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2025-03-21T14:48:31.663430
| 2020-08-01T19:12:19 |
368093
|
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"authors": [
"Denis Nardin",
"Donu Arapura",
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"Mikhail Borovoi",
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|
Stack Exchange
|
Non-abelian Ext functor and non-abelian $H^2$
Let $G$ be a group and
$$0\rightarrow K\rightarrow M\rightarrow N\rightarrow 0$$
a short exact sequence of groups. Now these are abelian groups, if I want to show that $\text{Hom}(G,M)\rightarrow \text{Hom}(G,N)$ is surjective, I would show that $\text{Ext}^1(G,K)=0$. However, if I'm studying the same question for non-abelian groups, then I do not have the tool of derived categories at my disposal. Can this be overcome with (non-abelian) cohomology?
Yes (in some sense).
@MikhailBorovoi How?
See the reference in my comment to this question.
@MikhailBorovoi, you can link to specific comments, not just questions. I think you mean this comment.
@LSpice: How can I do that?
See also: Flicker, Scheiderer, Sujatha, Grothendieck's theorem on non-abelian $H^2$ and local-global principles. J. Amer. Math. Soc. 11 (1998), no. 3, 731–750.
@MikhailBorovoi, the time stamp is a link to a specific anchor on the question page. For example, your reference to Flicker et al.: https://mathoverflow.net/questions/368093/non-abelian-ext-functor#comment928644_368093 .
I took the liberty to edit the title and to add the tag galois-cohomology. This is relevant to my answer. Roll back if you don't like this.
EDITED, taking into account the comments of Donu Arapura.
As JLA wrote, a homomorphism $f\colon G\to N$ gives an extension
\begin{equation}\label{e:E}
1\to K\to E\to G\to 1.\tag{E}
\end{equation}
This extension defines a homomorphism
$$b\colon G\to \operatorname{Out} K$$
called the band (lien, kernel) of \eqref{e:E}.
By definition, $H^2(G,K,b)$ is the set of isomorphism classes of extensions \eqref{e:E} bound by $b$.
A cohomology class $\eta(E)\in H^2(G,K,b)$ is called neutral if the extension \eqref{e:E} splits, that is, there exists a homomorphism $G\to E$ such that the composite homomorphism $G\to E\to G$ is the identity automorphism of $G$. In this case we obtain an action $\varphi$ of $G$ on the normal subgroup $K$ of $E$, and we obtain an isomorphism $E\overset{\sim}{\to}K\rtimes_\varphi G$ with the semidirect product.
There may be more that one neutral class in $H^2(G,K,b)$: they correspond to semidirect products with different actions $\varphi$ of $G$ on $K$. I have read that there may be no neutral elements, but I don't know examples. (In the Galois cohomology setting, for a connected reductive group, by Douai's theorem there always exists a neutral element in nonabelian $H^2$; see [2], Proposition 3.1).
If $K$ is abelian, then $\operatorname{Out} K = \operatorname{Aut} K$, so $b$ is just an action of $G$ on $K$, and $H^2(G,K,b)$ is the usual abelian group cohomology $H^2(G,K)$, where $G$ acts on $K$ via $b$.
The set $H^2(G,K,b)$ can be described in terms of cocycles. See Section 1.14 in Springer [1].
The band $b$ defines an action of $G$ on the center $Z=Z(K)$, and we may consider the usual (abelian) group cohomology $H^2(G,Z)$. From the cocyclic description of $H^2(G,K,b)$ it is clear that $H^2(G,Z)$ naturally acts on $H^2(G,K,b)$.
Moreover, if the set $H^2(G,K,b)$ is nonempty, then $H^2(G,Z)$ acts on it simply transitively; see Mac Lane, Homology, Theorem IV.8.8. The set $H^2(G,K,b)$ is nonempty if and only if a certain obstruction $\operatorname{Obs}(G,K,b)\in H^3(G,Z)$ vanishes; see Mac Lane, Theorem IV.8.7.
Note that we should not think that $H^2(G,K,b)$ "equals" $H^2(G,Z)$. First, $H^2(G,K,b)$ does not have a distinguished unit element. Secondly, $H^2(G,K,b)$ has a distinguished subset $N^2(G,K,b)$ of neutral elements. This is important because in many applications one uses nonabelian $H^2$ in order to determine whether a given extension \eqref{e:E} is split or not.
As far as I know, nonabelian $H^2$ is mostly used in the Galois cohomology setting.
Namely, if $k$ is an algebraic closure of a field $k_0$ of characteristic 0, $G=\operatorname{Gal}(k/k_0)$,
and $Y$ is a quasi-projective $k$-variety with additional structure (say, an algebraic group or a homogeneous space) such that for any $\sigma\in G=\operatorname{Gal}(k/k_0)$ there exists an isomorphism $\alpha\colon\sigma Y\overset{\sim}{\to}Y$, then it defines an extension
$$1\to \operatorname{Aut} Y\to E\to G\to 1,$$
where $E$ is the set of such pairs $(\alpha,\sigma)$
with a suitably defined composition law.
We obtain the cohomology class $\eta(Y)\in H^2(k_0,\operatorname{Aut} Y,b)$
of this extension for a suitable band $b$.
The variety $Y$ (with additional structure) admits a $k_0$-model if and only if $\eta(Y)$ is neutral, that is, the extension splits;
see this question.
For nonabelian $H^2$ in Galois cohomology see:
[1] T. A. Springer, Non-abelian $H^2$ in Galois cohomology, in: Algebraic Groups and Discontinuous
Subgroups, Proc. Sympos. Pure Math. 9, Amer. Math. Soc., Providence, 1966, 164-182.
[2] M. Borovoi, Abelianization of the second nonabelian Galois cohomology. Duke Math. J. 72 (1993), 217-239.
[3] Flicker, Scheiderer, Sujatha, Grothendieck's theorem on non-abelian $H^2$ and local–global principles.
J. Amer. Math. Soc. 11 (1998), no. 3, 731–750.
See also newer papers (they refer to these three), and this preprint.
If you are interested in nonabelian $H^2$ in Galois cohomology, please ask a separate question, and I will answer it when I have time
This is a very nice answer. I do have a couple of questions. (1) When you say $\eta(E)$ is neutral, does that mean there exists a lift of $b$ to $G\to Aut(K)$, such that $E$ is the semi direct product? So then it is clear, there may be more than one, or perhaps none. (2) What is the relationship to the old results of Eilenberg-Maclane "Cohomology theory in abstract groups. II." Annals (1947)? As I recall, they show that the set of extension classes is either empty (with obstructions in $H^3$) or parameterized by $H^2(G, Z)$, where $Z$ is the centre of $K$.
Let me know if you prefer that I ask these as separate questions, rather than as comments.
Of course it's your right not to name your paper if you don't want, and I apologise for an unwanted edit; but too many old MO comments become less useful when "here"-type links fade, so I'll put the name of "this preprint" here if it's all right: Borovoi - Extending the exact sequence of nonabelian $H^1$, using nonabelian $H^2$ with coefficients in crossed modules.
@LSpice: I think that my preprint is not quite relevant to this answer, and so I did not name it on purpose.
Thanks again for the interesting answer. Concerning an example. Let $F$ be a nonabelian free group. Choose a surjection $f:F\to G$, where $G$ is finite and nontrivial. Let $K$ be the kernel. $K$ is again nonabelian and free, so it must have trivial centre. By your answer, the nonabelian $H^2$ consists of a single element. This cannot be neutral, because the sequence can't split, as $F$ is torsion free.
Thank you, @DonuArapura, for the example.
If you have a morphism $f:G\to N\,,$ then you get a $K$-extension of $G$ by pulling back the $K$-extension of $N\,.$ The morphism $f$ lifts to a morphism into $M$ if and only if this extension is trivial. So you could show the map you want is surjective by showing that all $K$-extensions of $G$ are trivial.
If $K$ is abelian, then isomorphism classes of $K$-extensions correspond to classes in (abelian) group cohomology.
I think the general approach is sound, but all you need for the morphism to lift is that the extension is split (which doesn't mean it's trivial, since semidirect products exist), so you'll need done kind of pointed set of "extensions modulo splittings"
|
2025-03-21T14:48:31.664012
| 2020-08-01T19:46:54 |
368095
|
{
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],
"authors": [
"Fedor Petrov",
"ZUN LI",
"https://mathoverflow.net/users/136078",
"https://mathoverflow.net/users/4312"
],
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"url": "https://mathoverflow.net/questions/368095"
}
|
Stack Exchange
|
Estimating a combinatorial summation regarding multinomial distribution
We know that for $\sum_{i=1}^Kp_i=1, \forall i, p_i\ge0$, since $(\sum_{i=1}^Kp_i)^m=1, m\in\mathbb{N}^+$, we have multinomial equation $$\sum_{m_1+m_2+\dots+m_K=m}\frac{m!}{m_1!m_2!\dots m_K!}\prod_i p_i^{m_i}=1$$.
What can we say about the following quantity
$$\sum_{m_1+m_2+\dots+m_K=m}\frac{(Bm)!}{(Bm_1)!(Bm_2)!\dots (Bm_K)!}\prod_i p_i^{Bm_i}$$
for $B\in\mathbb{N}^+$? For example, its asymptotic lower bound/upper bound in terms of $B, K, p, M$, or even its closed-form.
By the roots of unity filter this is $B^{-K}\sum (p_1w_1+\ldots +p_Kw_K)^{Bm}$, where $w_j$ runs over roots of unity of degree $B$ for all $j=1, \ldots, k$.
Sorry could you please elaborate more about the roots of unity filter? Thanks.
See theorem 1 here http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf
|
2025-03-21T14:48:31.664105
| 2020-08-03T16:25:56 |
368237
|
{
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],
"authors": [
"Robert Israel",
"amateur",
"darij grinberg",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/50468"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368237"
}
|
Stack Exchange
|
Lagrange’s interpolation formula: Theoreme and Example
I would like to know where they come up with the formula of Lagrange interpolation (Lagrange’s interpolation formula),Lagrange_polynomial because I did some research, but I find a different definition of Lagrange interpolation. Beside would someone explain and elaborate more how they apply the formula 5.1 in example 5.6. because I don't know how to apply it. for example, what is $c_{i1}$
Thanks in advanced.
Theorem $5.5 ([26])$.
Let $f : k^{r} \to k$ be any function on a finite field $k$. Then there exists a unique polynomial $g : k^{r} \to k$, such that $\forall x \in k^{r}, f(x)=g(x).$
Any such mapping over a finite field can be described by a unique polynomial.
Using Lagrange interpolation, we can easily determine the polynomial. Let $f: k^{r}\to k$ be any function on $k$. Then
$$g(x)=\sum_{\left(c_{i1},\ldots,c_{ir}\right)\in k^{r}}f\left(c_{i1},\ldots,c_{ir} \right)\prod_{j=1}^{r}\left(1-\left(x_{j}-c_{ij} \right)^{p-1} \right)\quad \mbox{(5.1)}$$
is the unique polynomial that defines the same mapping as $f$.
Example $5.6$
Suppose $k=\mathbb{F}_{3},r=2,$ and the mapping $f$ is defined on $\mathbb{F}_{3}^{2}=\{0,1,2\}\times\{0,1,2\}$ as follows:
$$f(0,0)=0,$$
$$f(0,1)=1,$$
$$f(0,2)=2,$$
$$f(1,0)=1,$$
$$f(1,1)=2,$$
$$f(1,2)=0,$$
$$f(2,0)=2,$$
$$f(2,1)=0,$$
$$f(2,2)=1.$$
Then the polynomial $g$ that defines the same mapping as $f$ is constructed as follows:
$$\begin{align*}
g(x,y)&=0\\
&+1\left((1-x^{2})(1-(y-1)^{2}) \right)\\
&+2\left((1-x^{2})(1-(y-2)^{2}) \right)\\
&+1\left((1-(x-1)^{2})(1-y^{2}) \right)\\
&+2\left((1-(x-1)^{2})(1-(y-1)^{2}) \right)\\
&+0\\
&+2\left((1-(x-2)^{2})(1-y^{2}) \right)\\
&+0\\
&+1\left((1-(x-2)^{2})(1-(y-2)^{2})\right)\\
&=x+y.
\end{align*}$$
Reference:
Mathematical Concepts and Methods in Modern Biology 1st Edition Using Modern Discrete Models
It seems to me that the $c_i=(c_{i1},\ldots,c_{i,r})$ are the elements of $k^r$. Just need to list them in some order...
If by "they" you mean Lagrange, he was working over $\mathbb R$ rather than a finite field, but the principle is similar: to interpolate a function at a finite set of points, take a basis of functions that are $1$ at one of the points and $0$ at all the others.
Theorem 5.5 in the scan is wrong.
You would do well to study the idea of Lagrange interpolation. There is a unique polynomial of the minimal reasonable degree fitting data. In your setting, $t^4=t$ so any of $x+y,x^4+y,x+y^4,x^4+y^4$ or $(x+y)^4=x^4+x^3y+xy^4+x^4$ agree at all points.
Here are a few observations about your example: Consider the $9$ function values you specify, but as integers. The standard Lagrange interpolation would be
$$\begin{align*}
g(x,y)&=0\\
&+1 \frac {x \left( x-2 \right)
\left( y-2 \right) \left( y-1 \right) }{-2}
\\
&+2\frac {x \left( x-1
\right) \left( y-2 \right) \left( y-1 \right) }{4}
\\
&+1\frac {y
\left( x-2 \right) \left( x-1 \right) \left( y-2 \right) }{-2}
\\
&+2\frac{xy \left( x-2 \right) \left( y-2 \right)}{1} \\
&+0\\
&+2\frac {y \left( x-2 \right)
\left( x-1 \right) \left( y-1 \right) }{4}
\\
&+0\\
&+1\frac {xy \left( x-1
\right) \left( y-1 \right) }{4}
\\
&=\frac{9\,{x}^{2}{y}^{2}-15\,{x}^{2}y-15\,x{y}^{2}+21\,xy+4\,x+4\,y}{4}.
\end{align*}$$
And that is the unique polynomial of that degree over the reals, which does that. Then, $\mod 3,$ we have $4=1$ and the other coefficients are $0$, so indeed, $x+y.$
What you (and your book) wrote, treated as over the integers, comes out $9\,{x}^{2}{y}^{2}-18\,{x}^{2}y-18\,x{y}^{2}+6\,{x}^{2}+24\,xy+6\,{y}^{
2}-2\,x-2\,y-3
$
The values at the appropriate integer points are $$[0, 0, -3], [1, 0, 1], [2, 0, 17], [0, 1, 1], [1, 1, 2], [2, 1, -3], [0, 2, 17], [1, 2, -3], [2, 2, -11]$$ but again,$\mod 3,$ everything comes out fine.
|
2025-03-21T14:48:31.664331
| 2020-08-03T16:51:39 |
368239
|
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|
Stack Exchange
|
tame ramification and separability of the reduction
Let $K$ a local field, with ring of integer $\mathcal{O}_K$ or $\mathcal{O}$ ,of uniformizer $\pi$, residue field $k$ of caracteristic $p$.
Let $\varphi:\mathbb{P}^1_K\to\mathbb{P}^1_K$ finite separable morphism with tame ramifications: all the order of vanishing are prime to $p$. Let $\Phi:\mathbb{P}^1_{\mathcal{O}}\to\mathbb{P}^1_\mathcal{O}$ an extension of $\varphi$ to $\mathbb{P}^1_\mathcal{O}$ and let $\overline{\varphi}$ the reduction of $\Phi$ in $k$ that is $\overline{\varphi}=\Phi\times_\mathcal{O}\text{Id}_k:\mathbb{P}^1_k\to\mathbb{P}^1_k$.
Concretely $\varphi$ is associated with a fraction $\pi^k P/Q$ with $P,Q\in\mathcal{O}[t]$,$\Phi$ is $[\pi^kP:Q]$ or $[P:\pi^{-k}Q]$ (depending of the signe of $k$) and $\overline{\varphi}$ is the reduction of this fraction if it's possible (constant otherwise).
Question: can we tell that if all the ramifications of $\varphi$ are tame and $\overline{\varphi}$ is not constant (ie finite) then $\varphi$ is a separable morphism? If not can you give me an example. If not, is there a criterion for the reduction to stay separable? Thanks!
|
2025-03-21T14:48:31.664431
| 2020-08-03T17:25:23 |
368240
|
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"url": "https://mathoverflow.net/questions/368240"
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|
Stack Exchange
|
Isomorphisms mod nonstationary
Suppose $G \subseteq \mathrm{Add}(\omega_1)$ is generic over $V$. Let $X_i = \{ \alpha : G(\alpha) = i \}$. Is it true that $P(X_0)/\mathrm{NS} \cong P(X_1)/\mathrm{NS}$?
|
2025-03-21T14:48:31.664477
| 2020-08-03T17:56:28 |
368241
|
{
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"Math604",
"Willie Wong",
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"https://mathoverflow.net/users/66623"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368241"
}
|
Stack Exchange
|
improved Sobolev embedding
This is probably not a research level question but I am struggling with the geometry. My question is related to whether some monotonicity can increase the range of exponents in the Sobolev embedding.
For instance on the unit ball, nonnegative radially symmetric functions which are nondecreaing in the radial direction should satisfy a Sobolev embedding for an improved range of exponents: Indeed such functions must be large on the full boundary of the ball, yet the trace theorem prevents this from happening.
So I will ask the question on a finite cone; let $ S \subset \subset S^{N-1}$ be some nice spherical cap. For explicitness let us assume that the $x_1$ axis cuts through the center of $S$ (assume e.g. that $S$ is a ball in $S^{N-1}$ centered at some $y\in S^{N-1}$ lying on the $x_1$ axis). Let $ \Omega:=\{x=r \theta: 0<r<1, \theta \in S\}$ denote the finite cone. We now look at nonnegative functions which are zero on the side of the cone and moreover nondecreasing in the radial direction, i-e $ x \cdot \nabla u(x) \ge 0$ in $\Omega$.
So my question is:
Can we expect an improved Sobolev embedding for this class of functions? (here I am using the $H^1$ norm so I am asking about possible embedding of $H^1$ into some $L^p$ space for some improved exponents $p>2^*=\frac{2N}{N-2}$)
I suspect the answer is `no' and I am attempting to disprove it considering the first strategy that comes to mind: Take $ 0\le \phi \in C_c^\infty(B_1)$ a smooth radially nonincreasing function and then translate this function so that its support is centered at $y$, and scale it in order to concentrate its support at $y$. This sequence of functions will presumably violate any alleged improved Sobolev embedding, up to proving that these functions really have the correct monotonicity. Geometrically it looks like to me that this is so, but my geometric intuition almost always fails me now. Any comments would be great.
I am confused.... or maybe i don't understand your comment. On the unit ball if you just ask for radial functions then there is no improved range.
I misunderstood your question. Are you only interested in the $H^1 \to L^p$ embedding? // Also, when you say "radial increasing functions" do you mean functions that increase in the radial direction, or do you mean radially symmetric functions that also increase in the radial direction?
ya, i was not overly precise in my question. So let $X$ denote the functions with $H^1(\Omega)$ norm; the functions are zero on side of cone and the functions are nonnegative and satisfy $u_r \ge 0$ or $ x \cdot \nabla u(x) \ge 0$. So they are radial increasing but there is no radial symmetry assumption.
The answer is indeed "no." To see this, instead of considering translations of radial bump functions, one needs to use bump functions whose level sets are deformed e.g. to ellipsoids. More precisely, denote $x \in \mathbb{R}^n$ by $(x_1,\,x')$ and let $h$ be any decreasing function on $\mathbb{R}$. Then for
$$H(x) := h(\Lambda^2 x_1^2 + |x'|^2)$$
we have that $x \cdot \nabla [H(x-y)] \geq 0$ in the ellipsoid
$$\left\{\frac{(x_1 - 1/2)^2}{(1/2)^2} + \frac{|x'|^2}{(\Lambda/2)^2} \leq 1\right\},$$
which contains a neighborhood of $y$ in $B_1$ when $\Lambda$ is large (but not $1$). Now fix $\Lambda$ large and choose $h$ smooth with $h(s) = 1$ for $s \leq 0 $ and $h(s) = 0$ for $s \geq 1$. Then the family $$\lambda^{\frac{n-p}{p}}H(\lambda (x-y))$$
shows as $\lambda \rightarrow \infty$ that the exponent cannot be improved.
@ Connor Mooney. Thank you very much for your answer. Helpful as always.
|
2025-03-21T14:48:31.664723
| 2020-08-03T18:17:05 |
368243
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368243"
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|
Stack Exchange
|
Level sets of the polynomial function $^2+^2−+−$ over $\mathbb{F}_$
Let be an odd prime and assume $^2+ax+1$ is irreducible over the field $\mathbb{F}_p$. The polynomial function
$$\Psi:\mathbb{F}_p^2⟶\mathbb{F}_p,\quad (x,y)\mapsto ^2+^2−x+y-axy$$
is surjective, as proved here: Image of a polynomial function $x^2+y^2-x+y-axy$ over $\mathbb{F}_p$.
I would like to compute a set of representatives of the classes of the kernel of $\Psi$ (i.e., the relation $\ker(\Psi)=\{(x,y,t,w)\in \mathbb{F}_p^4, \, \Psi(,)=\Psi(t,w)\})$. So basically I would like to have an explicit set of $p-1$ elements of $\mathbb{F}_p^2$ that take on all the nonzero values in $\mathbb{F}_p$ when you apply $\Psi$. (Finding a solution to $\Psi(x,y)=0$ is obvious.)
Is there a way to do this in general, regardless of the value of $p$?
Here is a solution for $\frac{p-1}{2}$ of the level sets. Let $\varphi=x^2+y^2-x+y-axy$, and assume that $a\not=\pm 2$.
Following the suggestion in https://mathoverflow.net/q/356936 we first apply the change of variables given by $x= z+(a/2)y$ to obtain $$\varphi=z^2-z-b(y^2+(1+a/2)^{-1}y),$$ where $b=a^2/4-1$. Then letting $u=z-1/2$, $v=y+(2+a)^{-1}$, and $c=\frac{-1}{4}\left(\frac{1-a/2}{1+a/2}\right)$ we have $$\varphi=u^2-bv^2+c.$$
To compute the level sets you need to compute the fibres $\varphi^{-1}(d)$ for an image point $d$. If $d-c$ is a quadratic residue, then $d-c=e^2$ and so letting $s=u/e$ and $t=v/e$ we obtain: $$s^2-bt^2=1.$$
Having all solutions $(s,t)$ determines the solutions $(u,v)$, in turn the solutions $(y,z)$, and in turn the solutions $(x,y)$.
So it suffices to find the solutions to $s^2-bt^2=1.$ But such solutions are explicitly determined in:
Tekcan, Ahmet, The number of rational points on conics $C_{p,k}:x^2−ky^2=1$ over finite fields $\mathbb{F}_p$. Int. J. Math. Sci. 1 (2007), no. 2, 150–153.
MathSciNet Summary:
"Let $p$ be a prime number, $\mathbb{F}_p$ be a finite field, and let
$k\in \mathbb{F}_p^*$. In this paper, we consider the number of
rational points on conics $C_{p,k}: x^2−ky^2=1$ over $\mathbb{F}_p$.
We prove that the order of $C_{p,k}$ over $\mathbb{F}_p$ is $p−1$ if
$k$ is a quadratic residue mod $p$ and is $p+1$ if $k$ is not a
quadratic residue mod $p$....''
I am not sure what happens when $d-c$ is a non-residue (I haven't tried), but since $\varphi$ is onto and $(p-1)/2$ of the non-zero values are quadratic residues, this gets you roughly half way.
I am surprised that Tekcan could publish this result, which is a standard exercise in number theory (for any diagonal conic over $\mathbb{F}_p$, in any number of variables). This was also discussed multiple times on this site, see e.g. my response for https://mathoverflow.net/questions/65183/when-is-the-sum-of-two-quadratic-residues-modulo-a-prime-again-a-quadratic-resid
|
2025-03-21T14:48:31.664922
| 2020-08-03T19:17:22 |
368244
|
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|
Stack Exchange
|
Computable functions with limited domains
In the developments I've seen of primitive recursive and computable functions, the functions always have codomain $\mathbb{N}$, but are allowed to have domain $\mathbb{N}^{m}$ for any natural number $m$. This seems odd to me---treating the domains and codomains as fundamentally different.
One solution would be to allow functions $f\colon \mathbb{N}^m\to \mathbb{N}^n$ for any natural numbers $m,n$. Of course, such a function is really just an $n$-tuple of functions $(f_1,f_2,\ldots, f_n)$ where $f_i$ is just the $i$th coordinate of $f$, and computability for $f$ would amount to computability for each $f_i$.
However, I'm more interested in the opposite direction: limiting the domain to always be $\mathbb{N}$. This seems to match, more naturally, what an idealized machine is doing by taking in a single natural number and spitting out a single natural number (or not halting). Of course, a fortiori, one could develop the recursive functions as usual, look at the subclass of functions whose domains are $\mathbb{N}$, and called these the limited domain computable functions, and then show that from these we can reconstruct the non-limited functions in a simple way.
My question is if there is a more natural approach. Just as the recursive functions are built up from some starting functions, using very limited and natural operators, I wonder if there are ways to build up the "limited domain computable functions" similarly, in a non-"ad hoc" way. (For instance, it would be nice if we could do it without the need to develop a universal Turing machine first.)
In other words: Does the extra generality in the domain necessarily simplify the development of computable functions?
You might look-up abstract recursion theory. (I think Fenstad is a primary researcher.) Often one will deal with partial recursive functions as well, and there may be benefit to considering the domain as proper subset of N. One author who has likely addressed this issue in an accessible text is Oddifreddi. Gerhard "Recursion Theory Is My Weakness" Paseman, 2020.08.03.
If you are concerned about runtime complexity, then you should turn to computer science. There the exponents matter, and while the transformations in between are computationally equivalent, they may not be polytime or linear equivalent. Gerhard "Will Wait For Another Answer" Paseman, 2020.08.03.
@Zero Yes. More generally, given any sets $S,T$ with bijections $\varphi\colon S\to \mathbb{N}$ and $\psi\colon \mathbb{N}\to T$, we can pre-compose with $\varphi$ and post-compose with $\psi$, and then ask about "computable" functions from $S\to T$ (relative to the "encodings" afforded by $\varphi$ and $\psi$). This was one reason I wanted to define the computable functions using the limited domain---because up to the "nice recursive bijections" you mention there is no difference.
I think this is addressed in R. L. Goodstein's Recursive number theory, where he develops the entire theory of primitive recursive functions and recursive functions in an absolutely minimalistic manner. Primitive recursive functions $\mathbb{N}^2\to\mathbb{N}$ are unavoidable but he carefully isolates what is essential about these. IIRC, the key thing is to get to $\max(m,n) = (m \mathop{\dot{\smash-}} n) + n$, where ${\dot{\smash-}}$ denotes truncated subtraction.
Two papers of Julia Robinson seem to do the sort of thing you're looking for. Here are the MathSciNet data.
Robinson, Julia, General recursive functions, Proc. Amer. Math. Soc. 1, 703-718 (1950). ZBL0041.15101 MR0038912.
Robinson, Julia, A note on primitive recursive functions, Proc. Am. Math. Soc. 6, 667-670 (1955). ZBL0067.00204 MR0073536.
For the benefit of those not familiar with these two papers, let me summarize the connections I see with my question. Robinson shows that there are some significant simplifications by working with functions with limited domains. First, "substitution" can be replaced with the simpler "composition" operation. Second, "primitive recursion" can be replaced with the simpler "repeat the function" operation. (That is, $g(x) = f^{\circ x}(0)=f(f(\cdots(f(0))\cdots))$.) Third, "$\mu$-minimization" can be replaced with the simpler "first inverse" operation. [to be continued...]
With respect to the basic, or initial, functions, things get a little muddier, but still (I believe) simpler. Robinson shows that a finite set of initial functions suffices. She shows that one function won't suffice, but shows that two will suffice. She gives numerous different lists of such functions.
Moreover, one can replace "repeat the function" with "add two functions" if working with recursive functions of more than one variable. etc...
The paper by Julia's husband Raphael, entitled "Primitive recursive functions" may also be useful to people.
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2025-03-21T14:48:31.665244
| 2020-08-03T19:42:04 |
368245
|
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|
Stack Exchange
|
Lie monoids as monoids internal to the category of smooth manifolds?
This question can be thought as a complement to this one.
Lie groups can be defined as groups internal to the category of smooth manifolds. Lie monoids, however, as a particular case of Lie semigroups, seem to deserve a much more complicated definition (see, for instance, 'Lie semigroups and their applications', by Hilgert and Neeb, section 1.4).
Briefly, these are thought as closed subsemigroups of Lie groups, satisfying an extra property. This property, on its turn, is related to the infinitesimal counterpart of the notion of Lie semigroup (in the above reference, the notion of 'Lie wedge', whose definition, consequently, must precede that of a Lie semigroup).
What kind of difficulties appear if one tries to define a Lie monoid simply as a monoid internal to the category of smooth manifolds (or some related category)?
A LITTLE BIT OR FURTHER DISCUSSION
Lie groupoids, on their turn, can be defined as groupoids internal to the category of smooth manifolds. Is there an analogous notion of 'Lie category', in which morphisms are allowed not to be isomorphisms? Of course, the same question holds for its infinitesimal counterpart.
I tried to find some reference dealing with such a notion, but couldn't. Though, it seems to be a reasonable one to consider even within the realm of Lie groupoid theory. For example, if one wants to allow distinct objects to have distinct automorphism groups, but still be connected by morphisms, this notion seems to be a necessary step.
In particular, that's the case if one wants to allow morphisms between distinct objects to be not only isomorphisms between their automorphism groups, but also covering maps between them. I can't think right now of a concrete example coming, say, from Physics, but it sounds possible that the 'internal symmetries' of a system might 'collapse' in this particular way.
Besides that, exactly as Lie groupoids can be considered as natural generalizations of Lie groups (even if this shouldn't be considered the most appropriate point of view, for many reasons...), the 'Lie categories' would be natural generalizations of Lie monoids. Indeed, a 'Lie category' with one object would amount precisely to a Lie monoid.
Any references will be appreciated.
A commonly used "Lie monoid" is $[0,\infty)$, so you'd need to consider at least manifolds with boundary. Actually, since we also want $[0,\infty)^n$, you need manifolds with corners. That's the first complication. The other is that we know from the theory of semigroups in functional analysis that continuity at $0$ is very different from continuity at all the other points, so there's the potential for a lot of complication there.
There is indeed a notion of "Lie category", introduced a 1959 paper of Charles Ehresmann: Catégories topologiques et categories différentiables. This is accessible in his OEuvres Complètes, part I, pages 237–250. I can't say it's gained massive traction. Incidentally, I think this is the paper that introduced Lie groupoids, but under the name 'differentiable groupoids'. Most of the 19 citations recorded by MathSciNet seem to be citing it for this purpose. The name of Lie was originally attached to a special case, which you can still see if you read Mackenzie's book on Lie groupoids from the 1980s. (Note that there is a more recent notion of 'differentiable category' in a 2006 paper by Blute–Cockett–Seely that is unrelated, coming from a computer science perspective.)
One nice result is that the core of the underlying category—the maximal subgroupoid—is proved to be a Lie groupoid, though the proof is in a rather old style. This is analogous to the maximal subgroup of the Lie monoid of $n\times n$ matrices with multiplication is a Lie group.
I should add that in the meantime, I have learned that Grothendieck talked about internal categories and groupoids in more generality in the 1960/61 Seminaire Bourbaki, section 4 of Techniques de construction et théorèmes d'existence en géométrie algébrique III : préschémas quotients http://www.numdam.org/item/SB_1960-1961__6__99_0/ on page numbered 106. He called them $\mathbf{C}$-categories and $\mathbf{C}$-groupoids.
|
2025-03-21T14:48:31.665561
| 2020-08-03T21:18:31 |
368250
|
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|
Stack Exchange
|
Noncommutative torus as a von Neumann algebra
Le $\theta$ be irrational. One can define the noncommutative torus $A_{\theta}$ as a universal algebra generated by two unitaries $u,v$ satisfying the relation $vu=e^{2 \pi i \theta} uv$. This is an abstract defnition: however one can show that this algebra is simple and can be concretely represented as a $C^*$-subalgebra of $B(L^2(\mathbb{T}))$ generated by $U$ and $V$ where $Uf(x)=e^{2\pi i x}f(x)$ and $Vf(x)=f(x+\theta)$. Denote this concrete algebra as $\mathfrak{A}$ and consider $\mathfrak{A}''$ which is von Neumann algebra.
How to prove that $\mathfrak{A}''$ is a type $II_1$ factor (correct me if it isn't true)?
In context, ‘jest’ must be ‘is’.
Hi @JonBannon : I'm not too familiar with crossed products, is it clear that the "natural" vN completion of the abstract crossed product $C(S^1)\rtimes {\mathbb Z}$ coincides with the concrete representation in the question? Does this work by showing that the representation in the question is the GNS representation for the unique tracial state on ${\mathfrak A}$?
Thank you: it seems to me that this solves my problem-as one can construct the faithful tracial state on $A_{\theta}$ (and $A_{\theta}$ is infinite dimensional) then this factor has to be of type $II_1$. Or am I wrong?
No. This is irreducible. The commutant of $U$ is $L^\infty({\mathbb T})$ and its intersection with $V$-commutant is just ${\mathbb C}1$.
@Yemon Choi: I've deleted the comment containing the error. Thanks for catching this!
@NarutakaOZAWA Thanks! I was trying to derive a pedestrian argument to show that ${\mathfrak A}''$ had no faithful normal trace, using what we know about the unique tracial state on $A_\theta$, but your comment is much more incisive.
@NarutakaOZAWA many thanks for your comment-if you could rewirte it as an answer then I will accept it (by the way I'm pretty surprised that one obtain the whole $B(L^2)$ in this manner)
@truebaran: I have the unfortunate habit of speaking publicly first and then fixing what I said after the fact. I should have read your question carefully before commenting. I apologize.
@JonBannon it's fine, especially that your idea looked very plausible
@truebatan: nevertheless!
@YemonChoi, I just want to point out that there is no such thing as a "natural" vN completion of a C*-algebra. One could argue that the double dual is a candidate but it is rarely the object of interest. This is why vN algebras are almost always defined concretely as operators on a Hilbert space.
@Ruy You are right, of course. I was responding in haste to one of Jon's earlier comments (now deleted). What I meant was that in forming a reduced Cstar crossed product, there is (at least in general) a preferred space on which said crossed product is realised. So here I was imagining C(T) realised on L^2(T) as usual and then forming the crossed product by the appropriate action of Z resuts in a concrete Cstar algebra on a specific Hilbert space; what I meant by "the natural vN completion" was then the WOT-closure of that concrete Cstar algebra.
@YemonChoi, please see my answer below.
No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant:
$$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$
By the way, the invariant subspace problem for the Bishop operator $f(x)\mapsto xf(x+\theta)$ is still open in full generality.
https://mathscinet.ams.org/mathscinet-getitem?mr=353015
To support Ruy's answer: in my opinion the most natural representation of the quantum torus C*-algebra is the GNS representation coming from its tracial state. This can be explicitly described as the action on $l^2(\mathbb{Z}^2)$ given by $$Ue_{m,n} = e^{-i\hbar n/2}e_{m+1,n}$$ and $$Ve_{m,n} = e^{i\hbar m/2}e_{m,n+1}.$$ The von Neumann algebra they generate is indeed a $II_1$ factor.
I would even say this is the "quantum torus von Neumann algebra". There's more in Section 6.6 of my book Mathematical Quantization.
I wonder if people know what are the indices of finite index subfactors of this vN algebra. I was ready to guess that these should involve the rotation angle $\theta$, or $\hbar$ in your answer, but I now see that this is wrong. If I am reading it correctly you state in the 2nd paragraph of page 146 of your book that these algebras are all isomorphic to each other for all positive values of $\hbar$ (presumably just the irrational ones). Do you have a reference for this result?
I was wondering that too! It's been 20 years since I wrote that comment. I think the reason is because they're all hyperfinite (and there is only one hyperfinite $II_1$ factor). But I really don't remember how we know this ... I will think about it.
I guess it follows from Theorem 4 in "Elliott, George A.; Evans, David E., The structure of the irrational rotation C*-algebra, Ann. Math. (2) 138, No. 3, 477-501 (1993)".
Yes, that's it. Thanks!
As I was reading Yemon Choi's comment above it occurred to me that the situation of the crossed product $C(S^1)\times_\theta\mathbb{Z}$ is in fact a bit peculiar since the most standard representation of $C(S^1)$ one usually thinks of, namely as multiplication operators on $L^2(S^1)$, already comes equipped with a unitary representation of $\mathbb{Z}$ implementing the action by rotation.
This is not always the case for a general crossed product $A\times\mathbb{Z}$, so one usually starts with one's favorite representation of $A$ on some Hilbert space $H$ and builds the "regular representation" of the crossed product on the Hilbert space $H\otimes \ell^2(\mathbb{Z})$.
Even though that was not the representation the OP had in mind it is interesting to observe that, if the irrational rotation C*-algebra is completed in this other representation, one does indeed get a type $II_1$ factor, partly because the standard trace is a vector state in this representation and hence duly extends to a normal state on the weak closure.
PS: It was my original intention to reply to a comment by Yemon Choi, but I could not fit all of this within the 600 charactes size limitation. I therefore hope to be excused for shamelessly attempting to sidestep the rules and I am ready to delete this post should anyone complain!
|
2025-03-21T14:48:31.666139
| 2020-08-03T22:34:09 |
368257
|
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|
Stack Exchange
|
On the dual version of an isomorphism of spectral sequence term (from Cartan and Eilenberg)
I'm trying to take spectral sequences as a black box for application in commutative algebra and I admit that I haven't really gone through (or understand) all the proofs of all the isomorphisms between various terms of a spectral sequence and respective cohomologies assuming some term of the sequences are zero.
So in this post, my question is about the dual (homology) version of Proposition 5.6 (c) , Chapter XV, of Homological algebra by Cartan & Eilenberg. Let me first recall (for the case $r=2$ only) that it says that: if $E_2^{p,q}$ is a spectral sequence converging to
$ H^{p+q}$ and we have $E_2^{u,v}=0$ for $u+v=n-1, u\le p-2 $ and for $u+v=n, u<p$ and for $u+v=n+1, u\ge p+2$ and for $u+v=n, u>p$, then we have an isomorphism $E_2^{p,n-p}\cong H^n$ .
Now my question is: What is the dual version of this result (for first quadrant spectral sequence $E^2_{p,q}$ and $H_n$ ) ?
My naive idea: Since $E^2_{p,q}=E_2^{-p, -q}$ and $H_n=H^{-n}$, so if I just change all the $u,v,n,p,q$ in the conditions to it's negative and then rewrite the conditions, then do I get a valid dual result ?
Thanks
I think it is better to understand the main definitions and theorems you don't need the proofs.
the conditions in the theorem means that all the boundary maps coming to or going out of $E^{p,n-p}$ are zero and also all other elements on the diagonal $u+v=n$ are zero. just look at the definition of the boundary map in homology and rewrite the conditons
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2025-03-21T14:48:31.666276
| 2020-08-03T23:18:24 |
368260
|
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|
Stack Exchange
|
What is the explicit relationship between the shape parameters and the holonomy of a hyperbolic ideal triangulation?
Let $K$ be a hyperbolic knot in $S^3$. One way to describe the hyperbolic structure is to give a discrete, faithful representation $\pi_1(S^3 \setminus K) \to \operatorname{PSL}_2(\mathbb C)$, the holonomy representation.
This is hard to compute in practice.
A way that is easier is to decompose the knot complement into ideal tetrahedra, whose geometry is described by shape parameters.
It is then possible to write down equations for these shape parameters.
Solving them gives a description of the hyperbolic structure, and can be used to compute things like the volume.
[Warning: I know less about this paragraph, so I may be wrong.]
Is there a source that works out the details of how to go between these two descriptions? I would be especially interested in explicit examples of how to obtain the shape parameters from the matrices of the holonomy and vice-versa.
I think what you're looking for is in chapter 4 of Thurston's notes.
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2025-03-21T14:48:31.666378
| 2020-08-04T00:34:58 |
368264
|
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Stack Exchange
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On Galois' criterion for resolvents
Let $K$ be an algebraic number field, $f(x) = 0$ an algebraic equation over $K$ of degree $n \ge 2$ with only simple roots $x_1, \dotsc, x_n$, and $L \mathrel{:=} K[x_1, \dots, x_n]$ the splitting field of this equation over $K$. Any $y \in L$ then has a representation
$$y = P(x_1, \dotsc, x_n)\label{1}\tag{1}$$
with $P = P(X_1, \dotsc, X_n) \in K[X_1, \dotsc, X_n]$ a polynomial in the $n$ indeterminates $X_1, \dotsc, X_n$ over $K$. Such an element is also called a rational function of the roots — this being a priori a quotient $Q(x_1, \dotsc, x_n)/R(x_1,\dotsc, x_n)$, but such an expression can always be rewritten as a polynomial $P(x_1, \dotsc, x_n)$ —. A resolvent for the equation $f(x) = 0$ is then an expression $t = P(x_1, \dotsc, x_n)$ having, in classical parlance, the property that any rational function of the roots can be expressed as a rational function of $t$, or being, in modern terminology, a primitive element of the field extension $L:K$ (for this and what follows see [1]).
This notion goes back to Lagrange and Galois, and was taken by the latter as the point of departure of his theory of the solvability of equations (later vigorously exorcised by Dedekind and Artin, making the theory questionably more elegant and unquestionably much more impenetrable). The point of departure for obtaining, in return, such a resolvent, was the following observation.
Fix once and for all an enumeration of the roots as $x_1, \dotsc, x_n$.
Any permutation $\sigma \in \mathfrak{S}_n$ operates on the polynomials
$P \in K[X_1, \dotsc, X_n]$ via $P \mapsto P^{\sigma}$ with
$$P^{\sigma}(X_1, \dotsc, X_n) \mathrel{:=} P(X_{\sigma(1)}, \dotsc, X_{\sigma(1)}).$$
Thus to any $y \in L$ with a given representation \eqref{1} there
is attached a bunch of at most $n!$ different values $y^{\sigma} \mathrel{:=} P^{\sigma}(x_1, \dotsc, x_n)$. There are two basic claims:
(C1) If the values $t^{\sigma}$ are pairwise different, so that there are
exactly $n!$ different ones, then $t$ is a resolvent;
(C2) for every degree $d \ge 1$ (and even only $\mathbb{Z}$-coefficients),
the generic polynomial $P$ gives $t = P(x_1, \dots, x_n)$ satisfying
(C1), thus yielding a resolvent.
Here, "generic" means "not satisfying a certain nontrivial polynomial relation".
Now the condition (C1) is initially only sufficient, and the property of being
a resolvent does not make use of the special representation \eqref{1}, so it is a priori possible that there is a representation $t = Q(x_1, \dots, x_n)$ not
satisfying (C2) with $t$ still being a resolvent.
Question:
Can this happen?
[1] Edwards, H.M.,
Galois Theory
(Graduate Texts in Mathematics 101). Springer 1984
Despite your definition, I don't know what "generic" means. Which relation is "a certain nontrivial polynomial relation"? What does it mean to speak of whether a polynomial (which is what you are calling generic) satisfies a polynomial relation?
Also, since you edited it back in you obviously want it, so I apologise for an edit that went against your intentions; but "a bunch of at most $n!$ different values" is not usual English. Usually one would say something like "a bunch of different values, at most $n!$" or just "at most $n!$ different values".
Hopefully useful reference on non-modern Galois Theory, https://arxiv.org/abs/1301.7116
(C1) is only necessary for a primitive element when $n!$ is the degree of the extension $L/K$. But there are examples of degree $3$ Galois extensions, and $n!$ never equals $3$.
Let $K=L=\mathbb{Q}$ and take $f(x)=x(x-1)$. Thus, the two roots can be ordered as $x_1=0,x_2=1$. Any polynomial $Q(X_1,X_2)$ will yield a primitive element for this extension when we evaluate at $x_1,x_2$.
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2025-03-21T14:48:31.666642
| 2020-08-04T01:44:27 |
368266
|
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Stack Exchange
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Tits Reductive Groups over Local Fields Example 1.15 (Quasi-split special unitary groups in odd dimension)
I hope this question about Tits's paper "Reductive groups over local fields" in Algebraic groups and discontinuous subgroups ends up having an easy answer, but I'm a little stuck on the morass of notation. I have a couple questions about this example that will probably help me greatly.
I am considering Example 1.15, where $L$ is a separable quadratic extension of $K$. We take the standard Hermitian form, define our split torus etc.
We find root subgroups $U_{a_{ij}}(K)$ and $U_{2a_j}(K)$ in terms of matrices involving elements $c, c^{\tau}, d, d^{\tau}$ where all of these are in $L$, not $K$. See question 3 for an example.
Question 1: Why are these "$K$-points"? My guess is that I should let the Galois group act on this set of matrices in some fashion and these root subgroups $U_{a_{ij}}(K)$ and $U_{2a_j}(K)$ should be pointwise fixed under this action: is this the correct idea?
Question 2: I wish to compute the $\alpha(a_i u_i(c,d))$ and the $\alpha(2a_i, u_i(0,d))$. In order to do this, one supposedly first computes the $m(u_{ij}(c))$ and the $m(u_{i}(c,d))$. These elements should all be in $N(K)$, the normalizer of the torus $S(K)$. We must identify the apartment $A$ with $X_* \otimes \mathbb{R}$ and we do so as Tits says in (4) (I do not really understand the significance of (4)) and then … TaDa, we apparently have
$$
\alpha(a_i u_i(c,d))a_i+(1/2)\omega(d)
\quad\text{and}\quad
\alpha(2a_i, u_i(0,d))=2a_i+\omega(d).
$$
This second one at least seems to make sense: these should be affine functions which describe how conjugation by an element of the affine Weyl group acts on a coweight, so I generally expect the answer for $\alpha$ to be the vector part $a_i$ associated to the root and then a non-vector part associated to the valuation of $a$ for $u_i(a)$. But I do not see at all where we get $(1/2)\omega(d)$ in the first of these formulas.
Question 3: To compute the function $d$, let me take the example of $SU(3)$; here
$$
U_{a_1}(K)=
\left\{\begin{pmatrix} 1 & -c^{\tau} & d \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}\right\}.
$$
And if we quotient by $U_{2a_1}$, then we should be left with matrices like
$\begin{pmatrix} 1 & -c^{\tau} & 0 \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$. In computing $d$ are we merely checking the ramified degree of the extension $L/K$ for whatever field $c$ is permitted to live in?
I've got more questions for the rest of this example but I think the answers to these three questions can help me find my footing.
As to "why they are $K$-points", it's the same reason that the unitary group ${g \in \operatorname{GL}_n(\mathbb C) \mathrel: g g^* = 1}$ is the $\mathbb R$-points of an $\mathbb R$-group, not the $\mathbb C$-points. A very easy check: the Lie-algebra version of this gives a $K$-vector space, not an $L$-vector space. (Why does this seem to behave differently from more familiar situations? Because, unlike those more familiar situations, the Galois group doesn't simply act coördinatewise.)
What do you mean by "compute the function $d$"?
The $\omega(d)$ terms come from the fact that, similarly, the action of the normaliser of the split torus on the apartment corresponding to that torus is not what you might expect: see (4) on p. 41.
In section 1.6, Tits writes: "For every affine function $\alpha$ on $A$ whose vector part $a=v(\alpha)$ belongs to $\Phi$, one has an obvious inclusion $\overline{X}{2\alpha} \rightarrow \overline{X}{\alpha}$, and the quotient has a natural vector space structure over $\overline{K}$"..., the residue field. The dimension of this space is $d$.
Oh, $d$ like $d_\alpha$, not $d$ like $u_i(c, d)$. Yes, it is more or less as you say (although I'd argue we really care about the residual degree—but the two pieces of information are complementary). Note, by the way, that, though it's fine for casual discussion, your discussion of the quotient group is more like a discussion of a transversal of the collection of cosets: the set of matrices you describe isn't a group!
So I don't know how you can draw this 1/2 out of (4). In particular, (4) seems to be identical to the formula for example 1.14, at the bottom of page 39; we let the normalizer act on the apartment by the natural action on the vectors plus a term corresponding to the valuation. I am probably being slow here but I still don't see where the 1/2$\omega(d)$ comes from. I would simply expect $a_i+\omega(d)$.
You are right; I pointed to the wrong thing. Note $\nu(m(u_i(c, d))v = v$ means $-v_i = v_{-i} = v_i + \omega(d)$, i.e., $2v_i + \omega(d) = 0$. This can be expressed as the vanishing of $a_i + \frac1 2\omega(d)$ or, when $c = 0$ and so we are in $U_{2a_i}$, of $2a_i + \omega(d)$.
Also you make a very good comment about the fact that the matrices I wrote do not form a group; but nonetheless, we know $U_{a_{ij}}(c,d)$ are isomorphic to a 2-dimensional additive group over $K$ (is this correct?) corresponding to choices of $c$ and $d$, and that when we quotient as Tits asks, in this case I am checking the ramified degree of the field where $c$ lives?
You should not write $U_{a_{ij}}(c, d)$. Perhaps you mean $U_{a_i}(K)$, which is a set of matrices $u_i(c, d)$? This is not an additive group; it is not even Abelian.
By the way, as you go on, it may help to know that the $\delta$ that occurs later in this section is $0$ unless $p = 2$. (I thought that that was remarked somewhere, but I can't find it now.) Of course, Tits's Corvallis article is much easier reading than BT; but probably more pleasant than either is Yu's lovely article "Bruhat–Tits theory and buildings" in the Ottawa lectures on admissible representations of reductive, $p$-adic groups. It used to be on J.-K.'s web page, but anyway you can find it on Semantic Scholar.
Thank you so much for all your help! I'll take some time to digest (especially with your good point about the $U(K)$ not being nice additive groups as in the split case)
Incidentally, this is "as bad as (quasi-split but) non-split gets", which is why Tits does this particular example rather than, say, something not of type $\mathsf A$; in all other (quasi-split) cases at least the relative root system is reduced, so you do once again get additive groups (and $d_\alpha$ is an appropriate residual degree).
It's also probably worth your while (it was definitely worth mine!) to see @Marty's lovely discussion of this example from the point of view of parahoric subgroups.
It was that very post that inspired me to try out Tits again... I seem to be gradually piecing things together as I cycle through a couple of these excellent discussions. Thanks again!
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2025-03-21T14:48:31.667379
| 2020-08-04T03:19:55 |
368269
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"authors": [
"Jean Raimbault",
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Stack Exchange
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Reference for openness of subspace of PSL(2,R) representation variety corresponding to Teichmüller space
$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Hom{Hom}$Let $S$ be a compact oriented surface with nonempty boundary. There are two variants of Teichmuller space for $S$ you might consider:
The one that parameterizes finite-volume complete hyperbolic metrics on the interior of $S$. These correspond to discrete and faithful representations of the fundamental group of $S$ into $\PSL(2,\mathbb{R})$ that take the loops surrounding the punctures to parabolic elements.
The one that parameterizes finite-volume complete hyperbolic metrics on $S$ with geodesic boundary. These correspond to (certain, not all as in 1) discrete and faithful representations of the fundamental group of $S$ into $\PSL(2,\mathbb{R})$ that take the loops surrounding the punctures to hyperbolic elements.
Let $U \subset \Hom(\pi_1(S),\PSL(2,\mathbb{R}))$ be the set of representations in either 1 or 2, so you obtain Teichmüller space from $U$ by quotienting out by the conjugation action of $\PSL(2,\mathbb{R})$.
Question: What is a good reference for the fact that $U$ is open? I know many good sources for the corresponding fact when $S$ is a closed oriented surface, where in fact we can replace $\PSL(2,\mathbb{R})$ by an arbitrary Lie group (a theorem of Weil — here we require the representation to be discrete, faithful, and cocompact). But I don't know a source that does these variants.
In case 1, $U$ is not open in the space of all representations as you can deform the parabolic to an elliptic with infinite order and get a non-discrete group. It is probably open in the subspace of type-preserving representations (those that send peripheral elements to parabolics), but I don't know a reference for that.
In case 2 I also don't know a reference, but I think you can bootstrap an argument from the closed case: let $2S$ be the double along the boundary, all representations of $\pi_1(S)$ extend to representations of $\pi_1(2S)$ which are equivariant with respect to the involution given by the symmetry in the boundary. The map between representation spaces is an embedding and the result follows. (I did not check the details but they do not seem hard to write up.)
$\DeclareMathOperator\PSL{PSL}$Here is a complement to Jean Raimbault's first comment (I would have posted it as a comment, but I have not yet unlocked that privilege). Let $S$ be a thrice-punctured sphere, and let $\Gamma$ be the subgroup of $\PSL_2\mathbb{R}$ generated by $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$. We may identify $\Gamma$ with $\pi_1(S)$ and view the inclusion $\Gamma \hookrightarrow \PSL_2\mathbb{R}$ as a holonomy representation associated to the unique complete finite-area hyperbolic structure on $S$.
Here is a sequence of representations $\rho_n : \Gamma \rightarrow \PSL_2\mathbb{R}$ converging to the inclusion $\Gamma \hookrightarrow \PSL_2\mathbb{R}$ such that each of the $\rho_n$ is discrete but not faithful. We work in the upper half-plane model of $\mathbb{H}^2$. For $n\geq 3$, let $L_n$, $L_n'$ be the geodesics in $\mathbb{H}^2$ of equal Euclidean length passing through the real points $-1$, $1$, respectively, and intersecting at a purely imaginary point $z_n$ such that the angle facing $\infty$ formed at $z_n$ by $L_n$, $L_n'$ is $2\pi/n$. Let $\gamma_n \in \PSL_2\mathbb{R}$ be the clockwise "rotation" by $2\pi/n$ fixing $z_n$ (so $\gamma_n$ maps $L_n$ onto $L_n'$). Define $\rho_n$ by $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \mapsto \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \mapsto \gamma_n$. For each $n\geq 3$, the map $\rho_n$ is a holonomy representation $\pi_1(S) \rightarrow \PSL_2\mathbb{R}$ associated to an incomplete finite-area hyperbolic structure on $S$: namely, the holonomy representation associated to the thrice-punctured sphere obtained by removing the cone point from the orbifold $\rho_n(\Gamma)\backslash \mathbb{H}^2$.
You know, I also have been looking for a reference for openness in the type preserving setting and haven't been able to find it. (Actually, I'd like to have a theorem like this that works for orbifolds, even.)
At worst, I think you can prove it using the same arguments as in Weil's paper, though. Namely, Weil's proof (phrased in the $\mathbb H^2$ case) essentially involves taking a huge compact subset $K\subset \mathbb H^2$ that includes a fundamental domain of the image of a discrete, faithful representation $\rho$, letting $\Delta \subset \pi_1 S$ be a finite subset that includes all $\gamma\in \pi_1 S$ such that $\rho(\gamma)$ translates $K$ anywhere near itself, and then for $\rho'\approx \rho$, showing that $K / \rho'(\Delta)$ is a compact hyperbolic $2$-orbifold with $\rho'$-holonomy, implying that $\rho'$ is discrete and faithful. If $\rho$ has parabolics, one can instead take $K$ large enough so that it projects to a compact core of the quotient. Then $K / \rho'(\Delta)$ will be an (incomplete) hyperbolic $2$-orbifold with $\rho'$-holonomy, but since $\rho'$ is type preserving, one can ensure that the ends of this incomplete orbifold have parabolic holonomy, and hence the orbifold is contained in a complete finite volume orbifold.
Haven't really thought through the details, though.
Edit: Ah, it also follows from Theorem 1.1 in Bergeron and Gelander - A note on local rigidity, for example.
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2025-03-21T14:48:31.667716
| 2020-08-04T03:24:45 |
368270
|
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Stack Exchange
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Notion of distance between linear programs
Consider the linear programming problem
\begin{align}
\max_{x}&~c^Tx \\~s.t.~~a^Tx &\leq B~,~0\leq x_i \le1
\end{align}
where $c$ and $a$ are $n \times 1$ given non-negative vectors. $B$ is a positive constant (call it "Budget Constraint") and $x_i$'s are required to be in the interval $[0,1]$ (call it "Box Constraint"). Assume that the problem is always feasible. This is the mathematical formulation of an engineering problem we are trying to solve. Typically $n$ runs into tens of thousands for our problem. Let the dual variable corresponding to Budget Constraint be $\lambda$ and the dual variables ($n$ of them) corresponding to Box constraints be $\alpha_i$. What is important for us is the nature of $\lambda$ itself. Now, consider a slightly noisy version of the above LP problem
\begin{align}
\max_{x,\tilde{x}}&~c^Tx + \tilde{c}^T\tilde{x} \\~s.t.~~a^Tx + \tilde{a}^T\tilde{x}&\leq B~,~0\leq x_i \le1~,~0\leq \tilde{x}_i \le1
\end{align}
where $\tilde{c}$ and $\tilde{a}$ are non-negative $m \times 1$ vectors. Typically $m <<n$ and is often in hundreds. Let $\lambda_n$ denote the corresponding dual variable for Budget Constraint. In practice, we have often observed that $\lambda$ is close to $\lambda_n$. Intuitively, the problems seems to be close to each other depending on the nature of the additional terms.
Is there some literature which studies the notion of distance between such linear programs? We could only find literature that studies LP problems with noisy versions of $c$ and $a$ and not in terms of a number of variables. Our actual problem is a combination of both where the dimensionality, as well as the co-efficents, are noisy. Any pointers would be appreciated.
Why not consider the first LP as one with an expanded number of variables ${x,\tilde{x}}$ and setting some coefficients appropriately to zero? And then the second LP can be studied in the light of perturbation analysis of the first LP. Hope I got your question right.
@DSM Can you please point to references like that?
What I mentioned is just an extension of the variable space (so that the two LPs in your question are comparable), and the usual ideas in LP. You might want to look any standard references (Bertsimas) for perturbation/sensitivity analysis. As far as 'distance between LPs' is concerned, I have not come across such an idea.
@DSM most of such references assume that a small noise is added to a set of coefficients whereas in my case, they are literally zeroed out. Also, let's say $m$ is a random variable in itself, only very small in its values compared to $n$. Thus, the count of additional terms that can appear is randomized. Thus, I am not interested in a fixed $n + m$ which then reduces to $n$ variable LP via perturbation. I am interested in set of all LPs with $n+m$ variables and any notion of distance of them to the $n$ variable LP.
Okay. I think I did not note the random perturbation in the number of variables.
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2025-03-21T14:48:31.667950
| 2020-08-04T03:35:04 |
368271
|
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"Anthony Quas",
"Goulifet",
"Iosif Pinelis",
"Mark Schultz-Wu",
"R W",
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"sort": "votes",
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|
Stack Exchange
|
Discrete entropy of the integer part of a random variable
Let $X$ be a real valued random variable. Of course, the integer part $\lfloor X \rfloor$ of $X$ is a discrete random variable taking values in $\mathbb{Z}$. We can therefore define its discrete entropy
\begin{equation}
H(\lfloor X \rfloor) = - \sum_{n\in\mathbb{Z}} \mathbb{P}( \lfloor X \rfloor = n ) \log( \mathbb{P}( \lfloor X \rfloor = n ) ),
\end{equation}
which is in $[0,\infty]$ as a sum of nonnegative terms, since $- x \log x \geq 0$ for any $0 \leq x \leq 1$ (with the convention $0\log 0 = 0$).
I am looking for sufficient conditions such that $H(\lfloor X \rfloor ) < \infty$. For instance, is it sufficient to know that $X$ has a finite absolute moment $\mathbb{E}[|X|^p] < \infty$ for some $p>0$? Any condition of this type, possibly weaker, is welcome.
Motivation: There are strong connection between the differential entropy of $X$ (assuming $X$ has a pdf whose differential entropy is well-defined) and the discrete entropy of $\lfloor nX \rfloor$ when $n\rightarrow0$. This was the main topic of the 1959 paper from Alfred Rényi intitled On the dimension and entropy of probability distributions: I am questioning the assumptions under which the discrete entropy is well-defined.
Since $\lfloor X\rfloor$ has finite entropy if and only if $|\lfloor X\rfloor|$ has finite entropy, it suffices to consider random variables taking values in the natural numbers. Write $p_n$ for $\mathbb P(X=n)$ (so that $\sum_n p_n=1$). We have $X\in L^q$ if and only if $\sum p_n n^q<\infty$.
Suppose $X\in L^q$ so that $\sum p_n n^q<\infty$.
Then let $S_1=\{n\colon p_n<\frac{1}{n^2}\}$ and $S_2=\{n\colon p_n\ge \frac 1{n^2}\}$.
We have
$$
H(X)=\sum_n -p_n\log p_n=-\sum_{n\in S_1}p_n\log p_n-\sum_{n\in S_2}p_n\log p_n.
$$
Since $-t\log t$ is increasing on $[0,\frac 1e]$, the first sum is bounded above by
$$
\sum_{n\in S_1}\frac{2\log n}{n^2}<\infty.
$$
There exists an $n_0$ so that for $n\ge n_0$, $2\log n<n^q$. For $n\in S_2$ such that $n\ge n_0$, $-\log p_n<2\log n<n^q$, so that
$$
-\sum_{n\in S_2,\,n\ge n_0}p_n\log p_n\le
\sum_{n\in S_2,\,n\ge n_0}p_n n^q<\infty.
$$
Hence $H(X)<\infty$. (This trick appears in a couple of papers of mine: one with Ciprian Demeter in NYJM and another more recent preprint with Tamara Kucherenko and Christian Wolf).
That's a cute trick effectively, for a neat proof. Thanks!
@MattF: are you asking for examples where the entropy is infinite? My go-to example is $P(X=n)=c/[n(\log n)^2]$ for $n\ge2$ and with suitable normalization.
Using (say) decimal notation, ASCII encoding, and a delimiter symbol such as a space or comma, as well as the law of large numbers, one can almost surely encode $N$ independent copies of $\lfloor X \rfloor$ using $O( N {\bf E} \log( 2 + |X| ) ) + o(N)$ bits. Applying the Shannon source coding theorem, we conclude that
$$ {\bf H}( \lfloor X \rfloor ) \ll {\bf E} \log(2 + |X| )$$
which by Jensen's inequality also gives
$$ {\bf H}( \lfloor X \rfloor ) \ll_p \log(2 + {\bf E} |X|^p)$$
for any $0 < p < \infty$.
$\newcommand{\fx}{\lfloor X\rfloor}$ $\newcommand\Z{\mathbb{Z}}$
We shall prove more than requested: that $H(\fx)<\infty$ if $E\ln(1+|X|)<\infty$.
Indeed, let
$$p_n:=P(\fx=n),$$
so that
$$H(\fx)=-\sum_{n\in\Z}p_n\ln p_n.$$
Let $q\colon\mathbb R\to(0,\infty)$ be any function such that
$$\sum_{n\in\Z}q(n)=1\tag{1}$$
and
$$q(x)\le cq(\lfloor x\rfloor)\tag{2}$$
for some real $c>0$ and all real $x$.
Then by the Gibbs inequality for the Kullback–Leibler divergence between $(p_n)_{n\in\Z}$ and $(q(n))_{n\in\Z}$ we have
$$0\le KL((p_n)_{n\in\Z}||(q(n))_{n\in\Z})=\sum_{n\in\Z}p_n\ln\frac{p_n}{q(n)}=-H(\fx)+\sum_{n\in\Z}p_n\ln\frac1{q(n)},$$
whence, in view of (2),
$$H(\fx)\le\sum_{n\in\Z}p_n\ln\frac1{q(n)} \\
=\sum_{n\in\Z}\int_{[n,n+1)}P(X\in dx)\ln\frac1{q(n)} \\
\le\sum_{n\in\Z}\int_{[n,n+1)}P(X\in dx)\ln\frac c{q(x)} \\
=E\ln\frac c{q(X)}=\ln c+E\ln\frac1{q(X)}.$$
So,
$$H(\fx)<\infty\quad\text{if}\quad E\ln\frac1{q(X)}<\infty.$$
Taking here e.g. $q(x)=\frac C{(1+|x|)^2}$, where $C:=1/\sum_{n\in\Z}\frac1{(1+|x|)^2}[=\frac3{\pi ^2-3}]$, we have conditions (1) and (2) satisfied. So,
$$H(\lfloor X\rfloor)<\infty\quad\text{if}\quad E\ln(1+|X|)<\infty.$$
It follows that for any real $a>0$
$$H(\lfloor X\rfloor)<\infty\quad\text{if}\quad E|X|^a<\infty,$$
as was initially desired.
@MattF. If you restrict $\alpha$ to integers the answer is yes (Cauchy distribution). If you allow for fractional moments then this paper makes the answer seem to be "no".
@MattF. : If e.g. a random variable $X$ has the pdf $f$ given by $f(x)=\frac1{2|x|\ln^2|x|},1{|x|>e}$, then $EX_+^a=EX_-^a=\infty$ for all $a>0$, where $X_+:=\max(0,X)$ and $X_-:=\max(0,-X)$. Is this what your question is about?
Very nice. The question about the finiteness of the entropy for measures on $\mathbb Z$ with a finite logarithmic moment is more or less equivalent to that about the measures with a finite first moment on the set of words in a finite alphabet (which naturally arises in symbolic dynamics). The standard argument in the literature is the one quoted by Anthony - which I have always found somewhat artificial.
@IosifPinelis That's very nice, especially because the logarithmic moment's finiteness is a very mild condition. I have a question regarding the proof. The fact that $\sum_n p_n \log ( 1 / q(n))$, which is a sum of non-negative terms, is finite, is clear to be. But why is the Kullback-Leibler divergence well-defined, since it is the difference between two positive but possibly infinite quantities?
@Goulifet : The Kullback--Leibler divergence is well-defined (and nonnegative), because $p_n\ln\frac{p_n}{q_n}\ge p_n-q_n$ and $\sum_n(p_n-q_n)=0$, where $q_n:=q(n)$. Cf. https://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence#Properties and https://en.wikipedia.org/wiki/Gibbs%27_inequality.
@IosifPinelis I see, this makes sense. Thanks!
|
2025-03-21T14:48:31.668474
| 2020-08-04T05:02:35 |
368275
|
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|
Stack Exchange
|
How to mathematically characterize a feedback loop in ODEs?
I have a biological system that exhibits a feedback type of behavior. The diagram is a schematic of the system of ODEs. In this system, the total amount of $x_1, x_2, x_3$ is conserved; however, there are transitions between them at a rate of $m_{ij} > 0$ proportional to state $x_i$. During state $x_2$, the byproduct $y$ is produced at a rate of $\alpha > 0$ proportional to $x_2$. Additionally, $y$ degrades at a rate of $\gamma > 0$ proportional to itself. The amount of $y$ affects the transition rate from $x_2$ to $x_3$ positively. This is represented by a monotone function $f(y)$ that is bounded between $1$ and $m>1$.
Edit 2: the production of $y$ does not consume $x_2$, just the materials floating in the environment. $x_2$ acts like a machine that takes those materials to make $y$. The schematic is a simplified diagram of the transcription-translation processes for DNA-RNA-protein.
The standard biological description would categorize the process involved $y$ as a positive loop. In fact, the underlying biological process is known to exhibit a "positive feedback loop". But this does not make logical sense. My observation is that as $y$ increases, $f(y)$ increases, leading to a higher transition rate from $x_2$ to $x_3$; however, it is not clear that would lead to either an increase in $y$ or not. Furthermore, it seems that whether this is a positive or negative feedback loop relies on the relative ratios of the other transition rates. Numerically, I thought of characterizing the loop as the ratio between $\frac{dy/dt}{d^2y/dt^2}$, for instance, if $\frac{dy/dt}{d^2y/dt^2} > 0$, then I have a positive feedback loop and vice versa.
My question is: how do I characterize whether $y$ is involved in a negative or positive feedback loop?
I found a similar concept based on the Jacobian matrix that is used to characterize "qualitative stability", but I don't think it is quite the same. I have searched through many system biology/math-bio references, but I have not found such materials. If there is some standard method for such characterization, please help point me to the right references. Thanks in advance!
Edit 1: I believe the standard qualitative stability theory states that in general, stability will follow in a negative feedback loop (for instance, page 240 of Leah Edelstein-Keshet's Mathematical Models in Biology). Hence, I think it may be more precise for me to ask for a mathematical characterization of a "self-activating" or "self-inhibiting" loop involving the $y$ compartment.
Edit 3: As Alexandre Eremenko pointed out, I should explain the question and motivation from a mathematical point of view. The current theory of qualitative stability uses a linearized characterization of the feedback loop (either a self-inhibiting or a self-activating loop). In this example, this is equivalent to looking at $\partial Y'/\partial Y = -\gamma < 0$. Hence, $Y$ is concluded to be in a negative feedback loop. However, this is not the case (at least for some situations). A better description would involve the sign of $Y'/Y''$, which is better and allows for more possibilities but is still only an approximation. Thus I would like to ask if there is a way to mathematically characterize the feedback loop of $Y$ without the need for approximation.
This is a site for mathematical questions. Let me try to state your question in mathematical terms, and you tell us whether I translated it correctly or not.
Let $\mathbf{x}=(x_1,x_2,x_3,y)^T$ be a time dependent vector in $R^4$, satisfying the
differential equation
$$\mathbf{x}'=A(\mathbf{x})\mathbf{x},$$
where $A$ is the matrix
$$A=\left(\begin{array}{cccc}-m_{12}&m_{21}&0&0\\
m_{12}&-(m_{21}+m_{23}f(y)+\alpha)&m_{32}&0\\
0&m_{23}f(y)&-m_{32}&0\\
0&\alpha&0&-\gamma\end{array}\right),$$
and (I suppose) all $m_{ij}$ are positive and $f$ is some function $|f|\leq 1$. And you want to know whether $0$ is a stable equilibrium.
In fact the system decouples: the first 3 equations are independent of $y$.
Certainly the answer will depend on $m_{ij}$ and $k:=f(0)$, under some reasonable assumption about the behavior of $f$.
If $f$ is continuous, the answer depends first of all on the eigenvalues of the Jacobi matrix
$$J=\left(\begin{array}{cccc}-m_{12}&m_{21}&0&0\\
m_{12}&-(m_{21}+m_{23}k+\alpha)&m_{32}&0\\
0&m_{23}k&-m_{32}&0\\
0&\alpha&0&-\gamma
\end{array}\right).$$
A sufficient condition of stability (negative feedback) is that all these eigenvalues have negative real part, and if one of them has positive real part then the system is unstable (positive feedback).
Since one negative root of the characteristic polynomial, namely $-\gamma$
factors out, the question is reduced to stability of a cubic polynomial.
The stability criterion for a polynomial of third degree
$$\lambda^3+a_1\lambda^2+a_2\lambda+a_3$$
is $a_1>0,\; a_1a_2-a_3>0,\; a_3>0$.
Computation with Maple (if I made no mistake) shows that your system is indeed stable for all positive values parameters $m_{ij},\alpha,\gamma,k$ are positive. If some of them are allowed to be $0$, the question may depend on other properties of $f$.
Edit. In fact no computaton is necessary: the 3 times 3 upper left submatrix is a Jacobi matrix, and always has negative eigenvalues.
Since $x_1 + x_2 + x_3$ is a constant of motion (and I'd assume that the OP is interested in the domain $x_1>0, x_2>0, x_3>0$), I think that it makes sense to look for equilibria of the form $x = (a,b,c,0)$ and then stability of these, than at the origin only.
@Jaap Eldering: for the system I wrote, $H:=x_1+x_2+x_3$ is not a constant of motion since $dH/dt=-\alpha x_2$. For this reason I asked OP to confirm that I translated his problem to the system of differential equations correctly.
Thank you for your answer! My question is actually not about the stability of the system. I understand that general theory states that a negative feedback loop gives stability, while a positive one will not. My question is perhaps more biologically motivated in the sense that there is a lack of logical interpretation of the terms "positive" and "negative" feedback in a biological system. Perhaps, it is more precise for me to ask for a mathematical characterization of a "self-activating" or "self-inhibiting" feedback loop concerning $y$ in particular.
@Paichu: Let us first clarify your sttement that $x_1+x_2+x_3$ is conserved. If I translated your conditions to a system correctly, this statement is wrong. And it seems also clear from your picture that this sum decays at the rate $\alpha x_2$.
The creation of $y$ does not consume $x_2$, but rather the materials floating in the environment, which is assumed to be nondepletable. $x_2$ only serves to construct $y$. I apologize for not making this clear.
Well, I do not understand the last comment. Can you tell how to modify the system of equations (which I wrote from your explanation) to reflect what you want? You wish to remove $\alpha$ from the second line?
If we remove $\alpha$ from the second equation, then $x_1+x_2+x_3$ will indeed stay constant. But then there is evidently no "positive feedback": the orbits stay bounded for all times.
I agree with your statement and thank you for being patient. With $\alpha$ removed from the second equation, the orbit of $x_i$ is bounded. The positive feedback relates to the biological aspect of the problem. In the molecular biology community, $y$ is observed to be in a positive feedback loop (https://en.wikipedia.org/wiki/Tat_(HIV)#:~:text=In%20molecular%20biology%2C%20Tat%20is,acids%20depending%20on%20the%20subtype).
However, since the production rate of $y$ is $\alpha x_2$, a positive feedback loop would require $x_2$ to increase with $y$, which is not guaranteed in this system.
@Paichu: I do not understand what "biological aspect" is, or what is called positive feedback "in biological community". This site is for mathematicians, and if you cannot state the problem in mathematical terms, it is out of place here.
I apologize, I should have described it more clearly. I hope this will be clear. If species $x$ is involved in a positive feedback loop with itself, then increment in $x$ initiate more increment in $x$. Intuitively, I think this can be described as $x'/x'' > 0$. In the theory of qualitative stability (proof by Quirk and Ruppert, 1965), they classify the feedback using the Jacobian matrix, meaning if $\partial(x')/\partial x > 0$ then it is a positive feedback loop with itself. In a way, my question is whether there is a global theory used for this characterization (no local approximation).
|
2025-03-21T14:48:31.669037
| 2020-08-04T06:17:02 |
368279
|
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"Carlo Beenakker",
"Darth Vader",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368279"
}
|
Stack Exchange
|
Eigenvalues of product of operators
Let $A,B$ be two Trace class operators with spectral decomposition $\sum_{j\geq 1} \lambda_j \phi_j(\cdot)\otimes \phi_j(\cdot)$ and $\sum_{j\geq 1} \gamma_j \psi_j(\cdot)\otimes \psi_j(\cdot)$ respectively. Can one say anything about the eigenvalues and eigenvectors of $A^{1/2}BA^{1/2}$ maybe under further assumptions ? In particular I am curious if clean relationships exist between the eigenvalues of $A^{1/2}BA^{1/2}$ and $\lambda_j$, $\gamma_j$, for $j\geq 1$ in some special cases ?
the eigenvalues of $A^{1/2}BA^{1/2}$ are the same those of $AB$.
Are you assuming $A$ is a positive trace class operator? Otherwise $A^{1/2}$ doesn't make sense to me..
@Darth Vader Yes indeed
|
2025-03-21T14:48:31.669118
| 2020-08-04T06:41:55 |
368281
|
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"Adittya Chaudhuri",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368281"
}
|
Stack Exchange
|
When is the natural map of Tate cohomology an isomorphism?
First of all I want to say that I am not at all an expert in Group cohomology . Recently I attended a seminar where the speaker mentioned about something called Tate cohomology groups which in someway relate group homology and group cohomology in one sequence.
My question is the following:
Is there any result towards the characterization of the finite groups $G$ and $G$-modules $A$ such that the natural map $N:H_0(G,A) \rightarrow H^{0}(G,A)$ appearing in Tate cohomology is an isomorphism of groups?
(My reference is https://en.wikipedia.org/wiki/Tate_cohomology_group)
It seems interesting to me because when $N$ is an isomorphism it represents some sort of notion of duality between group homology and group cohomology at the zeroth level.
I asked this question to the speaker but did not get satisfactory answer . So I am asking here for an answer / partial answer to my question.
I apologise in advance if my question sounds stupid.
It holds, for example, if A is a projective Z[G]-module (III.1.1(c) in Brown's book 'Cohomology of Groups')
@DrewHeard Thanks for the Reference. I guess you mean Chapter 3, Section 1 in Brown's book. I have the first edition of the book with me but I could not find 1(c) in section 1 of chapter 3 there. Is it a different edition?
@DrewHeard Thanks I found it. It was in the Exercise.
|
2025-03-21T14:48:31.669237
| 2020-08-04T07:19:12 |
368282
|
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|
Stack Exchange
|
Continuity on the boundary of mixed elliptic problems
Let $B$ the interior of a Polyhedron in $\mathbb{R}^n$. Suppose $a^{ij} \in C^\infty (\overline{B})$ , $a^{ij}(x)y_i y_j \geq \lambda |y|^2$ $\forall x\in B, \ \ y\in \mathbb{R}^n$.
Set $\Gamma := \cup _i \Gamma _i $, where $\Gamma _i$ is the interior of a face of $\overline{B}$. Suppose also that $c\in L^q (B)$, $c_1 \in L^{q_1} (\partial B -\overline{\Gamma})$ for same $q$, $q_1 > n-1$. Consider the following problem
\begin{equation}
\left\{
\begin{array}{rl}
-(a^{ij}u_{x_i})_{x_j} + c u = f & \mbox{on} \ B\\
u=g & \mathrm{in\ } \Gamma\\
a^{ij}u_{x_i} \nu _j + c_1u=h & \mathrm{in\ } \partial B - \overline{\Gamma} \ \ \ \ (\nu \ \ \mbox{is a outward-pointing vector to } \ B)
\end{array}
\right.
\end{equation}
If $f\in C(B)\cap L^\infty (B) $ and $g,h \in C(\overline{B})$ then a solution of the mixed problem above is continous on $\overline{B}$ ?? Please, I would very much like references of papers or books that answer this question.
|
2025-03-21T14:48:31.669334
| 2020-08-04T07:40:59 |
368284
|
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"authors": [
"Alkan",
"Greg Hurst",
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"url": "https://mathoverflow.net/questions/368284"
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|
Stack Exchange
|
Is there a positive odd $n$ such that $\sigma(\sigma(n)) = \sigma(\sigma(n)-n)+\sigma(n)$?
Let $\sigma(n)$ denote the sum of the divisors of $n$. (https://oeis.org/A000203)
It is relatively easy to find numbers $n$ such that $f(g(n)) = g(f(n))$ where $f(n) = \sigma(n)$ and $g(n) = \sigma(n) -n$ (https://oeis.org/A291881) when $n$ is even.
Question. What is the smallest odd number $n \ge 1$ such that $\sigma(\sigma(n)) = \sigma(\sigma(n)-n)+\sigma(n)$ (if such an odd number exists) ?
If someone can show that there is no such odd $n$ or there is any reference to this spesific equation, that answer is also very welcome.
Thanks.
The only such $n$ less than $5.7\cdot10^9$ are all even: $2, 38040, 51888, 236644, 260880, 3097024, 5283852, 5667312, 11777472, 46120848, 52981252, 69128640, 121352208, 330364848, 485906400, 662736600, 769422720,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>,<PHONE_NUMBER>.$
Thanks for your computation. It is noted that it is greater than $2.10^{11}$, if such $n$ exists.
@Alkan: What is your motivation for pursuing a solution to this problem?
I am sorry for late response. I was curious about a possible connection to conjecture on existence of an odd perfect number and such nesting argument that question focuses on. So I wanted to share my observation on sequence entry and increase my knowledge on the subject. Best regards.
|
2025-03-21T14:48:31.669449
| 2020-08-04T08:12:06 |
368286
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Does the Skorokhod space with the uniform topology admit a smooth partition of unity?
Does the Skorokhod Banach space $D[0,1]$ (cadlag functions equipped with the uniform norm) admit a smooth partition of unity? I found Johanis - Smooth partitions of unity on Banach spaces, which provides several classes of Banach spaces with this property, but don't see how to relate the given criteria to the Skorokhod space.
The Skorohod space is not normed and thus not Banach.
@IosifPinelis Maybe it is a matter of notation: What I'm considering are the cadlag functions together with the topology given by the uniform norm. This is a non-separable Banach space (and as far as I know, non-separability was the main reason why Skorokhod introduced the four more common J and M topologies)
With the supremum norm topology the Skorohod space is linearly homeomorphic to $C([0,1])\oplus\ell^1([0,1])$ where neither or the factors has smooth bump functions by old results. Hence by reductio ad absurdum the answer is no, provided that the question is interpreted in the sense that every open cover has a subordinated smooth partition of unity..
@TaQ thanks for this point of view, apparently there is something I don't understand, as in the reference I quoted it is explicitly stated that $C([0,1])$ does admit a partition of unity. I will read some more and then comment further.
@TaQ I think that I understand now: A necessary condition for $C(K)$ to admit smooth partitions of unity is that $K$ be scattered (has no nonempty dense-in-itself subsets), which rules out intervals of the real line (so my previous comment was wrong). Would you mind extending your comment to an answer, so that I can accept it? In particular, could you provide some more details on the linear homeomorphism (the definition is clear, but continuity requires some work, doesn't it)?
My comment above was mistaken since $C([0,1])\oplus\ell^1([0,1])$ only continuously embeds in the Skorohod space $D$ as a proper nonclosed subspace. This in conjunction with the assumption of existence of smooth partitions of unity in $D$ does not (at least as I thought) lead to a contradiction with the results of Bonic and Frampton from 1965. See Section 14 (pp. 152−158, in particular 14.11(1)) in Kriegl and Michor's The Convenient Setting of Global Analysis (AMS 1997) for an account of bump functions.
@TaQ Thanks for your answer and the reference!
The Skorohod space $D$ has $C([0,1])$ as a Banach subspace that does not have smooth bump functions by results of Bonic and Frampton from 1965. If $D$ had smooth partitions of unity, it would posses smooth bump functions, hence also $C([0,1])$ which is false. So the answer is no.
See Section 14 (pp. 152−158, in particular 14.11(1)) in Kriegl and Michor's The Convenient Setting of Global Analysis (AMS 1997) for an account on bump functions.
|
2025-03-21T14:48:31.669637
| 2020-08-04T08:53:47 |
368287
|
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|
Stack Exchange
|
Looking for an algorithm that finds lowest cost linear subspace containing the vector
Let $v_1, v_2, \dots, v_N, u$ are vectors in $\mathbb{R}^n$, each defined by $n$ integer numbers. Typically, $n<N$, and each vector has only few (one to four) non-zero coefficients. Additionally, each of $v_1, v_2, \dots, v_N$ has a cost, an integer number $c_i > 0$. Need to find a subset $v_{i_1}, \dots, v_{i_k}$ such that
$u$ lies in the linear space spanned on $v_{i_1}, \dots, v_{i_k}$, and
The cost of the subset $\Sigma_{j=1}^kc_{i_j}$ is minimal.
Any ideas, including incomplete and/or non-optimal solutions, are very appreciated. Thanks in advance.
Given finite bounds $[L_i,U_i]$ on the multipliers $\lambda_i$, you can solve the problem via mixed integer linear programming as follows. Let binary decision variable $x_i$ indicate whether $\lambda_i \not= 0$. The problem is to minimize $\sum_i c_i x_i$ subject to
\begin{align}
\sum_i \lambda_i v_i &= u \tag1 \\
L_i x_i \le \lambda_i &\le U_i x_i &&\text{for all $i$} \tag2
\end{align}
Constraint $(1)$ forces $u$ to be in the span.
Constraint $(2)$ enforces $\lambda_i \not= 0 \implies x_i = 1$.
If $k$ is fixed, you can also impose cardinality constraint $\sum_i x_i = k$.
|
2025-03-21T14:48:31.669742
| 2020-08-04T09:00:01 |
368288
|
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"Andreas Blass",
"Taras Banakh",
"aduh",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368288"
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|
Stack Exchange
|
Is the Vietoris topology on compact subsets of $\mathbb R^n$ locally convex?
The title question says it all really.
If the question is negative for compact subsets of $\mathbb R^n$, is it affirmative for compact and convex subsets of $\mathbb R^n$? How about for all nonempty subsets of $\mathbb R^n$?
If the answer is affirmative, does the result also hold for all locally convex topological vector spaces?
Let me know if i misunderstood the question. Take a V shape $K$ on the plane, and engulf it with a thin open set $U$. The set $V={L\subset U}$ is open in the Vietoris topology. Assume that $K\in W\subset V$ is convex. But then $(K+K)/2\in W$. However, $(K+K)/2$ contains the middle between the ends of the V, and so $(K+K)/2$ is not even included in $U$.
I've encountered local convexity only as a property of topological vector spaces. Since the space of compact subsets of $\mathbb R^n$ isn't naturally a vector space, it's not clear to me what the question means. The comment by @bof uses a reasonable-looking interpretation of "locally convex" for this situation, but maybe some other interpretation was intended.
The hyperspace of compact convex sets indeed has a convexity structure as we can take Minkowski convex combinations $tA+(1-t)B={ta+(1-t)b:a\in A, b\in B}$ of compact convex sets. So, for the hyperspace of compact convex sets there is sense to speak about the local convexity. It suffices to check that the balls in Hausdorff metric around compact convex sets are convex in the sense of the Minkowski operation (which should be true).
I had in mind using Minkowski sums, as Taras Banakh suggested. I don't know why I didn't realize it when I was writing the question, but I guess the subsets should be compact and convex.
|
2025-03-21T14:48:31.669877
| 2020-08-04T09:28:17 |
368289
|
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"H A Helfgott",
"Jochen Wengenroth",
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"url": "https://mathoverflow.net/questions/368289"
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|
Stack Exchange
|
Is $A$ is small on bounded functions, is there a large subdomain on which $A$ is small?
Let $A$ be a symmetric linear operator of norm $\leq 1$ on the space of functions $f:S\to \mathbb{R}$, where $S$ is a set with $N$ elements. Define the inner product $\langle \cdot,\cdot\rangle$ by letting $\{x\}$ have measure $1/N$ for each $x\in S$.
Assume that, for every function $f$ such that $|f|_2=1$ and $|f|_\infty\leq K$, we have $|\langle f,A f\rangle|\leq \alpha$.
Does it follow that there must be a $Y\subset S$ of measure $O(1/K)$ (say) such that, for $X = S\setminus Y$ and any $f:S\to \mathbb{R}$ with $|f|_2=1$, $$|\langle f|_X, A(f|_X)\rangle|\leq 10 \alpha,$$ say?
Why doesn't this hold just by homoeneity for $Y=\emptyset $? And how do you define $A (f|_X) $? Do you put the restriction as $0$ outside $X $?
I'm not sure how homogeneity would be enough - $A$ could have eigenfunctions with enormous peaks. Sure, define $A(f|_X)$ that way if you wish -- it doesn't matter, since you are going to take the inner product with $f|_X$.
I must misunderstand something. For $t=K/|f|\infty$ one has $|tf|\infty\le K$ and hence $$\langle f, A(f)\rangle= t^{-2} \langle tf, A(tf)\rangle \le t^{-2}\alpha |tf|_2=\alpha|f|_2.$$
Yes, I stated the question wrongly. Just fixed it. Thanks!
|
2025-03-21T14:48:31.669992
| 2020-08-04T09:38:27 |
368291
|
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"Richard Garner",
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|
Stack Exchange
|
Can this ultrafilter convergence condition be expressed as a compactness condition?
Suppose that $X$ is a topological space. Let us say that an ultrafilter $\mathcal U$ on the Boolean algebra $C_X$ of clopen subsets of $X$ is partition-prime if whenever $X = \amalg_{i \in I} X_i$ is a clopen partition of $X$ we have $X_i \in \mathcal U$ for some $i \in I$.
The set $X'$ of partition-prime ultrafilters on $X$ forms a topological space under the usual Stone topology with basis $[V] = \{ \mathcal U \in X' \mid V \in \mathcal U\}$ for $V \in C_X$, and we have a continuous map $\eta \colon X \to X'$ sending $x$ to $\{ V \in C_X \mid x \in V \}$. This is clearly the reflector onto a certain full subcategory of topological spaces; the game is to figure out which one.
It is easy to see that $\eta$ is injective just when $X$ is totally disconnected, and that $\eta$ is a subspace inclusion just when $X$ is zero-dimensional Hausdorff. We would like to know when, moreover, $\eta$ is surjective, and hence a homeomorphism.
This will happen precisely when each partition-prime ultrafilter on $C_X$ is of the form $\eta(x)$ for some $x \in X$. It is easy to re-express this condition (under the assumption that $X$ is zero-dimensional Hausdorff) as saying that $\eta$ is surjective precisely when each partition-prime ultrafilter on the full powerset $\mathcal P X$ converges.
My question is whether there is a natural way of re-expressing this convergence condition as a compactness condition. The obvious guess is that this is the same as ultraparacompactness: every open cover of $X$ can be refined to a disjoint clopen cover.
It is not so difficult to show that our convergence condition is implied by ultraparacompactness. Indeed, if $X$ is ultraparacompact, and $\mathcal U$ is a completely partition-prime ultrafilter on $\mathcal P X$ with no convergent point, then the open sets not in $\mathcal U$ cover $X$; we can refine this cover to a disjoint cover $X = \amalg_i X_i$; and so $X_i \in \mathcal U$ for some $i$, contradicting the fact that $X_i \subseteq U$ for some open $U \notin \mathcal U$.
The other direction is much less clear. Suppose $X$ is zero-dimensional Hausdorff and ultraparacompact, and suppose towards a contradiction that $X = \bigcup_i U_i$ were an open cover with no disjoint refinement. Clearly it does no harm to assume that this is a cover by clopens, and the assumption of no disjoint refinement implies in particular that there is no finite subcover (since a finite cover by clopens can always be refined to a disjoint one). So the sets $V_i = U_i^c$ generate a proper filter $\mathcal F$ with no adherent point. Moreover, any clopen partition $X = \amalg_j X_j$ must fail to refine the cover $(U_i)_{i \in I}$, and so there must exist some part $X_j$ of the partition which meets each set in $\mathcal F$. Clearly, this is a necessary condition for $\mathcal F$ to be extendable to a partition-prime ultrafilter. If it were also sufficient we would be done: for then we could choose a partition-prime $\mathcal U$ extending $\mathcal F$ which, since $\mathcal F$ has no adherent point, would not converge: a contradiction.
So, my claimed characterisation of the property that every partition-prime ultrafilter converges as ultraparacompactness would follow if we could prove the following ultrafilter lemma:
If $\mathcal F$ is a proper filter such that for any clopen partition $X = \amalg_j X_j$
there is some $X_j$ which meets each set in $\mathcal F$, then there is a partition-prime
ultrafilter extending $\mathcal F$.
An obvious attempt at a transfinite argument works well at successor stages, but falls over at limit ordinals. So there remain three possibilities: this is true; this is false; this is independent of the axioms of set theory, and I don't know which of these it is.
Another way of looking at this ultrafilter lemma is via locale theory. Given any locale $L$, we can obtain a new locale $L'$ by taking sheaves on the Boolean algebra of complemented elements of $L$, for the topology whose covers are all clopen partitions. Now $L \to L'$ is a reflector into zero-dimensional ultraparacompact (aka "strongly zero dimensional") locales. Clearly the passage $X \mapsto X'$ described above sends $X$ to the space of points of the strongly zero-dimensional reflection $\mathcal O(X)'$: which would thus be ultraparacompact so long as $\mathcal O(X)'$ were spatial.
Now clearly not every strongly zero-dimensional locale is spatial (take any atomless complete Boolean algebra). But it's possible that the strongly zero-dimensional reflection of any spatial locale is spatial, and this is in fact exactly equivalent to the ultrafilter lemma I quote above.
Any ideas, either about the ultrafilter lemma, or about a different characterisation of the spaces of the form $X'$, would be gratefully received!
Thanks, yes, I meant clopen subsets. Edited to correct
Your condition amounts to saying that the uniformity $\mathcal{C}$ generated by all clopen partitions is complete: 'partition-prime' is the same as '$\mathcal{C}$-Cauchy'.
It is also implied by the condition that the space is $\mathbb{N}$-compact, which means that every clopen ultrafilter with the countable intersection property is fixed: if $\mathcal{U}$ is $\mathcal{C}$-Cauchy then it has the countable intersectionproperty; if $\{U_n:n\in\omega\}$ is a decreasing sequence in $\mathcal{U}$ with empty intersection (and $U_0=X$) then $\{U_n\setminus U_{n+1}:n\in\omega\}$ is a clopen partition with no member in $\mathcal{U}$.
In New properties of Mrowka's space $\nu\mu_0$ Kulesza showed the space $\nu\mu_0$ from the title is $\mathbb{N}$-compact; it is not strongly zero-dimensional, hance not ultraparacompact.
|
2025-03-21T14:48:31.670449
| 2020-08-04T09:51:43 |
368292
|
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"Anubhav Mukherjee",
"Henry",
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"Oğuz Şavk",
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|
Stack Exchange
|
Negative surgeries on negative knots
This question is two-fold.
The first question is rather specific: what are some small examples of negative surgeries on negative knots that give rise to the same 3-manifold? I know one class of examples coming from Borromean rings. By performing $-1/m$ and $-1/n$ surgery on two components of the Borromean rings, we get the double twist knot $K_{m,n}$ which is negative. Now, $-1/l$ surgery on $K_{m,n}$ is just $-1/l$, $-1/m$ and $-1/n$ surgery on the Borromean rings, and by the symmetry of the Borromean rings, it is the same as $-1/m$ surgery on $K_{l,n}$ and $-1/n$ surgery on $K_{l,m}$. I would like to know some other simple examples (preferably with knots with small number of crossings).
The second question is a bit vague: what is known about the class of 3-manifolds obtained as negative surgeries on negative knots? I am curious to know if there are some theorems saying this class of 3-manifolds are "nice" in some way. Any kind of input would be greatly appreciated.
what is negative knots? Also can you please elaborate your example, it is not clear to me.
@Anubhav A negative knot is a knot that has a knot diagram with only negative crossings.
To your convention, the right-handed trefoil is negative knot, right?
@Oguz Right-handed trefoil is positive.
$(-7)$-surgery on the left-handed trefoil yields the lens space $L(7,2)$ which is defined to be the $(-7/2)$-surgery along the unknot.
Similarly one can get more examples along negative torus knots producing lens spaces. Moser classified all surgeries along torus knots in [L. Moser, Elementary surgery along a torus knot, Pacific J. Math. 38 (1971), 737–745.].
In general, to find explicit examples for the first part of your question is a hard problem, sometimes impossible. Actually, it is related to the notion of cosmetic surgeries, see Ni and Wu's paper, and further articles.
You may predict conjectures or obtain obstructions due to Thurston's theorem: all but finitely many surgeries on a hyperbolic knot result in hyperbolic manifolds.
On the other hand, as Kegel said, L. Moser completely classified surgeries along torus knots as follows:
Theorem: Let $K$ be an $(r,s)$ torus knot in $S^3$ and let $Y$ be the $3$-manifold obtained by performing a $(p,q)$-surgery along $K$. Set $\sigma =rsp−q$.
(a). If $|\sigma|>1$, then $Y$ is the Seifert manifold $\Sigma(\alpha_1, \alpha_2, \alpha_3)$ over $S^2$ with three exceptional fibers of multiplicities $\alpha_1=s, \alpha_2=r$ and $\alpha_3=|\sigma|$.
(b). If $\sigma =±1$, then $Y$ is the lens space $L(|q|,ps^2)$.
(c). If $\sigma =0$, then $Y$ is the connected sum of lens spaces $L(r,s) \#L(s,r)$.
EDIT: Considering mirror symmetry of knots and following the common convention on surgeries, we have for $n \geq 1$,
$\Sigma(r,s,rsn-1)$ is obtained by $(-1,n)$-surgery along the left-handed $(r,s)$ torus knot.
$\Sigma(r,s,rsn+1)$ is obtained by $(-1,n)$-surgery along the right-handed $(r,s)$ torus knot.
Note that these are only integral homology spheres obtained by surgery on a torus knot in $S^3$.
Moser's notation is somehow non-standard. When she says $(p,q)$-surgery, she means gluing the meridian $\mu$ to $q\mu+p\lambda$, I think. Most people define this as the $(q,p)$-surgery. I am also not sure if the orientation of the lens spaces is really the one induced via the group action on $S^3$ or if she is working with the opposite one.
@MarcKegel thanks for your attention. You are right. To the common convention, what L. Moser did is $(q,p)$-surgery on the torus knot in $S^3$. I will edit the post thinking about the orientations.
|
2025-03-21T14:48:31.670705
| 2020-08-04T10:23:52 |
368297
|
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"Fedor Petrov",
"Sascha",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368297"
}
|
Stack Exchange
|
Limiting behaviour of elementary sequence
I am curious about the limiting behaviour of a certian sequence of functions
$$f_n:=\left(\sum_{k=1}^{\infty} 2^{-k} e^{i2^{k}/n}\right)^n$$
-where $i$ is the imaginary unit-to get a conjecture about the behaviour as $n$ tends to infinity I truncated the series at $1000$ (matlab does not really allow higher powers).
Then the modulus of the first $100$ items, i.e. $\{ \vert f_n \vert; n\in \{1,...,100\}\}$ of the sequence are shown here
So I find it hard to say whether the sequence tends to zero or not. Can one say analytically whether it converges or keeps on oscillating?
What is $i$ in this sum?
@FedorPetrov imaginary unit
Assume that $n$ goes to infinity so that $\lim 2^{m_0}/n=\alpha$ for certain integer $m_0$ which grows together with $n$. Then $|f_n|\to F(\alpha)$, where
$$
F(\alpha)=\exp\left(-\alpha^{-1}\sum_{t=-\infty}^\infty 2^{-t}\sin^2 2^t\alpha\right). \quad\quad(\star)
$$
Note that $F(2\alpha)=2F(\alpha)$ for $\alpha>0$, but $F$ is not constant. (Is there a short way to verify this being not constant without explicit accurate enough calculations?) So we may suppose without loss of generality that $2^{m_0}\leqslant n<2^{m_0+1}$.
Denoting $$z_n=\sum_{j=1}^{\infty} 2^{-j} e^{i2^{j}/n}$$
we get $$|f_n|^2=(z_n\overline{z_n})^n,\\z_n\overline{z_n}=\sum_{j,k=1}^\infty
2^{-(j+k)}e^{i(2^j-2^k)/n}=\sum_{j,k=1}^\infty
2^{-(j+k)}\cos\frac{2^j-2^k}n=\\
1-4\sum_{j>k}2^{-(j+k)}\sin^2\frac{2^{j-1}-2^{k-1}}n=1-\sum_{m=1}^\infty S_m,\quad S_m:=\sum_{\ell=0}^{m-1}2^{-m-\ell}\sin^2\frac{2^{m}-2^{\ell}}n.$$
Let's bound $S_m$. If $m\leqslant m_0$, i.e., $2^m\leqslant n$, we have $\sin \frac{2^m-2^l}n=\Theta(\frac{2^m-2^l}n)=\Theta(\frac{2^m}n)$ and $S_m=\Theta(\frac{2^m}{n^2})$. Thus $$\sum_{m\leqslant m_0} S_m=\Theta\left(\frac1{n^2}\sum_{m\leqslant m_0} 2^m\right)=\Theta\left(\frac1n\right).$$
If $m>m_0$, we may bound the sine by 1 that gives
$$
S_m\leqslant \sum_{\ell=0}^{m-1}2^{-m-l}=2^{1-m},
$$
thus
$$
\sum_{m> m_0} S_m\leqslant \sum_{m>m_0} 2^{1-m}\leqslant \frac4n.
$$
We get $z_n\overline{z_n}=1-\Theta(1/n)$ that is equivalent to $\log |f_n|=\Theta(1)$.
Now fix $\varepsilon>0$. We may find large $M$ such that $$\sum_{m:|m-m_0|>M} S_m<\varepsilon n^{-1}$$ from above estimates. Therefore with accuracy $\varepsilon$ the value of $n\sum_m S_m$ comes from $m$ near $m_0$. Denote $m_0=m+t$, and fix an integer $t$. We get
$$
S_m\sim 2^{1-m_0-t}\sin^2 \frac{2^{m_0+t}}n
\sim 2\alpha^{-1} n^{-1} 2^{-t} \sin^2 2^t\alpha.
$$
This gives $(\star)$.
actually the question was whether $f_n$ has a limit and I am sorry for not being more clear about that. If I understand you correctly, then you say this should not be the case, right?
Yes, it looks that the limit exists when $n/2^m$ tends to $\alpha$ and this function depending on $\alpha$ is not a constant function.
|
2025-03-21T14:48:31.670897
| 2020-08-04T10:31:42 |
368299
|
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"Hans-Peter Stricker",
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|
Stack Exchange
|
Time of peak of an SIR epidemic
I've learned some classical results on the peak and the attack rate of an idealized epidemic which evolves according to a SIR model
$\dot{s} = -\beta\cdot i \cdot s$
$\dot{i} = +\beta\cdot i \cdot s - i/\delta$
$\dot{r} = +i/\delta$
with infection rate $\beta$ and duration of infectiosity $\delta$, the basic reproduction number being $R_0 = \beta \cdot \delta$.
The classical results I have learned:
For $s(0) \approx 1$ the maximum of $i(t)$ is given by
$$i_{max} = \frac{R_0 - \ln{R_0} - 1}{R_0}$$
(see e.g. Hethcote's The Mathematics of Infectious Diseases, Theorem 2.1, p. 607)
and the attack rate $r_\infty = \lim_{t\rightarrow \infty}r(t)$ is given by
$$r_\infty = 1 + \frac{W(-R_0 e ^{-R_0})}{R_0}$$
with the Lambert W function (see e.g. here, Proposition 1.10, p. 3)
What I still haven't found is a reference for the time $t_{max}$ at which $i(t)$ reaches its maximum (i.e. $i(t_{max}) = i_{max}$) when $i(0) = i_0$ is given and $r(0) = 0$.
Find here some plots with a single patient 0 in a completely susceptible population of 10,000, i.e. $i_0 = 0.0001$. A period of 90 days is displayed. $[\delta] =$ 1 day, $[\beta] = $ 1/day.
The peaks for $\beta_1\delta_1 = \beta_2\delta_2$ – for example $[4,1.5]$ and $[6,1]$ or
$[2,1.5]$ and $[4,0.75]$ or $[2,1]$ and $[4,0.5]$ – have the same $R_0 = \beta\delta$ and thus roughly(1) the same $i_{max}$, but they differ in $t_{max}$. So even though $t_{max}$ decreases with increasing $R_0$ for fixed $\beta$ or $\delta$ (which is quite natural), it cannot depend on the product $\beta\delta$ alone (as $i_{max}$ does) but must depend also on the quotient $\beta/\delta$ or maybe the difference $\beta - 1/\delta$. (1) Roughly because of errors due to finite time steps $\Delta t$ instead of infinitesimal $dt$.
To get a better picture of the heights and timings of the peaks, these are all the curves above overlayed:
A good algebraic approximation for $t_{max}$ as a function of $\beta$ and $\delta$ would be welcome (assuming that no closed formula exists) – or simply a reference. (Consider $i_0$ a fixed parameter and $r(0) = 0$.)
Note that $\beta/\delta$ has a unit of 1/day2 and thus of an acceleration.
Seems like you fell in love with those equations and, especially, with the $I$ component of them:-). So let me try to show you how you can derive as many approximations as you want yourself, test them against simulations and (if you are lucky) find a few to your liking. Again, I'll put everything in the numerator: $\dot S=-\beta IS, \dot E=\beta IS-\lambda E, \dot I=\lambda E-\delta I, \dot R=\delta I$. I'll also normalize to $\lambda+\delta=1$ (time scaling) and denote $\rho=\lambda\delta\in(0,\frac 14]$ after such normalization. The total population will be normalized to $1$.
We shall assume that we are in the situation when the initially infected and exposed portions are very small and everyone is susceptible
Note that for a while (when $S\approx 1$), you run just a linear system on $I$ and $E$. Anything can happen here: for instance, if $E=0$, then $I$ initially goes down and there is no way its graph can match your idea of a peaking curve on that interval. We want to eliminate the decaying part of the solution from the initial data.
Fortunately, the linear theory is easy: you expect that all growth is determined by the eigenvector corresponding to the largest eigenvalue. The eigenvalues for the EI part are (under my normalization) $-0.5\pm\sqrt{0.25+(\beta-\delta)\lambda}$. So, you decompose the vector $(E,I)$ into the parts proportional to the eigenvectors and take the $I$-component of the part corresponding to the positive eigenvalie. That is your $I_{eff}$. You can now presume that starting with $I_{eff}$ and $E_{eff}$, you can run your approximate curve, whatever you decide it to be, infinite time in both directions. Thus, if you settled on some analytic curve $I(t)=F(t)$ approximating your solution that has the maximum at the origin and satisfies the equation on the whole line, then you just solve the equation $F(-t_{max})=I_{eff}$.
Now about how to find decent curves that describe the pandemic coming from $-\infty$ with $S=1,I=E=R=0$ there and going to $+\infty$ with $I=E=0$ there. First of all, determine the quantities that you know exactly. There is actually just one such quantity: the full integral $J=\int_{-\infty}^\infty I(t)dt$. It has two meanings. On one hand, $\delta J=R(+\infty)$. On the other hand, $e^{-\beta J}=S(+\infty)$. Then we get our first equation:
$$
e^{-\beta J}+\delta J=1,
$$
which can be solved in a unique way for $J>0$. So, from now on, I'll treat $J$ as a known quantity available for using in other formulae.
Next, generally speaking, we need to decide on some parametric family of curves $F_p(t)$, where $p$ is a set of parameters, that can be used for curve fitting. We need at least 2 free parameters since, after my normalization, we have $2$ degrees of freedom in the choice of $\beta,\lambda,\delta$. However, having just 2 formal parameters from the start seems too restrictive because we then need to guess everything just right. On the other hand, introducing too many parameters is also bad because we shall get too many complicated equations. So, perhaps, three or four would be a good choice. Note that we already have one nice to use relation for $p$:
$$
\int_{-\infty}^{\infty}F_p(t)\,dt=J\,.
$$
so ideally this integral should be at least approximately computable in terms of $p$.
Let's see what else we can discern before deciding on any particular kind of curve.
Write $I=\frac{I_0}{\psi}$ where $I_0$ is the maximal value attained at $0$. Then we can successively express $E$ and $S$ in terms of $\psi$ and its derivatives. This algebraic exercise yields
$$
\lambda E=\delta I+\dot I=I_0\left[\frac \delta\psi-\frac{\dot\psi}{\psi^2}\right]
\\
\beta\lambda SI=\lambda \dot E+\lambda(\lambda E)=I_0\left[\frac\rho\psi-\frac{\dot\psi}{\psi^2}-\frac{\ddot\psi}{\psi^2}+2\frac{\dot\psi^2}{\psi^3}\right]
\\
\beta\lambda S=\rho-\frac{\dot\psi}{\psi}-\frac{\ddot\psi}{\psi}+2\frac{\dot\psi^2}{\psi^2}
$$
(I used the normalization $\lambda+\delta=1$ in the process).
Now denote $B=\beta I_0$. The last equation that we haven't used (that for $S$) reads in this notation
$$
-\dddot\psi-\ddot\psi+5\frac{\dot\psi\ddot\psi}{\psi}-4\frac{\dot\psi^3}{\psi^2}=
-B\left[\rho-\frac{\dot\psi}{\psi}-\frac{\ddot\psi}{\psi}+2\frac{\dot\psi^2}{\psi^2}
\right]\,,
$$
which looks a bit scary until you realize that for every exponent $a\in \mathbb R$ (with one exception, which I leave to you to find), it has an asymptotic solution $ce^at+b+\dots$ at each infinity (the exponent is presumed to grow at the infinity here and $\dots$ stand for the decaying terms). To see it, just plug this form into the equation and find $b$ that eliminates all growth and constants (it depends on $a$, of course).
Now, it seems a good idea to presume that our actual solution of the IVP $\psi(0)=1,\dot\psi(0)=0$ also has this asymptotic form. Moreover, from the expression for $\beta\lambda S$, which can be rewritten as
$$
\rho-\frac{\dot\psi}{\psi}-\frac{d}{dt}\frac{\dot\psi}{\psi}+\frac{\dot\psi^2}{\psi^2}
$$
we can immediately see the equations for the exponents. Indeed, if $\psi\asymp e^{at}$, then $\frac{\dot\psi}{\psi}\to a$ and the derivative of it goes to $0$, so at $\pm\infty$, we have for the corresponding exponents
$$
\delta-a_{\pm}+a_{\pm}^2=\beta\lambda S(\pm\infty)
$$
and we know that $S(-\infty)=1$, $S(+\infty)=e^{-\beta J}$. So, these quadratic equations allow us to find $a_-$ as the unique negative root $0.5-\sqrt{0.25+(\beta-\delta)\lambda}$, which not surprisingly is just minus the "onset exponent" we found earlier. For $a_+$ we have two choices but it turns out that it is the smaller root that we really need (both are positive). Thus, our $F_p$ should be asymptotic to $e^{a_\pm t}$ at infinities at least approximately, which gives us two more equations for $p$. Finally, it would be nice to have our ODE satisfied at least at the point of the maximum to have the local behaviour near $0$ not too weird. This is the fourth equation. Thus, we should have at least 4 parameters not to sacrifice anything. We already have one: $I_0$, or, which is the same, $B$. It seems rather natural to take the two exponents $a_{\pm}$ as the other two, especially because we can force their values to be exactly right without much trouble. Thus, we need one more.
Here you have options. I played with a few and finally settled on the function of the kind
$$
\psi(t)=c+c_+e^{a_+t}+c_-{ea_-t}
$$
where $c\in[0,1)$ and $c_\pm$ can be immediately determined from $c$ and the conditions $\psi(0)=1,\dot\psi(0)=0$. The differential equation at the point $0$ of maximum becomes then a nice algebraic relation between $c$ and $B$ (quadratic, to be exact) once $a_{\pm}$ are known.
The nightmare equation then becomes the very first one: $\int F_p=J$. This requires to integrate $\frac 1{\psi}$ and there is no nice formula. However, there is a nice approximation: when $c=0$, we can find the full integral using residue technique, and when $a_-=-a_+$, we can find the dependence of $c$ exactly, so we just assume that it expands to the other values approximately. The quick numeric check shows that this assumption is not as idiotic as one can think, so we get the final relation, which, if you put everything together, reads
$$
B\frac{2}{\sin(\pi t)t^t(1-t)^{1-t}}
\frac 1{a_+-a_-}\frac {\arctan(\frac{\sqrt{1-2c}}c)}{\sqrt{1-2c}}=\beta J
$$
where $t=a_+/(a_+-a_-)$ (or something like that: when $c>\frac 12$, you need to replace the arctangent by the difference of logarithms: it is the same analytic function but the algebra is not programming-friendly here). Of course, you are more than welcome to experiment with other forms of the solutions.
Now, once you solve the resulting system, you can plug everythin in and see if the curve matches the simulation. Here are a few pictures: the black curve is the numeric solution of the ODE, the red curve is the fitting one (of the above kind), the green one is the best symmetric approximation, the blue line is the error in the equation (scaled in some reasonable way), the orange dot is the predicted time of maximum, the magenta horizontal line is the predicted maximum, etc.
Can one learn anything from this exercise? Honestly, I have no idea. I just wanted to show you how such things can be done, so you can try yourself. The approximate parametric form I suggested here is quite simple, but determining the parameters from $\beta,\lambda,\delta$ requires solving a few transcendental equations. The fit is pretty good though up to $\frac{\beta}{\delta}\approx 20$ uniformly in $\lambda$.
Oh my god, that's a lot of stuff to delve into. I'll do and try my best. Thanks a lot!
Still haven't found the time to digest your answer completely. Accepted it nevertheless. Thanks again.
|
2025-03-21T14:48:31.671589
| 2020-08-04T12:46:49 |
368303
|
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"Hans",
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|
Stack Exchange
|
A matrix inequality involving a singular matrix
I have three matrices $A \in \mathbb R^{n \times n}$, $B \in \mathbb R^{n \times n}$, and $X \in \mathbb R^{n \times n}$.
Suppose that $A$ is singular, $B = B^\top > 0$ and $X = X^\top > 0$.
Then, does the following inequality true?
$A^\top (A X^{-1} A^\top + B)^{-1} A \leq X$
My approach was decomposing $A$ into singular and non-singular part but, it was still unsuccessful...
By $<$ do you mean element-wise smaller or that the matrix difference is positive definite?
Both $B$ and $X$ are symmetric positive definite.
Using (say) the Jordan form, approximate $A$ by a nonsingular matrix $C_t$ so that $C_t\to A$ as $t\to0$. Then
$$C_t^\top (C_t X^{-1} C_t^\top + B)^{-1} C_t\le
C_t^\top (C_t X^{-1} C_t^\top)^{-1} C_t=X.$$
Letting now $t\to0$, we get
$$A^\top (A X^{-1} A^\top + B)^{-1} A \le X,$$
as desired.
|
2025-03-21T14:48:31.671672
| 2020-08-04T13:05:25 |
368305
|
{
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"authors": [
"David Roberts",
"https://mathoverflow.net/users/128235",
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"k.j."
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"sort": "votes",
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|
Stack Exchange
|
A morphism $F$ of algebraic stacks is quasi-compact iff $|F|$ is quasi-compact
This is 5.6.3. of Laumon, Moret-Bailly's "Champs Algeriques".
Let $S$ be a scheme, $\mathscr{X, Y}$ algebraic stacks over $S$, and $F : \mathscr{X \to Y}$ a morphism.
Then $F$ is quasi-compact (i.e., for every quasi-compact scheme $Y$ and every $Y \to \mathscr{Y}$, the stack $\mathscr{X} \times_\mathscr{Y} Y$ is quasi-compact) iff $|F|$ is quasi-compact.
($|F|$ is the canonical continuous map of topological spaces $|\mathscr{X}| \to |\mathscr{Y}|$.)
In this text, the authors assume that $S$ is quasi-separated and that all algebraic stacks are quasi-separated over the base scheme.
But I want to show this highlighted statement for non-qs base schemes and for non-qs algebraic stacks.
In the proof of "if" part, the authors says that, for quasi-compact algebraic stack $\mathscr{X, Y}$ and a morphism $F : \mathscr{X \to Y}$, $F$ is quasi-compact.
I think that this needs the quasi-separation of $S$ and $\mathscr{Y}$.
Taking a look at the Stacks Project, I couldn't find its references.
Is this holds for non qs stacks?
If so, how can I show it?
Can you remind us what $|F|$ is? Is it a map on coarse moduli spaces?
@DavidRoberts Thanks for pointing out. I edited it.
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2025-03-21T14:48:31.671786
| 2020-08-04T13:06:31 |
368306
|
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|
Stack Exchange
|
Forbidden structures for generalized hypertree width
Generalized hypertree width is a tree-width-like parameter for hypergraphs, which plays an important role in the study of constraint satisfaction problems and related areas. For its more well-known cousin, namely tree-width, there is a well-established theory detailing the "obstructions" for having small tree-width; for example, it is known that if a graph has tree-width at least $k$ then it contains as a minor the $f(k) \times f(k)$ grid, for $f(k)$ which grows with $k$ (this is the Excluded Grid Theorem).
My question is as follows: is there a similar theory for generalized hypertree width? In the paper Adler, I.; Gottlob, G.; Grohe, M. (2007). Hypertree width and related hypergraph invariants. European Journal of Combinatorics, Vol. 28(8), 2167-2181 I found some generalizations to the setting of hypertree decompositions of notions from graph minor theory (such as brambles and tangles), but nothing as developed as the Excluded Grid Theorem. What is the current state of research/knowledge on this issue?
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2025-03-21T14:48:31.671887
| 2020-08-04T13:23:03 |
368308
|
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|
Stack Exchange
|
Hilbert scheme with essential generators
Let $H$ be the Hilbert scheme of finite flat lci subschemes in $\mathbb{A}^n_k$ of degree $d$, where $k$ is some field. In other words the set of morphisms from an affine $k$-scheme $Spec(A)$ to $H$ consists of surjections $A[t_1, \dots, t_n] \to B$ (up to the evident notion of isomorphism) with kernel generated by $n$ elements, such that $A \to B$ is finite locally free of degree d. (One knows that $H$ is smooth of dimension $nd$.) Let $U \subset H$ denote the subfunctor consisting of those maps $A[t_1, \dots, t_n] \to B$ such that $A[t_1, \dots, t_{n-1}] \to B$ is surjective (i.e. the last generator is unnecessary). I believe this defines an open subscheme of $H$ (essentially by Nakayama's lemma).
Here is my question: Write $Z \subset H$ for the (reduced) closed complement of $U$ in $X$. What can be said about $Z$? In particular, what is its codimension?
Regarding the title: Naively, $Z$ consists of those maps $A[t_1, \dots, t_n] \to B$ such that $A[t_1, \dots, t_{n-1}] \to B$ is not surjective. In other words, the last generator is essential.
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2025-03-21T14:48:31.671996
| 2020-08-04T14:03:10 |
368309
|
{
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|
Stack Exchange
|
$H^{p,q}(X)$ versus $H^{q}(X, \bigwedge^p TX)$
Let $X$ be a Kahler manifold. To $X$ one can associate the cohomology groups $H^{p,q}(X)$, and $H^{(0,q)}(X, \bigwedge^p TX)$ with $TX$ being the holomorphic tangent bundle of $X$.
Is there a general relationship between these two cohomologies?
For instance, it is claimed in a footnote on page 30 in Hori, Iqbal, and Vafa - D-branes and mirror symmetry, that the Euler characters of the two agree up to a sign. It would be nice to know the origin of this.
Second we know that when $X$ is Calabi–Yau the two cohomologies match. For the more general case when $X$ is Kähler it would be good to know a precise way to measure the mismatch.
Just to check, you understand that $H^{p,q}(X)$ is $H^q(X, \bigwedge^q T^{\ast} X)$, where $T^{\ast} X$ is the cotangent bundle, and your question is what happens if we use the tangent bundle $TX$ instead of the cotangent bundle?
@David E Speyer Yes
By Serre duality $H^q(X,T)= H^{n-q}(X, \Omega^1\otimes \omega)^*$, where $\omega$ is the canonical bundle, and something similar holds for other exterior powers. So Euler characteristic match up to sign in the CY case, but I don't see how they would in general.
By the way, as @DavidESpeyer's comment indirectly indicates, the operator symbol for the alternating product is $\bigwedge$ \bigwedge, not $\Lambda$ \Lambda. I have edited accordingly.
If you combine those vector spaces in the correct way you get the Hochschild-Kostant-Rosenberg decomposition for Hochschild homology (resp. Hochschild cohomology), and the Euler characteristics of those agree (up to a sign); see e.g. Example 3.12 in Polishchuk's Lefschetz type formulas for dg-categories (where a direct geometric argument is also given in fact).
For $X=\mathbb{P}^1$, the only nonzero Hodge numbers are $h^{0,0}=h^{1,1}=1$; but $\dim H^0(TX)=8$.
@abx You mean $3$? And so? I know they don't agree.
Yes, sorry for the typo. If you know, what do you ask exactly?
@abx The Euler characteristic of Hochschild homology is 2 (there is only something in degree 0), the Euler characteristic of Hochschild cohomology is also (up to a sign) 2 (because 1-dimensional in degree 0, and 3-dimensional in degree 1).
@abx, my (admittedly vague) question is just asking if we know some sort of precise relationships between the two cohomologies. A spectral sequence maybe?
One reason that $H^p(X, \bigwedge^q TX)$ will not be as well behaved as $H^p(X, \bigwedge T^{\ast} X)$ is that it is not deformation invariant, and thus not topological. In other words, if we have a connected base $B$, and a flat projective family $\mathcal{X}$ over $B$ with smooth fibers, then will have the same Hodge numbers. This will not be true for $H^q(X, \bigwedge^p TX)$. Note that, in particular, this means that $H^q(X, \bigwedge^p TX)$ is not topological, since all fibers of such a family will be diffeomorphic.
Let $X_1$ be $\mathbb{P}^2$ blown up at $3$ general points, and let $X_2$ be $\mathbb{P}^2$ blownup at $3$ collinear points. I claim that $H^0(X_1, T) \cong \mathbb{C}^2$, but $H^0(X_2, T) \cong \mathbb{C}^3$. The surfaces $X_1$ and $X_2$ are diffeomorphic, and it is easy to make a smooth flat projective family over $\mathbb{A}^1$ with most fibers $X_1$ and some fibers $X_2$. This seems like bad news.
Recall that $H^0(\mathbb{P}^2, T)$ is $8$-dimensional. Writing $z_1$, $z_2$, $z_3$ for the homogenous coordinates on $\mathbb{P}^2$, it is spanned by $z_i \tfrac{\partial}{\partial z_j}$, modulo the relation $\sum z_k \tfrac{\partial}{\partial z_k}=0$. If we write a section of $T\mathbb{P}^2$ as $\sum A^i_j z_i \tfrac{\partial}{\partial z_j}$, then this section vanishes at the point $(x_1 : x_2 : x_3)$ if and only if $(x_1, x_2, x_3)$ is an eigenvector of the matrix $A^i_j$.
Let $U_j$ be the locus of $\mathbb{P}^2$ away from the points which are blown up in $X_j$. So $U_j$ is an open subset of both $\mathbb{P}^2$ and $U_j$, and we can restrict sections of the tangent bundle from the complete surfaces to $U_j$. Since $U_j$ is $\mathbb{P}^2$ remove a codimension $2$ locus, $H^0(\mathbb{P}^2, T) \cong H^0(U_j, T)$. On the other hand, I believe that a section $\sigma$ of $H^0(U_j, T)$ will extend to a section of $H^0(X_j, T)$ if and only if, considering $\sigma$ as a section on $\mathbb{P}^2$, that section vanishes at the blown up points (see computation below).
Thus, $H^0(X_j, T)$ will be sections $\sum A^i_j z_i \tfrac{\partial}{\partial z_j}$, subject to the condition that the matrix $A$ has eigenvectors at the blown up points, and modulo the relation $\sum z_k \tfrac{\partial}{\partial z_k}=0$.
If we blow up the points $(1:0:0)$, $(0:1:0)$, $(0:0:1)$, we are requiring that the matrix $A$ be diagonal. So $H^0(X_1, T)$ is matrices of the form $\left[ \begin{smallmatrix} \ast&0&0 \\ 0&\ast&0 \\ 0&0&\ast \\ \end{smallmatrix} \right]$ modulo scalar multiples of the identity. Three general points in $\mathbb{P}^2$ are equivalent, up to $PGL_3$, to these three points, so this is the general case.
On the other hand, if we blow up $(1:0:0)$, $(0:1:0)$ and $(1:1:0)$, then we are requiring that $(\ast:\ast:0)$ be an eigenspace of $A$. This means that we are looking at matrices of the form $\left[ \begin{smallmatrix} \lambda&0&\ast \\ 0&\lambda&\ast \\ 0&0&\ast \\ \end{smallmatrix} \right]$ , modulo scalar multiples of the identity. This vector space is one dimension larger.
I had trouble finding a reference for the claim I made about vector fields extending to the blow up iff they vanish at they blown up point, so here is a proof: The statement is local, so I'll check it in the affine plane. A vector field on $\mathbb{C}^2$ corresponds to a derivation from $\mathbb{C}[x,y]$ to itself. Namely, the vector field $f(x,y) \tfrac{\partial}{\partial x} + g(x,y) \tfrac{\partial}{\partial y}$ gives the unique derivation with $D(x) = f$ and $G(y) = g$.
Such a vector field extends to the plane blown up at $(0,0)$ if and only if the derivation maps the rings $\mathbb{C}[x,y/x]$ and $\mathbb{C}[x/y,y]$ to themselves. I'll do the computation for the first case; it is enough to compute the derivation on the generators of the ring.
We have $D(x) = f \in \mathbb{C}[x,y] \subset \mathbb{C}[x,y/x]$, so there is no problem there. We then have
$$D(y/x) = \frac{x D(y) - y D(x)}{x^2} = \frac{x g(x,y) - y f(x,y)}{x^2}.$$
Writing $f(x,y) = \sum f_{ij} x^i y^j$ and $g(x,y) = \sum g_{ij} x^i y^j$, we have
$$\frac{x g(x,y) - y f(x,y)}{x^2} = \frac{g_{00}}{x} - \frac{f_{00} y}{x^2} + \dots$$
where the ellipses are terms that are in $\mathbb{C}[x,y] \langle 1, y/x, y^2/x^2 \rangle \subset \mathbb{C}[x,y/x]$.
So the derivation takes $\mathbb{C}[x,y/x]$ to itself if and only if $f_{00} = g_{00} = 0$, as desired.
Maybe I should rewrite my first paragraph to stress lack of deformation equivalence more strongly.
@crispr See if you like this version better. Thanks for the feedback!
Like @crispr, I'm confused by the first paragraph. As I read it, it's phrased in a way that seems to imply that Hodge numbers are topological invariants. But they aren't ( https://mathoverflow.net/questions/42744/diffeomorphic-k%C3%A4hler-manifolds-with-different-hodge-numbers ). Your argument, if I understand it well, should be: while $H^p(X,\wedge^q T^*_X)$ is deformation invariant, $H^p(X,\wedge^q T_X)$ is not as well behaved because it's not deformation invariant.
|
2025-03-21T14:48:31.672824
| 2020-08-04T14:20:18 |
368312
|
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|
Stack Exchange
|
Conjugation of bosonic and fermionic
We use the notation from semi-infinite wedge formalism $\bigwedge^{\infty/2}V$ with vector space $V$ generated by $$\left\{\underline{s}\mid s \in \mathbb{Z}+\frac12\right\}$$,
we consider the charge zero subpsace where
$v_{\lambda}$ from the basis for partition $\lambda$.
$\psi$ is the fermionic operator and $\psi^{*}$ is conjugate. $:\psi_{k-n}\psi_{k}^{*}:$ is normal ordering.
We have the following operators on the charge zero subpace.
$$\mathcal{E}_{n}(z):= \sum_{k\in \mathbb{Z}+\frac12}\exp(z-\mu /2):\psi_{k-n}\psi_{k}^{*}: $$
$$\mathcal{F}_n := \sum_{k\in \mathbb{Z}+\frac12}\frac{k^n}{n!}:\psi_{k-n}\psi_{k}^{*}: $$
$$\alpha_n := \mathcal{E}_n (0)$$
Let $p_{r}(\lambda):=\sum_{i=1}^{\infty} (\lambda_i -i+\frac12)^r-(-i+\frac12)^r $ which are eigen value of $\mathcal{F}_r$
For $r=2$ It can be shown that
$$\exp({s\mathcal{F}_2})\alpha_{-n}\exp({-s\mathcal{F}_2})= \mathcal{E}_{-n}(-ns) $$
what will be the conjugation for
$$\exp({s\mathcal{F}_r})\alpha_{-n}\exp({-s\mathcal{F}_r})= ? $$
in terms of $\mathcal{E}_n$? How can it be computed?
What do ", $v_\lambda$ from the basis for partition $\lambda$" and "$\wedge^{\infty/2}V$" (which should probably be $\bigwedge^{\infty/2}V$, but that's not my issue), each on its own line, mean?
Yeah I have done some edit to make more sense.
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2025-03-21T14:48:31.672937
| 2020-08-04T15:23:57 |
368317
|
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|
Stack Exchange
|
Gelfand pair, weakly symmetric pair, and spherical pair
I am a bit confused with the relations among Gelfand pairs, weakly symmetric pairs, and spherical pairs defined in the book "Harmonic analysis on commutative spaces" written by professor Joseph A. Wolf.
For convenience, let me recall the definitions in this book, and just consider connected groups $G$.
Definition 1 Let $G$ be a connected Lie group, and $K$ a compact subgroup. If the algebra $L^1(K\backslash G/K)$ is commutative under convolution, then $(G,K)$ is called a Gelfand pair.
Definition 2 Let $G$ be a connected Lie group, and $K$ a compact subgroup. If there exists an automorphism $\sigma$ of $G$ such that $\sigma(g)\in Kg^{-1}K$ for all $g\in G$, then $(G,K)$ is called a weakly symmetric pair.
Definition 3 Let $G$ be a complex reductive linear algebraic group, and $H$ a reductive subgroup. Denote by $\mathfrak{g}$ and $\mathfrak{h}$ the Lie algebras of $G$ and $H$ respectively. If there exists a borel subalgebra $\mathfrak{b}$ in $\mathfrak{g}$ such that $\mathfrak{b}+\mathfrak{h}=\mathfrak{g}$, then $(G,H)$ is called a spherical pair.
Now on page 281 of Wolf's book, there are two results: Theorem 12.6.10 and Theorem 12.6.11.
Let $G_\mathbb{C}$ be a connected complex reductive algebraic group, and $H_\mathbb{C}$ a reductive algebraic subgroup. Suppose that $G$ is a real form of $G_\mathbb{C}$ such that $H:=G\cap H_\mathbb{C}$ is a compact real form of $H_\mathbb{C}$. Then $(G_\mathbb{C},H_\mathbb{C})$ is a spherical pair if and only if $(G,H)$ is a weakly symmetric pair (by Theorem 12.6.10) if and only if $(G,H)$ is a Gelfand pair (by Theorem 12.6.11).
Thus, suppose that we have a real reductive group $G$ with its compact subgroup $K$, then $(G,K)$ is a Gelfand pair if and only if $(G,K)$ is a weakly symmetric pair. But I do not think that the two definitions are equivalent. As far as I know, weakly symmetric pairs are Gelfand pairs, but there exist Gelfand pairs which are not weakly symmetric pairs.
Hence, I think that I am probably misunderstanding the definitions, the theorems, or the relations among three pairs. I shall be grateful if experts may give any comments.
@LSpice Thank you for the comment. So, do you mean that, for a real reductive group $G$ and its reductive subgroup $K$, $K=G\cap K_\mathbb{C}$ does not always hold?
I deleted some ill informed guesses (to one of which @Hebe's comment was responding). I think that the answer is that you have understood these implications correctly, and that the subtleties you expect occur only in the non-reductive case.
I like this question! I originally thought it was much less subtle, and so posted some ill informed guesses in the comments.
Although the notions of "weakly symmetric" and "Gelfand pair" differ in general (Lauret - Commutative spaces which are not weakly symmetric), they coincide for reductive groups. This is Theorem 7.3 of Akhiezer and Vinberg - Weakly symmetric spaces and spherical varieties.
Thank you so much for the answer and the references!
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2025-03-21T14:48:31.673134
| 2020-08-04T15:27:03 |
368318
|
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|
Stack Exchange
|
Integral convergence with two sequences of functions
I came across this theorem just stated but has not proved and marked by 'it is easy to see'.
Theorem If $u_m$ and $v_m$ converges to $u$ and $v$ in $L^2([0,T];H^1(\Omega))$ weakly and $L^2([0,T];L^2(\Omega))$ strongly, then for $w\in C(\overline{[0,T]\times \Omega})$,
$$\int_{0}^{T}\int_\Omega v_mwu_mdxdt\xrightarrow[m\rightarrow \infty]{}
\int_{0}^{T}\int_\Omega vwudxdt$$
where $\Omega\subset \mathbb{R}^n$ and $f\in L^2([0,T];L^2(\Omega))$ means that $$\int_{0}^T\|f(t)\|^2_{L^2(\Omega)}dt<\infty$$
and
$$\int_{0}^T\|f(t)\|^2_{L^2(\Omega)}dt=\int_{0}^{T} \int_\Omega |f(t,x)|^2dxdt.$$ And we use analogous definition for $L^2([0,T];H^1(\Omega))$.
The books says it is easy to see but I cannot come up with any useful tool to prove this. I guess the peculiar definition of $L^2([0,T],L^2(\Omega))$ would not be big matter. Does anybody know any idea on it?
I thought it is better to add the book where I saw this theorem. The question I asked is refined version of this following theorem.
The title of this book is 'Navier-Stokes Equations Theory and Numerical Analysis' written by Roger Temam. And it is page 289.
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2025-03-21T14:48:31.673248
| 2020-08-04T15:33:43 |
368321
|
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|
Stack Exchange
|
Uniqueness of solution to Cauchy problem with quadratic nonlinearity
Consider the non-linear differential operator
$$\mathfrak{L}: \ C^2((0,T)\times\mathbb{R}^2)\ni\varphi\equiv\varphi(t,x,y) \, \mapsto \, \partial_x^2\varphi + (\partial_x\varphi)^2.$$
For $U\subset\mathbb{R}^2$ open and bounded with smooth boundary $\Gamma:=\partial U$ and $T>0$, can you guarantee that the (smooth classical) solution to the Cauchy-problem
$$\left\{\begin{aligned}\label{lem:ContrastiveDiffusions:eq2}
\partial_t\varphi \ &= \ \mathfrak{L}\varphi\quad &&\text{in } \ (0,T)\times U,\\
\partial_x\partial_y\varphi \ &= \ 0 \quad &&\text{on } \ \{T\}\times\Gamma,
\end{aligned}\right.$$
with initial condition $\lim_{t\rightarrow 0+}\varphi = \delta(x-y)$, is unique?
(Given that the answer is not immediate, references to the literature are appreciated.)
Do you really want the boundary condition only at t=T? And I suppose you also want an initial condition?
Thanks @MichaelRenardy. The existing solution $\varphi$ that I found is a probability density (modelling the transition probability of a stochastic system to jump from state $x$ to state $y$ in $t>0$ units of time). Yes, I want the boundary at $t=T$ only.
Writing $\phi=\log v$ leads to the heat equation $v_t=v_{xx}$ but the boundary condition looks more complicate.
|
2025-03-21T14:48:31.673365
| 2020-08-04T17:11:44 |
368325
|
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|
Stack Exchange
|
Status of a conjecture of C.T.C. Wall?
In Wall's paper Unknotting tori in codimension one and spheres in codimension two, he states the following conjecture:
Any $h$-cobordism of $S^3 \times S^1$ to itself is diffeomorphic to $S^3 \times S^1 \times I$.
What is the status of this conjecture?
The conjecture has been solved. This is Theorem 16.1 in C.T.C. Wall. (1999). Surgery on Compact Manifolds, Second edition. Mathematical Surveys and Monographs,
Vol. 69. Here, two proofs of this theorem are given: the first one is an application of a more general method presented in the book, while the second one relies on some geometrical properties of $S^3 \times S^1$.
|
2025-03-21T14:48:31.673454
| 2020-08-04T17:18:03 |
368326
|
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|
Stack Exchange
|
$\omega_1$-approximation property for Sacks iteration— contradiction in literature?
The following is a folklore result.
Suppose $P$ is a countable support iteration of nontrivial forcings, $\langle P_\alpha, \dot{Q}_\alpha : \alpha < \omega_1 \rangle$. Then there is a complete embedding of $\mathrm{Add}(\omega_1)$ into $P$.
The forcing to add a Cohen subset of $\omega_1$ fails the $\omega_1$-approximation property, since it produces a “fresh” sequence— a sequence such that all initial segments are in the ground model.
In the 1979 paper, “Iterated perfect-set forcing,” Baumgartner and Laver seem to make a contrary claim. Lemma 6.2 states that the countable support iteration of Sacks forcing produces no fresh sequences of length some ordinal of uncountable cofinality. This is key to their argument that iterating Sacks forcing up to a weakly compact forces the tree property at $\omega_2$.
I do not see a flaw in their argument. Is the folklore claim correct? How is this resolved?
I've not seen that "folklore result" before - does it actually appear used in the literature?
@NoahSchweber It’s stated as Remark 0.3 in Goldstern’s “Tools for your forcing construction.”
Goldstern was an idiot. (But that was long ago, perhaps he has learned something since then?)
This is similar to "aren't the csi and the csp of ($\omega$-) Cohen forcing the same?". This holds true in the finite support case. But in the countable case, one is proper and the other collapses the continuum.
I don't see why the "folklore result" holds. The analogous result for finite support iteration is as follows.
$\textbf{Fact}$: If $\langle P_i, Q_j: i \leq \alpha + \omega, j < \alpha + \omega \rangle$ is a finite support iteration of non-trivial forcings, then $P_{\alpha + \omega}$ adds a Cohen real over $V^{P_{\alpha}}$.
Its proof goes as follows. WLOG, $\alpha = 0$. Fix $q_k, r_k \in V^{P_{\alpha + k}}$ such that $\Vdash_{P_{\alpha + k}} q_k, r_k \in Q_k \wedge q_k \perp_{Q_k} r_k$. Define $c \in V^{P_{\omega}} \cap 2^{\omega}$ by $c(k) = 1$ iff $(\exists p \in G(P_{\omega}))(p(k) = q_k)$. To see that $c$ is Cohen over $V$, suppose $D \in V$ dense in $2^{< \omega}$ and $p \in P_{\omega}$. Fix $n$ such that $p \in P_n$. Extend $p$ to $p' \in P_n$ in $n$ steps such that for each $k < n$, $p \upharpoonright k \Vdash_{P_k} (p(k) \perp_{Q_k} q_k) \text{ or } (p_k \leq_{Q_k} q_k)$. Let $s \in 2^n$ be determined by this $p'$. Choose an extension $s'$ of $s$ in $D$. The rest should be clear.
In the case of countable support iteration, the passage from $p$ to $p'$ could be problematic since one might have to extend $p$ infinitely many times. This also shows that the folklore result would work if we iterate countably closed forcings. In any case, Baumgartner-Laver result implies that the folklore result is false. The closest related result about countable support iteraton is the following. Any countable support iteration of non-trivial forcings of length $\omega_1$ collapse the continuum to $\omega_1$. This is the Sublemma inside Lemma 6.3 in the Baumgartner-Laver paper.
In fact this fails in the following example. Let $c\in 2^{\omega_1}$ be the name defined analogously to the case of finite support iteration. Let $Q_0$ be Cohen forcing, adding a Cohen real $x\in 2^\omega$. Now let $p\in P_\omega$ be defined as follows: $p(0)=$ trivial, and $p(n+1)$ is a name for ``if $x(n)=0$, then $q_n$, else $r_n$.'' -- Clearly [don't believe Goldstern if he says "clearly"!] $p$ forces that $c$ is not fresh, as its initial $\omega$-segment $x$ is not in $V$. - See also Johannes Schürz' more general answer.
Exactly. The rumour (not folklore ;) ) is even wrong if you iterate Cohen forcing (on $\omega$ !!) with countable support $\omega_1$ many times. Let $P$ denote the iteration of Cohen forcings. It follows that any complete embedding $F$ from $\text{Add}(\omega_1)$ into $P$ is an independent permutation of coordinates and $0$ and $1$'s. Indentify $2^\omega$ with the corresponding max antichain in $ \text{Add}(\omega_1)$ and wlog assume that for every $x \in 2^\omega$ it holds that $F(x)= \inf_{n \in \omega} F(x \restriction n)$. But then $F[2^\omega]$ has a uniform, countable support $A$. Wlog let $A=\omega$. But there is $p\in P$ such that $p$ codes the first Cohen-real horizontally into $p\restriction \omega\, (\cdot)\, (0)$. As the Cohen-real differs from any ground model real, the embedding cannot be complete.
Wanted to post this as a 'sketchy' comment, but turned out to be too long...
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2025-03-21T14:48:31.673776
| 2020-08-04T18:07:12 |
368328
|
{
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}
|
Stack Exchange
|
Model theory and logic of notions of programming methodology
Linear Programming corresponds to first order theory of reals with addition and order. What theory do notions such as semidefinite programming, second order cone programming and convex programming correspond to? (I am interested in quantified statements in addition to $\exists\forall$ statements).
You'll have to explain more clearly what you mean by "corresponds to". Should the answer be a first-order theory? Certainly there are first-order languages on the reals which are capable of expressing these kinds of programming problems, but it sounds like you're hoping for a tighter connection... can you make this more precise?
I am just looking for general framework to view these. And a more precise connection. If I know how to express my thoughts properly I would not be posting this. Sorry. It is unclear to me but I think there is something more deeper.
I'm torn between finding that this question needs details or clarity, and so joining the close voters, and finding it interesting even in its current incompetely posed form. I think MO has historically demonstrated tolerance for interesting "first give the right form of the question, and then answer it" questions of this type, so I came down on the side of upvoting—because I'd rather see if somebody can give an interesting answer to this question than keep MO formally tidy by removing it.
I've simultaneously upvoted and voted to close as unclear; I think there's probably a very good question lurking here, and so I want to encourage it, but at present I don't think it admits a reasonable answer.
@MattF. This is same as Alex Kruckman's comment.
(Perhaps) I am not worried about time complexity here but taken recommendation.
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2025-03-21T14:48:31.673924
| 2020-08-04T18:56:38 |
368330
|
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Stack Exchange
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Relation between $\mathbb{A}^1$-homotopy theory and derived algebraic geometry
I've often heard that one of the benefits of derived algebraic geometry, next to a cleaner intersection theory, is that "provides natural settings " for the $\mathbb{A}^1$-homotopy theory (the quote comes from Wikipedia's article on Derived Algebraic Geometry https://en.wikipedia.org/wiki/Derived_algebraic_geometry). However, I have not found an introduction to this online. Could someone explain how derived schemes show up in $\mathbb{A}^1$-homotopy theory and/or point me to a nice introduction to this?
The only source I know of which explains explicitly how to interpret derived geometry within motivic homotopy theory is the work of Adeel Khan quoted in the wikipedia article. This is not old enough to be part of a text book.
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2025-03-21T14:48:31.674014
| 2020-08-04T20:41:44 |
368337
|
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"warzasch"
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Stack Exchange
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Every immersion can be deformed to have only transverse self-intersections
I asked this question some time ago in MSE, but obtained no answer. Maybe this is the right place to post it.
Let $f : M^n \to \overline{M}^{n+k}$ be an immersion between smooth manifolds. Is it true that there exists a smooth map $F : M \times [0,1] \to \overline{M}$ such that the following conditions hold?
$F_0 = f$;
$F_t : M \to \overline{M}$ is an immersion for every $t \in [0,1]$;
$F_1(M)$ has only transverse self-intersections.
(Here, $F_t(p) = F(p,t)$ for every $(p,t) \in M \times [0,1]$.)
If this does not hold in this full generality, is it true for hypersurfaces ($k=1$)? For $\overline{M} = \mathbb{R}^{n+k}$?
Have you seen this, and the references therein? https://www.ams.org/journals/tran/1969-136-00/S0002-9947-1969-0234480-9/S0002-9947-1969-0234480-9.pdf . I’m on mobile right now...
The statement follows easily from Sard's lemma.
Suppose $k=\dim N-\dim N$.
Cover $M$ by a countable collection of charts $s_i\:U_i\to M$ such that $f\circ s_i$ extends to embeddings $U_i\times \mathbb{R}^k\to N$.
Applying Srad's lemma you can remove bad self-intersection from each chart.
Combining these perturbations you get the result.
One has to be bit careful not to make new bad self-intersection, but is can be easily achieved by making smaller-and-smaller perturbations in each chart.
The proof is very similar to Thom transversality theorem, and I am sure it is well known to someone.
Is it following by the same argument that these intersections can be taken to be double points?
@warzasch sure $ $ $ $ $ $ $ $ $ $
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2025-03-21T14:48:31.674142
| 2020-08-04T21:06:43 |
368338
|
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|
Stack Exchange
|
Basis for space of continuous, surjective monotone functions on $\mathbb{R}$
$\DeclareMathOperator\CM{CM}$
I recently came across Okhezin - Study of families of monotone continuous functions on Tychonoff spaces describing monotone functions on general topological spaces and I wonder the following more basic question. Let $\CM^+(\mathbb{R})$ be the set of continuous, surjective, and strictly monotone functions (actually these must be homeomorphisms) from $\mathbb{R}$ to $\mathbb{R}$.
Has anyone come across any literature describing a basis of the infinite-dimensional space $\CM^+(\mathbb{R})$?
$\operatorname{CM}^+(\mathbb R)$ is not a vector space. For example, it doesn't include $0$; but, even if you toss that in, it's not closed under addition.
$\newcommand\R{\mathbb R}$$\newcommand\LBV{\mathrm{LBV}}$As was noted in comments, the set of all monotone functions (or your version of it,
$\mathrm{CM}^+(\R)$) is not a linear space.
An appropriate linear space here is $\LBV$, the space of all (say) right-continuous functions from $\R$ to $\R$ of locally bounded variation — because any function of locally bounded variation is the difference between two nondecreasing functions.
Take now any $f\in\LBV$. Let $\mu_f$ be the corresponding signed Lebesgue–Stieltjes measure, defined by the formula $\mu_f((a,b])\mathrel{:=}f(b)-f(a)$ for all real $a$ and $b$ such that $a<b$. It is easy to check that for all real $x$ we have
$$f(x)=f(0)+\int_{(0,\infty)}\mu_f(du)\,1(x\ge u)-\int_{(-\infty,0]}\mu_f(dv)\,1(x<v);$$
that is,
$$f=f(0)\,1+\int_{(0,\infty)}\mu_f(du)\,1_{[u,\infty)}-\int_{(-\infty,0]}\mu_f(dv)\,1_{(-\infty,v)}.$$
Thus, any $f\in\LBV$ is the limit of linear combinations of the following, say basic, functions: (i) the constant $1$, (ii) the functions of the form $1_{[u,\infty)}$ for real $u>0$, and (iii) the functions of the form $1_{(-\infty,v)}$ for real $v\le0$. In this sense, one might want to say that these basic functions form a basis of $\LBV$.
$\big($Because $1_{(-\infty,v)}=1-1_{[v,\infty)}$ for all real $v$, we have another, equivalent set of the basic functions of $\LBV$, consisting of the following functions: (i) the constant $1$ and (ii) the functions of the form $1_{[u,\infty)}$ for all real $u$.$\big)$
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2025-03-21T14:48:31.674294
| 2020-08-04T23:47:00 |
368343
|
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|
Stack Exchange
|
Whitehead Theorem in $\mathbb{A}^1$-homotopy theory
I'm reading "UNSTABLE MOTIVIC HOMOTOPY THEORY" by Kirsten Wickelgren and Ben Williams (https://arxiv.org/pdf/1902.08857.pdf). There they have a version of Whitehead's Theorem, namely Prop 2.3, which says that $f:\mathcal{X}\rightarrow \mathcal{Y}$ is a weak equivalence if and only if the morphisms on homotopy sheaves $\pi_n^{\mathbb{A}^1}(\mathcal{X})\rightarrow \pi_n^{\mathbb{A}^1}(\mathcal{Y})$ are bijections for all basepoints of $\mathcal{X}$.
Naively, one would expect that Whitehead's Theorem should show that weak equivalence and equivalence are the same for a "nice" subset of spaces. This seems to show that the name "weak equivalences" for schemes up to $\mathbb{A}^1$-homotopy is well-choosen. This confuses me. Is the notion of weak equivalence in $\mathbb{A}^1$-homotopy a stronger property then in the classical topological case? For instance, do we have
$$[T,\mathcal{X}]_{\mathbb{A}^1}\cong [T,\mathcal{Y}]_{\mathbb{A}^1}$$
if $f$ is a weak equivalence of schemes up to $\mathbb{A}^1$-homotopy and $T$ is some scheme?
The condition you've stated implies that the homotopy sheaves are equivalent, and it is implied by the map being a weak equivalence, so they are equivalent. You're nullifying $\mathbb{A}^1$ in the $\infty$-category of Nisnevich sheaves of spaces on the category of affine schemes. This can be seen from the fact that $\mathbb{A}^1$-nullification is a left Bousfield localization.
The condition you've stated implies that the homotopy sheaves are equivalent, and it is implied by the map being a weak equivalence, so they are equivalent. You're nullifying $\mathbb{A}^1$ in the $\infty$-category of Nisnevich sheaves of spaces on the category of (affine) schemes (smooth over the base). This ultimately can be seen from the fact that $\mathbb{A}^1$-nullification is a left Bousfield localization.
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2025-03-21T14:48:31.674526
| 2020-08-05T00:17:37 |
368346
|
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|
Stack Exchange
|
Asymptotics of the number of minimal strongly connected digraphs
Is anything known about the number of minimal strongly connected digraphs on $n$ labeled nodes? (``Minimal’’ meaning that on the deletion of any arc, strong connectivity is lost.) Some values are given in sequence A130768 at OEIS.org, but no generating function or asymptotic estimate is given. This feels like something that should be known, being a natural digraph analog of the famous question of the number of trees on $n$ labeled vertices.
Context: I’m preparing lectures, and while I was writing notes on a theorem of the form “the following 5 statements are equivalent”, it occurred to me that it would be nice to discuss in class the question: “in how many ways can one show that $n$ statements are equivalent, without any redundant implications in the proof?”.
There is a webpage devoted to these digraphs, http://jglopez.etsisi.upm.es/MSC_Digraphs_Page/ but a quick look through it suggests that the asymptotics really are not known.
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2025-03-21T14:48:31.674626
| 2020-08-05T01:40:31 |
368351
|
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|
Stack Exchange
|
Chevalley restriction theorem: group vs lie algebra version
Let $G$ be a (split) reductive group over $k$, $T$ a split maximal torus, and W its Weyl group. I sometimes see the Chevalley restriction theorem stated as
(1) $k[G]^G \xrightarrow{\sim} k[T]^W$
and sometimes as
(2) $k[\mathfrak{g}]^G \xrightarrow{\sim} k[\mathfrak{t}]^W$.
I think both can be proved in essentially the same way, but is there a way to directly deduce them from each other?
I think (1) implies (2) directly: equip $k[G]$ and $k[T]$ with the filtrations by the powers of the maximal ideals at the origins. Then the map in (2) is the associated graded of the map in (1). Since the map in (1) is a strict map of filtered rings, it being an isomorphism implies its associated graded is an isomorphism.
It's also clear that the injectivity of (2) implies the injectivity of (1). However I'm not quite seeing what to do about surjectivity -- is there an adjective one can put in front of "filtered" to guarantee that a map is surjective if its associated graded is? (This is the case when the filtration is finite, but that is not satisfied here...)
There cannot be a completely formal way to prove equivalence of these two, because (1) is always true but (2) can fail in positive characteristic. (But I like this question, and hope that there is a good answer that avoids this in some sense shallow pitfall! It had never occurred to me to try to connect these two facts, rather than just proving both.)
Interesting, thanks for pointing that out. Although it seems that your objection applies to the implication (1) => (2) whereas I thought I was stuck on (2) => (1). Does my argument for (1) => (2) sound right in char 0? Certainly it breaks down in positive characteristic because formation of invariants doesn't commute with formation of associated graded.
It sounds plausible, but I wouldn't want to claim I've thought about it enough that I'd catch a subtle error if there were one.
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2025-03-21T14:48:31.674771
| 2020-08-05T02:39:21 |
368352
|
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|
Stack Exchange
|
Properties of differentiable functions on non-locally-bounded fields
I was reading some results on the structure of non-locally-bounded topological fields, and I was wondering what is known about differentiable functions on them. In particular, on the complex numbers with the usual topology, differentiable functions are infinitely differentiable. Is the same thing (or something similar) true of functions on fields with weird topologies?
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2025-03-21T14:48:31.674831
| 2020-08-05T02:51:38 |
368353
|
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"Alan C",
"Francois Ziegler",
"KConrad",
"LSpice",
"https://mathoverflow.net/users/133880",
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"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/3272"
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"sort": "votes",
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|
Stack Exchange
|
The $2\pi$ factor in the Fourier transform and dimensional analysis
I have been thinking about the $2\pi$ factor in the various conventions of the Fourier transform. For example, I was looking for a way to justify the following:
$(*)$ If we define $\hat f(\xi) = \int f(x) e^{-ix\xi} \, dx$, then the following equation is false: $f(x) = \int \hat f(\xi) e^{ix\xi} \, d\xi$
Obviously, one way to prove $(*)$ is to derive the correct form of the Fourier inversion formula, but I wanted a heuristic argument to convince myself quickly that $(*)$ is true. Here is what I came up with:
Suppose $x$ has units of time, and $f(x)$ is unitless. In the expression $e^{i\theta}$, the variable $\theta$ should have units of radians. Hence $\xi$ must have units of radians/time.
From $\hat f(\xi) = \int f(x) e^{-ix\xi} \, dx$, it follows that $\hat f(\xi)$ must have units of time (coming from the $dx$).
Combining 1 and 2, we can conclude $\int \hat f(\xi) e^{ix\xi} \, d\xi$ has units of time*radians/time = radians, which does not match the units of $f(x)$. Thus $f(x) = \int \hat f(\xi) e^{ix\xi} \, d\xi$ must be false, since the units on both sides do not agree. This "proves" $(*)$.
(On the other hand, if we define $\hat f(\xi) = \int f(x) e^{-2\pi ix\xi} \, dx$, then we can think of the $2\pi$ as having units of radians, so $\xi$ now has units of 1/time. As a result, there are no unit agreement issues with $f(x) = \int \hat f(\xi) e^{2\pi ix\xi} \, d\xi$.)
I have two questions:
Is there a way to make the above argument more precise? I would like to think of it as a dimensional analysis argument, but radians are dimensionless.
Is there a way to give a heuristic argument to see that the following is true?
$(**)$ Let $L > 0$. Suppose we define $\hat f(\xi) = \int f(x) e^{-Lix\xi} \, dx$, and suppose that $f(x) = \int \hat f(\xi) e^{Lix\xi} \, d\xi$ holds. Then $L$ must be $2\pi$.
(I asked this question on math.SE, but the responses there did not address my question.)
Another way to ask it: we incorrectly think that $[0, 1]$ has measure $1$ on both the time side and the spectral side. One or the other must have measure $1/(2\pi)$, or both might have measure $1/\sqrt{2\pi}$. Heuristically, why?
And, of course, any approach to this has to answer why $x \mapsto e^{-x^2/(2\pi)}$ is an eigenfunction (or maybe I've got my factors off; as you say, this is why one wants an argument like this, so as not to have to check every time!). To me that 'feels' more like the place for a heuristic argument than Fourier inversion.
The definition of what counts as an eigenfunction of the Fourier transform is tied up with how you define it. Letting $\hat{f}(y) = \int_{\mathbf R} f(x)e^{-2\pi ixy},dx$, $e^{-\pi x^2}$ is its own Fourier transform, but this is not true when the Fourier transform uses $e^{-ixy}$ in the integral.
Does this answer your question? The $2\pi$ in the definition of the Fourier transform
Specifically, from @FrancoisZiegler's reference, @WhatsUp's answer and the associated comments talk about "why $2\pi$?".
Thanks for the comments. WhatsUp's answer in the referenced question is the type of answer I'm looking for in my second question. I don't understand the comment by WhatsUp, which says that if we identify $\mathbb R \simeq \widehat {\mathbb R}$ with $y \mapsto (x \mapsto e^{iaxy})$, then the correctly scaled Haar measure on $\widehat {\mathbb R}$ is Lesbesgue measure scaled by the Lebesgue measure of $\mathbb R/(\ker(x \mapsto e^{iax}))$. Why is this the case?
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2025-03-21T14:48:31.675061
| 2020-08-05T03:54:25 |
368355
|
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|
Stack Exchange
|
Continuity of embeddings and systole as you vary a metric
Let $M$ be a smooth compact manifold and let $R(M)$ be the set of Riemannian metrics on $M$, topologized with the $C^\infty$ topology (viewing a metric as a section of an appropriate bundle).
I have two questions about $R(M)$, one that seems like it should be easy and the other hard.
Assume that $M$ has a nontrivial fundamental group and let $s:R(M) \rightarrow \mathbb{R}$ be the function that returns the systole of the metric, ie the length of the shortest non-nullhomotopic curve. Question: Is $s$ continuous? It feels like it should be, but I can’t seem to prove it.
Here is a crazier question. Can we find some large $n$ and a continuous function from $R(M)$ to the space of smooth maps from $M$ to $\mathbb{R}^n$ that takes a metric to an isometric embedding?
|
2025-03-21T14:48:31.675143
| 2020-08-05T04:54:30 |
368357
|
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|
Stack Exchange
|
Regularity of a singular Kaehler Einstein metric
On a manifold $X$ of general type i.e. $X$ is projective and $c_1(K_{X})$ semiample. One can construct a singular Kaehler Einstein metric $\omega_{\infty}$ in $-c_1(X)$. In particular, $\omega_{\infty}$ can be constructed as the limit of a normalized Kaehler Ricci flow, and in this case the potential $\varphi$ of $\omega_{\infty}$ is shown to be $L^{\infty}(X)$. I wonder whether it is possible to say something about $|\nabla \varphi|$ and higher derivatives of $\varphi$.
The gradient estimate is known, cf Proposition 4.1 here
https://arxiv.org/pdf/1404.0445.pdf . I am not aware of any higher order estimates.
Thank you and that is really helpful! Also the paper measures the gradient against the KE metric but I think it can be easily checked that it is actually true for any non singular metric.
no, it is crucial that the gradient be taken wrt to the KE metric (which has singularities). I see no reason why the potential should be Lipschitz.
|
2025-03-21T14:48:31.675240
| 2020-08-05T07:06:45 |
368359
|
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|
Stack Exchange
|
Projective Jacobian matrix and determinant
I was wondering if there is an interpretation of the projective Jacobian matrix in algebraic geometry. In affine space, this matrix represents the linear map on tangent spaces induced by a function. But if projective space, if $F$ if a endomorphism on, say, $P^2$, what geometrically would the Jacobian matrix represent? It is a 3 by 3 matrix. Is there some sort of natural "projective tangent space at a point"?
Supposing this was some sort of notion, then things still do not exactly behave as I would expect them to. For example, in the affine case, if F is an endomorphism on $A^n$, then the Jacobian determinant is zero at a point if and only if there is ramification at that point (and I believe this is regardless of characteristic). And, while it seems like this may be the case in characteristic 0, it is not in finite characteristics. The map $[yz:zx:xy]$ is a birational automorphism, but in fields of characteristic 2, its Jacobian determinat is identically $0$.
So does anyone know much about this, or know where I can find out more? The Jacobian determinant of a projective automorphism has been very useful to me in my research, and while I do not need to learn anything specifically, it would be nice to understand this tool I am using much more than I do now.
|
2025-03-21T14:48:31.675347
| 2020-08-05T07:28:06 |
368364
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368364"
}
|
Stack Exchange
|
On a degenerate SDE in the unit ball
This is a question about a diffusion process on the unit ball.
In this article J.S, the author considered the following SDE in the closed unit ball $E \subset \mathbb{R}^n$:
\begin{align*}
(1)\quad dX_t=\sqrt{2(1-|X_t|^2)}\,dB_t-cX_t\,dt,
\end{align*}
where $\{B_t\}_{t \ge 0}$ is an $n$-dimensional Brownian motion, $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^n$ and $c$ is a nonnegative constant. We define an elliptic operator $(\mathcal{A},\text{Dom}(\mathcal{A}))$ by $\mathcal{A}=C^2(\mathbb{R}^n)|_E$ and
\begin{align*}
\mathcal{A}f=(1-|x|^2)\Delta f-c x\cdot \nabla f,\quad f \in \text{Dom}(\mathcal{A}).
\end{align*}
Then, standard results from martingale problems show that there exists a diffusion process $\{X_t\}_{t \ge 0}$ on $E$ such that
$f(X_t)-f(x)-\int_{0}^{t}\mathcal{A}f(X_s)\,ds \quad(t \ge 0,\, x \in E)
$
is a martingale. Thus, the SDE $(1)$ possesses a solution. Furthermore, we can show that the solution is unique in the sense of distribution (by the way, the pathwise uniqueness for (1) is a very profound problem).
My question is as follows.
If $c=0$, then $\mathcal{A}$ is a weighted Laplacian on $E$ and the corresponding diffusion process $X$ is a time-changed absorbing Brownian motion on the interior of $E$. The boundary $\partial E$ is the cemetery of $X$. If $c>0$, the corresponding diffusion process is a time-changed standard Brownian motion with drift?
Thank you in advance.
As for the above question, I may solved it myself.
I think $\mathcal{A} f = (1-|x|^2) \nabla f - c x \cdot \nabla f$. Also, transforming to polar coordinates may shed insight into your questions.
@NawafBou-Rabee Thank you. $\mathcal{A}$ should be defined as you pointed out.
@NawafBou-Rabee I am rather at a loss in interpreting boundary conditions. Even if $c$ vanishes, the corresponding diffusion process is not a time changed reflected B.M., right?.
Clearly, if $c=0$, then the SDE has a fixed point on the unit sphere. That corresponds to an absorbed B.M., i.e., after the boundary is hit, the process will never leave the boundary for all subsequent times.
@NawafBou-Rabee I see. I asked a strange question. So, the solution to (1) is identified with a time-changed Brownian motion with drift?
It seems there is a slight transcription error in (1): the paper you mention considers an SDE with a quadratic diffusion coefficient. As the paper indicates, the classification of the boundary depends on $c$: if $c=0$ the process is a time-changed absorbed Brownian motion; if $0<c<2$, the process hits the boundary infinitely often and the solution can be identified with a time-changed reflecting Brownian motion with drift; and if $c \ge 2$, the process never reaches the boundary and can be identified with a time-changed Brownian motion with drift.
@NawafBou-Rabee I think (1) is correct as it is. (1) is interesting in that the diffusion coefficients is not Lipschitz continuous. The pathwise uniqueness for (1) holds if $c \ge 2(\sqrt{2}-1)$. If $c<2(\sqrt{2}-1)$, the uniqueness remains unknown.
Yes, you are right, (1) is correct. The html version of the paper did not display the square root for some reason.
|
2025-03-21T14:48:31.675578
| 2020-08-05T09:50:44 |
368373
|
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"sort": "votes",
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|
Stack Exchange
|
Spectral symmetry of a certain structured matrix
I have a matrix
$$ A= \begin{pmatrix} 0 & a & d & c\\ \bar a & 0 & b & d \\ \bar d & \bar b & 0 & a \\ \bar c & \bar d & \bar a & 0 \end{pmatrix} $$
As you can see, the matrix is always self-adjoint for any $a, b, c, d \in \mathbb C$.
But it has a funny property (that I found by playing with some numbers):
If $a,b,c$ are arbitrary real numbers and also $d$ is real, then the spectrum of $A$ is in general not symmetric with respect to zero. To illustrate this, we take $d := 2$, $a := 5$, $b := 3$, $c := 4$ then the eigenvalues are
$$\sigma(A):=\{10.5178, -6.54138, -3.51783, -0.458619\}$$
But once I take $d \in i \mathbb R$, the spectrum becomes immediately symmetric. In fact, $d := 2 i$, $a := 5$, $b := 3$, $c := 4$ leads to eigenvalues
$$\sigma(A)=\{-9.05607, 9.05607, -0.993809, 0.993809\}$$
Is there any particular symmetry that only exists for $d \in i\mathbb R$ that implies this nice inflection symmetry?
I am less interested in a brute-force computation of the spectrum than of an explanation of what symmetry causes the inflection symmetry.
Is it symmetric for all choices of $a$, $b$ and $c$ (real and or complex)?
@M.Winter good question, now I am tempted to say that it actually only works for $a,b,c$ real.
For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity:
$$X=\left(
\begin{array}{cccc}
0 & 0 & 0 & -i \\
0 & 0 & i & 0 \\
0 & -i & 0 & 0 \\
i & 0 & 0 & 0 \\
\end{array}
\right),\;\;XA+AX=0,\;\;X^2=I.$$
Hence the spectrum of $A$ has $\pm$ symmetry:
$$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$
so if $\lambda$ is an eigenvalue then also $-\lambda$.
how did you get $\det(\lambda I-A)=\det(\lambda X^2-XAX)$?
Since $X^2=1$, you can multiply $\det(\lambda-A)$ with $\det X^2=(\det X)^2$; and since the product of determinants is the determinant of the matrix product, you have $\det(\lambda-A)=\det[X(\lambda-A)X]=\det(\lambda X^2-XAX)$.
great! a useful trick in evaluating some determinants
An equivalent trick : Let $J:= \operatorname{diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.
|
2025-03-21T14:48:31.675757
| 2020-08-05T10:31:23 |
368377
|
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"Alexandre Eremenko",
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|
Stack Exchange
|
A holomorphic shrinking of a domain into a compact subset
This question is related to these two.
Let $X\subset \mathbb{C}^{n}$ be a bounded domain. I am interested in the following property: there is a sequence of continuous maps $\varphi_n:\overline{X}\to X$ which are holomorphic on $X$ and such that $\varphi_n(x)\to x$, for every $x \in X$. Every convex domain has this property.
If $n=1$ is this property equivalent to $X$ being a Jordan domain?
If $n>1$ and $X$ is pseudoconvex what are some sufficient conditions on $X$ (other than convexity) so that it has this property?
Can you explain what $\phi(x)$ is?
@AlexandreEremenko it was just an error, sorry
I guess it holds also for images of biholomorphic maps from the closed bounded starshaped domains into \mathbb{C}^n.
|
2025-03-21T14:48:31.675851
| 2020-08-05T12:46:52 |
368381
|
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"Pierre-Yves Gaillard",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368381"
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|
Stack Exchange
|
Isomorphism of hyperreal fields viewed as extensions of the field of reals
I asked this question on Mathematics Stackexchange but got no answer.
Question. Does $ZFC$ prove that there are non-principal ultrafilters $\mathcal U$ and $\mathcal V$ over $\mathbb N$ such that the respective ultrapowers $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ are not isomorphic when regarded as extensions of the field $\mathbb R$?
$ZFC+\neg CH$ proves that such ultrafilters do exist. More precisely, $ZFC+\neg CH$ proves the stronger statement that there are $\mathcal U$ and $\mathcal V$ as above such that $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ are not even isomorphic as fields. On the other hand, $ZFC+CH$ proves that $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ are always isomorphic as fields.
The question could be asked with $ZFC+CH$ instead of $ZFC$, but, seeing no reason to think that it cannot be settled in $ZFC$, the above formulation seemed more natural to me.
Since the fields $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ have $2^{2^{\aleph_0}}$ distinct structures of extensions of $\mathbb R$, as shown by Eric Wofsey, a field isomorphism between these two fields will not be $\mathbb R$-linear in general.
The transcendence degree of $\mathbb R^{\mathcal U}$ over $\mathbb R$ is $2^{\aleph_0}$. [Of course the same holds for $\mathbb R^{\mathcal V}$.] Indeed, for $x\in\mathbb R$ define $f_x\in\mathbb R^{\mathbb N}$ by $f_x(n)=e^{e^{nx}}$ [or $f_x(n)=\exp(\exp(nx))$ if you don't like small letters] and let $g_x$ be the image of $f_x$ in $\mathbb R^{\mathcal U}$. Then a straightforward argument shows that the subset $\{g_x\ |\ x>0\}\subset\mathbb R^{\mathcal U}$ is algebraically independent over $\mathbb R$.
As shown by tomasz, $ZFC$ proves that there are models $\mathcal A$ and ultrafilters $\mathcal U$ and $\mathcal V$ as above such that $\mathcal A^{\mathcal U}$ and $\mathcal A^{\mathcal V}$ are not isomorphic.
Rahman. M replaced the tag [tag:ultrapowers] with [tag:set-theory]. I rejected the edit because the question asks if certain pairs of ultrapowers are isomorphic. If I had to keep just one tag, I would choose this one. It seems to me the other tags I picked give a more precise idea of the question than the [tag:set-theory] tag. If you think the [tag:ultrapowers] tag is inappropriate, thanks for telling me why.
I don't like to argue about things in MO! I just explain you the reason: The question asks explicitly if you can prove a certain thing in ZFC while we know that under the failure of CH the answer is positive. There are certain properties of ultra-filters which are independent from ZFC. In particular an answer may show that there are something combinatotial behind the curation, and the concept of ultrapower is just an instance! This why I prefer set-theory to ultrapowers in such question.
By the way, I don't think that the ultrapowers is inappropriate, I just do not prefer it to set theory as already mentioned.
|
2025-03-21T14:48:31.676061
| 2020-08-05T14:13:27 |
368385
|
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"Alexandre Eremenko",
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"sort": "votes",
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|
Stack Exchange
|
Is this curve well known?
I consider the curve $c(t)=(x(t),y(t))$ in $\mathbb{R}^2$ such that
$\frac{d^2x(t)}{dt^2}=-(a\sin t+b)\frac{dy(t)}{dt}$
$\frac{d^2y(t)}{dt^2}=(a\sin t+b)\frac{dx(t)}{dt}$
$a,b\in\mathbb{R}$
Is the orbit curve of solution of above equations known?
What initial conditions? For example, all lines are solutions.
It is not clear what you mean by "known" but this system can be solved explicitly, in quadratures of elementary functions. Set $x'=u,\; y'=v,\; g(t)=a\sin t+b$. Then your system becomes
$$u'=-gv,\quad v'=gu.$$
Multiplying the first equation on $u$ and second on $v$ and adding, we obtain
$u'u+v'v=0,$ therefore $u^2+v^2=c$. Then the first equation becomes
$(u')^2=g^2(c-u^2)$, and this is a separable equation, with an explicit integral.
Thanks for your answer.
Excuse me for the wide implications of my remarks.
I wanted to know if it had a name or if there was a reference to be cited.
If this system is interesting for some reasons, perhaps it has a name, but you know better where it comes from.
|
2025-03-21T14:48:31.676166
| 2020-08-05T14:16:58 |
368386
|
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"Samir Canning",
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|
Stack Exchange
|
Use of flattening stratification (from Nitsure's construction of Hilbert and Quot schemes)
I study Nitin Nitsures paper on the Construction of Hilbert and Quot Schemes and not understand the propetry (F) completely:
In previous chapter (Embedding Quot into Grassmanian) it was
proved that there exist a morphism of functors
$\frak{Quot}$$^{\Phi, L}_{E/X/S} \to \frak{Grass}$$(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$
for appropriate positive integer $r$.
All used notation is explained rigorously in the linked paper on pages 7 and 24 but let me make some short remarks:
Here we deal with special case where $V$ and $W$ are
vector bundles on scheme $S$, $\pi: X := \mathbb{P}(V) \to S$
and $E = \pi^*(W)$ and $L =O_X(1)$. Futhermore $\Phi$ is the
Hilbert-polynomial.
Now let come to my actually question: The author intends to show
that given any locally noetherian $S$-scheme $T$
and a surjective homomorphism $f: W_T \otimes_{O_T} \operatorname{Sym}^r V_T \to
\mathcal{J} $ where $\mathcal{J}$ is a locally free $O_T$-module of
rank $\Phi(r)$, there exists a locally closed subscheme
$T' \subset T$ with universal property (F).
Question:
I not understand the universal property (F):
It says: Given any locally noetherian $S$-scheme $Y$ and
an $S$-morphism $\phi : Y \to T$, let
$f_Y$ be the pull-back of $f$, and let
$\mathcal{K}_Y = \operatorname{ker}(f_Y) =
\phi^* \operatorname{ker}(f)$. [...]
Why is $\operatorname{ker}(f_Y) =
\phi^* \operatorname{ker}(f)$? Indeed $\phi$ wasn't assumed to
be flat, therefore there is no reason with pullback
functor $\phi^*$ should be left exact.
We have exact sequence
$$0 \to \mathcal{K}:= \operatorname{ker}(f) \to
W_T \otimes_{O_T} \operatorname{Sym}^r V_T \xrightarrow{f}
\mathcal{J} \to 0$$
applying $\phi^*$ we obtain only
$$\phi^*\operatorname{ker}(f) \to
\phi^*(W_T \otimes_{O_T} \operatorname{Sym}^r V_T)
\xrightarrow{f_Y}
\phi^* \mathcal{J} \to 0$$
$\phi^*$ is right exact but in general not left exact
(this is only true with additional assumption ie when
$\phi$ is flat).
Why here nevertheless $\operatorname{ker}(f_Y) =
\phi^* \operatorname{ker}(f)$ is true?
Hint: the exactness on the left follows from $\mathcal{J}$ being locally free
I understand, locally by freeness the sequence splits and this property is preserved after application of $\phi^$ because $\phi^ J$ stays locally free and especially projective. And this preserves the exactness, yeah? Thank you!
|
2025-03-21T14:48:31.676352
| 2020-08-05T14:29:05 |
368387
|
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"Mohammad Golshani",
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|
Stack Exchange
|
How badly can the GCH fail globally?
It's known that we can have global failures of GCH---for example, where $\forall \lambda(2^\lambda = \lambda^{++})$---given suitable large cardinal axioms.
My question is whether we can have global failures of GCH where there is a weakly inaccessible cardinal between $\lambda$ and $2^\lambda$ for each $\lambda$. Similarly, whether we can have a cardinal fixed point between $\lambda$ and $2^\lambda$. I'd also be interested in whether $2^\lambda$ can be weakly inaccessible/a cardinal fixed point, for every $\lambda$.
Two questions: (1) What is weakly measurable? (2) Remind me what does the original Foreman-Woodin model satisfy?
oh, haha! (I use weaklymeasurable as a handle on instagram.) (1) is fixed. So, I think Foreman-Woodin was just $2^\lambda> \lambda^+$ and then Woodin got $2^\lambda = \lambda^{++}$.
In the Foreman-Woodin model The generalized continuum hypothesis can fail everywhere.
for each infinite cardinal $\kappa, 2^\kappa$ is weakly inaccessible.
This answers your last question. The answer to the first two questions can be yes as well. In the case of Foreman-Woodin model, they start with a supercompact $\kappa=\kappa_0$ and infinitely many inaccessibles $\kappa_n, n<\omega,$ above it. They first force to get $2^{\kappa_n}=\kappa_{n+1}$ preserving $\kappa$ supercompact, and this is reflected below for all cardinals. So if for example each $\kappa_n$ is measurable, then what you get in the final model is that for each infinite cardinal $\lambda, 2^\lambda$ has been measurable in $V$, in particular there are both weakly inaccessible and cardinal fixed points between $\lambda$ and $2^\lambda.$
See also the paper A model in which every Boolean algebra has many subalgebras
by Cummings and Shelah, where they build a model in which for each infinite cardinal $\kappa, 2^\kappa$ is weakly inaccessible and $Pr(2^\kappa)$ holds. Here $Pr(\lambda)$ is in some sense a large cardinal property (for example it holds if $\lambda$ is a Ramsey cardinal). For its definition see the paper.
Thanks, Mohammad! That answers my final question. Does it also answer my first two questions? Sorry if I'm missing something obvious!
@SamRoberts I added details.
Thanks so much, Mohammad!
|
2025-03-21T14:48:31.676627
| 2020-08-05T15:05:27 |
368389
|
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|
Stack Exchange
|
Bounded subsets of $\delta$-hyperbolic metric spaces
I was reading this book by Coorneart, Delzant, and Papadopoulos. I am stuck with this proposition in Chapter 1 (Proposition 1.5)
If $Y$ is a bounded $\delta$-hyperbolic subset of $X$, then $X$ is $\delta'$-hyperbolic with $\delta' = \delta + 6 \eta$ where $\eta = \sup_{x \in X} \operatorname{dist}(x,Y)$.
The authors have not given proof of the proposition. I have tried using the $\delta$-hyperbolicity condition of $Y$ but I am getting stuck. Please help.
@AmirSagiv Thanks for editing the question
Let $x,y,z,b \in X$ be given and assume that $\eta = \sup_{x\in X} d(x,Y) < \infty$. Fix $x' , y' , z' , b' \in Y$ such that $|x-x'|, |y-y'|$, etc. are all $\leq \eta$. (For a 'properly done' proof, use $|x-x'|<\eta + \varepsilon$ for some $\varepsilon>0$ and take $\varepsilon \to 0$ at the end).
Using the triangle inequality, one can show that
\begin{align*}
& (x'|y')_{b'} \leq (x|y)_b + 3\eta \\
& (x|z)_b \leq (x'|z')_{b'} + 3\eta \\
\text{and } & (y|z)_b \leq (y'|z')_{b'} + 3\eta .
\end{align*}
Then combining the above with the fact that Y is $\delta$-hyperbolic, we get
\begin{align*}
(x|y)_b & \geq (x'|y')_{b'} -3\eta
\\ & \geq \min\{ (x'|z')_{b'} , (y'|z')_{b'} \} - \delta - 3\eta
\\ & \geq \min\{ (x|z)_{b} -3\eta , (y|z)_{b} -3\eta \} - \delta - 3\eta
\\ & = \min\{ (x|z)_{b}, (y|z)_{b} \} - \delta - 6\eta .
\end{align*}
So $X$ is $\delta'$-hyperbolic with $\delta' = \delta + 6\eta$.
|
2025-03-21T14:48:31.676737
| 2020-08-05T15:17:04 |
368390
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368390"
}
|
Stack Exchange
|
Inverse mellin transform
Let $K_1(t)$ be the K-Bessel function, then we have
$\int_{0}^{\infty}K_v(y)y^s\frac{dy}{y}=2^{s-2}\Gamma(\frac{s+v}{2})\Gamma(\frac{s-v}{2})$ See page 106 of Bump's book Automorphic forms and representations.
I need a proof of the inverse Mellin transform
$$\frac{1}{2\pi i}\int_{C-i\infty}^{C+i\infty}\frac{\Gamma(z+3)\Gamma(z+2)}{z}\left(\frac{N}{4\pi^2nA}\right)^zdz=\frac{1}{8}G_1\left(4\pi\sqrt{\frac{nA}{N}}\right),$$
where $$G_1(x)=\int_{x}^{\infty}t^4K_1(t)dt$$
take the derivative with respect to $u=N/A$ and it becomes the usual inverse Mellin.
Thank you for saving the day!
|
2025-03-21T14:48:31.676804
| 2020-08-05T15:26:11 |
368391
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Ben Wieland",
"https://mathoverflow.net/users/4639"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368391"
}
|
Stack Exchange
|
Torus trick without surgery theory
It follows from surgery theory that in dimension $\geq 5$ every closed PL manifold homotopy equivalent to a torus has a finite cover which is PL homeomorphic to a torus. This is an important ingredient for Kirby's torus trick, and thus for the theory of topological manifolds.
In the introduction to The Topological Classification of Stratified Spaces, Weinberger refers to an alternative proof of this fact:
I should mention that nowadays using some amazing constructions of Edwards and Chapman, one can prove this result about tori by pure geometry, without invoking surgery.
However, he gives no further explanation or references. Can someone point me to a reference or give an explanation of this argument?
I believe that he returns to this in 9.4.1. I guess you should be suspicious of this answer because he doesn't give a forward reference in the introduction. There he cites Chapman, Approximation results in topological manifolds, 1981. See the introduction for the discussion of the avoidance of surgery.
|
2025-03-21T14:48:31.676908
| 2020-08-05T15:49:25 |
368396
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Guy Fsone",
"Hannes",
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"https://mathoverflow.net/users/51695",
"https://mathoverflow.net/users/85906",
"mlk"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368396"
}
|
Stack Exchange
|
Measure of the boundary of an BV-extension domain: do we have $|\nabla Eu|(\partial \Omega)=0?$
Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where
\begin{align}\label{eq:bounded-variation}
|u|_{BV(\Omega) }:= \sup\Big\lbrace \int_{\Omega} u(x) \operatorname{div} \phi(x) d x:~\phi \in C_c^\infty(\Omega, \mathbb{R}^d),~\|\phi\|_{L^\infty(\Omega)}\leq 1\Big\rbrace.
\end{align}
The vector $\nabla u =(\Lambda_1, \Lambda_2,\cdots, \Lambda_d)$ can be seen as a vector valued Radon measure on $\Omega$ such that
\begin{align*}
\int_\Omega u(x)\frac{\partial \varphi}{\partial x_i}(x) d x=
-\int_\Omega \varphi(x)d \Lambda_i(x), \quad \text{for all }\quad \varphi\in C_c^\infty(\Omega),~~i=1,\cdots, d.
\end{align*}
In particular if $u\in W^{1,1}(\Omega)$ then $u\in BV(\Omega)$, $\partial_{x_i} u(x)d x= d\Lambda_i(x)$ and
$ |u|_{BV(\Omega)}\asymp\|\nabla u\|_{L^1(\Omega)}.$
Let us recall that $\Omega \subset \mathbb{R}^d$ is called an $W^{1,p}$-extension (resp. $BV$-extension) domain if there exists a linear operator $E:W^{1,p}(\Omega)\to W^{1,p}(\mathbb{R}^d)$ (resp. $E: BV(\Omega)\to BV(\mathbb{R}^d)$) and a constant $C: = C(\Omega, d)$ such that
\begin{align*}
Eu\mid_{\Omega} &= u \quad\hbox{and} \quad \|Eu\|_{W^{1,p}(\mathbb{R}^d)}\leq C \|u\|_{W^{1,p}(\Omega)} \quad\text{for all}\quad u \in W^{1,p}(\Omega)\\
(\text{resp.}\quad
Eu\mid_{\Omega}& = u \quad\hbox{and} \quad \|Eu\|_{BV(\mathbb{R}^d)}\leq C \|u\|_{BV(\Omega)} \quad\, \text{for all}\quad u \in BV(\Omega)).
\end{align*}
Question:
1-If $\Omega$ is an $BV$-extension domain do we have $|EU|_{BV(\Bbb R^d)}(\partial\Omega)=0?$
2-If $\Omega$ is an $W^{1,p}$-extension domain do we have $|\nabla EU|_{W^{1,p}(\Bbb R^d)}(\partial\Omega)=0?$
Intuitively, the second question is more likely to hold if we assume in addition that the boundary $\partial \Omega$ has zero Lebesgue measure. Since $\nabla Eu$ is still a function when $Eu\in W^{1,p}(\Bbb R^d)$.
Or, does the fact that $\Omega$ is an extension domain implies that $|\partial\Omega|=0$?
To your last question: A Sobolev extension domain necessarily satisfies the measure density condition $|B(x,r) \cap \Omega| \geq c r^d$ from which $|\partial \Omega| = 0$ follows, see Sobolev embeddings, extensions
and measure density condition by Hajlasz, Koskela, Tuominen.
@Hannes Thank you for this reference.
For a domain with nice enough boundary (e.g. Lipschitz), you can always construct a $BV$-extension by simply extending with zero, which then does not fulfill your condition since possibly all of the boundary is a jump set. But that is a very specific extension out of many. So a better question might be, if on a $BV$-extension domain there is an extension for which the boundary is not part of the jump set. (Which sounds like it could be true.)
|
2025-03-21T14:48:31.677078
| 2020-08-05T15:55:04 |
368397
|
{
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"authors": [
"Mikhail Bondarko",
"curious math guy",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/2191"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368397"
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|
Stack Exchange
|
Stalk of motivic homotopy sheaves
In contrast to "classical" homotopy theory, in the motivic homotopy theory, we don't have homotopy group but rather homotopy sheaves in the Nisnevich topology, which is associated to the presheaf
$$\pi_n^{\mathbb{A}^1}(\mathcal{X}):Sm_k\rightarrow \text{Ab},\quad U\mapsto [S^{i,j}\wedge U_+,\mathcal{X}]_{\mathbb{A}^1}.$$
Morphisms of spaces $\mathcal{X}\rightarrow \mathcal{Y}$ induce morphisms of homotopy sheaves $\pi_n^{\mathbb{A}^1}(\mathcal{X})\rightarrow \pi_n^{\mathbb{A}^1}(\mathcal{Y})$. The Nisnevich topology has enough points,and thus an isomorphism of homotopy sheaves can be detected on stalks. So a natural question is what is the stalk of the homotopy sheaves? Is it enough to evalute $\pi_n^{\mathbb{A}^1}(\mathcal{X})$ on the spectra of local henselian rings?
Yes, sure.:) Moreover, in the motivic stable homotopy theory ($S^1$-stability is sufficient here; I don't know whether it is necessary) it suffices to evaluate homotopy groups at spectra of function fields.
@MikhailBondarko Would you happen to have a reference for this? Something I can quote?
Did you try Morel F., An introduction to A 1 -homotopy theory, in: Contemporary
Developments in Algebraic K-theory (M. Karoubi, A. O. Kuku, C. Pedrini
eds.), ICTP Lecture Notes, vol. 15, 2003, 357-441? It does not contain complete proofs, but the formulations are probably fine. Theorem 4.3.4 says something about the corresponding t-structure.:)
|
2025-03-21T14:48:31.677192
| 2020-08-05T16:57:21 |
368401
|
{
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"authors": [
"Federico Poloni",
"Wolfgang",
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"https://mathoverflow.net/users/29783"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368401"
}
|
Stack Exchange
|
Are there orthogonal Cauchy-like matrices with rational entries for any given size?
This is inspired by a recent question about the existence of orthogonal Cauchy-like matrices. It is proved that there are indeed such matrices, i.e. there are vectors $x,y,r,s\in\mathbb R^n$ such that for the matrix $C$ defined by
$$
C_{ij} = \frac{r_i s_j}{ x_i - y_j},
$$ we have $C C^T = I$.
In the pdf with the solution, the following fact is used: If $C$ is a $n \times n$ real Cauchy matrix $C$, i.e. there are vectors $x,y\in\mathbb R^n$ such that $$C_{ij} = \frac{1}{ x_i - y_j},$$ then its inverse admits the factorization
$$C^{-1} = D_aC^TD_b,$$
where $D_a,D_b$ denote the diagonal matrices with the entries of $a, b\in\mathbb R^n$, which can be computed by explicit formulae obtained by Lagrangian interpolation. BTW, it is not clear to me why the RHS features $C^T=(( \frac{1}{ x_j - y_i}))_{i,j=1}^n$, as the given formula $(7)$ in Theorem 1 of the original article (accessible here with different notation, as it calls $a,b$ what we call $x,y$ here) seems to contain rather $(( { x_j - y_i}))_{i,j=1}^n$ than the reciprocals.
In any case, my question is
whether the vectors $x,y,r,s\in\mathbb R^n$ defining a Cauchy-like matrix can all have rational entries, i.e. whether there is a Cauchy matrix $C$ such that each entry of the corresponding vectors $a,b$ in the factorization of $C^{-1}$ quoted above is a (rational) square?
Note that the given construction shows that there are Cauchy matrices where $a,b$ have only positive entries and uses their square roots at some point. But for rational square roots, I wouldn't know how to go about that even for $n=2$.
That denominator in the inverse is expected. If $C$ satisfies $D_x C - C D_y = \text{(rank 1)}$, then by multiplying by $C^{-1}$ on both sides you get that $C^{-1}$ satisfies $C^{-1}D_x - D_y C^{-1} = \text{(rank 1)}$, and by expanding you get those denominators.
@Federico yes I agree that that makes sense. In fact, I have a hard time interpreting the formula derived from Lagrange in the article in order to put it into a form with $D_a$ and $D_b$ for some vectors a,b.
Let us start with four rational numbers $x_1, x_2; y_1,y_2$ so that using the cross ratio
$$
r=(x_1,x_2,y_1,y_2)
$$
the numbers $-r$ and $1-r$ are rational squares.
(Exchanging the mid components $x_2, y_1$ bring $r$ into $1-r$. Exchanging the first and/or last two components we obtain the multiplicative inverse, etc. - so we want in the following to set in evidence squares times cross ratio values known to be squares.)
For instance for $0,1;-1,8/17$ we are producing $r=-9/16$. It may be simpler to follow the construction based on this example.
Let $C$ be the associated Cauchy matrix. In the example:
$$
C=\begin{bmatrix}1 & -17/8\\1/2 & 17/9\end{bmatrix}\ .
$$
Let $a_1,a_2;b_1,b_2$ be the rational squares:
$$
\begin{aligned}
a_1 &= (x_1-y_1)^2 \; (x_1, y_1, y_2, x_2) \ ,\\
a_2 &= (x_1-y_2)^2 \; (x_1, y_2, y_1, x_2) \ ,\\[2mm]
b_1 &= 1\ ,\\
b_2 &= (-1)\; \left(\frac{x_2-y_1}{x_1-y_1}\right)^2 \;(x_1, x_2, y_1, y_2) \ .
\end{aligned}
$$
Then we have $C^{-1}=D_a\; C^T\; D_b$.
Let now $L$ be a the Cartan-like matrix of the shape $L=D_r\; C\; D_s$. Then:
$$
\begin{aligned}
L &=D_r\; C\; D_s\ ,\\
L^T &=D_s\; C^T\; D_r\ ,\\
L^{-1}
&=D_s^{-1}\; C^{-1}\; D_r^{-1}\\
&=D_s^{-1}\; D_a C^T D_b\; D_r^{-1}
\ ,\\[3mm]
&\qquad\text{ and we want $L^T=L^{-1}$, i.e.}\\[3mm]
D_s\; C^T\; D_r &= D_s^{-1}\; D_a C^T D_b\; D_r^{-1}\text{ i.e.}\\
C^T &= D_s^{-2}\; D_a C^T D_b\; D_r^{-2}\ .
\end{aligned}
$$
Recall that $-r$ and $1-r$ both squares
implies $a,b$ squares, as wanted in the OP, so we can arrange for $s,t$ with rational entries.
To make the above easy to test, here is some sage code making the computations.
var('x1,x2,x3,y1,y2,y3');
def r(s,t,u,v):
return (s-u)/(s-v)/(t-u)*(t-v)
a1 = (x1-y1)^2 * r(x1, y1, y2, x2)
a2 = (x1-y2)^2 * r(x1, y2, y1, x2)
b1 = 1
b2 = (-1) * (x2-y1)^2 / (x1-y1)^2 * r(x1, x2, y1, y2)
C = matrix([ [1/(x1-y1), 1/(x1-y2)] , [1/(x2-y1), 1/(x2-y2)] ])
Da = diagonal_matrix( [a1, a2] )
Db = diagonal_matrix( [b1, b2] )
print("Is C^-1 = Da * C^T * Db? %s"
% bool(C^-1 == Da * C.transpose() *Db))
And we get:
Is C^-1 = Da * C^T * Db? True
We use now instead of general variables the special values:
x1, x2, y1, y2 = 0, 1, -1, 8/17
(just replace the first var line with the above, keep the next lines of the used code) and ask for the values of $a$, $b$:
sage: a1, a2, b1, b2
(16/25, 576/7225, 1, 9/4)
Now consider the matrix $L$ given by
sage: L = diagonal_matrix([sqrt(b1), sqrt(b2)]) * C * diagonal_matrix([sqrt(a1), sqrt(a2)])
sage: L
[ 4/5 -3/5]
[ 3/5 4/5]
Which is an orthogonal matrix. (I found this problem while searching for the tag elliptic-curves, but the above solution is maybe closer to K-theory.)
|
2025-03-21T14:48:31.677479
| 2020-08-05T18:31:31 |
368404
|
{
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],
"authors": [
"Gerald Edgar",
"M. Reza. K",
"Wilberd van der Kallen",
"erz",
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"url": "https://mathoverflow.net/questions/368404"
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|
Stack Exchange
|
If all points of a real function with positive values would be local minimum, can one say it is constant function?
During my studies I faced a function $f:\mathbb{R} \to \mathbb{R}^+ $ with the property: for all $x \in \mathbb{R} $ and all $y$ in open interval $(x-\frac{1}{f(x)} ,x+\frac{1}{f(x)}) $ we have $f(x) \leq f(y)$.
At first I guessed maybe it is necessarily constant function but I could not show this. I tried to generate a counter example for my guess and unfortunately I could not again. would anyone please help me?
The answer to the question in the title is (of course) no. [Define $f(x) = 2$ for $x \ne 0$ and $f(0)=1$.] But the question in the text is more complicated.
What is $\mathbb{R}^+$? Is $f(x)=0$ allowed?
$\mathbb{R}^+ $ is denoted for the open interval $(0,+\infty)$
My answer is: Such a function must be constant. Suppose (for puoposes of contradiction) $f$ is a nonconstant function $f : \mathbb R \to (0,+\infty)$ such that
$$
\forall x\in\left(a-\frac{1}{f(a)},a+\frac{1}{f(a)}\right),\quad
f(x) \ge f(a) .
$$
For $m > 0$, define $U_m := \{x : f(x)>m\}$.
Lemma 1: For all $m \in \mathbb R$, the set $U_m$
is open.
Proof. Let $x \in U_m$. Then $(x-1/f(x),x+1/f(x)) \subseteq U_m$. So $U_m$ is open.
Define
$\mathcal G = \{(a,b,m) :
a < b, f(a)\le m, f(b)\le m, \text{ and }\forall x\in(a,b),f(x)> m\}$.
Lemma 2: Let $a,b \in \mathbb R$ with $f(a) \ne f(b)$, and $m>0$.
Then there exists $a_1, b_1$ with $a < a_1 < b_1 < b$ and $m_1 \ge m$ such that
$(a_1,b_1,m_1) \in \mathcal G$.
Proof. One of $f(a), f(b)$ is smaller, assume WLOG that $f(a) < f(b)$.
Now $U_{f(a)}$ is open,
$a \notin U_{f(a)}$, and $b \in U_{f(a)}$. The maximal open interval
$(c,d) \subseteq U_{f(a)}$ with $c<b< d$ is nonempty, in fact $(c,b) = (c,d) \cap (a,b) \ne \varnothing$. By maximality, $f(c) \le f(a)$. Choose $e$ with $c < e < b$. For $x \in (c,e)$
we have $f(c) < f(x)$ so $c \le x-1/f(x)$ and thus
$f(x) \ge 1/(x-c)$. Thus $f(x)$ is unbounded on $(c,e)$.
Let $m_1$ be such that $m_1 \ge m, m_1 > f(c), m_1 > f(e)$.
The open set $U_{m_1}$ has $U_{m_1} \cap (c,e) \ne \varnothing$
and $c,e \notin U_{m_1}$. So let $(a_1,b_1)$ be a maximal open subinterval
of the open set $U_{m_1} \cap (c,e)$.
Lemma 3: Let $(a_1, b_1, m_1) \in \mathcal G$, and let $m > m_1$.
Then there exist $a_2, b_2$ with $a_1 < a_2 < b_2 < b_1$ and $m_2 \ge m$
such that
$(a_2, b_2, m_2) \in \mathcal G$.
Proof: If $f(a_1) \ne f(b_1)$, apply Lemma 2 directly. Otherwise, pick any $c \in (a_1,b_1)$ and apply Lemma 2 to $a_1, c, m$.
Main Proof: We assume $f$ is not constant. So there exist $a_1 < b_1$ with
$f(a_1) \ne f(b_1)$. By Lemma 2 there exist $a_2, b_2$ with
$a_1 < a_2 < b_2 < b_1$ and $m_2 > 2$ so that $(a_2,b_2,m_2) \in \mathcal G$.
Then by Lemma 3, there exist $a_3, b_3$ with
$a_2 < a_3 < b_3 < b_2$ and $m_3 > 3$ so that $(a_3,b_3,m_3) \in \mathcal G$.
Continuing recursively, we get sequences $(a_k), (b_k), (m_k)$ so that
$$
\forall k:\quad a_k < a_{k+1} < b_{k+1} < b_k,\quad m_k > k,\quad\text{and }
(a_k,b_k,m_k)\in\mathcal G .
$$
The sequence $(a_k)$ is increasing and bounded above, so it converges.
Let $a = \lim_{k\to\infty} a_k$. Then $a_k < a_{k+1} \le a \le b_{k+1} < b_k$.
From $(a_k,b_k,m_k) \in \mathcal G$ we conclude $f(a) > m_k > k$.
This is true for all $k$, so $f(a)$ is not a real number, a
contradiction.
Your answer was enjoyable and very nice to me. It really engaged me and I am thinking if instead of $\mathbb{R}$ in domain of the $f$, replace a complete metrically convex metric space, does your answer still work?
@M.Reza.K for every $x,y$ in the complete metrically convex space there is an isometry $g:[a,b]\to X$ such that $g(a)=x, g(b)=y$. Apply the result to $h=f\circ g$. It follows that $f(x)=h(a)=h(b)=f(y)$. I am interested if we can drop the completeness: the proof seems to work for rational numbers... Perhaps it also works for intrinsic metric spaces?
@erz My proof certainly fails for rational numbers. A decreasing sequence of closed intervals may have empty intersection in $\mathbb Q$.
This is not an answer but it greatly narrows down the class of functions $f$ with the described property.
If $c>0$, then the sets $f^{-1}([c,+\infty))$ and $f^{-1}((c,+\infty))$ are both open.
Indeed, every $x$ in any of these sets comes with an open ball of radius $\frac{1}{f(x)}$.
Consequently, the sets $A_c=f^{-1}((-\infty, c))$ and $B_c=f^{-1}((-\infty, c])$ are both closed. In particular, $f$ is lower semi-continuous.
Let $x\in B_c$. We know that if $d(y,x)<\frac{1}{c}\le \frac{1}{f(x)}$, then $f(x)\le f(y)$. On the other hand, if $y\in B_c$ and $d(y,x)<\frac{1}{c}$, then $d(y,x)<\frac{1}{f(y)}$, from where $f(y)\le f(x)$. Hence, if $y\in B_c$ and $d(y,x)<\frac{1}{c}$, we get $f(y)= f(x)$.
Not only this means that $f$ is locally a constant on $B_c$, but also that $f^{-1}(c)$ is closed.
Let us prove the following strengthening of the Baire's theorem: let $X$ be a complete metric space, and let $B_n$ be a sequence of closed sets such that $X=\bigcup B_n$. Then $X=\overline{\bigcup int B_n}$.
Indeed, if $U\subset X\backslash \bigcup int B_n$ is open and nonempty, it is metrizable with a complete metric, and $B_n\cap U$ is closed in $U$. Since $U=\bigcup (B_n\cap U)$, from the usual Baire's theorem, there is $m$ such that $V=int (B_m\cap U) \ne \varnothing$. But then $V=int (B_m\cap U)=int B_m\cap U\subset int B_m \backslash \bigcup int B_n=\varnothing$. contradiction.
So, combining these observation we get the following picture: there is a dense open set $U$ such that $f$ is locally constant on $U$; every level set of $f$ is closed, and the distance between elements of $f^{-1}(c)$ and $f^{-1}(d)$ is at least $\frac{1}{\max \{c,d\}}$.
The only way it's not a constant function is when the components of $U$ are like the gaps in the Cantor set. So the possible counterexample may be constructed like the Cantor's staircase, but the smaller the component, the larger is the value on it. I was unable to perform this fine tuning though, maybe it's impossible.
|
2025-03-21T14:48:31.677866
| 2020-08-05T19:30:38 |
368407
|
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|
Stack Exchange
|
Unit group of octonions over finite fields
One can define the algebra $A(K)$ of octonions over an arbitrary field $K$, see for example the command OctaveAlgebra in GAP: https://www.gap-system.org/Manuals/doc/ref/chap62.html .
When $K$ is a finite field, this is a finite dimensional $K$-algebra and thus has finitely many elements. Let $A_q$ denote the octonions over a field with $q$ elemetns.
Question 1: What is the number of units in $A_q$? Can one even describe the (possibly non-associative) group of units up to isomorphism?
For $q=2$ the order is 120 and for $q=3$ the order is 4320. In both cases it is indeed a group according to GAP.
Question 2 is motivated by Will Sawin's comment (I forgot the unit "group" might not be associative):
Question 2: For which $q$ is the unit "group" of $A_q$ associative?
It would be interesting to see what the smallest $q$ is such that the unit "group" is not associative.
I guess it should be a (Moufang) loop of units instead of a group, because it's not associative (except maybe in characteristic 2 it is). I don't know how easy those are to describe...
@WillSawin Thanks, I forgot about that. For $q=2$ and $q=3$ it is indeed a group according to GAP.
Units are exactly octonions with norm different from $0$, so you need to count the number of solutions of $\sum_{k=1}^8x_i^2\neq 0$, where $x_i\in \mathbb{F}_q$.
Just to add that GAP has a bug in dealing with the non-associativity here. This will be fixed.
This is all worked out in the article "A class of simple Moufang loops" by L.J. Paige. The short answer is that the loop of units has size $q^3(q^4-1)(q-1)$, and is not associative for any $q$. The example given by Paige (lemma 3.5) is given in terms of Zorn vectors as
$$\left[\begin{pmatrix}
1 & (0,0,1)\\
(0,0,0) & 1 \\
\end{pmatrix}\begin{pmatrix}
1 & (1,0,0)\\
(0,0,0) & 1 \\
\end{pmatrix}\right]\begin{pmatrix}
0 & (0,1,0)\\
(0,-1,0) & 1 \\
\end{pmatrix}=\begin{pmatrix}
0 & (1,1,1)\\
(-1,-1,1) & 2 \\
\end{pmatrix}$$
and
$$\begin{pmatrix}
1 & (0,0,1)\\
(0,0,0) & 1 \\
\end{pmatrix}\left[\begin{pmatrix}
1 & (1,0,0)\\
(0,0,0) & 1 \\
\end{pmatrix}\begin{pmatrix}
0 & (0,1,0)\\
(0,-1,0) & 1 \\
\end{pmatrix}\right]=\begin{pmatrix}
1 & (1,1,1)\\
(-1,0,1) & 1 \\
\end{pmatrix}$$
so these two products cannot be equal over any characteristic. For $q=2$ we obtain the smallest simple nonassociative Moufang loop, which has order 120.
The article actually shows that a certain subloop modulo its center is a simple Moufang loop. At the time before Paige's result, the only simple Moufang loops known where the simple groups. Liebeck later proved the converse: Every finite simple Moufang loop that is not a group corresponds to such a subloop of octonions over some $\mathbb F_q$. In particular we shouldn't expect a simple classification.
In fact Paige loops are not associative for $q\in {2,3}$ either (despite the claim in the OP), as stated in Theorem 4.1 of the linked article.
@pregunton Thanks for this comment. I obtained the units with the GAP command U:=units(A); and then it returns true when I check IsGroup(U); So maybe it is a GAP bug?
The article of Liebeck: The classification of finite simple Moufang loops (MSN).
@Mare, it's a little hard to tell from the documentation whether Units is meant to handle a non-associative ring. The definition of a ring allows it, but then the documentation says that Units always returns a group, so it's not so clear.
I edited the answer to correctly reflect that Paige loops are always nonassociative. Some of the lemmas in the paper require being careful in the cases $q\in {2,3}$, but the part that shows nonassociativity works over all $q$.
@LSpice using the command StructureDescription(U) seems to calculate forever (even though U has only order 120 for q=2). So the developers also might forgot about the fact the units do not need to be associative.
|
2025-03-21T14:48:31.678380
| 2020-08-05T20:14:54 |
368410
|
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|
Stack Exchange
|
Arithmetic statement which is independent, and whose independence is independent, and so on?
My vague intuition is that not only it is common for a simple arithmetic proposition $p$ to be independent of ZFC, but it is common for the statement "$p$ is independent of ZFC" to be independent, and so on. If we let $I(p)$ be the statement that $p$ is independent of ZFC, then this "iterative independence" property is $I^*(p) = \forall n \ge 0, I^n(p)$.
Of course, it would be impossible to prove $I^*(p)$ for any particular statement $p$, at least under ZFC. So the best one could do is find some set $P$ of propositions s.t. we can prove at least one $p \in P$ has $I^*(p)$, but where we don't know which one it is.
Is it possible to find such a set $P$ containing at least one iteratively independent statement?
if $p$ is independent, isn't $I(p)$ automatically independent, because $I(p)$ is true, therefore not disprovable, because we accept soundness of ZFC in the meta-theory, and proving $I(p)$ would require proving consistency of ZFC, which is not possible in ZFC, because we accept consistency of ZFC in the meta-theory?
@WillSawin : That should certainly be an answer.
If we fix things to avoid Will Sawin's observation, then the answer is yes under any reasonable interpretation I can think of.
For example, consider the following: let $J(p)$ be the sentence "If $\mathsf{ZFC}$ is consistent then $p$ is independent over $\mathsf{ZFC}$." We can indeed prove in $\mathsf{ZFC}$ many instances of $J$ (e.g. that $J$ holds of the Godel-Rosser sentence for $\mathsf{ZFC}$), so this is nontrivial.
However, note that if we assume that $\mathsf{ZFC}$ is sound (this is overkill but let's do it anyways), then we have $\mathsf{ZFC}\vdash J^n(p)$ for some finite $n$ only if $p$ really is independent of $\mathsf{ZFC}$. Since the set of sentences which are independent over $\mathsf{ZFC}$ is not c.e., there must be many $p$s such that $\mathsf{ZFC}$ doesn't prove any of the $J^n(p)$s.
The non-c.e.-ness claim above is an instance of a more general fact: if $T$ is any theory to which Godel's first incompleteness theorem applies, then the set of $T$-independent sentences is not c.e. This is because if it were, then the set of $T$-theorems would be both c.e. (because it obviously is) and co-c.e. (because every non-$T$-theorem is either $T$-disprovable, which is obviously a c.e. condition, or is $T$-independent which is a c.e. condition by assumption). But that would make the set of $T$-theorems computable, which can't be the case (e.g. if it happened we'd be able to whip up a computable complete extension of $T$).
Basically, any reasonable iterated independence property can't fully capture independence.
Thanks for the $I \to J$ fix and the simple positive answer! However, I don't quite understand the last sentence, since I don't know what you mean by "fully capturing independence". The thing I'm after is just that generically $J^*(p)$ should often be true, even if we can't ever know it, which is what you've established.
If $p$ is independent of ZFC, and ZFC is sound, then $I(p)$ is independent of ZFC.
Indeed, because ZFC is sound, it can't disprove $I(p)$ because $I(p)$ is true.
However, $I(p)$ implies consistency of ZFC, which isn't provable in ZFC by Godel's theorem (using soundness again), so $I(p)$ is not provable.
Thus $I(p)$ is independent.
Yep, this is of course correct; I should have gone with Noah's fix below.
|
2025-03-21T14:48:31.678733
| 2020-08-05T20:15:26 |
368411
|
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|
Stack Exchange
|
Reference to log-transition-density of a diffusion process
Consider the scalar diffusion $X=(X_t)_{t\geq 0}$ given by
$$\mathrm{d}X_t \ = \ b(X_t)\,\mathrm{d}t \ + \ \sigma(X_t)\,\mathrm{d}B_t, \qquad X_0=\xi,$$
with $b$, $\sigma$ smooth, $\xi$ absolutely continuous and $B=(B_t)_{t\geq 0}$ a standard Brownian motion (w.r.t. some prob. measure $\mathbb{P}$).
Suppose that the transition function $(t,x,\mathrm{d}y)\mapsto\mathbb{P}(X_t\in\mathrm{d}y\mid X_0=x)$ of $X$ admits a positive Lebesgue-density $\rho\equiv\rho(t,x,y)\in C^{1,2}(G)\cap C(\bar{G})$ with $G:=(0, T)\times\mathbb{R}^2$ for some $T>0$.
I'm interested in (reasonable) conditions on $b$ and $\sigma$ which guarantee that $D_T:=\{x\in\mathbb{R}\mid \partial_x\, \partial_y\, \log(\rho(T,x,x))\neq 0\}$ is dense in $\mathbb{R}$.
Question: Can you recommend any literature where the log-density $\log(\rho)$ of a general diffusion (of the above type) is analysed in enough detail to potentially derive (reasonable) sufficient conditions on $b$ and $\sigma$ for $D_T\subseteq\mathbb{R}$ to be dense?
This question combines two different differentiable dependences. One for the initial condition see here:
Differentiable dependence on the initial condition of the solution of a SDE
and the other for the integration variable in the law of $X_{t}$, see theorem 2.3.3 in Nualart's Malliavin calculus and related topics:
|
2025-03-21T14:48:31.678848
| 2020-08-05T20:57:31 |
368414
|
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|
Stack Exchange
|
Is there a Kauffman bracket invariant of colored links?
I want to distinguish between links where the components have different (or same) colors.
In the Alexander polynomial we can assign a different variable to each component, but what about a Kauffman bracket? Is there a generalization so that we could distinguish between multi-component links where some of the components are colored?
The Jones polynomial can be understood as coming from representations of quantum groups. The regular Jones polynomial comes from the defining representation of $U_q(\mathfrak{sl}_2)$. It is possible to color/label components of the link with different representations of $U_q(\mathfrak{sl}_2)$ to obtain various colored Jones polynomials.
However, this is a little different than the multivariable Alexander polynomial because the invariant you end up with is still a polynomial in a single variable.
Thanks! Do you have any (readable) references for 1) The Jones obtained from the representation of $U_q(\mathfrak{sl}_2)$ and 2) The coloring/labeling of components?
I have heard great things about "An Introduction to Quantum and Vassiliev Knot Invariants" by Jackson and Moffatt and at a glance it looks nice. I think Chapter 9 would have the details you are asking about
I should also say, because of how "nice" representations of $U_q(\mathcal{sl}_2)$ are, there are deep relations between the colored Jones polynomial and Jones polynomials of cables. Sometimes the cabling construction is used as a definition of the colored Jones polynomial to make working with it easier
|
2025-03-21T14:48:31.678996
| 2020-08-06T00:03:17 |
368422
|
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|
Stack Exchange
|
Explanation for why an ideal fluid doesn't have increasing entropy?
The equations of motion for a very simple ideal fluid (specifically a calorically perfect, monatomic, ideal gas) are \begin{align*}\dot{\rho}+\nabla \cdot (\rho u)=0 \;&\text{(mass conservation)} \\ \dot{(\rho u)}+\nabla \cdot (\rho u u) + \nabla p=0 \;&\text{(momentum conservation)} \\ \dot{(\rho e)} +\nabla \cdot (\rho ue+\rho p)=0 \;&\text{(energy conservation)} \\ e=\frac{1}{2}u^2+\frac{3}{2}p \;&\text{(equation of state)}\end{align*} where $\rho$ is the density, $u$ the velocity, $p$ the pressure, and $e$ the total energy (including the internal energy).
I've noticed that these equations are time reversible, i.e. if we have a solution on the time interval $[0,T]$, then by simply sending $u \to -u$, $t \to -t$ we get a solution on $[-T,0]$. From the point of view of thermodynamics, specifically the fact that total entropy is (weakly) increasing, this only makes sense if entropy is constant.
EDIT: In response to some of the comments I've deleted the example of gas expansion, since as pointed out, this wasn't strange. However I'd like to mention that a very simple model of particle collisions in a gas gives rise to the above equations:
Assume particles are interacting through collisions only (i.e. not through 'long range' forces), and there are sufficient collisions occuring that to a good degree of accuracy, the distribution of velocities of particles at any point is isotropic (after subtracting the mean velocity). For example if the velocity distribution of all particles at a given point is always a spherical gaussian this would be the case. Under just this assumption, the above equations follow.
I'm not denying that they are time reversible, and so must have constant entropy. It's just that I have no intuitive explanation for this, other than simply computing the equations. The statistical model mentioned is not time reversible, in fact it relies heavily on frequent collisions, and so time reversibility is a rather suprising fact.
Is there some other explanation, perhaps more intuitive than simply computing the equations, that explains time reversibility?
Your equations don't seem to have any mechanism that would make them thermodynamic. There is no reference to heat or temperature, nor to any sort of statistical distribution. Just describing a macroscopic matter flow doesn't yet imply the generation of entropy.
About the specific question, yes, expansion of a gas can be reversible (and apparently is under your assumptions). Personally, I do not see anything weird about it: the reverse process looks like a gas moving initially at some speed inwards decelerating and compressing due to inertia.
Reversibility only holds for classical solutions. For weak solutions, an entropy conditions needs to be imposed; without it, uniqueness is lost.
I am having trouble seeing the logical difference between your scenario and the following (and if there is a difference, understanding that might also be instructive): A metal block, warmer than its surroundings, sitting on a tabletop. You simply give Newton's first law as the equation of motion, and it correctly describes the block just sitting there. Of course, in reality, the block is also cooling down irreversibly, but your equation of motion doesn't know anything about that.
@MichaelEngelhardt So if I understand you, you're saying the issue is that I haven't included any equation for how heat evolves. That's because for this statistical model heat is the same thing as (exchange of) pressure: the internal energy is (proportional to) the pressure. So infact we do have an evolution equation for heat, which is what makes this strange - since if heat is being exchanged there really ought to be entropy increase
@MichaelRenardy that would explain it, thank you, is there any further reading I could do on that subject?
Hmm yes, at the level of the statistical model you may have a connection between heat and pressure, but it seems to me this is lost at the level of the equation of motion. Pressure is just a form of force, and it will get volume elements to move, changing their energy. The connection to the concept of heat is lost. It seems to me this is also apparent from the lack of any mention of temperature, the conjugate to entropy.
This is a very important issue, to which an answer must be made in mathematical terms, rather than by waving hands.
Yes, the Euler system (conservation of mass, momentum and energy) is time-reversible. So where is the error when we say Entropy is non-decreasing, but the system is time-reversible, therefore the entropy must be constant along trajectories ?
The point is that the Cauchy problem (i.e. solving the PDEs together with imposing an initial data) is not uniquely solvable. It is so when the initial data is smooth enough, but only for some finite time interval. For rather general smooth initial data, the smooth solution exists only for a finite time interval $(0,T_{\max})$. As $t\rightarrow T_\max$, some first derivative becomes infinite somewhere (generic behaviour). Beyond $T_\max$, the solution is not any more smooth ; it is at best piecewise smooth, with $\rho,u,e,p$ being discontinuous accross hypersurfaces. These discontinuities are known as shock waves and contact discontinuities.
It turns out that once shock waves develop, the Euler system is not any more sufficient to select a unique solution. There are actually infinity many, among which only one has a physical sense. The way to recognize that one, and to select it from a mathematical perspective, is to add a so-called entropy criterion. This is nothing more than saying that when a particle crosses a shock, then its entropy increases.
This entropy criterion is expected to guaranty the uniqueness of the solution (this is still an open problem). But, being an inequality, is not compatible with the time reversal. This is why the Cauchy problem, at far as physicall meaningful solutions are concerned, is irreversible, despite the apparent time-reversibility of the Euler system.
Edit. To answer Michael's concerns, the constancy of entropy along smooth solutions is just the well accepted fact in thermodynamics that smooth flows are time-reversible. Of course, if you have in mind a finer description, the situation will be different. The mesoscopic level is described by a kinetic equation, say that of Boltzmann, which is irreversible: the entropy does increase whenever the local distribution of the gas deviates from the Gaussians (= Maxwell's equilibria). Thus the only reversible model is at the microscopic level, where particles obey to Newton's law and interact through short range forces (or hard spheres dynamics).
To come back to the Euler system, the boundary conditions at discontinuities are not ad hoc. They just express that the mass, momentum and energy are conserved.
Thank you, this makes sense. Where should I look if I wanted to do more reading into this?
There are many good texts. The best exposition I know of is that by C. Dafermos. But you could have a look to my book as well Systems of conservation laws I, Cambridge UP.
This explains well what goes wrong with the macroscopic flow equations - as it should, it would be worse if it didn't! Physically, it raises questions still. For one, until $T_{max} $, entropy indeed doesn't increase - can this really be faithful to the underlying statistical description in general? Secondly, nothing becomes infinite in real life - so shouldn't one stop trusting the macroscopic equations well before $T_{max} $? Is it realistic to assume that a boundary condition at discontinuities can do justice to the physics, or is one attaching golden buttons to rotten coats?
@DenisSerre - Thank you for the additional discussion!
Q: Explanation for why an ideal fluid doesn't have increasing entropy?
A: The entropy will in fact increase for the most probable initial conditions.
The question in the OP refers to the socalled irreversibility paradox (or Loschmidt paradox): The statistical evolution of the ideal gas is irreversible even though the equations of motion are reversible. The evolution of the statistical distribution $f(r,v,t)$ is governed by the Boltzmann equation,
$$\frac{\partial f}{\partial t}+v\cdot\frac{\partial f}{\partial r}=\int dv_2 dv'_1 dv'_2 w(v'_1,v'_2;v,v_2)[f(v'_1,r,t)f(v'_2,r,t)-f(v,r,t)f(v_2,r,t)],$$
for some collision rate function $w$. This equation describes the relaxation of $f$ towards the equilibrium Maxwell-Boltzmann distribution, in accordance with the second law of thermodynamics.
From a modern perspective, a resolution of the irreversibility paradox is discussed by Freddy Bouchet in arXiv:2002.10398: The Boltzmann equation only holds for a subset of microscopic initial conditions compatible with a certain $f(r,v,0)$. Not every microscopic state of a macroscopic system will evolve in accordance with the second law, but only the majority of states, a majority which however becomes so overwhelming when the number of atoms in the system becomes very large that irreversible behavior becomes a near certainty.
I'll endeavor to read the paper, but forgive me for now: "The Boltzmann equation only holds for a subset of microscopic initial conditions compatible with a certain f(r,v,0)". Are you saying the boltzman equation doesn't have solutions (or perhaps uniqueness) for every initial data f? i.e. is this in the same as @MichaelRenardy's comment on the OP?
the Boltzmann equation describes a unique solution of increasing entropy for any initial condition on $f_0(r,v)=f(r,v,0)$; but there exists microscopic initial data (positions and velocities of molecules) that would give the same $f_0$ but would evolve with decreasing entropy. We ignore this initial data because it is of measure zero in the limit of a large number of molecules.
Oh I see, yes, and for example this explains why the Boltzmann equation isn't itself time reversible. What I still find strange is that the equations I've given above can be (I believe) derived from the Boltzmann equation, and so the whole 'ignore improbable initial data' spiel should have already been taken care of. Yet they are time reversible. I'm guessing therefore there are some 'more probable' initial conditions (in terms of u,p,rho, not the microscopic discription), and for these we get increasing entropy? Thanks for entertaining my ignorance
the conservation laws can be derived from the Boltzmann equation, but not the other way around; one way to see this, is that it is not possible to derive $dS/dt\geq 0$ from the conservation laws.
|
2025-03-21T14:48:31.679701
| 2020-08-06T00:35:22 |
368424
|
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|
Stack Exchange
|
A number similar to Landau's Totient constant
The real number given by the absolutely convergent series
$$\displaystyle A = \sum_{k=1}^\infty \frac{|\mu(k)|}{k \phi(k)}$$
is known as Landau's Totient Constant. It can be explicitly evaluated to be $\frac{\zeta(2)\zeta(3)}{\zeta(6)}$. Indeed, we see that $A$ can be expanded into an Euler product
$$\displaystyle A = \prod_p \left(1 + \frac{1}{p(p-1)} \right).$$
We then have
\begin{align*} 1 + \frac{1}{p(p-1)} & = \frac{p^2 - p + 1}{p(p-1)} \\
& = \frac{p^3 + 1}{p(p^2 - 1)} \\
& = \frac{(p^3 - 1)(p^3 + 1)}{p(p^2 - 1)(p^3 - 1)} \\
& = \frac{p^6 - 1}{p(p^2 - 1)(p^3 - 1)} \\
& = \frac{1 - p^{-6}}{(1 - p^{-2})(1 - p^{-3})},
\end{align*}
and from here we see that
$$\displaystyle \prod_p \left(1 + \frac{1}{p(p-1)} \right) = \prod_p \left(\frac{1 - p^{-6}}{(1 - p^{-2})(1 - p^{-3})} \right) = \frac{\zeta(2) \zeta(3)}{\zeta(6)}$$
as desired.
I am looking to evaluate the related number
$$\displaystyle B = \sum_{k=1}^\infty \frac{\mu(k)}{k \phi(k)} = \prod_p \left(1 - \frac{1}{p(p-1)} \right),$$
where there is no absolute value in the numerator. The "complete the cube" trick used above obviously will not work, as one gets the quadratic polynomial $x^2 - x - 1$ instead in the numerator. Is there a nice expression for $B$? In lieu of that, is there a reasonable interpretation for $B$?
That number has a "well-known" interpretation: it is the proportionality factor in the Artin primitive root conjecture. I suspect there is no tidy formula for this number.
The decimal expansion, along with many references to the literature, is given at http://oeis.org/A005596
This constant is known as Artin's Constant. The book Finch S. R. Mathematical constants (section 2.4) gives the following information.
A rapidly convergent expression for Artin's constant is as follows [12-18]. Define Lucas' sequence as $$
l_{0}=2, \quad l_{1}=1, \quad l_{n}=l_{n-1}+l_{n-2} \quad \text { for } n \geq 2
$$
and observe that $l_{n}=\varphi^{n}+(1-\varphi)^{n},$ where $\varphi$ is the Golden mean [1.2]. Then
$$
\begin{aligned}
C_{\text {Artin }} &=\prod_{n \geq 2} \zeta(n)^{-\frac{1}{n} \sum_{k \mid n} l_{k} \cdot \mu\left(\frac{n}{k}\right)} \\
&=\zeta(2)^{-1} \zeta(3)^{-1} \zeta(4)^{-1} \zeta(5)^{-2} \zeta(6)^{-2} \zeta(7)^{-4} \zeta(8)^{-5} \zeta(9)^{-8} \cdots.
\end{aligned}
$$
|
2025-03-21T14:48:31.679881
| 2020-08-06T04:21:53 |
368429
|
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|
Stack Exchange
|
When does a group and its pro-algebraic completion have equivalent categories of arbitrary representations?
In the following everything is over some field $k$.
Let $G$ be a discrete group. We write $G^{\text{alg}}$ for its pro-algebraic completion. The latter is a pro-affine pro-algebraic group which arises as the Tannakian dual to the category $\mathrm{Rep}_{\text{fd}}(G)$ of finite-dimensional $G$-representations. By definition there is an equivalence of categories
$$
\mathrm{Rep}_{\text{fd}}(G)\simeq \mathrm{Rep}_{\text{fd}}(G^\text{alg}).
$$
Under what conditions does this lift to an equivalence of categories of possibly-infinite dimensional representations?
For the usual story, I think you want finite dim representations such that $G^{alg}\to Gl_n(k)$ is a morphism of group schemes. I sort of doubt there is a useful statement for infinite dim reps, unless $G$ is finite of corse.
Is it possible that finiteness is necessary? Certainly Rep(G) (infinite dim) recovers G, and Rep(G^alg) recovers G^alg (it's just the ind-completion I think), so if the two are equivalent doesn't that say G and G^alg are iso as group schemes?
I agree $Rep(G{alg})$ should be the ind-completion, What concerns me is that this may not contain the regular representation of G, when it’s infinite.
Right, but the finite dimensional representations are already enough to recover G^alg, by definition, so what's the issue? Maybe I'm missing something obvious
The issue is whether $Rep(G^{alg})=Rep(G)$. I thought that’s what you wanted.
If the definition of infinite-dimensional representation of $G^{alg}$ is what I think it is, then it sounds to me like the answer is "iff $G$ is finite" precisely for the reason Donu points at: the regular representation of $G$ won't be a representation of $G^{alg}$ because it will fail to be a filtered colimit of finite-dimensional representations. Right?
|
2025-03-21T14:48:31.680044
| 2020-08-06T05:56:49 |
368435
|
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|
Stack Exchange
|
Mod $N^2$ evaluation of a polynomial defined by first $N-1$ roots
Given a prime $N$ and integer $g$, where $g$ is able to generate the multiplicative subgroup $(\mathbb{Z}/N^2\mathbb{Z})^*$, I am interested in any results simplifying or evaluating $f\in (\mathbb{Z}/N^2\mathbb{Z})[x]$ where
$$f(x) := \prod_{k=0}^{N-2} (x-g^k)$$
I know that this polynomial in $\mathbb{Z}/N\mathbb{Z}[x]$ is equivalent to $(x^N-1)$ because they are both monic polynomials of the same degree with the same roots.
Similarly in $\mathbb{Z}/N^2\mathbb{Z}[x]$, $$\prod_{k=0}^{N(N-1)-1} (x-g^k)=(x^{N(N-1)}-1)$$
I have tried generating a few small examples of this product but have not seen an obvious pattern emerge. This polynomial also seems different from cyclomatic polynomials because the roots are close to each other rather than separated by by a gcd relation.
Ideally, I want to evaluate $f(x)$, but I am also curious what research has been done on these partial products. My preliminary searches have failed to find a name for this polynomial much less any results, so any information would be informative.
Note: This equation and question are very similar to the second quantity asked about in https://math.stackexchange.com/questions/3398653/polynomial-whose-roots-are-some-of-the-nth-roots-of-unity. But not much by way of follow up has been given for possible ways to evaluate or simplify the polynomial
When you say "evaluating $f$", do you mean computing the coefficients of $f$, or computing $f(\alpha)$ for $\alpha$ in $\mathbb{Z}/N^2\mathbb{Z}$ (or some $\mathbb{Z}/N^2\mathbb{Z}$-algebra)?
The latter, computing $f(\alpha)$.
Pollard has an algorithm which can be used to evaluate $f(x)$ in $O(\sqrt{N})$ operations in $\mathbb{Z}/N^2\mathbb{Z}$. Pollard's paper ("Theorems on factorization and primality testing") is not available freely online, but there's a convenient description of the algorithm in Section 2.1 of Bostan's "Computing the N-th Term of a q-Holonomic Sequence".
Thank you. That is a noticeable improvement over just calculating all the terms!
|
2025-03-21T14:48:31.680210
| 2020-08-06T06:31:48 |
368439
|
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|
Stack Exchange
|
Consistency strength of lifting through a lot of collapsing
What is the consistency strength of the following situation?
$j : V \to M$ is an elementary embedding definable from parameters in $V$, with critical point $\kappa$.
$\mathbb P$ is a forcing that collapses all ordinals between $\kappa$ and $j(\kappa)$.
$j$ can be lifted through $\mathbb P$.
One can deduce that $j(\kappa)$ is regular in $V$. The only examples I know of such liftings involve almost-huge cardinals.
If the answer is not known, you can define a new notion "pretty huge", and argue that since almost huge is pretty huge, is it always the case that a pretty huge cardinal is almost huge?
You can deduce stronger hypotheses. Let $G$ be $\mathbb P$-generic. Then $\kappa$ is regular in $V[G]$ since it is the critical point of an elementary embedding of $V[G]$ in some outer model. Similarly, $j(\kappa) = (\kappa^+)^{V[G]}$. By assumption, $j(\kappa)\leq (\kappa^+)^{V[G]}$. For the other direction, $j(\kappa)\geq (\kappa^+)^{V[G]}$ since $j(\kappa)$ is the target of the critical point of an elementary embedding of $V[G]$ in some outer model. (The target model $Q$ of this embedding must contain $P^{V[G]}(\kappa)$ and therefore $j(\kappa)\geq\kappa^{+Q}\geq \kappa^{+V[G]}$.)
$\text{AD}^{L(\mathbb R)}$ suffices. The situation actually holds in the model $H = \text{HOD}^{L(\mathbb R)}$. We will have $\kappa = \omega_1$ and $j : H\to \text{Ult}(H,U)$ equal to the ultrapower of $H$ by the club measure $U$ over $\omega_1$ as computed in $L(\mathbb R)$ (using all functions in $L(\mathbb R)$).
For any number $n$, the $\Sigma_n$-satisfaction predicate of $L(\mathbb R)$ with ordinal parameters is definable over $H$ from its restriction to ordinals less than $\Theta$, so any subclass of $H$ that is ordinal definable over $L(\mathbb R)$ is definable from parameters over $H$. In particular, $j$ is definable from parameters over $H$.
Let $N$ be a $\mathbb P_\text{max}$-extension of $L(\mathbb R)$.
Note that $H = \text{HOD}^N$ by the homogeneity and definability of $\mathbb P_\text{max}$. Let $\mathbb P\in H$ be the Vopenka algebra of $N$ for adding a subset of $\omega_2$ to $H$. There is a set $A\subseteq \omega_2$ such that $N= L[A]$, and so $N = H[G_A]$ where $G_A\subseteq \mathbb P$ is the $H$-generic ultrafilter associated to $A$.
In $N$, $\text{NS}_{\omega_1}$ is saturated. Let $G\subseteq P(\omega_1)\setminus\text{NS}_{\omega_1}$ be $N$-generic, and in $N[G]$ let $i : N\to \text{Ult}(N,G)$ be the generic ultrapower embedding associated to $G$ (using functions in $N$).
Now as usual, we cite a theorem due to Woodin: $j = i\restriction H$. This follows from Theorem 4.53 in The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal.
Now in $H$, we have the situation you were looking for with $\kappa = \omega_1.$ Note that $i(\omega_1) = (\omega_2)^N$ by saturation, which means that all $H$-cardinals between $\kappa$ and $j(\kappa)$ are collapsed to $\kappa$ in $N$. Moreover $j$ lifts through the forcing $\mathbb P$ (to $i$) by construction.
What happens if you require a "super" kind of modification, namely that $j(\kappa)$ can be unbounded when ranging over possible $j$s, how far does this blow up the consistency strength?
Very interesting. How does the extender in $H$ corresponding to $j$ relate to standard large cardinal notions? I wonder how it interacts with other forcings.
The model $H$ can be presented as a fine structure model. It's an open question whether $j$ is the branch extender of an iteration tree by the sequence of this model. The short part of $j$ is just the $\omega_2$-length linear iteration of the unique normal ultrafilter over $\omega_1$ in $H$.
@AsafKaragila I have no idea, I don't see how to get something like that from AD.
Oh, I thought it was obvious you can't. But maybe it's not that obvious...
I’m guessing there’s little hope of analyzing this $\mathbb P$ combinatorially within $H$. Dumb question: Is it in $HOD^H$?
Yeah, $\text{HOD}^H = H$ by another of Woodin's theorems, using the fine structural analysis of $H$.
You're right, $\mathbb P$ is probably very complicated. Viewing $\mathbb P$ as an order on $\Theta$, I am not sure that it is even ordinal definable over $H\cap V_\Theta$. I think you can show $\mathbb P$ is $\Theta$-cc in $H$ using Bukovsky's theorem. (By yet another theorem of Woodin, $\Theta$ is the unique Woodin cardinal of $H$.)
|
2025-03-21T14:48:31.680537
| 2020-08-06T07:59:09 |
368443
|
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|
Stack Exchange
|
Are the character degrees determined by the conjugacy class sizes?
The computation below (part 1) shows that if two finite groups of order at most $100$ have the same (ordered) list of conjugacy class sizes, then they also have the same (ordered) list of (irreducible) character degrees.
Question: Is it true in general?
If so, is there an explicit way to determine the character degrees from the conjugacy class sizes?
The converse is false, SmallGroup(64,42) and SmallGroup(64,134)) are counterexamples. At order at most $100$, there are exactly four types of counter-examples, three ot order $64$, and one of order $96$, see the computation below (part 2).
Computation
gap> BL:=[]; for d in [1..100] do n:=NrSmallGroups(d);; for r in [1..n] do g:=SmallGroup(d,r);; if not IsAbelian(g) then SC:=CharacterDegrees(g);; CC:=ConjugacyClasses(g);; L:=List(CC,c->Size(c));; Sort(L); Add(BL,[SC,L]); fi; od; od;
Part 1
sage: LLL=[[] for i in range(100)]
....: for l in BL:
....: LLL[len(l[1])].append(l)
....: for ll in LLL:
....: S=[]
....: for l1 in ll:
....: if not l1[1] in S:
....: S.append(l1[1])
....: SS=[l1[0]]
....: for l2 in ll:
....: if l1[1]==l2[1]:
....: if l2[0] not in SS:
....: SS.append(l2[0])
....: if len(SS)>1:
....: print(l1[1]); print(SS)
sage:
Part 2
sage: LLL=[[] for i in range(100)]
....: for l in L:
....: LLL[len(l[1])].append(l)
....: for ll in LLL:
....: S=[]
....: for l1 in ll:
....: if not l1[0] in S:
....: S.append(l1[0])
....: SS=[l1[1]]
....: for l2 in ll:
....: if l1[0]==l2[0]:
....: if l2[1] not in SS:
....: SS.append(l2[1])
....: if len(SS)>1:
....: print(l1[0]); print(SS)
....:
[[1, 8], [2, 6], [4, 2]]
[[1, 1, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 8, 8, 8, 8], [1, 1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8], [1, 1, 1, 1, 2, 2, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8]]
[[1, 8], [2, 10], [4, 3]]
[[1, 1, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 12, 12, 12, 12], [1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 8, 8, 12, 12]]
[[1, 16], [2, 4], [4, 2]]
[[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4], [1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]]
[[1, 8], [2, 14]]
[[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 8, 8, 8, 8], [1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]]
gap> S:=List([42,134],n->SmallGroup(64,n));;
gap> for g in S do Print(CharacterDegrees(g)); od;
[ [ 1, 8 ], [ 2, 6 ], [ 4, 2 ] ]
[ [ 1, 8 ], [ 2, 6 ], [ 4, 2 ] ]
gap> for g in S do L:=List(ConjugacyClasses(g),c->Size(c));; Sort(L);; Print(L); od;
[ 1, 1, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 8, 8, 8, 8 ]
[ 1, 1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8 ]
From experience, I would say that a conjecture like that has very little chance of being true! You could ask the converse question of whether the character degrees determine the conjugacy class sizes, but I am sure the answer to that is also no.
@DerekHolt: Your second sentence is answered in the post.
There is quite a lot of literature trying to relate class sizes and character degrees, by authors such as S. Dolfi. I think it is fair to say that the relationship between the two is unclear.
According to the body of the question, I guess you mean "determined by" in the title.
@FrançoisBrunault: Ok, I fixed the title.
SmallGroup(128,227) and SmallGroup(128,731)) are counterexamples.
gap> S:=List([227,731],n->SmallGroup(128,n));;
gap> for g in S do L:=List(ConjugacyClasses(g),c->Size(c));; Sort(L);; Print(L); od;
[ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8 ]
[ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8 ]
gap> for g in S do Print(CharacterDegrees(g)); od;
[ [ 1, 16 ], [ 2, 12 ], [ 4, 4 ] ]
[ [ 1, 8 ], [ 2, 22 ], [ 4, 2 ] ]
Here is a general comment related to my answer to a previous MO question. If $\chi$ is a complex irreducible character of a finite group $G$, and $\chi$ takes a root of unity value at $x \in G$, then $\chi(1)$ divides $[G:C_{G}(x)]$, since $\frac{[G:C_{G}(x)] \chi(x)}{\chi(1)}$ is an algebraic integer. Many irreducible characters will take a root of unity value somewhere, since a theorem of J.G. Thompson states that any irreducible character $\chi$ takes value $0$ or a root of unity on at least $\frac{1}{3}$ of the group elements.
However, notice that if $G$ is a $p$-group, then no non-linear irreducible character of $G$ ever takes a root of unity value anywhere.
Which previous MO question are you talking about?
I had to look it up. It was MO108406
|
2025-03-21T14:48:31.680893
| 2020-08-06T08:02:40 |
368444
|
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|
Stack Exchange
|
Existence of model of morphism and good reduction
In this question someone help me to understand why there is not always a model for a morphism $\mathbb{P}^1_K\to\mathbb{P}^1_K$. He says to me that the existence of model is equivalent of good reduction. Now I would like to understand this sentence ie take the "good" definitions and find a proof.
Let $K$ be a local field, of integer $\mathcal{O}$ and uniformizer $\pi$. Take $\varphi:\mathbb{P}^1_K\to\mathbb{P}^1_K$ a finite morphism and $f\in K(t)$ its associated rationnal function.
I propose the following definition for model: a model of $\varphi$ is an extension of $\varphi$ to $\mathbb{P}^1_\mathcal{O}$ that is a morphism $\Phi:\mathbb{P}^1_\mathcal{O}\to\mathbb{P}^1_\mathcal{O}$ with $\varphi$ the extention of scalar of $\Phi$ to $K$ that is $\varphi=\Phi\times_\mathcal{O}\text{Id}_K$ ie the following diagram commute,
$\require{AMScd}$
\begin{CD}
\mathbb{P}^1_K @>{\varphi}>> \mathbb{P}^1_K\\
@VVV @VVV\\
\mathbb{P}^1_\mathcal{O} @>{\Phi}>> \mathbb{P}^1_\mathcal{O}
\end{CD}
I'm not sure of the "good" definition for good reduction, I propose: as $f\in K(t)\setminus K$, $K=\text{Frac}(\mathcal{O})$ and $\mathcal{O}$ is factorial, then one can write $f=P/Q$ with $P,Q\in\mathcal{O}[t]$ coprimes (no common irreducible divisor). Then one says that $\varphi$ has a good reduction if $\overline{P}$ and $\overline{Q}$ are coprimes in $k[t]$ and the degree of $P$ and $Q$ are preserve under reduction (I take this because $\pi t^2+1$ and $\pi t+1$ reducing to coprimly but I don't want to tell that this is a good reduction (not finit morphism), but I don't need it in the proof below...)
Question: with these definitions do we have that the existence of model is equivalent to good reduction?
I have elements of proof:
I think that $P,Q\in\mathcal{O}[t]$ will define a model if $(P,Q)=\mathcal{O}[t]$ that is $(P,Q)\cap\mathcal{O}=(1)$.
In general $(P,Q)\cap\mathcal{O}=(\pi^q)$ with $q\in\mathbb{N}$. So there is a model if $q=0$.
If there is a model then $q=0$ ie $1\in(P,Q)$ so reducing the Bezout relation one has $1\in(\overline{P},\overline{Q})$ so $\overline{P}$ and $\overline{Q}$ are coprimes (but how to prove that degree is preserved, or that the reduction stay a finite morphism??)
If $P$, $Q$ are coprimes then $(\overline{P},\overline{Q})=1$. Here I have a problem: I'd like to show that $(P,Q)\cap\mathcal{O}=(\pi^q)$ with $q>0$ is not possible. If $(\overline{P},\overline{Q})\cap k=(\overline{\pi^q})$ the conclusion comes shortly, but I have only $(\overline{P},\overline{Q})\cap k\supseteq(\overline{\pi^q})$
Thnaks for your help!
|
2025-03-21T14:48:31.681070
| 2020-08-06T08:21:32 |
368445
|
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|
Stack Exchange
|
Example of BV vector field $c$ without bounded divergence such that $u$ is bounded where $u_t + div(cu) = 0$
What is an example of vector field $c: \mathbb R_+ \times \mathbb R^N \to \mathbb R^N$ with $c \in L^1(\mathbb{R}_+, BV(\mathbb R^N))$ without bounded divergence $div_x c$ but such that there exists a distributional solution $u$ of $$u_t + div(cu) = 0$$ with
$$\alpha \le u \le \beta, $$
$\alpha, \beta >0$?
Would a local statement make you happy? In other words are you interested in large $t$, large $|x|$, or is it really important for you that the bounds on $u$ hold globally in $R_+\times R^N$? Would you be happy e.g. with $c\in L^1_{loc}(R_+,BV)$?
@leomonsaingeon Yes, I'd be interested in a local example too.
|
2025-03-21T14:48:31.681146
| 2020-08-06T08:41:10 |
368446
|
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"Alkan",
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|
Stack Exchange
|
On $\sum_{k=1}^nk^3 = x^3 + y^3$ with $x,y \ge 1$
My question is related to https://oeis.org/A269839.
It is well-known that there are parametric families of solutions for cubes that are sums of consecutive cubes: https://arxiv.org/pdf/1603.08901.pdf. Famous and simplest one is $6^3 = 3^3 + 4^3 + 5^3$ which Euler noted.
I have also curiosity about solutions to $\sum_{k=1}^nk^3 = x^3 + y^3$ with $x,y \ge 1$, although it seems that there is no any reference on it. Largest solution that I found as below. $$\sum_{k=1}^{524^{3}}k^3 =<PHONE_NUMBER>34081017703266616960000 =<PHONE_NUMBER>4^3 +<PHONE_NUMBER>6^3$$
In particular, I am not sure about that there are additional probably interesting values of $n$ such as $7^{2}$, $61^{3}$, $293^{3}$, $440^{3}$ and $524^{3}$.
Question. Can be infinitely many solutions to diophantine equation $\sum_{k=1}^nk^3 = x^3 + y^3$ with $x,y \ge 1$ ?
Any comment is welcome in order to suggest helpful ideas on question. Also any additional term is very welcome.
Thanks.
I think you are wolcome to study solutions of $ 2(n^4+n^2)+4n^3= 8(x^3 + y^3)$
@zeraouliarafik: Yes, we can rewrite the equation based on $\sum_{k=1}^nk^3 = T_{n}^{2}$ as you did, $n^2 + 2n^3 + n^4 = 4(x^3+y^3)$. Best regards.
@Alkan: $\sum_{k=1}^nk^3 = x^3 + y^3$ can be transformed to $v^2 = -3s^4+(3n^4+6n^3+3n^2)s$.
We get $(n,s,x,y)=(2112278, 237469827, 70918018, 166551809).$
@Tomita: Many thanks for your comment. Yes, I agree with your transformation. Is this larger solution that you found yet? May you look further ranges and can you confirm $291^{3}$ ? I am also not sure that there are other solution(s) between $2112278$ and $291^{3}$. Best regards.
@Tomita: Thanks, I understand. Your example is also confirmation of last term of OEIS entry. Best regards.
Sorry. I have a typo, what I mean is $293^{3}$. Best regards.
|
2025-03-21T14:48:31.681280
| 2020-08-06T09:51:50 |
368450
|
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|
Stack Exchange
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Two kinds of generating functions
Sorry for a possibly off-the-topic question, but I am afraid to gain the necessary overview to give an answer (supposed the question is not ill-posed) is beyond my capabilities.
In the course of creating random graphs I got to know the concept of generating functions generators for statistical distributions (as thoroughly described in Luc Devroye's book Non-uniform Random Variate Generation). Such functions (as algorithms) are necessary to create given degree sequences to produce configuration model random graphs.
On the other side I learned to know that generating functions (in the sense of generatingfunctionology) can help to calculate combinatorial properties of graphs, especially random graphs.
So I wonder if there is some "deep" connection between both kinds of "generating functions". Both have to do with statistics resp. combinatorics of graphs. Can knowing one help to know the other?
Your link seems to point to probability generation functions on the natural numbers. These are related to those described in generatingfunctiology and techniques from the latter help with the former. It is more that being able to manipulate of power series is useful. In practice, moment generating functions and the related characteristic functions are used more often
Can you give an example or link a definition of the first kind of "generating function"?
@MarcusM: You caught me! Devroye only talks about "generators" (of distributions), not literally about "generating functions". In fact a generator is more of an algorithm (a recipe) than of a mathematical function: it takes as input some parameters and gives as output a such-and-such distributed number. Only in my mind I called these generators "generating functions". I adjusted my question accordingly.
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2025-03-21T14:48:31.681417
| 2020-08-06T10:45:20 |
368454
|
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|
Stack Exchange
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Extending Sobolev function on Riemannian manifold
Let $(M, \mu, d)$ be a geodesically complete non-compact Riemannian manifold such that measure $\mu$ is volume doubling, i.e. \begin{equation}\label{VD}\mu(B(x, 2r))\leq C\mu(B(x, r))\end{equation} for some constant $C>0$ and also $M$ satisfying the $L^{2}$-Poincare inequality $$\frac{1}{\mu(B(x, r))}\int_{B(x, r)}{|u-u_{B(x, r)}|^{2}d\mu}\leq cr^{2}\frac{1}{\mu(B(x, \delta r))}\int_{B(x, \delta r)}{|\nabla u|^{2}d\mu}$$
for all $u\in W^{1, 2}(B(x, r))$ with $\delta>1$ and $\nabla u$ being the weak gradient of $u$.
For a fixed point $x_{0}\in M$ and $\alpha, \beta\in \mathbb{R}_{+}$ consider the annuli $$P_{\alpha, \beta}=\{x\in M:\alpha<d(x, x_{0})<\beta\}.$$
Question: Is there a statement known, such that under these assumptions (or even stronger), any function $u\in L_{loc}^{2}(P_{\alpha, \beta})$ with $$\int_{P_{\alpha, \beta}}{|\nabla u|^{2}d\mu}<\infty$$ can be extended to a function $\widetilde{u}\in L_{loc}^{2}(M)$ such that $$(*)\int_{M}{|\nabla \widetilde{u}|^{2}d\mu}\leq C\int_{P_{\alpha, \beta}}{|\nabla u|^{2}d\mu}?$$
The closest statement I found is from "On extensions of Sobolev functions defined on regular
subsets of metric measure spaces" by P. Shvartsman. In this paper he proofs that considering a regular set $S$, i.e. a set such that there are constants $\theta_{S}\geq 1$ and $\delta_{S}>0$ such that for every $x\in S$ and $0<r\leq\delta_{S}$ $$\mu(B(x, r))\leq \theta_{S}\mu(B(x, r)\cap S),$$ then any function $u\in L^{2}(S)$ such that $u_{1, S}^{\#}\in L^{2}(S)$ where $$u_{1, S}^{\#}(x):=\sup_{r>0}\frac{r^{-1}}{\mu(B(x, r))}\int_{B(x, r)\cap S}{|u-u_{B(x, r)\cap S}|d\mu}$$ can be extended to a function $\widetilde{u}\in CW^{1, 2}(M)$ such that $$\|\widetilde{u}\|_{CW^{1, 2}(M)}\leq C(\|u\|_{L^{2}(S)}+\|u_{1, S}^{\#}\|_{L^{2}(S)}),$$ where $CW^{1, 2}(M)$ is the Calderon-Sobolev space which coincides with the classical Sobolev space $W^{1, 2}(M)$ if one assumes volume doubling and $M$ satisfying the $L^{2}$-Poincare inequality.
Thanks in advance for your help!
If the constant $C$ is allowed to depend on the manifold $M$ and the annulus $P_{\alpha,\beta}$, then, since $\overline{P}_{\alpha,\beta}$ is compact, it suffices to do this locally. Therefore, an extension theorem for a domain in $\mathbb{R}^n$ suffices. Stein's book, Singular Integrals and Differentiability Properties of Functions, contains such a theorem,
If the annulus is small, then it is basically an Euclidean annulus and there is an extension operator for Sobolev spaces. However, if the annulus is large is may happen that it goes around a "neck" in a manifold and meets along the boundary as the picture shows.
Then a smooth function on the annulus can approach to $0$ from one side of the common part of the boundary and to $1$ from the other side. Such a function has no Sobolev extension at all.
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2025-03-21T14:48:31.681587
| 2020-08-06T12:33:22 |
368459
|
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|
Stack Exchange
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Approximation of vector field by vector fields with all trajectories closed
Let $\vec j$ be a smooth compactly supported vector field in $\mathbb{R}^3$ such that $div(\vec j)=0$. Are there known either necessary or sufficient conditions such that there exists a sequence of smooth compactly supported vector fields $\{\vec j_i\}$ which converges to $\vec j$ say in $C^2$ topology, $div(\vec j_i)=0$, and such that all the trajectories of each $j_i$ are closed.
If $C^2$ topology can be replaced by something else, that might be of interest too.
|
2025-03-21T14:48:31.681650
| 2020-08-06T12:46:23 |
368461
|
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|
Stack Exchange
|
Hedetniemi for pseudo-chromatic number $\psi(G)$
Let $G=(V,E)$ be a finite simple graph. We say a map $p:V\to [n]:=\{1,\ldots,n\}$ is a pseudo-coloring if for all $a\neq b\in[n]$ there is $v\in\psi^{-1}(\{a\})$ and $w\in\psi^{-1}(\{b\})$ such that $\{v,w\}\in E$. We denote the maximal number $m$ such that there is a pseudo-coloring $p:V\to [m]$ by $\psi(V)$.
An easy argument shows that every coloring in the traditional sense is also a pseudo-coloring, which implies that $\psi(G) \geq \chi(G)$.
Let $G, H$ be finite simple undirected graphs.
Do we necessarily have $$\psi(G\times H) \leq \min\{\psi(G),\psi(H)\}?$$
(By $G\times H$ we denote the categorical product, sometimes also referred to as the tensor product of graphs.)
"every coloring in the traditional sense is also a pseudo-coloring" - why so? I see it only for colorings in the minimal number of colors.
If I understand correctly, the difference between the pseudo-chromatic number and the achromatic number is that a pseudo-coloring is not required to be a proper coloring, i.e., adjacent vertices may have the same color. Have I got that right?
So for instance $K_{2,2}$ has chromatic number $2$ and achromatic number $2$ but it has pseudo-chromatic number $3$, is that right?
Unfortunately $\psi(G)$ seems to be a usual notation for the achromatic number of a graph $G$.
It is not true. Let $G$ and $H$ be graphs, and let $p_{max}$ be maximal pseudo-coloring of the graph $H$. Show that the map $p((x,y))=p_{max}(y)$ is pseudo-coloring of graph $G\times H$. Fix some $\{u,v\}\in E(G)$. For arbitrary distinct colors $a,b$ there exist $\{k,l\}\in E(H)$ such that $p_{max}(k)=a$ and $p_{max}(l)=b$. Then the edge $\{(u,k),(v,l)\}$ connects colors $a$ and $b$ in $G\times H$. Hence $\psi(G\times H)\geq \max\{\psi(G),\psi(H)\}$.
Pedantically: $\psi(G\times H)\ge\max{\psi(G),\psi(H)}$ assuming each graph has at least one edge.
Here is a very simple example for $\psi(G\times H)\gt\max\{\psi(G),\psi(H)\}$.
The graph $G=H=K_3$ has achromatic number and pseudo-chromatic number equal to $3$. The tensor product $K_3\times K_3$ has achromatic number and pseudo-chromatic number $5$, as shown by the following coloring: assuming $V(K_3)=[3]$, define $p(1,1)=p(1,2)=1$, $p(1,3)=p(2,3)=2$, $p(3,3)=p(3,2)=3$, $p(3,1)=p(2,1)=4$, $p(2,2)=5$.
|
2025-03-21T14:48:31.681925
| 2020-08-06T12:52:02 |
368463
|
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Stack Exchange
|
Suggestions for special lectures at next ICM
(I am posting this in my capacity as chair of the ICM programme committee.)
ICM 2022 will feature a number of "special lectures", both at the sectional and plenary level, see last year's report of the ICM structure committee. The idea is that these are lectures that differ from the traditional ICM format (author of a recent breakthrough result talking about their work). Some possibilities are
a Bourbaki-style lecture where a recent breakthrough result (or series of results) is put into a broader context,
a "double act" where related results are presented by two speakers,
a survey lecture on a subfield relevant for some recent development,
a lecture that doesn't fit into any of the existing sections,
a lecture creating new connections between different areas of mathematics,
but these are not meant to be exhaustive in any way. So what special lecture(s) would you like to see at the next ICM?
(Unless it is self-evident, please state what makes the lecture you would like to see "special". If you would like to nominate someone for an "ordinary" plenary lecture instead, please do so by sending me an email.)
How about a presentation of twenty three or so open problems to guide the next century of mathematics research. :)
Although not directly a "research question", I think this is a great use for mathoverflow!
@SamHopkins We're 20 years late for this century ;-) More seriously, assuming we do have an "open problems" lecture, would you have someone in mind for delivering it?
Great question. I think having a lecture on crisp, un-hyped research questions at the interface of math+AI, math+quantum computing, math+X in general will be quite valuable. Math+AI may be too broad and diffuse to be useful, so being more focused, e.g., on math+optimization-in-AI (my personal bias) could be more broadly useful. Similarly, topics such as: the role of "geometry of polynomials" and its wider connections can be a great topic. Perhaps, injecting "diversity" not only of people but of topics is going to be very helpful.
Part of the Bourbaki style is that the recent breakthrough is not presented by the author(s), and sometimes describes several related works, or the emergence of a new subdomain.
@Sam my understanding is that Hilbert only presented ten problems at the meeting. The other thirteen were included in his write-up.
I think that an important debate should be if there is a crisis in physics what importance have this for the mathematics (I've read a book, I disagree with part of content but it seems to me very valuable and brave; my belief is that there is a crisis in theoretical physics): I would like to know if one or two physicists could explain if there is a crisis in physics and if mathematicians can help about it (these talks / conferences can be also inspiring as reflections for mathematicians, because maybe it can be also a subtle crisis in mathematics in relation to the cited crisis in physics).
@user142929 What book?
The Spanish edition of the book Lost in Math: How Beauty Leads Physics Astray by professor Sabine Hossenfelder , @MonroeEskew
@user142929 My understanding is that physicists who believe there is a crisis in physics tend to have very diverging views about what the solution is, which I think lead to even more divergent views of what mathematicians could do to help. So there would be a big issue of who to pick for such a talk - choosing the most respected / distinguished physicist would seem to mean choosing whoever's ideas are already the most influential, which would be against the spirit.
@user142929 I imagine Sabine Hossenfelder, who as the title suggests things theoretical physicists rely on math too much, would not think there is much we can do to help and probably would not want to give such a talk. But certainly others with a similar view on the existence of a crisis might.
I can’t imagine it would be helpful to have a talk about purported or potential crises in physics at a math conference, particularly so for the ICM.
Many thanks @crispr I appreciate your comment and the feedback of previous professors in comments. I do not have much more to say (I'm not a professional mathematician), if ICM-MathOverflow wants to consider my question as a proposal or modulate this in the future I'm agree (it's in their hands).
Sabine Hossenfelder is criticizing certain physical theories motivated by recent math, not the math behind them. While she has interesting things to say (and I greatly enjoy her writing), I don't see how ICM would be the right venue for this. "See how your research is getting misapplied" is a dog-bites-man story to us mathematicians.
@MartinHairer thank you for asking this question here. I'm not game enough to suggest n answer, and it would be buried in the scores of comments on the existing answer most related. But I offer to the committee something along the lines of what is covered in this review. It seems more constructive than the discussion below.
Other idea could be a lecture about the site MathOverflow itself: brief history with comments/anecdotes/curiosities from senior users, papers that born/inspired in MathOverflow, Mathoverflow's community, statistics about the site, MathOverflow in social network, extracts (if it's possible) about answers and/or unsolved problems and a selection of extracts, importance of Meta, a call to recruit more new contributors, future perspective or projects ... I don't know what of these could be suitable and legitimate, if some professor wants to edit a post with this proposal I accept it. Best week
Other brave critics that I know, were by Renate Loll from last section of the article George Musser, La paradoja más famosa de la física se acerca a su fin, Investigación y Ciencia ( the journal Investigación y Ciencia is the Spanish edition of Scientific American), Febrero 2021, Nº 533. The article was also published in Quanta Magazine, but I know the Spanish edition of the cited article.
How about a lecture on proof assistants/formal proofs?
Most mathematicians are still skeptical of the value of proof assistants, and it's certainly true that proof assistants are still very difficult for the average mathematician to use. However, I think that much of the skepticism stems from a lack of understanding of what proof assistants have to offer. A popular misconception is that proof assistants just give you a laborious way of increasing your certainty of the correctness of a proof from 99% to 99.9999%. But that's not where their primary value lies, IMO.
For example, having a large body of formalized mathematics available could help machine learning algorithms figure out what constitutes "interesting" mathematics and help them autonomously discover interesting new definitions and concepts—something that seems beyond what computers can do now. For another example, there are increasingly many cases where editors can't find a referee for a complicated and potentially important paper because the referees are skeptical and don't want to waste time studying something that might be wrong. If proof assistants become sufficiently easy to use that authors are routinely required to formally verify their proofs before submission, then referees can focus on the more rewarding work of assessing whether a result is interesting and important instead of spending the bulk of their time checking correctness.
A good lecture on this topic could give the subject a valuable boost. Incidentally, if you want to poll people to assess interest, I would recommend polling younger people. This is one topic where I would value the opinion of younger mathematicians and students more than the opinion of senior mathematicians.
EDIT: Kevin Buzzard ended up giving precisely such a talk at ICM 2022: The rise of formalism in mathematics.
Personally I would love to see an interactive session with a proof assistant. I think few mathematicians have ever used one (I haven't). This might help in explaining why and how proof assistants can be useful to the mathematical community.
@FrançoisBrunault If you want to watch such an interactive session now, you can choose a lecture here: https://www.youtube.com/playlist?list=PLVZep5wTamMmvdvczjrLctDM9T4nBse1M (Of course such a session designed for the ICM would be a bit different).
@WillSawin So great, thanks! On a more general note, computers are used more and more to prove theorems but we don't really "see it" during the talks.
"If proof assistants become sufficiently easy to use that authors are routinely required to formally verify their proofs before submission"? And that such a requirement would not be considered inhumanely cruel by those authors? This would be great but are there many mathematicians expecting this to happen in, say, half a century? Personally, I am not among them.
@AlexGavrilov : I don't know, but a lot can happen in half a century. In 1960, it might have been considered inhumanely cruel to expect authors to produce camera-ready copy of their papers, but by 2010 pdflatex was standard.
Assia Mahboubi and Enrico Tassi just published a very relevant book. From their intro: Mathematical Components is the name of a library of formalized mathematics for the Coq system. It covers a variety of topics, from the theory of basic data structures to advanced results in various flavors of algebra. This library constitutes the infrastructure for the machine checked proofs of the Four Color Theorem and of the Odd Order Theorem. I think Assia and Enrico would both be terrific candidates for a lecture on proof assistants for mathematics.
Prof. Kevin Buzzard has been working on bringing the Lean proof assistant to the mathematics community, and has already given a lecture on proof assistants for mathematics at Microsoft Research: https://www.youtube.com/watch?v=Dp-mQ3HxgDE. I think he would also be an excellent candidate for such a lecture at ICM.
Speaking as a young mathematician, this kind of topic worries me. Aside from the obvious issues that can come with automation, an important part of doing mathematics is that the ideas are actually understandable to other mathematicians, and it’s not clear to me that: 1) if we make program that can infallibly verify a sequence of mathematical ideas, then there won’t be a trend toward just showing that the program says it works and not actually explaining (giving intuition, etc.) and 2) if we make a program that can prove things for us, then the proofs will be comprehensible to a human being.
@RobinGoodfellow : Your concerns are legitimate. But regarding #1, I think proof assistants will actually help, for 2 reasons. First, they'll force the author to understand the proof well enough to explain it to a computer. You'd be surprised how often authors don't completely grasp their own proofs, yet get away with it. I expect a decrease in excessively sketchy and vague papers. Second, if the computer is checking correctness, I expect increased social pressure on the humans to explain their proofs. What use is the human otherwise?
(continued) As for #2, this is already a problem with more conventional computer-assisted proofs, but I don't think that this is the fault of technology. Rather, it's just a fact of life that some theorems are probably only provable by means of a long and not particularly enlightening computation. Surely we'd rather have a computational proof than no proof at all? I also think that it's human nature to seek conceptual proofs, and I don't see human nature changing just because our ability to find computational proofs increases.
@RobinGoodfellow: Computer-verified proofs still require a human to come up with the proof. This might change somewhat with AI, but probably not in foreseeable time to an extent that would uproot mathematical practice (more likely, AI will play the role of a better Omega tactic, unpacking "straightforward induction" or "analogous to the previous proof" arguments that commonly get left to the reader to unpack). As to intuition or explaining why a proof works: I'd struggle to do that with my own (non-formalized) proofs, so I think that ship has already sailed. Mathematics is weird. ...
... John von Neumann said 70 years ago that "in mathematics, you don't understand things; you get used to them". Much of the time, you discover a new phenomenon by experimentation, then get used to it, and the habits you form become a kind of intuition that you cannot explain to someone who hasn't done comparable work. What is the intuition behind the tensor product being left-exact? Behind the Schur functions being symmetric? Behind the Witt vectors existing and classifying various things? Behind the RSK correspondence for semistandard tableaux?
@TimothyChow — There is wisdom in what you are saying, though (regarding your last point) I think that the human tendency toward laziness may overcome the human inclination toward conceptual proofs. We already often use theorems in the literature that we don’t ourselves know how to prove, so it isn’t entirely inconceivable that we could end up moving toward just asking software to prove everything for us without ourselves understanding (if research-level mathematics continues at all) after proof-making is automated.
@darijgrinberg — I think this might be more of a field-dependent thing. In topology and geometry (especially in low-dimensions), intuition tends to play a more important role.
@darijgrinberg you explained why I do/did (non-algebraic) combinatorics!
It would be great to have Dr Z.
@GilKalai : Who is Dr Z?
I imagine it's Doron Zeilberger.
I always enjoy Doron Zeilberger's talks, but I wasn't aware that he knew much about proof assistants.
@TimothyChow, Dear Tim, I suppose I consider the larger area of computerized proofs and computer assisted proofs that may include or relate to your suggestion of proof assistants/formal proofs (I am not sure what precisely you refer to). In any case it will be great to have Dr. Z.!
I think this would be a great topic for a special lecture. Perhaps there is also a case for a session in which this is explored in greater depth.
I suggest lectures on big and transformative ideas. For example, it would be great to have a lecture by Tim Gowers about the future of mathematics publishing, and getting away from the issues with our current model. He has spoken and written on topics like this before, e.g., in this blog post. Another option in the same vein might be an update on the Polymath project.
I'd too love to hear from Gowers about publishing, but my impression is that a panel would be even better, since the experience is currently distributed across various people (e.g., the editorial board of Algebraic Combinatorics will have something to say about zombie journals; whoever is behind fixing the K-Theory fiasco will probably know a lot about good and bad contracts; I'd also like to hear some failures, like OA journals folding, as told by their participants).
One of the tangible outcomes of Gowers’ foray into the crisis in publication was the creation of two journals (Sigma, Pi) where the author (or their grant/university) pays a publisher in order to have their article appear. I’m not convinced that this is a desirable outcome.
@Gordon that was merely the first stage. Gowers' more recent, arXiv-overlay, efforts, Algebraic Combinatorics and Advances in Combinatorics, would be more to your liking perhaps. Check this blog post for more recent comments about FofM https://gowers.wordpress.com/2018/06/04/a-new-journal-in-combinatorics/
@DavidRoberts Thanks for that link, pretty much exactly mirrors how I feel about it. Although small fry in the journal world, I was an editor of the Australasian Journal of Combinatorics when we decided to go electronic with a free-to-write, free-to-read model. With APC there is always an obvious potential conflict of interest.
@GordonRoyle I'd go further and say I am convinced that this is not a desirable outcome (Speaking as someone working in the UK, which produced the Finch report, and who has his name on a paper that appeared in Sigma for reasons not of his choosing)
A topic worthy of a special lecture, and with no obvious other place to go, is ways we as mathematicians can make our field more diverse, equitable, and inclusive. As we know, women and minorities are underrepresented in math. This has less to do with differences in talent and more to do with structural inequality in society, different access to mathematics as students, and perceptions from individuals in underrepresented groups that the mathematical community is not welcoming to them. A special lecture at ICM, drawing attention to these issues and including concrete suggestions for improving the situation, might go a long way towards making math more diverse in the future.
In addition to being the ethically correct thing to do (as being a mathematician is generally among the top jobs in terms of life satisfaction, and hence should be open to all), making math more diverse would also lead to better mathematics, as a diversity of thought and background will lead to new approaches to problems we care about. For example, lack of diversity has contributed to bad and biased algorithms, e.g., in mathematics related to criminal justice. There is already a large literature about concrete strategies to make math more diverse, including work of Uri Treisman, the book Whistling Vivaldi, the book Successful STEM Mentoring Initiatives for Underrepresented Students, and the Harvard implicit bias research. Sadly, many mathematicians are unaware of this body of research, and it doesn't neatly "fit" within our existing silos.
A great speaker for such a special lecture would be Francis Su, who has served in the leadership of both the AMS and MAA, who has worked on these issues for years, and who recently published Mathematics for Human Flourishing, a book which describes itself as "An inclusive vision of mathematics—its beauty, its humanity, and its power to build virtues that help us all flourish." Another great speaker would be Dave Kung.
In the same vein, one could imagine a special lecture on how to use mathematics for social good. Several texts and resources have recently appeared on this topic, including this book, this compendium, and these curricular guides. Mathematicians might appreciate a survey of work in this direction, including pointers on how to pivot their research and/or teaching in a direction of social justice.
This response has produced an extended discussion - the comments have been moved to chat.
Additional discussion is active at a different chat room: https://chat.stackexchange.com/rooms/111634/discussion-about-suggestion-of-icm-talk-focusing-on-making-math-more-diverse-inc Sorry about the potential confusion!
The Weapons of Math Destruction would make an interesting and timely topic for such a lecture.
Is the book that good?
What is "good"? It is a relevant topic for mathematicians and society, perhaps much more so than most math lectures.
More generally, mathematicians have very little ethics training compared to any other type of scientist (probably in part because theoretical mathematics does not function like a science). I think this is something that we need to be talking about more, and Cathy's book is one example of that.
The topic is certainly relevant. Yet, that was not my question.
I found the book interesting and learned many things from it.
I think one lecture topic should be devoted to (some aspects of) the communication and dissemination of mathematics. Even though it is like fitting a mini conference into one hour, aspects of bringing the subject to more people is important and current practitioners and presenters should be made aware of good practices in communication.
It might be useful to invite Matt Parker or Kelsey Houston-Edwards to speak about some of their process for emphasizing and explaining a topic. We as a group might shift our perspective on what goals are important to present (by lecture, Youtube video, blog post, or preprint) a subject. Even if we cannot all become great communicators, we can try to make our areas of study accessible to those who are.
Gerhard "Is My Point Coming Across?" Paseman, 2020.08.06.
Great idea. It strikes me that the communication of current research in mathematics to the general public is light years behind that which is taking place in theoretical physics and cosmology for example.
Especially since we lost Michael Atiyah in 2019, I would like to see a talk dedicated to the unity of mathematics. The idea of addressing the "tower of Babel" tendency of increased specialization is always needed, I think. This can be accomplished in several ways already suggested. Perhaps by giving an overview, or a list of visionary questions, or imagining new ways to accomplish a sense of unity in the diversity of the subject. Maybe a lecture entitled "the unity and diversity of mathematics". Such a title may even bring in topics mentioned such as inclusiveness, etc.
Just as aside comment (I hope don't disturb) I add the following reference. The article (in Spanish) La joven promesa española de las Matemáticas by Laura Moreno from the web page of the newspaper El Mundo (date 2 OCT. 2017) refers in its third paragraph a quote of the words that Sir Michael Atiyah provided as advice to Ismael Sierra.
Particularly in memory of John Conway, whose creations were mathematically interesting and nontrivial, while of potential appeal to a wide audience: a lecture on developments in accessible mathematics. The idea would be to present progress in solving old problems and new challenges in areas that could be reported by the nonspecialist media, to give the public a taste of what mathematicians do.
I'd suggest a lecture discussing when and how a computer can be useful to prove or disprove conjectures. As a first example, think about Euler's sum of powers conjecture. In 1769, Euler proposed a generalisation of Fermat's last theorem: for all integers $n$, $k$ greater than $1$, the equation
$$ a_1^k + a_2^k + \cdots + a_n ^k = b^k $$
implies that $n \geq k$. The conjecture is true for $k=3$ (this follows from Fermat's last Theorem). However, it has been first disproven for $k=5$ in 1966 via a direct computer search by L. J. Lander and T. R. Parkin. The couterexample they found was:
$$ 27^5 + 84^5 + 100^5 + 133^5 = 144^5 $$
Moreover, combining some results on elliptic curves, N. Elkies restricted the variables in the case $k=4$ and was able to find a counterexample using a computer:
$$ 2682440^4 + 15365639^4 + 18796760^4 = 20615673^4 $$
Here, it is intersting to notice that a computer search had not been able to find it (this is due to the fact that many parameters were involved): it was also necessary some work to restrict the situation to a more suitable case.
As a second example, consider the search for some kinds of primes: it has been conjectured that there exist infinitely many Wall-Sun-Sun primes; however, thanks to some computer searches, we now know that, if any such a prime exists, it must be bigger than $9.7 \cdot 10^{14}$.
As a third example, I will cite the search for lower bounds of de Bruijn–Newman constant: before the proof by Brad Rodgers and Terence Tao that $\Lambda \geq 0$, computer searches had established some bounds on this constant. Note also the relation with the searches for counterexamples to Riemann Hypothesis.
EDIT: Some examples of important results whose proofs required, at some steps, the help of a computer can be found, for instance, here. In some cases (e.g. Erdos discrepancy problem), a first (partial) proof involved the use of a computer, but later the conjecture has been completely proven without it. I think it may be also interesting to discuss the fact that many mathematicians, at least when the first cases of computer-assisted proofs appeared, did not accept the solutions as they were 'infeasible for a human to check by hand'.
I like the idea, however I don't particularly like your examples. Sure there was some interesting work involved in reducing the search space, but this still comes down to just checking if solutions in this space work. I feel similarly about the Wall-Sun-Sun primes - has the computer work brought us any results of theoretical interest, or just told us some finite set doesn't work? The last example seems more interesting - it is by no means obvious any of this can be achieved with a finite computation, and yet here we are.
Nice point. I have added a link with some examples of conjectures whose proofs required, at some point, the use of a computer. In these cases, a computer has indeed led to results of theoretical interest.
In my opinion some examples like the Erdos discrepancy problem and the de Bruijn-Newman constant essentially form arguments against the importance of computers, because the eventual proofs did not use any ideas generated by the computer experiments but instead totally different ones. Of course there are other areas where computers have been helpful (either in the sense of helping proofs, or in providing information about areas where proofs are not likely to be forthcoming) but one has to be careful in choosing examples.
A recent case where computer calculations were helpful to an eventual non-computerized proof is the recent proof of the sphere packings in dimension 24 https://arxiv.org/abs/1603.06518 where IIRC computer calculations that had been previously been done by Cohn and Elkies, Cohn and Kumar, Cohn and Miller, ... suggested properties of the key function used in the proof, which were used to rapidly find the right function once the general principles for how to construct such a function were found by Viazovska. (These works were discussed at the 2018 ICM.)
Similar is the progress on expressing numbers as sums of three cubes. There's a lot of work to trim down the search space and be clever about the algorithm.
A great recent example of a computer-assisted proof is the Kaluba--Kielak--Nowak proof that $\mathrm{Aut}(F_n)$ has Kazhdan's property (T) for $n\geq 5$. This seemingly completely abstract statement relies on an intensive computer calculation, but this came as a complete surprise to the community.
There are also examples where the computer plays a decisive role in finding the statement of the theorem (or the conjecture). This is different from computer-aided proofs.
I agree with @Wojowu -- this needs better examples. I think significant parts of the classification of finite simple groups (including, as I've learnt today, the first construction of the Janko group $J_4$!), rely on computer calculations. I'd love to know how often computer-relying proofs appear these days, and how often they get replaced by human-readable proofs within 10 years of their publication, ideally with a comparison between mathematical disciplines.
@WillSawin If we didn't have computers, people would still think there are sequences with discrepancy at most $2$.
@mathworker21 This is not true at all. The proof that there are no sequences with discrepancy at most $n$ for any $n$ would still have been found (except for the importance of computers in email, blogging, etc. to its proof).
@WillSawin You can't say that. There probably wouldn't have been so much attention placed on the problem if people thought there were sequences with discrepancy at most $2$ (which they would have thought -- you need to go up to length 1161 for heaven's sake!).
@mathworker21 This is not true. There was interest before 2014 - there was a Polymath in 2010, and at the time I believe most people believed the discrepancy was infinite. Why would someone believe there is a long discrepancy 2 sequence without finding one?
@WillSawin ok, you win. I am curious why people didn't think there would be an infinite sequence with discrepancy $2$. Like, who would have expected $L(d)$, the largest sequence with discrepancy at most $d$, to grow so quickly!!
@mathworker21 I'm not sure. An easy argument shows it grows at least exponentially, and you might expect that if an easy argument attains a certain growth then cleverer constructions can attain much higher growth. Anyways, I think we still don't know if the overall growth is asymptotically exponential.
During the lockdown I've seen an online talk by Pierre Pansu about persistent homology. Roughly (I'm not the right person to explain it) this is a robust and recent computational way to compute homology, at several scales, with the aim to ignore "noise". It's for instance used in shape recognition. Pansu's talk (which was in a geometric group theory seminar) was explicitly to advertise its used in pure math, and precisely in geometric topology / group theory, where it ought to bring new computational methods, more powerful than naives ones (e.g., if one wishes to under the shape, e.g., computing homological invariants, of small pieces of Cayley graphs). The talk was great and motivating (more than my poor summary!)
PS MathSciNet search for "persistent homology" (anywhere) yields papers: 0 in years $\le 2004$, 25 in 2005-2010, 100 in 2010-2015, and 200 in 2015-2020.
Absolutely agree that we need to focus on the pure-math applications, particularly after seeing the applied-math applications being hyped to death.
I didn't claim that we need to focus on the pure math applications, and even less said anything negative about the applied math applications. Pansu advertised the applied math applications, and suggested possible new uses in pure math. He was then talking to an audience in pure math and more specifically in geometric group theory. I'm not suggesting one should do exactly the same. Actually these two aspects address respectively "a survey lecture on a subfield relevant for some recent development and a lecture creating new connections between different areas of mathematics.
Was it this talk? https://www.youtube.com/watch?v=XMrsMlF73jc
@JCK no this was in April 2020, but the second half of the talk was close to your link. The first half of the talk was really a quick introduction to persistent homology. His slides (in French) are here.
I thought "pure math" when applied becomes applied math, no ?
@PiyushGrover you're playing with the word "applied", and I actually rather said "use". One might use/apply/adapt methods coming from rather applied math (traditionally applied, say, image recognition), to, say, compute Betti numbers of some manifolds defined by arithmetic means.
Maybe a panel lecture on tools for online collaboration.
A lot of people now know about and attend online seminars (as listed on researchseminars.org), and there has been some panel discussions already (e.g. this one). But as time goes by, probably more maturity is developing.
One could also be insterested by other aspects:
Machine-Learning-inspired live subtitles, which could help Alice and Bob collaborate when neither speak well enough the languages that the other speaks ;
prospects for automatic speech-to-$\LaTeX$ for taking live notes, or writing a draft
ordering equipement for a whole bunch of universities together, to get a better deal from providers
Indeed, these tools make positions at smaller universities perhaps more attractive than they used to, since daily collaboration/interactions is not restricted to departmental colleagues. They even make collaboration between academics and people from other places more possible (e.g. people working in public agencies, or the private sector).
How about a survey lecture on the impact of algebraic geometry in mathematical physics? Second proposal: A survey about the impact of mathematical algorithms for computational simulation in science and engineering.
In their recent ICM paper, Numbers, germs and transseries, Proceedings of the International Congress of Mathematicians, Rio de Janeiro, 2018, Volume 2, edited by B. Sirakov, P. N. de Souza and M. Viana, World Scientific Publishing Company, Singapore, pp. 19-42, Aschenbrenner, van den Dries and van der Hoeven discussed the ambitious program they are engaged in for extending asymptotic differential algebra to all of the surreals. During the last decade, there have been a wide array of advances in the theory of surreal numbers. I'd like to see a talk discussing those advances as well as the future prospects of Conway's theory.
Some meager developments on this theme can be found in the articles Surreal substructures, especially Section 8, and Defining a surreal hyperexponential.
Empirical processes are key to certain subfields such as high dimensional statistics, compressed sensing,... Even though the field of empirical processes is far from being new, I believe that presenting recent results by Naor, Latawa, van Handel or others, while having a view on recent applications could be beneficial to many.
Further, challenges arise both in applications and in theory and a talk (with two speakers?) could have its place at the ICM. It could either be a survey lecture or a lecture presenting connections, or even a survey of the connections. It could help more 'applied people' dig into some theoretical aspects or the other way round.
A lecture by Ken Ribet similar to his talk as the outgoing AMS president, updating the work on the Taniyama-Shimura-Weil and Serre conjectures, and modern proofs of Fermat’s Last Theorem.
https://youtu.be/mq9BS6S2E2k
Frank Calegari gave a (wonderful!) special lecture entitled "30 years of modularity: number theory since the proof of Fermat's Last Theorem" which very much fits with your suggestion. See https://www.youtube.com/watch?v=L0Z4Ng6ZJbY for the recording.
@SamHopkins this version has better audio https://youtu.be/HDRqU6oW--4
Other idea could be a lecture about the site MathOverflow itself. This is my proposal for a special lecture, the slides could be about
presentation and what's MathOverflow;
brief history with comments/anecdotes/curiosities from senior users;
papers that were borned/inspired in MathOverflow;
MathOverflow's community;
statistics about the site;
MathOverflow in social networks;
extracts (if it's possible) about answers and/or unsolved problems;
importance of Meta and critics;
a call to joint/recruit more new contributors;
perspectives or future projects...
underrepresented subcategories of mathematics;
policy of moderation (adding a greeting/thanking for the moderator team from the community of users of MathOverflow;
inherent problems of the site, for example downvotes without comments/feedback;
I don't know what of these could be suitable and legitimate for the owners of the site, moderators and community of the site.
I would like to dedicate with all respect this proposal to the Ukrainian casualties of the war started previous week.
I hope that this post is nice for all professors and users here, and I hope that in some ICM in next years, could be a talk/lecture about the importance of the site for professional mathematicians and other persons. If my post is suitable as answer in this thread of posts, feel free to add comments about improvements/critics of my proposal of special lecture for a ICM
This is already an answer.
Many thanks for your feedback @Carl-FredrikNybergBrodda I didn't known/remembered the answer that you mention (I see now the content and the dates of the cited answer and my comments), I think that my proposal is much more general (with all respect to the cited post that I've upvoted), but if there are more comments in the same way than yours I can to delete my answer in next few days Many thanks again.
Other: underrepresented subcategories of mathematics; policy of moderation (adding a greeting/thanking for the moderator team from the community of users of MathOverflow; downvotes without comments/feedback.
I would like to dedicate with all respect this proposal to the ukrainian casualties of the war started today (I think that isn't polite edit this dedicatory in these moments and situation, and I don't know what is the prespectives of next ICM)
@user142929: Just letting you know: I upvoted, because I share your views.
It may we worthwhile for the community to debate the following.
Do either (or both) of the following tend to diminish the importance of aesthetics in mathematics?
The "unreasonable" usefulness of mathematics in "real" life.
The pursuit of mathematical research as a career.
In particular, as a consequence, has the overall aesthetic quality of mathematics diminished over the last century or so? One could make the case that the aesthetic quality of mathematical work has become inaccessible, not only to the common man, but also to the lay man and (to a surprising degree) to the working mathematician from a different area of mathematics.
The former influences mathematics by defeating Weyl's famous claim that given a choice between what is useful and what is beautiful, he would choose the second.
The latter leads mathematicians to relentlessly tunnel down rabbit holes and not come out to meet and to exchange notes with other rabbits!
Of course, there are counterpoints! (Else this would not be worth debating.)
Weyl's quote is about truth and beauty. Not utility and beauty.
@Lucia Thanks for the correction. I will leave it as it is begging poetic licence since looking for truth is halfway to looking for utility!
No; I don't think you capture Weyl's meaning at all with that misquote. Weyl certainly did not have applications in mind. More relevant would be to quote Oscar Wilde from "The picture of Dorian Gray": "We can forgive a man for making a useful thing as long as he does not admire it. The only excuse for making a useless thing is that one admires it intensely."
@Kapil I just wanted to say, I like your suggestion. I agree that the need to get a job, and to be seen as doing "useful" work (grant organizations might be to blame here) has probably reduced the amount of mathematics done for aesthetic reasons. But, there are also WAY more mathematicians now than 50 years ago, so we probably haven't seen a drop in aesthetic quality, at least for the papers that "rise to the top." Still, I like that you draw attention to this issue. You might enjoy the Journal of Humanistic Mathematics.
@DavidWhite Thanks for the suggestion of reading JHM.
An expository debate between Peter Scholze and Shinichi Mochizuki on the veracity of the latter's claimed proof of the abc conjecture.
Why debate? A fistfight would probably be more fun for the spectators?
A public debate is probably not the best way to go about this. However, I would like to hear a talk outlining Mochizuki's proof and its criticisms, specifically on whether the doubts expressed by Schloze have been explained away in the version of the paper which will be published soon.
@D.S.Lipham to my knowledge, and I have been following rather closely, Mochizuki's papers have not changed in any substantial way. He has been updating them continuously on his website and providing close commentary on the changes (from saying when italics have been removed from a single word, to when specific subsubsections have been "clarified"). Nothing has been done about the complaints of the community as focused through Scholze and Stix, apart from saying on his blog they don't get basic logic.
I would suggest a short lecture on the usefulness of MathOverflow in Mathematical Research. The very question is indicative of the importance that ICM has given to MO. It would be better if some great problems and answers of MO and their impact in the larger body of Mathematics be lectured upon.
In addition, I suggest a lecture on Influence of Combinatorics in Mathematics. This is on the basis of my observation that recently the number of papers on arXiv is maximum in Combinatorics (the second is, I think Number Theory). Along with this, there are several papers in other topics wich crosslist to Combinatorics as a secondary topic. This clearly shows the wide influence of Combinatorics on all of Mathematics.
why the downvotes?
How does "most arXiv papers are in combinatorics" show the "wide influence of combinatorics on all of mathematics"? For all I know from this it could equally be the case that combinatorics is a very large but very isolated field of maths. (Of course I know this is not the case, but again, I don't see how cited arXiv data support that)
@Wojowu I mean several papers in other fields are also crosslisted in combinatorics along with their primary subject ( like many number theory papers are also cross listed in math.CO tag along), and number theory is as usual widely regarded as the'queen of mathematics'
Different fields have different publishing habits.
@FrançoisBrunault yes, but why not combinatorics wield an influence on Mathematics (and Mathematicians)?
I see your point, but I don't find this to be a particularly strong case in this point. Anyway, if there is much to be said in how combinatorics impacts other areas, I would be interested in it, since combinatorics doesn't usually strike me as a thing with terribly many connections to other things.
@Wojowu you mean, would something like category theory be good?
I hope you are in good health, I've edited a closely related proposal than yours, let me to know if it is a problem. I think that my proposal is more general. Any case feel free to edit my post to edit to add your post as reference or adding improvements of my proposal. Many thanks.
No need for so many downvotes on this.
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2025-03-21T14:48:31.685483
| 2020-08-06T13:38:18 |
368468
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Stack Exchange
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Name for vertex of a digraph reachable from a directed cycle
I'm wondering if there is an established name for vertices of a finite directed graph that are reachable from a directed cycle. These also can be described as endpoints of arbitratily long directed paths or even as endpoints of directed paths that are as long as the number of vertices in the graph. They are also the endpoints of a left infinite directed paths in the graph.
These crop up a lot in what I am working on now and so I need to name them. I was going to call them essential vertices but I prefer to use established nomenclature if possible.
In my context the digraphs have no sinks and so these vertices are on bi-infinite paths.
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2025-03-21T14:48:31.685608
| 2020-08-06T14:34:47 |
368471
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Stack Exchange
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A relation between norm and spectral radius for some matrix operators on Banach spaces $\ell^{p}$
Let $A=(a_{i,j})_{i,j=1}^{\infty}$ be a semi-infinite matrix with real entries. Suppose further that $A$ and $A^{T}$ (matrix transpose) represent bounded operators on $\ell^{p}$ for $p\geq1$. Denote further the spectral radius of $A^{T}A$ as
$$
r_{p}(A^{T}A):=\sup\{|\lambda| \mid \lambda\in\sigma(A^{T}A)\},
$$
where $\sigma(A^{T}A)$ is the spectrum of $A^{T}A$ regarded as an operator on $\ell^{p}$.
It is well-known that, if $p=2$, then $r_{2}(A^{T}A)=\|A\|_{\ell^{2}\to\ell^{2}}^{2}$ since $A^{T}A$ is self-adjoint and $\|A^{T}A\|_{\ell^{2}\to\ell^{2}}=\|A\|_{\ell^{2}\to\ell^{2}}^{2}$.
Is there an $\ell^{p}$-variant of the equality for general $p\geq1$? That is, can be the spectral radius $r_{p}(A^{T}A)$ related to the norm $\|A\|_{\ell^{p}\to\ell^{p}}$ or perhaps
to $\|A\|_{\ell^{p}\to\ell^{p}}$ and $\|A^{T}\|_{\ell^{q}\to\ell^{q}}$, where $q$ is the dual conjugate to $p$ ($1/q+1/p=1$)?
A modification of the claim or imposing additional conditions on $A$ is possible. I do not have a clear picture about the relation. I just would like to know if there are certain results of this kind. Thank you.
If $A,A^T=\ell^p\rightarrow\ell^p$, then the adjoints $A^T,A$ map $\ell^{p'}$ into itself. By interpolation (Riesz-Thorin), they map $\ell^2$ into itself. It will be often the case that the spectrum of $A^TA$ is not sensitive to the exponent $p$, in which case one will have $r_p(A^TA)=r_2(A^TA)=\|A\|_2\ne\|A\|_p$.
In general, the spectral radius of a bounded operator is bounded by its norm, i.e.
$$ r_p(A^T A) \leq \Vert A^T A\Vert_{\ell^p\to\ell^p},$$
therefore you have the inequality
$$ r_p(A^T A) \leq \Vert A^T \Vert_{\ell^p\to\ell^p}\ \Vert A \Vert_{\ell^p\to\ell^p}. $$
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2025-03-21T14:48:31.685762
| 2020-08-06T14:42:51 |
368472
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Stack Exchange
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Is $T(n)=\sum_{k=1}^{n}\frac{\lambda(k)\Lambda(k)}{k} \geq 0$ and what is the upper bound of $T(x)=\sum_{n\leq x} \lambda(n)\Lambda(n) $?
Let $\Lambda(n)$ denote the von Mangoldt function: $\Lambda(n)=\log p$ when $n=p^e$ is a prime power ($e\ge 1$) and $\Lambda(n)=0$ otherwise.
and $\lambda(n)$ be Liouville Function, , I'm interested to know about average growth of the following function $T_j(x)=\sum_{n\leq x}\frac{\lambda(n)\Lambda(n)}{n^j} $ such that I want to determine its upper bound for large $n$ , Probably it is known that positivity of $T(n)=\sum_{k=1}^{n}\frac{\lambda(k)}{k} $ for large $n$ is an open problem such that its confirmation led to proof of the Riemann hypothesis according to Pál Turán ,My two dependents questions are :
Question:
a) Is $T(n)=\sum_{k=1}^{n}\frac{\lambda(k)\Lambda(k)}{k} \geq 0$ (Does this sum also behave like related sum to Pólya conjecture )
b) What is the upper bound of $T_j(x)=\sum_{n\leq x}\frac{\lambda(n)\Lambda(n)}{n^j} $ , take the case $j=0$?
Note: The motivation of this question is to look if the positivity of the titled sum also led to proof of Riemann hypothesis
This question makes no sense: If $n =p^e$ is a prime power (so usually $e=1$) then $\lambda(n) =(-1)^e$ which is usually just $-1$.
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2025-03-21T14:48:31.685886
| 2020-08-06T16:31:00 |
368478
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Stack Exchange
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Hadwiger number of a graph: Question about the original article from 1943
I am analyzing Hadwiger's original article (Hadwiger, Hugo (1943), "Uber eine Klassifikation der Streckenkomplexe", Vierteljschr. Naturforsch. Ges. Zurich, 88: 133–143) for my work related to the Hadwiger Conjecture in graph theory.
This article is in German, and Hadwiger's terminology from 1943 is very different from current graph theory terminology. For example, he says "Komplex" for what we now call a graph, and he says "Simplex S(n)" for what we now call the complete graph $K_n$ over $n$ vertices. Also, when he says "K(k)", now we would call this a graph with Hadwiger number $k$ (for him, K(k) is not the complete graph).
In this article, he defines what we now call the Hadwiger number ${\rm had}(A)=k$, the size of the largest complete graph $K_k$ that can be obtained by contracting edges of $A$ (where $A$ is an undirected graph).
His original definition is
Ein Komplex A heisst ein K(k), wenn er sich auf einen S(k),
aber nicht auf einen S(k+1) zusammenziehen lässt. Ein K(k) ist notwendig ein zusammenhängender Komplex.
I am translating this into current terminology "An undirected graph $A$ has Hadwiger number $k$, if the complete graph $K_k$, but not $K_{k+1}$, can be obtained by contracting edges of $A$. It is necessarily a connected graph."
In the next sentences he says
Die Eigenschaft, ein K(k) zu sein, kennzeichnet eine Art des höheren Zusammenhangs, die durch die natürliche Zahl k gegeben ist.
What does "höheren Zusammenhangs" mean here? My current terminology translation would be "The property to have Hadwiger number $k$ characterizes a higher degree of connectedness, which is given by the natural number $k$".
But my second, more deliberate translation sounds more logical to me: "The property to have Hadwiger number $k$ characterizes a more abstract idea of connectedness, which can just be expressed by the natural number $k$". But it is more of an interpretation than the first version.
I have asked two German native speaker scientists, but they were not sure as they are no experts in graph theory. It would be great if someone could help me with this.
If that helps: In algebraic topology one would say the same about vanishing of homotopy groups up to degree k: Das Verschwinden der Homotopiegruppen bis zur Ordnung k charakterisiert eine höhere Art des Zusammenhangs charakterisiert eine Art des höheren Zusammenhangs.
The meaning of „höher“ is as in „in einem höheren Sinn“, which (I think) translates into English as „in a higher sense“.
@ThiKu great comment. It is true that Zusammenhang might not refer to connectedness here. Let me give it a thought or two, it is interesting.
Actually I meant that Zusammenhang would mean connectedness but that höherer Zusammenhang might mean „connectedness in a higher sense“ as well as a ‚numerically higher‘ connectedness.
Let me give it a try. As a disclaimer, English is not my mother tongue, so my translation might have linguistic flaws.
First of all, I would say the sentence is hard to translate and it is a bit informal, i.e. it is not a rigorous mathematical statement. In my view, this sentence gives an informal motivation why it it is useful to look a "K(k)" graphs, i.e. graphs with Hadwiger number $k$. Secondly, I think your 2nd version is relatively close to what this sentence means, but I think the meaning goes beyond connectedness.
My suggestion is to translate it in the context of the beginning of the whole paragraph where you took the quotes from. The first sentence in this paragraph reads:
"Im folgenden sprechen wir von einer Möglichkeit der Klassifikation der
Streckenkomplexe, die besonders im Hinblick auf das Problem der chromatischen Zahl von besonderem Interesse zu sein scheint."
(My translation: In the following, we speak of a possible classification of graphs, which seems to be of particular interest regarding the problem of chromatic number.)
So this is setting the scene really broadly, and announcing that the following definition could be relevant for the chromatic number problem, and in particular for the famous 4-color-problem.
Next, he defines the Hadwiger number (his "K(k)" graphs), and then there is the sentence you are asking about. Here is my suggested translation:
The property of having Hadwiger number $k$ ($k$ a natural number) characterizes a deeper connection between those graphs. (in the sense of: deeper than the ordinary notion of connectendness)
I don‘t think that it is a statement about connections between the graphs. It is a statement about one graph for the moment.
@ThiKu I would tend to think "connection between these graphs" is actually ok. Grammatically, Hadwiger refers to one object, but what he really refers to is one property that all these graphs (with Hadwiger parameter k) have in common. I think, his term "ein K(k)" is representing a whole class of graphs, the defining property of this class.
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2025-03-21T14:48:31.686279
| 2020-08-06T17:54:56 |
368483
|
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|
Stack Exchange
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Geometric interpretation of the Weyl tensor?
The Riemann curvature tensor ${R^a}_{bcd}$ has a direct geometric interpretation in terms of parallel transport around infinitesimal loops.
Question: Is there a similarly direct geometric interpretation of the Weyl conformal tensor ${C^a}_{bcd}$?
Background: My understanding is that the Weyl conformal tensor is supposed to play a role in conformal geometry analogous to the role of the Riemann curvature tensor in (pseudo)Riemannian geometry. For instance, it is conformally invariant, and (in dimension $\geq 4$) vanishes iff the manifold is conformally flat, just as the Riemann curvature tensor is a metric invariant and vanishes iff the manifold is flat. The two tensors also share many of the same symmetries. So it would be nice to have a more hands-on understanding of the Weyl tensor when studying conformal geometry.
Notes:
I'd be especially happy with a geometric interpretation which is manifestly conformal in nature, referring not to the metric itself but only to conformally invariant quantities like angles.
I'm also keen to understand any subtleties which depend on whether one is working in a Riemannian, Lorentzian, or more general pseudo-Riemannian context.
One aspect which is important to have in mind when dealing with the Weyl tensor is that due to its algebraic symmetries it identically vanishes if the manifold dimension is less than four. In three dimensions there is another conformal tensor built from the metric and its derivatives, called the Bach tensor, which shares many of the properties of the Weyl tensor. In Lorentzian geometry, the importance of the Weyl tensor also lies on the fact that, on space-times solving the Einstein equations, the Weyl tensor encodes all dynamically propagating degrees of freedom (e.g. gravitational waves).
Penrose has some way of thinking about the Weyl tensor geometrically, but only for Lorentzian signature 4-manifolds, if I remember correctly, but I think his approach splits into self-dual and anti-self-dual parts first, and then gives each one an interpretation.
There is such an interpretation, with a few caveats. Essentially, there is a canonical connection on a certain vector bundle for which the "principal part" of the curvature is the Weyl tensor in dimensions $n\geq4$, and the Cotton tensor when $n=3$. I will describe this from the point of view of the tractor calculus, but avoid introducing unnecessary bundles where needed. This can also be described using the Fefferman–Graham ambient metric or using Cartan connections. This summary mostly follows Bailey–Eastwood–Gover, though Armstrong and articles written by Gover are also good references. I use abstract index notation throughout.
First, we define conformal densities. Given a conformal manifold $(M,c)$, a conformal density of weight $w\in\mathbb{R}$ is an equivalence class of pairs $(g,f)\in c\times C^\infty(M,c)$ with respect to the equivalence relation $(g,f)\sim(e^{2\Upsilon}g,e^{w\Upsilon}f)$. Let $\mathcal{E}[w]$ denote the space of conformal densities of weight $w$. We similarly define $\mathcal{E}^i[w]$ as the space of equivalence classes of pairs $(g,v^i)\in c\times\mathfrak{X}(M)$ with respect to the equivalence relation $(g,v^i)\sim(e^{2\Upsilon}g,e^{w\Upsilon}v^i)$. Here $\mathfrak{X}(M)$ is the space of vector fields on $M$.
Next, we define the space of sections of the standard tractor bundle. Fix a metric $g\in c$. Define $\mathcal{T}_g^A=\mathcal{E}[1]\oplus\mathcal{E}^i[-1]\oplus\mathcal{E}[-1]$. Given another metric $\hat g \mathrel{:=} e^{2\Upsilon}g\in c$, we identify $(\sigma,v^i,\rho)\in\mathcal{T}_g^A$ with $(\hat\sigma,\hat v^i,\hat\rho)\in\mathcal{T}_{\hat g}^A$ if
$$ \begin{pmatrix} \hat\sigma \\ \hat v^i \\ \hat\rho \end{pmatrix} = \begin{pmatrix} \sigma \\ v^i + \sigma\Upsilon^i \\ \rho - \Upsilon_j v^j - \frac{1}{2}\Upsilon^2\sigma \end{pmatrix} . $$
(Recall these are densities, so exponential factors are suppressed.) The space of sections $\mathcal{T}^A$ is the result after making this identification. Note that the top-most nonvanishing component is actually conformally invariant modulo multiplication by an exponential factor. Because of this, we call the top-most nonvanishing component the projecting part.
There is a canonical connection on (the vector bundle whose space of sections is) $\mathcal{T}^A$, the standard tractor connection, which, given a choice of metric $g\in c$, is given by the formula
$$ \nabla_j \begin{pmatrix} \sigma \\ v^i \\ \rho \end{pmatrix} = \begin{pmatrix} \nabla_j\sigma - v_j \\ \nabla_j v^i + \sigma P_j^i + \delta_j^i\rho \\ \nabla_j\rho - P_{ji}v^i \end{pmatrix} . $$
Here $P_{ij}=\frac{1}{n-2}\left( R_{ij} - \frac{R}{2(n-1)}g\right)$ is the Schouten tensor and $n=\dim M$. It is straightforward to check that this is well-defined, in the sense that it is independent of the choice of matrix $g\in c$.
Given a metric $g\in c$, it is straightforward to compute that
$$ (\nabla_i\nabla_j - \nabla_j\nabla_i)\begin{pmatrix} \sigma \\ v^k \\ \rho \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ C_{ij}{}^k & W_{ij}{}^k{}_l & 0 \\ 0 & -C_{ijl} & 0 \end{pmatrix} \begin{pmatrix} \sigma \\ v^l \\ \rho \end{pmatrix} . $$
This is conformally invariant by construction. The "3-by-3" matrix is the tractor curvature, and its projecting part is $W_{ij}{}^k{}_l$ when $n\geq4$ and $C_{ij}{}^k$ when $n=3$. Standard interpretations of holonomy then give the interpretation of the Weyl tensor in terms of parallel transport around infinitesimal loops that I indicated in the first paragraph.
Finally, given your bullet points, let me emphasize that the signature of $c$ plays no role here, and everything is manifestly conformally invariant.
Added in response to a comment. There are many geometric motivations for introducing the standard tractor bundle. One is that the conformal group of the sphere is $\operatorname{SO}(n+1,1)$, so it makes sense that the right replacement of the tangent bundle of a conformal $n$-manifold should be a bundle of rank $n+2$, as is the standard tractor bundle. Note that the metric on $\mathcal{T}$ has signature $(n+1,1)$, assuming we start with a conformal manifold of Riemannian signature (if $c$ has signature $(p,q)$, the metric on the standard tractor bundle has signature $(p+1,q+1)$).
Another motivation comes from the ambient metric. First, note that the flat conformal sphere $(S^n,c)$ (i.e., the conformal class of the round $n$-sphere) can be identified with the positive null cone $\mathcal{N}$ centered at the origin in $\mathbb{R}^{n+1,1}$. This is done by noting that the projectivization of $\mathcal{N}$ is $S^n$ and identifying sections of $\pi\colon\mathcal{N}\to S^n$ with metrics in the conformal class $c$ by pullback of the Minkowski metric. (Incidentally, this leads to a proof that $\operatorname{SO}(n+1,1)$ is the conformal group of $S^n$.) In this case, a fiber $\mathcal{T}_x$ of the standard tractor bundle is identified with $T_p\mathbb{R}^{n+1,1}$ for some $p\in\pi^{-1}(x)$; this is made independent of the choice of $p\in\pi^{-1}(x)$ by identifying tangent spaces at points subject to a homogeneity condition matching that of $\mathcal{E}^i[-1]$ above. The standard tractor connection is then induced by the Levi–Civita connection in Minkowski space, after making some identifications.
For a general conformal manifold $(M^n,c)$ of Riemannian signature, Fefferman and Graham showed that there is a "unique" Lorentzian manifold $(\widetilde{\mathcal{G}},\widetilde{g})$ which is "formally Ricci flat" and in which $(M^n,c)$ isometrically embeds as a null cone. Here formally Ricci flat means that the Ricci tensor of $\widetilde{g}$ vanishes to some order, depending on the parity of $n$, along the null cone, and I write unique in quotes because the metric is only determined as a power series to some order along the cone, and this up to diffeomorphism. One recovers the standard tractor bundle and its canonical connection from that of $(\widetilde{\mathcal{G}},\widetilde{g})$ as in the previous paragraph. See Fefferman–Graham for details, or Čap–Gover for a detailed description of the relation between the tractor calculus and the ambient metric, including the identifications I didn't detail. A similar construction for other signatures works, consistent with what is described in the previous paragraph.
Thanks! It's great to learn that to any conformal manifold is naturally associated this tractor bundle and a connection on it of which the Cotton and Weyl tensors are components of the curvature -- just as the Riemann curvature tensor is the curvature of the Levi-Civita connection on the tangent bundle. Right now the tractor bundle and its connection look much more mysterious to me than the tangent bundle and the Levi-Civita connection. Is there anything to be said about what they "mean geometrically" which might make them seem as natural as the tangent bundle and Levi-Civita connection?
I have added some comments which give one possible interpretation. I recommend you read the introduction of Čap--Gover, cited in my answer, for further details or interpretations.
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2025-03-21T14:48:31.686880
| 2020-08-06T18:29:07 |
368489
|
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|
Stack Exchange
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Sign of the function $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$
It is well-known that the Mertens function $M(n)=\sum_{k=1}^n\mu(k)$ changes sign infinitely many times when $n\rightarrow +\infty$. Let $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$, then $\lim\limits_{n\rightarrow +\infty}f(n)= 0$.
Question: Does the function $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$ also change sign infinitely many times when $n\rightarrow +\infty$ ?
Yes it does. To see this, note that by partial summation,
$$\frac{1}{\zeta(s + 1)} = s\int_{1}^{\infty}\sum_{n \leq x} \frac{\mu(n)}{n} x^{-s} \, \frac{dx}{x}$$
for all $\Re(s) > 0$. Now let $\Theta$ denote the supremum of the real part of the zeroes of $\zeta(s)$, and suppose in order to obtain a contradiction that there exists some $\varepsilon > 0$ and $x_{\varepsilon} > 1$ such that
$$\sum_{n \leq x} \frac{\mu(n)}{n} < x^{-1 + \Theta - \varepsilon}$$
for all $x > x_{\varepsilon}$. Then Landau's lemma (Lemma 15.1 of Montgomery-Vaughan) states that if $\sigma_c$ is the infimum of $\sigma \in \mathbb{R}$ for which
$$\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{n}\right) x^{-\sigma} \, \frac{dx}{x}$$
is convergent, then
$$\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{n}\right) x^{-s} \, \frac{dx}{x}$$
is holomorphic in the right half-plane $\Re(s) > \sigma_c$ but not at the point $\sigma_c \in \mathbb{R}$. On the other hand, this integral is equal to
$$\frac{1}{s + 1 - \Theta + \varepsilon} - \frac{1}{s\zeta(s + 1)}$$
for $\Re(s) > 0$ and hence for $\Re(s) > \sigma_c$ by analytic continuation. However, this expression has a pole at $s = -1 + \Theta - \varepsilon$ and no other poles on the real line segment $\sigma > -1 + \Theta - \varepsilon$, yet by the definition of $\Theta$, there are poles in the strip $-1 + \Theta - \varepsilon < \Re(s) \leq -1 + \Theta$. Thus a contradiction is obtained, and so it follows that
$$\sum_{n \leq x} \frac{\mu(n)}{n} = \Omega_{+}\left(x^{-1 + \Theta - \varepsilon}\right).$$
The same method shows that
$$\sum_{n \leq x} \frac{\mu(n)}{n} = \Omega_{-}\left(x^{-1 + \Theta - \varepsilon}\right),$$
which implies an infinitude of sign changes. Moreover, with more work, one can show that
$$\sum_{n \leq x} \frac{\mu(n)}{n} = \Omega_{\pm}\left(\frac{1}{\sqrt{x}}\right),$$
and if one is sufficiently enthusiastic, then under the assumption of the Riemann hypothesis and the linear independence hypothesis,
$$\limsup_{x \to \infty} \sqrt{x} \sum_{n \leq x} \frac{\mu(n)}{n} = -\liminf_{x \to \infty} \sqrt{x} \sum_{n \leq x} \frac{\mu(n)}{n} = \infty.$$
The "true" rate of growth is probably the following:
$$0 < \limsup_{x \to \infty} \frac{\sqrt{x}}{(\log \log \log x)^{5/4}} \sum_{n \leq x} \frac{\mu(n)}{n} < \infty, \quad -\infty < \liminf_{x \to \infty} \frac{\sqrt{x}}{(\log \log \log x)^{5/4}} \sum_{n \leq x} \frac{\mu(n)}{n} < 0.$$
See also these two answers of mine.
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2025-03-21T14:48:31.687068
| 2020-08-06T18:35:53 |
368490
|
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"Eoin",
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|
Stack Exchange
|
What can be said about the Chow rings of classifying spaces of semi-direct products of groups?
For instance, what can we say about the Chow ring of the classifying space of a semi-direct product $CH^*(B(G\ltimes H))$, in terms of the Chow rings of $CH^*(BG)$, $CH^*(BH)$, and the singular cohomology groups $H^*(BG,\mathbb{Z})$, $H^*(BH;\mathbb{Z})$ and $H^*(B(G\ltimes H);\mathbb{Z})$?
You might find some references in the book
Group Cohomology and Algebraic Cycles by Burt Totaro.
E.g. Lemma 2.12 handles some wreath products.
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2025-03-21T14:48:31.687127
| 2020-08-06T18:37:25 |
368491
|
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"url": "https://mathoverflow.net/questions/368491"
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|
Stack Exchange
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What is the importance of singularities of type II in the Mean Curvature Flow?
I am reading the Mean Curvature Flow and Isoperimetric Inequalities by Manuel Ritoré, Carlo Sinestrari, Vicente Miquel and Joan Porti and I am curious to know what is the importance in understand the singularities of type II in the Mean Curvature Flow.
Thanks in advance!
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2025-03-21T14:48:31.687190
| 2020-08-06T18:44:01 |
368493
|
{
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"authors": [
"Leo Moos",
"Pedro Lauridsen Ribeiro",
"https://mathoverflow.net/users/103792",
"https://mathoverflow.net/users/106263",
"https://mathoverflow.net/users/11211",
"user7868"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368493"
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Stack Exchange
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Reference for sets of locally finite perimeter on Riemannian manifolds
I am looking for a reasonably complete reference for Ennio De Giorgi's theory of sets of locally finite perimeter (also christened by him as Caccioppoli sets, after Renato Caccioppoli's pioneering work on the subject) and the associated theory of functions of locally bounded variation in Riemannian manifolds.
I am aware there are quite a few excellent references on the subject in Euclidean space, such as:
L. Ambrosio, N. Fusco, D. Pallara, Functions of Bounded Variation and Free Discontinuity Problems (Oxford University Press, 2000);
L. C. Evans, R. F. Gariepy, Measure Theory and Fine Properties of Functions (CRC Press, 1992);
H. Federer's classic, Geometric Measure Theory (Springer-Verlag, 1969 - particularly Section 4.5);
M. Giaquinta, G. Modica, J. Soucek, Cartesian Currents in the Calculus of Variations I (Springer-Verlag, 1998);
E. Giusti, Minimal Surfaces and Functions of Bounded Variation (Birkhäuser, 1984);
F. Maggi, Sets of Finite Perimeter and Geometric Variational Problems (Cambridge University Press, 2012);
W. F. Pfeffer, The Divergence Theorem and Sets of Finite Perimeter (CRC Press, 2012);
L. Simon, Lectures on Geometric Measure Theory (Australian National University, 1984);
W. P. Ziemer, Weakly Differentiable Functions (Springer-Verlag, 1989).
However, the extension of this theory to Riemannian manifolds lies at the very basis of important topics in the latter such as isoperimetric inequalities, positive mass theorems and Penrose inequalities, which are also of importance in General Relativity. Most contemporary papers in the subject quote one or more of the above references and (implicitly or explicity) dismiss the work needed in performing such an extension as straightforward and thus left to the reader. A few papers do spell a few details of what should change in general Riemannian manifolds, but essentially restricting themselves to the very basics - i.e. the definition of functions of (locally) bounded variation and sets of (locally) finite perimeter therein, but no more. This makes these papers quite hermetic to newcomers.
At the other extreme, we do have an extension of De Giorgi's theory to metric measure spaces (which of course encompass Riemannian manifolds), but usually the treatments of such a far-reaching extension focus on aspects absent even on Riemannian manfolds. There is considerable interest in the particular case of sub-Riemannian manifolds, but these are also sufficiently far removed from Riemannian manifolds so as to be of little applicability to the latter.
I am particularly interested in the treatment of topics such as Part Two (Chapters 12-20) of F. Maggi's book cited above in the context of Riemannian manifolds. The closest thing I could find to what I am looking for is the Diplomarbeit of Alexander Volkmann (2010) (see particularly Chapter 2), but it stops short of going beyond lower semicontinuity of perimeter and the ensuing closure of the class of sets of (locally) finite perimeter under finite unions and intersections. De Giorgi's structure theorem and Federer's theorem on the measure-theoretic boundary of sets of locally finite perimeter, for instance, seem to find no discussion at all in the literature of Riemannian manifolds, despite their obvious importance. Other references that do discuss a few aspects of the theory in the context of Riemannian manifolds are
M. Miranda, D. Pallara, F. Paroneto and M. Preunkert, Heat semigroup and functions of bounded
variation on Riemannian manifolds, J. reine angew. Math. 613 (2007) 99-119;
A. Carbonaro, G. Mauceri, A note on bounded variation and heat semigroup on Riemannian manifolds, Bull. Austral. Math. Soc. 76 (2007) 155-160;
B. Güneysu, D. Pallara, Functions with bounded variation on a class of Riemannian manifolds with Ricci curvature unbounded from below, Math. Ann. 363 (2015) 1307-1331,
but all of the above are rather focused on the extension of De Giorgi's heat kernel characterization of the variation of $L^1$ functions to Riemannian manifolds and do not go much further than that.
I wish there was more work in this area. Thanks for sharing what you have - the Volkmann reference saved me a lot of work. (The link above doesn't work for me, but I found it at https://alexandervolkmann.files.wordpress.com/2014/10/diplom.pdf).
Thanks for letting me know about the broken link (though gmvh has already edited my post with the active link). It has been a while...
@PedroLauridsenRibeiro Apologies for commenting on an old question; in case you are still interested, I wanted to write a quick comment. At first sight Leon Simon's book seems to be set in Euclidean space, but the statements actually apply to subsets of Riemannian manifolds. For the general theory - including Allard regularity for example -, you can either work in charts or with a Nash embedding. In fact, for some chapters (including the 37th, 'Codimension 1 Minimizing Currents'), Leon Simon's book explicitly works with subsets of Riemannian manifolds.
@LeoMoos thanks for the comment. I have no doubt that the main results of De Giorgi's theory should still hold for Riemannian manifolds, but the actual proofs in the correct generality cannot be easily found in a systematic form in a reference that can be cited - they rather remain either as scattered, incomplete pieces or as folklore. Moreover, you seem to refer to results on Federer's theory of rectifiable sets and currents (which is admittedly more general) rather than De Giorgi's approach. Simon's book is no different - the discussion in Section 14 is restricted to Euclidean space.
@PedroLauridsenRibeiro It's difficult to make an absolute statement ('the proofs in Simon's book all apply in Riemannian manifolds'), but I think the truth is not far off. Since you mention Section 14, the results given there - notably Thm. 14.3 - do generalise immediately, because you can work in bilipschitz charts. The image of a set with finite perimeter in $(M,g)$ has finite perimeter with the Euclidean metric in the chart. Last, what you say about currents is not completely accurate. In the top dimension, currents (with boundaries representable by integration) are Caccioppoli sets.
@LeoMoos ok, let me elaborate on why I don't think Simon's book qualifies as a reference in the terms posed in my question. I do understand these results all generalize to Riemannian manifolds, but the point is that there is nontrivial (albeit straightforward) work to be done to get there and most of it isn't done in any reference I know of. As I said in the OP, "details are left to the reader" doesn't qualify - what you said of Simon's book can be said of most of the references I quoted in Euclidean space.
(continued) Volkmann's diploma thesis linked above illustrates what a reference I'm asking for should look like. Unfortunately, as I said above, it discusses only the very basics of De Giorgi's theory in the setting of Riemannian manifolds and key results are missing. Finally, regarding rectifiable sets and currents, by "more general" I meant they go beyond top dimension - due to that, the language used by Federer to discuss them is somewhat different from De Giorgi's. The latter's structure theorem forms precisely the bridge between both setups in top dimension.
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2025-03-21T14:48:31.687766
| 2020-08-06T18:48:25 |
368494
|
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|
Stack Exchange
|
Sum of indicator functions of binomial random variables
Let $x_1, x_2,..., x_m$ be iid binomial random variables (each with a number of trials n and probability of success in each trial p). Define a list of binary indicator variables $y_1,y_2,...,y_m$ for at least $K$ successes i.e.,
$ y_i =
\begin{cases}
1 & \text{if } x_i\geq K\\
0 & \text{else}
\end{cases}$
What is the variance of $z=\sum_1^m y_i$ ?
We have $Var\,z=m\,Var\,y_1$,
$$Var\,y_1=P(x_1\ge K)(1-P(x_1\ge K))=P(x_1\ge K)P(x_1<K))=(1-F(K-1))F(K-1)
=(1-I_{1-p}(n-K+1,K))I_{1-p}(n-K+1,K)$$
according to this, where $F$ is the cdf of the binomial distribution with parameters $n$ and $p$, and $I_\cdot(\cdot,\cdot)$ is the regularized incomplete beta function. Thus,
$$Var\,z=m\,(1-I_{1-p}(n-K+1,K))I_{1-p}(n-K+1,K).$$
|
2025-03-21T14:48:31.687841
| 2020-08-06T18:58:15 |
368496
|
{
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"authors": [
"Ali Taghavi",
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}
|
Stack Exchange
|
A kind of holomorphicity of maps on Hilbert space
Let $H$ be an infinite dimensional seperable Hilbert space. Is there an Irreducible involutive sub algebra $D$ of $B(H)$ with the following properties?:
1)For every open set $U\subset H$ and every Frechet differential map
$f:U \to H$ with $Df(x)\in D,\; \forall x\in U$, the mapping $f$ is automatically $C^{\infty}$
2)For every open set $U\subset H$, a uniform limit $f$ of a sequence of Frechet differential maps $f_n:U \to H$ with $Df_n(x)\in D$ is Frechet differentiable which satisfies $Df(x)\in D,\;\forall x\in U$.
The motivation comes from the concepet of holomorphic functions when we put (the real analogy) $ H=\mathbb{R}^2$ and $D=\left \{\begin{pmatrix} a&b\\ -b&a\end{pmatrix} \mid a,b\in \mathbb{R} \right \}$
In this book, Thm 7.19 (1) $\Longleftrightarrow$ (9) answers this for $D$ the complex linear bounded maps, but you have to assume that $f$ is Gateaux differentiable with the derivative locally Lipschitz.
From Thm 7.17 (7) you can deduce an answer, but you have to adapt the definitions and strengthen the convergence at least at one point.
There is also the book:
Dineen, Seán
Complex analysis in locally convex spaces. (English) Zbl 0484.46044
North-Holland Mathematics Studies, 57. Notas de Matematica (83). Amsterdam - New York - Oxford: North-Holland Publishing Company. XIII, 492 p.
Thank you very much for your answer
|
2025-03-21T14:48:31.687961
| 2020-08-06T19:29:04 |
368498
|
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"Nate",
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"url": "https://mathoverflow.net/questions/368498"
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|
Stack Exchange
|
What characterizes the incidence matrix of a tripartite hypergraph?
The incidence matrix of a graph $G = (V,E)$ is a matrix with $|V|$ rows and $|E|$ columns, in which element $v,e$ is $1$ if node $v$ is incident to edge $e$, and $0$ otherwise.
In bipartite graphs, the incidence matrix has the following property (called "total unimodularity"): the determinant of every square submatrix is either $-1$ or $0$ or $1$. This property is very useful, because it implies that every linear program in which the constraints are based on this matrix (for example: for finding a maximum cardinality matching, maximum weight matching or minimum vertex cover) has a solution in which all values are integers.
I am now researching tripartite hypergraphs. Their incidence matrix need not be totally unimodular. For example, consider the hypergraph on the vertices $\{1,4,2,5,3,6\}$ with hyperedges: $\{ \{4,5,3\}, \{4,2,6\}, \{1,5,6\} \}$. Its incidence matrix is 6 by 3:
\begin{pmatrix}0& 0& 1\\0 &1 &0\\1& 0& 0\\1& 1& 0\\1& 0& 1\\0& 1& 1\end{pmatrix}
The bottom submatrix is a 3-by-3 square matrix whose determinant is $-2$.
Is there another property, weaker than total unimodularity, which characterizes the incidence matrices of tripartite hypergraphs?
Cross-posted from math.SE
I believe this is more commonly referred to as an "incidence matrix", not that it matters hugely.
|
2025-03-21T14:48:31.688089
| 2020-08-06T19:41:39 |
368500
|
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"Eduardo Longa",
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"Joseph O'Rourke",
"M. Winter",
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|
Stack Exchange
|
Minimal pizza cutting
Given a circle, we want to divide it into $n$ connected equally sized pieces. In such a way that the total length of the cutting is minimal. What can we say about the solution for each $n$. Are they unique (up to some symmetry). Do all cuttings arise from the intersection of three straight segments with angle 120º?
These are conjectured to be the best solutions for $n \in \{2, 3, 4, 7\}$
Case $n= 2$ goes as follows: For simplicity the radius is $1$. If there is only one or no points on the circumference, then we know the minimal curve is the circumference, which has length $\frac{2\pi}{\sqrt{2}}$ bigger than 2 and we are done. Hence, we have at least two points in the circumference. If they are in the same diameter, the shortest curve is the line, so the cutting has at least length 2.
Now, pick a diameter parallel to $\overline{AC}$, which leaves both points on the same side. Since the curve connecting both needs to have area $\pi/2$ in needs to contain at least one point on the other side of the diameter.
Now, $\overline{AB}$ is bounded below by a straight line, same with $\overline{BC}$. And the sum of those two is smaller than $\overline{AO} + \overline{OC}$ where $O$ is the center of the circle.
For $n\in\{3, 4, 7\}$ I don't have a proof and they might not be minimal.
As for the conditions on the cuttings, probably something like picewise $C^2$ is enough.
Why skipping $n=5,6$?
@WhatsUp I have no guess of what they might look like. Case 7 is specially symmetric
Your solution for $n=7$ generalizes to having a regular $(n-1)$-gon in the center instead of a hexagon. Might these be solutions for $n=5,6$? And have you already checken whether having a triangle in the center is worse than your proposed solution for $n=4$? Except for the case $n=7$, these solutions do not have $120^\circ$ angles.
Cox and Flikkema have some candidate solutions from small n here: https://doi.org/10.37236/317, and the reference therein will lead you to some literature on the existence and regularity of solutions to this problem (not sure about uniqueness though). In general the arcs of constant curvature and meet in threes at 120º, but not necessarily straight.
@M.Winter my trigonometry is a bit rusty, so please double check the following, for a triangle to have area $pi/4$ it requires side $\frac{\sqrt{\pi}, \sqrt[4]{3}}$. Just three of those is more than $4$ so it is worse than taking four radii
Another condition (that I believe rules out your n=4 candidate) is that the arcs meet the boundary perpendicularly.
@M.Winter A triangle in the center is definitely worse, with a total of about $4.9718$. The proposed one is about $3.9624$, which is slightly better than taking $4$ radii.
The cuttings have to be either line segments or arcs of circles. The meet three at a time forming 120 degree angles.
MSE posting: Unit circle is divided into equal pieces, what is the least value of the perimeters of the parts?.
Here is recent paper coauthored by Cox (whom Yoav Kallus cited),
with a different focus:
Headley, Francis, and Simon Cox. "Least-perimeter partition of the disc into $N$ regions of two different areas." arXiv:1901.00319 (2019).
arXiv abstract.
Their first figure illustrates Yoav's points that
the arcs "meet in threes at $120^\circ$, but not necessarily straight,"
and that "the arcs meet the boundary perpendicularly":
|
2025-03-21T14:48:31.688357
| 2020-08-06T20:26:57 |
368507
|
{
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"authors": [
"Balazs",
"Denis Nardin",
"Igor Makhlin",
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|
Stack Exchange
|
Reference for the multiprojective Nullstellensatz?
Didn't get a single comment in over a day at math.SE, so maybe the question is more appropriate here.
I'm looking for a reference to a generalization of Hilbert's Nullstellensatz to the multiprojective setting. This would appear to be a rather basic fact but I'm having trouble finding a statement online or in the literature.
Consider the product $$\mathbb P=\mathbb P(\mathbb C^{n_1})\times\dots\times \mathbb P(\mathbb C^{n_k}).$$ Let $R$ be the ring of polynomials in variables $X_i^j$ with $j\in[1,k]$ and $i\in[1,n_j]$. Let $I\subset R$ be a multihomogeneous ideal, meaning that $I$ is homogeneous with respect to degree in the variables $X_1^j,\dots,X_{n_j}^j$ for every $j$. Then $I$ is seen to define a subvariety $V(I)\subset\mathbb P$ of points in which all $p\in I$ vanish. Question: where in the literature can I find a necessary and sufficient condition for $I$ to contain all polynomials vanishing on $V(I)$? (Preferably in a standard reference such as Hartshorne.)
I suspect that the condition is that $I$ is (a) multihomogeneous, (b) radical and (c) saturated with respect to the irrelevant ideals $\langle X_1^j,\dots,X_{n_j}^j\rangle$ for all $j$. I think I know how this can be proved but I'm looking for a precise reference.
Update. Here's a brief summary of what has been found so far. First of all, Balazs found this book, Theorem 2.14 there contains a condition of this sort for the case of the product $\mathbb P^1\times \mathbb P^1$. However, I'm afraid their condition is wrong, see comments. Next, I found a paper where on page 8 it says that "the assignment $V \mapsto I(V)$ is a one-to-one correspondence between..." However, I think that this claim is also wrong by the same counterexample. Finally, the latter paper cites Chapter 5 of this book which is in French but Proposition 2.17 there says that there is a correspondence between closed subschemes in $\mathbb P$ and ideals $I\subset R$ satisfying conditions (a) and (c) above. My French is nowhere near enough to tell whether a condition for subvarieties is also found in this chapter (i.e. the statement that adding radicality will ensure the subscheme being reduced).
Update 2. Shameless self-promotion time! The statement is now proved as Theorem 1.8.1 in this paper.
I found a reference in "Arithmetically Cohen-Macaulay Sets of Points in P^1 x P^1" by Guardo and Van Tuyl for the bigraded case, their Theorem 2.11 and thereabouts (google will show this if you search for "bigraded nullstellensatz"). But they only say "the proof is as in the graded case, so will be omitted"...
As for the case of projective space you can see $\mathbb{P}$ as the quotient of $\mathbb{A}^{n_1+\cdots+n_k}$ minus the "coordinate axes" (i.e. those subschemes obtained by replacing one $\mathbb{A}^{n_i}$ factor by 0) by an action of $\mathbb{G}_m^k$. Therefore the closed subschemes in $\mathbb{P}$ are in bijection with the closed subschemes of $\mathbb{A}^{n_1+\cdots+n_k}$ that are stable by the $\mathbb{G}_m$-action and contain the "coorodinate axes". From this the usual Nullstellensatz yields the description you're after. Probably there aren't many references because the proof is so short.
@Balazs, thanks, a reputable source would be good enough for me to quote with or without a proof. But, if I'm reading this correctly, I think there is an issue here (aside from it being bigraded rather than multigraded). I'm not sure their Theorem 2.14 is correct: I don't think their notion of "projective relevance" is what is needed here. For instance, the ideal $\langle y_0(x_0-x_1),y_1(x_0-x_1)\rangle$ is projectively relevant as per Definition 2.12 but it is not the vanishing ideal of its zero set, that would be $\langle x_0-x_1\rangle$. Saturatedness is a stronger condition.
@DenisNardin, thanks, that's more or less the argument I had in mind. There is a nuance, however. You want to obtain the ideals that are the vanishing ideals of their zero sets and those are not the ideals corresponding to subvarieties in affine space containing the $k$ coordinate subspaces. Instead, you want the subvariety to contain as little from these subspaces as possible, i.e. for it to be the closure of itself minus said subspaces. This is where saturatedness comes into play.
So the argument, although not hard, is not exactly obvious. I agree, however, that it is very much the norm to sweep stuff like that under the rug and, by the looks of it, I might have to do just that.
|
2025-03-21T14:48:31.688692
| 2020-08-06T20:44:44 |
368508
|
{
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"Ali Enayat",
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|
Stack Exchange
|
When do two ultrafilters yield isomorphic ultrapowers?
Fix a cardinal $\lambda$$\newcommand{\cU}{\mathcal U}\newcommand{\cV}{\mathcal V}$. Consider the equivalence relation on $\beta\lambda$ given by $\cU\sim \cV$ when for all first-order structures $M$ we have $M^{\cU}\cong M^{\cV}$.
By considering the structure $(\lambda,A)_{A\subseteq \lambda}$, one can show that the $\sim$-classes have at most $2^\lambda$ elements (arguing as here).
If we consider the action of the symmetric group $S(\lambda)$ on $\beta\lambda$, it is easy to see that for each $\cU\in \beta\lambda$ and $f\in S(\lambda)$, we have $\cU\sim f(\cU)$.
Is the converse also true, namely, if $\cU\sim \cV$, are they necessarily conjugate by a permutation of $\lambda$? If not, is it true for $\lambda=\aleph_0$ (or any $\lambda$)?
Edit: Given that the answer is positive (for signature of size $2^\lambda$), it would also be interesting to know whether the equivalence still holds if we define $\cU\sim \cV$ when $M^{\cU}\cong M^{\cV}$ for all first-order structures $M$ whose signature is either
strictly smaller than $\lambda$, or
of size at most $\lambda$, or
strictly smaller than $2^{\lambda}$.
For countably complete ultrafilters $U$ and $W$, this is true, because for any set $x$, it $y$ is its transitive closure, one has $j_U(y,x,\in)\cong j_W(y,x,\in)$, and by wellfoundedness, we may act as though these two structures are actually equal. Hence $j_U$ and $j_W$ agree on all sets, so they are Rudin-Keisler equivalent, which is what you're asking about. For the countably incomplete case, I will need some aspirin.
Do you want to restrict to signatures that are finite or maybe countable or maybe of size less than the continuum? Otherwise there is an unsatisfying positive answer.
@GabeGoldberg: What do you mean by $j_U, j_W$? Regarding signatures: the question I was pondering was without any restriction on the signature, though I guess it would be interesting to ask about the restriction to smaller signatures (although I was more inclined to think of signatures of size $\lambda$, $2^{\lambda}$ or $2^{<\lambda}$, but if you say it already trivializes for continuum, then maybe that does not make sense.).
No you're right, I meant in the countable case. In large cardinal theory, if $U$ is countably complete, then $j_U : V\to M_U$ denotes the transitive collapse of the ultrapower of the universe of sets by $U$.
@GabeGoldberg: I see. My knowledge of the Rudin-Keisler order is rather superficial --- I suppose it is a standard fact that $j_U=j_W$ implies that $U$ and $W$ are equivalent. But if they are not $\sigma$-complete, then $j_U$ does not make sense, as $V^U$ is not well-founded, no? Anyway, a definitive answer even in the case of $\lambda=\aleph_0$ would be interesting, I guess.
If one allows an arbitrary signature, the answer to this question is fairly well-known. Consider the structure $(\lambda,A)_{A\subseteq \lambda}$. Let $(M,R_A)_{A\subseteq \lambda}$ be the common ultrapower by $\mathcal U$ and $\mathcal V$. Note that $A\in \mathcal U$ if and only if $[\text{id}]_U \in R_A$ by the definition of an ultrapower. Fix functions $f,g:\lambda\to \lambda$ such that $[f]_\mathcal V = [\text{id}]_\mathcal U$ and $[g]_\mathcal U = [\text{id}]_\mathcal V$. We have $f_*(\mathcal V) = \mathcal U$: for any $A\subseteq \lambda$,
$A\in \mathcal U$ if and only if $[\text{id}]_\mathcal U\in R_A$ or equivalently
$[f]_\mathcal V\in R_A$, which by definition means $\{\alpha < \lambda : f(\alpha) \in A\}\in \mathcal V$ or $f^{-1}[A]\in \mathcal V$; that is, $A\in f_*(\mathcal V)$.
Similarly $g_*(\mathcal U) = \mathcal V$. It
follows that $(g\circ f)_*(\mathcal V) = \mathcal U$.
A fundamental theorem (discovered independently by Rudin, Keisler, Blass Katetov, Frolík, and maybe others, proof below) states that for
any ultrafilter $\mathcal W$ over $X$ and any $h :X\to X$, if $h_*(\mathcal W) = \mathcal W$
then $[h]_\mathcal W =[\text{id}]_\mathcal W$
Therefore $[g\circ f]_\mathcal V = [\text{id}]_\mathcal V$. Hence there is a set $A\in \mathcal V$ such that $g\circ f\restriction A$ is the identity. In other words, $f$ is one-to-one on a set in $\mathcal V$. It is then not hard to modify $f$ on a null set to make it a permutation.
Proof that $h_*(\mathcal W) = \mathcal W$ implies $[h]_\mathcal W= [\text{id}]_\mathcal W$: Assume not, and so without loss of generality $h(x) \neq x$ for all $x\in X$. Consider the graph $G$ with vertex set $X$ and edge set $E = \{\{x,y\}\in [X]^2 : h(x) = y\}$. We claim $G$ is $3$-colorable. Any finite connected induced subgraph $H$ of $G$ with $n$ vertices contains at most $n$ edges (as $x\mapsto \{x,f(x)\}$ is a partial surjection), and hence contains at most one cycle. Therefore removing at most one edge of $H$ yields an acyclic and hence 2-colorable graph, and this easily implies $H$ is $3$-colorable. By compactness, $G$ is $3$-colorable. Therefore there is a partition $\{A_0, A_1, A_2\}$ of $X$ such that $G\restriction A_n$ is discrete for $n =0,1,2$. This means that $h^{-1}[A_n]\cap A_n\neq \emptyset$ for $n =0,1,2$. Therefore if $A_n\in \mathcal W$, then $A_n\notin h_*(\mathcal W) =\mathcal W$, contradiction.
The question for countably incomplete ultrafilters and finite signatures is far more interesting and seems to be sensitive to set theoretic hypotheses. The answer seems to be yes assuming Woodin's HOD Conjecture. Details on request.
Details: I can actually answer the finite signature question positively without the HOD Conjecture, although my proof looks like overkill. I need a lemma.
Lemma. Suppose $i,j : V_\alpha\to N$ are elementary embeddings that are continuous at regular cardinals $\delta_0 < \delta_1 < \alpha$, and suppose there is a partition $\vec S$ of $\{\alpha < \delta_1 : \text{cf}(\alpha) = \delta_0\}$ into $\delta_1$ stationary sets such that $i(\vec S) = j(\vec S)$. Then $i\restriction \delta_1 = j\restriction \delta_1$.
Given this, we proceed as follows. Consider the structure $M = (V_\alpha,\in,\delta_0,\delta_1,\vec S,f)$ where $\delta_0 > \lambda$ is regular, $\delta_1 \geq 2^\lambda$ is regular, $\vec S$ is a stationary partition as in the lemma, and $f$ is a surjection from the cardinal $2^\lambda$ onto $P(\lambda)$. The ultrapowers of this structure by $\mathcal U$ and $\mathcal V$ coincide, and so we can identify them. Let's say the ultrapower is $(N,E,d_0,d_1,\vec T,g)$. (The signature has one relation symbol, interpreted as $E$, and four constant symbols.)
Let $i,j:V_\alpha \to N$ be the ultrapower embeddings associated to $\mathcal U$ and $\mathcal V$. These embeddings are continuous at $\delta_0$ and $\delta_1$ since these cardinals are regular and above the underlying set $\lambda$ of $\mathcal U$ and $\mathcal V$. Since $i(\vec S) = \vec T = j(\vec S)$, the hypotheses of the lemma are true, so $i\restriction \delta_1 = j\restriction \delta_1$. In particular, these embeddings agree on the ordinal $2^\lambda$.
Now $i[P(\lambda)] = i(f)[i[2^\lambda]] = g[i[2^\lambda]] = g[j[2^\lambda]] = j[P(\lambda)]$. Inverting the transitive collapse shows $i\restriction P(\lambda) = j\restriction P(\lambda)$. This suffices to run the easy argument from the arbitrary signature case.
Proof of Lemma. Let $d_0 = i(\delta_0)$, $d_1 = i(\delta_1)$, $\vec T = i(\vec S) = \langle T_a : a < d_1\rangle$. We run an argument due to Solovay to show that $j[\delta_1]$ is equal to the set $\{a < d_1 : T_a\text{ meets every $\delta_0$-club in $d_1$}\}$. By symmetry (and since $j(\vec S) = \vec T$), we have the same characterization of $j[\delta_1]$, and this proves $i[\delta_1] = j[\delta_1]$, which easily implies the lemma.
(For the record, a set $C\subseteq d_1$ is $\delta_0$-club if it is cofinal in $d_1$ and any increasing $\delta_0$-sequence of elements of $C$ has a supremum in $N$ and this supremum belongs to $C$. Two(?) examples of $\delta_0$-clubs : $i[\delta_1]$ and $j[\delta_1]$.)
First assume $T_a$ meets every $\delta_0$-club in $d_1$, and we will show $a\in i[\delta_1]$. The point is that $T_a$ meets $i[\delta_1]$. Take $\xi < \lambda$ with $i(\xi)\in T_a$. Note that $\text{cf}(\xi) = \delta_0$, so there is some $\alpha < \lambda$ with $\xi \in S_\alpha$. Hence $\xi \in i(S_\alpha) = T_{i(\alpha)}$. Since $\xi \in T_a\cap T_{i(\alpha)}$ and the $T_b$ are pairwise disjoint, $a = i(\alpha)$.
Conversely let us show that $T_{i(\alpha)}$ (i.e., $i(S_\alpha)$) meets every $\delta_0$-club in $\delta_1$. Towards this fix such a $\delta_0$-club $C$. Then the usual argument shows that $C\cap i[\delta_1]$ is a $\delta_0$-club in $d_1$. This means $i^{-1}[C]$ is a $\delta_0$-club in $\delta_1$ in the usual sense. Therefore there is some $\xi \in S_\alpha\cap i^{-1}[C]$. Now $i(\xi)\in i(S_\alpha)\cap C$, so $i(S_\alpha)$ meets $C$, as desired. This proves the lemma.
Request${{{}}}$
Thanks! :-)${}$
@AndrésE.Caicedo: Thanks for getting me to write the details. There is an interesting technical problem here: one can prove (schematically) in ZF that if $j_0,j_1 : V\to M$ are definable elementary embeddings, then $j_0\restriction \text{Ord} = j_1\restriction \text{Ord}$ (even if $M$ is illfounded). Yet I do not see how to prove that if $j_0,j_1 : V\to M$ are (isomorphic to) ultrapower embeddings, then $j_0\restriction \text{Ord} = j_1\restriction \text{Ord}$ without assuming the HOD Conjecture.
Two minor comments: (1) In the penultimate sentence of the last paragraph $=[\mathrm{id}]_{\mathcal{W}}$ is missing at the end of the sentence. (2) The result attributed in the first paragraph to Rudin and others was also proved in 1967 by Miroslav Katětov in the paper accessible through: https://dml.cz/handle/10338.dmlcz/105124
@AliEnayat Thanks! Apparently the theorem was also proved by Frolík
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