example).
The notion of stability is closely related to the stability of the Dirichlet problem, and to the problem of uniform approximation by harmonic functions.
My question is as follows: is there a relation between Hausdorff dimension and (1,2) stability?
For example, is it true that if $\partial E$ has Lebesgue measure 0, then $E$ is (1,2) stable?
Or more generally, can you give some references on the relation between the geometry of a given space and it's (1,2) stability? I am interested in the following question as well: is it true that a Jordan domain is (1,2) stable?
|
2025-03-21T14:48:29.903582
| 2020-02-18T16:51:20 |
353007
|
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|
Stack Exchange
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Reference for the following flow equation
I'm looking for reference on the following partial differential equation. $\partial_tF(t,x) = G(t)((\partial_xF(t,x))^2 + \partial_x^2F(t,x))$, where G(t) is a fixed Schwartz function. If possible, I also want to know if there's reference on the following stochastic partial differential equation, namely the previous deterministic equation with a Brownian white noise $\eta$: $\partial_tF(t,x) = G(t)((\partial_xF(t,x))^2 + \partial_x^2F(t,x)) + \eta$.
I'm a theoretical physicist. So I apologize in advance if the question is trivial for people working in PDE. For me, the origin of these two equations comes from a version of 0-dimensional exact renormalization group flow in the sense of Polchinski, where x in the above equation is to be understood as dynamical fields in usual QFT, t is to be understood as log time scale of renormalization group flow, and F(t,x) is to be understood as a generating functional of all multi-point interaction. For those of you who know RG, all issues related to non-locality, anomalous dimension, etc. are suppressed for the sake of simplicity of toy model.
Also, I think the following comment is important. My approach so far is that this equation has structural similarity with Ricci flow in the sense that the evolution operator contains a square of 1st order derivative (the product of two Christoffel symbols in Ricci curvature) and a 2nd order derivative (the derivative of Christoffel symbol in Ricci curvature). Therefore, I try to massage my equation a bit and see if I can use some technique from Ricci flow to study it, although many of the complexity from Ricci flow is not present here. I'm not an expert so I am not confident that I am on the right path.
Thank you very much for the help.
Modulo some signs, your first equation is a Hamilton-Jacobi-Bellman equation for a stochastic dynamical system, for a special case. This equation is used for computed an optimal control for a system governed a system of ODEs.
You can consult e.g., these notes (section 2.1) https://homes.cs.washington.edu/~todorov/courses/amath579/Todorov_chapter.pdf
and references therein.
Also wikipedia has a decent discussion: https://en.wikipedia.org/wiki/Hamilton%E2%80%93Jacobi%E2%80%93Bellman_equation
See the last example.
Hello thanks for the response. I'm aware that my equation is a HJB equation since this is built from HJB perspective (something called HJB approach to holographic reconstruction in physical language). What I'm not certain is whether there is previous analysis of my particular form of HJB equation. As I understand, HJB equation is a framework where one may fill in different forms of Hamiltonian to drive the flow. So I'm wondering if anything has been said about my form of HJB equation in particular.
Ok, I guess one can write the optimal control problem from where this comes from, and think about it. Are there any specific properties you want to know about the solution of this equation ? What do we know about G?
If G(t) is positive, then at least for the deterministic version it can be absorbed on the left hand side by redefining time. The equation then becomes equivalent to the Burgers equation. There are enough papers on the Burgers equation even to satisfy Carl Sagan, so that should give you a start.
@MichaelRenardy: Thanks a lot. That seems exactly what I'm looking for.
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2025-03-21T14:48:29.903834
| 2020-02-18T17:05:34 |
353008
|
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|
Stack Exchange
|
Homotopy Gerstenhaber algebras: description via operads vs derivations
There are at least a couple of definitions in the literature for an $E_2$-algebra, also known as a homotopy Gerstenhaber algebra, also known as $G_{\infty}$-algebra.
Suppose $V$ is a graded vector space. Let $S(V), \text{Lie}(v)$ denote the graded symmetric algebra and the graded free Lie algebra of $V$ respectively and $S_+(V)$ be the symmetric algebra in strictly positive degree.
$\textbf{Definition (Tamarkin)}$ A $G_{\infty}$-algebra structure on $V$ is a degree $+1$ map $$ \delta: S_+(\text{Lie}(V^*[1])[1]) \rightarrow S_+(\text{Lie}(V^*[1])[1]) $$ which behaves as a derivation of both $\cdot$ and $[\,,]$ such that $\delta^2 = 0$.
Unpacking this leads to saying that we have a collection of maps $$ m_{k_1, \dots, k_n}: V^{\otimes k_1} \otimes \dots \otimes V^{\otimes k_n} \rightarrow V$$ of degree $3-(k_1 + \dots k_n + n)$ obeying appropriate symmetry and associativity relations.
The second definition pertains to algebras over the little disk operad. Let $D_2(k)$ denote the configuration space of $k$ little disks inside a big disk and let $\text{Chains}_{\bullet}(D_2(k))$ be the singular chain complex. Letting $\mathcal{P}(k) = \text{Chains}_{\bullet}(D_2(k))$, one can define an operadic structure on this collection of vector spaces. This leads one to the
$\textbf{Definition (Getzler-Jones?)}$ An $E_2$-algebra is an algebra over the operad $\text{Chains}(D_2)$.
How can one show that these two definitions are equivalent?
Short of a full proof of equivalence, it would be nice to understand a description of the cycle in $D_2(k_1 + \dots +k_n)$ which corresponds to the map $m_{k_1, \dots, k_n}$. For example: if one were working in $H_{\bullet}(D_2)$, the homology operad, one associates to the point class in configuration space of two disks the operation $\cdot = m_2$, and to the cycle involving one little disk going around the other the bracket $[\,,] = m_{1,1}$ in the Gerstenhaber algebra. Is there an explicit description of the cycle corresponding to $m_{k_1, \dots, k_n}$?
The equivalence between these two notions is nontrivial, since it amounts to a choice of formality isomorphism for the operad of little disks.
Let $D_2$ be the little disks operad. The easy part of the equivalence is to show that $G_\infty$ algebras are equivalent to algebras over the operad $H_*(D_2)$. If you take the operad of graded $\mathbb Q$ vector spaces, $H_*(D_2)$, its algebras are Gerstenhaber algebras-- this can be seen by computing the homology of the configuration spaces. Then Koszul self-duality for the Gerstenhaber operad gives the relationship between $G_\infty$ algebras and Gerstenhaber algebras. I.e. every $G_\infty$ algebra may be rectified to an algebra over $H_*(D_2)$.
Let $C_*(D_2)$ be chains on the little disk operad. To obtain an equivalence between $C_*(D_2)$ algebras and $H_*(D_2)$ algebras, one needs to choose a zig-zag of quasi-isomorphisms between the two operads. These formality isomorphisms are difficult to construct, and there are many possible choices, because these operads admit a large automorphism group. See Kontsevich's paper, https://arxiv.org/abs/math/9904055, for a discussion.
Thanks for the answer. Please take a look at the edits I made.
Here are the operads that are involved in that game:
the operad $D_2$ of little disks, which is a topological operad.
its chain operad $C_{-*}(D_2,\mathbb{k})$, which is an operad in cochain complexes (of $\mathbb{k}$-modules).
its homology operad $H_{-*}(D_2,\mathbb{k})$, which is known to be isomorphic to the Gerstenhaber operad $G^{\mathbb{k}}$, which is itself a binary quadratic operad satisfying the Koszul property.
the minimal resolution of $G^{\mathbb{k}}$ is the operad $G^{\mathbb{k}}_\infty$ governing ($\mathbb{k}$-linear) $G_\infty$-algebras.
Being a resolution, $G^{\mathbb{k}}_\infty$ is obviously quasi-isomorphic to $G^{\mathbb{k}}$. As Phil Tosteson mentions in his answer, the difficult part relies on proving that $C_{-*}(D_2,\mathbb{k})$ is formal. It's only proven over a field $\mathbb{k}$ of characteristic zero, and it's actually not formal for $\mathbb{k}=\mathbb{F}_p$ (see e.g. the introduction of https://arxiv.org/pdf/1903.09191.pdf). To my knowledge, there are essentially two different proofs:
one by Tamarkin: https://sites.math.northwestern.edu/~tamarkin/Papers1/Formality.pdf.
another one by Kontsevich, which generalizes to higher dimension: https://arxiv.org/abs/math/9904055 (see also https://arxiv.org/abs/0808.0457 for more details).
Note that the formality quasi-isomorphisms from these two proofs happen to coincide, after the correct "choice of associator" has been made. See https://arxiv.org/abs/0905.1789.
Finally, I don't think it is meaningful to ask which cycle corresponds to the map $m_{k_1,\dots,k_n}$. The reason is that $m_{k_1,\dots,k_n}$ is not closed. You may ask if they are represented by nice chains... I don't know the answer to that question (and I suspect that it is close to be as hard as proving the formality itself), but for the $m_{1,\dots,1}$ the answer is known:
first observe that $D_2$ is weakly homotopy equivalent to the operad of compactified configuration spaces of points in the plane.
$m_{1,\dots,1}$ (with $n$ "$1$"s) can be represented by the top-dimensional/fundamental cell of the compactified configuration space $\overline{C}_n\simeq D_2(n)$ of $n$-points in the plane.
Thanks! In $\overline{C}_n$ is the group of dilations and overall translations implicitly modded out? So that the fundamental cell of $\overline{C}_n$ is $2n-3$ dimensional?
Yes, the group we use to mod out is the semi-direct product of translations of (real, positive) dilations. hence, yes, the dimension of the top dimensional cell is 2n-3.
Great. It might be worth asking this as a separate question, but what is the argument that there is a morphism of operads from the $L_{\infty}$-operad to the suboperad of top dimensional cells in the compactified configuration space? If you know any references I'd appreciate it.
I don't have a reference in mind, but I think the argument is not so difficult. As a graded operad (ie without the differential), the $L_\infty$-operad is free. So, you define a morphism of graded operads sending $m_{1,\dots,1}$ to the top dimensional cell in $\overline{C}_n$. Then you check that it is compatible with the differentials (intuitively, the boundary of the top dimensional cell is a union of products of top-dimensional cells of $\overline{C}_m$'s for $m<n$.
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2025-03-21T14:48:29.904252
| 2020-02-18T18:38:19 |
353013
|
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|
Stack Exchange
|
3-uniform hypergraphs and their circuit space
So, I'll break this post into two questions. Both concern 3-uniform hypergraphs. A 3-uniform hypergraph $H=(V,E)$ consists of a set of vertices $V$ and a set of edges $E$, where each edge $e\in E$ is a set of exactly three vertices from $V$.
1) Here is a well-defined notion of a circuit in a 3-uniform hypergraph. Construct the incidence matrix $B\in\{0,1\}^{|V|\times|E|}$ where entry $B_{ve}=1$ if and only if $v\in e$. So every column of $B$ has exactly three 1s. Then $\text{ker}(B)=\{x\in\{0,1\}^{|E|}:Bx=0\}$ describes the circuit space. In words, a circuit is a subset of edges such that every vertex is included in an even number of edges from the subset.
This seems like a natural generalization of the circuit space for normal graphs, which is also the kernel of the incidence matrix. However, I can't find any work using this definition of circuit space for hypergraphs! Is there a name for this type of hypergraph circuit or some literature I should check out?
2) It is efficient (polynomial time in $|V|$ and $|E|$) to find a basis $\{x_i\}_i$ for the circuit space of a 3-uniform hypergraph since it's just the null space of B. However, can we find a "minimal" circuit basis efficiently? A minimal circuit basis is a circuit basis that uses the minimal number of hyperedges, i.e. minimize, over all bases, $\sum_i|x_i|$ where $|x_i|$ is the Hamming weight of vector $x_i$.
The corresponding problem of finding a minimum circuit basis (often called minimum "cycle" basis because the basis consists of cycles) for normal graphs is efficient -- Horton's algorithm being the first polynomial-time algorithm. Is there any way to find a minimal circuit basis for 3-uniform hypergraphs efficiently? What if we put additional conditions on the 3-uniform graph? Such conditions may be similar to those in arxiv.org/abs/1002.3331, where an efficient algorithm for counting spanning trees of certain classes of 3-uniform hypergraphs is given even though it's hard in general.
Thanks for your help!
If you don't get any responses, I'd suggest deleting this question after you copy it over to the theoretical computer science stack exchange at https://cstheory.stackexchange.com/, where it may receive more attention. Cross-posting is frowned upon, hence my suggestion to delete it here.
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2025-03-21T14:48:29.904430
| 2020-02-18T18:42:57 |
353015
|
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Stack Exchange
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Are $\pm f\sqrt{1+g^2}$ and $\pm fg\sqrt{1+g^2}$ smooth if $f,fg,fg^2$ are smooth?
This is a follow-up on the previous question.
Suppose that $f$ and $g$ are functions from $\mathbb R$ to $\mathbb R$ such that the functions $f,fg,fg^2$ are smooth, that is, are in $C^\infty(\mathbb R)$. Does it then necessarily follow that, for some function $h\colon\mathbb R\to\{-1,1\}$, the
functions $hf\sqrt{1+g^2}$ and $hfg\sqrt{1+g^2}$ are smooth?
Of course, the problem here is that the function $g$ does not have to be smooth, or even continuous, at zeroes of the function $f$; if $f$ has no zeroes, then one can obviously take $h=1$.
One may also note that the continuity of the
functions $hf\sqrt{1+g^2}$ and $hfg\sqrt{1+g^2}$ (at the zeroes of $f$ and hence everywhere) follows easily from the inequalities $|hf\sqrt{1+g^2}|\le|f|+|fg|$ and $|hfg\sqrt{1+g^2}|\le|fg|+|fg^2|$.
Still no. Consider the function $f(x)=(x^2+r^2)^2$, $g(x)=\frac x{x^2+r^2}$ with small $r>0$. Then $f$, $fg$, $fg^2$ are polynomials of degree $\le 4$ with bounded coefficients but $f\sqrt{1+g^2}$ is very close to $x^2|x|\sqrt{1+x^2}$ as close to $0$ as you wish when $r$ is small enough, so the maximum of the fourth derivative in an arbitrarily small neighborhood of the origin can be forced to be very large by choosing $r$ small enough. Now just take your favorite $C^\infty$ function $\psi$ that is $1$ on $[-1,1]$ and is supported on $[-2,2]$ and use $\psi f$ and $\psi g$ instead of $f$ and $g$. You'll get a compactly supported building block that you can scale and translate with the possibility to blow up the fourth derivative by choosing $r$ last. So just scale to some disjoint intervals $I_j$ accumulating to $0$ with sufficiently fast decaying heights to make individual multiplications by controlled polynomials irrelevant after which choose $r_j$. Near the center of each interval $I_j$ the function $f$ is strictly positive, so $h$ is of no use there. What may really help (no guarantee though) is to assume that $g$ is continuous, but that is, probably, too much for your purposes.
This is very good! I do see the explosion $H^{(4)}(0)=24-3/r^4$ for $H:=f\sqrt{1+g^2}$ with your $f$ and $g$, which should be enough. Yet, I don't see in what exact sense $H(x)$ is close to $x^2|x|\sqrt{1+x^2}$; also, if that is/were so, I think already $H'''$ would be bad, but it is quite tame. This phenomenon seems pretty unusual: the derivatives of $H$ up through the 3rd order are quite tame, but the 4th one explodes very fast! It would be useful to me (and I hope others) if you could explain how you found such a phenomenon.
$\newcommand{\de}{\delta}$This is to provide a detalization/formalization of fedja's answer.
For $r\in(0,1]$ and real $x$, let
\begin{equation*}
f_{0,r}(x):=(x^2+r^2)^2,\quad g_{0,r}(x):=\frac x{x^2+r^2},
\end{equation*}
\begin{equation*}
H_{0,r}:=f_{0,r}\sqrt{1+g_{0,r}^2},\quad F_{0,m,r}:=f_{0,r}g_{0,r}^m,
\end{equation*}
\begin{equation*}
f_r:=f_{0,r}\psi,\quad g_r:=g_{0,r}\psi,
\end{equation*}
\begin{equation*}
H_r:=f_r\sqrt{1+g_r^2},\quad F_{m,r}:=f_r g_r^m,
\end{equation*}
where $m\in\{0,1,2\}$ and $\psi$ is any function in $C^\infty(\mathbb R)$ such that $\psi=1$ on the interval $[-1/2,1/2]$ and $0$ outside the interval $[-1,1]$.
Then for each $k\in\{0,1,\dots\}$ we have $\max_{m=0}^2\sup_{0<r\le1}\|F_{0,m,r}^{(k)}\|_\infty<\infty$ and hence
\begin{equation*}
C_k:=\max_{m=0}^2\sup_{0<r\le1}\|F_{m,r}^{(k)}\|_\infty<\infty. \tag{1}
\end{equation*}
However,
\begin{equation*}
H_r^{(4)}(0)=H_{0,r}^{(4)}(0)=24-3/r^4\sim-3/r^4\to-\infty
\end{equation*}
as $r\downarrow0$; this crucial fact can be verified either by a direct calculation or by using (with $r$ fixed) the Maclaurin expansions $\sqrt{1+v}=1+v/2-v^2/8+o(v^2)$ (with $v=\dfrac u{(r^2+u)^2}$) and then $\dfrac1{(r^2+u)^2}=\dfrac1{r^4}\,\Big(1-\dfrac{2u}{r^2}\Big)+o(u^2)$ (with $u=x^2$).
For real $x$ and $m\in\{0,1,2\}$, let now
\begin{equation*}
f(x):=\sum_{j=1}^\infty a_j f_{r_j}\Big(\frac{x-c_j}{\de_j}\Big), \quad
g(x):=\sum_{j=1}^\infty a_j g_{r_j}\Big(\frac{x-c_j}{\de_j}\Big),
\end{equation*}
\begin{equation*}
F_m(x):=f(x)g(x)^m=\sum_{j=1}^\infty a_j F_{m,r_j}\Big(\frac{x-c_j}{\de_j}\Big)
\end{equation*}
\begin{equation*}
H(x):=f(x)\sqrt{1+g(x)^2}=\sum_{j=1}^\infty a_j H_{r_j}\Big(\frac{x-c_j}{\de_j}\Big),
\end{equation*}
where
\begin{equation*}
r_j:=a_j:=e^{-j}, \quad c_j:=\tfrac12\,(x_j+x_{j+1}), \quad x_j:=1/j,\quad \de_j:=\tfrac12\,(x_j-x_{j+1})\sim1/(2j^2)
\end{equation*}
as $j\to\infty$.
Then, by (1) and dominated convergence, the functions $f=F_0$, $fg=F_1$, $fg^2=F_2$ are in $C^\infty(\mathbb R)$, with
\begin{equation}
\|F_m^{(k)}\|_\infty\le\sum_{j=1}^\infty a_j C_k/\de_j^k<\infty
\end{equation}
for all $k\in\{0,1,\dots\}$.
However, $hH\notin C^\infty(\mathbb R)$, and even $hH\notin C^4(\mathbb R)$, for any function $h\colon\mathbb R\to\{-1,1\}$. Indeed, assume the contrary. Then, for each natural $j$ and all $x\in[c_j-\de_j/2,c_j+\de_j/2]$, we have
$h(x)H(x)=a_j h(x)H_{r_j}\big(\frac{x-c_j}{\de_j}\big)$, which will be continuous in $x$ at $x=c_j$ only if $h$ is the constant $1$ or the constant $-1$ in some neighborhood of $c_j$. Hence,
\begin{equation}
|(hH)^{(4)}(c_j)|=a_j|H^{(4)}(0)|/\de_j^4\sim 3a_j/(r_j^4\de_j^4)\to\infty
\end{equation}
as $j\to\infty$. Since $c_j\to0$ as $j\to\infty$, we see that $(hH)^{(4)}$ is unbounded in any neighborhood of $0$. So, $hH\notin C^4(\mathbb R)$, as claimed.
This time the answer is yes. Edited: the answer is no, as shown by fedja in his excellent answer. However, the following proves that with an appropriate choice of $h$, the functions $h f \sqrt{1 + g^2}$ and $h f g \sqrt{1 + g^2}$ have Taylor expansion (of an arbitrary order) at every real point. This is weaker than smoothness, but at least some positive result.
It is sufficient to consider a fixed zero of $f$. When trying to define the square-root of $h_1 = f^2 + f^2 g^2$ and $h_2 = f^2 g^2 + f^2 g^4$ in a neighbourhood of $a$, we need to choose among two options:
$$ f(x) \sqrt{1 + (g(x))^2} \qquad \text{or} \qquad f(x) \sqrt{1 + (g(x))^2} \operatorname{sign} (x - a) $$
for $h_1$, and similarly
$$ f(x) g(x) \sqrt{1 + (g(x))^2} \qquad \text{or} \qquad f(x) g(x) \sqrt{1 + (g(x))^2} \operatorname{sign} (x - a) $$
for $h_2$.
Suppose first that $a$ is a zero of $f$ of finite multiplicity $a$, say, $$ f(x) \sim C (x - a)^n $$ as $x \to a$ for some $C \ne 0$. Since $f g$ and $f g^2$ are smooth at $a$, and $g = (f g) / f$, $g^2 = (f g^2) / f$, we have the following possibilities:
$g$ has a pole at $a$ of degree $-m < \tfrac{1}{2} n$;
$g$ is smooth and non-zero at $a$ (and we set $m = 0$);
$g$ has a zero at $a$ of multiplicity $m > 0$ (possibly infinite).
It is thus easy to see that $a$ is a zero of $h_1 = f^2 + f^2 g^2$ and $h_2 = f^2 g^2 + f^2 g^4$ of multiplicity $2 \min\{n, n + m\}$ and $2 \min\{n + m, n + 2 m\}$, respectively (watch out when $m = 0$!). Consequently, one can define smooth square-roots of $h_1$ and $h_2$ by an appropriate choice of sign — and it is easy to see that $f \sqrt{1 + g^2}$ and $f g \sqrt{1 + g^2}$ is the right choice if and only if $m$ is even or $m \geqslant 0$; otherwise, choose $f g \sqrt{1 + g^{-2}}$ and $f g^2 \sqrt{1 + g^{-2}}$ instead. (Note that in the latter case we have $|g^{-2}| < 1$ in a neigbourhood of $a$.)
Now consider a zero $a$ of $f$ of infinite multiplicity: suppose that $$f(x) = o(|x - a|^n)$$ as $x \to a$ for every $n$. If $f g^2$ also has a zero at $a$ of infinite multiplicity, then $h_1 = f^2 + f^2 g^2$ and $h_2 = f^2 g^2 + f^2 g^4$ have a zero at $a$ of infinite multiplicity, too, and so any choice of signs near $a$ leads to square-roots of $h_1$ and $h_2$ with a zero at $a$ of infinite multiplicity.
However, it is possible that $f g^2$ has a zero at $a$ of finite multiplicity $k$. In this case $a$ is still a zero of $h_1 = f^2 + f^2 g^2$ of infinite multiplicity, and hence both choices of sign of the square-root of $h_1$ are smooth at $a$. On the other hand, $f^2 g^2 + f^2 g^4$ has a zero at $a$ of finite multiplicity $2 k$. As in the first part of the proof, it is again possible to choose a square-root of $f^2 g^2 + f^2 g^4$ which is smooth at $a$; namely, $f g^2 \sqrt{1 + g^{-2}}$ is smooth at $a$. (As before: note that we necessarily have $|g^{-2}| < 1$ in a neigbourhood of $a$.)
One has to deal with some technicalities when $a$ is not an isolated zero of $h_1$ or $h_2$. Fortunately in this case it is necessarily a zero of infinite multiplicity, and hence the choice of sign of the square root does not matter at all.
Ok, after reading fedja's answer I see an error in my solution. Still, I'll leave it here for a while, in case anyone feels like solving a puzzle and trying to identify the flaw on their own.
|
2025-03-21T14:48:29.904874
| 2020-02-18T18:44:05 |
353016
|
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Stack Exchange
|
Is there a term for a not-necessarily-convex set whose non-extreme points can be expressed as a linear combination of two other points in the set?
This question was asked on Math.SE here, but received no replies after several months. So I have posted it here, though with somewhat revised structuring of the question.
Let $V$ be a real vector space, and for any $S \subset V$ write $\mathrm{hull}(S)$ for the convex hull of $S$.
Given $S \subset V$, it is easy to see that for any $\mathbf{x} \in S$ the following statements are equivalent:
$\mathbf{x}$ is not an extreme point of $\mathrm{hull}(S)$;
there exists $n \geq 2$, points $\mathbf{x}_1,\ldots,\mathbf{x}_n \in S \setminus \{\mathbf{x}\}$ and values $\lambda_1,\ldots,\lambda_n \in [0,1]$ such that $\sum_{i=1}^n \lambda_i=1\,$ and $\,\sum_{i=1}^n \lambda_i\mathbf{x}_i=\mathbf{x}$.
Is there a term for a set $S \subset V$ with the property that for all $\mathbf{x} \in S$, if $\mathbf{x}$ is not an extreme point of $\mathrm{hull}(S)$ then there exist $\mathbf{x}_1,\mathbf{x}_2 \in S \setminus \{\mathbf{x}\}$ and $\lambda \in [0,1]$ such that $\lambda\mathbf{x}_1 + (1-\lambda)\mathbf{x}_2=\mathbf{x}$?
In other words, is there a term for sets $S \subset V$ with the property that in the further-above characterisation of non-extremity, one can simply fix $n=2$?
Obviously, by definition, any convex set has this property; but there are also non-convex sets with this property.
[Some trivial examples that are nonetheless interesting to observe: An arbitrary union of open line segments has this property. A set consisting of the three corners of a triangle has this property; but a set consisting of the three corners of a triangle plus one interior point of the triangle does not have this property. Nonetheless, a set consisting of the boundary of a triangle has this property, and so does a set consisting of the boundary of a triangle plus any subset of the interior of the triangle.]
Motivation from ergodic theory. Let $(X,\mathcal{X})$ be a measurable space and let $f \colon X \to X$ be a measurable map. An $f$-invariant measure is a probability measure $\mu$ on $X$ such that $\mu(A)=\mu(f^{-1}(A))$ for all $A \in \mathcal{X}$. An $f$-ergodic measure is an $f$-invariant measure $\mu$ with the additional property that the following equivalent statements hold:
for any $A \in \mathcal{X}$ with $f^{-1}(A)=A$, we have $\mu(A) \in \{0,1\}$;
for any $A \in \mathcal{X}$ with $\mu(f^{-1}(A) \triangle A)=0$, we have $\mu(A) \in \{0,1\}$;
the only $f$-invariant measure that is absolutely continuous with respect to $\mu$ is $\mu$ itself.
Given the equivalence of these three statements, the following well-known result is not difficult to show:
Proposition 1. The $f$-ergodic measures are precisely the extreme points of the convex set of $f$-invariant measures.
But sometimes, when we have a map with an invariant measure, we like to consider the map as only defined modulo null sets under this measure. So it might be useful to have some kind of convex-geometric characterisation of ergodicity that doesn't require measure-preserving maps to be defined more precisely than in this "mod null sets" sense.
With this goal in mind, I have obtained (by essentially the same proof as Proposition 1) the following generalisation of Proposition 1:
Proposition 2. Let $S$ be a set of probability measures on $(X,\mathcal{X})$, and suppose we have an $S$-indexed family $(f_\mu)_{\mu \in S}$ of measurable maps $f_\mu \colon X \to X$ such that for each $\mu \in S$, $\mu$ is $f_\mu$-invariant. Suppose furthermore that for every $\mu \in S$ and every probability measure $\nu$ on $(X,\mathcal{X})$ that is absolutely continuous with respect to $\mu$, we have
$$ \nu \in S \ \ \Leftrightarrow \ \ \text{$\nu$ is $f_\mu$-invariant.} $$
Then $S$ has precisely the property I am asking about in this MO question, and for each $\mu \in S$, $\mu$ is $f_\mu$-ergodic if and only if $\mu$ is an extreme point of $\mathrm{hull}(S)$.
(In my Math.SE question I gave a version of this proposition for general Markov operators.)
|
2025-03-21T14:48:29.905128
| 2020-02-18T19:56:20 |
353020
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Alexander Kalmynin",
"Sylvain JULIEN",
"https://mathoverflow.net/users/101078",
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"url": "https://mathoverflow.net/questions/353020"
}
|
Stack Exchange
|
Is the abscissa of convergence of $s\mapsto\sum_{n>0}(ng_{n}/2)^{-s}$ known?
The famous Polignac conjecture posits that the $n$-th prime gap $g_{n}:=p_{n+1}-p_{n}$ attains all even positive integral values infinitely many times, which implies $\displaystyle{\zeta_{Pol}:=s\mapsto\sum_{n>0}(ng_{n}/2)^{-s}}$ has an abscissa of convergence $\sigma_{Pol}$ less or equal to $1$.
Is this abscissa of convergence known unconditionally or under some widely believed conjecture other than Polignac's?
Well, the abscissa of convergence is $1$, though I see no way to deduce anything about it from the Polignac conjecture. $\sigma_P \le 1$ is obvious from $|\zeta_P(s)| \le \zeta(\Re s)$, while for $\sigma_P \ge 1$ it is enough to prove that $\zeta_P(1) = \infty$:
$$\zeta_P(1) \ge \sum_{k=1}^\infty \frac{1}{2^{k+1}}\sum_{n = 2^k}^{2^{k+1}}\frac{1}{g_n} \gg \sum_{k = 1}^\infty \frac{1}{2^k} \frac{2^k}{k} = \infty,$$
where in the second step we used that $\sum_{n = 2^k}^{2^{k+1}} g_n \le p_{2^{k+1}} \ll 2^k \log(2^k)$ and, say, AM-HM inequality.
Obviously, abscissa of convergence is at most 1, because the series is dominated by the series for $\zeta(s)$. In fact, it is equal to 1. Indeed, assume that your series converges for some $s=1-\varepsilon$, $\varepsilon>0$. Then the sum
$$
\sum_{X<n<2X} (ng_n)^{-s}
$$
is bounded by constant independent of $X$. Let us prove this is not the case.
Observe that for large integer $X$ we have
$$
\sum_{X<n<2X} g_n=p_{2X}-p_{X}\sim X\log X
$$
By Cauchy-Schwarz we get
$$
\left(\sum_{X<n<2X} g_n\right)\left(\sum_{X<n<2X} (ng_n)^{-1}\right)\geq \left(\sum_{X<n<2X} \frac{1}{\sqrt{n}}\right)^2\geq ((X-1)/\sqrt{2X})^2\gg X,
$$
so that
$$\sum_{X<n<2X} (ng_n)^{-1}\gg \frac{1}{\log X}$$
therefore
$$\sum_{X<n<2X} (ng_n)^{-s}\geq \sum_{X<n<2X} n^\varepsilon (ng_n)^{-1}\gg \frac{X^\varepsilon}{\log X},$$
which contradicts boundedness.
Many thanks to the both of you. I guess no non trivial upper bound for $g_n$ can be deduced from this result?
@SylvainJULIEN, you see, the abscissa of convergence in this case is more about lower bounds. You can have $g_{n\ln n}=2$ and all other values as large as you want and still get the same result. There are conjectures on moments of $g_n$, maybe they can give us something about the behaviour of $\zeta_{Pol}$ but probably not vice versa
|
2025-03-21T14:48:29.905432
| 2020-02-18T20:50:10 |
353027
|
{
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],
"authors": [
"Anton Petrunin",
"Bumblebee",
"David Handelman",
"Jim Belk",
"Martin Sleziak",
"Nate Eldredge",
"Nik Weaver",
"Peter LeFanu Lumsdaine",
"Piotr Hajlasz",
"Timothy Chow",
"Will Brian",
"YCor",
"https://mathoverflow.net/users/105652",
"https://mathoverflow.net/users/121665",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1441",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/42278",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/54507",
"https://mathoverflow.net/users/6514",
"https://mathoverflow.net/users/70618",
"https://mathoverflow.net/users/8250",
"nikola karabatic"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353027"
}
|
Stack Exchange
|
Existence of a strange measure
The answer to this question must be known, but I do not know where to find it. It is related to the Ulam measures I believe.
Question. Is there a finitely additive measure defined on all subsets of positive integers $\mathbb{N}$, with values into $\{0,1\}$, (only two values) that is
$$
\mu:2^{\mathbb{N}}\to \{0,1\}
$$
such that $\mu(\{n\})=0$ for every $n\in\mathbb{N}$ and $\mu(\mathbb{N})=1$ ?
I believe such a result is needed in a proof of the co-area inequality that is stated in Coarea inequality, Eilenberg inequality. I am working with my student on some generalizations of that result so this is a research related question.
This is called non-principal ultrafilter (namely $\mu^{-1}({1})$)
@YCor What would be the best (easy to read) reference to this construction?
I don't know (it's extensively documented anyway), but here's a proof (of its existence on any infinite set $X$): the set of ultrafilters is closed in the pointwise convergence topology, hence is compact, and its subset of principal ultrafilters (=Dirac measures) is an infinite discrete subset, hence is not compact and hence is a proper subset. Verifications are straightforward.
@YCor Thank you so much. Following your suggestion about the non-principal ultrafilters I managed to complete all details as a direct consequence of Zorn's lemma. I will write a short proof soon. I actually needed this result for the co-area formula for mappings into metric spaces.
An approach that functional analysts might like is to observe that $l^\infty/c_0$ is a C*-algebra. So let $\phi: l^\infty/c_0 \to \mathbb{C}$ be a unital *-homomorphism, and then define $\mu(S) = \phi(1_S + c_0)$.
Without using ultrafilters: Let $B$ be $2^N$ as a boolean algebra (viewed as a ring in the usual way), and let $I$ be the ideal generated by projections onto singletons; then $I$ consists of the projections onto finite subsets, and is not equal to $B$. Let $M$ be any maximal ideal of $B$ containing $I$ (exists since $B$ is unital). Then $B/M $ is just the two-element field, and define $\mu$ as the corresponding homomorphism $ M \to B/M$. [Of course, $B \setminus M$ is a non-principal ultrafilter, but it isn't necessary to bring in such a complicated notion.]
Hmm, I didn't see the answer below (my comment is essentially the same) ...
A very similar question was already discussed on MO: https://mathoverflow.net/q/95954/105652
@DavidHandelman: A non-principal ultrafilter isn’t any kind of “complicated notion” — it’s just spelling out the details of what “maximal ideal” means in this special case (or rather, the dual conditions).
@Peter LeFanu Lumsdaine It depends on your notion of what "complicated" is. When I asked my 4th year and graduate algebra class if they had seen ultraproducts (for a proof of the existence of algebraic closures of fields), only a few of them responded affirmatively. But all had seen maximal ideals, and frequently.
@DavidHandelman : Maybe "unfamiliar terminology" is a better expression than "complicated notion".
@DavidHandelman: Sure, I wouldn’t quibble with describing ultrafilters as, say, less familiar.
A related post on [math.se]: How can an ultrafilter be considered as a finitely additive measure?
The answer is yes. I wrote a proof using YCor's comment.
Theorem. There a finitely additive measure defined on all subsets of positive integers $\mathbb{N}$, with values into $\{0,1\}$, (only two values) that is
$$
\mu:2^{\mathbb{N}}\to \{0,1\}
$$
such that $\mu(\{n\})=0$ for every $n\in\mathbb{N}$ and $\mu(\mathbb{N})=1$.
Proof.
Let
$$
\mathcal{F}_0=\{A\subset\mathbb{N}:\, \mathbb{N}\setminus A\ \ \text{is finite}.\}
$$
$\mathcal{F}_0$ is a filter. By a filter we mean here a family $\mathcal{F}\subset 2^\mathbb{N}$ with the following properties
$\mathbb{N}\in \mathcal{F}$,
$\emptyset\not\in\mathcal{F}$,
$A\in\mathcal{F}$, $A\subset B$ $\Rightarrow$ $B\in\mathcal{F}$,
$A,B\in\mathcal{F}$ $\Rightarrow$ $A\cap B\in\mathcal{F}$.
Lemma. There is a family $\mathcal{F}\subset 2^{\mathbb{N}}$ such that $\mathcal{F}_0\subset\mathcal{F}$ and $\mathcal{F}$ has properties 1.-4. and also
property
$A\subset\mathbb{N}$ $\Rightarrow$ $A\in\mathcal{F}$ or $\mathbb{N}\setminus A\in\mathcal{F}$.
Remark. A family $\mathcal{F}\subset 2^\mathbb{N}$ satisfying properties 1.-5. is called an ultrafilter.
Proof.
Filters containing $\mathcal{F}_0$ are ordered by the inclusion. The union of a chain of filters is a filter (this is obvious). Therefore by the Kuratowski-Zorn lemma there is a maximal filter $\mathcal{F}$ that contains $\mathcal{F}_0$. It remains to show that $\mathcal{F}$ satisfies property 5. (it has properties 1.-4. since it is a filter).
Suppose to the contrary that there is $A\subset\mathbb{N}$ such that
$$
(*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
A\not\in\mathcal{F}
\quad
\text{and}
\quad
\mathbb{N}\setminus A\not\in \mathcal{F}.
$$
Define
$$
\widetilde{\mathcal{F}}=
\{ E\subset \mathbb{N}:\, \exists B\in\mathcal{F}\ \ A\cap B\subset E\}.
$$
Note that $\mathcal{F}\subset\widetilde{\mathcal F}$, because for $B\in\mathcal{F}$, $A\cap B\subset B$ so $E=B\in\widetilde{\mathcal{F}}$.
Also $\mathcal{F}\subsetneq\widetilde{F}$ is a propert subset, because $\mathbb{N}\in\mathcal{F}$, $A\cap\mathbb{N}\subset A$ so $A\in\widetilde{\mathcal F}$ by definition and hence $A\in\widetilde{\mathcal F}\setminus\mathcal{F}$.
It remains to prove that $\widetilde{\mathcal F}$ is a filter (has properties 1.-4.) to reach a contradiction with the maximality of $\mathcal{F}$.
Clearly $\widetilde{\mathcal F}$ has properties 1., 3., and 4. (by property 4. for $\mathcal{F}$).
It remains to prove property 2. Suppose to the contrary that $\emptyset\in \widetilde{\mathcal F}$. Then $A\cap B=\emptyset$ for some $B\in\mathcal{F}$. Then $B\subset\mathbb{N}\setminus A$ so $\mathbb{N}\setminus A\in\mathcal{F}$ by property 3. for $\mathcal{F}$ which contradicts $(*)$. The proof is complete.
$\Box$
Now we define the finitely additive measure
$$
\mu:2^{\mathbb{N}}\to \{0,1\}
$$
such that $\mu(\{n\})=0$ for every $n\in\mathbb{N}$ and $\mu(\mathbb{N})=1$ as follows
$$
\mu(A)=
\begin{cases}
1 & \text{if $A\in \mathcal{F}$},\\
0 & \text{if $A\not\in\mathcal{F}$}.
\end{cases}
$$
Finite-additivity of $\mu$ follows from the fact that no two subsets with measure $1$ can be disjoint.
Note: Countable additivity fails because
$$
\mathbb{N} = \bigcup_{n=1}^\infty \{n\}
\quad
\text{but}
\quad
\mu(\mathbb{N}) \neq \sum_{n=1}^\infty \mu(\{n\}) \ .
$$
$\Box$
This interpretation of ultrafilter was used by Kleiner and Leeb in “Rigidity of quasi-isometries ...”
Just about any form of the axiom of choice can be used to prove this. I like topology, so here's a proof using Tychonoff's theorem.
Consider the space $2^{2^\mathbb{N}}$ of all functions $2^{\mathbb{N}}\to\{0,1\}$ under the product topology. By Tychonoff's theorem this is compact, and it is easy to check from the definition of the product topology that the set $\mathcal{F}$ of all finitely-additive $\{0,1\}$-valued measures on $\mathbb{N}$ is closed in $2^{2^\mathbb{N}}$. Then the sequence of measures $\mu_k\in\mathcal{F}$ defined by
$$
\mu_k(S) = \begin{cases}1 & \text{if }k\in S,\\ 0 & \text{if }k\notin S\end{cases}
$$
must have a limit point $\mu\in\mathcal{F}$. Then $\mu(\mathbb{N})=1$ since $\mu_k(\mathbb{N})=1$ for all $k$, and $\mu(\{n\}) = 0$ for all $n\in\mathbb{N}$ since $\mu_k(\{n\})=0$ whenever $k>n$.
I like this proof! Here's a fun exercise to go with it: even though the closure of your sequence is compact, no subsequence of it converges! (This seems wrong at first, especially if you're accustomed to working with metric spaces -- but there's no contradiction here.)
This is basically the detailed version of the 3-line proof I gave in a comment.
@YCor Sorry, I hadn't noticed your comment, but you're right that it's the same proof as the one you suggest.
My comment seems to be buried so I'd like to repeat it here. There is a simple C*-algebra construction that answers the question. The quotient space $l^\infty/c_0$ is a unital commutative C*-algebra, so there is a $*$-isomorphism $\Phi: l^\infty/c_0 \cong C(X)$ for some compact Hausdorff space $X$. For any $x \in X$, define $\mu_x(S) = \Phi(1_S)(x)$.
Every such $\mu_x$ works. In fact, this establishes a 1-1 correspondence between the set of non-principal ultrafilters on $\mathbb{N}$ and the spectrum of $l^\infty/c_0$.
And $X$ is in fact the Stone-Čech remainder $\beta \mathbb{N} \setminus \mathbb{N}$, right? So there's another connection.
Right, the elements of $\beta\mathbb{N}\setminus\mathbb{N}$ correspond to free ultrafilters.
@NikWeaver: What is $c_0$?
@Bumblebee: it is one of the standard sequence spaces.
This can be proved without introducing ultrafilters by name, by doing "finitary measure theory" and using Zorn's lemma.
An algebra $A$ on a set $X$ is just a $\sigma$-algebra without the $\sigma$, i.e. $\newcommand{\powerset}{\mathcal{P}}A \subseteq \powerset(X)$ and is closed under finite unions and complements (and therefore all other Boolean operations).
Let $\newcommand{\N}{\mathbb{N}}F \subseteq \powerset(\N)$ be the set of finite sets and their complements (so-called cofinite sets). This is an algebra. Furthermore, we can define a 2-valued finitely-additive measure $\mu : F \rightarrow \{0,1\}$ to be $0$ on the finite sets and $1$ on the cofinite sets. The existence of the required 2-valued finitely-additive measure on $\powerset(\N)$ then follows from:
Proposition For any algebra $A \subseteq \powerset(X)$ and finitely-additive 2-valued measure $\mu : A \rightarrow \{0,1\}$, there exists a finitely-additive measure $\overline{\mu} : \powerset(X) \rightarrow \{0,1\}$ extending $\mu$.
Proof: Most of the difficulty is in believing that it's true. We use Zorn's lemma. The poset consists of pairs $(B,\nu)$ where $B \supseteq A$ is an algebra of sets, and $\nu : B \rightarrow \{0,1\}$ is a finitely-additive measure extending $\mu$. The order relation $(B_1,\nu_1) \leq (B_2,\nu_2)$ is defined to hold when $B_1 \subseteq B_2$ and $\nu_2$ extends $\nu_1$. Every chain in this poset has an upper bound - we just take the union of algebras (this is the step that fails for $\sigma$-algebras) and define the measure on the union in the obvious way.
Let $(B,\nu)$ be a maximal element in the poset. Suppose for a contradiction that $B \neq \powerset(X)$, so there is some $U \in \powerset(X) \setminus B$. We contradict the maximality of $B$ by extending $\nu$ to a larger algebra $B'$ including $U$. Define $B' = \{ (U \cap S_1) \cup (\lnot U \cap S_2) \mid S_1, S_2 \in B \}$. It is clear that $B \subseteq B'$ and $U \in B'$, and with a little Boolean reasoning we can prove that for all $S_1,S_2,T_1,T_2 \in B$:
$$
((U \cap S_1) \cup (\lnot U \cap S_2)) \cup ((U \cap T_1) \cup (\lnot U \cap T_2))\\ = (U \cap (S_1 \cup T_1)) \cup (\lnot U \cap (S_2 \cup T_2))
$$
and
$$
\lnot ((U \cap S_1) \cup (\lnot U \cap S_2)) = (U \cap \lnot S_1) \cup (\lnot U \cap \lnot S_2)
$$
This proves that $B'$ is an algebra.
Now, define $d \in \{0,1\}$ to be the "outer measure" of $U$, i.e. $d = 0$ if there exists $S \in B$ such that $U \subseteq S$ and $\nu(S) = 0$, otherwise $d = 1$. Without loss of generality we can take $d = 1$, because we can exchange the roles of $U$ and $\lnot U$. We define $\nu'((U \cap S_1) \cup (\lnot U \cap S_2)) = \nu(S_1)$. This is well-defined because if $(U \cap S_1) \cup (\lnot U \cap S_2) = (U \cap T_1) \cup (\lnot U \cap S_2)$, then $U \cap S_1 = U \cap S_2$, so $U \subseteq \lnot (S_1 \triangle S_2)$, so as $d = 1$, $\nu(\lnot (S_1 \triangle S_2)) = 1$, and therefore $\nu(S_1) = \nu(S_2)$. The identities we used to prove that $B'$ is an algebra can then be used to prove that $\nu'$ is finitely additive, and it follows directly from the definition that it extends $\nu$. So we successfully contradicted the maximality of $B$. $\square$
Of course, I actually think ultrafilters are a good thing to know about, both in topology and logic. There is also no metamathematical benefit in doing it this way - over ZF the existence of a non-principal ultrafilter on $\N$ and a finitely-additive 2-valued measure on $\powerset(\N)$ are equivalent. The above proof is based on something I came up with while reproving Stone duality in the case where the points of the Stone space are defined to be Boolean homomorphisms into $2$, rather than ultrafilters. The proposition above is a special case of the fact that complete Boolean algebras (such as $2$) are injective objects in the category of Boolean algebras.
|
2025-03-21T14:48:29.906235
| 2020-02-18T21:48:58 |
353029
|
{
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"authors": [
"Fernando Muro",
"https://mathoverflow.net/users/12166"
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|
Stack Exchange
|
$\ell$-adic Eilenberg-MacLane space and Brown representability
I posted the following question on MathStackexchange, where it was suggested that I should move my question to Mathoverflow, which do here (https://math.stackexchange.com/questions/3550741/algebraic-geometric-analogue-of-browns-representability). In topology, the Brown representability theorem is very useful to show for instance that the functor $X\rightarrow H^i(X,A)$ is representable. My question is whether or not there exists an algebraic geometric analogue of the (pointed) homotopy category of CW complexes and of Brown's representability theorem. I would like to use this to show the existence of an $\ell$-adic analogue of the Eilenberg-MacLane space, i.e. to show the functor
$$X\rightarrow H^i_{ét}(X,\mathbb{Q}_{\ell})$$
is representable.
https://mathoverflow.net/questions/242660/representability-of-weil-cohomology-theories-in-stable-motivic-homotopy-theory
|
2025-03-21T14:48:29.906322
| 2020-02-18T22:05:44 |
353030
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"https://mathoverflow.net/users/150135",
"sdigr"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353030"
}
|
Stack Exchange
|
Transversality for algebraic spaces
$\DeclareMathOperator\dim{dim}$I want to apply EGA IV 4, Proposition 17.13.2 to a cartesian diagram
of algebraic spaces over a fixed scheme $S$.
I know the relative dimensions $\dim(\mathfrak{X}'/S)$, $\dim(\mathfrak{Y}/S)$ and $\dim(\mathfrak{X}/S)$ and I want to use the Proposition to determine the relative dimension of $\mathfrak{Y}'$ by showing that the map on tangent spaces is surjective.
The motivation for my question is L. Lafforgue, Proposition 1, page 29 where he obviously applies the Proposition from EGA to a diagram of algebraic spaces.
The problem that I am concerned about is, that the Proposition is formulated and proven only for the case of schemes, so a priori it is not immediately applicable to the diagram.
I expect at least one of the following ideas to work:
State and prove the Lemma for algebraic spaces. (If it still holds?)
How is the tangent space of an algebraic space even defined? Is this the tangent space of a functor as defined in deformation theory?
Use the fact that every algebraic space has an atlas and apply the lemma to the resulting diagram.
I could not figure any of these ideas out yet and would appreciate, if someone could explain me what the author had in his mind.
I found an answer myself. Actually it is very easy. All one has to do is to repeat the reduction step in precisely the same way as one can reduce from stacks to spaces. I can write this as an answer, if appreciated.
|
2025-03-21T14:48:29.906717
| 2020-02-18T22:07:28 |
353031
|
{
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"Abdelmalek Abdesselam",
"JustWannaKnow",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353031"
}
|
Stack Exchange
|
Thermodynamic limit and Gaussian measures
Let $\Lambda \subset \mathbb{Z}^{d}$ be finite and fixed and consider $\mathbb{R}^{|\Lambda|}$ be the vector space of all sequences $\varphi = (\varphi_{x})_{x\in \Lambda}$. We equip $\mathbb{R}^{|\Lambda|}$ with its Borel $\sigma$-algebra $\mathbb{B}(\mathbb{R}^{|\Lambda|})$. We denote by $\nu$ the Lebesgue measure restricted to $\mathbb{B}(\mathbb{R}^{|\Lambda|})$.
Now, in statistical mechanics, one usually consider an action $S_{\Lambda}: \mathbb{R}^{|\Lambda|}\to \mathbb{R}$ which is assumed to be a measurable function. This action gives rise to a finite volume Gibbs measure $\mu_{\Lambda}$, defined by means of its density with respect to $\nu$:
\begin{eqnarray}
d\mu_{\Lambda}(\varphi) := \mbox{const.} e^{-S_{\Lambda}(\varphi)}d\nu(\varphi) \tag{1}\label{1}
\end{eqnarray}
where the $\mbox{const.}$ term in (\ref{1}) is a normalization factor so that $\mu_{\Lambda}$ is a probability measure on $\mathbb{R}^{|\Lambda|}$.
From the physics point of view, we are interested in studying the behavior of the system in the infinite volume limit $\Lambda \nearrow \mathbb{Z}^{d}$. Thus, we have to first ensure that the infinite volume measure $\mu_{\mathbb{Z}^{d}} \equiv \mu$ exists in some sense. We usually take this limit as the weak-limit:
\begin{eqnarray}
\int d\mu e^{i\langle f, \varphi\rangle} := \lim_{n\to \infty}\int d\mu_{\Lambda_{n}}e^{i\langle f, \varphi\rangle} \tag{2}\label{2}
\end{eqnarray}
provided this limit exists. Here, $\Lambda_{n}$ is a sequence of increasing sets converging to $\mathbb{Z}^{d}$.
A particular case of the above scenario is when the action $S_{\Lambda}$ comes from a strictly positive quadratic form $Q_{\Lambda}: \mathbb{R}^{|\Lambda|}\times \mathbb{R}^{|\Lambda|} \to \mathbb{R}$. If we set $S_{\Lambda}(\varphi) = Q_{\Lambda}(\varphi, \varphi)$, the measure (\ref{1}) becomes Gaussian.
Now, it is a known fact that Gaussian measures are consistent in the sense of Kolmogorov Theorem, so that this very same Theorem implies the existence of a measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. (*)
My Question is whether we can interpret this measure $\mu$ obtained using Kolmogorov's Theorem as a weak limit of the form (\ref{2}) in some sense. My point here is: it seems legit to consider this $\mu$ as a infinite volume measure, but the path taken to define this limit was not by means of a limit such as (\ref{2}). Besides, the inner product $\langle f,\varphi \rangle$ only makes sense on $\mathbb{R}^{\mathbb{Z}^{d}}$ if we restrict it to some subspace, say $\mathcal{l}^{2}(\mathbb{Z}^{d})$. How to connect these two scenarios?
EDIT: Let me elaborate more on (*). Simon's book states Kolmogorov's Theorem as follows.
Theorem [Kolmogorov]: Let $\mathcal{I}$ be a countable set and let a probability measure $\mu_{|I|}$ on $\mathbb{R}^{|I|}$ be given for each finite set $I\subset \mathcal{I}$, so that the family of $\mu_{I}$'s is consistent (i.e. $\mu_{I}(A) = \mu_{I'}(A\times \mathbb{R}^{|I'|-|I|})$ if $|I'|\ge |I|$). Then there is a probability measure space $(X, \mathcal{F}, \mu)$ and random variables $\{f_{\alpha}\}_{\alpha \in \mathcal{I}}$ so that $\mu_{I}$ is the joint probability distribution of $\{f_{\alpha}\}_{\alpha \in \mathcal{I}}$.
Now, the proof of this theorem shows that $X$ is actually $\dot{\mathbb{R}}^{|\mathcal{I}|}$ where $\dot{\mathbb{R}}:=\mathbb{R}\cup \{+\infty\}$ is a compactification of $\mathbb{R}$. Thus, if we take $\mathcal{I}=\mathbb{Z}^{d}$, and for each finite $\Lambda \subset \mathbb{Z}^{d}$ the associate $\mu_{\Lambda}$ to be a Gaussian measure (say, nondegenerate and associated to some positive-definite matrix $C_{\Lambda}$), the above Theorem proves the existence of a (Gaussian) measure on $\mathbb{R}^{\mathbb{Z}^{d}}$.
EDIT 2: The measures $\mu_{\Lambda_{n}}$ are, in fact, defined in $\mathbb{R}^{\mathbb{Z}^{d}}$ with the product topology. It can be constructed from (\ref{1}) by taking, instead of $S_{\Lambda}$, an action $S:\mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ with free boundary conditions outside each $\Lambda_{n}$.
The question does not quite make sense yet. To talk about weak convergence, you need your measures to be on a fixed topological space. I guess here $\mathbb{R}^{\mathbb{Z}^d}$ with the product topology. So you have to decide what $\varphi$ is doing outside $\Lambda_n$, for each finite $n$. Also, your $\mathcal{Q}$ should be a $\mathcal{Q}_{\Lambda}$. You mention an alternate direct construction of $\mu$ in infinite volume. Explain precisely what you mean. Does it come from a $\mathcal{Q}$ in infinite volume?
related https://mathoverflow.net/questions/277678/weak-convergence-for-discrete-time-processes-using-characteristic-functions
Edited! Thanks!!
What do you mean by free boundary conditions?
Not quite an answer (because I still don't understand the question) but too long for a comment.
Take the following very simple example where the $\varphi_{x}$ are iid $\mathscr{N}(0,1)$ random variables. There is no problem constructing their joint law $\mu$ as a probability measure on $\mathbb{R}^{\mathbb{Z}^d}$ with the product topology. The way to do that is using the Daniell-Kolmogorov Extension Theorem.
Now what about the physicists' beloved notion of action? Well, it just does not make sense.
In this case, a physicist may want to write formally
$$
d\mu(\varphi)=\frac{1}{Z}\ e^{-S(\varphi)}\ \prod_{x\in\mathbb{Z}^{d}} d\varphi_x
$$
with the action $S(\varphi)=\frac{1}{2}Q(\varphi,\varphi)$, (BTW please put the 1/2)
where
$$
Q(\varphi,\varphi)=\sum_{x\in\mathbb{Z}^d} \phi_x^2\ .
$$
Now if the random variable $\varphi\in\mathbb{R}^{\mathbb{Z}^d}$ is sampled according to the (well defined) probability measure $\mu$, then almost surely
$$
Q(\varphi,\varphi)=\infty
$$
as can be shown say by Kolmogorov's Three Series Theorem.
This is all to say the question would make more sense if, rather than talking about actions and the matrix $Q$, it talked instead about the covariance matrix/bilinear form $C=Q^{-1}$.
That's a very nice comment. The point is: how to systematically define what we mean by a infinite volume limit, in order to study the system in the thermodynamic limit? The Gaussian measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$ constructed using Kolmogorov's Theorem is a legit infinite volume limit but, in general, the infinite volume measure is assumed to be that satisfying (2). It then seems that the limit is taken conveniently, depending on the model rather than a systematic approach.
although one could, one does not construct an infinite volume Gaussian measure using (2) but directly. (2) is used for constructing more physically interesting non-Gaussian measures.
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2025-03-21T14:48:29.907170
| 2020-02-19T01:04:19 |
353040
|
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|
Stack Exchange
|
Are there any necessary conditions of lacunary functions known?
On the internet, most theorems about lacunary function only give the sufficient conditions. For example, Ostrowski-Hadamard Gap Theorem concerns the asymptotic length of null Taylor coefficients, while a theorem of Carlson states that a Taylor series with integral coefficients is either rational or lacunary.
However, it seems like that a Taylor series not satisfying the conditions in these theorems may also be lacunary. Are there any theorems about the necessary conditions of lacunary functions?
Incidentally, Carlson's theorem also assumes the radius of convergence is 1. For instance, $\sum {2n\choose n}x^n$ is neither rational nor lacunary.
Edit. I added two references to the work of Miloš Kössler I found recently: he, by using the variable transformation $z\mapsto \zeta + \eta e^{-i\theta} \zeta^2$, $0< \eta \le {3/ 4}$, gives two necessary and sufficient condition for the power series \eqref{1} to have the point $e^{i\theta}$ as a singular boundary point. These condition, which do not rely on the one of Fabry described below, are briefly described in the final notes.
Terminology: by singular boundary point I mean a boundary point across which you cannot analytically continue an analytic function defined on the interior of the give domain, while this is always possible at a regular boundary point.
A necessary and sufficient condition.
I am not aware of any necessary condition for the "lacunarity". However, a conceptually simple necessary and sufficient condition for a boundary point to be a singular point is known: and as a corollary, you can use it and check every boundary point of the domain of your function is singular or not.
Without any restriction on generality, let's consider a power series
$$
f(z)=\sum_{n=0}^{\infty} a_n z^n,\label{1}\tag{1}
$$
whose radius of convergence $R_f$ is 1 and suppose we want to check if $z=1$ is singular or not.
If the expansion of $f(x)$ as a power series centered at any point on the real segment $]0,1[$ converges on a disk which includes $z=1$ then this point is regular, otherwise it is singular.
This form of the condition is fully given by Markushevich ([2] chapter IX §IX.7 pp313-314) and Titchmarsh ([3] chapter 7, §7.23, p. 216): however, this latter author follows Landau in simplifying the calculations by introducing the following function $F(\zeta)$ ([3] chapter 7, §7.23, pp. 216-217, [1] chapter 5, §19, p. 76-77). Let
$$
F(\zeta)=\frac{1}{1-\zeta}f\left(\frac{\zeta}{1-\zeta}\right)=\sum_{n=0}^{\infty} b_n z^n,
$$
where $b_n=\sum_{m=0}^{n}\binom{n}{m} a_m$. Then $z=1$ is a singular boundary point for $f$ if and only if
$$
R_F=\limsup_{n\to\infty}|b_n|^{-\frac{1}{n}}=\frac{1}{2}, \label{2}\tag{2}
$$
and thus we get the following
Corollary. Let $f:\Bbb C\to\Bbb C$ be an analytic function whose power series expansion at $0\in\Bbb C$ is \eqref{1}. Then $f$ is lacunary on the open unit disk $\Bbb D$ (i.e. $\partial\Bbb D$ is the "natural boundary" for $f$) if and only if
$$
\limsup_{n\to\infty} |a_n|^\frac{1}{n}=1\;\wedge\;\limsup_{n\to\infty}{\left|\sum_{m=0}^{n}\binom{n}{m} a_me^{im\theta}\right|^\frac{1}{n}}\!\!=2\quad\forall \theta\in[0,2\pi]
$$
Notes
According to Landau ([1] chapter 5, §19, p. 76), this criterion is due to Fabry.
Reference [2] have been translated in English as Markushevich A.I.
The theory of analytic functions:
a brief course, Moscow: MIR: however, I do not have access to a copy of that book, therefore I refer to the Italian edition listed below.
Miloš Kössler, by using the variable transformation $z\mapsto \zeta + \eta e^{-i\theta} \zeta^2$, transforms \eqref{1} in the power series $F(\zeta)=\sum_{n=0}^{\infty} A_n(\eta, e^{i\theta}) \zeta^n$. Then proves that ([A1], p. 27) a necessary and sufficient condition for the point $e^{i\theta}$ to be a singular boundary point for \eqref{1} is that
$$
\limsup_{n \to \infty} {|A_n(\eta, e^{i\theta})|^{1\over n}}=\frac{2\eta}{\sqrt{1+4\eta}-1}\qquad 0< \eta \le {3\over 4}\label{3}\tag{2b}
$$
Obviously, if the above value is independent from $\theta\in [0,2\pi]$ then the all the boundary of the unit disk is made of singular points for \eqref{1}. Kössler's condition may be interesting due to the fact that while the coefficients $\{b_n\}_{n\in \Bbb N}$ above are customarily polynomials of order $n$ respect to $e^{i\theta}$, the coefficients $A_n$ are polinomials at most of degree $\big[{n \over2}\big]$ (respect to the same variable). In the paper [A2] (pp. 528-529), with the same title, he further simplifies condition \eqref{3} assuming directly $\eta= {3/4}$ and defining a sequence $\{B_n\}_{n\in \Bbb N}$ by which \eqref{3} becomes
$$
\limsup_{n \to \infty} {|B_n(e^{i\theta})|^{1\over n}}= {3\over 4}.\label{4}\tag{2c}
$$
Each $B_n$ is a polynomial whose (variable, depending on an arbitrarily small positive constant $\mu$) number of terms is howhever less than $\big[{n \over2}\big]$, thus again \eqref{4} could be easier to veryfy respect to conditions \eqref{2} and \eqref{3} for some power series $f$.
References
[1] Landau, Edmund; Gaier, Dieter, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie. 3 erw. Auflage (German), Berlin-Heildelberg-New York: Springer-Verlag, pp. XI+201 (1986), ISBN: 3-540-16886-9, MR0869998, Zbl 0601.30001.
[2] Markushevich, Alekseĭ Ivanovich, Elementi di teoria delle funzioni analitiche. Translated from the Russian by Ernest Kozlov, (Italian)
Nuova Biblioteca di Cultura, Serie Scientifica. Roma: Editori Riuniti; Moscow: Edizioni Mir. pp. 384 (1988), ISBN: 88-359-3284-X, MR1011460, Zbl 0694.30002.
[3] Titchmarsh, Edward Charles, The theory of functions 2nd ed., (English), Oxford: Oxford University Press, pp. X+454 (1939), JFM 65.0302.01, MR3728294, Zbl 0336.30001.
Addendum reference
[A1] M. Kössler, "Sur les singularités des séries entières" (French), Atti della Reale Accademia dei Lincei, Rendiconti, Classe di Scienze Fisiche Matematiche Naturali, Serie V, 32, No. 1, pp. 26-29 (1923), JFM 49.0236.01.
[A2] M. Kössler, "Sur les singularités des séries entières" (French), Atti della Reale Accademia dei Lincei, Rendiconti, Classe di Scienze Fisiche Matematiche Naturali, Serie V, 32, No. 1, pp. 528-531 (1923), JFM 49.0236.03.
To the downvoter: why the downvote? Is there a particular reason for which my answer in not useful?
|
2025-03-21T14:48:29.907694
| 2020-02-19T01:47:56 |
353042
|
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Stack Exchange
|
Number of regions formed by $n$ points in general position
Given $n$ points in $\mathbb{R}^d$ in general position, where $n\geq d+1$. For every $d$ points, form the hyperplane defined by these $d$ points. These hyperplanes cut $\mathbb{R}^d$ into several regions. My questions are:
(1) is there a formula in terms of $d$ and $n$ that describes the number of regions?
(2) the same question for the number of bounded regions?
I tried many key words on google but found nothing helpful. Any reference or ideas will be appreciated. Thanks
General position as defined in https://en.wikipedia.org/wiki/General_position
Can you maybe elaborate? Both for the general position arrangement where all 4 points lie on the boundary of their convex hull, as well as for the arrangement where one point lies in the interior, I count 6 bounded regions.
The maximum number of cells in an arrangement of $k$ hyperplanes in dimension $d$ is $O(k^d)$. Your $k$ is $\binom{n}{d}$.
@JosephO'Rourke Thanks for the info. Can you direct me to a reference where I can find the results?
@GHfromMO I think you are right, the number of bounded regions may vary from case to case. I am wondering whether the total number of regions would also vary or just be the same?
So an upper bound is $O(n^{d^2})$, which accords with Richard Stanley's exact count for $d=2$.
This is more of a long comment than an answer. It should be possible
to compute the number of regions and number of bounded regions using
Whitney's theorem for the characteristic polynomial $\chi(t)$ (Theorem
2.4 of these
notes), and Zaslavsky's theorem that the number of regions is
$(-1)^d \chi(-1)$, and the number of bounded regions is (in this
situation) $(-1)^d\chi(1)$ (Theorem 2.5 of the previous link). We need more
than the usual definition of "general position." We want the
position to be generic enough for the argument below (generalized to
$d$ dimensions) to hold.
Here is the computation for $d=2$. First, the empty intersection (the
ambient space $\mathbb{R}^2$) contributes $t^2$ to $\chi(t)$. The
${n\choose 2}$ lines will contribute $-{n\choose 2}t$. Now we must
consider all subsets of the lines that intersect in a point $p$. Let
$p$ be one of the original $n$ points. Then ${n-1\choose 2}$ pairs of
lines intersect in $p$, ${n-1\choose 3}$ triple of lines intersect in
$p$, etc., giving a contribution to $\chi(t)$ of
$$ {n-1\choose 2}-{n-1\choose 3}+{n-1\choose 4} -\cdots = n-2. $$
We have to multiply this by $n$ since there are $n$ choices for
$p$. There are now $3{n\choose 4}$ choices of two lines that don't
intersect in one of the original $n$ points, but they still intersect
by genericity. Thus we get an additional contribution of $3{n\choose
4}$. It follows that
$$ \chi(t) = t^2-{n\choose 2}t+n(n-2)+3{n\choose 4}. $$
The number of regions is
$$ \chi(-1) = \frac 18(n-1)(n^3-5n^2+18n-8). $$
The number of bounded regions is
$$ \chi(1) = \frac 18(n-1)(n-2)(n^2-3n+4). $$
Can someone extend this argument to $d$ dimensions?
Addendum. I worked out $d=3$. Here are the details. Let $X$ be an
$n$-element "generic" subset of $\mathbb{R}^3$. Thus $X$ determines
a set $\mathcal{A}$ of ${n\choose 3}$ hyperplanes. We need to find
all subsets of $\mathcal{A}$ with nonempty intersection. A
$j$-element subset that intersects in an $e$-dimensional affine
space contributes $(-1)^jt^e$ to the characteristic polynomial
$\chi(t)$.
Case 1: $e=3$. We take the intersection over the empty set to get
$\mathbb{R}^3$. This gives a term $t^3$.
Case 2: $e=2$. Each hyperplane contributes $-t^2$, giving a term
$-{n\choose 3}t^2$.
Case 3: $e=1$. (a) Any two hyperplanes intersect in a line (by
genericity), giving ${{n\choose 3}\choose 2}t$.
(b) Any $j\geq 3$ hyperplanes containing the same two points
$p,q\in X$ meet in a line. There are ${n\choose 2}$ choices for
$p,q$ and ${n-2\choose j}$ for the remaining element of $X$ in the
hyperplanes. Thus we get a contribution
$$ {n\choose 2}\sum_{j\geq 3}(-1)^j {n-2\choose j}t =
{n\choose 2}\left[-1+(n-2)-{n-2\choose 3}\right]t. $$
Case 4: $e=0$. (a) Any three hyperplanes intersect at a point,
except when all three contain the same two points $p,q\in X$. There
are ${{n\choose 3}\choose 3}$ ways to choose three hyperplanes, and
${n\choose 2}{n-2\choose 3}$ ways to choose them so that they
intersect in two points of $X$. Hence we get a contribution
$$ -\left[ {{n\choose 3}\choose 3}-{n\choose 2}{n-2\choose 3}
\right] $$
to the constant term of $\chi(t)$ (the minus sign because the
number of hyperplanes is odd).
(b) Any $j\geq 4$ hyperplanes meeting at a point $p\in X$. We can
choose $p$ in $n$ ways. We then must choose $j$ two-element subsets
of $X-p$ whose intersection is empty. There are ${{n-1\choose
2}\choose j}$ ways to choose $j$ two-element subsets of $X-p$. If
their intersection is nonempty, then they have a common element $q$
which can be chosen in $n-1$ ways, and then we can choose the
remaining elements in ${n-2\choose j}$ ways. This gives the
contribution
$$ n\sum_{j\geq 4}(-1)^j\left[ {{n-1\choose 2}\choose j}-
(n-1){n-2\choose j}\right] $$
$$ \ = n\left[ -1+{n-1\choose 2}-{{n-1\choose 2}\choose 2} +
{{n-1\choose 2}\choose 3}-(n-1)\left(-1+(n-2)
-{n-2\choose 2}+{n-2\choose 3}\right)\right]. $$
(c) Any $j\geq 3$ hyperplanes meeting at $p,q\in X$, together with
one additional hyperplane not containing $p$ or $q$. There are
${n\choose 2}$ choices for $p,q$ and ${n-2\choose j}$ ways to
choose $j$ hyperplanes containing $p,q$. There are then
${n-2\choose 3}$ ways to choose the additional hyperplane not
containing $p$ or $q$. Thus we get the contribution
$$ -\left[ \sum_{j\geq 3}(-1)^j{n-2\choose j}\right]
{n-2\choose 3} $$
$$ -{n-2\choose 3}{n\choose 2}\left[-1+(n-2)-{n-2\choose 2}
\right]. $$
Putting all this together gives the characteristic polynomial
$$ t^3-{n\choose 3}t^2+
\frac{1}{72}n(n-1)(n-3)(n^3-2n^2-16n+68)t $$
$$ -\frac{1}{1296}n(n-2)(n-3)(n^6-4n^5-74n^4+698n^3-2129n^2
+2276n-120). $$
The number of regions is
$$ \frac{1}{1296}(n-2)(n^8-7n^7-62n^6+938n^5-4295n^4+8429n^3
-4932n^2-2016n-648). $$
The number of bounded regions is
$$ \frac{1}{1296}(n-1)(n-2)(n-3)(n^6-3n^5-77n^4+603n^3
-1508n^2+1056n+216). $$
Conceivably there could be an error in the computation, but I checked
it for $n=4,5$ by a brute force computation.
This method should extend to any $d$, but the computation will be
more complicated, and I am too lazy to work out the details.
A note on the stronger condition required, here is an example of what can go wrong: Consider the simple case $6$ points in the plane, no $3$ on a line. There is a special degenerate case where we can partition the $6$ points into three pairs $(A, A')$ $(B, B')$ and $(C,C')$ where the lines $AA'$, $BB'$, and $CC'$ all intersect at a point. This will have one less region than the general case.
Theorem 2.2 of Ardila-Billey https://arxiv.org/abs/math/0605598 gives a combinatorial description of the relevant matroid (described dually, as the intersections of generic hyperplanes rather than the spans of generic points). I don't see an easy way to extract the characteristic polynomial from it, but maybe you do.
I replicated your result for $d=2$ based on my answer below. I proved there that $x$ generic lines produce $2x$ unbounded regions and $(x-2)(x-1)/2$ bounded ones. Then replaced $x$ with $n(n-1)/2$. Last, at each of the $n$ points the bounded regions that would exist in a generic $(n-1)$-lines case are collapsed to a point. So the total number of regions collapsed to a point is $n(n-3)(n-2)/2$. This leaves $(x-2)(x-1)/2$-$n(n-3)(n-2)/2$ bounded regions, which matches your calculation. I think $d=3$ could be tacked in a similar way, but much more messy.
Perhaps it is worth quoting this theorem, even though
it does not distinguish bounded from unbounded cells,
and is phrased in terms of the number of hyperplanes
rather than the number of points determining the hyperplanes.
Theorem. Let $H$ be a set of $n$ hyperplanes in $\mathbb{R}^d$.
The maximum number of $k$-dimensional cells in the arrangement
${\cal A}(H)$ formed by $H$, for $0 \le k \le d$, is
$$ \sum_{i=0}^k \binom{d-i}{k-i} \binom{n}{d-i} \;.$$
The maximum is attained exactly when ${\cal A}(H)$ is simple.
An arrangement is simple if every $d$ hyperplanes meet in a point, and no $d+1$ hyperplanes have a point in common.
In the OP's situation, the arrangement is non-simple.
For example, in $d=2$, $4$ points determine $6$ lines,
but each point has $3$ lines through it.
For $d=2$ and $k=2$, the above equation reduces to
the familiar expression
$$\binom{n}{2} + \binom{n}{1} + \binom{n}{0}
= \frac{1}{2}( n^2 + n + 2) \;.$$
Handbook of Discrete and Computational Geometry, 3rd ed. Chapman and Hall/CRC, 2017.
CRC link.
Chapter 28, Thm.28.1.1, p.724.
This non-answer completes Joseph O'Rourke's nice non-answer, for the case of $n$ hyperplanes in $\mathbb{R}^d$ in general position. But it also suggests that the OP situation may also well have unique answers.
Define:
$U_{d,n}=$ number of unbounded regions cut by $n$ hyperplanes in $\mathbb{R}^d$
$B_{d,n}=$ number of bounded regions
$T_{d,n}=$ total number of regions $=U_{d,n}+B_{d,n}$
$S_{d,n}=$ number of regions cut on the sphere $S^d$ by $n$ great $S^{d-1}$-circles
Then $U_{d,n}, B_{d,n}$, $T_{d,n}$ and $S_{d,n}$ are unique with these formulas:
$U_{d,0}=1$, $\quad B_{d,0}=0$, $\quad T_{d,0}=1$, $\quad S_{d,0}=1$
$U_{1,n}=2$, $\quad B_{1,n}=n-1$, $\quad T_{1,n}=n+1$, $\quad S_{1,n}=2n$
and for $n>0$
$U_{d+1,n}=S_{d,n}$
$S_{d,n}=U_{d,n}+2B_{d,n}$
$T_{d,n+1}=T_{d,n}+\sum_{i=0}^{i=d-1}{n\choose i}$
Proof.
In $\mathbb{R}^{d+1}$ take a huge and growing $S^d$ sphere, so that all the bounded regions zoom down to a point at the center of the sphere, the hyperplanes become great circles on the sphere and the unbounded regions corresponds to regions cut by the circles on the sphere. Therefore if the numbers are unique (as will be proved at the end) 1. follows.
Centrally project $\mathbb{R}^d$ onto a half $S^d$ (tangent to it). Complete the semisphere to a sphere by central symmetry. Then the hyperplanes become great circles, the $B_{d,n}$ bounded regions in $\mathbb{R}^d$ become $2B_{d,n}$ regions in $S^d$ and the $U_{d,n}$ unbounded ones become $U_{d,n}$ regions stretching across the suture line (equator) of the sphere. Again by unicity 2. follows.
Start with $\mathbb{R}^d$ and $n$ hyperplanes in generic position inside it. Now add a new hyperplane in generic position the following way:
first chose a point inside one region: no matter how that point is eventually stretched to a hyperplane, to it will split the region in two, for a gain of 1, or $n \choose 0$. Now stretch that point to a line: since it is a generic line it will meet each of the $n$ hyperplanes once and at each meeting the line will cross into one one new region and split it - with a gain of $n \choose 1$ new regions. Next stretch the line to a generic 2-plane, which will meet once each of the $(d-2)$-dimensional intersections of two hyperplanes; at each meeting the growing plane will arrive from having already crossed 3 of the 4 regions, to cross into the fourth and cut it; this a gain of another $n \choose 2$ regions.
In general as a generic $m-1$-plane grows to a generic $m$-plane it will meet all the $n \choose m$ $(n-m)$-dimensional intersections of $m$ hyperplanes, and each time it will go from cutting $2^m-1$ regions before crossing the intersection to cutting all $2^m$ after crossing, for a total gain of $n \choose m$ regions. This continues up to $m=d-1$, proving 3.
Proof of Unicity.
By induction:
$U_{d,0}$, $\quad B_{d,0}$, $\quad T_{d,0}$, $\quad S_{d,0}$ are unique;
$T_{d,n}$ unique $\implies$ $T_{d,n+1}$ unique (by the proof of 3.);
$U_{d,n}$ and $B_{d,n}$ unique $\implies$ $S_{d,n}$ unique (by the proof of 2. and the fact that the construction can be reversed in a non-unique way to show that $S_{d,n}=U_{d,n}+2B_{d,n}$ for some values of $U_{d,n}$ and $B_{d,n}$);
$S_{d,n}$ unique $\implies$ $U_{d+1,n}$ unique (by the proof of 1.);
$U_{d+1,n}$ and $T_{d+1,n}$ unique $\implies$ $B_{d+1,n}$ unique (as $B=T-U$).
I have a feeling that jiggling the $n \choose d$ hyperplanes of the OP into generic positions while carefully keeping track of the corresponding increase of regions (both bounded and unbounded) one may be able to reach an answer to the OP's question too. For a start this should be relatively easy for $d=2$ (but I haven't done it).
|
2025-03-21T14:48:29.908423
| 2020-02-19T03:21:50 |
353043
|
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"GTA",
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|
Stack Exchange
|
gluing Berkovich spaces
In his paper Etale cohomology for non-Archimedean analytic space (IHES), Berkovich explained how to glue $k$-analytic spaces (Prop. 1.3.3) and show its uniqueness using the Prop 1.3.2 (gluing morphism).
In Prop 1.3.3, it has two cases that we can glue the $k$-analytic spaces $X_i$ along $X_{ij}\subset X_i$.
(a) all $X_{ij}$ are open in $X_i$
(b) all $_{ij}$ are closed in $X_i$ and for each $i$, $X_{ij}$ is non empty for finitely many $j$.
Q1. Do we need, in (b) of Prop 1.3.3, the condition that $X_{ij}$ are closed in $X_i$? I'm not sure where he used this condition in the construction.
Prop 1.3.2 says if $X$ is $k$-analytic space with covering $\{X_i\}$ by analytic domains that is quasi-net (i.e. for each $x\in X$ there are $X_{i_1},\cdots,X_{i_n}$ such that $x\in \cap X_{i_k}$ and $\cup X_{i_k}$ is neighborhood at $x$), then we can glue the morphisms from $X_i$ to $Y$ that are compatible as the morphism from $X$ to $Y$. Here the quasi-net condition is necessary for example to make sure the continuity of glued morphism. He used this to prove the uniqueness of glued analytic spaces.
Q2. When we glue $\{X_i\}$ as in Prop.1.3.3, does the covering $\{X_i\}$ of $X$ (glued analytic space) becomes quasi-net?(so that we can apply the Prop 1.3.2)
I think this can fail when the $X_{ij}$ are not open, as the quotient map $\prod X_i \to X$ is not a open map.
If $X_{ij}$'s are not closed then the quotient topology might be weird, so that $X$ might not be even Hausdorff. You need some sort of compatibility between topologies of $X_{i}$. For your second question in the case of 1.3.3(b), indeed it forms a quasinet, because $x\in X$ can be contained in only finitely many $X_{i}$'s. Namely pick one $i$ such that $x\in X_{i}$, then any $x\in X_{j}$ would mean $x\in X_{ij}$ so there are possibly only finitely many $X_{j}$'s that can contain $x$. It is a neighborhood because $\cup \left(X_{j}-\cup X_{jk}\right)\subset\cup X_{j}$ is open.
I think the last question on the relation with adic spaces should be asked separately.
@GTA I guess you mean X_j - union of X_jk where X_jk's do not contain x and X_jk are closed so X_j - union of X_jk is open and take union of them? I think this is correct and gives me more sense why we require X_ij being closed.
@JakobWerner I see. I removed it now
|
2025-03-21T14:48:29.908611
| 2020-02-19T04:55:21 |
353046
|
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|
Stack Exchange
|
The mean of a running maximum
Suppose $$ is a one-dimensional standard Brownian motion defined on some probability space $(\Omega, \mathcal F, P)$ and let $():=\exp\{()−\frac{1}{2}−\frac{1}{+1}\}$ for $\ge 0$. Note that $(\infty):=\limsup_{t\to\infty}()=0$ a.s. because $\lim_{t\to\infty}\frac{()}{}=0$ a.s.
My question is: How to show that $E[\sup_{0\le t\le \infty}()]=\infty$? Many thanks. (I posed this question in stackexchange earlier today. Not sure if this is the right place to ask this question.)
Many thanks.
The supremum of a Brownian motion with drift -1/2 is distributed as an exponential random variable of mean 1.
This seems to me a standard exercice in a probability course: Ignore the $1/(t+1)$ term as $X(t)\geq e^{-1}\exp(W(t)-t/2)$. The term $M_t:=\exp(W(t)-t/2)$ is well known to be a martingale so $\mathbb{E}(M_t)=1$ for all $t$. If $\mathbb{E}(\sup_{0\leq t\leq \infty}X_t)<\infty$ then by dominated convergence $$1=\mathbb{E}(M_t)\rightarrow_{t\rightarrow \infty} \mathbb{E}(M_\infty)=0$$
Great! Many thanks.
|
2025-03-21T14:48:29.908742
| 2020-02-19T05:24:43 |
353047
|
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"Dabed",
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"Steven Stadnicki",
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|
Stack Exchange
|
Regularizing the sum of all primes
In the spirit of a similar question for the harmonic series, is there a way to regularize the (divergent) sum of all primes?
$$ \sum_{p \text{ prime}} p $$
Neither of these questions obtained a successful regularization:
Is it possible to assign a value to the sum of primes?
Is it possible to sum the divergent series with prime coefficients?
Is it possible to sum the divergent series with prime coefficients?
The prime zeta function, unfortunately, has a natural boundary on the imaginary line that prevents analytic continuation by the usual means.
On the other hand, we know that
$$ \prod_{p \text{ prime}} p = 4\pi^2 $$
To be specific: Is there a different kind of analytic continuation, such as the technique referred to by Gammel in this question, that can be used to continue the prime zeta function beyond its natural boundary?
Another vague idea that occurred to me is, instead of trying to "tunnel through" the natural boundary on the imaginary line, we could try to reach the other side by going around the Riemann sphere in the opposite direction: i.e. through infinity, popping out on the other side of the boundary. My search for something like this yielded Confinement as Analytic Continuation Beyond Infinity by Yamazaki and Yonekura. The image below shows what the function looks like around infinity (on the positive side). Could it be extended to the negative side?
Edit: Möbius inversion yields
\begin{align}
P(-1)
&= \sum_{p \text{ prime}} p \\
&= \sum_{n \geq 1} \frac{\mu(n)}{n} \log \zeta(-n) \\
&= \sum_{n \geq 1} \frac{\mu(2n)}{2n} \log \zeta(-2n) + \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log \zeta(-(2n-1))
\end{align}
Because the zeta function is zero at the negative even integers, the summands of the first series are undefined, strictly speaking. However, we notice that
\begin{align}
\sum_{n \geq 1} \frac{\mu(2n)}{2n} \log \zeta(-2n)
&= \sum_{n \geq 1} \frac{\mu(2n)}{2n} \log 0 \\
&= \left(\sum_{n \geq 1} \frac{\mu(2n)}{2n}\right) \log 0 \\
&= 0 \log 0 \\
&= \log 0^0 \\
&= \log 1 \\
&= 0 \\
\end{align}
Therefore we can get rid of these problematic terms and end up with
\begin{align}
P(-1)
&= \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log \zeta(-(2n-1)) \\
&= \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log \frac{(-1)^{2n-1} B_{(2n-1)+1}}{(2n-1)+1} \\
&= \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log \left(-\frac{B_{2n}}{2n}\right) \\
&= \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log (-1) - \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log 2n + \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log B_{2n} \\
&= 0 \log (-1) - \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log 2n + \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log B_{2n} \\
&= - \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log 2n + \sum_{n \geq 1} \frac{\mu(2n-1)}{2n-1} \log B_{2n} \\
\end{align}
where the first series converges but the second diverges.
see https://math.stackexchange.com/q/84642/87355
@CarloBeenakker The same link is in the OP.
The sum of the first $n$ primes is tabulated at http://oeis.org/A007504 – maybe there's something useful in one of the links/references given there.
The cited question from stackexchange was also asked on Overflow: https://mathoverflow.net/questions/179511/is-it-possible-to-sum-the-divergent-series-with-prime-coefficients (the comments there are not encouraging).
@GerryMyerson Good catch, I'll add it to the question.
Isn't this just a duplicate of the linked MO question?
Does this answer your question? Is it possible to sum the divergent series with prime coefficients?
@StevenStadnicki That's the same question Gerry Myerson already linked to.
@user76284 Sorry; that second message was autogenerated by the 'flag as dupe' option. The broad question (is there anything in this question that wasn't in the other?) still stands, though...
@StevenStadnicki I've highlighted a specific question I have about a technique referred to by Gammel that he says can continue functions beyond natural boundaries.
Maybe, related: https://mathoverflow.net/questions/389255/what-is-the-regularized-numerocity-of-prime-numbers
Looks like the regularized sum is $0$, intuitively.
@Anixx What's the intuition behind that?
Looking at the graphic. It's oscillations go to zero visually, even though there are infinitely many logarithmic singularities. I think, one can smooth out those logarithmis singularities somehow. I am talking about the prime zeta function graphic (left image here: https://mathworld.wolfram.com/PrimeZetaFunction.html)
@Anixx That's $P(0)$, not $P(-1)$.
Ah, then it is more related to my linked question, I somehow thought they are similar.
It was too long for a comment and overall as far as possible can be from rigorous so if it isn't helpful just tell me and I'll delete it.
In his blog John Baez talks of particular approach that he learn of to shown that $\sum_{n=1}^{\infty}n=-\frac{1}{12}$ he says that we have the formal series
$f(0)+f(1)+f(2)+...=[(1+e^{D}+e^{2D}+...)f(x)](0)=[\frac{1}{1-e^{D}}f(x)](0)=[\frac{D}{1-e^{D}}F(x)](0)=[(-1+\frac{D}{2}-\frac{D^2}{12}+...)F(x)](0)$
And letting $f(x)=x$ he gets $f(0)+f(1)+f(2)+...=0+1+2+...$ while on the other side it gives $[(-1+\frac{D}{2}-\frac{D^2}{12}+...)\frac{x^2}{2}](0)=[(-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}+0)](0)=\frac{-1}{12}$.
Now the divergence of harmonic series can be shown be using the Mercator series $\ln(\frac{1}{1-x})=\sum\frac{x^n}{n}$ and letting $x\to1$ which gives $\infty=\ln(\infty)=\sum\frac{1}{n}$
Euler used this to show also the divergence of primes by taking another logarithm of the mercator series and letting again $x\to1$ obtaining $\infty=\ln(\ln(\infty))=\ln(\sum\frac{1}{n})=\ln(\prod_p \frac{1}{1-p^{-1}})=\sum\ln(\frac{1}{1-p^{-1}})=\sum_p\sum_k\frac{1}{kp^k}=\sum_p\frac{1}{p}+\text{constant}$.
Just as $\ln(\frac{1}{1-x})=\sum\frac{x^n}{n}$ for $|x|<1$ we also have that $\ln(\ln(\frac{1}{1-x}))=\sum_p\frac{x^p}{p}$ but only when $x\to 1$ yet we could try to repeat the argument given before.
If for $\ln(\frac{1}{1-x})=\sum\frac{x^n}{n}$ we take the derivative and let $x=e^{D}$ then we get the formal series from above $\frac{1}{1-e^{D}}=\sum e^{nD}$.
So doing the same for $\ln(\ln(\frac{1}{1-x}))=\sum_p\frac{x^p}{p}$ gives $\frac{1}{(e^{D}-1)\ln(1-e^{D})}=\sum_p e^{pD}$(+), that is our formal series should be:
$f(2)+f(3)+f(5)+...=[(e^{2D}+e^{3D}+e^{5D}...)f(x)](0)=[\frac{1}{(e^{D}-1)\ln(1-e^{D})}f(x)](0)=[\frac{D}{(e^{D}-1)\ln(1-e^{D})}F(x)](0)=[\text{Taylor series}(\frac{D}{(e^{D}-1)\ln(1-e^{D})})F(x)](0)$
While before we had a finite second derivative $\frac{1}{2!}(\frac{y}{1-e^y})''|_{y=0}=-\frac{1}{12}$
Now we have a undefined second derivative$\frac{1}{2!}(\frac{y}{(e^y-1)\ln(1-e^y)})''|_{y=0}=\text{undefined}$
Were you able to obtain a concrete value with this approach?
@user76284 I updated the post so now instead of asking wolfram for the taylor series for which it was hard to tell what it was giving I asked for the second derivative and in one hand we get $-\frac{1}{12}$ but on the other it says is undefined so it diverges and that is why the taylor series looked weird, I ignore if further can be done but I suppose at the end it boils down to what you mentioned in your question that there is a natural border that prevents analytic continuation but at least we can try to gain some insight while trying to do it even if at the end we can't escape it.
It might be possible to overcome the boundary through generalized analytic continuation, by matching limits from the right side of the boundary to those of a function that is analytic to the left of it.
@user76284 Thanks for mentioning the topic of generalized analytic continuation that is nice to hear and something I would like to learn about yet now there is no more I can contribute for example I haven't even seen how the argument to $\prod p = 4\pi^2$ and $\prod n = \sqrt{2\pi}$ go along which I suppose would be useful but I would also have a value for the series so if I ever have an idea I will come back here.
You said $\sum_{n=1}^{\infty}\frac{1}{n}=-\frac{1}{12}$. This is wrong. The regularized value of this series is $\gamma$. The value $-\frac{1}{12}$ is the regularized value of $\sum_{n=1}^{\infty}n$.
@Anixx Oops edited, thank you very much.
Alas, it looks like the imaginary axis is the natural analyticity boundary for $P(1/s)$:
Flatten[Table[{x,y,N@Abs[PrimeZetaP[1/(x+I y)]]},{y,-.5,.5,.0025},{x,.001,1,.0025}],1]
//ListContourPlot
gives
|
2025-03-21T14:48:29.909252
| 2020-02-19T08:29:23 |
353049
|
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"M A Sofi",
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|
Stack Exchange
|
Lipschitz choice of "norm attaining maps'
I am not very sure if the following problem has been treated in the literature and if so, whether it always holds:
A Banach space $X$ is isomorphic to a Hilbert space if the 'norm attaining' map $F$ from $S_{X^*}$ into $S_X$ satisfying $x^*(F(x^*)) = 1$ is a bilipschitz equivalence. Here $S_Z$ denotes the unit sphere of the Banach space $Z$.
I think this will be correct, if I interpret your assumption correctly.
The assumption implies that the duality mapping on $X$ with gauge $t\mapsto t^2$, $$j_X(x) := \bigl\{ x^* \in X^* \colon x^*(x) = \|x^*\|\,\|x\|,~\|x^*\| = \|x\|\bigr\},$$ is bijective and Lipschitz continuous via $j_X(x) = \|x\|\, F^{-1}(x/\|x)$. Hence $X$ is $2$-smooth [Remark 5, XR]), so of type $2$ [Prop. IV.5.10, DGZ]. In particular, it is reflexive and $F$ induces the duality mapping $j_{X^*}$ on $X^*$ in the same way as $F^{-1}$ did on $X$. Analogously, $X^*$ is $2$-smooth, and this implies that $X$ is of cotype $2$ [Prop. IV.5.12, DGZ]. Finally, a Banach space which is both of type and cotype $2$ is isomorphic to a Hilbert space by Kwapien's theorem.
[XR] Xu, Zong-Ben; Roach, G. F., Characteristic inequalities of uniformly convex and uniformly smooth Banach spaces, J. Math. Anal. Appl. 157, No. 1, 189-210 (1991)
[DGZ] Deville, Robert; Godefroy, Gilles; Zizler, Václav, Smoothness and renormings in Banach spaces, Pitman Monographs and Surveys in Pure and Applied Mathematics. 64. Harlow: Longman Scientific & Technical. New York: John Wiley & Sons, Inc.. 376 p. (1993).
The argument looks plausible. Shall have to check more carefuly.
|
2025-03-21T14:48:29.909378
| 2020-02-19T09:44:51 |
353052
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353052"
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|
Stack Exchange
|
Coloring $\mathbb{N}$ such that $\{a, b, a+b\}$ is not monochromatic
Let $\mathbb{N}:=\omega \setminus \{0\}$. Is there $k\in \mathbb{N}$ and a map $c:\mathbb{N}\to \{1,\ldots,k\}$ such that for all $a,b\in\mathbb{N}$ the restriction $c|_{\{a,\,b,\,a+b\}}$ is non-constant?
No, this is the content of Schur's Theorem.
|
2025-03-21T14:48:29.909562
| 2020-02-19T10:33:41 |
353055
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Gerry Myerson",
"Giuseppe",
"Iosif Pinelis",
"https://mathoverflow.net/users/152578",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/3684"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353055"
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|
Stack Exchange
|
First-order non-linear differential equation and transcendental equation
I'm trying to solve this differential equation :
$$ \frac{dy}{dx}= \frac{-2 y^3}{(y+1)^2(y+2)^2} $$
with the boundary condition $y(x_0)=x_0$, $x>0$, and $y(x)$ being a positive function.
The integration of the equation is straightforward, however after integration, one gets a transcendental equation of the form $$a y(x)+by^2(x)+c \log(y)+\frac{d}{y}+\frac{e}{y^2}+ g(x_0)= z (x-x_0)$$
where $a,b,c , d, e,z $ are constants, and $g(x_0) $ is a function of $x_0$.
I tried to solve it with Lagrange inversion theorem, however due to the non triviality of the LHS, the computation of the $n$'th derivative is very complicated, is there any other way to solve it ?
Also posted to m.se https://math.stackexchange.com/questions/3552459/first-order-non-linear-differential-equation-and-transcendental-equation without notification to either site, an abuse of the system.
Calculating the $n$th derivative for arbitrary $n$ is not easy even for simple elementary functions. See e.g. this Mathematica answer for $\tan^{(n)}x$:
or the Faà di Bruno formula.
What can be done in your case is as follows: You have $y'=R_0(y)$, where $R_0(y)$ is a certain rational function of $y$. Hence, $y''=R_0'(y)y'=R_0'(y)R_0(y)=:R_1(y)$. Generally, for any natural $n$, recursively you have
$$y^{(n)}(x)=R_{n-1}(y(x)),$$
where $R_n:=R_{n-1}'R_{n-1}$. To find $R_{n-1}'$, you may want to first decompose $R_{n-1}$ into partial fractions. As for the value of $y(x)$, you compute it by solving numerically your transcendental equation or by solving numerically your original differential equation.
The Mathematica notebook image below shows such a numerical calculation (done in about 2.2 sec) of the set $\{(n,y^{(n)}(0))\colon n\in\{1,\dots,9\}\}$ for the solution $y$ of your differential equation with $y(0)=1$:
The calculation of $R_{10-1}$ takes apparently too much memory.
Dear Iosif Pinelis, Thank you a lot for your answer ! I may indeed ressort to a Taylor Expansion untills some order to approximate the solution, however don't you think there could be a way to get a closed form in terms of some special function for example?
@Giuseppe : I very much doubt that a closed form in terms of some special function is possible here. Mathematica's commands FunctionExpand and FullSimplify cannot do anything with this inverse function.
If you're trying to find the solution as a power series about $x=x_0$, that's fairly straightforward. Don't use Lagrange, though. Instead, substitute $y = x_0 + \sum_{i=1}^n a_i (x - x_0)^i$ in to the differential equation, expand to the desired order, equate coefficients of each power, and solve for the $a_i$. Here's the answer to order $7$, according to Maple:
$$y \left( x \right) =(x_{{0}}-2\,{\frac {{x_{{0}}}^{3}}{ \left( x_{{0}}
+1 \right) ^{2} \left( x_{{0}}+2 \right) ^{2}}} \left( x-x_{{0}}
\right) -2\,{\frac {{x_{{0}}}^{5} \left( {x_{{0}}}^{2}-3\,x_{{0}}-6
\right) }{ \left( x_{{0}}+1 \right) ^{5} \left( x_{{0}}+2 \right) ^{5
}}} \left( x-x_{{0}} \right) ^{2}-{\frac {4\,{x_{{0}}}^{7} \left( 3\,{
x_{{0}}}^{4}-18\,{x_{{0}}}^{3}-35\,{x_{{0}}}^{2}+36\,x_{{0}}+60
\right) }{3\, \left( x_{{0}}+1 \right) ^{8} \left( x_{{0}}+2 \right)
^{8}}} \left( x-x_{{0}} \right) ^{3}-{\frac {2\,{x_{{0}}}^{9} \left(
15\,{x_{{0}}}^{6}-135\,{x_{{0}}}^{5}-203\,{x_{{0}}}^{4}+753\,{x_{{0}}}
^{3}+1170\,{x_{{0}}}^{2}-396\,x_{{0}}-840 \right) }{3\, \left( x_{{0}}
+1 \right) ^{11} \left( x_{{0}}+2 \right) ^{11}}} \left( x-x_{{0}}
\right) ^{4}-{\frac {4\,{x_{{0}}}^{11} \left( 105\,{x_{{0}}}^{8}-1260
\,{x_{{0}}}^{7}-1062\,{x_{{0}}}^{6}+12528\,{x_{{0}}}^{5}+15889\,{x_{{0
}}}^{4}-22824\,{x_{{0}}}^{3}-37848\,{x_{{0}}}^{2}+2880\,x_{{0}}+15120
\right) }{15\, \left( x_{{0}}+1 \right) ^{14} \left( x_{{0}}+2
\right) ^{14}}} \left( x-x_{{0}} \right) ^{5}-{\frac {4\,{x_{{0}}}^{
13} \left( 945\,{x_{{0}}}^{10}-14175\,{x_{{0}}}^{9}-552\,{x_{{0}}}^{8}
+205254\,{x_{{0}}}^{7}+167497\,{x_{{0}}}^{6}-768099\,{x_{{0}}}^{5}-
1044390\,{x_{{0}}}^{4}+571608\,{x_{{0}}}^{3}+1258368\,{x_{{0}}}^{2}+
66960\,x_{{0}}-332640 \right) }{45\, \left( x_{{0}}+1 \right) ^{17}
\left( x_{{0}}+2 \right) ^{17}}} \left( x-x_{{0}} \right) ^{6}-{
\frac {8\,{x_{{0}}}^{15} \left( 10395\,{x_{{0}}}^{12}-187110\,{x_{{0}}
}^{11}+161979\,{x_{{0}}}^{10}+3503880\,{x_{{0}}}^{9}+688353\,{x_{{0}}}
^{8}-21504726\,{x_{{0}}}^{7}-21815219\,{x_{{0}}}^{6}+37940508\,{x_{{0}
}}^{5}+61133076\,{x_{{0}}}^{4}-9402048\,{x_{{0}}}^{3}-44133840\,{x_{{0
}}}^{2}-5866560\,x_{{0}}+8648640 \right) }{315\, \left( x_{{0}}+1
\right) ^{20} \left( x_{{0}}+2 \right) ^{20}}} \left( x-x_{{0}}
\right) ^{7}+O \left( \left( x-x_{{0}} \right) ^{8} \right) )
$$
It appears that the coefficient $a_n$ of $(x-x_0)^n$ is a polynomial of degree $2n-2$ in $x_0$ times $x_0^{2n+1}/((x_0+1)(x_0+2))^{3n-1}$
|
2025-03-21T14:48:29.909819
| 2020-02-19T10:37:11 |
353057
|
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|
Stack Exchange
|
Prime/irreducible elements in certain (integral) ring extensions
The answer to this question says the following:
Let $R$ be a finitely generated $k$-algebra, where $k$ is a field.
If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde{R}$, the integral closure of $R$ in its fraction field $Q(R)$.
Now, let $S$ be a $k$-algebra such that $R \subseteq S \subseteq \tilde{R} \subset Q(R)$. Clearly, $R \subseteq S$ is integral.
When the following property is satisfied? If $p \in R$ is a prime element, then $p$ is a prime element in $S$.
A condition that may be relevant: $R^{\times}=S^{\times}$ (= same invertible elements).
More generally (see the following questions: i and ii):
Let $A \subseteq B$ be two commutative integral domains which are $\mathbb{C}$-algebras.
Question 1: Is it possible to find mild conditions on $A, B, A \subseteq B$ such that the following property P is satisfied:
Property P: Every prime element in $A$ remains prime in $B$.
An empty example: $A=\mathbb{C}[x^2,x^3]$, $B=\mathbb{C}[x]$. There are no prime elements in $\mathbb{C}[x^2,x^3]$, so the property is satisfied.
A nice example: If $B$ is the integral closure of $A$ in the fraction field of $A$, then the property is satisfied, as was quoted above.
Non-examples:
(i) $A=\mathbb{C}[x^2]$, $B=\mathbb{C}[x^2,x^3]$.
$x^2$ is prime (= irreducible in a UFD) in $\mathbb{C}[x^2]$, but $x^2$ is not prime in $\mathbb{C}[x^2,x^3]$ (it remains irreducible), since $x^2$ divides $x^6=x^2x^2x^2=x^3x^3$ but it does not divide $x^3$.
(ii) $A=\mathbb{Z}$, $B=\mathbb{Z}[i]$. $2$ is prime in $\mathbb{Z}$ (= irreducible in a UFD), but $2$ is not prime in $\mathbb{Z}[i]$ (also, it is not irreducible), since $2=(1+i)(1-i)$ divides the product $(1+i)(1-i)$ but it does not divide $1+i$ or $1-i$; this example was presented in an answer to this question.
Question 2: Is it possible to find mild conditions on $A, B, A \subseteq B$ such that the following property I is satisfied:
Property I: Every irreducible element in $A$ remains irreducible in $B$.
Notice that the first, empty example, is not valid anymore since $x^2,x^3$ are irreducibles in $\mathbb{C}[x^2,x^3]$, but reducibles in $\mathbb{C}[x]$.
Remark: I do not mind to further assume that $B$ is a finitely generated $A$-algebra, or even that $B=A[b_1,b_2]$ or $B=A[b]$, for some $b_1,b_2,b \in B$.
Thank you very much!
You probably need something like everything in sight being Noetherian and catenary.
@NeilEpstein, thank you very much! You probably mean that the above propery is satisfied if $R,S$ are Noetherian and $S$ is catenary, as R. van Dobben de Bruyn has suggested (for example, if $R$ is a finitely generated $k$-algebra). So, for example, if $R$ is a finitely generated $k$-algebra, then if $R \subseteq S$ is integral and $Q(R)=Q(S)$, then every prime element in $R$ remains prime in $S$?
@NeilEpstein, please what if we remove one of the following two conditions: (1) $R \subseteq S$ is integral. (2) $Q(R)=Q(S)$. For example, $R=k[x^2] \subsetneq k[x^2,x^3]=S$ with $Q(R)=k(x^2) \subsetneq k(x)=Q(S)$ and $x^2$ is prime in $k[x^2]$ but not prime in $k[x^2,x^3]$. I guess other conditions should be added in order to guarantee the above property (flatness instead of (2) is not enough since here the extension is free). What about being etale? https://math.stackexchange.com/questions/3548524/when-every-prime-element-remains-prime-in-a-ring-extension?noredirect=1#comment7313893_3548524
Also, what if we replace "prime" by "irreducible"?
|
2025-03-21T14:48:29.910025
| 2020-02-19T10:39:17 |
353058
|
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|
Stack Exchange
|
Asymptotic expansion of hypergeometric 2F2
I would like to find an asymptotic expansion for the hypergeometric function
$$
_{2}F_{2}\left(a,b;c,d;z\right),\quad a,b,c,d\in\mathbb{R}.
$$
The parameters are fixed. $z$ is real and $z\rightarrow +\infty$.
Could someone shed light on it?
$$_{2}F_{2}\left(a,b;c,d;z\right)=\frac{\Gamma (c) \Gamma (d)}{\Gamma (a) \Gamma (b)}e^z z^{a+b-c-d}\left(1+{\cal O}(z^{-1})\right)$$
As an example, the plot shows $_{2}F_{2}\left(a,b;c,d;z\right)$ (blue) and the asymptotics (gold) for $a=1,b=2,c=3,d=4$.
Thank you! Could you provide me with some reference book?
for example https://www.researchgate.net/publication/263796676_A_note_on_the_asymptotic_expansion_of_generalized_hypergeometric_functions
|
2025-03-21T14:48:29.910103
| 2020-02-19T10:48:49 |
353060
|
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|
Stack Exchange
|
Marcenko Pastur law when the dimensionality/sample size ratio $p/n \to 0, \infty$? Lack of resources?
Let $X: \Omega \to \mathbb{R}^{p \times n}$ be a random matrix so that each entry $X_{ij}$ is a random variable with $\mathbb{E}X_{ij}=0, \mathbb{E}X_{ij}^2=\sigma^2$
I was wondering what would happen if we keep every hypothesis in the Marcenko-Pastur theorem, but change only one: assume that $p/n \to 0$ or $p/n \to \infty$. Below I'll refer time and again to the above link and the symbols used therein.
From the numerical experiements that I'm doing and from the expression of $\lambda_{+, -}$ in the link above, it looks like:
(1) When $p/n \to 0$, the empirical spectral ditribution of $Cov := \frac{1}{n}XX^{*}$ approaches the Dirac measure at $\sigma^2$. This is apparent from the numerical experiments, and also from the fact that in this case, when $\lambda:= lim_{p,n\to \infty}p/n < 1$, there's no isolated mass, but the continuous part defined using $\nu$ has support $\lambda_{+, -}$, and as $\lambda \to 0$, this support shrinks to $\sigma^2$.
(2) When $p/n \to \infty$, he empirical spectral ditribution of $Cov := \frac{1}{n}XX^{*}$ approaches the Dirac measure at $0$. This is because when $\lambda:= lim_{p,n\to \infty}p/n > 1$, then the Marcenko Patur distribution has two parts: one isolated mass at $0$ with weight $1-\frac{1}{\lambda}$, and the other is the contonuous distribution with density $\nu$, and with support $[\sigma^2(1-\frac{1}{\lambda})^2, \sigma^2(1+\frac{1}{\lambda})^2]$. Clearly the first isolated pass approach the Dirac measure and the second part approaches $0$ as its support moves infinitely away.
But the above two explanations are heuristics. Is there a mathematically rigorous proof where they actually prove these statements, say perhaps using the Stieltjes' transform method, so proving that for example when $p/n \to \infty$, the Stieltjes' transform ($ST$) of $ESD(cov X)\to -\frac{1}{z}= ST(\delta_0)$
N.B. I tried to look this up on the internet, but was a bit surprised by the lack of resources available. I wonder why there isn't much on these extreme cases?
I am not sure that I understand why there is a difficulty: first of all, it is helpful to subtract $|n-p|\delta(\lambda)$ from the eigenvalue density and consider only the continuous part $\delta\nu(\lambda)$; we may then without loss of generality assume $p\leq n$, because $XX^\ast$ and $X^\ast X$ have the same $\delta\nu$. Then the usual derivations of the MP law may be applied, which hold in the limit $n\rightarrow\infty$ for any fixed $p/n\in(0,1]$.
@CarloBeenakker Thanks for your comment! I'm not sure I understand what exactly you're doing to the random matrix $X$ in question to boil things down to the case where $p/n\to \lambda \in (0, \infty)$, so that usual derivations of the MP law may be applied. From what you wrote it looks like you're transforming the limiting spectral density of the random matrix for which $p/n \to {0, \infty}$, but how does that transform the random matrix itself from $X$ to say $Y$ so that the for $Y$, $p/n\to \lambda \in (0, \infty)$? It'd be great if you write a detailed answer.
It is enough to understand the case $p/n\to 0$, as the eigenvalues of $XX^*$ match those of $X^*X$ up to an extra atom at $0$. In that case you can argue as follows, with $\sigma=1$: writing $W=n^{-1}XX^*$, set $Y=W_I$. Compute $Q:=p^{-1}Etr( YY^*)=p^{-1}\sum_{i,j} E(|Y_{ij}|^2)$. An easy calculation (at least if say $EX_{ij}^4<\infty$) reveals that $Q\to_{n\to\infty} 0$, which implies by standard interlacing of eigenvalues the conclusion you sought.
|
2025-03-21T14:48:29.910382
| 2020-02-19T10:59:05 |
353061
|
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|
Stack Exchange
|
The meaning and purpose of "canonical''
This question is jointly formulated with Neil Barton. We want to know about the significance of canonicity in mathematics broadly. That is, both what it means in some detail, and why it is important.
In several mathematical fields, the term 'canonical' pops up with
respect to objects, maps, structures, and presentations. It's not
clear if there's something univocal meant by this term across
mathematics, or whether people just mean different things in different
contexts by the term. Some examples:
In category theory, if we have a universal property, the relevant
unique map is canonical. It seems here that the point is that the map
is uniquely determined by some data within the category. Furthermore, this kind of scheme can be used to pick out objects with certain properties that are canonical in the sense that they are unique up to isomorphism.
In set theory, L is a canonical model. Here, it is unique and definable. Furthermore, its construction depends only on the ordinals-- any two models of ZF with the same ordinals construct the same version of L.
In set theory, other models are termed 'canonical' but it's not
clear how this can be so, given that they are non-unique in certain
ways. For example, there is no analogue of the above fact for L with respect to models of ZFC with unboundedly many measurable cardinals. No matter how we extend the theory ZFC + "There is a proper class of measurables," there will not be a unique model of this theory up to the specification of the ordinals plus a set-sized parameter. See here.
Presentations of objects can be canonical: The most simple being
that of fractions, whose presentation is canonical just in case the
numerator and denominator have no common factors (e.g. the canonical
presentation of 4/8 is 1/2). But this applies to other areas too; see here.
Sometimes canonicity seems to be relative. Given a finite-dimensional vector space, there is a canonical way of defining an isomorphism between V and its dual V* from a choice of a basis for V. This determines a basis for V*, and thus the initial choice of basis for V yields a canonical isomorphism from V to V**. But two steps can be more canonical than one: The resulting isomorphism between V and V** does not vary with the choice of basis, and indeed can be defined without reference to any basis. See here.
Other examples can be found here.
Our soft questions:
(a) Does the term 'canonical' appear in your field? If so what is the
sense of the term? Is it relative or absolute?
(b) What role does canonicity play in your field? For instance, does it help to solve problems, help set research goals, or simply make results more interesting?
Does this answer your question? What is the definition of "canonical"?
@FrancoisZiegler Thanks for the link, it is very helpful. However I don't think it addresses the "purpose" part of our question, that is part (b). We want more than a survey of how people generally use the word; we want moreover to know what the "right" answer is based on the underlying motivations.
I echo the helpfulness comment! That material is super-useful. I think especially interesting is the conjecture that it might help direct research goals. For instance, one of the key research goals in set theory is to build a canonical model for a supercompact cardinal. It's important that such a model have some sort of canonicity properties (good reasons to think that supercompacts are consistent and so there should be some model just don't satisfy the same goal). But it's very unclear what that is meant to mean given that such a beast wouldn't be unique.
It's therefore a real mathematical problem to isolate what we mean in certain contexts by canonical; that's how you know when you've solved the problem.
It seems you are not using an appropriate definition of canonicity in set theory. I even mentioned that in the linked question! (By the way, the link does not work.)
@AndrésE.Caicedo Thanks for pointing out the error. I am not proposing a definition but just pointing out that the way in which things can be canonical seems to get weaker as you move to larger cardinals. So I'm trying to see if there is a solid notion underneath that accords with broader mathematical practice, which could clarify the goals of the "inner model program."
First you ask for canonicity. Then you'll ask for canonical canonicity. Then you'll ask about canonicity of canonicity and for a canonical canonicity of canonicity. Where will it end???
"...to pick out objects with certain properties that are canonical in the sense that they are unique up to isomorphism," better to say "unique up to unique isomorphism".
Good question. I had a professor make me pin down what I meant by a canonical object one time when I was a junior. It was a worthwhile exercise.
Should this have the tag [big-list]?
Looking at the answers, I wonder if maybe there is no canonical meaning to the word canonical.
@AsafKaragila, that is the canonical conclusion to such questions.
The term "canonical basis" is used in representation theory. One of the fundamental examples is the Kazhdan–Lusztig basis of the Hecke algebra of a Coxeter group. Why is the term "canonical" used? Well, vaguely speaking, the thought process begins by trying to categorify the Hecke algebra. There's no mechanical recipe for categorification, and there are various categories that may be regarded as categorifying the Hecke algebra, such as (in the crystallographic case) the Bernstein–Gelfand–Gelfand category of representations of the associated complex simple Lie algebra, or a certain category of perverse sheaves on the corresponding flag variety. But once you have a category, you can get a distinguished basis by considering the classes of simple objects in the Grothendieck group. These give rise to the KL basis in a natural way.
Regarding your question of whether this use of "canonical" is absolute or relative, I would say that it is relative. On the other hand, the KL basis exudes an uncanny odor of being the "right" basis in some sense. Consider for example the positivity of the coefficients of the KL polynomials. The search for an explanation of this positivity has led to the discovery of all kinds of unexpected algebraic and (sometimes) geometric structure. For more information, see for example The Hodge theory of Soergel bimodules and The $p$-canonical basis for Hecke algebras.
In my experience "canonical" means "the simplest way possible" within some context. Often it turns out that this way is uniquely determined, at least when one puts some "natural" restrictions.
For instance, when we want to embed a domain $R$ into its field of fractions $Q(R)$, the simplest way to do that is to map $r$ to $\frac{r}{1}$. All other formulas either don't work, or they don't define a ring homomorphism, or they are not injective. And this is actually the only embedding which can be defined for every domain $R$ and is a natural transformation.
Another example is the projection map $X \times Y \to X$, $(x,y) \mapsto x$. Again this is the simplest way to produce an element of $X$ out of an element of $X \times Y$. And this is actually the only choice which is natural in $X$ and $Y$.
The canonical basis of $K^n$ is another example. Here the simplicity is measured by the number of zeroes in each basis vector, and zeroes should be considered to be simple of course.
In order to illustrate that uniqueness is not required in general, for example one says that for sets $X$ there are two canonical maps $X \to X \sqcup X$. Likewise, there are two canonical maps $X \times X \to X$.
In some cases there is no canonical solution. For example, I would argue that there is no canonical bijection $\mathbb{N}^2 \to \mathbb{N}$, and in fact I don't see a clear measure for simplicity here. Cantor's pairing function is a polynomial bijection which can therefore be considered to be quite simple, but this is just one choice among many others. And one could argue that $(n,m) \mapsto 2^n \cdot (2m+1)-1$, even though it's not polynomial, is actually much simpler since here bijectivity is trivial.
One purpose of canonical maps, structures etc. is to focus on what is relevant and useful.
For Bourbaki, the adjective "canonical" is simply a way to assign a name to
a well-defined object. For instance, when the quotient map $E\rightarrow E/R$ is defined ($E$ is a set, $R$ an equivalence relation), Bourbaki says "it is called the canonical map from $E$ to $E/R$"; then each time a quotient map occurs, Bourbaki will say "let us consider the canonical map $E\rightarrow E/R$ ... ", and it should be clear for the reader what it is.
I assume you mean "object defined before"? Every object is well-defined, otherwise we wouldn't call it an object, right?
Maybe "object" is not the right term. For me a ring is an object, but it is not canonical. The ring of integers is.
Ok, so you mean a specific object.
I believe that there is no definition of canonical which covers all of its uses in mathematics. Heuristic evidence for this is that the definition of a canonical map in Wikipedia is vague.
The word appears in number theory and in some cases it can be made rigorous, and in other cases it is more of a heuristic issue.
Sometimes people say "canonical isomorphism" when they just mean "natural (in the sense of category theory) isomorphism".
In algebraic geometry there is a convention where canonically isomorphic objects are declared equal and the $=$ symbol is used to denote the canonical isomorphism. Strictly speaking this is abuse of the $=$ sign but it has never seemed to cause any problems in algebraic geometry. To give an explicit example: if $R$ is a commutative ring and $M$ is an $R$-module then there is a canonical identification of the abelian groups $M_f:=M\otimes_R R[1/f]$ and $M_g:=M\otimes_R R[1/g]$ when the set of primes containing $f$ equals the set of primes containing $g$ -- both of these are the value on the open subset $D(f)=D(g)\subseteq Spec(R)$ of the quasicoherent sheaf associated to the $R$-module $M$. In EGA1, (1.3.3) Grothendieck says that the map between these modules is "un homomorphisme canonique fonctoriel" and then uses the $=$ symbol and writes $M_f=M_g$. They are not equal in the strict sense, but they both satisfy the same universal property so as long as mathematicians only ever do things to these modules which can be done using the universal property -- and this is exactly what they do -- then this abuse of the $=$ symbol will not cause problems. But one could in theory argue that many many uses of the $=$ symbol in EGA and elsewhere in algebraic geometry should really say "is canonically isomorphic to". In Milne's book on etale cohomology he explicitly admits this, saying at the beginning of his book that canonical isomorphisms will be denoted $=$, without ever explicitly saying what a canonical isomorphism is. Here, what appears to be going on is that the isomorphisms are sufficiently natural that every conceivable sensible diagram that one can come up with will commute. Note that there is an extra subtlety here -- $A = B$ is usually in mathematics a true-false statement. In Milne's book it is sometimes actually implicitly encoding a map from $A$ to $B$, and the usual proof that equality is an equivalence relation is being beefed up to assertions of the form "composing two canonical isomorphisms gives a canonical isomorphism" and so on. This phenomenon occurs all over the place and is very well-hidden. However there is no doubt that this convention "works".
Finally, in the Langlands philosophy one can find assertions saying that the local or global Langlands correspondences are canonical. This is a use of the term which I find more unsettling because the situation seems to be that people believe that there is one "special" correspondence, and they can list a bunch of properties which it should have, but nobody has proved the theorem that there is at most one correspondence with these properties and indeed it may well not be true that the current list of properties that we have defines the correspondence uniquely. One can try and wriggle out of this by asking that the correspondence commutes with functoriality, however functoriality is a big open question and the statement of functoriality is not completely understood in full generality (what happens for non-classical groups at bad primes). My current thinking about this use of the word is that it is more expressing a hope that one day in the future, people will figure out what we were talking about. However if one works with general connected reductive groups, I am pretty sure that this day is not yet here. It is for me a very interesting usage of the word.
Finally it might be worth mentioning that even in the case of GL_1, there are two canonical Langlands correspondences! Class field theory is the statement that two abelian groups are canonically isomorphic, but these groups contain elements of order bigger than 2, and so one can apply inversion on one side but not the other and change an isomorphism into a different one. They are distinguished by having different names -- one is "the canonical isomorphism sending uniformisers to arithmetic Frobenii" and the other is "the canonical isomorphism sending uniformisers to geometric Frobenii".
Perhaps you didn't think it was sufficiently relevant to mention, but I personally think it's worth pointing to your Xena project blog post for further examples and observations.
This seems broader than the other question which was interpreted mainly in the “category theory” sense (1.).
An early, maybe earliest,a case of sense (4.) “normal form” is Jacobi (1837) calling canonical the “Hamilton form”b of the equations of mechanics, and also any variables, coordinates or “elements” in which they take this form; today we would speak of Darboux coordinates or Darboux normal formc of a symplectic structure. Remarks:
A big difference with the case of fractions is that the normal coordinates (or isomorphism to normal form) are far from unique.
Sylvester in another context (1851, p. 190) attributes the phrase to Hermite, presumably (1854):
I now proceed to the consideration of the more peculiar branch of my inquiry, which is as to the mode of reducing Algebraical Functions to their simplest and most symmetrical, or as my admirable friend M. Hermite well proposes to call them, their Canonical forms.
Similar: the Jordan canonical formd and Frobenius rational canonical forme of a matrix.
So far as I can tell, Jacobi may well have originated the phrase “normal form” too (1845, 1850).
a one can find “ad formam canonicam” at least once in Euler: De reductione formularum integralium ad rectificationem ellipsis ac hyperbolae (1766, p. 28); also often aequatio canonica.
b notoriously used before by Lagrange and Poisson.
c called canonical by Frobenius (1877), canonical or normal by Darboux (1882).
d called canonical by Jordan (1870, p. 114); Kronecker (1874) bitterly deplored the terminology.
e called normal by Frobenius (1879, pp. 207-208).
Do I misunderstand, or are you suggesting that Sylvester in 1851 attributed to Hermite a usage that the latter used in 1854?
@LSpice Either I missed an earlier publication, or they first talked (or corresponded) about it. Everything here suggests “canonical” has a longer oral tradition — much as we might say but not write “God-given”.
The term aequatio canonica occurs already in Harriot's Artis analyticae praxis, published in 1631, 10 years after the death of its author. Moritz Cantor (vol II, p 791) surmises that this is the earliest occurrence of an expression in this family.
@AnstenKlev Oh yes. Even earlier analogia canonica in Vieta (1579, §§XI, XIX), doctrina canonica in Snellius (1627). I found quite a bit more history that I’ll add here when I have time: Napier, Leibniz, Bernoulli, Cotes, Euler, Lagrange, Gibbs, Milankovic,... also used it; I think it was always more or less tongue-in-cheek for from the Book.
(a) Does the term 'canonical' appear in your field? If so what is the sense of the term? Is it relative or absolute?
Practical algorithms for the graph isomorphism problem solve the graph canonization problem instead of merely checking two given graphs for isomorphism. Normally this is done by computing a canonical labeling, but other canonical forms would serve the same purpose, namely to allow efficient set/map operations (in the sense of C++ std::set/std::map) after the given graphs have been canonized.
The sense of the term here seems to be to single out a unique representation or hash for a given graph. I would say the term is relative here, i.e. it is not important to determine the "best canonization," and it isn't even clear what "best" would mean in this context.
(b) What role does canonicity play in your field? For instance, does it help to solve problems, help set research goals, or simply make results more interesting?
The graph canonization problem is slightly more difficult than the graph isomorphism problem. So after László Babai showed that graph isomorphism can be solved in quasipolynomial time, it was clear that one should try to also find a corresponding canonization algorithm. So László Babai wrote a follow up report Canonical form for graphs in quasipolynomial time: preliminary report and others tried to address the problem from a more general point of view (like Pascal Schweitzer and Daniel Wiebking in A unifying method for the design of algorithms canonizing combinatorial objects)
Let me also mention a different example where I used the term canonical myself. The straightening theorem for vector fields says that around a point where the vector field is non-zero, there exist a local coordinate system in which the vector field is constant. But this coordinate system is not unique. To get a unique straightening, I extended the vector field by an additional coordinate, and set the vector field there to 1. This allows me to use a unique straightening, which I call the canonical straightening. I used this canonical straightening for short proofs of some theorems, and I believe that it made the arguments more transparent.
I also extended that trick to the case of multiple vector fields whose Poisson brackets (or maybe it is called Lie bracket in this case) form a nilpotent Lie-algebra. Here I extend with more additional coordinates, set the vector field there to a simple presentation of the nilpotent Lie-algebra, and get a coordinate system (in a "canonical" way) where the vector fields are given by that simple presentation. That was done in a German text. I never came around to investigate whether that trick could also be made to work for solvable Lie-algebras, or whether something similar could even be done for arbitrary Lie-algebras.
My use of canonical in the last example above (where I use some simple presentation of a nilpotent Lie-algebra) doesn't feel significantly different to me than the use of canonical in the graph canonization problem. So it is neither a "best" nor "universal" construction, but just a sufficiently unique construction.
Every group (actually I believe the same holds for rings and modules) admits the following (silly) canonical presentation. If $G$ is a group, consider the free group $F(G)$ on the (underlying) set $G$. Then
$$G = F(G)/R,$$
where $R$ is the subgroup normally generated by words of the form $ghk^{-1}$, where $g$, $h$ and $k$ are in $G$ satisfying $gh=k$.
@MattF. I believe that it is a theorem that there is no general algorithm to determine whether a given finite group presentation presents the trivial group, so I think there is no hope for an arbitrary group to have a canonical presentation in the way you describe. I choose to call it canonical because the presentation can be described without making any choices involving $G$ or its elements.
More generally, when $T$ is a monad and $(X,h)$ is a $T$-algebra, then $(X,h)$ has a canonical presentation $T^2(X) \rightrightarrows T(X) \to X$. The two parallel maps on the left are $T(h)$ and $\mu_X$, the map on the right is $h$. I would not call this presentation silly, it is quite useful at least in the general theory.
@RyleeLyman That it is undecidable whether a finite presentation defines the trivial group is one of the consequences of the Adian-Rabin theorem, which (very informally) says that "all we can really say about a group given by a finite presentation is that it is finitely presented". But it is important to note that your "canonical presentation" is a finite presentation iff $G$ is finite, so Adian-Rabin is a bit of a sideline.
Given Francois Ziegler's answer, my interpretation here may be ahistorical, but here's how I think about the use of "canonical" in the context of symplectic geometry/mechanics.
If you have a phase space (what we'd now call a symplectic manifold) and you pick half of your coordinates (say those corresponding to position $q_1,\ldots,q_n$) then the other half come for free (the canonically conjugate momenta): the
functions $q_1,\ldots,q_n$ generate Hamiltonian vector fields $X_{q_1},\ldots,X_{q_n}$ and the "$p_k$ momentum axis" is simply
the flowline of $X_{q_k}$. Indeed, this is one way you construct Darboux coordinates (e.g. in Arnold's book on classical mechanics).
Most people who study mathematics beyond high school level would be introduced to the word canonical in Algebra I. At first, a set of vectors $\{e_1,e_2,\ldots ,e_n\}$ is defined as basis if it is linearly independent and spans the whole space. How ever, that set is not unique to a specific space, so we introduce canonical basis with Kronecker delta
$$e_i=(\delta_{1i},\delta_{2i}, \ldots ,\delta_{ni}).$$
What does the word canonical mean in this context? I would say a way to pick out a unique basis, that is standard and can be applied in practice. More interestingly, is it chosen arbitrarily or is it inherent for us, humans, to think of space in terms of up, left, right?
I think that any Linear Algebra course that introduces the definition of a basis before discussing the standard (not canonical, at least not by most people's definition) example of a basis of $k^n$ is poorly organised.
Some of the examples given in the question are a careless misuse of the word. Who writes "canonical basis" for $K^n$ when they mean "standard basis", and who writes of a "canonical presentation of a fraction" when they mean a "fraction in its lowest terms", which isn't even canonical unless everything is positive.
In my field, arithmetic geometry, "canonical" has a well-understood meaning even if it is difficult to write down a precise definition. In his 1980 book, Milne was comfortable assuming that his readers would know what it meant (in his later writings, he has switched to using $\simeq$ for "canonically isomorphic"). Roughly, it means that the object can be constructed without making any arbitrary choices. There is a huge difference between saying two objects are isomorphic and saying they are canonically isomorphic. Barr botched this in his translation of Grothendieck's Tohoku paper by replacing "=" (meaning canonically isomorphic) with isomorphic.
I agree that the use of "canonical" is problematic in the Langlands program. There are major conjectures saying that some set (of representations) is bijective to some other set. After Serre pointed out that this only means that the two sets have the same cardinality, the word "canonical" was added. It is part of the problem to figure out what that means.
Well, Bourbaki, who is known for careful use of words, writes canonical basis for what you call standard basis.
@FredRohrer, the problems is what kind of object $K^n$ is—or, perhaps to phrase it differently, how many automorphisms we allow it. As a bare vector space (with a full $\operatorname{GL}_n(K)$ of automorphisms), it's probably not reasonable to call the standard basis canonical, since it's not preserved by automorphisms. As a product of copies of $K$ indexed by the cardinal $n$, it has no non-trivial automorphisms, and the standard basis can reasonably be identified as a canonical choice. Bourbaki is careful, but can still make mistakes, or—if that is too strong—infelicities.
|
2025-03-21T14:48:29.912358
| 2020-02-19T11:20:34 |
353064
|
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|
Stack Exchange
|
Matroids with no relaxations (~ weak maps)
There's an operation in matroid theory which is called "relaxation".
To keep things simple, let's consider a matroid $M$ with set of bases $\mathcal{B}$. If $M$ has a subset $H$ of $M$ that is both a circuit and a hyperplane, then one may construct a matroid $\widetilde{M}$ whose set of bases is $\widetilde{\mathcal{B}} = \mathcal{B} \cup \{H\}$.
Notice that both $M$ and $\widetilde{M}$ are matroids on the same ground set and of the same rank. This says that the identity map $i:\widetilde{M} \to M$ is indeed a weak map (every independent set in $M$ is independent in $\widetilde{M}$).
I want to know the following:
Do connected matroids with no relaxations have a name? Have they been studied? Do they have any nice properties?
Of course, many matroids are of that type: for instance uniform matroids are connected and do not admit any relaxation (they are just as big as possible: they don't have any circuit+hyperplane).
The matroids that do not admit any relaxation are exactly those matroids that do not have a circuit-hyperplane. This is a very diverse class, and it's hard to think of properties they have in common, other than the property in the definition. I do not know of any name for this class of matroids, and I do not think they have been studied as a class.
|
2025-03-21T14:48:29.912502
| 2020-02-19T14:01:33 |
353074
|
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|
Stack Exchange
|
Elementary equivalence between $n\mapsto n+1$ and its inverse on the Stone-Čech remainder?
Consider structures $(A,f)$ encoding a Boolean algebra $A$ endowed with an automorphism $f$. There is an obvious notion of isomorphism between such structures.
Consider the endomorphism $\hat{\Phi}$ of the Boolean algebra $2^\omega$ of subsets of $\omega$ given by $A\mapsto \{a\in\omega:a+1\in A\}$. It induces an automorphism $\Phi$ of the quotient Boolean algebra $2^\omega/\mathrm{fin}$, where $\mathrm{fin}$ is the ideal of finite subsets. (Under Stone duality, this corresponds to the self-homeomorphism of the Stone-Čech remainder of $\omega$ induced by $n\mapsto n+1$.)
Whether $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^{-1})$ are isomorphic is essentially unknown (see my previous related question for more details).
My question is
Are $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^{-1})$ elementary equivalent?
That is, do they satisfy the same first-order sentences (in the language of Boolean algebras endowed with an automorphism)?
Note that $\Phi^n$ has exactly $2^{|n|}$ fixed points for $n\neq 0$. In particular, $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^n)$ are not not equivalent for $|n|\ge 2$.
If the question has a positive answer, it is tempting to ask whether one can characterize simply those $(A,f)$ having the same first-order theory as $(2^\omega/\mathrm{fin},\Phi)$.
If they are elementary equivalent, then a back and forth argument shows that they are isomorphic under CH. (And whether they are isomorphic under CH is unknown.) So I think this is open.
@PaulMcKenney this was indeed my hope to pass from EE to isomorphism, but at the same time if they're not EE, it should be for some "concrete"reason. So I hope this approach leads to something.
@PaulMcKenney how does the back-and-forth argument work? is $(2^\omega/\mathrm{fin},\Phi)$ $\omega_1$-saturated? since it seems to reduce to my previous question, I think this would be worth an answer.
That's what I was thinking, but it's no longer clear to me that $(2^\omega / \mathrm{fin}, \Phi)$ is saturated.
@PaulMcKenney: I was thinking a bit more about this question today and, for what it's worth, I can show that $(2^\omega/\mathrm{fin},\Phi)$ is not $\omega_1$-saturated. I have a somewhat tedious proof of this assertion, but this comment is too short to contain it . . . but feel free to email me if you'd like some details.
@WillBrian I'd be interested, and probably some other readers too (you might post it as an answer here even if it's an extended comment).
@WillBrian Cool! I'll send you an email, I am quite interested.
Yes, these two structures are elementarily equivalent.
This is proved as a corollary to another theorem, which states
Theorem: CH implies that $\Phi$ and $\Phi^{-1}$ are conjugate to each other in the automorphism group of $\mathcal P(\omega) / \mathrm{fin}$.
You can find a proof of this on the arXiv (here), and a short overview of the proof in my answer to this MO question.
Let me add a few words explaining why the theorem quoted above implies that the structures $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$ are elementarily equivalent (even without assuming CH).
The short version is: we can force CH without changing the theory of $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$, so if they're isomorphic in the forcing extension, and their theories have not changed, they must have been elementarily equivalent to begin with.
In more detail, let $\mathbb P$ denote the poset of countable partial functions $\omega_1 \times \omega \to \omega$, ordered by extension, i.e., the usual poset for forcing CH with countable conditions. Let $G$ be a $V$-generic filter on $\mathbb P$.
Because $\mathbb P$ is a countably closed notion of forcing, no new subsets of $\omega$ are added by $\mathbb P$, which means that $V \cap \mathcal P(\omega) / \mathrm{fin} = V[G] \cap \mathcal P(\omega) / \mathrm{fin}$. Furthermore, because $\Phi$ is definable by the simple formula $\Phi([A]) = [A+1]$, the action of $\Phi$ is the same in $V$ and in $V[G]$. Thus the structures $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle^V$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle^{V[G]}$ are one in the same.
It follows (by the ``absoluteness of the satisfaction relation'') the theory of $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ is the same in $V$ and in $V[G]$.
The same argument applies to $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$, so the theory of $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$ is also the same in $V$ and in $V[G]$.
Because $V[G] \models$ CH, the theorem quoted above implies $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$ are conjugate in $V[G]$.
In particular, they are elementarily equivalent in $V[G]$.
Because the theories of these two structures is the same in $V$ and in $V[G]$, this means they have the same theory in $V$ as well.
|
2025-03-21T14:48:29.912878
| 2020-02-19T14:12:03 |
353077
|
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|
Stack Exchange
|
Brown representability in slice category
Brown's representability theorem gives us a very nice set of conditions to check that a (contravariant) functor $Hot^{op}\rightarrow Set$ is representable. Choose an object $X$ in $Hot$. Then it is seems natural to ask whether or not an analogue of the Brown representability theorem is true for the slice category, i.e. if there exists a nice set of conditions to check whether or not a contravariant functor
$$F:(Hot/_X)^{op} \rightarrow Set$$
is representable. Is there such an analogue? I've searched online, but couldn't find an comments on the subject. My hope would be that this functor is representable if $F$ respects coproducts and sends homotopy pushouts to weak pullbacks.
One way I would hope to recover such a criterion is to consider the functor $$Hot\rightarrow Hot/_X, Y\mapsto (Y\rightarrow \{*\})$$
which by composition gives us a functor $Hot^{op}\rightarrow Set$, for which we know we can apply Brown's Theorem. I don't know however if this is enough to test representability of the functor $F$.
Neeman proves a version of Brown representability which holds in an arbitrary compactly generated triangulated category $\mathcal C$: if $H\colon\mathcal C^{\mathrm{op}}\to \mathit{\mathcal Ab}$ is a functor sending coproducts to products and exact triangles to long exact sequences, then $H$ is representable by an object in $\mathcal C$. Hopefully that applies to this setting!
Please define the notation $Hot$ in the question. Note that the stable homotopy category is triangulated, but the unstable one is not.
The naive slice category $Hot_{/X}$ is not very meaningful homotopically: it is usually more useful to consider the version of the slice category where a morphism includes a homotopy making the triangle commute; in other words, the homotopy category of $Top_{/X}$. As pointed out by others you also need the words "pointed connected" somewhere. For the slice one can either look at $(Top_^{\geq 1}){/X}$ or $(Top{/X})_^{\geq 1}$ (the latter means: spaces over $X$ with a section which is surjective on $\pi_0$). Both of their homotopy categories satisfy Brown representability.
A modern version of Brown's representability theorem is Thm <IP_ADDRESS> in Lurie's Higher Algebra. While it does not subsume Brown's original theorem, it can be used instead in most cases of interest (such as the two cases from my previous comments).
Yes, sliced homotopy categories of pointed connected spaces satisfy Brown representability. (We had better be using $Hot$ to denote the homotopy category of pointed connected spaces, as Brown representability is false for the homotopy category of unbased or non-connected spaces.)
The abstract version of Brown’s representability theorem has the following requirements on a category $C$: it must have coproducts, weak pushouts and a “compact generating set”. This is a set of objects, maps out of which jointly detect isomorphisms and commute with some choice of sequential weak colimits. Note this is more general than Neeman’s version for compactly generated triangulated categories, as $C$ need not be triangulated and indeed $Hot$ is not. That said, Neeman has proved vastly more general theorems in the well-generated and perfectly generated triangulated case which cannot be extended unstably. (At least not yet!)
As is common in these situations, the coproducts and weak pushouts in the slice category are no problem, while weak sequential colimits can be chosen, as in $Hot$, as the homotopy colimit/Milnor’s infinite mapping telescope.
As to the compact generators, we may take the set of all spheres over $X$. This is analogous to results on, say, local presentability of slice categories that may be familiar, but let’s go ahead and see the proof. Compactness is immediate from the case in $Hot$. For generation, if $f:A\to B$ is a map over $X$, then a map $S^n \to A$ is killed by $f$ equivalently whether we’re over $X$ or not, while if $S^n\to B$ fails to factor through $A$ in $Hot$, then it certainly fails to do factor over $X$. Thus the set of spheres over $X$ see $f$ as an isomorphism over $X$ if and only if the set of spheres sees $f$ as an isomorphism in $Hot$, if and only if $f$ is really as isomorphism.
Edgar Brown wrote two papers on this topic: one in the Annals in 1962 that focused on the category of topological spaces, and a second one "Abstract homotopy theory"
Trans. Amer. Math. Soc. 119 (1965).
His second paper is very axiomatic, and I believe your situation is easily checked to satisfy his properties. (Check this!)
This paper obviously predated Quillen's work on model categories (it likely partially inspired Quillen), so of course he doesn't use that language. I confess that I have always been a bit disturbed by the number of papers on Brown Representability written by authors who show no indication that they have ever looked at the original papers. Yes, people have written about more general versions (and less general, when they assume a triangulated category!) but I encourage folks to use their library resources to look at Brown's own work.
Brown’s abstract theorem is precisely the one I explained how to apply. Thanks for bringing it up, as I neglected to cite my results. I agree that these papers are read less than they should be, at least the second one.
To explain the connection with Brown’s paper, the compact generator is his $C_0$, without the superfluous assumption of closure under finite sums, while Brown spells out “there are weak sequential colimits with respect to with $C_0$ is compact” explicitly. Other than that the assumptions are precisely his, and have not been improved for unstable homotopy 1-categories since 1964.
|
2025-03-21T14:48:29.913325
| 2020-02-19T14:52:20 |
353080
|
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|
Stack Exchange
|
Positively curved manifold with collapsing unit balls
Can we find a complete connected noncompact Riemannian manifold $(M^n,g)$ such that the curvature operator $Rm>0$ and
$$
\inf_{p \in M} \text{Vol}_gB(p,1)=0?
$$
You should add connected otherwise there are obvious counter-examples.
@MoisheKohan right.
I wonder, are there such examples if we only require the sectional curvature to be positive?
You should take a look to Example 1 of the paper "VOLUMES OF SMALL BALLS ON OPEN MANIFOLDS: LOWER BOUNDS AND EXAMPLES" by Croke and Karcher (you can find it here https://www.ams.org/journals/tran/1988-309-02/S0002-9947-1988-0961611-7/S0002-9947-1988-0961611-7.pdf), it seems to answer your question positively with an explicit construction. However (as proved in their Theorem 1) the answer is negative for $2$ dimensional manifolds.
No, they attach the cones to a paraboloid and then smooth. The resulting hypersurface is connected. I’ve tried to detail more in my answer below.
The answer is negative if $\dim M=2$ and positive otherwise, as shown in the paper:
Croke, C. B., & Karcher, H. (1988). VOLUMES OF SMALL BALLS ON OPEN MANIFOLDS: LOWER BOUNDS AND EXAMPLES. AMERICAN MATHEMATICAL SOCIETY (Vol. 309). https://www.ams.org/journals/tran/1988-309-02/S0002-9947-1988-0961611-7/S0002-9947-1988-0961611-7.pdf
The negative result is an immediate consequence of Theorem A, while the positive one follows from Example 1. Their construction for the latter is as follows:
Consider the $d$-dimensional hypersurface given by the paraboloid $\mathcal P = \{x\in \mathbb R^{d+1}\mid x_{d+1}=x_1^2+\ldots+x_d^2\}$. At any point $A$ of $\mathcal P$ one can consider its tangential cone $\mathcal C$, i.e., the Euclidean cone of vertex $V$ with axis along the line $AV$ and that intersects $\mathcal P$ tangentially. One then can consider $\tilde {\mathcal P}$ to be $\mathcal P$ with $\mathcal C$ "attached" (see the Figure at p.765).
This is a piecewise $C^\infty$ hypersurface, with a conical singularity at $V$ and that is $C^1$ at the intersection $\mathcal P\cap \mathcal C$. Outside of these regions, the curvature of $\tilde {\mathcal P}$ is $K>0$, since the curvature of $\mathcal P$ is positive and the one of the cone is zero. Moreover, it is always possible to smooth a conical singularity preserving the $K>0$ bound, and the same is true for the $\mathcal P\cap \mathcal C$ part. (You can maybe look at this thesis, e.g., Lemma 3.2.4.) So we have constructed a smooth hypersurface with $K>0$, and where we can estimate the volume of balls contained in the smoothed conical region by the volumes of the corresponding regions in the non-smoothed cones. Moreover, this operation can be done with a sequence of tangent cones $\mathcal C_n$, as soon as they are sufficiently distant one from the other.
The estimation of the volumes of the balls on the tangent cones is the object of the claim at the end of p.765. In particular, there the authors show that if the vertex of the cone is $V = (k+1,\ldots, 2k+1)$ for $k\ge 5$ (I'm fixing $r=1$), this volume is bounded by the volume of a "spherical spindle" :
$$
\operatorname{vol}(B(V,1))\le C_d\sin^{d-2}\theta ,
$$
where $\theta$ is an angle bounded by $\tan^2\theta\le 6/k$. (Here I chose $\varepsilon =1\le k^2/(2k+2)$.)
This shows that it is possible to attach to $\mathcal P$ a sequence of cones $\mathcal C_n$ with vertices $V_n\to +\infty$, and smooth them out so that
$$
\lim_{n\to +\infty}\operatorname{vol}(B(V_n,1))\le (C_d+1) \lim_{n\to +\infty} \sin^{d-2}\left(\theta_n\right) = 0.
$$
This proves that the obtained hypersurface (which has positive curvature) satisfies your requirement.
That's a really cool example! In the nutshell the idea seems to be the following. If we take a round sphere $S^{d-1}$ in $\mathbb R^d$, take a point $p$ far from it and construct the cone with the centre in $p$ that is tangent to $S^{d-1}$, the unit ball centred at $p$ in the surface of the cone will have a very small volume, provided $d>2$. If $d=2$, then this will be a segment of length $2$, of course. That's why the example doesn't work in dimension $2$ :).
Exactly! Very good explanation of the idea! :)
|
2025-03-21T14:48:29.913682
| 2020-02-19T16:14:34 |
353085
|
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|
Stack Exchange
|
Anti-concentration inequalities: lower bound on realized second moment
Let $X:(\Omega,\Sigma)\rightarrow (\mathbb{R}^d;\mathbb{B}(\mathbb{R}^d))$ be a Borel-measurable random vector. What are some general classes of such random vector for which one can give a "lower concentration inequality" of the form:
$$
\mathbb{P}(\|X\|^2>\lambda) \geq \mbox{(insert non-trivial lower bound)}
$$
where $\lambda>0$.
Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Wikipedia) yields
$$P(\|X\|>u)\ge1- e^{-(E\|X\|-u)^2/(2n)}$$
for $u\le E\|X\|$.
A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following:
$$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$
for $u>0$, where $X$ is just as above.
Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then
$$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u)
\ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$
for $u>0$.
Hmm. The Hoeffding-Azuma inequality you quote is not the one in the reference, and is actually wrong as the example $n=1$ and $u=E|X|$ shows. I believe there is a typo also in the reference you give (Bartlett's notes) - it should be an upper bound, not a lower bound, as the example t=0 (and the proof given there) show.
@oferzeitouni : Thank you for your comment. This is now corrected.
How come the Azuma inequality becomes an anti-concentration one? The bound should be a concentration bound around the mean, not a lower bound on the probability to deviate from the mean. Am I wrong?
@CuriousGuy : The lower bound here is not on the probability to deviate from the mean -- notice the condition $u\le E|X|$.
Thanks for the answer: what is the considered martingale then?
@CuriousGuy : The martingale is $(E(|S_k|,|,X_1,\ldots,X_{k-1}))$, where $S_k:=X_1+\cdots+X_k$.
Well, in your case notice that $\|X\|^2>0$ and so $\mu\triangleq \mathbb{E}[\|X\|^2]>0$. Thus, for any $\lambda \in \left(0,\mu\right)$ the Cantelli Inequality gives
$$
\Pr(X\ge\lambda)
\ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2},
$$
where $\sigma\triangleq \mathbb{E}\left[\left(\|X\|^2 - \mu\right)^2\right]$.
This is interesting, but it only holds for small $\lambda$...not sure I'm I'm asking for too much...
$0<\lambda<\mu$?
|
2025-03-21T14:48:29.913897
| 2020-02-19T16:47:35 |
353087
|
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|
Stack Exchange
|
Identity involving dimensions and contents of partitions
Let $d_\lambda$ with $\lambda\vdash n$ be the dimensions of irreducible representions of the permutation group $S_n$. Let $C^{\nu}_{\lambda,\mu}$ be the standard Littlewood-Richardson coefficients.
I have been led to define the quantity
$$ F_n=\frac{1}{n!(n+1)!}\sum_{\lambda\vdash n}\sum_{\nu\vdash n+1}C^{\nu}_{\lambda,(1)} d_\lambda d_\nu M_\nu^2,$$
where $M_\nu$ is the product of all non-vanishing contents in the partition $\nu$, i.e.
$$ M_\nu=\prod_{(i,j)\in\nu, i\neq j}(j-i).$$
By comparing two different ways of computing the same result, I was expecting that $F_n$ should be equal to 1 for all $n$. Instead, when I actually calculated the values, I got $F_0=F_1=F_2=1$, fine, but $F_3=\frac{37}{36}$, which is curious.
Does anyone know how to evaluate $F_n$ in general, or something very similar to it (maybe by some mistake I ended up with a slightly wrong definition of $F_n$) that could shed some light into this situation?
This is a partial observation:
The Littlewood-Richardson coefficient in your case is simply 1,
if and only if $\nu/\lambda$ is a skew shape, and 0 otherwise.
Hence, (by using more standard notation for dimensions)
$$
F_n = \frac{1}{n!(n+1)!}\sum_{\lambda \vdash n} \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda}} (f^\lambda) (f^\nu) (M_\nu)^2.
$$
Moreover, you also have that for $\nu \vdash n+1$,
$$
f^\nu = \sum_{\substack{\mu\vdash n \\ \nu \supset \mu}} f^\mu,
$$
since every SYT of shape $\nu$ can be constructed by adding a box to a SYT of shape $\mu$.
Rearranging a bit,
$$
F_n = \frac{1}{n!(n+1)!}\sum_{\lambda\vdash n}
f^\lambda
\sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda}}
\left(
\sum_{\substack{\mu\vdash n \\ \nu \supset \mu}} f^\mu\right)
(M_\nu)^2.
$$
Further rearranging,
$$
F_n = \frac{1}{n!(n+1)!}\sum_{\lambda,\mu\vdash n}
f^\lambda f^\mu
\sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda,\\ \nu \supset \mu}}
(M_\nu)^2.
$$
Now it might be a good idea to consider the separate cases $\lambda=\mu$, and $\lambda \neq \mu$.
This might help a bit.
I don't see why $\lambda = \mu$. If they differ by "moving" a single box, eg. $\lambda = (n)$ and $\mu = (n-1, 1)$, then there is a $\nu$ that contains both of them, in the example $\nu = (n, 1)$.
@lambda Yes, you are correct! I had a feeling it was simplifying a bit too much... It has a oscillating tableau feel to it now, with the removal + addition.
I realize now that, in the original quantity
$$ F_n(\mu)=\frac{1}{n!(n+m)!}\sum_{\lambda\vdash n}\sum_{\nu\vdash n+m}C^{\nu}_{\lambda,\mu} d_\lambda d_\nu M_\nu^2 \delta_{D(\nu),D(\mu)},$$
I should have imposed the condition that the Durfee square of $\nu$ is equal to the Durfee square of $\mu$.
In the particular case $\mu=(1)$, mentioned in the question, this implies that $\nu$ is a hook, which of course simplifies things enormously. Once this condition is imposed, we indeed get $F_n((1))=1$, as I had expected.
|
2025-03-21T14:48:29.914252
| 2020-02-19T18:34:36 |
353094
|
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|
Stack Exchange
|
Metric and particular system of PDE
I have a big problem to solve this system:
$\Delta f−hf^2=0$
$p|\nabla f|^2+hf^3=0$
where $h$ and $p$ are constants (with $h \neq 0$ and $p \neq 0$, $p \neq -1$), $f$ is a scalar function defined on a 4-manifold ($f:M \rightarrow \mathbb{R}$) where $M$ is a 4-manifold not compact and where $\Delta f$ is the Laplacian of $f$ (trace of Hessian of $f$, with positive sign, not negative), and $\nabla f$ is the gradient of $f$ for the metric $g$ (where $g$ is the metric of
$M$).
Should I find $f$ and $g$ excluding cases of flat metric $g$.
1) Are there solutions?
2) Can the metric $g$ admit a scalar curvature ($S$) equal to $-p(p+1)x$, i.e., $S=-p(p+1)x$, for some negative constant $p$ ?
As in my previous solution in the 3-dimensional case (discussed here), we can set $f=-(p/h)x$ for a function $x$ that satisfies
$$
\Delta x + p\,x^2 = |\nabla x|^2 - x^3 = 0.\tag 1
$$
Conversely, if $x$ satisfies this system for a metric $g$, then $f = -(p/h)x$ will satisfy the original equations.
Again, the same argument as before shows that the $1$-form $\omega_1 = x^{-3/2}\, \mathrm{d}x$ is a $1$-form of $g$-norm equal to 1, so one can write $g$ locally in the form
$$
g = {\omega_1}^2 + {\omega_2}^2 + \cdots + {\omega_n}^2.
$$
Again, fixing the orientation so that $\omega_1\wedge\cdots\wedge\omega_n$ is the oriented volume form, we have $\ast_g\omega_1 = \omega_2\wedge\cdots\wedge\omega_n$, and the definition of the Laplacian gives us $\mathrm{d}(\ast_g\mathrm{d}x) = -\Delta x\,\,{\ast_g}1$. Just as before, we now compute, using $\mathrm{d}x = x^{3/2}\,\omega_1$ that
$$
\mathrm{d}(x^{3/2-p}\,\omega_2\wedge\cdots\wedge\omega_n) = 0,
$$
and hence, locally, one can write
$$
x^{3/2-p}\,\omega_2\wedge\cdots\wedge\omega_n = \mathrm{d}y^2\wedge\cdots\wedge\mathrm{d}y^n\tag2
$$
for some functions $y^2,\ldots,y^n$ that, with $x$, gives a local coordinate system. Moreover, (2) implies that
$$
{\omega_2}^2 + \cdots + {\omega_n}^2 = x^{(2p-3)/(n-1)}\,g_{ij}(x,y)\,\mathrm{d}y^i\mathrm{d}y^j,
$$
for some function $g_{ij}$ satisfying $\det(g_{ij}) = 1$.
Conversely, given functions $g_{ij}$ on a domain in $\mathbb{R}^n$ so that $(g_{ij})>0$ and $\det(g_{ij}) = 1$, we have that the metric
$$
g = x^{-3}\,(\mathrm{d}x)^2 + x^{(2p-3)/(n-1)}\,\bigl(g_{ij}(x,y)\,\mathrm{d}y^i\mathrm{d}y^j\bigr)
$$
and the function $x$ satisfy the equation (1).
Finally, the equation $S =-p(p+1) f$ is a second order PDE for the functions $g_{ij}$, and the existence of solutions should be fairly straightforward as long as $n>2$. It's not clear that it's worth writing out the details.
|
2025-03-21T14:48:29.914454
| 2020-02-19T19:27:08 |
353098
|
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|
Stack Exchange
|
Real root of the derivative of a prime cyclotomic polynomial
Consider the graph of $y=x^n$ with $n$ odd, and draw a tangent to its negative arc that crosses the graph at $(1,1)$. Equating the slopes gives us $\frac{x^n-1}{x-1}=nx^{n-1}$ at the point of tangency. Since
$$nx^{n-1}-\frac{x^n-1}{x-1}=(x-1)\left(\frac{x^n-1}{x-1}\right)'$$ this point has to be the only real (negative) root of $$\left(\frac{x^n-1}{x-1}\right)'=(n-1)x^{n-2}+(n-2)x^{n-3}+\dots+2x+1=0.$$ When $n=p$ an odd prime this is just the derivative of the $p$-th cyclotomic polynomial $\Phi_p'$ (not sure if $\frac{x^n-1}{x-1}$ has a special name in general). For $n=3$, $x_3=-\frac12$, and for $n=5$ one can express $x_5$ by Cardano's formula, but it is not very illuminating.
Since cyclotomic polynomials are a classical topic I thought it would be easy to find information on the roots of their derivatives. But my searches in the usual places (MathSciNet and Google Scholar) came up short, only the values of $\Phi_p'$ at $0$ and $1$ are typically discussed. I am mostly interested in some good asymptotics for the negative root $x_p$, but information on complex roots is welcome too, if there is any. From geometry, $x_n$ is obviously monotone decreasing, so the asymptotics is the same for all large odd $n$, and, since $y=x^n$ approaches a vertical step, $x_n\to-1$. But I'd like something more precise, an expansion ideally. Same for $y_n=x_n^n\to0$. I'll appreciate even a hint in the right direction, the typical asymptotic methods for roots seem to have the parameter enter the equation differently.
|
2025-03-21T14:48:29.914602
| 2020-02-19T20:00:35 |
353102
|
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|
Stack Exchange
|
Equivalent definitions of the ring $B_{\mathrm{cris}}$
I'm reading Laurie's note about Fargues-Fontaine Curve and I think he uses a different definition of $B_{\mathrm{cris}}$. Usually when $R$ is a perfect ring of characteristic $p$, $A_{\mathrm{cris}}(R)$ is defined as $p$-adic completion of divided power envelope of the map $W(R)\to R$ and $B_{cris}=A_{\mathrm{cris}}[1/t]$.
But in these notes when $R$ is ring of valuation of an algebraically closed perfectoid field $B$ is defined as completion of $\mathrm{Frac}(W(R))$ with respect to all Gauss norms and defined Fargues-Fontaine curve using $B$ in place of $B_{\mathrm{cris}}$.
I want to know the relation between $B$ and $B_{\mathrm{cris}}$ in general. Is it true that they are isomorphic if R is the valuation ring of a perfectoid field?
If somebody could fix English in "and defined Fargues-Fontaine curve by it"? I don't now the relevant math enough to do so.
Some relations between analogous rings $B, B^+$ and $B^+{\mathrm{cris}, \rho}$ are discussed in Section 1.2 of Fargues' and Fontaine's paper "Vector bundles on curves and p-adic Hodge theory", https://webusers.imj-prg.fr/~laurent.fargues/Durham.pdf . But it might require some care to see that/whether those agree with $B, B^+$ from Lurie's notes (and for which $\rho$ is $B{\mathrm{cris}, \rho}^+=B^{+}_{\mathrm{cris}}$).
It is not true that $B_{cris} = B_{cris}^+$. What is true is that $B_{cris} = B_{cris}^+[1/t]$. In addition, what you define in the first paragraph is not $B_{cris}^+$ but $A_{cris}$.
@LaurentBerger you are right I edited the question.
@PavelČoupek thanks it's helpful.
|
2025-03-21T14:48:29.914747
| 2020-02-19T20:13:02 |
353104
|
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|
Stack Exchange
|
What is the etale homotopy type of the Witt group of braided fusion categories?
The Witt group $\mathcal{W}$ of braided fusion categories (see also the sequel paper) can be defined over any field; I am happy to restrict to characteristic $0$ if it matters.
Is $\mathbb k \mapsto \mathcal W(\mathbb k)$ an (affine?) algebraic group scheme?
Assuming $\mathcal W$ is sufficiently scheme-like for the following question to make sense, what I really want to know is:
What is the etale homotopy type of $\mathcal{W}$?
The answers to your questions are essentially in the first paper you cite. The second paper has more information on finer structure, like torsion, but the basic properties are all we need.
In the first paragraph of the introduction, the authors mention that any embedding of algebraically closed fields of characteristic zero yields an isomorphism on rational points, so it is geometrically a zero dimensional object. In particular, it is perhaps more profitable to think of it as a big Galois module. By Corollary 5.23, $\mathcal{W} \otimes \mathbb{Q}$ is a vector space of countably infinite dimension, so (assuming one manages to define $\mathcal{W}(R)$ for commutative $\mathbb{Q}$-algebras $R$) it is in fact an ind-affine group ind-scheme, rather than an affine group scheme.
|
2025-03-21T14:48:29.914881
| 2020-02-19T21:29:15 |
353110
|
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|
Stack Exchange
|
The mean along the eccentric anomaly of an ellipse log distance to a point within the ellipse
Conjecture.
Let
$$ f(r,\alpha,p, \theta) = \ln\left(\left(r\sin\alpha-\sin\theta\right)^{2}\left(1-p\right)^{2}+\left(r\cos\alpha-\cos\theta\right)^{2}\left(1+p\right)^{2}\right). $$
Then for any real $|r|<1,\ |p|<1$ and $\alpha$
$$ \int_{0}^{2\pi}f(r,\alpha,p, \theta)d\theta\ =\ 0 .$$
The picture is the following: we have an ellipse with center in $(0,0)$ and semi-axes adding up to two and we have a point $P = \left((1-p)r\sin\alpha,\ (1+p)r\cos\alpha\right)$ within this ellipse. Then we measure the distance from from a point on the ellipse $\left((1-p)\sin\theta,\ (1+p)\cos\theta\right)$ to the point $P$ and realize that on average the log of this distance is zero.
Also here is my sketch on desmos.
This fact sounds so simple to me that I'd thought that someone else has already proven it.
Some of my thoughts on this:
For $r=p=0$ this is trivial.
For just $r=0$ I could prove it the following way: consider the derivative $\int_{0}^{2\pi}\partial_pf(0,\alpha,p, \theta)d\theta$ and through the substitution $t= \tan\theta$ get an integral of a rational function
$$
\int\limits_{-\infty}^\infty \frac{1+p-(1-p)t^2}{((1+p)^2+(1-p)^2t^2)(1+t^2)}dt
$$
that one can show to be 0 for any $|p|<1$ and therefore the integral of the function $f(0,\alpha,p,\theta)$ itself must not depend on $p$, namely also be 0 as for $p=0$.
For $p=0$, just a circle and a point inside, the Pythagorean trigonometric identity applies and under the integral we basically have $\ln(1+r^2-2r\cos(\theta-\alpha))$, which after double angle $\theta=2\eta+\alpha-\pi$ substitution reduces to $\ln\left((1+r)^2\cos^2\eta+(1-r)^2\sin^2\eta\right)$, i.e. the previous case.
Here comes the feeling that the complete proof is somewhere very close, yet I completely stuck. Please share any ideas.
Is there some reason that you used the "elliptic-curve" tag? An elliptic curve is not an ellipse. The integral you've written is not an elliptic integral. If you don't know what an elliptic curve is, you should remove that tag. If you do, please edit your question to indicate how it is related to an elliptic curve.
presumably you want not only $|p|$ but also $|r|$ to be less than 1?
Otherwise $r=2$, $p=0$ gives $f=5-4\cos(\alpha-\theta)$ and the integral of $\log f$ over $\theta$ from 0 to $2\pi$ equals $4\pi\log 2$.
This is trivial. Just write the argument of logarithm as $$\left(1-p r e^{i a+i t}-r e^{i t-i a}+p e^{2 i t}\right) \left(1-p r e^{-i a-i t}-r e^{i a-i t}+p e^{-2 i t}\right),$$ where $a=\alpha$, $t=\theta$. From Fourier series expansion it is obvious that the integral will be $0$.
|
2025-03-21T14:48:29.915106
| 2020-02-19T23:09:46 |
353114
|
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|
Stack Exchange
|
Is the union of l^p a Banach space under some norm?
As a set of sequences, take the union of $\ell^p$, $p\geq 1$. As $p$ increases, the $\ell ^p$ space is larger, with strict inclusion.
However, this infinite union is strictly contained in $c_0$, consider
$x_n = 1/\log (n+1)$, so the usual $c_0$ norm will not yield a closed space, as $c_{00}$ is dense within any $\ell ^p$.
Is there some norm which turns the union of l^p into a Banach space, presumably with $c_{00} $dense?
This probably isn’t the kind of answer you were hoping for. As a vector space, it has a basis of the same cardinality as the cardinality of a basis for $l^2$. You can therefore “copy” the l^2 norm onto your space.
Are you talking about a Hamel basis? In which case one needs AC and necessarily can't write it down explicitly. Or a Schauder basis, where we can't be certain one exists until we actually have a norm? And what do you mean by "copy"? Can you be a bit more specific?
Hamel basis. There is a vector space isomorphism $\phi:X\to l^2$ between the two spaces. Then you can define a norm on $X$ by $|x|=|\phi(x)|$. Of course this is not explicit as you say, and depends on the axiom of choice.
As pointed out above, the question of whether a given vector space has a Banach space structure is one about cardinality. However, your space, as a union of a sequence of Banach spaces, has a natural and perfectly respectable locally convex structure (the inductive limit) for which the finite sequences are dense.
@AnthonyQuas: it's not explicitly stated in the question, but I suspect the OP wants the canonical embeddings of the $\ell^p$-spaces into the union to be continuous.
So, I agree that the cardinality of the 2 Hamel bases would be the same. However I am not convinced that the union would be complete under the norm derived from this. Further, I would like to know if one could actually write a norm down, which almost definitionally would require not using AC...
If two vector spaces are isomorphic (which they are if they have Hamel bases of the same cardinality), and one is a Banach space, then you can transfer the norm to the other and this normed space is isometrically isomorphic to the first one and so also Banach. In the other direction if you could put a norm on your space to make it a Banach space in such a way that the coordinate functionals are continuous, then by a suitable version of the closed graph theorem its topology would have to agree with the one I described above and this is clearly not the case.
By the way, to look at your question from the positive side, namely that of searching for a well behaved functional analytic point of view, then the lc topology I described is one of a type investigated by Komatsu and has many useful properties. For example, it is a complete, Hausdorff, bornological $(DF)$-space, its dual is naturally identifiable with the (Fréchet space) intersection of the $\ell^p$-spaces and there are nice descriptions of its bounded sets and weakly convergent sequences.
Maybe take the dual of some modular sequence space $\ell{M_n}$ with $M_n$ chosen wisely. For instance if $M_n(t)=t^{1+\delta_n}$ with $\delta_n\downarrow 0$, then the uniform $\Delta_2$ condition is satisfied. I'm worried it will just turn out to be $\ell_1$, but maybe not.
Many thanks, I think the modular space suggestion is a good place to look! Will try
Although such a norms exists by the axiom of choice, it is an interesting question what nice properties the unit vector basis $(e_n)_n$ could possibly have. From my point of view, it cannot have very nice ones.
For example, $(e_n)_n$ cannot be a Schauder basis. It cannot even be the spatial component of a Markushevich basis. Let's see why not.
Let $X = \cup_{1\leq p<\infty}\ell_p$, $\|\cdot\|_X$ be a complete norm for $X$, and let us assume that $(e_n,f_n)_n$ is an M-basis for $(X,\|\cdot\|_X)$.
For every $1\leq p < \infty$ the inclusion map $I_p:\ell_p\to X$ is well defined. By the closed graph theorem it is also bounded. Indeed, if $(x_n)_n$ converges to zero in $\ell_p$ and $(x_n)_n$ converges to some $x\in X$ (with respect to $\|\cdot\|_X$) then $x$ must be zero. Here we used the fact that $(e_n)_n$ is the spatial component of an $M$-basis and completeness. Both will be used again later.
But it is also the case that for no $1\leq p<\infty$, $\ell_p$ the map $I_p$ is an isomorphism, otherwise your space $X$ would be just $\ell_p$. A small argument (using automatic continuity of linear maps on finite dimensional spaces) tells us that for any $n\in\mathbb{N}$ if we restrict $I_p$ to the tail-space $Y_n$ of $\ell_p$ (i.e., the subspace of all vectors in $\ell_p$ with the first $n$ coordinates zero), then it cannot be an isomorphism. The conclusion is that for any $1\leq p<\infty$, $\varepsilon >0$, and $n\in\mathbb{N}$, there exist $k\in\mathbb{N}$ and a scalar sequence $(a_i)_{i=n+1}^{n+k}$ so that if we define $x = \sum_{i=n+1}^{n+k}a_ie_i$ then $\|x\|_p = 1$ while $\|x\|_X\leq \varepsilon$.
Now we are almost finished. Take $p_n = n$ and inductively pick $(a_i)_{i=k_{n-1}+1}^{k_n}$ so that if $x_n = \sum_{i=k_{n-1}+1}^{k_n}a_ie_i$, then $\|x_n\|_{p_n} = 1$, yet $\|x_n\|_X\leq 2^{-n}$.
We use completeness again and put $x = \sum_{n=1}^\infty x_n$. For every $i\in\mathbb{N}$, the $i$'th coordinate of $x$ (i.e., $f_i(x)$) is $a_i$. By assumption, there must exist some $1\leq p<\infty$ so that $(a_i)_i\in\ell_p$. This is absurd because for every $1\leq p<\infty$, if we pick $n$ with $p_n\geq p$ then
$$\|(a_i)_i\|_p \geq \|(a_i)\|_{p_n} \geq \|(a_i)_{i\geq k_{n-1}+1}\|_{p_n} = (\sum_{j=n}^\infty \|x_j\|_{p_n}^{p_n})^{1/p_n} \geq (\sum_{j=n}^\infty \|x_j\|_{p_j}^{p_n})^{1/p_n} = \infty.$$
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Subsets and Splits
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