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2025-03-21T14:48:31.183563
| 2020-06-06T14:47:54 |
362335
|
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"Ali Taghavi",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/36688",
"leo monsaingeon"
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}
|
Stack Exchange
|
Riemannian metrics on matrix space for which the restriction of trace function to each complete geodesic is a bounded function
Edit: According to comment by Leo Monsaingeon I revise my question:
Is there a Riemannian metric on $M_n(\mathbb{R})$ for which the function $trace$ is a bounded function on every complete(whole) geodesic? that is for every complete geodesic $\gamma$, the restriction $trace$ to $\gamma$ is a bounded function? What about the same question if we replace $M_n(\mathbb{R})$ with $GL_n(\mathbb{R})$?
Remark By a complete geodesic we DO NOT mean exclusively a geodesic whose parameter $t$ can be varied in whole $\mathbb{R}$, but we mean $\{\gamma(t)\mid t\in I\}$ where $I$ is the maximal interval of definition of $\gamma$. So we mean the whole trajectory of a non parametrized geodesic.
Do you mean complete geodesics? Or just two-points geodesics between arbitrary matrices $M,N$? In the latter case the Hilbert structure induced by the Frobenius scalar product should do, right?
@leomonsaingeon I think it is the Eucliean metric so the geodesic $\begin{pmatrix}t&0\0&0\end{pmatrix}$ has an unbounded trace, right?
well, not if you mean "geodesics between arbitrary pairs of points", in which case your time parameter cannot run across the whole real line. But I guess from your comment that you really mean "complete geodesics", right? If so perhaps it would be worth editing your question (also, there is a typo in your title "o"->"of")
@leomonsaingeon I mean complete geodesic, otherwise not only trace but also every continuous function on a compact set9geoesic from M to N) is a bounded function
right, but the continuity of the trace function depends on the choice of the metric (although I agree that pretty much all the Riemannian structures should give an equivalent topology). So it was not completely clear for me what you meant. Now it is ;-)
@leomonsaingeon Thank you for your comment. I revise my question.
It is possible to find such metrics. However, they are a bit unnatural, in that they aren't really constructed from any inherent properties of matrices. For concreteness, I'm going to restrict my attention to $M_2$ and $Gl_2$, but this construction can be modified to work for any larger $n$ as well.
We consider $2 \times 2$ matrices $$
\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]. $$
To construct the desired metric, we start by changing coordinates on $M_2$ as follows.
$x = a+d$, $y= \exp(a-d)$, $z=b$, $w=c$
This gives a map from $M_2$ to a half-space in $\mathbb{R}^4$. For a Riemannian metric, we use $$ds^2 = \frac{dx^2+dy^2}{y^2} +dz^2+dw^2.$$
Notice that the $x$-coordinate is simply the trace of the original matrix, so what's left to show is that $x$ is bounded on any geodesic. Different geodesics will have different bounds, however.
To show that $x$ is bounded on any geodesic, observe that as a Riemannian manifold, $(M_n, g)$ is simply the metric product $\mathbb{H} \times \mathbb{R}^2$. As such, the geodesics are of the form $\gamma(t) = (c(t), \ell(t))$.
Here, $c(t)$ forms an arc of a semi-circle in $\mathbb{H} $ whose center is on the axis $y=0$. Meanwhile, $\ell(t))$ is a line in $\mathbb{R}^2$. As such, it follows the $x$ coordinate is bounded by the center and radius of $c(t)$.
When we try to do this for $Gl_2$, we have to deal with the fact that the space is disconnected. However, we can use a similar idea on each connected component. Namely, we change coordinates to
$x = a+d$, $y= \exp(a-d)$, $z=b$, $w=ad-bc$ and take the metric
$ds^2 = \frac{dx^2+dy^2}{y^2} +\frac{dz^2+dw^2}{w^2}.$
This gives $Gl_2$ the metric structure of $\mathbb{H} \times \mathbb{H}$ on each of its connected components, which also has the desired geodesic property.
|
2025-03-21T14:48:31.183838
| 2020-06-06T14:57:34 |
362336
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362336"
}
|
Stack Exchange
|
Identity on convolution with Gaussian measure
I've came across an identity once (I don't remember where) concerning convolutions of Gaussian measures. If I'm not mistaken, this identity was
\begin{eqnarray}
(\mu_{C}*f)(y) = \exp\bigg{[}\frac{1}{2}\bigg{(}\frac{\partial}{\partial x}, C\frac{\partial}{\partial x}\bigg{)}\bigg{]}f(x)\bigg{|}_{x=y} \tag{1}\label{1}
\end{eqnarray}
where $C=(C_{ij})$ is a $n\times n$ Gaussian covariance, i.e. a positive-definite matrix and
$$\bigg{(}\frac{\partial}{\partial x},C\frac{\partial}{\partial x}\bigg{)} := \sum_{i,j}\frac{\partial}{\partial x_{i}}C_{ij}\frac{\partial}{\partial x_{j}}$$
is a differential operator.
I recently discovered that D. Brydges states a very similar result in his notes on functional integrals (page 34, exercise 4.3.3). He mentions that this identity is related to Wick's theorem for Gaussian measures. Note, however, that Brydges states (\ref{1}) in the case $f$ is a polynomial.
I checked some references I know to look for this version of Wick's Theorem but I have found nothing and I'd like to know more about this identity. If $f$ is assumed to be a polynomial, it seems to me that the natural way to prove it is to consider the Fourier transform of $\mu_{C}$ in order to obtain its moments. But does this identity hold in more general cases? If this is the case, how to address it and prove it? Also, any reference is welcome here. Thanks!
Upon Fourier transformation the convolution becomes a product of the Fourier transform ${\cal F}[f]$ of the function $f$ and the Fourier transformed Gaussian measure, which is again a Gaussian with covariance matrix $C^{-1}$,
$${\cal F}[\mu_{C}*f](k) = \exp\left(-\tfrac{1}{2}\sum_{n,m}k_n C_{nm} k_m\right){\cal F}[f](k).$$
Upon inverse Fourier transformation $k_n\mapsto i\partial/\partial x_n$, hence
$$(\mu_{C}*f)(x) = \exp\left(\tfrac{1}{2}\sum_{n,m}\frac{\partial}{\partial x_n} C_{nm} \frac{\partial}{\partial x_m} \right)f(x).$$
This holds irrespective of whether $f$ is polynomial or not.
(Not a complete answer really, but too long for a comment.)
I believe the answer depends on the way you define the operator $$A = \exp(\tfrac12 \sum_{n,m} \tfrac\partial{\partial x_n}C_{nm}\tfrac\partial{\partial x_m}).$$
In the sense of functional calculus, $A$ is the exponential of the elliptic operator $$L = \tfrac12 \sum_{n,m} \tfrac\partial{\partial x_n}C_{nm}\tfrac\partial{\partial x_m}$$ (that is, $A = \exp(L)$). Since $L$ generates a strongly continuous semigroup of operators $\exp(t L)$ on various function spaces (for example, on $L^p(\mathbb{R}^n)$), $A$ is well-defined on these function spaces, and it is a classical result that $A$ is the convolution with the Gauss–Weierstrass kernel (up to change of variables $x \mapsto C^{-1/2} x$). Thus, the identity (1) works for $f$ in these function spaces. (This is essentially what Carlo Beenakker wrote in his answer.)
If one understands $A$ formally as a series of differential operators, then convergence is an issue. For polynomials, however, this series becomes a finite sum, and it is not difficult to verify this formally, either by appealing to the Fourier transformation (again as in Carlo's answer), or by integration by parts (as in Ander Aguirre's answer, if I understand it correctly). The same should be true if $f$ is a real-analytic function with power series coefficients decaying sufficiently fast, but I did not attempt to work out the exact condition.
|
2025-03-21T14:48:31.184316
| 2020-06-06T15:14:11 |
362338
|
{
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"authors": [
"Cubikova",
"Jakob Werner",
"LSpice",
"https://mathoverflow.net/users/112369",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/8145"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362338"
}
|
Stack Exchange
|
What is the definable functor associated to an algebraic scheme (model theory of valued fields)
I have a very basic question regarding algebraic model theory. I am trying to read Espaces de Berkovich, polytopes, squelettes et théorie des modèles (MSN) by Antoine Ducros. The relevant section is Section 0.31. Let me give the setup: Let $K$ be a valued field and let $\mathscr{C}_K$ be the category of non-trivially valued, algebraically closed field extensions of $K$. We write $\mathscr{L}_\mathrm{val}$ for the language of valued fields, having three sorts (the valued field, its residue field and the value group).
Let $\mathscr{S}$ be a product of sorts. For each model of the theory of non-trivially valued, algebraically closed field extensions of $K$, i.e. for each $F \in \mathscr{C}_K$, we can then make sense of $\mathscr{S}(F)$, and this gives a functor $\mathscr{S} \colon \mathscr{C}_K \to \mathrm{Set}$. If $\Phi(x)$ is a formula of $\mathscr{L}_\mathrm{val}$ (with parameters in $K$), whose free variables $x$ live in $\mathscr{S}$, then the subfunctor of $\mathscr{S}$ mapping $F$ to $\{x \in \mathscr{S}(F) \mid \Phi(x)\}$ is called a definable subfunctor of $\mathscr{S}$.
In general, an abstract functor from $\mathscr{C}_K$ to $\mathrm{Set}$ is called definable if it is isomorphic to a definable subfunctor of some product of sorts.
Now it is claimed that a scheme of finite type $\mathscr{X}$ over $K$ gives rise to a definable functor on $\mathscr{C}_K$. I don't understand why this is true. If $\mathscr{X}$ is affine, then I believe that the associated functor should simply be the “functor of points”. Namely, if $\mathscr{X}$ is isomorphic to $\mathrm{Spec}(K[T_1, \dots, T_n] / \langle f_1, \dots, f_r\rangle)$, then its associated functor is isomorphic to the definable subfunctor of $F \mapsto F^n$, given by the equations $f_i$. I have no idea, what is meant in the non-affine case. I would be very glad about some clarification. Also, if some of my writing above hints at some misunderstanding on my side, please let me know.
This follows by elimination of imaginaries in algebraically closed fields. Given a finite affine cover of your scheme, one obtains a functor to Set by gluing the affine charts. This is indeed not definable, but a quotient of a definable set by a definable equivalence relation. Elimination of imaginaries tells you precisely that such a quotient is in definable bijection with a definable set. You can find the details in Chapter 4 (A. Pillay, Model theory of algebraically closed fields) of the following book:
Bouscaren, Elisabeth, Introduction to model theory, Bouscaren, Elisabeth (ed.), Model theory and algebraic geometry. An introduction to E. Hrushovski’s proof of the geometric Mordell-Lang conjecture. Berlin: Springer. Lect. Notes Math. 1696, 1-18 (1998). ZBL0925.03160.
Chapters 1 and 2 also contain an introduction to general model theoretic concepts including elimination of imaginaries (Chapter 2).
I edited in links to the book and the individual chapters. Since you later cite Chapter 1 separately, I think you didn't mean your reference and corresponding ZBL link to refer only to the first chapter, but rather to the whole book. That's ZBL0920.03046.
Thank you very much.
If I understand “gluing affine charts” correctly, then you are saying that also for general $\mathscr{X}$, the associated functor is the functor of points, sending $F$ to $\mathscr{X}(F) = \mathrm{Hom}_K(\mathrm{Spec}(F), \mathscr{X})$, right? Could you maybe indicate for some simple non-affine scheme like $\mathbf{P}^1_K$, how the bijection that you get to a definable functor would look like?
By the way, thank you for the hints to the literature, I will definitely have a look at it.
Indeed, the functor is isomorphic to the functor of points. In the case of $\mathbf{P}_K^n$, the functor of points is isomorphic to the functor sending $F$ to the set of $n$-dimensional subspaces of $F^{n+1}$. The usual equivalence relation on $F^{n+1}$ given by $(x_0,\ldots, x_n)\sim(y_0,\ldots,y_n)$ if and only if there is $\lambda\in F^*$ such that $\lambda x_i=y_i$ for each $i\in {0,\ldots,n}$ is the corresponding definable equivalence relation.
In the special case of $\mathbf{P}_K^1$, you can identify such a quotient with the definable functor which sends $F$ to the (definable) set ${(x,y)\in F^2 : y=1} \cup {(1,0)}$. For $\mathbf{P}_K^n(F)$ you can cook up a similar description. For general schemes of finite type, the fact that you can find this definable (set) functor is precisely the content of elimination of imaginaries.
Note that the choice is made up to definable bijection (or up to isomorphism of definable functors in your terminology). Indeed, I could also define the functor for $\mathbf{P}_K^1$ as sending $F$ to ${(x,y) \in F^2 : y=1}\cup{(2,0)}$. In any case, the existence of such a functor is granted by elimination of imaginaries.
@Cubikova I see. So this is really only an abstract existence statement “there exists some isomorphism onto a definable functor”, and this isomorphism is in no way canonical or natural, not even after the choice of an affine covering. Thank you for your help.
|
2025-03-21T14:48:31.184654
| 2020-06-06T15:22:43 |
362339
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362339"
}
|
Stack Exchange
|
strict convexity of the Legendre-Fenchel transform
Let $d$ be a positive integer.
Let $L:\mathbb{R}^d\to\mathbb{R}$ be a differentiable function with continuous derivatives.
Assume that the Legendre-Fenchel transform of $L$ exists everywhere, is denoted by $H:\mathbb{R}^d\to\mathbb{R}$ and defined by:
$$ H(p)=\sup_{\alpha\in\mathbb{R}^d} p\cdot\alpha - L(\alpha).$$
Assume that there exists a unique $\alpha\in\mathbb{R}^d$ achieving the maximum in the definition of $H(p)$ for any $p\in\mathbb{R}^d$.
Is it possible to conclude that $L$ is strictly convex? We did not assume that $L$ is convex.
A quick drawing makes me think that the latter conclusion holds in dimension $d=1$.
Any help would be appreciated. Please let me know if you think about a reference which may contain the answer.
Best.
|
2025-03-21T14:48:31.184733
| 2020-06-06T15:40:56 |
362340
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362340"
}
|
Stack Exchange
|
Is there a $3\times 3$ matrix over a Dedekind domain not similar to a matrix with zero top right entry?
Let $R=\mathbb{Z}[\sqrt{-5}]$, which is well known to be a Dedekind
domain but not a PID. Let $\mathrm{M}_{3}(R)$ be the set of $3\times3$
matrices over $R$. Does there exist a matrix $A\in\mathrm{M}_{3}(R)$
such that for all $g\in\mathrm{GL}_{3}(R)$ the $(1,3)$-entry of
$gAg^{-1}$ is not $0$? In other words, does there exist a matrix
$A\in\mathrm{M}_{3}(R)$ such that the $R$-similarity class of $A$
does not contain any matrix whose top right entry is zero?
I have asked the question in the simplest case where I don't know the answer, but one can of course generalise to $n\times n$ matrices and other rings.
If $R$ is a PID it is known that any $A\in\mathrm{M}_{3}(R)$
is $R$-similar to a matrix whose $(1,3)$-entry is zero. Namely, if
$A=(a_{ij})$, $a_{ij}\in R$, then one can verify by direct computation
that if $\begin{pmatrix}x & y\\
z & w
\end{pmatrix}\in\mathrm{GL}_{2}(R)$ and
$$
g=\begin{pmatrix}1 & 0 & 0\\
0 & x & y\\
0 & z & w
\end{pmatrix},
$$
then the first row of $g^{-1}Ag$ is $(a_{11}\ \ \ a_{12}x+a_{13}z\ \ \ a_{12}y+a_{13}w)$.
Thus, if $R$ is a PID, we can take $y,w\in R$ such that $y\gcd(a_{12},a_{13})=a_{13}$
and $w\gcd(a_{12},a_{13})=-a_{12}$ and $x,z\in R$ such that $xw-yz=1$
(which exist since $\gcd(y,w)=1$). This approach breaks down when
$R$ is not a PID. On the other hand, if $R$ is an arbitrary commutative
ring with identity and $(a_{12},a_{13})=(1)$, that is, if there exist
$b_{1},b_{2}\in R$ such that $b_{1}a_{12}+b_{2}a_{13}=1$, then $\begin{pmatrix}x & y\\
z & w
\end{pmatrix}=\begin{pmatrix}b_{1} & -a_{13}\\
b_{2} & a_{12}
\end{pmatrix}$ will do. Thus, for this question, a starting point might be to consider
matrices $A$ whose entries all lie in some maximal ideal of $\mathbb{Z}[\sqrt{-5}]$.
|
2025-03-21T14:48:31.184869
| 2020-06-06T16:34:32 |
362349
|
{
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],
"authors": [
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/90871",
"user3653831"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362349"
}
|
Stack Exchange
|
A Mathieu-like equation with a quadratic term
I have the following equation:
$$y''(x)+\left[a-bx^2-c\cos(2x)\right]y(x)=0.$$
This equation is the Schrodinger equation for a quantum pendulum with a quadratic term. If we set $b=0$, the solutions are Mathieu functions.
What are the solutions in the case $b\neq 0$? I'm not sure where to look for this type of equations. Mathematica seems to not know how to solve it.
this is an anharmonic oscillator, which has no closed form solution; even the small-$c$ case where the potential has a term $x^2$ and a term $x^4$ cannot be solved exactly.
@CarloBeenakker Thanks for the comment. It does save some time.
|
2025-03-21T14:48:31.184939
| 2020-06-06T16:35:58 |
362350
|
{
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"authors": [
"Filip",
"https://mathoverflow.net/users/114985"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362350"
}
|
Stack Exchange
|
Nakajima reflection functors and the flavour/framing group action
Nakajima has constructed so-called reflection functors that are isomorphisms between different quiver varieties that have the same framing $\mathbf{w}:$
$$\Phi_{\sigma}:\mathfrak{M}_\zeta(Q,\mathbf{v},\mathbf{w})\rightarrow \mathfrak{M}_{\sigma \cdot \zeta}(\sigma*_\mathbf{w} \mathbf{v},\mathbf{w}),$$
Here $\sigma\in W$ is an element of the Weyl group for Kac-Moody Lie algebra $\mathfrak{g}_Q.$
The question is: Is it known/written somewhere how do these maps combine with $\operatorname{GL}(\mathbf{w})$-actions on the both sides? E.g. is it equivariant with respect to them? Moreover, is it equivariant with respect to the $\mathbb{C}^*$-action that acts by weight $1$ on all edges of the quiver $Q$? I am interested in ADE types only, but I guess the answer does not depend on that.
Just to add now: There is a paper "Etingof conjecture for quantized quiver varieties" by Bezrukavnikov-Losev which deals with the related question. They claim rather a twisted-equivariant property of reflection functor, with respect to a subgroup of $GL(\mathbf{w})\times \mathbb{C}^*.$
|
2025-03-21T14:48:31.185039
| 2020-06-06T17:24:52 |
362355
|
{
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],
"authors": [
"Karim KHAN",
"https://mathoverflow.net/users/152650"
],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362355"
}
|
Stack Exchange
|
$ \{x\in X:h(x)\leq r\} $ is sequentially compact subset of $X$?
Let $(X,\|.\|)$ be a reflexive Banach space and $(D,\|.\|)$ be a Suslin subspace of $X$ such that $D$ is weakly closed subset of $X$.
Take $h:X\to [0,+\infty]$ such that $h(x)=\|x\|$ if $x\in D$ and $h(x)=+\infty$ otherwise.
Can we say that for all $r\in\mathbb{R}$
$$
\{x\in X:h(x)\leq r\}
$$
is sequentially compact subset of $X$?
$\overline{B}(0,r)$ is weakly compact thus ${x:h(x)\leq r}$ is relatively weakly compact, it follows from Eberlein - Šmulian theorem that ${x:h(x)\leq r}$ is sequentially compact subset of $X$
|
2025-03-21T14:48:31.185105
| 2020-06-08T13:17:30 |
362505
|
{
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"authors": [
"Gerry Myerson",
"Zach Hunter",
"https://mathoverflow.net/users/130484",
"https://mathoverflow.net/users/3684"
],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362505"
}
|
Stack Exchange
|
Does there always exist a matrix satisfying certain tracial conditions
Given odd integers $0<a<b$, I want to know if there exists an $n$ by $n$ real valued square matrix $M$ such that
$$ M_{ij} = M_{ji} \quad \forall i,j \in \{1,2\dots n\}$$
$$ \sum_{i=1}^n M_{ij} = 0\quad \forall j \in \{1,2\dots n\} $$
$$\operatorname{Tr}(M^a)\operatorname{Tr}(M^b) = \sum_{i=1}^n \lambda_i^a \sum_{i=1}^n\lambda_i^b < 0 $$
for some finite $n$.
If there always exists such an $M$, this would resolve a case which may lead to improving a result of
Locally common graphs by Csóka, Hubai, and Lovász. In the paper's language, I am trying to prove that there always exists a balanced graphon $U$ such that $t(C_a,U)t(C_b,U)<0$. Proposition 4.3 proves a similar special case, where the two cycles are connected. Since they ignored the disconnected case, I suspect my problem may be trivial. I am not too familiar with manipulating the trace function, so I'm not sure.
Maybe you could accept your answer, Zach, so this question stops being bumped by the Community Bot three times every year.
done. sorry, I don’t think I was aware of how the bumping system worked.
In hindsight this was rather easy. If $G$ is an edge-weighted graph with adjacency matrix $M$, then the count of weighted homomorphisms of $C_k$ onto $G$ is equal to $\operatorname{Tr}(M^k)$.
It is sufficient to be able to construct graphs $G_{k-}$ (resp. $G_{k+}$) such that $C_k$ has only one embedding, which has negative (resp. positive) weight, with all other cycles being arbitrarily long.
To do this, you essentially just need to glue an arbitrarily long path to each vertex of $C_k$, and then glue all the ends of the paths together. The weighting is straightforward to handle afterwards. (give all but one edge of $C_k$ unit weight, and give the last edge weight $k-3$ for the positive case, or weight $-(k-1)$ in the negative case, and then alternate the path weights appropriately)
We have that $C_a$ cannot embed in $G_{b+}$, and $C_{b+}$ can embed only finitely many times into $G_{a-}$. Thus, for $N$ sufficiently large, defining $G$ to be one copy of $G_{a-}$ and $N$ copies of $G_{b+}$, the weighted embeddings of $C_a$ are negative while those of $C_b$ are positive, as desired.
|
2025-03-21T14:48:31.185285
| 2020-06-08T13:19:10 |
362507
|
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"dohmatob",
"https://mathoverflow.net/users/35520",
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"ofer zeitouni"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362507"
}
|
Stack Exchange
|
Upper bound for $\mathbb P(|f(A+XX^T)-f(A)| > \epsilon)$, where $A$ is a fixed pd matrix and $X$ has random iid entries
Let $A$ be a fixed $n$ by $n$ real symmetric positive definite matrix with eigenvalues $\lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_n > 0$, and let $f(A):=\sum_{i=1}^n\log\lambda_i$, and let $X$ be a random $n$ by $k$ matrix with real iid copies distributed according to $N(0,\sigma^2/k)$.
The regime
$n$ is fixed (in particular, $n \not \to \infty$).
$k \to \infty$ (in particular, $k \gg n$).
Question
How close is $f(A+XX^T)$ to $f(A)$ in espectation ?
What is an upper-bound for $\mathbb P(|f(A+XX^T)-f(A)| > \epsilon)$?
Observations
I've observed that $f(A+XX^T)$ is approximately $\mathcal N(\mu,s^2)$, for some $\mu \in \mathbb R$, and $s > 0$.
Back-of-envelop calculation
By triangle inequality, one has
$$
\begin{split}
|f(A+XX^T) - f(A)| \le &|f(A+\sigma^2 I_n) - f(A)|\\
&\quad + |f(A+XX^T) - f(A+\sigma^2 I_n)|.
\end{split}
\tag{*}
$$
Note that as $k\rightarrow \infty$, $XX^T \rightarrow \sigma^2 I_n$ in probability. Thus, by the delta method, we know that $f(A+XX^T) - f(A+\sigma^2 I_n) \longrightarrow \mathcal N(0,s^2/k)$, where
$$
s^2 := n\sigma^4\|(A+\sigma^2 I_n)^{-1}\|_F^2 \le \sigma^4(\sum_{j=1}^n\lambda_j((A + \sigma^2)^{-1})^2 \le (\sqrt{n}\sigma^2\eta(A))^2,
$$
where
$$
\begin{split}
\eta(A) &= \eta(A; \sigma^2) := \text{trace}(A+\sigma^2 I_n)^{-1} = \sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1}
\le n\min(\sigma^{-2},\lambda_n(A)^{-1}).
\end{split}
$$
On the other hand,
$$
|f(A + \sigma^2 I_n) - f(A)| = \sum_{i=1}^n\log(1 + \sigma^2/\lambda_i) \le \sigma^2\sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1} = \sigma^2\eta(A).
$$
Putting everything together then gives
$$
\begin{split}
E_X|f(A+XX^T) - f(A)| &\le \sigma^2\frac{n}{\lambda_n} + E_X|f(A+XX^T)-f(A+\sigma^2 I_n)|\\
&\le \sigma^2\eta(A) + \sqrt{\frac{n}{k}} \sigma^2 \eta(A) \to \sigma^2\eta(A).
\end{split},
$$
Thus it appears that,
To have $E|f(A+XX^T) - f(A)|$ small, it is sufficient to have $\sigma^2 \eta(A) \ll 1$ and $k \rightarrow \infty$.
This doesn't solve my problem, but it raises suspicion to what the important problem parameters could be; here, $\sigma$, $k$, and $\eta(A)$ (or $n/\lambda_n$, for an even cruder analysis).
Any reasonable answer must depend on how $\lambda_n$ is close to $0$. This is because the minimal eigenvalue of $A+XX^T$ will be of order $\sqrt{n}$ and least. So if for example $\lambda_n=e^{-n}$ then there is a huge difference between $f(A)$ and $f(A+XX^T)$ if $n$ is large.
Sure, $n$, $k$, $\lambda_1$, and $\lambda_n$, and $\sigma$ are all problem data, and so I'd expect any reasonable answer to depend on them.
The short answer is that if $\lambda_n>>n$ then you have concentration, and quantitative answers depend on regimes (for example, if all eigenvalues of $A$ are much larger than $n$ then you have good concentration). The question is too open ended for me to seriously answer. If there is a specific asymptotic regime that you care about then please write it.
$A$ is a fixed $n$ by $n$ matrix, where $n$ is small (say, $ n = 100$). In particular, $n$ doesn't go to $\infty$. As for $k$, one may limit onself to the regime $k \rightarrow \infty$, if that helps. If this clarifies the situation, please let me know (so I can update the question with these poiints).
In the regime $k\to \infty$ , $XX^T\sim kI_n$. In particular, there is no hope for what you want, as $f(A+XX^T)\to\infty$. Did you mean to normalize $XX^T$? Also, what are entries $N(0,\sigma^2 I)$? are the entries real? complex?
@oferzeitouni Thanks for the comments. $X$ has real iid entries which are distributed according to $N(0,\sigma^2/k)$. See corrected question.
@oferzeitouni I've added an answer based on Berry-Esseen. I'd be grateful if you could take a look and tell me what you think. Thanks in advance!
I am not sure why one needs Berry Esseen. For all I can tell, a simple estimate based on variance would do. Namely, the difference between the eigenvalues of $XX^T$ and $1$ is, with high probability, bounded by $n$ times $1/\sqrt{k}$ (just bound the individual off-diagonal entries of $XX^T$; each such entry is a sum of independent random variables). As for your proof, sorry but I don't have the time now to check it.
@oferzeitouni Thanks for pointing out that Berry-Esseen is a sub-optimal overkill for my problem. I've updated my answer with a "direct" computation which leverages sub-Gaussianity, combined with Weyl's inequality.
Also, one can work out the details of your suggestion above by constructing a Gershgorin confidence set for the eigenvalues of $XX^T$. My guess is that something like this is done under the hood in the "book proof" of Weyl's inequality.
Below, I provide a "high-probability" non-asymptotic bound (see (+) below) based on non-linear Berry-Esseen theory developed by Iosif Pinelis. I'd be grateful if someone would kindly check that I didn't screw up anything. Thanks in advance!
Main tool: non-linear Berry-Esseen theory
Let $ \mathcal H$ be a (possibly infinite-dimensional) Hilbert space with topological dual $\mathcal H^*$, and let $g:\mathcal H \rightarrow \mathbb R$ be a function such that
$g(0)=0$,
$g$ has linear approximant $L \in \mathcal H^*$ valid on a nonzero neighborhood of the origin, i.e there exist $\epsilon > 0,M>0$ such that
$$
|g(z) - L(z)| \le \frac{M}{2}\|z\|^2,\;\forall \|z\| \le \epsilon. \tag{1}
$$
Note that under such conditions, we automatically have that $g$ is (Fréchet) differentable at $0$ and $L=\nabla g(0)$.
Let $Z$ be a random vector on $\mathcal H$ such that $E Z = 0$ (i.e $Z$ is centered) and suppose $\tilde{\sigma} := \|L(Z)\|_{Z,2} := (E_Z|L(Z)|^2)^{1/2} < \infty$. The quantity $\tilde{\sigma}^2/k$ will play the role of a "proxy variance" for the random variable $g(Z)$. Finally, let $Z_1,\ldots,Z_k$ be iid copies of $Z$ and set $\overline{V} := (1/k)\sum_{i=1}^kZ_i$.
Theorem (Corollary 3.7 of reference paper). Let $p \in (2,3]$ such that $\|Z\|_{Z,p} := (E_Z\|Z\|_p)^{1/p} < \infty$. Then
$$
\sup_{t \in \mathbb R}\left|\mathbb P\left(\frac{g(\overline{V})}{\tilde{\sigma}/\sqrt{k}}\ge t\right)-\Phi(t)\right| \le Ck^{1-p/2},
$$
where $C$ is a constant which only depends on the distribution of $Z$ (in particular, $C$ is independent of the sample size $k$ and $p$).
Application: our problem
For our own business, we let
$\mathcal H$ be the euclidean space of $n$ by $n$ matrices equipped with the Frobenius trace inner product $\langle B,\tilde{B}\rangle_{Fro} := \text{trace}(B^T\tilde{B})$. We denote induced norm by $\|B\|_{Fro} := \langle B,B\rangle_{Fro}^{1/2}$. Note that this space is nothing but euclidean $\mathbb R^{n^2}$ in disguise.
Define
$$g(B) := \begin{cases}f(B + A + \sigma^2 I_n) - f(A+\sigma^2 I_n),&\mbox{ if }B + A + \sigma^2 I_n \succ 0,\\0,&\mbox{ else.}\end{cases}
$$
where $f(B) := \sum_{j=1}^n\log\lambda_j(B)$.
It's not hard to see that $g$ is differentiable at $0$ with derivative $L:=g'(0) := (A+\sigma^2 I_n)^{-1}$. Moreover, $g$ has Lipschitz continuous gradient at $0$ and and so (1) is satisfied.
For our random vector, we take $Z = RR^T-\sigma^2 I_n$, where $R \in \mathbb R^n$ is a random vector with iid entries from $N(0,\sigma^2 I_n)$. Thus $XX^T-\sigma^2 I_n = (1/k)\sum_{i=1}^k Z_j =: \overline{V}$, where $Z_j = X_jX_j^T$ are iid copies of $Z$ (where $X_i$ is the $i$th column of the random $n$ by $k$ matrix $X$ in our original problem). Of course, $\|Z\|_{Z,p} < \infty$ for all $p$ (Gaussians have finite moments).
By Cauchy-Schwarz inequality, one computes
$$
\begin{split}
\tilde{\sigma}^2 &= \|L(Z)\|_Z = \|L\|_{Fro}^2\|Z\|_{Z,2}^2 = \text{trace}((A+\sigma^2 I_n)^{-2})E_Z\|Z\|_2^2\\
& \le (\sum_{j=1}^n(\lambda_j(A) + \sigma^2)^{-1})^2\cdot n \sigma^4 \le n\sigma^4\eta(A)^2 = (\sqrt{n}\sigma^2\eta(A))^2,
\end{split}
$$
where
$$
\begin{split}
\eta(A) &= \eta(A; \sigma^2) := \text{trace}(A+\sigma^2 I_n)^{-1} = \sum_{i=1}^n(\lambda_i(A)+\sigma^2)^{-1}
\le n\min(\sigma^{-2},\lambda_n(A)^{-1}).
\end{split}
$$
Putting everything together, and applying the above theorem, we obtain that for every $t \ge 0$, w.p at least $\Phi(t\sqrt{k})+\mathcal O(k^{-1/2})$,
$$
|f(A+XX^T) - f(A + \sigma^2 I_n)| = |f(\overline{V}+A+\sigma^2 I_n) - f(A + \sigma^2 I_n)| \le t\sqrt{n}\sigma^2\eta(A).
$$
Combining with (*) then gives: for all $t \ge 0$, it holds w.p at least $\Phi(t\sqrt{k/n})+\mathcal O(k^{-1/2})$ that
$$
|f(A+XX^T) - f(A)| \le \sigma^2\eta(A) + t\sigma^2\eta(A) = (1+t)\sigma^2\eta(A). \tag{+}
$$
Edit: Improved non-asymptotic bound for the term $|f(A+XX^T)-f(A+\sigma^2 I_n)|$
It has been noted in the comments section of the question that Berry-Esseen sounds like an over-kill (and might produce sub-optimal bounds for my particular problem in which everything is sub-Gaussian...). To remedy this, I'll use well-known concentration results for estimation of covariance matrices, and then Weyl's inequality. Interestingly, just as in the Berry-Esseen method above, the key requirement is that the entries of $X$ be iid with finite moments of order $2 + \epsilon$.
By Proposition 2.1 of HOW CLOSE IS THE SAMPLE COVARIANCE MATRIX TO
THE ACTUAL COVARIANCE MATRIX?, we know that for any $t \ge 0$, it holds with probability $1-2e^{-ct^2}$ that
$$
\|XX^T-\sigma^2I\| = \left\|\frac{1}{k}\sum_{j=1}^kR_iR_i^T-\sigma^2 I_n\right\| \le t\sigma^2\sqrt{\frac{n}{k}}.
$$
Thus by Weyl's inequality, it holds with probability $1-2e^{-ckt^2/n}$ that
$$
\max_{1 \le i \le n} |\lambda_i(A+XX^T)-\lambda_i(A+\sigma^2)| \le \|XX^T-\sigma^2I_n\| = \left\|\frac{1}{k}\sum_{j=1}^kR_iR_i^T-\sigma^2 I_n\right\| \le t\sigma^2,
$$
and so by the triangle inequality, it holds with probability $1-2e^{-ckt^2/n}$ that
$$
\begin{split}
|f(A+XX^T)-f(A+\sigma^2 I_n)| &= \left| \sum_{i=1}^n \log\left(\frac{\lambda_i(A+XX^T)}{\lambda_i(A + \sigma^2 I_n)}\right)\right|\\
& \le \sum_{i=1}^n \left|\log\left(\frac{\lambda_i(A+XX^T)}{\lambda_i(A + \sigma^2 I_n)}\right)\right| \\
& \le \sum_{i=1}^n\log\left(1 + \frac{t\sigma^2}{\lambda_i(A)+\sigma^2}\right)
\\
&\le t\sigma^2\sum_{i=1}^n (\lambda_i(A)+\sigma^2)^{-1}\\
&= t\sigma^2\eta(A).
\end{split}
$$
This is what I had in mind. Thumbs up.
|
2025-03-21T14:48:31.185945
| 2020-06-08T13:25:43 |
362508
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362508"
}
|
Stack Exchange
|
If a stochastic flow is Fréchet differentiable in the spatial parameter, does the induced transition semigroup preserve differentiability?
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(E,\mathcal E)$ be a measurable space, $X:\Omega\times[0,\infty)\times E\to E$ be $(\mathcal A\otimes\mathcal B([0,\infty))\otimes\mathcal E,\mathcal E)$-measurable and $$X^x_t:=X(\;\cdot\;,t,x)\;\;\;\text{for }(t,x)\in[0,\infty)\times E.$$ It's easy to see that $$\kappa_t(x,B):=\operatorname P\left[X^x_t\in B\right]\;\;\;\text{for all }(x,B)\in E\times\mathcal E$$ is a Markov kernel and $$(\kappa_tf)(x):=\int\kappa_t(x,{\rm d}y)f(y)=\operatorname E\left[f(X^x_t)\right]\tag1$$ for all $\mathcal E$-measurable $f:E\to\mathbb R$ with $(\kappa|f|)(x)<\infty$ for all $(t,x)\in[0,\infty)$.
(1): If $X(\omega,t,\;\cdot\;)$ is Fréchet differentiable, are we able to show that $\kappa_tf$ is Fréchet differentiable for all Fréchet differentiable $f:E\to\mathbb R$?
(2): What can we conclude (and how is this related to the former assumption) if $$E\ni x\mapsto C^0([0,\infty),L^p(\operatorname P,E))\;,\;\;\;x\mapsto X(\;\cdot\;,\;\cdot\;,x)\tag2$$ is Fréchet differentiable?
|
2025-03-21T14:48:31.186045
| 2020-06-08T13:26:30 |
362509
|
{
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],
"authors": [
"Abdelmalek Abdesselam",
"Jochen Wengenroth",
"JustWannaKnow",
"Nate Eldredge",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/7410"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362509"
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|
Stack Exchange
|
Strong topology on a topological vector space
I'm not sure this is an appropriate question for this site but I've tried math stack exchange and got no answers. Also, this problem arose in one of my research problems, so I'm stating it here.
The strong operator topology is defined on Simon and Reed's book as follows. It is the weakest topology on $\mathcal{L}(X,Y)$ such all the maps $E_{x}: \mathcal{L}(X,Y) \to Y$ defined by:
$$E_{x}(T) := Tx $$
are continuous for all $x \in X$. Here, $X$ and $Y$ are supposed to be Banach spaces and $\mathcal{L}(X,Y)$ is the space of all bounded linear operators from $X$ to $Y$. A neighboorhood basis for this topology, in Simon's words, is given by the sets of the form:
$$ \{S: \hspace{0.1cm} S\in \mathcal{L}(X,Y), \hspace{0.1cm} ||Sx_{i}||_{Y}<\epsilon, \hspace{0.2cm} i=1,...n\}$$
where $x_{1},...,x_{n}$ is any finite collection of elements of $X$ and $\epsilon > 0$.
I know the notion of strong topology can be extended to more general spaces such as topological vector spaces, but I don't want to get too deep into the theory. However, I'm interested in the case where $X$ is not Banach but $Y = \mathbb{C}$ is Banach.
My question is: In my setup, if $X$ is a Fréchet space and $Y=\mathbb{C}$ is Banach, the above definition seems to work just fine if I replace $\mathcal{L}(X,Y)$ the space of bounded linear operators to its analogue, the space of all continuous linear maps. The same properties seem to hold in this case. Is it a correct definition of a strong topology to my particular case? In other words, if I was to consider $X$ as a topological vector space and $X^{*}$ its topological dual, would the strong topology defined on $X$ be the same topology I'm proposing?
If $Y = \mathbb{C}$ then $\mathcal{L}(X,\mathbb{C}) = X^$ and the strong operator topology on $\mathcal{L}(X, \mathbb{C})$ is the same as the weak- topology on $X^*$. It's more usual to use the latter name.
@NateEldredge thanks for the comment! Simon's terminology 'strong operator topology' for $\mathcal{L}(X,\mathbb{C}) = X^{*}$ is defined when $X$ is Banach. Here, you are defining this topology as I proposed?
Yes, the definitions still make perfect sense for any topological vector space. Reed and Simon probably add the Banach assumption because that's the only case they care about, and because they want to prove theorems that may require that assumption.
Oh, right. My bad. Gonna fix it right now! Thanks
I consider both names strong operator topology as well as weak$^$ topology* quite unfortunate (although they are of course standard). A good name, IMHO, would be topology of pointwise convergence.
@JochenWengenroth this is because, in this topology, a net of operators $T_{\alpha}$ converges to $T$ iff $||T_{\alpha}x-Tx||\to 0$ for every $x \in X$, right?
Yes, that's right.
This is a clash of two cultures which use the adjective "strong" for a topology with completely different meanings. I agree with Jochen's that this choice of terminology is quite unfortunate. I believe the question the OP is after is what is the correct topology on spaces of distributions like $\mathscr{D}'$, $\mathscr{S}'$ and their sequence space concrete realization like $s'$, etc. The answer is the strong topology in the sense of the topological vector spaces literature. The more precise (Jochen comment compliant) terminology would be the topology of uniform convergence on bounded sets.
First review the basic definitions given in
https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-d-omega/3511753#3511753
to which one can add the following. For a LCTVS $V$, and a subset $A\in V$, we say that $A$ is bounded iff for every continuous seminorm $\rho$ on $V$,
$$
\sup_{v\in A}\rho(v)\ <\ \infty\ .
$$
The strong dual $V'$ is the space of continuous linear forms $L:V\rightarrow\mathbb{C}$
with locally convex topology defined by the collection of seminorms
$$
\rho_A(L)=\sup_{v\in A}|L(v)|
$$
indexed by the (nonempty) bounded sets $A$ in $V$.
Take the space of sequences $s$. To a nonempty subset $A\subset s$, one can associate an envelope ${\rm env}(A)$ which is the sequence $(a_n)$ given by
$$
a_n=\sup_{x\in A}|x_n|\ .
$$
Exercise 1: Show that a $A$ is bounded iff ${\rm env}(A)\in s_+$ (sequences in $s$ with nonnegative entries).
Exercise 2: Consider $s'$ realized as a space of sequences. Show that the previous strong topology is the same as the locally convex topology defined by the seminorms
$$
||y||_{\omega}=\sum_n \omega_n\ |y_n|
$$
indexed by $\omega\in s_+$.
Note that one can take $\ell^{\infty}$ or $\ell^p$ norms instead of $l^1$ with the same result. The relation to the envelope is more immediate with the $\ell^{\infty}$ variant. However, the $\ell^1$ choice allows a more immediate comparison with the weak-$\ast$ topology
which is defined by the seminorms
$$
||y||_x=\left| \sum_n x_n y_n \right|
$$
indexed by $x\in s$. A glance at the last formula, versus the one for $||y||_{\omega}$,
should be enough to see that the $||y||_x$ are very bad seminorms to work with.
Exercise 3: Repeat the process for $V=s'$ instead of $s$ and show that the strong dual is $s$ with its original topology. Namely, $s$ is reflexive, just like a finite-dimensional space.
Moral of the story: The strong and weak-$\ast$ topologies are not that far apart because a bounded set "behaves like a single vector", however, it allows one to put the absolute values where they should be, i.e., "inside" the sum for the duality pairing. For more on this, see my answer to the MO question: Is the tensor product of distributions a continuous bilinear map with respect to the weak topology?
sorry lots of typos in my first draft of the post. I hope it is fixed now.
@AbdelmalekAbdesselam thanks again for the answer! Indeed, the problem is related to these spaces of sequences you mentioned. I'm working out the details of your previous answers and, in special, the realization of $s'$ as the dual of $s$. I realized that when you said the dual was equiped with the strong topology, I automatically assumed that this was the strong operator topology on Simon & Reed's book, but the underlying spaces there are both Banach. I tried to look for equivalences on the internet but I didn't find anything closely related, only alternative constructions.
BTW, I also did a very extensive search on the internet about materials discussing these sequence spaces $s$ and $s'$ but I didn't find anything really helpful. I think this is because of the identifications of $s$ and $s'$ to $\mathcal{S}$ and $\mathcal{S}'$, and people prefer to work with the latter. Also, I don't think these sequence spaces have a proper name on the literature, so it is difficult to search. Just out of curiosity, do you know any material on these sequences?
There is material see https://mathoverflow.net/questions/361048/on-k%c3%b6the-sequence-spaces/361062#361062 but I do not recommend that for beginners. The point of sequence spaces is that they are so simple compared to S, S' etc that you can prove things you need by yourself.
Nice! I'm going to take a look at that! Thanks!!
|
2025-03-21T14:48:31.186485
| 2020-06-08T13:29:32 |
362510
|
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|
Stack Exchange
|
Euclidean embedding of the mesh
$M$ is a topological mesh, i.e. triple $M=(V,E,F)$, where $V$ is the vertex, $E$ is the edge and $F$ is the face, such that $M$ is homeomorphic to the sphere.
Suppose that we have a metric $l :E\to\mathbb R^+$( on each triangle of $F$, $l$ satisfies the triangle inequality) with positive sectional curvature.
Q Is there a method to embed $M$ to the three-dimensional Euclidean space $\mathbb R^3$, such that two vertices' distance coincides with the distance of metric $l$.
Is there any reference about the question.
|
2025-03-21T14:48:31.186552
| 2020-06-08T13:51:48 |
362511
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"https://mathoverflow.net/users/143783",
"https://mathoverflow.net/users/36721",
"lrnv"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362511"
}
|
Stack Exchange
|
Can this be translated to a truncated multivariate moment problem?
Fix $\mathbf t \in \mathbb{R}_+^d$. I am looking for a r-atomic measure $\nu$ on $\mathbb{R}_{+}^{d}$ that solves the following 'moment' conditions.
$$\forall\, \mathbf i \in \mathbb{N}^{d} \, s.t \, \lvert\mathbf i \rvert \le n \in \mathbb N, \; \mu_{\mathbf i}^{\mathbf{t}} = \int\limits_{\mathbb R_+^d} \frac{\prod\limits_{j=1}^d r_j^{i_j}}{\left(1+\sum\limits_{j=1}^{d} r_j t_j \right)^{\lvert \mathbf i \rvert}}\nu\left(d\mathbf r\right)$$
For given values of $\mu_{i}^{t}$.
Is there a transformation of my equations (by a push-forward change on $\nu$ for exemple) that would translate to a more 'classical' multivariate truncated moment problem ? i.e, i want an expression without the bottom part of the fraction..
Edit : My idea is currently to increase the dimension by one, by the change of variable :
$$\mathbf p = \left(r_1,...,r_d,\frac{1}{\left(1+\sum\limits_{j=1}^{d} r_j t_j \right)}\right)$$
Since $\mathbf t$ is fixed, this change of variable might be bijective, and it makes $\mu_{\mathbf{i}}^{\mathbf t}$ the $\left(i_1,...,i_d,\lvert\mathbf i \rvert\right)$th moment of an other random variable.. Will this work ? I am not very confident about which algorithm exists to solve such a r-atomic moment problem.
Changing the variables from $\mathbf r=(r_j)$ to $\mathbf s=(s_j)$, where
$$s_j:=g(\mathbf r)_j:=\frac{r_j}{1+\sum_i t_ir_i},$$
we have
$$r_j:=\frac{s_j}{1-\sum_i t_is_i}.$$
So, the transformation $g$ is a homeomorphism of $\mathbb R_+^d$ onto
$$\Sigma:=\Big\{\mathbf s\in\mathbb R_+^d\colon \sum_i t_is_i<1\Big\}.$$
Therefore, the problem can indeed be rewritten as the multivariate moment problem to find a measure $\rho$ such that
$$\mu_{\mathbf i}^{\mathbf{t}}=\int_\Sigma \rho(d\mathbf s)\,
\prod_{j=1}^d s_j^{i_j}\quad
\forall\, \mathbf i \in \mathbb{N}^{d} \ s.t. \ \lvert\mathbf i \rvert \le n \in \mathbb N. $$
The measure $\nu$ can then be obtained by the formula $\nu(B)=\rho(g(B))$ for all Borel $B\subseteq\mathbb R_+^d$.
Waouh. Thanks, i was sure that it could be done but i did not found the $g_j$ functions. You made my day. I now have to found how the moment problem can be solved numericaly, but this is another story.
I am glad this helped.
|
2025-03-21T14:48:31.186725
| 2020-06-08T14:56:59 |
362516
|
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|
Stack Exchange
|
Is every extension of ZFC interpretable in a finite extension of ZC + rank?
Let's speak of the theory $\sf ZC + rank$ as the first order set theory with axioms of Extensionality, Separation, infinity, and choice (written as usual), plus iterative powers and foundation, those are:
Iterative Power: $\forall \text { ordinal } \alpha \exists x : x=P^\alpha(\emptyset)$
where $P^\alpha(\emptyset) = \bigcup \{P(P^\beta(\emptyset)) : \beta < \alpha\} $
Foundation: $\forall x \exists \text { ordinal } \alpha: x \in P^\alpha (\emptyset)$
Now ordinal is meant to be a Von Neumann ordinal.
is every extension of $\sf ZFC$ interpretable in a finite extension of $\sf ZC + rank$?
is every finite extension of ZFC interpretable in a finite extension of ZC + rank?
Definitely not every extension - ZFC has uncountably many pairwise incompatible extensions (e.g. for any sequence $(a_n)\in{1,2}^\omega$ consider ZFC+"$2^{\aleph_n}=\aleph_{n+a_n}$" for all $n$), but ZC+rank has only countably many finite extensions, and each interprets only finitely many theories.
@Wojowu, then it might be more plausible to ask if every finite extension of ZFC is interpretable in a finite extension of ZC + rank?
@Wojowu, I just want to ask about the theories "ZFC + for all $n:2^{\aleph_n}=\aleph_{n+a_n}$" now each of these must be "effectively generated" so it must have a clear definition of its terms, it appears to me that $a$ is a sequence (so its domain is N) of binary strings which would code some reals I suppose. But this sequence must be definable otherwise the theory "ZFC + for all $n:2^{\aleph_n}=\aleph_{n+a_n}$" is not well defined, and so not effectively generated. So we only have countably many such extensions.
@EmilJeřábek, yes I assume consistency and that ZC+rank is $\Sigma_1$-sound.
@Wojowu, I see where I missed, I should have said any recursively axiomatizable extension of ZFC. I noticed that after reading Emil's answer, although my comment preceded this answer and it was clear that I was meaning effectively generated extensions of ZFC.
$\let\res\restriction\def\N{\mathbb N}$By the discussion in comments, it is missing from the question that all the theories are assumed consistent (otherwise the answer is trivially yes, as everything is interpretable in the inconsistent extension) and recursively axiomatizable (otherwise the answer is trivially no, as there are $2^\omega$ pairwise incompatible complete extensions of ZFC, while only countably many of these are interpretable in any given consistent extension of ZC + rank).
With these caveats, the answer to both questions is yes, for general reasons that work for most other pairs of natural theories. In what follows, if $T$ is a theory with a fixed r.e. set of axioms, let $\Box_T$ denote the formalized provability predicate for $T$, and $T\res n$ the theory axiomatized by the axioms of $T$ with Gödel number below $n$.
Proposition. Let $S$ be an extension of $I\Delta_0+\mathrm{EXP}$, and $T$ be a recursively axiomatized theory such that
$$\tag{$*$}S\vdash\Box_{T\res n}\phi\implies T\vdash\phi$$
for all $n\in\N$ and all $T$-sentences $\phi$. Then every consistent r.e. extension of $T$ is interpretable in a consistent finite extension of $S$.
Observe that $(*)$ holds whenever $S\subseteq T$ (or just $S$ is $\Sigma_1$-conservative over $T$) and $T$ is a locally essentially reflexive theory, for example $S=\mathrm{ZC+rank}$ and $T=\mathrm{ZFC}$. But in fact, $(*)$ holds whenever $S$ is $\Sigma_1$-sound, which pretty much any natural theory is. Also, it is not difficult to show that if $S$ is locally essentially reflexive, then condition $(*)$ is necessary.
Proof. Let $U\supseteq T$ be recursively axiomatized and consistent. Then $(*)$ implies that
$$V=S+\{\mathrm{Con}_{U\res n}:n\in\N\}\equiv S+\{\neg\Box_{T\res n}\neg\phi:n\in\N,\phi\in U\}$$
is consistent; since it is an extension of $S$ by a r.e. set of $\Pi_1$-sentences, there exists a $\Pi_1$-sentence $\psi$ such that $S+\psi$ is a $\Sigma_1$-conservative extension of $V$ by a theorem of Lindström [1]. In particular, $S+\psi$ is consistent, and since it proves $\mathrm{Con}_{U\res n}$ for all $n\in\N$, it interprets $U$ by the usual interpretation existence lemma. QED
Reference:
[1] Per Lindström: On partially conservative sentences and interpretability, Proceedings of the American Mathematical Society 91 (1984), no. 3, pp. 436–443, doi: 10.2307/2045318.
|
2025-03-21T14:48:31.187009
| 2020-06-08T15:24:33 |
362519
|
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|
Stack Exchange
|
Which fields of mathematics are most straightforward for a statistician to do research in, other than probability theory?
I have a Ph.D. in Statistics and have always been interested in pure mathematics, but never had a chance to really pursue it. My mathematics background includes real analysis, linear algebra, functional analysis, and measure theory. These were taken at the graduate mathematics Ph.D. level. I am wondering how many classes are necessary for me to take in pure mathematics in order to do research in pure mathematics. Does it require me to go back to school or are there certain areas of pure mathematics that would be easier for someone like me to dive into? Thanks in ahead.
I am not qualified enough to fully answer such a broad question. You could perhaps employ your expertise in statistics to help bridge the gap with applied category theory: an example is this preprint by Rémy Tuyéras.
I suggest looking for research on rigorous mathematical foundations of statistics that you find interesting and are not too far of a leap from what you already know. From there you can probably gradually migrate into research in areas such as probability and the others mentioned in the answers. But it's best to start with something that has connections to your current area of expertise.
Perhaps ergodic theory as a bridge from statistics to dynamical systems (or many other fields, ergodic theory has numerous applications and links).
I have PhD in Statistics and work in Machine learning. I encountered many gaps in the study of positive definite functions/kernels. As a statistician I have very specific questions which can move my field forward if answered but mathematicians never even thought of these questions. What is even more frustrating - I cannot interest any of my colleagues to look into my specific question because they are only interested in working in their own very narrow fields. So I am educating myself in all these math fields you have just listed.
@MattF. thank you.
|
2025-03-21T14:48:31.187175
| 2020-06-08T17:01:09 |
362524
|
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|
Stack Exchange
|
Classicality theorems for Hilbert eigenforms of low weight
Let $p$ be a prime, $k \geq 2$ an integer and $f$ a normalized overconvergent, cuspidal eigenform of finite slope, weight $k$, and level divisible by $p$. The classicality theorem of Coleman gives says that $f$ is classical exactly if $\mathrm{val}_p(a_p(f))<k-1$ or if $\mathrm{val}_p(a_p(f))=k-1$ and $f$ is not in the image of the $k-1$ iteration of the operator $\theta = qd/dq$.
Now suppose instead that $f$ is a Hilbert modular form of partial weight $1$. For simplicity, let's say $F$ is a totally real field quadratic over $\mathbb{Q}$, and $f$ is normalized overconvergent, cuspidal Hilbert eigenform for $F$ of finite slope, weight $[k_1,k_2,w]=[k,1,w]$ where $k\geq 2,w\in \mathbb{Z}$, and level divisible by $p$. In this context there are theta operators constructed by Goren and Kassaei, which correspond to the different weights, so let's say $\theta$ is the one corresponding to the first weight $k_1=k$. Is there a similar classicality theorem in this context? (Say, if the slope is small enough compared to an expression involving $k$ and $w$, then $f$ is classical? And perhaps the boundary case has something similar to do with the $\theta$ operator?)
Such a criterion is conjectured by Breuil in section 4 of "conjectures de classicité sur les formes de Hilbert surconvergentes de pente finie" but classicality results I am familiar with (such as in the work of Tian and Xiao) always assume the weight is cohomological.
Thanks!
I think Stephane Bijankowski's article called Classicite de formes modulaires de Hilbert is maybe what you want. I can't see any restriction on the weight in his theorem and since it does not appear to be a cohomological construction, so it should probably work (note this I think is for geometric hmfs, so you might need to argue a bit more to get your arithmetic hmfs statement). If not, then I expect you can find it somewhere in the "Arithmetique $p$-adique des formes de Hilbert".
|
2025-03-21T14:48:31.187323
| 2020-06-08T17:04:16 |
362526
|
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|
Stack Exchange
|
Representability result
Let $X$ and $S$ be schemes over a field $k$.
Reading this paper, there is a result on the representability of a morphism (proposition 3.1, page 4).
Which result or reference on representability is used in this proposition?
The forgetful morphism $ {\mathcal C(r, X)\to\mathcal M(r,X) \times \Bbb A^1} $ is representable.
Where $\mathcal C(r,X)$ is a stack over $ \operatorname{\textbf{Aff}} /\Bbb A^1 $ with objects the vector bundles on $X\times_k S$, and $ \mathcal M(r,X) $ is the stack of rank-$ r $ vector bundles of degree zero on $ X $.
The authors seem to construct an $S$-category of triples(which reminds me of a $2$-fiber product) and prove that it is represented by a geometric object but I don't see which theorems or definition they're using.
Many Thanks for any insights or references.
|
2025-03-21T14:48:31.187421
| 2020-06-08T17:14:31 |
362527
|
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"Anthony Quas",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362527"
}
|
Stack Exchange
|
Chromatic number of distance graphs over the integers
Let $D\subseteq\mathbb{N}^+$, and consider the graph $G_D$ with vertices set $\mathbb{N}$ and edges set $\{(x,y)\in\mathbb{N}\times\mathbb{N}\;s.t.\;|x-y|\in D\}$. I expect that if $D$ is dense enough in $\mathbb{N}^+$, then the chromatic number of $G_D$ is large. As Wojowu pointed out in the comments, positive density does not guarantee infinite chromatic number. Hence, one can ask the following question:
if $D$ has density (say, e.g., lower asymptotic density) one in $\mathbb{N}^+$, is it true that the chromatic number of $G_D$ is infinite?
Thank you for any suggestion.
If $D$ contains no numbers divisible by $k$, then the graph is $k$-colorable, by coloring each residue class modulo $k$ differently. Hence for no $d<1$ does density $d$ guarantee infinite chromatic number. On the other hand, I believe density equal to $1$ is sufficient to guarantee that.
Sure, thank you a lot for pointing it out. I edited the question, focusing on the case of density one.
Following up on this comment, if the density is big, you’ll have big cliques (and thus big chromatic number). That is, if you insist there are no $k$-element cliques, then in particular $D$ has no sets of the form $x, 2x, ..., kx$. But this means $D$ has at most $(1-\varepsilon)N$ elements less than $N$ (for some $\varepsilon > 0$ depending on $k$).
(E.g., $\epsilon = 1/k^2$ can be proven by noting there are at least $n/k$ such sets the complement must intersect, and any element meets at most $k$ of these. So the complement has size at least $n/k^2$)
@PatDevlin: you should post an answer, this solves the question. Actually, you only need to say that if $D$ has density $1$, then so does $D_2:={x\in\mathbb{N}|2x\in D}$, and so do $D_3$, ..., $D_k$. Then $D\cap D_2\cap \dots\cap D_k$ has density one, and is non empty. You get a $k$-clique.
One gets this way, with a crude estimate, that if the lower density of $D$ is at least $1-\frac{2}{k(k+1)}$ then $G_D$ has a $k$-clique, and in particular has chromatic number at least $k$. There might be lots of interesting follow-up questions in this.
I believe infinite chromatic number also follows from the weaker assumption of $D$ having upper Banach density 1. However, the argument is different from what @BenoîtKloeckner has written (based on the comments of @PatDevlin) since two sets of upper Banach density 1 need not intersect.
Thank you everybody for the interesting comments! If all of you prefer, I can edit the post by summarizing what you said so far, and by posing more refined questions. Otherwise, I keep the actual post and accept the answer of who want to write it.
I'll post a proof of what I've claimed. But if one of the earlier commenters posts theirs then it should be accepted over mine.
Maybe you're familiar with the following somewhat-related paper, showing that if $D$ grows exponentially, then the chromatic number is finite. Katznelson, Y.(1-STF)
Chromatic numbers of Cayley graphs on $\mathbb Z$ and recurrence. (English summary) Paul Erdős and his mathematics (Budapest, 1999).
Combinatorica 21 (2001), no. 2, 211–219.
@Anthony Quas, thank you for the reference
This post, subsequent comments, and answers all seemed to go quite well with everybody being very cordial! Yay community! :-) [PS, I was getting groceries, so I forgot to reply for a bit, but I didn’t feel like writing an answer since it didn’t feel like the right proof anyway! What’s there now feels much better.]
I'll show that if $G_D$ has chromatic number $k$ then $D$ has upper Banach density at most $(k-1)/k$.
So suppose $G_D$ has chromatic number $k$. Let $\mathbb{N}$ be partitioned into $P_1,\ldots,P_k$, where each $P_i$ is independent with respect to $G_D$. Without loss of generality, $P:=P_1$ has upper Banach density at least $1/k$. Let $Q=\{|x-y|:x,y\in P\text{ are distinct}\}$. Then $Q\subseteq \mathbb{N}^+\backslash D$ since $P$ is $G_D$-independent. We claim that $Q$ has lower Banach density at least $1/k$, which implies the desired result for $D$.
(The proof of the claim is an adaptation of Ruzsa's Covering Lemma and/or the well-known fact that if a set $A$ of integers has positive upper Banach density then $A-A$ is syndetic.)
Call a set $X\subset\mathbb{N}$ $P$-separating if $(x+P)\cap (y+P)=\emptyset$ for all distinct $x,y\in X$. Since $P$ has upper Banach density at least $1/k$, it follows that any $P$-separating subset of $\mathbb{N}$ has size at most $k$. So we may choose a $P$-separating set $X$ of maximal size. Now fix $a\in\mathbb{N}^+$ such that $a>\max X$. By maximality, there is some $x\in X$ such that $(a+P)\cap (x+P)\neq\emptyset$. So there are $p,q\in P$ such that $a+p=x+q$. Since $a>x$ it follows that $a\in x+Q$.
Altogether, we have shown that $X+Q$ is cofinite in $\mathbb{N}^+$. Since $|X|\leq k$, it follows that $Q$ has lower Banach density at least $1/k$.
Remark. The proof actually shows that if $G_D$ has chromatic number $k$ then there are $k$ translates of the complement of $D$ whose union is cofinite in $\mathbb{N}^+$, which I suppose is stronger than saying $D$ has upper Banach density at most $(k-1)/k$.
Nice answer, thank you!
The upper bound is tight (as suggested in Ilya Bogdanov's answer and observed earlier in the comments by Wojowu). If $D$ is the set of natural numbers not divisible by $k$ then $D$ has density $(k-1)/k$, and $G_D$ is $k$-colorable via the partition into equivalence classes modulo $k$.
Assume that the chromatic number is $k$. Among the numbers $1,2,\dots, N$ there are at least $N/k$ numbers of the same color, say $a_1,\dots, a_t$. Then $D$ does not contain at least $t-1\geq N/k-1$ numbers not exceeding $N-1$, namely $a_i-a_1$, $i\geq2$. Thus the density of $D$ is at most $1-1/k$. Surely, this estimate is tight...
Thank you for your answer!
|
2025-03-21T14:48:31.187949
| 2020-06-08T17:23:50 |
362528
|
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|
Stack Exchange
|
Non-homogeneous Poisson process with intensity function $\lambda\cdot f_{X_1}$
I want to show that there is a non-homogeneous Poisson process with a certain intensity function, but I have some problems while showing that this Poisson process satisfies the axioms(?). I am using the axioms as follows:
$N(0) = 0$
if $s\leq t$, then $N(s)\leq N(t)$
etc
as usual. Now the problem is given by:
Let $X_1,X_2,\dots$ be i.i.d. continuous random variables with common probability density function $f_{X_1}$ and let $Z \xrightarrow{d}Po(\lambda)$ be independent of $X_1,X_2,\dots$. Now define the random point set $\mathcal{P} = \{X_1,\dots,X_Z\}$ ($\mathcal{P}$ = $\emptyset$ if $Z = 0$).
Show that $\mathcal{P}$ is an inhomogeneous Poisson point process with intensity function $\lambda\cdot f_{X_1}$.
Sorry if my question is not clear enough, and thank you!
Indeed, to me the problem is not clear. What do you want: a Poisson point process or a Poisson process? Please try to clarify the assumptions.
@DieterKadelka If I am right, either Poisson point process or Poisson process means the same. If not, I was just meaning a normal Poisson process which is defined by the set of random points
I see you are done with this, but I would like to point out that in case the $X_i$ are uniform on (0,1) , this is the well know connection between the times of the poisson process and the order statistics of uniforms
$\newcommand{\la}{\lambda}$
Take any $t_0,\dots,t_n$ such that $0=t_0<\dots<t_n=\infty$. For each $j\in[n]:=\{1,\dots,n\}$, let $I_j:=[t_{j-1},t_j)$ and
\begin{equation}
\nu_j:=\#\{i\in[Z]\colon X_i\in I_j\}.
\end{equation}
Then, for each nonnegative integer $z$, the joint conditional distribution of $(\nu_1,\dots,\nu_n)$ given $Z=z$ is the multinomial distribution with parameters $z,p_1,\dots,p_n$, where
\begin{equation}
p_j:=P(X_1\in I_j).
\end{equation}
So, for any nonnegative integers $k_1,\dots,k_n$
\begin{align}
P(\nu_1=k_1,\dots,\nu_n=k_n)
&=\sum_{z=0}^\infty P(Z=z)P(\nu_1=k_1,\dots,\nu_n=k_n|Z=z) \\
&=\sum_{z=0}^\infty \frac{\la^z e^{-\la}}{z!}
\frac{z!}{k_1!\cdots k_n!}\, p_1^{k_1}\cdots p_n^{k_n}\,1\{k_1+\dots+k_n=z\} \\
&=\prod_{j\in[n]}
\frac{(\la p_j)^{k_j}e^{-\la p_j}}{k_j!}.
\end{align}
So, $\nu_1,\dots,\nu_n$ are independent random variables and $\nu_j\sim Poisson(\la p_j)$ for each $j\in[n]$.
This means that indeed $\mathcal{P}$ is an inhomogeneous Poisson point process with intensity function $\la f_{X_1}$.
I am curious because there was a comment that pointed out the difference between Poisson process and Poisson point process and you said that $\mathcal{P}$ is a Poisson point process. Are they actually different or is it just a more precise name of it?
@MathislikeFriday : The Poisson point process (say on the real line) is a random integer-valued measure, which may be identified with the random set of points, whose realizations are the support sets of the corresponding realizations of the random measure; this is what your random set $\mathcal P$ is. Also, the random integer-valued measure, say $\mu$, that is the Poisson point process may be identified with its counting function $N_\cdot$ defined by $N_t:=\mu([0,t])$ for $t\ge0$.
Previous comment continued: The counting function $N_\cdot$ may also be considered as the stochastic process $(N_t)$, which is the Poisson process corresponding the Poisson point process.
Thank you for very detailed answer! It's clear now
I have another question about your answer. Why/How did you set up the parameter of multinomial distribution as $p_j := \mathbb{P}(X_1\in\mathcal{I}_j)$? I don't get where $X_1$ came from
Given $Z=z$, we have $z$ independent trials, each trial with $n$ possible outcomes. The probabilities of the outcomes in each trial are $p_1=P(X_1\in I_1),\dots,p_n=P(X_1\in I_n)$. Here instead of $X_1$ you can write $X_i$ with any natural $i$, since the $X_i$'s are iid.
Thank you again sir :)
Basically a Poisson point process can be defined on the real line by considering the number of points of the process in the interval (a.b], which is \lambda*(F(b)-F(a)) (\X_i s are iid), then use the definition of the intensity…
|
2025-03-21T14:48:31.188220
| 2020-06-08T17:47:32 |
362530
|
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|
Stack Exchange
|
Hopf algebra structure on Frobenius algebras
It was shown by Abrams (see https://www.sciencedirect.com/science/article/pii/S0021869399979012 ) that every Frobenius algebra has a canonical coalgebra structure.
Question 1: Has it been studied when the Frobenius algebra together with the coalgebra structure is a Hopf algebra?
It seems that this can happen nearly never but I am not so experienced with Hopf algebras.
The question sounds also rather natural so probably it has been studied already or has a trivial answer (that there are no non-trivial cases?)
Question 2: Is there a program that can test whether a given Frobenius algebra (given by quiver and relations) is a Hopf algebra ? (for example finding all possible coalgebra structures and testing for the Hopf algebra conditions).
Concerning question 2, do you know the work of C. Cibils and M. Rosso? I don't hace the precise reference, but look for them on Google together with Hopf quivers, or Hopf path algebras/ Hopf path coalgebras. They look when a path algebra/coalgebra is Hopf. I don't remember if they consider relations. There are very few quivers that works
The compatibility conditions on the product/coproduct for Frobenius algebras and Hopf algebras are very different. It's almost never the case that the same coalgebra/algebra satisfies both the Frobenius axioms and the Hopf algebra axioms. (Probably it's not hard to prove a result in that direction, but I don't know it off the top of my head. The trivial example of just the base field will satisfy both.)
One weird complication is that finite dimensional Hopf algebras always have a Frobenius algebra structure, but the coproduct for the Hopf algebra structure and the coproduct for the Frobenius structure are different! In particular, the counit for the Frobenius structure is given by an integral, not by the counit.
In case $A$ is a non-semisimple quiver algebra, the trace $\lambda$ of the Frobenius structure satisfies $\lambda(1)=0$ (since it is essentially given so that it maps only elements of the socle of the algebra to non-zero elements) and thus it can not be an algebra map. Thus in the case of non-semisimple quiver algebras, this coalgebra structure indeed never gives rise to a Hopf algebra structure.
|
2025-03-21T14:48:31.188398
| 2020-06-08T17:56:26 |
362531
|
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|
Stack Exchange
|
Derivations of differential operators
For a smooth affine variety $\operatorname{Spec} A$ over a ring $R$ we have an algebra of differential operators $\mathcal{D}_A$ (here I mean not the Grothendieck differential operators but PD-ones). Is it possible to describe $\operatorname{Der} \mathcal{D}_A$? As far as I understand, over $\mathbb{C}$ there is an explicit relation with $\Omega_{A}^1$ but I am mostly interested in the case of $R = \mathbb{Z}/p^n \mathbb{Z}$ and $A = \mathbb{Z}/p^n \mathbb{Z} [x]$. How to describe it in this case?
In case $R$ is a field $k$ of characteristic $p$, I can answer this question.
For any $k$-algebra $D$, set $HH^1(D) = Der_R(D)/Inn(D)$ to be the space of outer derivations. Thus we have a short exact sequence
$$ 0\to Inn(D) \to Der_R(D) \to HH^1(D)\to 0.$$
If $D = \mathcal D_A$ for a smooth $k$-algebra $A$, then the center of $D$ is just $k$, so $Inn(D) = D/k$. The outer derivations $HH^1(D)$ are as follows:
Theorem: $HH^1(\mathcal D_A) = \varprojlim\limits_r (A / A^{(p^r)}) / A$, where $A^{(p^r)}$ is the $k$-span of $\{a^{p^r} \mid a \in A\}$.
In case $A = k[x]$, you can think of this limit as the space of power series $\sum_i c_i x^{\alpha_i}$ where for any $r \geq 0$, all but finitely many $\alpha_i$ are divisible by $p^r$; then one goes modulo polynomials. An example of a nontrivial such series is $x + x^p + x^{p^2} + x^{p^3} + \cdots$. Taking the commutator with this series gives a derivation because every differential operator on $k[x]$ commutes with some $x^{p^r}$.
The displayed theorem appears in my recent preprint "Hochschild cohomology of differential operators in positive characteristic," https://arxiv.org/abs/2303.07373. See especially Example 5.1.
|
2025-03-21T14:48:31.188540
| 2020-06-08T18:13:44 |
362533
|
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|
Stack Exchange
|
Character table of the symmetric group $S_n$ according to James
In James, "The representation theory of the symmetric groups" an algorithm is described to produce the character table of a symmetric group. The proof involves the equation (pp. 22,23)
$$\lvert S_\lambda\rvert \lvert S_\nu\rvert \langle \text{ind}_{S_\lambda}^{S_n} 1_{S_\lambda}, \text{ind}_{S_\mu}^{S_n} 1_{S_\mu}\rangle = \sum_{\mu} \frac{n!}{\lvert S_\mu \rvert} \lvert S_\lambda \cap S_\mu \rvert \lvert S_\nu \cap S_\mu\rvert$$
which is verified by first applying Frobenius reciprocity and then getting to the intermediate (apparently equal) term
$$\lvert S_{\lambda} \rvert \sum_{\mu} \text{ind}_{S_\lambda}^{S^n}(g^\mu) \lvert S_\nu \cap S_\mu \rvert$$
(where $g^\mu$ is some element of type $\mu$). Looking at James, Kerber, "The representation theory of the symmetric group" I find that this part of the statement is removed and instead there is a general discussion about the values of $\langle \text{ind}_{S_\lambda}^{S_n} 1_{S_\lambda}, \text{ind}_{S_\mu}^{S_n} 1_{S_\mu}\rangle$. Could someone explain to me, what the status of those few lines in the first book is? Is it easy and I'm just missing something easy? Or does it in fact use more machinery than is available at that point?
Edit: a formulation of the question that does not involve characters is how to see
$$\lvert S_\lambda \backslash S_n / S_\nu \rvert = \frac{\lvert S_n \rvert}{\lvert S_\lambda \rvert \lvert S_\nu \rvert }\sum_{\mu} \frac{1}{\lvert S_\mu \rvert} \lvert S_\lambda \cap S_\mu \rvert \lvert S_\nu \cap S_\mu \rvert\text{.}$$
Edit2: I think he is saying that $\lvert (g^\mu)^{S_n} \cap S_\nu \rvert = \lvert S_\mu \cap S_\nu \rvert$ (the first is the conjugacy class of $g^\mu$ in $S_n$).
Solution: In the end this is about handwriting: James writes a gothic S for the symmetric groups and a curly C for conjugacy classes. Just, both of these are little squiggles and I only realize now that they are supposed to be distinct. So every $S_\mu$ has to be a $C_\mu = (g^{\mu})^{S_n}$. I apologize to everyone who spent time on this!
P.S: James himself gets confused about this: The fraction $n!/\lvert C_\mu \rvert$ is the order of the centralizer of $g^\mu$ so the denominator has to be $C_\mu$ rather than $S_\mu$. However, what James writes is very clearly (once you get sensitive to the distinction) an $S_\mu$.
Is Mackey's formula mentioned in the intermediate lines?
No, that's the thing. It appears that it should be used but he specifically does not. He applies Frobenius reciprocity (which would also be the first step toward Mackey) but then he writes the resulting scalar product out with $1_{S_\mu}$ as a sum. Sorry, I should probably add that intermediate step to the question.
@GeoffRobinson: It's certainly not mentioned but mabye it is Mackey's formula after all (and then some more steps so it's not apparent).
Is the sum ranging over all partitions $\mu$ of $n$ or over all compositions $\mu$ of $n$ ?
Also, which part of James pp. 22-23 are you referring to?
@darijgrinberg: Partitions. It's a sequence of displayed equations around the page break: the last three lines on page 22 and the first few on page 23.
And $S_\mu$ is the Young subgroup (so, in particular, $S_{\left(n\right)} = S_n$ and $S_{\left(1,1,\ldots,1\right)} = \left{\operatorname{id}\right}$)? In this case, this is false for $n = 2$ and $\lambda = \nu = \left(1,1\right)$ unless I'm mistaken.
@darijgrinberg: You are saying the left hand side is 1 while the right hand side is 1 + 1, yes? So perhaps there is a restriction on \mu (that \nu has to dominate \mu?). (I suppose it's even more apparent in the formulation of edit2: the conjugacy class of (1 2) does not meet $S_\nu$ at all, so $\mu = (2)$ should not be allowed.)
I'm talking about the equality in your first Edit; conjugacy classes don't even appear there.
@darijgrinberg: right, there the right hand side is not even an integer if $\mu = (2)$ is allowed. So I think it's safe to assume that if $n = 2$ and $\lambda = \nu = (1,1)$ then $\mu$ is supposed to only range over $(1,1)$.
Hah, I got confused by James's Fraktur too. Someone should really republish the book in 21th century typesetting.
@darijgrinberg: In any case, thanks for pointing out that it doesn't work on small examples! After making apologies about your counterexample I looked at $\mu = (2,1)$ and $\nu = (3)$ and there it just made no sense at all.
|
2025-03-21T14:48:31.188845
| 2020-06-08T20:04:02 |
362540
|
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|
Stack Exchange
|
Fredholmness of elliptic operator on Hölder spaces
Let $(M,g)$ be a smooth oriented closed Riemannian manifold, $E\to M$ a smooth vector bundle, and $C^{k,\alpha}(E)$ the Banach space of sections of $E$ that are $k$-times differentiable (with respect to the Levi-Civita connection for $g$) and whose $k^{th}$ derivatives are all $\alpha$-Holder, $0<\alpha<1$. Let $L: C^\infty(E)\to C^\infty(E)$ be a linear elliptic partial differential operator of order $m$. Is it true that $L$ extends to a Fredholm operator from $C^{k+m,\alpha}\to C^{k,\alpha}$?
I'm under the impression the answer is yes, and I came up with a proof. It's just a slight modification of the proof of Fredholmness for the Sobolev spaces $W^{k,p}$. We want to show the (obviously continuous) extension has (1) finite dimensional kernel , (2) closed range, and (3) finite dimensional cokernel.
The key observation is that for any $r\leq k, \beta> \alpha$, the identity map $C^{k,\alpha}\to C^{r,\beta}$ is compact.
(1) If $u_n$ is a $C^{k+m,\alpha}$-bounded sequence contained in $\ker L$, then by passing to $C^{k,\alpha}$ we extract a subsequence along which it converges in $C^{k,\alpha}$ to a $C^{k,\alpha}$ section $u\in \ker L$. By the Schauder estimates we can control the $C^{k+m,\alpha}$-norm of $u_n$ uniformly by the $C^{k,\alpha}$ norm. This gives convergence in $C^{k+m,\alpha}$. This implies the unit ball of the kernel is compact, and hence this kernel is finite dimensional.
(2) First we prove a Poincare type inequality: let $(\ker L)^\perp$ be the set of $C^{m+k,\alpha}$-sections that are $L^2$-orthogonal to the kernel. Then for any $u\in (\ker L)^\perp$, $$\|u\|_{C^{k,\alpha}}\leq C\|Lu\|_{C^{k,\alpha}}$$ Indeed, if not, we can find a sequence of sections $u_n$ with $\|u_n\|_{k,\alpha}=1$ and $\|Lu_n\|_{k,\alpha}\to 0$. The Schauder estimates then give you a uniform upper bound on $\|u_n\|_{m+k,\alpha}$. Passing to $C^{m+k,\beta}$, for $\beta<\alpha$, the $u_n$ converge along a subsequence in $C^{m+k,\beta}$ to a section that is $L^2$-orthogonal to $\ker L$. However, $Lu=0$, so $u=0$, which is impossible.
With this in mind, let $y_n = Lx_n \in C^{k,\alpha}$ converge to $y\in C^{k,\alpha}$. We may assume $x_n$ is $L^2$ orthogonal to the kernel. By the Schauder estimates and the inequality proved above, $$\|x_n-x_m\|_{m+k,\alpha}\leq C\|y_n-y_m\|_{k,\alpha}$$ and hence the $x_n$ are Cauchy. The result follows.
(3) By compactness, $C^{k,\alpha}$ embeds into $W^{k,p}$ for any $p\geq 1$. $L$ is Fredholm on $W^{k,p}$, so there is a splitting $$W^{k,p}=\ker L^* \oplus \textrm{Im}(L)$$ where the second summand denotes the image of $L: W^{k+m,p} \to W^{k,p}$. We write $u$ uniquely as $$u=u_1+Lu_2$$ with $u_1\in \ker L^*$, $u_2\in W^{k+m,p}$. If $M$ has dimension $n$, then Morrey's inequality gives $u_2\in C^{r,\beta}$ if $$k+ m - \frac{n}{p}> r + \beta$$ Since we can choose $p$ as large as we like, we see $u_2\in C^{k+m-1, \alpha}$. Also, $Lu_2 = u-u_1$ implies $Lu_2\in C^{k,\alpha}$, because $u_1$ is smooth by the Weyl lemma. Appealing to the Schauder estimates once more we see $u_2\in C^{k+m,\alpha}$. This implies we have a splitting $$C^{k,\alpha} = \ker L^* \oplus \textrm{Im}(L)$$ where now the second summand is the image of $L: C^{k+m,\alpha}\to C^{k,\alpha}$. Hence, the cokernel identifies isomorphically with $\ker L^*$, and this is finite dimensional.
Is something in my work incorrect? Is there some obstruction I've missed? I can't see anything wrong with it--I'm just fairly agitated that I've spent a long time looking in texts and online and haven't found what should be a basic fact in any source at all. Thanks.
|
2025-03-21T14:48:31.189072
| 2020-06-08T20:41:25 |
362544
|
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"Carlo Beenakker",
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|
Stack Exchange
|
How do I join the dots between the formal definition of exterior derivative and the intuition
I am self-studying differential forms, and I'd like to understand the proof of Stoke's theorem.
I have read through the document here by Dan Piponi and really liked his explanation of the exterior derivative as finding the boundary in the picture (section 4, page 6), but I have a hard time to connect this with the formal definition of the exterior derivative.
Reading through the comments of his answer here, I am not sure if the concept of 'finding the boundary in the picture' applies to just xdy or any differential forms?
two earlier MO questions are similar: one and two --- is there something left to answer here?
well, I think joining the dots is essentially proving, at least some local version of Stoke's theorem. I understand that there is a gap between the definition, via essentially a formula, of the exterior differential, and the intuition, which is essentially the local version of Stoke's theorem. Historically, it came from generalizing several lower dimensional special cases, and is thus non-trivial. Many books discuss it. I would guess you could find it in some volume of Spivak, one of Lee's books, possibly in Boothby's book and many other places.
Perhaps my recent answer in Carlo's reference link "one" may join your dots.
|
2025-03-21T14:48:31.189197
| 2020-06-08T21:10:48 |
362546
|
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"Anurag Sahay",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362546"
}
|
Stack Exchange
|
Moments of Dirichlet $L$-functions on the critical line
I'm looking for a reference for some questions related to the moments of Dirichlet characters on the critical line,
$$ M_k(T;\chi) = \int_T^{2T} |L(1/2+it,\chi)|^{2k}\,dt, $$
where $\chi$ is a Dirichlet character and $k > 0$ is a fixed real number (though I mostly care about $k \in \mathbb{N}$).
Conjecturally, one expects that $M_k(T;\chi) \sim c_k(\chi) T (\log T)^{k^2}$ for fixed $k,\chi$ and $T \to \infty$.
I am looking for two results that probably exist in the literature, because analogous results exist for the Riemann zeta function, and the same tools should generalize, but I'm having trouble finding them on mathscinet or google scholar.
Question 1
Does the sharp estimate $M_k(T;\chi) \ll_{k,\chi} T (\log T)^{k^2}$ conditionally on GRH for $L(s,\chi)$ appear in the literature?
This estimate was shown on RH for the Riemann zeta function by Harper. For $L$-functions, I found this paper of Milinovich and Turnage-Butterbaugh where they show almost sharp conditional upper bounds on products of automorphic $L$-functions. Specializing to a single Dirichlet $L$-function, their results give on GRH that $M_k(T;\chi) \ll_{k,\chi,\epsilon} T (\log T)^{k^2 + \epsilon}$ which is just shy of what I think can be proved.
Question 2
In his paper on applying the Montgomery-Vaughan mean value theorem to the fourth moment of the Riemann zeta function, Ramachandra states that the methods of his paper can prove the following theorem:
"Theorem 4. Let $\chi_1$ and $\chi_2$ be two characters mod $q_1$ and $q_2$ respectively. Then for $T \geq 2$, the mean value
$$ \frac{1}{T} \int_0^T \left|L\left(\frac{1}{2}+it,\chi_1\right)L\left(\frac{1}{2}+it,\chi_2\right)\right|^2\,dt $$
is $C(\log T)^4 + O((\log T)^3)$ or $D(\log T)^2 + O(\log T)$ according as $\chi_1$ and $\chi_2$ are equivalent characters or not. (Here $C$ and $D$ are constants depeneding on $\chi_1$ and $\chi_2$."
He doesn't provide a proof of this, but notes that the details of the proof will be published elsewhere.
My question here is: did Ramachandra ever publish the details of the proof? I couldn't find it.
I am especially interested in a reference with a precise value for $C$ and $D$ as functions of the characters, especially if this exists in the literature already.
Thanks in advance!
With regards to Question 1: I'm pretty sure a detailed proof of the sharp estimates under GRH aren't written down anywhere, but they should be a pretty straightforward modification of Harper's result.
Thanks - I was afraid that would be the case. I found a few analogous results for classes of $L$-functions associated with automorphic forms (for example, this), but since I don't know much automorphic forms I was having trouble telling whether this contains the result I wanted.
I guess I'll put a pin on this and come back to it when I understand Harper's method.
Since I discovered the answer to my Question 2 in the literature, I figured I would update this in case someone stumbles upon it in the future.
It does not appear to be the case that that Ramachandra published the details of his proof anywhere [though it's a straightforward modification of the earlier results in the same paper]. However, Topacogullari [1] has recently proved a full asymptotic formula with power savings in the error term. His result is also uniform in the conductors of the characters involved.
For primitive characters $\chi,\chi_1,\chi_2$, $\chi_1 \neq \chi_2$ with conductors $q,q_1,q_2$ respectively, the constants $C(\chi)$ and $D(\chi_1,\chi_2)$ are given by
$$ C(\chi) = \frac{1}{2\pi^2} \frac{\varphi(q)^2}{q^2} \prod_{p \mid q} \left(1 - \frac{2}{p+1}\right), $$
and
$$ D(\chi_1,\chi_2) = \frac{6}{\pi^2} |L(1,\overline{\chi_1}\chi_2)|^2 \frac{\varphi(q_1)\varphi(q_2)}{\varphi(q_1 q_2)} \prod_{p \mid q_1 q_2}\left(1 - \frac{1}{p+1}\right). $$
As Peter Humphries mentions in the comments to the OP, the answer to Question 1 is no, this result does not appear to have been written down yet.
[1] Topacogullari, Berke. "The fourth moment of individual Dirichlet L-functions on the critical line." Mathematische Zeitschrift (2020): 1-48.
|
2025-03-21T14:48:31.189465
| 2020-06-08T22:37:54 |
362551
|
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|
Stack Exchange
|
Formally real non-Jordan algebras
Jordan, von Neumann and Wigner [1] showed that for any finite-dimensional real vector space $A$ with a bilinear commutative power-associative operation $\circ : A \times A \to A$, the formal reality condition
$$ a_1 \circ a_1 + \cdots + a_n \circ a_n = 0 \implies a_1, \dots, a_n = 0 $$
implies the remaining Jordan algebra axiom: for all $a \in A$, multiplication by $a \circ a$ commutes with multiplication by $a$.
Thus, in the definition of a formally real Jordan algebra (which is the kind of Jordan algebra important in quantum mechanics), we can leave out the law $(a \circ a) \circ (a \circ b) = a \circ ((a \circ a) \circ b)$ without harm, since it follows from the other conditions... in the finite-dimensional case.
But their proof uses finite-dimensionality. So, my question is:
Do you know an example of an infinite-dimensional vector space with a commutative bilinear power-associative operation obeying the formal reality condition that is not a Jordan algebra?
[1] P. Jordan, J. von Neumann and E. Wigner, On an algebraic generalization of the quantum mechanical formalism, Ann. Math. 35 (1934), 29-64.
I'd try the free commutative power-associative $\mathbf{R}$-algebra on 2 generators.
Nice question. +1 Still no answers.
|
2025-03-21T14:48:31.189569
| 2020-06-08T23:04:06 |
362553
|
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|
Stack Exchange
|
Mapping class groups of $T^2 \times [0, 1]$ and $T^2 \times S^1$
Are the mapping class groups of $T^2 \times [0, 1$] and $T^2 \times S^1$ explicitly known?
Do your diffeomorphisms and isotopies fix the boundary pointwise or no? For the latter the answer is $GL_3 \Bbb Z$, due to Waldhausen, who gives the answer for all Haken manifolds.
Yes, for the former case, I am thinking about the diffeomorphisms which fix the boundary point-wise.
Then the mapping class group is $\Bbb Z^2$, generated by Dehn twists around each factor of the $T^2$ separately. This all requires a little bit of work but not too much from Waldhausen's result. In fact, you can compute using a result of Ivanov here the homotopy type of the entire diffeomorphism group; $GL_3(\Bbb Z) \rtimes T^3$ for the 3-torus, just $\Bbb Z^2$ for $T^2 \times I$ rel boundary, and $\Bbb Z/2 \times (GL_2(\Bbb Z) \rtimes T^2)$ for $T^2 \times I$.
Thanks a lot! What does the T^3 stand for in $GL_3(\mathbb Z) \rtimes T^3$? I think I might be misunderstanding the notation but you said that the mapping class group for the 3-torus is just $GL_3(\mathbb Z) $. Also, by ``rel boundary'' do you mean in the case in which you fix diffs on the boundary pointwise?
$T^3$ is the group given by the 3-torus, aka $\Bbb R^3/\Bbb Z^3$, and the action of $GL_3(\Bbb Z)$ descends from the action on $\Bbb R^3$. When you take connected components (the mapping class group) you are just left with $GL_3(\Bbb Z)$. I just thought I would mention these slightly stronger results as well which include information on the higher homotopy groups of Diff in addition to the mapping class group $\pi_0$ Diff, because they are accessible thanks to Ivanov.
Thanks! Just for clarification regarding the latter case, do you indeed mean that ``rel boundary'' is the case in which you fix diffs on the boundary pointwise and the $\mathbb Z/2 \times (GL_2(\mathbb Z) \rtimes T^2)$ when you don't fix the boundary pointwise?
Yes, sorry for not clarifying!
|
2025-03-21T14:48:31.190093
| 2020-06-08T23:05:04 |
362554
|
{
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"Alessio",
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|
Stack Exchange
|
Algebraic vector bundles on the punctured spectrum: an exact reference for a result
Let $(R, \mathfrak m)$ be a Noetherian local ring of depth at least $2$. Let $X=Spec(R)$ denote the affine- scheme with structure sheaf $\mathcal O_X$ and $U=Spec(R)\setminus \{\mathfrak m\}$ be the punctured spectrum and write $\mathcal O_U=\mathcal O_X|_U$. Then it is known that $\Gamma_U(\mathcal O_U)\cong R$. Let $\mathfrak Vect(U)$ be the category of Algebraic vector bundles on $U$ and $\mathcal C$ denote the category of finitely generated reflexive $R$-modules that are locally free on the punctured spectrum. Let $F: \mathcal C \to \mathfrak Vect(U)$ be the functor which sends a module $M$ to $\tilde M |_U$ (here $\tilde M$ is the sheaf defined by $M$ on $X$) .
From Horrock's famous paper https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s3-14.4.689 , I think it does follow that $\Gamma_U :\mathfrak Vect(U) \to \mathcal C$ is an equivalence of categories and in fact we have an isomorphism of functors $\Gamma_U \circ F \cong Id_{\mathcal C}$ and $F \circ \Gamma_U \cong Id_{\mathfrak Vect(U)}$. Is this indeed true ? And if it is, what is a precise explicit reference for this (because the paper of Horrocks does not explicitly state this result) ?
One modern account can be found in this thesis of Majidi-Zolbani, especially Chapter 1 and Appendix A. The point is that if $E$ is a vector bundle on $U$, then the global sections $\Gamma_U(E)$ is a finite $R$ module that is locally free on $U$. Sheafifying gets you back $E$. To prove that this induced a true equivalence between categories may require a bit more details, but for many practical purposes this plus some facts about local cohomology is enough. For example you can deduce Horrocks splitting criterion.
Nice reference! A compact version of the equivalence can be found also in Section 3 of the following paper by Burban and Drozd: https://arxiv.org/pdf/0803.0117.pdf
|
2025-03-21T14:48:31.190246
| 2020-06-08T23:16:33 |
362555
|
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|
Stack Exchange
|
Poisson reduction in odd/graded Poisson geometry?
I would like to know whether there is any literature on Poisson reduction of $\mathbb Z$- or $\mathbb Z_2$-graded Poisson algebras.
A $\mathbb Z$-graded Poisson algebra with degree $p\in\mathbb Z$ Poisson structure $(-,-)$ is a $\mathbb Z$-graded algebra $P$ so that $(-,-):P\times P\to P$ is a degree $p$ linear map, on top of appropriately graded versions of antisymmetry and Jacobi identities etc. These arise in e.g. the Batalin-Vilkovisky formalism (for $p=+1$) in mathematical physics.
By Poisson reduction I mean the following, in the non-graded case, which is in some sense dual to Marsden-Weinstein symplectic reduction, when the Poisson structure arises from a symplectic structure on a manifold. (Summary to follow taken from this n-Cat cafe post, thanks Urs.) If $I$ is an algebra ideal in $P$ with
$$
(I,I)\subseteq I,
$$
the quotient algebra $P/I$ does not inherit a Poisson structure, but the smaller algebra
$$
(P/I)^I
$$
of elements $p\in P/I$ with $(p,I)=0$ does:
$$
(p+I,q+I)=(p,q)+(p,I)+(I,q)+(I,I)=(p,q)+I.
$$
Evidently this argument generalises to the graded case (and it's certainly implicit in the physics literature) but I have not been able to find any (explicit) discussions.
Graded symplectic reduction is treated in this paper by Mehta! (https://arxiv.org/abs/1009.1280.) I might write an answer when I've processed it
|
2025-03-21T14:48:31.190355
| 2020-06-08T23:28:22 |
362556
|
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|
Stack Exchange
|
The number of uniformly finite subgroups of some Lie groups
Let $G$ be the group $SO(n)$, $SU(n)$ or $Sp(n)$.
Let $F_m$ be the collection of finite subgroups of $G$ such that its order is bounded by $m$. Two elements in $F_m$ are identified if they are conjugate in $G$.
Can we prove that $\#F_m \le C(n,m)$ for some finite number $C(n,m)$?
I am sure this was asked earlier; indeed, there are only finitely many finite subgroups of the given order in each connected Lie group up to conjugation.
It amounts to showing that for every finite group $F$ there are finitely many conjugacy classes of homomorphisms $F\to G$. This is a standard fact, at least in the OP's setting. For instance this is explicit in [D.H. Lee, T.S. Wu. On conjugacy of homomorphisms of topological groups. Illinois J. Math. 13 1969 694-699], but already follows from Weil's rigidity (1964), and was possibly known earlier. (For the general case: use conjugacy of maximal compact subgroups —due to Iwasawa— to reduce to the compact case.)
|
2025-03-21T14:48:31.190466
| 2020-06-08T23:33:48 |
362560
|
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|
Stack Exchange
|
Existence of algebraic integer with absolute value equal to reciprocal of maximum of $1$ and absolute value of a given algebraic number
Consider a number field $K$, and let $v_1, \cdots v_n$ ($n \in \mathbb N$) be some finite (i.e. non-archimedean) places of $K$. Is the following true?
For every $\alpha \in K^\times$ there exists $\beta \in \mathcal{O}_K$ for which $\alpha\beta \in \mathcal{O}_K$ and
$$|\beta|_{v_j} = \frac{1}{\max \{1, |\alpha|_{v_j} \}} \text{ for every }j \in \{1, \cdots ,n\} \hspace{10mm} \cdots (1)$$
I could see that this was immediate if $\mathcal{O}_K$ is a UFD (or equivalently a PID) for in that case, I could simply write $\alpha$ uniquely as $\alpha_1/\alpha_2$ where $\alpha_1$ and $\alpha_2$ are algebraic integers sharing no common prime factor and let $\beta:=\alpha_2$. However, I couldn't verify this in the general case. Two things I tried are the following:
I tried to naturally generalize the above approach for UFD's, by factoring the principal fractional ideal $\alpha \mathcal O_K$ uniquely into prime ideals and consider the "denominator" ideal of $\alpha \mathcal O_K$ (that is, if $\alpha \mathcal O_K = \prod_{i=1}^k \mathfrak{p}_i^{a_i} \prod_{j=1}^l \mathfrak{q}_j^{-b_j}$ where $\mathfrak{p}_i$ and $\mathfrak{q}_j$ are all distinct prime ideals and $a_i, b_j \in \mathbb N$ for all $i \in [k], j \in [l]$, then the ideal I'm talking about is $\mathfrak{a} := \prod_{j=1}^l \mathfrak{q}_j^{b_j}$). This need not be a principal ideal but I could raise it to the power of its order in the ideal class group. However that would disturb the exponents, violating my requirement (1). Not sure if there's a work-around.....
I tried using the Strong Approximation Theorem, in an attempt to obtain $\beta$ so as to make the $v$-adic absolute values of the difference $\beta-\alpha^{-1}$ sufficiently small for $v \in \{v_1, \cdots v_n\}$ (so the set of absolute values $w$ for which I'm trying to make $|\beta-\alpha^{-1}|_w$ sufficiently small are supersets of $\{v_1, \cdots v_n\}$), but that hasn't worked out so far.....
I haven't had any luck in finding a counterexample either. I would really appreciate any help, and would also like to know if there is any similar result along these lines.
Edit 1: Another thing I tried along the lines of Approach 1 was to write $\alpha$ as $\beta / \gamma$ (where $\beta$ and $\gamma$ are algebraic integers) and compare the aforementioned prime factorization of $\alpha \mathcal{O}_K$ with those of $\beta \mathcal{O}_K$ and $\gamma \mathcal{O}_K$. What I obtained (after some careful exponent comparison) was the following:
$$\beta\mathcal{O}_K = \big(\prod_{i=1}^k \mathfrak{p}_i^{\alpha_i} \big) \mathfrak{a_1}\mathfrak{a_2}\mathfrak{a_3}$$
$$\gamma \mathcal{O}_K = \big( \prod_{j=1}^l \mathfrak{q}_j^{b_j} \big) \mathfrak{a_1}\mathfrak{a_2}\mathfrak{a_3}$$
where $\mathfrak{a_1}$ is made up of prime factors among the $\mathfrak{p}_i$'s, $\mathfrak{a_2}$ is made up of prime factors among the $\mathfrak{q}_j$'s and $\mathfrak{a_3}$ is made up of primes not in the set $\{\mathfrak{p}_i : 1 \leq i \leq k\} \cup \{\mathfrak{q}_j : 1 \leq j \leq l\}$. Of course, one or more of the $\mathfrak{a}_i$ could be trivial (i.e. the unit ideal) but I don't think that's necessary.
And there seems to lie the source of my problem in Approach 1 - I couldn't seem to get rid of the common prime factors between $\beta\mathcal{O}_K$ and $ \gamma \mathcal{O}_K$, in order to be able to generalize the PID approach. Anyway I don't expect that the "numerator" and "denominator" ideals $\prod_{i=1}^k \mathfrak{p}_i^{a_i}$ and $\prod_{j=1}^l \mathfrak{q}_j^{b_j}$ of $\alpha \mathcal{O}_K$ to be principal, if they did, then I should've been able to obtain algebraic integers $\beta$ and $\gamma$ for which $\alpha = \beta / \gamma$ and the principal ideals $\beta\mathcal{O}_K$ and $\gamma \mathcal{O}_K$ would factor as $\prod_{i=1}^k \mathfrak{p}_i^{a_i}$ and $\prod_{j=1}^l \mathfrak{q}_j^{b_j}$ respectively, and life would've been a lot easier.
Edit 2 (More details on Approach 2): As asked in a comment by @Arno Fehm, here are some more details on my second approach. We know that the number of places $w$ of $K$ for which $|\alpha|_w>1$ or $|\alpha|_w<1$ are both finite. As such, I can fix some $\epsilon \in (0, \min\{1, |\alpha|_w^{-1} : w \in N_K\})$ (where I use $N_K$ to denote the set of non-archimedean places of $K$), and then use SAT to obtain a $\beta \in K$ such that $|\beta - \alpha^{-1}|_w < \epsilon$ for all $w \in S:= \{v_1, \cdots , v_n\} \cup \{w \in N_K: |\alpha|_w>1\}$ and $|\beta|_w \leq 1$ for all the other (remaining) non-archimedean places $w$ of $K$. This ensures that $|\alpha\beta|_w, |\beta|_w \leq 1$ for all the places $w \in N_K \setminus S$, whereas for places $w \in S$, I have
$$|\beta - \alpha^{-1}|_w < \epsilon < \min\{1, |\alpha|_w^{-1}\} = \frac{1}{\max\{1, |\alpha|_w\}} \hspace{2mm} \cdots (2)$$
Now for the places $w \in S$ for which $|\alpha|_w \geq 1$, I could show by means of the ultrametric inequality that (2) forces
$$|\beta|_w = \frac{1}{|\alpha|_w} = \frac{1}{\max\{1, |\alpha|_w\}} \leq 1$$
Problems start occurring for those places $w \in S$ for which $|\alpha|_w<1$ (so $w$ has to be one of $v_1, \cdots , v_n$). In this case, (2) yields $|\beta - \alpha^{-1}|_w<1$ which in fact again forces $|\beta|_w = |\alpha|_w^{-1}>1$, for otherwise the ultrametric inequality leads to the following contradiction
$$1>|\beta - \alpha^{-1}|_w = \max\{|\beta|_w, |\alpha^{-1}|_w\} \geq |\alpha^{-1}|_w = |\alpha|_w^{-1} > 1$$
This means that $\beta$ in fact cannot be an algebraic integer if $|\alpha|_{v_j}<1$ for one of the $j \in [n]$ (not to mention that (1) clearly fails for such $j$ as well).
If this approach looks promising, I would really like to know how I should choose my parameters $\epsilon$, $S$ etc. to make it work.
I guess it doesn't work to take $\beta$ to be the smallest positive integer $n$ such that $n\alpha$ is an algebraic integer?
I would think that your approach 2. with strong approximation should work. Why didn't it work out?
@Gerry Myerson Thanks for the suggestion. Unfortunately, I've already tried writing $\alpha$ as the ratio of two algebraic integers and comparing the factorizations of all the principal ideals involved (I have added details in Edit 1 above). Even if I use the stronger result and restrict $\gamma$ to be a natural number $n$, I don't see a way of going around looking at the factorization of $n \mathcal{O}_K$. I would really like to know if there's an alternative approach besides looking at factorizations or if there's something that I am missing.
@Arno Fehm Thanks for your suggestion. I also added my approach using SAT and where it runs into problems. Is there some way of making it work?
Thanks for the details. I might miss something subtle, but as far as I understand you simply want to get $\beta\in\mathcal{O}_K$ with $|\beta|_v=(\max{1,|\alpha|_v})^{-1}$ for all $v$ in the set $S$. So you want $|\beta-\alpha^{-1}|_v$ to be small for $v\in S$ except at those $v$ where $|\alpha|_v<1$, where instead you demand that $|\beta-1|_v$ is small. Strong approximation gives that.
You're right! All I had to do was to instead require that $\beta$ be $w$-adically close to $1$ rather than $\alpha^{-1}$ for those places $w$ in $S$ (i.e. among ${v_1, \cdots v_n}$) for which $|\alpha|_w < 1$ and the same argument works. Thanks a lot!
The answer is already in the comments, but here again for completeness: It indeed follows easily from strong approximation:
Let $S=\{v_1,\dots,v_n\}\cup\{v:|\alpha|_v>1\}$. By the strong approximation theorem one can find $\beta\in\mathcal{O}_K$ that is close to $1$ at those $v\in S$ with $|\alpha|_v<1$ and close to $\alpha^{-1}$ at the other $v\in S$. In particular, $|\beta|_v=1$ for those $v\in S$ with $|\alpha|_v<1$ and $|\beta|_v=|\alpha|_v^{-1}$ at the other $v\in S$.
Thus $|\beta|_v=\max\{1,|\alpha|_v\}^{-1}$ for $v\in S\supseteq \{v_1,\dots,v_n\}$,
and $|\alpha\beta|_v\leq 1$ for all finite $v$, hence $\alpha\beta\in\mathcal{O}_K$.
|
2025-03-21T14:48:31.190941
| 2020-06-09T01:01:23 |
362561
|
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|
Stack Exchange
|
Prerequisites/Preparation for understanding a research paper - global solutions to Einstein field in Bondi Coordinates
I would like to read this paper:
João L. Costa, Filipe C. Mena, Global solutions to the spherically symmetric Einstein-scalar field system with a positive cosmological constant in Bondi coordinates arXiv:2004.07396
and understand it thoroughly but I feel I lack many prerequisites. I have taken a General Relativity course based on Sean Carroll’s “Spacetime and Geometry”, but that’s about it.
I have no experience with reading research papers and this would be my first. I am currently reading “A first course in GR” by Schutz to refresh my GR skills, but I feel it would be inefficient.
I need some help in coming up with a reading list which would at least get me started, and which I would come back to refer to from time to time.
Thanks.
It's not clear to me where you feel your prerequisites are lacking. Is it mathematical background (i.e. differential geometry) or physics background (GR itself, and unfamiliarity with the literature there)?
You would need to learn some PDE, the initial value problem and the characteristic problem in GR.
Thank you @MBN! This gives me a direction.
|
2025-03-21T14:48:31.191072
| 2020-06-09T02:14:27 |
362563
|
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|
Stack Exchange
|
Conditions under which $\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n}$
I was working with some Dirichlet series and I realized that I have never seen any general conditions under which
\begin{equation}
\sum_{n=1}^{\infty}\frac{a_n}{n}=\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}\label{1}\tag{1}
\end{equation}
holds. This is obviously not true in a general case since if so there would be a very simple proof of the PNT from just applying this to $a_n=\mu(n)$. My question is: under what conditions does \eqref{1} hold?
I can show that if $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges then \eqref{1} must hold using a very simple proof, but I can't find any broader statements. The ideal condition that I would like to show is that if the partial sums $\sum_{n=1}^{N}\frac{a_n}{n}$ are bounded then \eqref{1} must hold. I don't know how I would go about proving this though, and any insights on this general area would be greatly appreciated.
If you take $a_n=\frac{\Lambda(n)}{n^{it}}$ for a fixed $t \ne 0$, RHS converges to $-\zeta'(1+it)/\zeta(1+it)$ but LHS oscillates finitely (as $LHS_N-RHS$ ~$-N^{-it}/{it}$) so the partial sums are bounded but the series doesn't converge; for real $a_n$ Hardy-Littlewood proved that $a_n \ge -C/\log n$ is a sufficient condition for 1 above to hold - conditions like this are part of Tauberian theory for which there is a standard reference by J Korevaar - https://www.springer.com/gp/book/9783540210580 where you will find lots of stuff about this
Please use a high-level tag like "nt.number-theory". I added this tag now. Regarding high-level tags, see https://meta.mathoverflow.net/q/1075/
Here are analogous Tauberian theorems for power series and Dirichlet series that involve a condition of analytic continuation to a boundary point plus one extra condition that is necessary for the series to converge at some point on that boundary.
Power series: If $c_n \to 0$ then $\sum_{n \geq 0} c_nz^n$ converges for $|z| < 1$. Fatou showed this series converges at each number $z$ with $|z| = 1$ to which the series admits an analytic continuation from inside the disc $\{z : |z| < 1\}$. Note that "$c_n \to 0"$ is a necessary condition for $\sum c_nz^n$ to converge at some number $z$ where $|z| = 1$, so including it as a hypothesis is reasonable.
Dirichlet series: If $(a_1 + \cdots + a_n)/n \to 0$ then $\sum a_n/n^s$ converges for ${\rm Re}(s) > 1$. Riesz showed this series converges at each $s$ with ${\rm Re}(s) = 1$ to which the series admits an analytic continuation from the half-plane ${\rm Re}(s) > 1$. The condition $(a_1 + \cdots + a_n)/n \to 0$ is necessary for $\sum a_n/n^s$ to converge at some number $s$ where ${\rm Re}(s) = 1$, so including that as a hypothesis is reasonable.
For your motivating example with $a_n = \mu(n)$, the hypothesis $(a_1 + \cdots + a_n)/n \to 0$ is known by completely elementary methods to be equivalent to the Prime Number Theorem (in the form "$M(x) = o(x)$" for $M(x) = \sum_{m \leq x} \mu(m)$), so this theorem of Riesz isn't really a good approach to proving the Prime Number Theorem even though the analytic continuation hypothesis for $\sum \mu(n)/n^s = 1/\zeta(s)$ from ${\rm Re}(s) > 1$ to ${\rm Re}(s) = 1$ is valid.
I'm not sure if the answer to this is known, but would it be sufficient to show that $\liminf_{n\to\infty}(a_1+...+a_n)/n\to0$ for the interchange to be valid?
I don't know, but I think it is unrealistic to think you're going to find a very simple proof of the Prime Number Theorem this way. Even if a simple-sounding condition is sufficient, surely the proof of that is going to involve serious work or the verification that an example actually fits the condition is going to involve serious work.
Actually, the main example I have in mind is not proving the prime number theorem (the methods I am using have already led me to a new all-elementary proof of the PNT). I am actually trying to prove a different conjecture of mine which would follow nicely from a simple condition to let me interchange limit. If you want to discuss this further I would love to, and we can move this conversation to a private chat.
If you are genuinely sure you have a new all-elementary proof of PNT, try to reach out to one of the number theorists near you (well, could be hard to meet "in person" these days). All conditions you have wished to be true here so far are not true, so you had better be really sure your proof of PNT does not involve a subtle misunderstanding of what is legitimate with limit processes. In particular, simple conditions that justify interchanging of limits could themselves have proofs as hard as that of PNT! See Korevaar's book on Tauberian theorems for a survey of what is known in this area.
I made sure that all interchanges of limits were very justified, and I used no high-level theorems. It is not a full proof of the prime number theorem since the approach taken was showing that the average order of the Mobius function is zero but I only showed that if it must exist it must be zero without showing that it exists (I actually prove something stronger but that's the gist of it). Do you wish to discuss this further? I do not know who to reach out to about it.
Oh! That is a huge difference from PNT itself. There are numerous summation or mean value calculations where the only possible value they could have is 0 and the fact that that they do have those values has long been known to be equivalent to PNT (proof of equivalence relying on methods much more basic that what goes into all known proofs of PNT), which just emphasizes why the actual existence of those limits is really, really hard. Analytic number theory books discuss this kind of stuff, e.g., Theorems 4.14 and 4.15 in Apostol's "Introduction to Analytic Number Theory".
Oh, okay. The specific statement of what I prove is that $\lim_{x\to1^-}(1-x)\sum_{n=1}^{\infty}\mu(n)x^n=0$, which is stronger but probably does not imply the PNT.
Let us continue this discussion in chat.
It does imply PNT. One version of the Hardy-Littlewood Tauberian theorem for power series says that if $\sum_{n \geq 0} a_nx^n$ converges for $|x| < 1$, $(1-x)\sum_{n \geq 0} a_nx^n \to A$ as $x \to 1^{-}$, and (Tauberian condition) $a_n \geq -C$ for some $C > 0$ then $(a_0 + \cdots + a_n)/n \to A$ as $n \to \infty$. Taking $a_n = \mu(n)$, we get that if $(1-x)\sum \mu(n)x^n \to A$ as $x \to 1^{-}$ then $(\mu(1) + \cdots + \mu(n))/n \to A$ as $n \to \infty$. The only possible limit for that is $0$, and it's equivalent to PNT. The proof of the HL Tauberian theorem is not easy at all.
In D. J. Newman's paper
A simple analytic proof of the prime number theorem
published in The American Mathematical Monthly
87 (1980) 693-696, Newman proved a result of this type which may also be useful to you (this is a minor variant of the criterion given by KConrad in his answer).
It says the following:
Let $|a_n| \leq 1$ and suppose that the Dirichlet series
$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$
admits a holomorphic continuation to the line $\mathrm{re}\, s = 1$. Then
$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$
converges for all $\mathrm{re}\, s \geq 1$.
actually assuming that the Dirichlet series converges for $\Re s >1$, it is enough to have only one-sided boundness $a_n \ge -C$ in the result above (note that $|a_n| \le C$ implies the convergence of the Dirichlet series in $\Re s >1$, but one-sided boundness doesn't necessarily, so we need this extra hypothesis - the issue though in this and @KConrad above is that one needs strong analytic information about the function on the whole boundary line, while the results of Littlewood-Hardy (and there are other sufficient conditions too) are just coefficient results, no analytic assumptions
@Conrad the analytic information for the Fataou and Riesz theorems in my answer doed not need analytic information to the whole boundary (unit circle or ${\rm Re}(s) = 1$) but just to a specific point of interest. For example, consider the series $\sum z^n/n$, which has coefficients tending to $0$ and has an analytic continuation to each point on the unit circle except $z = 1$, since we know about the function $-\log(1-z)$. At each particular point on the unit circle other than $z = 1$, the Fatou theorem implies convergence of the series to $-\log(1-z)$. You don't need to know (contd.)
analytic continuation of the function inside the unit disc to all of the unit circle minus ${1}$ to be able to say at a particular number $z$ on the unit circle, like $z = i$, that the series converges there to $-\log(1-z)$. By partial summation it's easy to see the series converges at each $z$ with $|z| = 1$ other than $z = 1$, but the partial summation does not make it evidence that the series is converging to $-\log(1-z)$.
@KConrad - you are right and I stand corrected from the point of view of the full boundary, though the condition of analytic continuation even at a point is quite strong; Fatou's Theorem is indeed optimal as noted in your post since $a_n \to 0$ is obviously necessary while being sufficient under analytic continuation; the Hardy-Littlewood Tauberian theorems have stronger conditions on the coefficients (and just about applies to the $\log(1-z)$ example as $n|e^{int}/n| =1$) but require only radial convergence; same for the Dirichlet series and Riesz vs H-L,
@Conrad what are a few Dirichlet series of practical interest to which the H-L Tauberian theorem applies and the Dirichlet series does not admit analytic continuation to a boundary point of interest?
Sorry to disturb an old question, but I noticed that the other answers, having a number theory focus, only give a rather complicated condition for convergence of the series in the $s\to 1^+$ limit (existence of an analytic continuation). Maybe this is enough in the context of Dirichlet series (I don't know the topic well enough), but it leaves the original question unanswered as written.
Here's a straightforward direct condition that allows one to interchange $\sum_n$ and $\lim_{s \to 1^+}$ (or any other limit). It is obviously sufficient if the sequences $(a_n(s) := a_n/n^s)_{n=1}^\infty$ converge in the topology of the $\ell^1$ norm (absolute value sum). Vitali's convergence theorem is an iff condition for convergence in general $L^p$ spaces. Specializing to $\ell^1$ ($p=1$, counting measure on $\mathbb{N}$), there are exactly two conditions: (a) pointwise convergence $a_n(s) \to a_n(1)$ (already implicit in the question) and (b) uniform smallness of the summation tails (for given $\varepsilon > 0$, there is a sufficiently large $N$, so that for all $n\ge N$ the bound $\sum_{n=N}^\infty |a_n/n^s| < \varepsilon$ is uniform as $s\to 1^+$).
|
2025-03-21T14:48:31.191725
| 2020-06-09T03:09:53 |
362567
|
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|
Stack Exchange
|
Vanishing sequence and subsequence with particular decay
Assume I have a sequence $\{a_m\}$ that is vanishing and strictly positive:
$$
0<a_{m+1}\leq a_m\leq\ldots\leq a_1<\infty, \quad \lim_{m\to \infty}a_m = 0
$$
Is it true or false that this has a subsequence $\{a_{m_n}\}$ such that
$$
a_{m_n}\asymp n^{-1}
$$
which is to say, there exists a constant $C>0$ such that for all $n$:
$$
\frac{1}{Cn} \leq a_{m_n} \leq \frac{C}{n}
$$
Isn't this false for any sequence that goes to zero "quickly", e.g., $1/n^2$, or $1/2^n$?
No, if $a_m=1/m^2$, take $m(n) = \lceil \sqrt{n}\rceil$
What would the subsequence be for $1/2^n$?
So you're allowing, say, $m(10)=m(11)=m(12)=\cdots=m(16)$? A subsequence can contain repeated values, even if those values aren't repeated in the source sequence?
Yes. The subsequence can have repeated values.
For $1/2^n$, take $m(n) = \lceil \log_2 n\rceil$. In fact, if we can write $a_m=e^{-\phi(m)}$ and $\phi$ is strictly increasing (i.e., invertible), then $m(n) = \lceil \phi^{-1}(\log n)\rceil$ is satisfactory.
The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have
$$a_{k+1}<\frac1{n_k}<a_k$$
for all $k$.
So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|=\ln(k+1)$. Similarly, for all $m\ge k+1$ we have
$1>n_k a_{k+1}\ge n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_{k+1})|=\ln(k+1)$. So, $|\ln(n_k a_m)|\ge\ln(k+1)$
for all $k,m$.
So, for any $(m_n)$
$$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\ln(k+1)\to\infty$$
as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.
|
2025-03-21T14:48:31.191978
| 2020-06-09T04:10:46 |
362570
|
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|
Stack Exchange
|
Differentiable dependence on the initial condition of the solution of a SDE
Let
$b,\sigma:\mathbb R\to\mathbb R$ be differentiable and Lipschitz continuous
$(\Omega,\mathcal A,\operatorname P)$ be a probability space
$(\mathcal F_t)_{t\ge0}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
$(W_t)_{t\ge0}$ be an $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
$(X^x_t)_{(t,\:x)\in[0,\:\infty)\times\mathbb R}$ be a real-valued continuous $\mathcal F$-adapted process on $(\Omega,\mathcal A,\operatorname P)$ with $$X^x_t=x+\int_0^tb(X^x_s)\:{\rm d}s+\int_0^t\sigma(X^x_s)\:{\rm d}W_s\tag1$$ for all $t\ge0$ almost surely for all $x\in E$
Are we able to show that $\mathbb R\ni x\mapsto X^x_t(\omega)$ is differentiable for all $(\omega,t)\in(\Omega\setminus N)\times[0,\infty)$ for some $\operatorname P$-null set $N$?
Note that the hypotheses are sufficient to guarantee the existence of the process $X$ above. However, I'm not sure if they are sufficient to guarantee the existence of a real-valued continuous $\mathcal F$-adapted process $(Y^x_t)_{(t,\:x)\in[0,\:\infty)\times\mathbb R}$ with $$Y^x_t=\int_0^tb'(X^x_s)Y^x_s\:{\rm d}s+\int_0^t\sigma'(X^x_s)Y^x_s\:{\rm d}W_s\tag2$$ for all $t\ge0$ almost surely for all $x\in\mathbb R$ as well. However, if this process $Y$ exists, then it's clearly a candidate for the desired process obtained by differentiating $X$ with respect to the spatial parameter.
Maybe it's useful to remember how the existence of $X$ is proved: For fixed $T>0$ and $\xi\in L^2(\operatorname P)$, we know that the map $$\Xi_T(Z):=\xi+\left(\int_0^tb(s,Z_s)\:{\rm d}s\right)_{t\in[0,\:T]}+\left(\int_0^t\sigma(s,Z_s)\:{\rm d}W_s\right)_{t\in[0,\:T]}\in\mathcal B_T$$ for $Z\in\mathcal B_T$, where $$\mathcal C_T:=\left\{Z:Z\text{ is a continuous }(\mathcal F_t)_{t\in[0,\:T]}\text{-adapted process on }(\Omega,\mathcal A,\operatorname P)\right\}$$ is equipped with $$\left\|Z\right\|_T:=\left\|\sup_{t\in[0,\:T]}Z_t\right\|_{L^2(\operatorname P)}\;\;\;\text{for }Z\in\mathcal C_T$$ and $$\mathcal B_T:=\left\{Z\in\mathcal C_T:\left\|Z\right\|_T<\infty\right\},$$ has a unique fixed point and there is a constant $c\ge0$, depending only on $b$ and $\sigma$, with $$\left\|\Xi_T(Z)\right\|_T^2\le3\left(\operatorname E\left[|\xi|^2\right]+c(T+4)\left(T+\int_0^T\left\|Z\right\|_t^2\:{\rm d}t\right)\right)\tag3$$ for all $Z\in\mathcal B_T$.
Maybe this can be used to show that $$\left\|\frac{X^{x+h}-X^x}h-Y^x\right\|_T\le c_1|h|\;\;\;\text{for all }T>0\text{ and }h\in\mathbb R\tag4$$ for some $c_1\ge0$, depending only on $b$ and $\sigma$, which should even allow for a stronger conclusion then the one asked for in the question.
Have you looked at Kunita's st flour lecture notes on flows of diffeomorphism?
@oferzeitouni Yes, sure. But Kunita is hard to read and I think the presented theory is overkill for this rather simple scenario.
I looked at Kunita's lecture and the subject does not seem addressed. Indeed Kunita is hard to read. To further your question, I was wondering if a kind of Taylor's expansion can be used like : $X_t^{x_1}(\omega) = X_t^{0}(\omega) + (x_1 - 0) \frac{\partial X_t^x}{\partial x}(x_1)$
As mentioned in the comments, Kunita's lectures cover this eg. Lectures on
Stochastic Flows And Applications, i.e. depending on the regularity of the coefficients we have analogous differentiability in initial conditions
Another source is in Varadhan's notes too Stochastic1
As you mentioned, indeed the proof goes through creating new systems satisfied by the derivative processes and then showing existence-uniqueness for the entire system.
|
2025-03-21T14:48:31.192250
| 2020-06-09T08:21:27 |
362578
|
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|
Stack Exchange
|
A variation of the law of large numbers for random points in a square
I uniformly mark $n^2$ points in $[0,1]^2$. Then I want to draw $cn$ vertical lines and $cn$ horizontal lines such that in each small rectangle there is at most one marked point. Surely, for a given constant c it is not always possible.
But it seems that for $c=100$ when n tends to infinity, the probability that such a cut exists should tend to one, as a variation of the law of the large numbers.
Do you have any idea how to prove this rigorously?
What's you heuristics for the suggestion that the probability goes to one?
As a side comment, there are results describing the asymptotics of the side of the largest empty square with sides parallel to those of [0,1]^2. A version of rectangles instead of squares also exist, I think. The proof is, however, is not a LLN-type one.
@Vysotsky not a heuristics, but a somewhat weaker statements can be observed experimentally and it would follow from the positive answer to my question.
The statement is that the degree of a tropical curve through n^2 random points in a square is concentrated near n.
@Vysotsky note also that we cut AFTER we marked points randomly, so there is more freedom.
@Nikita Kalinin Of course, this is the main point (and difficulty)! Otherwise the claim will not work, as per the answer attempts below. It is unclear to me why your problem should in a sense correspond to some law of large numbers.
@Nikita Kalinin It is a nice problem! But somehow I share Iosif Pinelis's scepticism that cn would not suffice. You do have a reasonably quick algorithm how to compute the cuts for a given realisation of points, do not you?
Given $n^2$ i.i.d. uniform points in $[0,1]^2$, the goal is to draw a configuration of $cn$ vertical lines and $cn$ horizontal lines such that in each small rectangle there is at most one marked point.
We show below that $c$ must satisfy $c=\Omega(n^{1/3})$ for this to be typically possible:
In fact, $\Theta(n^{4/3})$ lines are necessary and sufficient for such a configuration to exist with substantial probability.
More precisely, denote by $p_n(k)$ the probability that a configuration of $k$ vertical lines and $k$ horizontal lines separate $n^2$ i.i.d. uniform points $\{x_j\}_{j=1}^{n^2}$ in $[0,1]^2$.
Claim: For suitable constants $0<c_1<c_2<\infty$, we have (omitting integer part symbols):
(a) $\; p_n(c_1 n^{4/3}) \to 0$ as $n \to \infty$, and
(b) $\; p_n(c_2 n^{4/3}) \to 1$ as $n \to \infty$.
This is proved below with $c_1=1/20$ and $c_2=3/2$; no attempt has been made to optimize these constants.
Proof: Consider an auxiliary grid of $L:=n^{4/3}$ uniformly spaced vertical lines and
$L$ uniformly spaced horizontal lines in the unit square. This grid defines $L^2$ grid squares of side length $1/L$.
(a) Call a grid square $Q$ nice if it contains exactly two of the $n^2$ given points $\{x_j\}$. Observe that for two distinct grid squares, the events that they are nice are negatively correlated. Call a nice grid square $Q$ good if there is at most one other nice square in its row and at most one other nice square in its column. The probability that a specific grid square $Q$ is nice is
$${n^2 \choose 2}L^{-4}(1-L^{-2})^{n^2-2}=(1/2+o(1))L^{-1}.$$
Given that $Q$ is nice, The conditional expectation of the number of nice squares (other than $Q$) in the row of $Q$ is $1/2+o(1)$ Thus, given that $Q$ is nice, Markov's inequality implies that the conditional probability that there are two or more additional nice squares in the row of $Q$ (besides $Q$ itself) is at most $1/4+o(1)$.
The same applies to the column of $Q$, and we deduce that
$$P(Q \; {\rm is \; good}\; | Q \; {\rm is \; nice}) \ge 1/2+o(1) \, ,$$ so
$$P(Q \; {\rm is \; good} ) \ge (1/4+o(1))L^{-1} \, .$$
Let $G$ denote the number of good grid squares. Then the mean satisfies
$$E(G) \ge (1/4+o(1))L \,.$$
Observe that if we replace one point $x_i$ by $x_i'$ then $G$ will change by at most 5, so Mcdiarmid's inequality, see [1, Theorem 3.1] or [2], implies that for $n$ large enough,
$$P(G \le L/5) \le \exp(-\frac{(L/21)^2}{25n^2}) \to 0 \,. {\rm as} \; n \to \infty \,.$$
(Alternatively, one could invoke the Efron-Stein inequality or estimate the variance directly to verify this.)
Now suppose that $S$ is a set of vertical and horizontal lines that separate the points
$\{x_j\}_{j=1}^{n^2}$.
For each good grid square $Q$, a line of $S$ is required to separate the two points $x_i, x_j$ in the square, and each such line can be used for at most two good squares. Thus $|S| \ge G/2$ so
$$p_n(L/20) \le P(\exists \; {\rm separating } \; S \; {\rm with } \; |S| \le L/10) \le P(G \le L/5) \to 0
$$.
(b) Denote by $M$ the number of pairs $(i,j)$ such that $1 \le i<j \le n^2$ and $x_i,x_j$ fall in the same grid square. Then
$E(M) = {n^2 \choose 2}L^{-2} \le L/2$, and another application of McDirarmid's inequality implies that $P(M \ge L) \to 0$ as $n \to \infty$.
Finally, construct a separating set of lines $S$ by combining the $2L$ lines of the auxiliary grid with one separating line for each pair $(i,j)$ counted in $M$
(we can take half of these lines vertical and half horizontal). Then $P(|S| \ge 3L) \to 1$
as $n \to \infty$ and $p_n(3L/2) \to 1$ as well.
[1] McDiarmid, Colin. "Concentration." In Probabilistic methods for algorithmic discrete mathematics, pp. 195-248. Springer, Berlin, Heidelberg, 1998.http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=8B1FFFE4553B63543AFEA0706E686E65?doi=<IP_ADDRESS>.5794&rep=rep1&type=pdf
[2] McDiarmid, C. (1989). "On the method of bounded differences". Surveys in Combinatorics. London Math. Soc. Lectures Notes 141. Cambridge: Cambridge Univ. Press. pp. 148–188. MR 1036755
Do you think the constant $k$ for which $p_n(k·n^{4/3}) → r ∈ (0,1)$ as $n→∞$ is rational or irrational? And what do you think it is?
Very nice proof. I initially read the question wrong, and thought the grid was assumed uniform. But rather it can be actively chosen. A couple of points that I had to struggle with: the first estimate can be seen by realizing the binomial factor is 1 + o(1). For the constant 5, note removing a point can promote at most 4 grids to being good (2 on the same row and 2 on the same column), and adding a point can promote at most 1 grid, namely the grid where the point lands. Lastly for part b, two points are not on the same x or y cord with probably 1. How likely is it to get sharp estimate for c?
A similar reasoning with the bound of $n^{4/3}$ also appears in the paper 'On Separating Points by Lines' by Har-Peled and Jones
Interestingly, if we allow the lines to have arbitrary directions, it still requires roughly n^{4/3} (up to a log correction) lines to separate all the points.
https://www.cambridge.org/core/journals/proceedings-of-the-london-mathematical-society/article/economical-covers-with-geometric-applications/486374A93F4351DF26C155F6C3FE35AE
That is beautiful. Thanks for the reference!
Label your $N$ points as $(x_i,y_{\sigma(i)})$ with $x_1 < \cdots < x_N$ and $y_1 < \cdots < y_N$ ; this defines a uniform random permutation $\sigma \in \mathfrak{S}_N$, and all the information about the problem is encoded in $\sigma$.
Let $C$ be the number of axis-parallel cuts needed to scatter all the points. An easy lower bound is $C \geq L-1$, where $L$ is the length of the longest monotone subsequence of $\sigma$. It is well known that $L \sim 2 \sqrt{N}$ in probability, so if we could show that the above lower bound is typically sharp, this would solve the problem.
Interesting! Could you comment on how to find the cuts based on the permutation?
This is an interesting connection. However, I am afraid that the longest monotone subsequence will account only for a rather small subset of the set of $n^2$ uniformly chosen points in the square.
The idea is that maybe some of the techniques used for longest increasing subsequence can apply here. If the answer to the OP is positive, it implies that the longest increasing subsequence is $O(\sqrt{N})$, which is not a trivial fact. Also, I'd like to see a permutation with $C \gg L$.
@S.Surace I don't undestand the question. I don't know how to find the cuts, this is just a lower bound.
I was thinking that since all the information is encoded in $\sigma$, one should be able to construct the cuts from that. But this is a separate issue.
Here is an example with $C \gg L$. Split ${1,\dots,N}$ into $\sqrt{N}$ blocks (=intervals) of size $\sqrt{N}$ and let $\sigma$ be the permutation which reverses the order inside each block. Then $L=\sqrt{N}$ and $C \geq \sqrt{N}(\sqrt{N}-1)$, since a cut can only scatter within one block, and each block requires $\sqrt{N}-1$ cuts to be scattered.
@Guillaume Aubrun This means that the points are placed very close to either of the diagonals, if I got it correctly. However, this is very unlikely.
This is to show rigorously that the uniform rectangular grid does not work -- cf. the answer by mike. As in the answer by Dieter Kadelka, suppose that the $cn$ vertical lines and the $cn$ horizontal lines partition the unit square into $$N:=(cn+1)^2$$ small squares of equal area $1/N$, where $c\ge1$. Let
$$K:=n^2.$$
Then the probability that each small square contains at most one random point is
\begin{aligned}
\frac{N(N-1)\cdots(N-K+1)}{N^K}&=\Big(1-\frac1N\Big)\cdots\Big(1-\frac{K-1}N\Big) \\
&<\exp\Big\{-\sum_{k=1}^{K-1}\frac kN\Big\} \\
&=\exp\Big\{-\frac{(K-1)K}{2N}\Big\}\to0\ne1
\end{aligned}
as $n\to\infty$, as claimed.
(I don't think any choice -- even a random one, depending on the random points -- of $cn$ lines in both the vertical and horizontal directions is enough, for any real $c>0$. I think, instead of $cn$, one needs at least something like $n\ln^a n$ for some real $a>0$.)
I think there remains the possibility that the grid is chosen dependent on the $n^2$ random points, but I would not bet on this.
@DieterKadelka : Yes, the very fast convergence to $0$ (rather than $1$) for the uniform grid strongly suggests that even a $cn\times cn$ grid dependent on the $n^2$ random points will not be enough.
yes, sure, that is why I wrote that we mark points and only then we draw lines...
@Nikita Kalinin: Do you have an algorithm to do this effectively?
@DieterKadelka : If I were going to try to make such an algorithm (which I am not, have no interest in doing, and refuse to even think through enough details to see if it could plausibly be made to work), I would start with a partitioning $k$-d tree having the defect that non-root interior nodes partition the entire space, not just the space associated with their parent. Depth control of that tree might give a useful bound on the number of partitioning lines.
If we draw the horizontal lines to have $p$ rows with $n^2/p$ points each, then finding where to put the vertical lines becomes a birthday problem : we take the points from left to right and find in wich row they belong. When the row is already taken, we must draw a vertical line and start a new column. Set $X_i$ the number of points in column $i$. The repartition function of $X_1$ converges to $\exp(-x^2/p)$.
If we can suitably control the law of the other $X_i$, this would give an order of $n^2/\sqrt{p}$ columns. For $p\sim n^{4/3}$, this means around $n^{4/3}$ columns.
This provides as upper bound an order of $n^{4/3}$ horizontal lines and vertical lines. If some form of concentration occur for the size of the columns, we would get a critical value at $c_{crit}n^{4/3}$, with proba 1 of having a solution for constants larger than $c_{crit}$.
To control $X_i$ (so that the columns are big enough in law), note that while there are still enough points, we can use modified birthday problems to find lower bounds in law on the size of the colums. For the right-most columns, just start again from the right. Set $(Y_i)_{1\leq i\leq N}$ the size of each column (starting from the right). $Y_i$ is distributed as $X_i$, and there is a relation between the columns from the left and from the right (the bounds of a column from the right should be in two adjacent columns from the left, and vice-versa). So, a column is always of order $\sqrt p$.
Nikita Kalinin, you were right, this problem indeed reduces to a rather general form of the law of large numbers, namely subadditive ergodic theorem! This theorem is used, e.g., to prove the result on the length of the largest increasing subsequence mentioned by Guillaume Aubrun (thanks for hinting the direction). So I just mimic its proof due to Hammersley (see, e.g. Example 7.5.2 in Durrett's "Probability: Theory and Examples").
First of all, as noted above, instead of the original setting with $n^2$ points in $[0,1]^2$ we can consider your question for a Poisson point process (PPP) of unit intensity on the positive quadrant restricted to $[0,n]^2$.
For any integer $0 \le m <n$, let $C_{m,n}$ be the minimal positive integer number such that it is possible to cut the rectangle $[[m,n]]:=\{(x,y) \in [0, \infty)^2: m \le x, y \le n\}$ with the points of the PPP using $C_{m,n}$ horizontal and $C_{m,n}$ vertical cuts to satisfy the condition. The key observation is that
$$C_{0,m} + C_{m, n} \le C_{0,n}.$$
Indeed, just use optimal cuts to cut $[[0,m]]$ and $[[m,n]]$; then more cuts may be needed.
By the subadditive ergodic theorem,
$$\lim_{n \to \infty}\frac{C_{0,n}}{n} = \sup_{n \ge 1} \frac{E C_{0,n}}{n}:=c, \qquad a.s.$$
Now the problem is to figure out whether $c$ is finite or not, which is the most interesting bit. I did not succeed so far.
One way to prove finiteness is to show that $E C_{0, n+1} - E C_{0,n}$ is bounded. This appears to be wrong on a first sight, but apparently this is not so trivial. On the other hand, finiteness would mean that cutting $[[0,n]]$ and $[[n, 2n]]$ in a nearly optimal way would automatically fix the two remaining subsquares of $[[0,2n]]$. This should not be true, but not simple to prove as well.
The last comments are that the same argument goes through if replace $C_{m,n}$ by the total number of horizontal and vertical cuts. And the sequence $E C_{0,n}/n$ is strictly increasing.
Not and answer, but a long comment: I don't think a uniform rectangular grid works. Take a planar poisson process with intensity 1 and look at (0,n)x(0,n). This is your setup except I have replaced the uniforms with a poisson process with an expected n^2 events in the space. Divide it into a rectangular grid with (cn)^2 rectangles of area 1/c^2 each. The number of points in each rectangle is poisson with parameter 1/c^2,and different rectangles are independent. You wonder if any have of the (cn)^2 rectangles have two points in them. As the parameter is fixed there is a probability going to 1 that at least one of them does.
It is sufficient to assume that the $cn$ vertical and $cn$ horizontal lines are at $\frac{1}{cn}, \frac{2}{cn+1}, \ldots$, so that we have $N := (cn+1)^2$ small rectangles each with area $p^{(n)} := \frac{1}{N}$. Let $X_1,\ldots,X_N$ be random variables with $X_i$ the random number of points in rectangle $i$. Then $X := (X_1,\ldots,X_N)$ has the multinomial (sometimes called polynomial) distribution $\cal{M}(n; p^{(n)},\ldots,p^{(n)})$ ($N$ parameters $p^{(n)}$). As far as I understand your problem you want an estimation for $\mathbb{P}(X_1 \leq 1, \ldots, X_N \leq 1)$. But $q^{(n)} := 1 - \mathbb{P}(X_1 \leq 1, \ldots, X_N \leq 1) \leq \sum_{i=1}^N \mathbb{P}(X_i \geq 2)$, where each $X_i$ has Binomial distribution $Bin(n,p^{(n)})$. Thus $q^{(n)} \leq N \cdot \left(1 - (1 - p^{(n)})^n - n \cdot p^{(n)} \cdot (1 - p^{(n)})^{n-1}\right)$, which goes to $0$ when $c > 1$.
Edit: The estimation above only works for $n$ points, not for $n^2$ points, as required.
It looks like your bound on $q^{(n)}$ is $\sim1/(2c^2)\ne0$.
Thank you @Iosif Pinelis, you are right. The estimation has to be sharpened (if it is possible).
|
2025-03-21T14:48:31.193393
| 2020-06-09T08:31:57 |
362579
|
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|
Stack Exchange
|
An observation on the Riemann $\xi$ function
Anyone seen these conclusions about the Riemann xi function or see any errors here?
With $\xi(s)$ the entire Landau Riemann xi function
defined by the Hadamard product representation
$$\xi(s) = (1/2) \prod_{\rho_m} [1 - \frac{s}{\rho_m}]$$
where the product ranges over the nontrivial zeros of the Riemann zeta function, let
$$\vartheta(z) = 2\xi(\frac{1}{1-z}) $$
$$= \Xi(\frac{1}{1-z}) $$
$$= \prod_{\rho_m}[1- \frac{1}{\rho_m} \frac{1}{(1-z)}].$$
Let $u = 1/(1-z)$, then $z = (u-1)/u$ and
$$\vartheta(\frac{u-1}{u}) = \Xi(u)$$
$$= \Xi(1-u) = \vartheta(\frac{u}{u-1}),$$
so
$$\vartheta(z) = \vartheta(\frac{1}{z})$$ $$ = \Xi[\frac{1}{1-(1/z)}]= \Xi[\frac{z}{z-1}]$$
$$= \prod_{\rho_k} [1 - \frac{1}{\rho_k} \frac{z}{z-1} ].$$
(Edit June 10: Corrected transcription error. At last moment I changed notation from $\Omega$ in my notes to $\Xi$ but missed the change in this last set of equalities since I read them in my mind as the same Hadamard product. Sorry for the confusion.)
We then have the two equivalent expressions
$$\vartheta(z) = \prod_{\rho_m}[1- \frac{1}{\rho_m} \frac{1}{(1-z)}] $$
$$ = \prod_{\rho_k} [1 - \frac{1}{\rho_k} \frac{z}{z-1} ].$$
The zeros of the last product are given implicitly by
$$1 = \frac{1}{\rho_k} \frac{z_k}{z_k-1},$$
so
$$z_k= \rho_k/(\rho_k-1)$$
and the other product gives a null factor when
$$1 = \frac{1}{\rho_m}\frac{1}{1-z_k},$$
so
$$z_k = (\rho_m-1)/\rho_m.$$
Each zero is paired with its complex conjugate so that both equations are simultaneously satisfied only if $\rho_k = \bar{\rho}_m$ giving
$$\frac{\rho_m-1}{\rho_m} = \frac{\bar{\rho}_m}{\bar{\rho}_m-1}$$
implying
$$(\rho_m-1)(\bar{\rho}_m-1) = \rho_m \bar{\rho}_m$$
and, therefore, that
$$Real(\rho_m) = 1/2.$$
Edit to address some comments:
If $\vartheta$ were an entire function with real zeros only, e.g., a polynomial with real zeros only, or the reciprocal of the gamma function written as it's Weierstrass factorization, the result would not apply since the necessary pairing that each zero's complex conjugate is also a zero, which is distinct but related to $\rho = 1 - \bar{\rho}$ since $ 3 + i \alpha$ and $3 - i\alpha$ are a conjugate pair that do not satisfy that relation, nor does the symmetry relation $f(s) = f(1-s)$ apply. The symmetry relation and factorization are indeed enough (explicit values for the zeros are not needed) to ensure the critical line--that's the point. Check that $\vartheta(z) = \vartheta(1/z)$ is a severe restriction that when evaluated at $z =0$ giving $\vartheta(0)) = \vartheta(\infty) = 1 = \prod_{\rho_k}[1+(1/\rho_k]$ enforces that $\rho_m = 1/2 + i\alpha$ and $ \rho_k = 1/2 - i\alpha$ are indeed solns.
(See, I believe Bump and others on the local Riemann hypothesis of other functions with critical lines.)
The two conditions of the general form of the factorization and the symmetry are necessary and sufficient to prove the critical line hypothesis without resort to calculating the actual values of the imaginary parts of the zeros (regardless of how interesting they are and the associations to the PNT) is the crux of the matter. (Anticlimactic, if true.)
Edit to address a comment:
From "Remark on Dirichlet Series Satisfying
Functional Equations" by Eugenio P. Balanzario:
"By a theorem of H. Hamburger (see [6], page 31) the zeta function of Riemann is determined by its
functional equation (FE). Hence, if we want to produce other Dirichlet series satisfying a functional equation, then it is necessary to change (FE) somehow."
[6] Titchmarsh, E. C. The theory of the Riemann zeta function, Oxford, 1986.
I should say, since I use this in the formal analysis, that the FE, the Hadamard factorization, and the fact that if $\rho$ is a zero then so is its complex conjugate are necessary (but not necessarily independent) conditions. I do not assume that $\rho = 1 - \bar{\rho}$, that is to be determined from the conditions.
Edit Jun 11;
Prompted by the initial responses, I thought more carefully about my assumptions and concluded before looking at the extended and revised responses that I indeed made a mistake by omitting potential solutions of the form of multiple factors, say $$(1-\frac{1}{p_k}\frac{z}{z-1}) (1- \frac{1}{p_m}\frac{z}{z-1}) = 0,$$ so, in effect, tantamount to assuming the RH is true, as LI did, to arrive at his criteria. Thanks to the responders for the input, the attention, and effort to show me the errors of my way--I learned some things along the way. I'll look over your last responses to see if I can learn some more.
note that leaving aside questions of convergence as pretty much all those products are conditional convergent, we know that $y_k=1-\rho_k$ is a root for each $\rho_k$ and obviously $\frac{y_k-1}{y_k}=\frac{\rho_k}{\rho_k-1}$ so that is the required pairing not that $\rho_k=\bar \rho_m$ which happens precisely when $\Re \rho =1/2$ so the above is the tautology $\Re z =1/2$ iff $\bar z =1-z$ which is true but not particularly helpful
note also that the manipulations above have nothing to do with zeta or xi, only with the symmetry $f(s)=f(1-s)$ so you can pick any entire function of small enough order that satisfies that (so we have a representation without say exponential terms) and conclude that its roots are on the critical line and that is manifestly untrue as there are lots of counterxamples
@Conrad, well that's interesting that it has nothing to do with the Riemann xi and zeta since it is precisely the notation and function defined in Li's classic "The Positivity of a Sequence of Numbers and the Riemann Hypothesis" and explored by Coffey in "Relations and positivity results for the derivatives of the Riemann ξ function"
I meant in the sense that the manipulation above applies to any function that has a symmetry like the zeta and which is nice enough so it is factorable (even conditionally only), so one cannot conclude anything particular about the zeroes from it - now if one adds other conditions etc, who knows, but my remark was strictly about the formal manipulation above
For some conditions that the symmetry condition alone imposes, see https://math.stackexchange.com/questions/28737/does-the-functional-equation-f1-r-rfr-have-any-nontrivial-solutions-besi/145159#145159
where does the symmetrized version of the Davenport-Heilbronn function fit in? It definitely has symmetry, conjugate invariance and Hadamard product on the zeroes (entire order $1$) and lots of non-critical zeroes
Last I heard, fractional linear (Moebius) transformations preserve analyticity.
Okay, good comment. Let me check that.
@Conrad, please check the recent edit.
There is the following error: Given a zero $\rho$ of $\xi(s)$, by the functional equation $\xi(s)=\xi(1-s)$, also $1-\rho$ is a zero of $\xi(s)$. Because $\xi$ is a real${}^*$ function, $\overline{\xi(s)}=\xi(\bar s)$, then also $\bar\rho$ is a zero of $\xi(s)$. The error in the deduction above is the implicit assumption that these two different operations of producing a new zero from a known one, yield the same result: $\bar\rho = 1-\rho$. But there is no reason for that. Instead, the four zeros $\rho$, $1-\rho$, $\bar\rho$, and $1-\bar\rho$ can all be mutually different.
${}^*$(Edit) This shall mean here: A function $f:\Omega\subset\mathbb{C}\to\mathbb{C}$ is real, if its domain $\Omega$ is invariant under complex conjugation, i.e. $\overline{\Omega} = \Omega$, and if it has the property $\overline{f(\bar{z})} = f(z)$ for each $z\in\Omega$ (where $z\mapsto\bar{z}$ denotes complex conjugation in both the domain and range of $f$). If we define the complex conjugate $\bar f$ of the function $f$ (regardless whether it is real or not) by $\bar f:\overline{\Omega}\to\mathbb{C}, z\mapsto \overline{f(\bar{z})}$, then $f$ is real if $\bar f = f$.
(Edit) Some more detailed explanation:
The functional equation $\vartheta(z^{-1}) = \vartheta(z)$ is only a reformulation of $\xi(s)=\xi(1-s)$. So, the corresponding respective pairings $(z, z')$ and $(\rho, \rho')$ of zeros of $\vartheta$ and $\xi$ with $z'= z^{-1}$ and $\rho'=1-\rho$ are essentially the same:
With $z=(\rho-1)/\rho$ and $z'=(\rho'-1)/\rho'$ we have
$$z= z'^{-1}\Leftrightarrow\frac{(\rho-1)}{\rho} = \frac{\rho'}{(\rho'-1)} \Leftrightarrow\rho=1-\rho'.$$
On the other hand, there is the pairing $(\rho,\bar\rho)$ of zeros of the Riemann zeta function $\zeta$ with their complex conjugates, which comes from the fact that $\zeta$ (besides $\xi$) is real in the sense defined above, i. e. $\overline{\zeta(s)}=\zeta(\bar s)$, which is easily seen from its defining sum in the half plane ${\rm Re}\,s>1$ and its unique analytic continuation.
Now, in the sentence justifiying the equation in the Question post
$$\frac{\rho_m−1}{\rho_m}=\frac{\bar\rho_m}{\bar\rho_m−1}$$
both pairings were implicitly assumed to be identical. Instead, $\vartheta(z^{-1}) = \vartheta(z)$ implies for each zero $\rho$ of $\xi$ only the existence of a zero $\rho'$ with
$$\frac{\rho−1}{\rho}=\frac{\rho'}{\rho'−1}.$$
But there is no a priori reason why $\rho'$ should be $\bar\rho$.
First, $\xi(s)$ is not a real function. See Mathworld where you can see separate plots of the real and imaginary parts over a region and a Taylor series in a complex variable. But more importantly your line of reasoning does not reflect mine. I'll add more detail in the question in the morning to hopefully convince you and others of that, but nothing to contradict what I have already written.
@Tom, I added the definition of my use of "real function". In this sense, $\xi$ is a real function, which can also be seen from the (conditionally convergent) product representation you gave, because in this product each zero $\rho$ is paired with its complex conjugate $\bar\rho$, which leads to factors $(1-s/\rho)(1-s/\bar\rho)=1-s (1/\rho + 1/\bar\rho) + s^2/|\rho|^2$ where $s$ has only real coefficients.
Regarding your reasoning, I tried to exhibit just the core of your argument, because neither the product representation nor the introduction of $\vartheta$ seem to be essential.
This is an extended comment based on my original comments and edits in the OP - the three conditions for an entire function $F$ mentioned in the OP:
1.symmetry (symmetrized functional equation) : $F(1-s)=F(s)$
2.$F$ real (various equivalent definitions being: conjugate invariant, $F(\bar z)=\bar F(z)$, real values on the real line, real Taylor coefficients at real values etc)
3.$F$ has a (conditional) Hadamard factorization of the given type (in terms of zeroes only with no exponential factors)
are very general and there are fairly arbitrary functions that satisfy them.
Basically one needs a set of complex numbers $\rho_k=\sigma_k+it_k, \sigma_k \ge \frac{1}{2}, t_k \ge 0, \rho_k \ne 1$ with a density condition of the type: $A_R$ is the number of $\rho_k, |\rho_k| \le R$, then the cardinality of $A(R)$ is $O(R^{2-\delta})$ for some fixed $\delta >0$ (which implies by standard stuff that their convergence exponent which is obviously the same as the convergent exponent of the set obtained from them under $s \to 1-s, s \to \bar s$ is at most $2-\delta$ so the sum of $1/|\rho_k|^2$ converges) and then one can construct a function like that by taking for each $\rho_k$ the polynomial (omitting repeated factors if any)
$P_k(z)=(1-\frac{z}{\rho_k})(1-\frac{z}{1-\rho_k})(1-\frac{z}{\bar \rho_k})(1-\frac{z}{1-\bar \rho_k})$
which is generally of degree $4$ but it can be of degree $2$ if either $\sigma_k=1/2$ or $t_k=0$ and even of degree $1$ if for some $k$'s, we have $\rho_k=1/2$
Obviously $P_k(1-z)=P_k(z)$ and $P_k$ is real (and it trivially satisfies Hadamard factorization of order zero) while if $|z| \le M$ and $|\rho_k|$ large, one clearly has $|\log P_k(z)| \le C_M/|\rho_k|^2$ (since $|(1-\frac{z}{\rho_k})(1-\frac{z}{1-\rho_k})| \le 1+C_M/|\rho_k|^2) $ which means that under the mentioned density hypothesis on the $|\rho_k|$, we get an entire function $F(z)=\Pi_{k}P_k(z)$ where the product is actually normally convergent and which has zeroes precisely at $\rho_k$ and its transforms under $s \to 1-s, s \to \bar s$, while it has the required (conditional as we need to group the zeroes as noted like in the usual $\xi$ case for that matter) Hadamard factorization
This being said, obviously most $F$ as above won't be symmetrizations of Dirichlet series like for $\xi$ or the symmetrizations of $L_{\chi}$ functions (all those have extra properties which are quite stringent - for example since the symmetrization involves the $\Gamma$ function which is of order $1$ and maximal type (in the extended Nevanlinna sense for meromorphic functions), all those have order $1$ and maximal type too as their zeroes satisfy the usual Riemann-von Mangoldt density $O(R\log R)$ and even among symmetrizations of Dirichlet series, we have such that come from Dirichlet series that do not have multiplicative coefficients so they are without Euler products
(eg the symmetrization of the Davenport-Heilbronn function is one such - for reference, I include the definition of both:
$f(s)= 1/1^s+\tan \theta/2^s-\tan \theta/3^s-1/4^s+1/6^s+\tan \theta/7^s...$, where $\theta$ is the unique angle in $(0, \pi/2)$ for which $\tan 2\theta =\frac{\sqrt 5 -1}{2}$
$F(s)=(\frac{5}{\pi})^{s/2}\Gamma(\frac{1+s}{2})f(s), F(s)=F(1-s)$
and a very detailed analysis of its zeroes is in the Karatsuba-Voronin RZ bible starting in chapter 6, page 212)
.
In particular, the OP's hypothesis that those three conditions by themselves allow one to draw conclusions about the zeroes is not correct
Also, and more immediately, Landau's $\zeta(2s)\zeta(2s-1)$...
|
2025-03-21T14:48:31.194517
| 2020-06-09T08:37:20 |
362580
|
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"Marouani Samir",
"YangMills",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362580"
}
|
Stack Exchange
|
Kähler-hyperbolic and Kobayashi-hyperbolic complex manifold
I look for a clear proof of the implication Kähler-hyperbolic in the sense of Gromov implicate Kobayashi-hyperbolic for compact complex manifolds.
Corollary 4.2 here https://arxiv.org/pdf/1609.06918.pdf
Thank you for the reference Is it the same proof as Gromov's theorem
|
2025-03-21T14:48:31.194575
| 2020-06-09T08:38:08 |
362581
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362581"
}
|
Stack Exchange
|
Tropical abelian variety as a limit
A tropical abelian variety is given by a quotient of a real vector space $V \cong \mathbb{R}^g$ with a fixed integral structure $\Gamma_2$, by a lattice $\Gamma_1$, equipped with some aditional structure (polarization). How precisely can this be seen as a limit of a family of complex abelian varieties (in terms of their lattices)?
There is such a limit sketched by the end of the first page of this paper:
https://arxiv.org/pdf/2002.02347.pdf
, but I cannot see what this means (if, for example, $\gamma$ generates $\Gamma_1$ and $\Gamma_2=\mathbb{Z} \subset \mathbb{R}$ than $X_\epsilon = \mathbb{C}^*/{\epsilon^{-1} e^{\gamma\mathbb{Z}}}$ and $\epsilon^{-1}e^{\gamma \mathbb{Z}}$ is not even a subgroup of $\mathbb{C}^*$ for most of $\epsilon$.) On the other hand this does not seem complicated at all so I have probably missed something.
I also read in the literature that this degeneration goes back to Mumford but his paper seems a bit involved and uses different language. I asked this question on Math Stack Exchange but maybe it is more suitable here.
|
2025-03-21T14:48:31.194690
| 2020-06-09T08:42:55 |
362582
|
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"authors": [
"Jean Van Schaftingen",
"Mr Pie",
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"https://mathoverflow.net/users/42047",
"https://mathoverflow.net/users/58793",
"tituf"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362582"
}
|
Stack Exchange
|
A conjecture of De Giorgi on weighted Sobolev spaces
Let $\mu$ be a probability measure on $\mathbb{R}^d$ which is absolutely continuous with respect to the Lebesgue measure with density $\rho$. Assume that, for all $t>0$,
\begin{align*}
\exp \left(t \rho\right), \exp \left(t \rho^{-1}\right) \in L^1_{loc}.
\end{align*}
Then, a conjecture of De Giorgi asserts that the Meyers-Serrin theorem holds for the weighted Sobolev spaces associated with $\mu$, namely $H=W$.
The only reference I know speaking of this conjecture is this paper
https://iopscience.iop.org/article/10.1070/SM1998v189n08ABEH000344/meta
Thus, I would like to know if there have been recent advances in proving or disproving this conjecture.
Thanks in advance.
It might make sense making the title for specific with something like “A conjecture of De Giorgi on weighted Sobolev spaces”.
I did some diggings and some readings and found out that the conjecture has been solved here
https://link.springer.com/article/10.1134/S1064562413060173
and extended recently to a wider context in
https://www.degruyter.com/view/journals/crll/2019/746/article-p39.xml
in his paper Zhikov deals with a bounded domain $\Omega$, have you understand how to pass to the whole $\mathbb R^N$ ?
On De Giorgi’s conjecture: Recent progress and open problems by Chan and Wei reviews the status of the problem in 2017.
A second attempt to locate this elusive conjecture: De Giorgi gave a talk in Lecce (Italy) in 1995 entitled "Congetture sulla continuità delle soluzioni di equazioni lineari ellittiche autoaggiunte a coefficienti illimitati" (Conjectures on the continuity of solutions of selfadjoint elliptic linear equations with unbounded coefficients"). The talk was typed out but never published.$^\ast$ Progress on these conjectures is discussed in
Xiao Zhong, Discontinuous solutions of linear, degenerate elliptic equations (2008).
Sungwon Cho, A certain example for a De Giorgi conjecture (2014).
$^\ast$ De Giorgi's Bibliography says:
"De Giorgi, in particular in the last years of his life, used to
circulated his writings among friends and colleagues, asking for
opinions. We plan in the future to collect and to make available all
these unpublished writings." I tried to contact professor Zhong for a copy, but the email bounced.
It appears De Giorgi has more than a single conjecture.
|
2025-03-21T14:48:31.194880
| 2020-06-09T09:52:57 |
362587
|
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"authors": [
"Jamie Gabe",
"Math Lover",
"Nik Weaver",
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"https://mathoverflow.net/users/129638",
"https://mathoverflow.net/users/23141"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362587"
}
|
Stack Exchange
|
Need reference of books which deals with ideal theory of tensor product
Is there any book which deals with Ideal theory of tensor product of $C^{\ast}-$ algebras
Is there an "ideal theory" of C*-algebras? Doesn't seem like much can be said in general.
@NikWeaver: I was looking for a book which discusses for instance ideals of min, max or projective tensor product?
In the textbook by Brown-Ozawa, Corollary 9.4.6 shows that two-sided closed ideals in the spatial tensor product $A \otimes B$ are generated by tensor products of two-sided closed ideals, provided $A$ or $B$ is exact. Consequently, $Prime(A\otimes B) \cong Prime(A) \times Prime(B)$ if $A$ or $B$ is exact, where $Prime(-)$ is the prime ideal space (the space of prime ideals with the hull-kernel topology). Also, Section 2.8 of the Blanchard-Kirchberg JFA 2004 paper deals with the ideal structure of spatial tensor products, even if none of the $C^\ast$-algebras are exact.
@JamieGabe: I will have a look. Thank you.
|
2025-03-21T14:48:31.194989
| 2020-06-09T10:57:05 |
362589
|
{
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"Ben McKay",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362589"
}
|
Stack Exchange
|
On the smoothness of transition functions
Let $p:E \longrightarrow M$ be a smooth fibre bundle, with standard fibre space $F$ and $G$ a Lie group acting effectively on $F$ as a structure group.
Then, are the transition functions always smooth?
I mean, if for every open $U\subseteq M$ and function $g:U \longrightarrow G$ such that the function:
$$ H:U\times F \longrightarrow U \times F$$
$$ H(x,s)=(x,g(x)\cdot s)$$
It's a diffeomorphism. Then, does it follow that $g$ is smooth?
PS: In Michor's book, natural operators in differential geometry, I read a similar fact in the step 5 of 9.11, but I don't see why it's true. https://www.mat.univie.ac.at/~michor/kmsbookh.pdf
Do you mean $f=g$?
Yes, my fault sorry. Thanks. I edited it
It is true but tricky to prove; I can't think of a reference off hand. You want to know that that an effective Lie group action allows one to smoothly determine each element $g$ by knowing ``how it acts''. You can prove this by thinking about the action of $G$ on the bundle of jets of local coordinates. The effectiveness, and analyticity of a Lie group acting on a homogenous space, ensures that if you take high enough jets, the stabilizer is trivial. That ensures that you can solve for an element $g$ uniquely and smoothly if you are giving the information of what $g$ does to a high enough jet.
Thanks. I'm not familiar with Jets enough to undesrtand it, but I think I catch the idea. Is it like calculate the "Taylor expansion" on $s$ of $g(x)\cdot s$ and recover $g(x)$?
That is the idea. But you only need to compute Taylor series up to some finite order to get this to work. The group action cannot fix a point and fix an analytic coordinate system at that point without fixing all nearby points. The Hilbert basis theorem can prove that the subgroup fixing that point can only fix a finite number of Taylor coefficients of a coordinate system without fixing all points in the connected component of that point. At least when there are finitely many components to the homogeneous space, this should do the trick. I think there is a way to deal with infinite components.
Mmm, I see that I have to study more to completely understand it. The analyticity property of the action means that the action can be recovered by the Taylor coefficients? Is it necessary the action to be transtitive? You've mentioned that $(G,F)$ is a homogeneus space.
Do you have any recommendation to read these properties about Jets and group actions?
About step 5 in 9.11:
The first line of the display in step 5 gives a formula for the transition functions involving mappings $x\mapsto c^x_{\alpha}$ which are smooth enough
(see 4 lines above this), and parallel transport which is smooth even in the choice of curves (if they depend on finitely many parameters) by theorem 8.9, 5 pages before.
|
2025-03-21T14:48:31.195204
| 2020-06-09T12:20:26 |
362596
|
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"Alok",
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|
Stack Exchange
|
Maximal ideal space of $\ell^\infty(A)$, $A$ a commutative unital Banach algbera
Let $A$ be a commutative unital complex Banach algebra with norm $\|\cdot\|_A$, and let $\ell^\infty(A)$ denote all bounded sequences $(a_n)_{n\in \mathbb{N}}$ with $a_n\in A$, $n\in \mathbb{N}$, with pointwise operations and
the supremum norm:
$$
\|(a_n)_{n\in \mathbb{N}}\|_{\ell^\infty(A)}=\sup_{n\in \mathbb{N}} \|a_n\|_{A}.
$$
Then $\ell^\infty(A)$ is itself a Banach algebra with this norm and pointwise operations. Each element in the product of the maximal ideal space of $\ell^\infty$ with the maximal ideal space of $A$ gives rise to an element of the maximal ideal space of $\ell^\infty(A)$.
Question: Can the maximal ideal space of $\ell^\infty(A)$ ever be bigger than the product of the maximal ideal space of $\ell^\infty$ with the maximal ideal space of $A$?
What happens if $A=\ell^\infty$?
Does this show that it is strictly bigger? Can't see it yet... could you give an outline please?
The maximal ideals of $l^\infty$ correspond to ultrafilters on $\mathbb{N}$. The algebra $l^\infty(l^\infty)$ is isomorphic to $l^\infty(\mathbb{N}^2)$. Is every ultrafilter on $\mathbb{N}^2$ a product of two ultrafilters on $\mathbb{N}$?
Many thanks for this!
|
2025-03-21T14:48:31.195312
| 2020-06-09T13:21:43 |
362602
|
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"Hollis Williams",
"Yellow Pig",
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|
Stack Exchange
|
Coulomb branch varieties and symplectic singularities
I was recently looking at the survey article of Fu on symplectic resolutions which has a number of open questions and conjectures at the end. (I think one of these was existence of a classification of when symplectic singularities admit symplectic resolutions, which is discussed on other MathOverflow questions).
Another perhaps simpler question which was mentioned was whether there is a proof that Coulomb branch varieties are always symplectic singularities. Has any progress been made towards a proof of this or is it of similiar difficulty to the other question I mentioned?
Edit: I'm slightly surprised that no-one has answered this question. It seems like people who work on these types of thing are not very active on Math Overflow for some reason, I'm not sure how that is meant to help people who wish to learn more about the subject.
Sorry, which survey article by Fu do you mean? The one I found https://arxiv.org/abs/math/0510346 from 2005 does not seem to mention Coulomb branches (I think they had not been introduced yet by BFN at that time).
This new paper https://arxiv.org/abs/2005.01702 by A. Weekes answers your second question in the case of quiver gauge theories.
Perhaps you would have got more answers if you had answered Yellow Pig's question.
Hi abx, I corresponded privately with Yellow Pig when they contacted me and did answer their question: I did not then feel the need to post the same answer here.
There is a very recent paper by G. Bellamy that shows that every Coulomb branch is a symplectic singularity.
The paper is very nice. You can find it here
|
2025-03-21T14:48:31.195456
| 2020-06-09T13:40:29 |
362604
|
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"url": "https://mathoverflow.net/questions/362604"
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|
Stack Exchange
|
Multi-simplicial generalization of $\Gamma$-spaces
Is there a generalization of Segal's theorem that the inclusion of $X_1$ into $\Omega|X_*|$ is a weak equivalence for a $\Gamma$-space $X_*$ if $X_1$ is group like? Specifically, I am looking for a result something like $\hat{X} \rightarrow \Omega ^n |X_{*,\dots,*}|$ is a weak equivalence where $X_{*,\dots,*}$ is a multi-simplicial space subject to some conditions like a $\Gamma$-space and $\hat{X}$ is a space derived from the spaces of $X_{*,\dots,*}$, subject to some condition on $\pi_0,\dots,\pi_{n-1}$
The paper "Iterated monoidal categories" by Balteanu, Fiedorowicz, Schwanzl, Vogt gives an analogous result for functors $(\Delta ^{op})^n \rightarrow Top$ and n-fold loop spaces.
|
2025-03-21T14:48:31.195524
| 2020-06-09T13:48:46 |
362607
|
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"url": "https://mathoverflow.net/questions/362607"
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|
Stack Exchange
|
Is there a list of all real unital subalgebras of M(2,C)?
Is there a complete classification of all real unital subalgebras of $M(2,\mathbb C)$ up to isomorphism? The list should include $M(2,\mathbb C)$, the quaternions, complex numbers, split-complex numbers, dual numbers, 2x2 real matrices, $\mathbb C \oplus \mathbb C$, the tensor product of the dual numbers with $\mathbb C$, the dual-complex numbers and so on.
|
2025-03-21T14:48:31.195590
| 2020-06-09T13:50:49 |
362609
|
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"Asaf Shachar",
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|
Stack Exchange
|
When is the optimum of an optimization problem affine in the constraint parameter?
While working on a variational problem I have reached to the following question:
Let $f:(0,\infty) \to [0,\infty)$ be a $C^1$ function satisfying $f(1)=0$. Suppose that $f(x)$ is strictly increasing on $[1,\infty)$ and strictly decreasing on $(0,1]$. Define $F:(0,1) \to [0,\infty)$ by
$$
F(s)=\min_{xy=s,x,y\in(0,\infty)} f(x)+ f(y).
$$
Question: For which functions $f$, $F(s)$ has an affine part? Can we
characterize such functions?
The motivation is that I am applying Jensen inequality with $F$, and an affine part (in contrast to strict convexity) gives some flexiblity.
The only example that I know of is when $f(x)=(x-1)^2$, and
$$
F(s) =
\begin{cases}
1-2s, & \text{ if }\, 0 \le s \le \frac{1}{4} \\
2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4},
\end{cases}
$$
is affine on $[0,\frac{1}{4}]$.
Is this $f$ the only choice which makes $F$ affine?
For qubic and quartic penalizations this is not the case; if $f(x)=(1-x)^3$, then
$$
F(s)=\begin{cases}
1 - 3 s - 2s^{3/2} &\text{ if } 0<s\le1/9, \\
2 + 6 s - 2(3 + s)s^{1/2} &\text{ if } 1/9\le s<1.
\end{cases}
$$
and similarly for $f(x)=(x-1)^4$.
Here is an attempted analysis:
Assume that there is a $C^1$ map $s \to (x(s),y(s))$ giving a minimizer to the problem, i.e. for any $s \in (0,1]$
$$
F(s)=f(x(s))+f(y(s)), \, \, \,x(s)y(s)=s. \tag{1}
$$
Lagrange's multipliers give
$$
f'(x(s))=\lambda(s) y(s),f'(y(s))=\lambda(s) x(s). \tag{2}
$$
$F'(s)=\lambda (s)$, so $F''(s)=0$ if and only if $\lambda(s) < 0$ is constant. ($\lambda < 0$ since $f'|_{(0,1)} < 0$ by our assumption.)
I don't see how to proceed from here.
For $f(x)=(x-1)^2$ we have $\lambda(s)=-2$: The minimum (for $s \in [0,\frac{1}{4}]$) is obtained at $x(s)+y(s)=1$, so $
f'(x(s))=2(x(s)-1)=-2y(s).
$
Looks to me like it takes a "perfect storm" to overcome the non-convexity of $xy=s$.
Thanks, I think you might be right. I tried to solve a "easier " question: When is the optimum $F(s)$ affine? This miracle seems even more surprising than convexity. I have edited the question to include an attempted proof for that subproblem. (It reduces to a coupled ODE+a functional equation. which I don't know how to analyze).
Say that $F$ is affine on an interval $I$, with $F'\equiv\lambda$, a constant. Then for $x=x(s)$ and $y=y(s)$, one has $f'(x)=\lambda y$ and $f'(y)=\lambda x$, thus
$$xf'(x)\quad (=\lambda xy)\quad=yf'(y).$$
This implies a functional equation
$$xf'(x)=\frac1\lambda f'(x)f'(\frac1\lambda f'(x)).$$
Simplifying, one finds that the function $\frac1\lambda f'=:g$ is a functional square root of the identity:
$$g\circ g={\rm id}_I.$$
Notice that if $g$ is not the identity itself, then it tends to be a dcreasing function: suppose $g(x)\ne x$, say $z:=g(x)>x$, then $g(z)=x<g(x)$.
Conversely, suppose now that $g$ is a decreasing square root of the identity. Then choose a constant $\lambda$ and define $f$ by $f'=\lambda g$. Suppose in addition that $x\mapsto xg(x)$ is strictly convex (perhaps too strong a hypothesis). Then the level sets of $xf'(x)$ consist in pairs $(x,y)$, which turn out to be the $(x(s),y(s))$ above, where $s:=xy$. Then $f$ answers your query.
Now, to construct a functional square root of identity, one proceeds as follows. Choose arbitrarily a point $a>0$ and $g:[0,a]\rightarrow{\mathbb R}$ a decreasing function such that $g(a)=a$. Let $b:=g(0)$, so that $g([0,a])=[a,b]$. Then extend the definition of $g$ to $(a,b]$ by $g(x):=g^{-1}(x)$. Then $g\circ g$ over $[0,b]$.
Thank you, that is a very interesting answer. Just to make sure that I understand the 'forward' direction: Suppose that you start with a decreasing square root of the identity $g$ such that $xg(x)$ is strictly convex, and set $f'=\lambda g$ as you described.Then if $x(s),y(s)$ are minimizers, then $xg(x)=yg(y)$ (by the Lagrange multiplier argument) - so are uniquely determined up to order by the convexity assumption on $xg(x)$. (I guess we could also require $xg(x)$ to be strictly concave, right?...
This is what happens in the example described in the question, where $f(x)=(x-1)^2$, and $g(x)=1-x$.
|
2025-03-21T14:48:31.196179
| 2020-06-09T14:31:31 |
362613
|
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"Ariel Weiss",
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|
Stack Exchange
|
Are the representations attached to nonclassical $p$-adic eigenforms of weight 1 Hodge-Tate?
Let $f$ be a nonclassical finite slope overconvergent $p$ adic eigenform of weight 1, and let $\rho_f$ be the associated $p$-adic Galois representation of $Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$. Is $\rho_f|_{Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)}$ necessarily Hodge-Tate? When $f$ is classical or when the weight is larger than 1 this is always true, but in weight $1$ this can perhaps fail because the Sen operator may not be semisimple.
If it had Hodge-Tate weights all 0, wouldn’t the image of inertia be finite? So it would be de Rham.
|
2025-03-21T14:48:31.196264
| 2020-06-09T14:46:12 |
362616
|
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|
Stack Exchange
|
Alternative proof of existence of absolute value of a functional on a C*-algebra
The usual proof of the existence of an absolute value of a functional on a C*-algebra $A$ uses the polar decomposition of normal functionals on $A^{**}$, which relies on the compactness of the unit ball of $A^{**}$ in the weak*-topology.
Is it possible to derive the existence of an absolute value of a bounded linear functional on a C*-algebra via a compactness argument in $A^*$? For a definition of the absolute value of $\varphi \in A^*$, I mean a $\tau \in A^*_+$ such that $\|\tau\| = \|\varphi\|$ and $|\varphi(a)|^2 \leq \|\varphi\| \tau(a^* a)$.
Furthermore, is it possible to use this to show that $\varphi$ can be represented as $\varphi(a) = \langle \pi_\tau(a) \xi \,\vert\, \eta\rangle$ where $\pi_\tau$ is the GNS representation associated with $\tau$ and $\|\varphi\| = \|\xi\| \|\eta\|$?
I haven't worked out the details, but another way to think about absolute values is to use a $2\times 2$ matrix construction. What I have in mind is how you move from completely bounded maps to completely positive maps, by putting the cb map on the off diagonal, and using a Hahn-Banach argument to find cp maps on the diagonal which make the whole matrix CP. This allows you to use Stinespring to prove to representation result for CB maps. I think the same works for linear functionals, but I haven't written this out to check there is not a lurking circular argument...
That does work, although you need to embed a proof that the CB norm of a linear functional is its norm, so it's not that much different than the full CB representation theorem. One other approach I tried that doesn't work is proving representation for self-adjoint functionals using Jordan Decomposition, and then trying to lift that to a self-adjoint functional in the 2x2 case. You do better than the naive estimate of $4 |\varphi|$, but you end up with a $\sqrt{2}$ factor in the norms of the vectors.
This is maybe a "backwards" answer to what you might have been hoping for...
An affirmative answer to the 2nd question would give the 1st question as well. Indeed,
if there is a $*$-representation $\pi:A\rightarrow B(H)$ and $\xi,\eta\in H$ with $\|\varphi\| = \|\xi\| \|\eta\|$ and $\varphi(a) = \langle \pi(a)\xi, \eta \rangle$, then by rescaling we may suppose that $\|\xi\|=\|\eta\|=\|\varphi\|^{1/2}$, and then $\tau:a\mapsto \langle \pi(a)\xi,\xi\rangle$ is a positive functional with $\|\tau\| = \|\xi\|^2 = \|\varphi\|$. Further,
$$ \|\varphi\| \tau(a^*a) = \|\eta\|^2 \|\pi(a)\xi\|^2
\geq | \langle \pi(a)\xi, \eta \rangle|^2 = |\varphi(a)|^2. $$
I do not know a "compactness" argument which can show this. However, a while ago I wrote up an argument of how to prove Kaplansky Density using Arens products; see Notes on GitHub. To do this in a non-circular way, you need to use Hahn-Banach, and need to find a way to prove exactly this representation result, in a "simple" way. The way I did this was as follows.
Let $H$ be a Hilbert space. Let $B(H)_*$ be the trace-class operators on $H$, the predual of $B(H)$. By Hahn-Banach (Goldstine's Theorem) for $\mu\in B(H)^*$ there is a net $(\omega_i)$ in $B(H)_*$ converging weak$^*$ to $\mu$, and with $\|\omega_i\|=\|\mu\|$ for each $i$. Thus, for an ultrafilter $\mathcal U$ refining the order filter, the natural map $(B(H)_*)_{\mathcal U} \rightarrow B(H)^*$ (given by "take weak$^*$-limit") is a metric surjection. With $K = \ell^2(H)$, and $\pi_0:B(H)\rightarrow B(K)$ the "diagonal" map, for each $\omega\in B(H)_*$ we can find $\xi,\eta\in K$ with $\omega = \omega_{\xi,\eta}\circ\pi_0$. Now let $K = (\ell^2(H))_{\mathcal U}$ the ultrapower, a Hilbert space. We can find $\xi = (\xi_i), \eta=(\eta_i)\in K$ with $\omega_i = \omega_{\xi_i,\eta_i}\circ\pi_0$ for each $i$. Thus, with $\Pi:B(H)\rightarrow B(K)$ the diagonal of $\pi_0$, we have that
$$ \langle \Pi(x)(\xi), \eta \rangle = \lim_{i\rightarrow\mathcal U} \langle \pi_0(x)\xi_i, \eta_i \rangle = \lim_{i\rightarrow\mathcal U} \omega_i(x) = \mu(x). $$
Given a $C^*$-algebra $A$, by the GNS construction, we can exhibit $A$ as a subalgebra of $B(H)$ for some $H$. For $\varphi\in A^*$ take a Hahn-Banach extension to $\mu\in B(H)^*$. Then the previous paragraph, with $\pi$ the restriction of $\Pi$ to $A$, gives the required representation.
I was going to update my question when I fully convinced myself I didn't understand, but there's a proof like the one I suggested in the middle of Theorem 1 in [http://ousar.lib.okayama-u.ac.jp/en/list/authors/T/Minoru,Tomita/item/33396](Spectral theory of operator algebras I) by Tomita. He says that since every bounded functional is a linear combination of states, you can find such a $\tau$, but maybe not one of the desired norm. Why can you find such a $\tau$ to begin with? It's also strange that he picks one of minimal norm, but never uses minimality...
That aside, your proof seems like the morally correct one, so I'm going to accept it. There's a variant of this idea in Theorem 1.10.8 of Ilijas Farah's book Combinatorial Set Theory of C*-Algebras, which is actually attributed to Ozawa in the acknowledgements. However, it uses a Banach limit of functionals on the diagonal rather than a Hilbert space ultrapower, and hence needs a further step in the form of an inequality that is rather difficult to motivate.
I don't know your motivation for an alternative proof. But maybe Bishop-Phelps theorem helps?
|
2025-03-21T14:48:31.196740
| 2020-06-09T14:46:39 |
362618
|
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|
Stack Exchange
|
Dense linear span implies closed convex hull has non-empty interior
Let $X$ be a Fréchet space and let $Y\subseteq X$ such that $\overline{\operatorname{span}(Y)}=X$. It seems intuitive to me that $\operatorname{int}\big(\overline{\operatorname{co}(Y)}\big)$ is a non-empty open subset of $X$. But how to show this?
What happens if $X=\ell^2$ and $Y$ consists of the vectors $2^ne_n$, where the $e_n$ are an orthonormal basis for $X$? It seems to me that, if there are counterexamples, they'd look something like this. (Iosif's answer, which appeared as I was typing mine, indicates that I got it backward; $2^{-n}$ instead of $2^n$.)
The answer is no. E.g., let $X=\ell^2$ and
$$Y=\{x\in\ell^2\colon\sum_n nx_n^2\le1\}.$$
Then $Y$ is a closed convex set, spanning $X$, but the interior of $Y$ is empty.
Details: $Y$ is convex because the function $\mathbb R\ni u\mapsto u^2$ is convex. To show that $Y$ is closed one may use the Fatou lemma.
If the interior of $Y$ were not empty, then, by the symmetry of $Y$ (that is, by the property $-Y=Y$), $0$ would be in the interior of $Y$. So, for some real $h>0$ and all natural $n$, we would have $he_n\in Y$, where $(e_1,e_2,\dots)$ is the standard basis of $\ell^2$. But $he_n\notin Y$ for $n>1/h^2$, a contradiction.
what would be a reasonable condition on $Y$ (or $X$ for that matter) such that this doesn't happen?
@Zorn'sLama : I suspect $X$ would have to be finite dimensional for this to hold for all $Y$, but don't know at the moment whether this is true.
If for example, $co(Y)$ is dense in $Ball(0,1)$ in $X$, then everything should work no?
@Zorn'sLama : Concerning your last comment: yes, of course.
ok, fantastic, then my theorem still holds ;) Thanks Iosif! (Had a small hard palpatation when I tried to extend if naively I see).
@Zorn'sLama: Your previous comment is just saying something trivial: if $\operatorname{co}(Y)$ is dense in the ball then its closure contains the ball, so of course the interior of the closure is nonempty.
|
2025-03-21T14:48:31.196908
| 2020-06-09T15:37:39 |
362622
|
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|
Stack Exchange
|
Constrain representation of tempered distribution
This is a follow-up to this question.
Let $T$ be a tempered distribution on $\mathbb{R}^d$.
Then there is a multiindex $\alpha \in \mathbb{N}_0^d$, an $n \in \mathbb{N}_0$ and a bounded continuous function $f$ on $\mathbb{R}^d$ such that
\begin{equation}
T = \partial^\alpha \left( 1 + \left \Vert x \right \Vert^2 \right)^n f
\end{equation}
Suppose now, that for all $\phi \in C_c^\infty \left( \mathbb{R}^d \right)$
\begin{equation}
\phi \ast T \in L^\infty \left( \mathbb{R}^d \right)
\end{equation}
can we then constrain $\alpha, n$ or $f$?
Intuitively, I would expect to conclude that we can choose $n = 0$ which would be in line with
\begin{equation}
\begin{aligned}
\left \vert \phi \ast T \left( x \right) \right \vert
&\le \int \left \vert \left( \partial^\alpha \phi \right) \left( x - y \right) \right \vert
\left( 1 + \left \Vert y \right \Vert^2 \right)^n \left \vert f \left( y \right) \right \vert \mathrm{d} y \\\\
&\le A^n \left \Vert f \right \Vert_{L^\infty} \left( 1 + \left \Vert x \right \Vert^2 \right)^n \int \left \vert \left( \partial^\alpha \phi \right) \left( x - y \right) \right \vert
\left( 1 + \left \Vert x - y \right \Vert^2 \right)^n \mathrm{d} y \\\\
&\le A^n C_{\partial^\alpha \phi,n} \left \Vert f \right \Vert_{L^\infty} \left( 1 + \left \Vert x \right \Vert^2 \right)^n
\end{aligned}
\end{equation}
where $0 < C_{\partial^\alpha \phi,n} < \infty$ and $A > 0$ is a constant chosen such that
\begin{equation}
1 + \left \Vert y \right \Vert^2
\le
A \left( 1 + \left \Vert x \right \Vert^2 \right)
\left( 1 + \left \Vert x - y \right \Vert^2 \right)
\end{equation}
for all $x, y \in \mathbb{R}^d$.
For e.g $f = 1$, it is easy tow show that we can find a $\phi$ producing the above asymptotics, that is "saturating" the inequality - up to a constant factor.
But generally $f$ could be wildly oscillating and do all sorts of weird things.
Hence, using any specific $\phi$ seems out of the question - but then I really struggle with ideas to move further...
(I would really like to be able to show $n = 0$, so my perspective is probably biased. On the other hand it seems extremely natural.)
|
2025-03-21T14:48:31.197120
| 2020-06-09T16:33:50 |
362627
|
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Stack Exchange
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Request for examples: verifying vs understanding proofs
My colleague and I are researchers in philosophy of mathematical practice and are working on developing an account of mathematical understanding. We have often seen it remarked that there is an important difference between merely verifying that a proof is correct and really understanding it. Bourbaki put it as follows:
[E]very mathematician knows that a proof has not really been “understood” if one has done nothing more than verifying step by step the correctness of the deductions of which it is composed, and has not tried to gain a clear insight into the ideas which have led to the construction of this particular chain of deductions in preference to every other one.
[Bourbaki, ‘The Architecture of Mathematics’, 1950, p.223]
We are interested in examples which, from the perspective of a professional mathematician, illustrate this phenomenon. If you have ever experienced this difference between simply verifying a proof and understanding it, we would be interested to know which proof(s) and why you did not understand it (them) in the first place. We are especially interested in proofs that are no longer than a couple of pages in length. We would also be very grateful if you could provide some references to the proof(s) in question.
We are sorry if this isn’t the appropriate place to post this, but we were hoping that professional mathematicians on MathOverflow could provide some examples that would help with our research.
Cantor's diagonal argument (or in general Cantor's theorem) is a very popular example of a theorem whose proof is extremely easy to verify, but it takes some students (and amateurs) quite some time to fully digest and understand it. This is one of the reason "counterexamples to Cantor's theorem" are so frustrating, because it's always so easy to throw the "counterexample" against the proof and see it fails.
I think that Charles Fefferman said that he checked Kuranishi's proof of CR embedding line by line, but could never understand what made it work.
Conversely, it's possible to understand a proof without verifying all the details! I think this situation is much more common for working mathematicians.
I am sorry for an unhelpful comment, but literally every proof is like that. I think you can even find good examples among answers to MathOverflow questions.
I would view a proof like a path from A to B. Verifying the proof behaves like walking along that path and understanding the proof is more like understanding at each intersection, why one chooses a specific direction and whether there are any short-cuts.
Many of Ramanujan's formulas on denesting radicals and diophantine equations can be proven with simple algebra - but how he found those in the first place is only guesswork.
@HenrikRüping I like the path idea. A metaphor I've used in the past is that sometimes stepping though a proof is like being a spelunker. You have made it though the cave, but the path that you took is still dark and hidden from your sight. Other times it is like climbing a (sufficiently gentle) mountain. From the summit you can look down and see at a glance the path that you took.
There might be some interesting arguments here about machine proving... since machines can verify (and sometimes even create) proofs, but as far as current technology goes they understand nothing.
I want to emphasize that is also happens frequently that understanding a proof is easy (due to conception, in particular when it comes to things like geometry or similar) but the technical execution is hard and therefore also verifying the correctness of a proof.
What makes you think that the proofs by themselves can constitute examples of this phenomenon? And not, for example, a combination of the proof, the person (the mathematician trying to check/understand it), and their state of knowledge and experience at the time they are trying to understand it?
@R.vanDobbendeBruyn "When I was young, I was able to understand any proof, whether it was correct or not!" (Paraphrasing Mark Twain, "When I was young, I remembered everything, whether it had happened or not.")
This distinction between verifying and understanding a proof seems quite similar to the P vs NP problem in theoretical computer science. To make the analogy clear, verifying a proof could be equated to verifying an answer to an NP problem, while coming up with an answer to an NP problem - in general believed to be much harder than the verification - could be equated to understanding a proof (as in, you know how to get there). If I gave you a 3-colorable graph and a valid coloring of it, you could easily check the color assignment, while having no idea of the process I used to get it.
Technically, a mathematician could obfuscate all her results in such a way that everyone would believe she actually proved what she proved, by verifying her proofs, while gaining very little insight in how she actually proved it (her creative process before the obfuscation). Fortunately such practices are frowned upon today, but similar practices are not unheard of in history (it reminds me of mathematicians solving the degree 3 polynomials their colleagues were sending them, to show that they had found a general solving method, which they were not sharing).
I'm not a professional mathematician thus I don't know if my reference fits with your need. I know a video by Carlos M. Madrid Casado from the official channel of YouTube fgbuenotv that is the official channel of the Fundación Gustavo Bueno. The video that I refer is in Spanish and has title Carlos Madrid - ¿Qué es la filosofía de las matemáticas? with date Marh, 5th 2018. In my view/belief (I don't know about your subject but the conferences of the author are excellents) it can be interesting in the context of your question from the minute $\approx 1:19$ until the minute $\approx 1:33$.
Don Zagier has a well-known paper, A one-sentence proof that every prime $p\equiv 1\pmod 4$ is a sum of two squares. An undergraduate mathematics major should be able to verify that this proof is correct. But as you can see elsewhere on MathOverflow, most professional mathematicians are unable to "understand" this proof just by studying it in isolation. By lack of "understanding" is meant, for example, the inability to answer questions such as, "Where did those formulas come from? How did anybody ever come up with this proof in the first place? Is there some general principle on which this proof is based, that is not being presented explicitly in the proof?"
I am one of those at MO who has complained about this proof. However, I was shown the light about what is really going on there by this pretty wonderful Mathologer video on youtube: https://www.youtube.com/watch?v=DjI1NICfjOk. (Oh, and now I see that Moritz Firsching explained the same thing in the other thread.)
Ivan Niven has published A simple proof that $\pi$ is irrational. Verifying that the proof is correct requires only elementary calculus. On the other hand, to "understand" it, a professional mathematician will probably need to study some general theory (what sort of general strategies are there for constructing an irrationality or transcendentality proof?) and/or some of the history of the subject. Otherwise, it looks like some complicated formulas are being pulled out of nowhere.
"In the remaining sections of this paper we briefly discuss various occurrences of the stability and pinching phenomena in differential geometry. The results we present are, for the most part, not new and we do not provide the detailed proofs. (These can be found in the papers cited in our list of references). What may be new and interesting for non-experts is an exposition of the stability/pinching philosophy which lies behind the basic results and methods in the field and which is very rarely (if ever) presented in print. (This common and unfortunate fact of the lack of an adequate presentation of basic ideas and motivations of almost any mathematical theory is, probably, due to the binary nature of mathematical perception: either you have no inkling of an idea or, once you have understood it, this very idea appears so embarrassingly obvious that you feel reluctant to say it aloud; moreover, once your mind switches from the state of darkness to the light, all memory of the dark state is erased and it becomes impossible to conceive the existence of another mind for which [the] idea appears non-obvious.)" --Mikhail Gromov, "Stability and Pinching" pp. 64-65. Bold emphasis added.
I find this same phenomenon sometimes makes it difficult to write papers (I'm a physicist, not a mathematician though), because after having worked on a problem for a long time one loses perspective since after the fact the results may seem almost trivial
@Kai Indeed! I also find it to be an important aspect to grapple with when teaching: we are often too familiar with the given subject and need to actively empathize with out students, to remind ourselves that we too did not always find a subject trivial.
From a logic viewpoint, verifying a proof is a syntactic business, also known as symbol pushing, whereas understanding a proof is a semantic matter. I cannot resist giving a layman analogy as an appetizer: Many people can follow a recipe to bake a cake, but not as many can design the recipe or know how to tweak it to make something else.
Many of the best examples involve the proof of an existential statement that requires constructing a complicated witness for it. You may be familiar with the construction of the reals via Dedekind cuts of rationals or via Cauchy sequences of rationals, and these proofs can be easily checked step by step by any student who understands some basic mathematics, but how many students truly understand these constructions? Do they know that the Dedekind cut approach extends to completion of linear orders, while the Cauchy sequence approach extends to completion of metric spaces? Do they get some sense of wonder that both ways happen to produce the same result in the case of completing the rationals?
In a similar vein, one can construct the complex numbers by defining them to be the set $C = \mathbb{R}^2$ with $+,·$ defined by $(a,b)+(c,d) = (a+c,b+d)$ and $(a,b)·(c,d) = (ac-bd,ad+bc)$, and then checking that $(C,+,·,(0,0),(1,0))$ is a ring and that $(a,b)·(\frac{a}{\sqrt{a^2+b^2}},-\frac{b}{\sqrt{a^2+b^2}}) = (1,0)$ for every $(a,b) ∈ C ∖ \{(0,0)\}$. But in my view this proof ought to feel mysterious and unsatisfying unless you actually understand the motivation for these definitions and know the construction via field extension (i.e. $R[X]/(X^2+1)R[X]$). This is because there is a priori no reason at all for the above definitions of $+,·$ to make $·$ associative and distributive over $+$, unless you already understood that the reals can be extended to a field with some element $i$ such that $i^2+1 = 0$, and that the resulting field must be a vector space over the reals with dimension $2$, which together force $+,·$ to necessarily obey those definitions! Otherwise, you would be totally in the dark as to why those definitions should work, even though you can plainly see that they do.
These are examples from basic mathematics, but I hope it demonstrates how an existential statement might have a proof that does not in itself provide any understanding of the proof, in the same way you might observe that the cake mixture rises upon baking without having a clue why it does...
As for personal experience, there is one particular proof that I have never felt I really understood, even though I have a solid grasp of the formal proof itself:
$
\def\pa{\text{PA}}
$
Theorem: Let $T = \pa + \{ \ c>1 \ , \ c>1+1 \ , \ c>1+1+1 \ , \ \cdots \ \}$, where $c$ is a fresh constant-symbol. Then $T$ is conservative over $\pa$.
Proof: Take any arithmetical sentence $Q$ such that $\pa ⊬ Q$. Then $\pa+¬Q$ is consistent and so has a model (by completeness), which clearly finitely satisfies $T+¬Q$, and hence $T+¬Q$ has a model (by compactness). Thus $T ⊬ Q$.
Although I completely understand the completeness and compactness theorems, I somehow cannot intuitively understand how this proof works. Why should we have to invoke models? Can we instead directly show that every proof of any arithmetical $Q$ over $T$ can be converted into a proof of $Q$ over $\pa$?
No models are needed for the conservativity of T over PA. Given a proof in T of an arithmetical Q, let $n$ be larger than all the (finitely many) numbers $m$ such that $c>1+1+\dots+1\ \ (m$ summands) was used in the proof. Throughout the proof, replace $c$ with $1+1+\dots+1\ \ (n$ summands), and you have (modulo trivialities) a proof of Q in PA.
@AndreasBlass: Oh that's so simple, thanks!
Doron Zeilberger's proof of the alternating sign conjecture is perhaps an extreme case of this. He enlisted a large team of referees (88 apparently) to each verify a small part of the long but rather modular proof. A shorter and presumably more conceptual proof was later given by Kuperberg, and Zeilberger then used Kuperberg's methods to prove a refined version of the conjecture. No one of the referees could be said to understand the proof. Arguably, Zeilberger didn't have a good understanding of why the conjecture was true, past the nearly 100 elementary steps, since the second proof seems to have illuminated the conceptual concepts, leading to a proof of a stronger statement.
Here are the referee reports, which make for interesting reading! https://sites.math.rutgers.edu/~zeilberg/asm/REPORTS
This doesn't meet the criterion of being just a couple of pages long. If we allow this type of example then there are tons of examples of extremely computational proofs that only a computer can verify, and nobody "understands" except at a high level (four color theorem, Flyspeck, checkers is a draw, various results in extremal combinatorics, etc.).
Here the distinction is between proof P and proof Q of the same resultThe question involves the distinction between verifying and understanding proof P. It can’t always be said that understanding proof Q increases understanding of proof P.
@AaronMeyerowitz : Yes, that's a very good point. Kuperberg's proof gives additional insight into the theorem, but I don't think it helps anyone "understand" Zeilberger's proof; it's a completely different proof.
OK, I didn't see that requirement. But it's an interesting corner case. I agree that the proofs are different, as Zeilberger himself explains (his proof gives something stronger that Kuperberg's doesn't). But there are cases where an old elementary proof is known to be essentially the same as a new high-powered but more conceptual proof. Consider for instance things in number theory dating back to Fermat that are implicitly about constructing points on elliptic curves. Or proving no such points exist by infinite descent.
One such (rather unremarkable) example is the proof at https://mathoverflow.net/a/239931/36721, which I wrote myself, but now don't remember how it was devised. So indeed, I don't really understand it. :-)
Sergey Pinchuk gave a counterexample to the Real Jacobian Conjecture (that every polynomial map from $\mathbb{R}^n$ to itself whose jacobian determinant is everywhere nonvanishing is invertible). The verification is by high school level algebraic manipulations and takes 2 pages (the whole paper is less than 4 pages). But of course it does not give a clue about how he came up with the map. As he states in the paper, "The most difficult (and invisible) part of the present construction was to find polynomials $t$, $h$, $f$ with nontrivial Newton diagrams and satisfying Abhyankar's condition."
One example in logic is the proof that $A \Longrightarrow A$ in a "Hilbert-style" deduction system, where our only inference rule is Modus Ponens and we have the two axiom schemata
K: $\beta \Longrightarrow (\alpha \Longrightarrow \beta)$
S: $(\alpha \Longrightarrow (\beta \Longrightarrow \gamma)) \Longrightarrow ((\alpha \Longrightarrow \beta) \Longrightarrow (\alpha \Longrightarrow \gamma))$
What gibberish are these? Well, it turns out those are precisely what you need to prove the Deduction Theorem for these systems — that if $\alpha \vdash \beta$ (using $\alpha$ as hypothesis, one can prove $\beta$) then $\vdash \alpha \Longrightarrow \beta$ (with no hypotheses, one can prove $\alpha \Longrightarrow \beta$) — by providing an algorithm that transforms any proof of $\alpha \vdash \beta$ into a (three times as long) proof of $\vdash \alpha \Longrightarrow \beta$. The idea is that for any step $\gamma$ in the original proof, you have a step $\alpha \Longrightarrow \gamma$ in the transformed proof, and then you have a bunch of extra steps to make it all fit together. Axiom scheme S is precisely what you need in order to do a modus ponens $\beta \Longrightarrow \gamma, \beta \vdash \gamma$ when there is a $\alpha \Longrightarrow$ prefix on everything. Axiom scheme K lets you put that $\alpha \Longrightarrow$ prefix on things that are theorems anyway, to import them into the hypothetical realm. However, that part is just definitions / an axiom system, not the mysterious theorem.
To complete the proof transformation required for the Deduction Theorem, you do however also need to prove that $\vdash \alpha \Longrightarrow \alpha$, since this is what you get as the transformation of the step where your proof uses the hypothesis for the first time. Frege had this as a separate axiom ($\alpha \Longrightarrow \alpha$ is a tautology, as he did for K and S, rather than prove it from earlier theorems as he did for everything else (in the implicational fragment of propositional calculus).">at least sort of), but it turns out it can be proved from schemata K and S alone, in just five steps:
$(A \Longrightarrow ((A \Longrightarrow A) \Longrightarrow A)) \Longrightarrow ((A \Longrightarrow (A \Longrightarrow A)) \Longrightarrow (A \Longrightarrow A))$
instance of axiom scheme S, with $\alpha = \gamma = A$ and $\beta = A \Longrightarrow A$
$A \Longrightarrow ((A \Longrightarrow A) \Longrightarrow A)$
instance of axiom scheme K, with $\alpha = A \Longrightarrow A$ and $\beta = A$
$(A \Longrightarrow (A \Longrightarrow A)) \Longrightarrow (A \Longrightarrow A)$
from 1 and 2 by modus ponens
$A \Longrightarrow (A \Longrightarrow A)$
instance of axiom scheme K, with $\alpha = \beta = A$
$A \Longrightarrow A$
from 3 and 4 by modus ponens
Checking that this is a proof is trivial. Explaining how it works… is a different matter entirely.
Enter the Curry–Howard correspondence
The names K and S are not from formalised logic, but from the combinatory calculus: these are the standard names for the two higher order functions satisfying the identities
$K(x)(y) = x$
$S(f)(g)(x) = f(x)(g(x))$
TeXheads may prefer to think of those as the TeX macros that would be defined as
\def\K#1#2{#1} % A.k.a. \@firstoftwo
\def\S#1#2#3{#1{#3}{#2{#3}}} % Not a common utility, but should be
The fun thing about those two is that they generate the whole combinatory calculus — any lambda-term whatsoever can be mechanically translated to a composition of $K$ and $S$. In particular the identity function $I(x)=x$ can be so expressed, in the sense that
$S(K)(K)(x) = K(x)(K(x)) = x$ for all $x$, hence $I = S(K)(K)$.
Now the untyped combinatory calculus, like the untyped lambda calculus, is too powerful for many purposes (it lets you do anything), so a major activity in these parts of logic and foundations of computer science is to tame it by stamping types on everything. The basic system is the simply-typed theory, where you have some set of atomic types and the ability to make function types; $f : \alpha \to \beta$ means $f$ is a function that takes an argument of type $\alpha$ and returns an argument of type $\beta$. To get functions of several variables, you use currying, so instead of some $f : \alpha \times \beta \to \gamma$ you have $f : \alpha \to (\beta \to \gamma)$. For example the $K$ combinator has type $\alpha \to (\beta \to \alpha)$ since its result has the same type ($\alpha$, say) as the first argument whereas its second argument can have any type ($\beta$).
Similarly analysing the defining identity $S(f)(g)(x) = f(x)(g(x))$ of $S$, we may arbitrarily let $x : \alpha$ (we pick the name $\alpha$ for the type of $x$). Then $f$ and $g$ must both have some $\alpha \to$ type, as they both take $x$ as their first argument. $f(x)$ take $g(x)$ as its argument, but the type of $g(x)$ is not constrained, so let's call that $\beta$, making $g : \alpha\to\beta$. The type of $f(x)(g(x))$ is likewise not constrained, so let's call that $\gamma$. Then we get
$f : \alpha \to (\beta \to \gamma)$
$S : (\alpha \to (\beta \to \gamma)) \to ((\alpha\to\beta) \to (\alpha\to\gamma))$
Simply typed terms are too restrictive for proper programming — you can only write trivial programs — so there is an entire industry of designing more complicated type systems for doing more (though still less than the untyped calculus can) while still keeping the functions tame; at least this gives the theorists something to do. However we will only need to consider the $S(K)(K)$ expression for the identity, which is perfectly possible to type with simple types, provided one observes that we can't use the same types for the two $K$s (they will be two different instances of the same untyped combinator). If we take $A$ to be the concrete type of some $x$ that our $S(K)(K)$ should apply to, then the second $K$ must fit the pattern for a $g$ argument of $S$ and have a type of the form $A \to (B \to A)$ for some (so far not constrained) type $B$. The first $K$ must take $x$ as its first argument and $g(x) : B \to A$ as its second, so the first $K$ rather has type $A \to ((B \to A) \to A)$. This means $\alpha = \gamma = A$ and $\beta = B \to A$ in the typing of $S$, so our typed identity of type $A \to A$ is in fact
$S_{(A \to ((B \to A) \to A)) \to ((A \to (B \to A)) \to (A \to A))} ( K_{A \to ((B \to A) \to A)} )( K_{A \to (B \to A)} )$
where the indices show the exact type instance of the three combinators that is at hand.
As it turns out, that combinatory term also serves as a blueprint for the above proof that $A \Longrightarrow A$, because the instances of the axiom schemes K and S have exactly the same structure as the possible types of the combinators $K$ and $S$, if one substitutes the implication arrow $\Longrightarrow$ for the function type arrow $\to$ (and for simplicity replaces $B$s by $A$s as well). Modus ponens has the same structure as the type inference "if $x$ has type $\beta$ and $f$ has type $\beta \to \gamma$, then $f(x)$ has type $\gamma$". This is (an elementary instance of) the Curry–Howard correspondence.
Opinions vary as to whether this is is a Deep Insight into the way logic really works, or just a funny coincidence. People can get ideological about it. For me, the only way I can reproduce that 5 step proof is to derive it by assigning types in $S(K)(K)$.
Perfect example. I just read this and immediately I fall back to not understanding the point of the Deduction Theorem. How could it not be the case? How would you even think that it is a thing to doubt?
@Mitch : Surely as soon as one grasps the distinction between $\vdash$ and $\implies$ one is led to ask what the relationship between them is?
@TimothyChow. Of course. $\rightarrow$, $\Rightarrow$, $vdash$, $\vDash$ all so distinct. But sometimes you just... it's all so... just look at them... side by side... it's so obvious!
@Mitch: It's obvious on the semantic side, but formalised logic is about what you can establish on the syntactic side. Gödel couldn't arithmetise the concept of truth, because that is semantic, but he could arithmetise the concept of provability because that is (in the modern view) strictly syntactic.
Maybe the fact that the Jones polynomial is a well-defined invariant of knots qualifies. If you're willing to accept Reidemeister theorem, whose proof is itself somewhat easy to follow though more delicate than one might think, then using the skein relation this is really straightforward and can be explained to someone with no mathematical background and basically boils down to drawing pictures. So checking that it works is easy, but the fact that it works is fairly miraculous, and understanding this involves some deep mathematics.
Isn't the same true of most knot invariants? Especially the ones like this one which are given in terms of the knot diagrams
@Wojowu Not, e.g., the Alexander polynomial.
@Wojowu the same case could be make about the HOMFLY polynomial, but apart from that all the "classical" knot invariants are either super elementary or have a clear topological definition (or both) e.g. knot group, linking number, crossing number,... On the other hand, others "quantum" invariant do not have such a simple rule to compute them, the only exception being the Alexander polynomial which in a way is both classical and quantum, and as Will say does have a clear topological meaning.
@Adrien Thank you, I'm showing off my ignorance in the field of knot theory; my little experience with the field is through the more elementary invariants presented via knot diagrams. Sometimes I'm forgetting knot theory is a subfield of topology and not combinatorics :)
For myself, I would mention the Lee-Yang theorem (see a mathoverflow post here or the book of David Ruelle "STATISTICAL MECHANICS Rigorous Results".
I admit that I have checked the proof many times. But I still don't really understand what happens there nor have any physical insight why this result should be true.
A variation on this is the "not rediscoverable proof". This is a proof of a result which you can verify, and perhaps even understand why it is a proof, but which you could not discover on your own just given the theorem statement. I think this arises from studying an area, and then finding a surprising and unexpected consequence. Perhaps this speaks more to the nature of understanding, in that a deductive consequence come as such a surprise.
The amazing example of this I have found is Lovasz's cancellation theorem for finite structures, from one of his first published papers. Even after having reviewed the proof many times, I still could not imagine how I could come up with it. More technical detail can be found in an answer of Eric Wolsey, with some of my commentary nearby. https://mathoverflow.net/a/269545/ .
Gerhard "Understands Understanding Isn't Very Understandable" Paseman, 2020.06.12.
A good proof is a proof that makes us wiser. If the heart of the proof is a voluminious search or a long string of identities, it is probably a bad proof. If something is so isolated that it is sufficient to get the result popped up on the screen or a computer, then it is probably not worth doing. Wisdom lives in connections. If I have to calculate the first 20 digits of π by hand I certainly become wiser afterwards because I see that these formulas for π that I knew take too much time to produce 20 digits. I will probably devise some algorithms which minimize my effort. But when I get two millions of digits of π from the computer using somebody else's library program I remain as stupid as I was before.
Yuri I. MANIN, Good Proofs are Proofs that Make us Wiser
While I agree very dearly with erz's comment that every mathematical result/proof/... is like that, I also think that there are some examples that showcase this better than other ones.
A large portion of commutative algebra would make for a good example. It is not too hard to verify many of the standard textbook results but to really understand the arguments, one might need to think about related theories like algebraic geometry.
It is not too bad to verify proofs of Noether Normalization or Lying Over or Insert More Examples but what do they really mean?
It is easy to verify that affine varieties induce morphisms of algebras and vice versa algebra-homomorphisms between coordinate rings induce morphisms of varieties. And it is easy to prove/verify that they are compatible in a natural way. But when I studied these things in my second year at university I didn't understand why we wanted to prove this or why we should even expect this to work.
After learning just the fundamentals of category theory and bits of geometry, however, one would realize that there is some algebro-geometric correspondences going on and that this is just the most natural way of defining an equivalence of categories, so of course it had to work.
Some of this might not even be considered "verifying a proof" but rather "verifying a definition" - which clearly should happen even more often, since it is oftentimes easy to "verify" a definition but years have gone into the development of the definition. (I'm thinking of the definition of a topology, or e.g. definitions of model category and $\infty$-categories, and so on and so on.)
It is rather frequent that results on algebra get illuminated by geometry. For me, this is "understanding" of the commutative algebra facts. But, for some people the geometric view just adds another layer of confusion.
A Robbins algebra is an algebra including a single binary relation satisfying associativity, commutativity, and the "Robbins axiom" $\neg \left( \neg \left(a \lor b \right) \lor \neg \left(a \lor \neg b \right) \right) = a$.
McCune's proof that Robbins algebras are synonymous with Boolean algebras is a relatively famous application of computer-assisted automated theorem proving; the output of the computer was small enough to be checked "by hand". This is on the "flip-side" of, for example, the Four-Color Theorem, wherein the output is likely not human-readable and more trust must, in a sense, be placed in the computer.
My limited understanding is that, although human-readable, the proof per se lacks intuition - it isn't clear how a human could have come up with the proof of the Robbins conjecture.
I expect that a human, or group of humans, could have come up with a proof of this by doing reams and reams of inferences from axioms. It's not so much the presence of the computer that induces the lack of intuition, but the manner in which the proof was constructed.
George Boolos wrote a paper entitled Gödel's Second Incompleteness Theorem Explained in Words of One Syllable wherein, indeed, he does explain the second incompleteness theorem using only words of one syllable. I am not sure if the monosyllabic explication offered in the beginning of the paper is an example of verifying a proof, as the explication more or less just states results as opposed to proving them or even sketching what the proofs look like. However, I believe that the second half of the paper fills in enough of the details for any mathematician to follow the proof and verify the result, but lacks enough of the formal definitions and technical machinery to allow someone to fully understand the proof. A mathematician who has no background in mathematical logic reading this paper could easily verify the proof on the third page starting from "We may prove Gödel's second incompleteness theorem," but I highly doubt they could give an explanation of what is really going on 'under the hood' of the proof, or any of the corollaries and consequences of the theorem, nor do I think would they be able to use the same techniques used in this proof to prove other results in computability theory or proof theory just from reading this paper alone.
To add to this, I have never seen one of these "In words of one syllable..." or "In words of two syllables or less..." things that did not end up acting as a reductio ad absurdum of requests to express things in layman's terms. That is to say, putting such restrictions only makes things harder to understand for beginners, more opaque and less motivated. The problem is that when people don't understand something, they don't know how to ask for help with understanding it that they can actually use.
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2025-03-21T14:48:31.199512
| 2020-06-09T16:48:29 |
362628
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What does hyperbolicity of curves buy us in the arithmetic context?
This is going to be a fairly vague question but hopefully it will have concrete answers:
There is this recurrent phenomenon in the arithmetic of curves (possibly stacky,affine) where there is a "phase transition" associated with the euler characteristic being positive, zero or negative.
For instance, Mordell's conjecture (Falting's theorem) states that there are finitely many integral points if the euler characteristic is negative. In fact, this theorem is true not only for affine curves but also for stacky curves.
Another example is the failure of local-global principles for integral points for euler characteristic $\leq 0$.
Question: Is there some key set of features in the "hyperbolic" case that come up often in proofs?
Let me illustrate the kind of answer I am looking for with reference to the complex case. In this case, negative euler characteristic corresponds exactly to having the hyperbolic disc as the universal covering space. So a very common argument is that given a map to a hyperbolic curve, we can lift it in some appropriate sense to the universal cover and then by Liouville, this lift will often be constant.
Is there an analogous collection of arithmetic hyperbolic techniques to keep in mind?
I am not entirely sure what you are looking for but a powerful tool in arithmetic hyperbolicity is Diophantine approximation and Nevanlinna theory. Vojta has some nice notes about (https://math.berkeley.edu/~vojta/cime/cime.pdf). Another useful theorem which is used in Diophantine approximation is the Schmidt Subspace Theorem. Yuri Bilu has a wonderful Bourbaki survey about this theorem and its applications (http://www.numdam.org/article/AST_2008__317__1_0.pdf). Finally, Ariyan Javanpeykar has a nice survey about hyperbolicity (https://arxiv.org/pdf/2002.11981.pdf). Hope this is helpful!
I am not entirely sure what I am looking for either but your links seem to be along the right lines! Thank you!
|
2025-03-21T14:48:31.199699
| 2020-06-09T17:10:41 |
362631
|
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|
Stack Exchange
|
Ext between a module and its higher Auslander-Reiten translate
Let $A$ be a representation-finite algebra and $M$ an indecomposable module with finite projective dimension $g >0$.
Question 1: Do we have $dim(Ext_A^g(M, \tau_g(M)))=1$? Here $\tau_g(M)=\tau ( \Omega^{g-1}(M))$.
Question 2: Could this even be true when $A$ is not represenation-finite? Probably not, but the computer did not find a counterexample yet.
Question 1: I think that the answer is no. We have
$$\operatorname{Ext}^g_A(M,\tau_g(M)) \simeq \operatorname{Ext}^1_A(\Omega^{g-1}_A(M), \tau(\Omega^{g-1}_A(M)).$$
Look at the following example in QPA:
gap> Q := DynkinQuiver( "A", 5, [ "l", "l", "r", "r" ] );
<quiver with 5 vertices and 4 arrows>
gap> KQ := PathAlgebra( GF(17), Q );
<GF(17)[<quiver with 5 vertices and 4 arrows>]>
gap> arrows := ArrowsOfQuiver( Q ) * One( KQ );
[ (Z(17)^0)*a_1, (Z(17)^0)*a_2, (Z(17)^0)*a_3, (Z(17)^0)*a_4 ]
gap> a1 := arrows[ 1 ];
(Z(17)^0)*a_1
gap> a2 := arrows[ 2 ];
(Z(17)^0)*a_2
gap> a3 := arrows[ 3 ];
(Z(17)^0)*a_3
gap> a4 := arrows[ 4 ];
(Z(17)^0)*a_4
gap> rels := [ a2 * a1, a3 * a4 ];
[ (Z(17)^0)*a_2*a_1, (Z(17)^0)*a_3*a_4 ]
gap> A := KQ / rels;
<GF(17)[<quiver with 5 vertices and 4 arrows>]/<two-sided ideal in <GF(17)[<quiver with 5 vertices and 4 arrows>]>,
(2 generators)>>
gap> S3 := SimpleModules( A )[ 3 ];
<[ 0, 0, 1, 0, 0 ]>
gap> g := ProjDimensionOfModule( S3, 4 );
2
gap> OmegaS3 := NthSyzygy( S3, g - 1 );
<[ 0, 1, 0, 1, 0 ]>
gap> DTrOmegaS3 := DTr( OmegaS3 );
<[ 1, 0, 0, 0, 1 ]>
gap> ext := ExtOverAlgebra( OmegaS3, DTrOmegaS3 );
[ <<[ 1, 0, 0, 0, 1 ]> ---> <[ 1, 1, 0, 1, 1 ]>>,
[ <<[ 1, 0, 0, 0, 1 ]> ---> <[ 1, 0, 0, 0, 1 ]>>, <<[ 1, 0, 0, 0, 1 ]> ---> <[ 1, 0, 0, 0, 1 ]>> ], function( map ) ... end ]
gap> Length( ext[ 2 ] );
2
|
2025-03-21T14:48:31.199808
| 2020-06-09T17:11:26 |
362632
|
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|
Stack Exchange
|
Question on induction of unramified representations
$\def\anonabs{\lvert\cdot\rvert}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\SO{SO}
$Let $F$ be a $p$-adic local field of characteristic zero.
Let $\chi$ be an unramified character of $F^{\times}$ and $\anonabs$ the absolute value.
Then it is well known that the normalized induced representation $\Ind_{B_2}^{\GL_2}(\chi\anonabs^{1/2} \boxtimes \chi\anonabs^{-1/2})$ has a sub-representation $\chi\circ \det_{\GL_2}$, where $B_2$ is a Borel subgroup of $\GL_2$.
Let $\pi$ be an unramified admissible smooth representation of $\SO_n$.
Then I am wondering that whether the set of all irreducible unramified constituents of $\Ind_{(\GL_1 \times \GL_1) \times \SO_n}^{\SO_{n+4}}((\chi\anonabs^{1/2} \boxtimes \chi\anonabs^{-1/2}) \boxtimes \pi)$ are the same as those of $\Ind_{\GL_2 \times \SO_n}^{\SO_{n+4}}(\chi\circ \det_{\GL_2} \boxtimes \pi)$. If is it true, how does the unramified condition play a role?
(Here, $(\GL_1 \times \GL_1) \times \SO_n$ and $\GL_2 \times \SO_n$ are Levi subgroups of parabolic subgroups of $\SO_{n + 4}$, respectively.)
Surely you mean to consider parabolic induction (thus requiring you to pick two parabolic subgroups of $\operatorname{SO}_{n + 4}$), not induction directly from the Levi?
first, it seems that the one dimensional representation of GL(2) is a quotient of your given induced representation if I am not mistaken. The twisted steinberg representation should be the sub. Second, that assertion is true. Here unramified plays a role because the representation induced from an unramified character has only one unramified component. in the GL2 case in your example, that is the one dimensional representation if the given character is unramified. The steinberg representation is not unramified, and thus the representation induced from St has no unramified component.
@LSpice, Yes! I mean parabolic induction!
@QZ0, Thank you very much! Right! I should have said that the one dimension representation is a quotient of the induced representation. For the second, you mean the set of all irreducible unramified constituents of $\pi_1=\text{Ind}{(GL_1 \times GL_1) \times SO_n}^{SO{n+4}}((\chi |\cdot|^{1/2} \boxtimes \chi|\cdot|^{-1/2}) \boxtimes \pi)$ are the same as that of $\pi_2=\text{Ind}{GL_2 \times SO_n}^{SO{n+4}}(\chi\circ \det_{GL_2} \boxtimes \pi)$. But the set of irreducible ramified representations of $\pi_2$ might be properly contained in that of $\pi_1$. Am I right?
|
2025-03-21T14:48:31.199977
| 2020-06-09T19:28:35 |
362641
|
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|
Stack Exchange
|
Symmetric powers and Frobenius
Let $X$ be a reduced affine scheme of characteristic $2$.
The symmetric square is $S^2 X := (X \times X)/\Sigma_2$. In studying the map $X^2 \to S^2 X$, one quickly observes the following. We have $X^2 = \Delta \coprod (X^2\setminus \Delta)$, where $\Delta \simeq X$ is the diagonal. This is a $\Sigma_2$-equivariant stratification and hence induces a stratification of the quotient $S^2 X = Z \coprod U$, where $Z$ is the image of $\Delta$ and $U$ is the image of $X^2 \setminus \Delta$. Moreover (and this is the main point for me), the map $X^2 \setminus \Delta \to U$ is étale, $Z \simeq X$ and the map $X \simeq \Delta \to Z \simeq X$ is the absolute Frobenius.
My question is: Does this generalize the higher symmetric powers and higher characteristics? Namely, if $X$ is a reduced affine scheme of characteristic $p>0$, is there a (natural) stratification of $X^n$ such that along the strata, the map $X^n \to S^n X$ can be described in terms of étale maps and Frobeniusse?
(I assume that, if true, this should not be particularly hard to establish. So I suppose this is mainly a reference request.)
$\coprod$ \coprod is meant as a large operator (like $\prod$ \prod), and so is both too large and badly spaced. There appears to be no universally agreed notation for the infix version, but $\sqcup$ \sqcup and $\amalg$ \amalg are common choices. Would you be amenable to replacing $\coprod$ by either of those? (Also, is 'Frobeniusse' a typo, or an exotic plural of 'Frobenius'?)
The ramification divisor of the map $X^n\rightarrow S^nX$ is the union of the diagonals $\Delta _{ij}$ defined by $x_i=x_j$, it has degree 2 onto its image. If $p>2$ I don't see how it could fit into your picture.
@abx That’s only a divisor if $X$ is a curve, right?
For example if A has characteristic p then the image of the map $(A^{\otimes n})^{\Sigma_n} \to A$ is $A^p$, so the restriction to the diagonal is just the Frobenius again.
|
2025-03-21T14:48:31.200153
| 2020-06-06T20:09:51 |
362366
|
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|
Stack Exchange
|
Upper-triangular matrices as union of centralizers of cyclic elements
Let $p$ be a prime number and $G=GL_n ( \mathbb{Z} / p \mathbb{Z}
)$ such that $n\leq p$. Consider the set $U$ of upper-triangular matrices of $G$
having entries of $1$ on the diagonal. The cardinality of $U$ is $p^{\frac {n(n-1)} 2}$ and $U$ is a subgroup of $G$, in particular $U$ is a Sylow $p$-subgroup of $G$. Recall that an element $M$ in $G$ is said to be a cyclic matrix if the characteristic polynomial of $M$ is equal to its minimum polynomial. Let $C$ be the set of cyclic matrices of $U$.
Question: Can $U$ be written as $U=\bigcup_{M\in C}C_{U}(M)$?
Any help would be appreciated so much. Thank you all.
In other words, does every element of $U$ commute with a matrix in $M\in U$ such that $M-I_n$ has rank $n-1$?
For $n=3$ the centralizer of $1+E_{12}$ in $U$ consists of matrices $1+aE_{12}+bE_{13}$. No such matrix is cyclic.
In fact, for $n\leq p$, centralisers of cyclic matrices cover all the elements in $GL(n,p)$. See Proposition 5.1 in [][https://sci-hub.tw/10.1007/s10801-011-0288-2].
Can you give the reference to your link? it's blocked where I am.
Anyway, my argument is correct. It's true that $1+E_{12}$ belongs to the centralizer to a cyclic matrix $M$, but $M$ cannot be chosen in $U$ (which is your requirement).
See Theorem 1.4 in "Azizollah Azad · Mohammad A. Iranmanesh · Cheryl E. Praeger · Pablo Spiga, Abelian coverings of finite general linear groups and an application to their non-commuting graphs, J Algebr Comb (2011) 34:683–710
DOI 10.1007/s10801-011-0288-2".
OK thanks, so I confirm the assertion from this paper doesn't rule out my (trivial) counterexample. That proposition doesn't deal with $U$, but with the whole linear group. Computations of centralizers in $U$ for $n=3$ are quite trivial.
Thank you so much for the time and effort taken to answer the question.
Anyway Marc Wildon already commented your previous question about the centralizer of $I_3+E_{12}$.
|
2025-03-21T14:48:31.200308
| 2020-06-06T20:42:23 |
362367
|
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|
Stack Exchange
|
Fourier analysis and fractional calculus
Do Fourier transform properties still hold in the case of fractional derivatives ?
i.e I have seen many times that some lectures define fractional derivative as :
$$\frac{d^{\alpha}}{dx^{\alpha}}f=\mathscr{F}^{-1}\big[\mathscr{F}[f(x)](w)\cdot w^{\alpha}\big](x)$$
Indeed fractional derivatives of exponentials do not really look like exponentials ...
Thanks
Too long for a comment. Your spelling of the name of the mathematician Joseph Fourier is incorrect. Also your formula is almost impossible to read: your fractional derivative on the lhs is the Fourier multiplier $(i\xi)^\alpha$ and thus formally you find
$$
\left(\left(\frac{d}{dx}\right)^\alpha f\right)(x)=\int e^{i x\xi} (i\xi)^\alpha \hat f(\xi) d\xi/(2π).
$$
You may also say that the inverse Fourier transform of the homogeneous $(i\xi)^\alpha$, say for $\Re\alpha>-1$, is also an homogeneous distribution of degree $-1-\alpha$ and that the above fractional derivative is the convolution with
$$
c_+(\alpha) x_+^{-\alpha-1}+c_-(\alpha) x_-^{-\alpha-1}=w_\alpha(x),
$$
where $c_\pm(\alpha)$ are constants so that the formal formula is
$$
\left(\left(\frac{d}{dx}\right)^\alpha f\right)(x)=\int f(y) w_\alpha (x-y) dy.
$$
Note that if $-\alpha-1$ happens to be a negative integer $-k$, then we have
$
x_+^{-k}=c_k \delta_0^{(k-1)},
$
where $\delta_0^{(k-1)}$ stands for the $(k-1)$th derivative of the Dirac mass at 0.
|
2025-03-21T14:48:31.200419
| 2020-06-06T20:49:11 |
362368
|
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|
Stack Exchange
|
Two $\mathbb Z$-algebra structures on $\mathbb Z\otimes_{\mathbb S} R$
$\newcommand{\Sph}{\mathbb S} \newcommand{\Z}{\mathbb Z} \newcommand{\F}{\mathbb F}$
In this question I will abuse notation by writing $A$ for the (generalized) Eilenberg-MacLane spectrum associated to the abelian group (chain complex) $A$.
Let $R$ be a usual (associative, unital) ring. My question is about the two natural $\Z$-algebra structures on $\Z\otimes_\Sph R$ : the first one "by inclusion to the left", and the second one by $\Z\to R$ followed "by the inclusion to the right".
A priori, one shouldn't expect these to be equivalent $E_1-\Z$-algebra structures, and in fact in general they aren't - I think I have an argument that proves it for $R=\mathbb F_2$ (I'll sketch the argument below)
They are however equivalent in certain cases, e.g. when $R$ is a generalized monoid algebra (over $\Z$), and so my (perhaps too broad) question is :
What are some interesting conditions on $R$ for them to be equivalent ? inequivalent ?
(to make the statement precise, I'm mainly interested in the $E_1-\Z$-algebra structure; but I'm also interested in the $E_n$-structure when $R$ is commutative, $n\in\mathbb N_{\geq 2}\cup\{\infty\}$)
Perhaps a first step would be to try to generalize the fact for $\mathbb F_2$ to odd primes :
Is it true that they're inequivalent for $R=\mathbb F_p$ for all odd primes $p$ ?
I suspect this should be true : indeed suppose they were equivalent as $E_1-\Z$-algebras; then by tensoring with $\F_p$ over $\Z$ we get $\F_p \otimes_\Sph \F_p \simeq \Z\otimes_\Sph (\F_p\otimes_\Z \F_p) \simeq \Z\otimes_\Sph \F_p[\epsilon]$ where $\epsilon^2= 0$ and $|\epsilon|=1$.
Then we get, by the Thom isomorphism and using a description of $\F_p$ and $\Z$ as Thom spectra as described e.g. in A simple universal property of Thom ring spectra (Antolin-Camarena and Barthel) : $$\F_p\otimes_\Sph\Sigma^\infty_+\Omega^2S^3 \simeq \F_p[\epsilon]\otimes_\Sph \Sigma^\infty_+\Omega^2S^3\langle 3\rangle$$ as $E_1$-algebras (where $S^3\langle 3\rangle$ is the $3$-connected cover of $S^3$)
So essentially this would be saying that "removing" the degree $1$ class on the left and "adding it freely, with a zero square" wouldn't change the algebra structure, which seems rather unlikely.
For $p=2$, we know (by a description of $\pi_*(\F_2\otimes_\Sph \F_2)$) that the degree $1$ class on the left is not nilpotent, so it indeed fails badly, which gives the result I mentioned earlier.
For odd $p$, such a simple argument does not work (as the degree $1$ class on the left obviously squares to $0$), but I still suspect that the equivalence does not hold. In fact the term on the left has $\pi_*$ the dual mod $p$ Steenrod algebra so I think a careful analysis of that one could lead to a proof, but I don't know enough about it.
I'm also interested in other examples - I've been told that there might be examples of inequivalence when $R$ is a ring of algebraic integers, but unfortunately I don't know how to attack that question so :
Are there nice examples of rings of algebraic integers where the inequivalence holds ? Is there some characterization of those ?
|
2025-03-21T14:48:31.200652
| 2020-06-06T20:59:52 |
362369
|
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|
Stack Exchange
|
Is the class of power-associative binars finitely axiomatizable?
A binar is simply a set $S$ equipped with a single binary operation $*$. A power-associative binar is a binar where the subalgebra generated by a single element is associative. Equivalently, they can be axiomatized with the infinite set of equations, $\{(x*x)*x=x*(x*x), ... \}$. Is there some finite set of axioms in the signature $*$ that can axiomatize power-associativity?
I can think of some two variable identities that would imply power associativity, but would be strictly stronger. Are you sure a finite basis exists? Gerhard "Trying Not To Get Hyper" Paseman, 2020.06.06.
Try the following for a counterexample. Let an algebra be generated by x and have standard power arithmetic up to n. Then, whenever two items have exponents which sum to a prime bigger than n, let the kth power times the jth power be equal to a symbolic term x to the power (k,j). In multiplying these, you can collapse the products as you like, as long as you keep (k,j) distinct from (j,k). Gerhard "Always Look For Prime Examples" Paseman, 2020.06.06.
No.
Indeed, let $\mathcal{V}_n$ be the variety of magmas generated by the relating identities with variable $y$ saying that for every $k\le n$, all products of $k$ copies of $y$ are equal. Since the variety of power-associative magmas is $\bigcap_n \mathcal{V}_n$, a negative answer to the question is equivalent to showing that for every $n$ the relatively free magma $M_n$ on 1-generator $x$ in $\mathcal{V}_n$ is not associative.
Write by induction $x^1=x$, $x^k=xx^{k-1}$. I claim that in $M_n$ we have $x^{n+1}\neq x^nx$. Indeed, $x^nx$ can be rewritten in all ways $(ab)x$ with $a,b$ products of $k,\ell$ copies of $x$ (in some order) for $k+\ell=n$. No such $(ab)x$ is a relator, and no nontrivial relator has the form $(x=\dots)$. Hence $(ab)x$ can only be transformed by a relator substitution into another $(a'b')x$.
(This actually shows that $\mathcal{V}_{n+1}$ is properly contained in $\mathcal{V}_n$.)
In case it helps, here is a pedestrian definition of $M_k$. Let $J_1(x)$ be the free magma on 1 generator $x$ with chosen total order; enumerate pairs $(P_i,Q_i)$ of distinct elements of $J_1$ with $P_i<Q_i$ and $P_i,Q_i$ of the same size; choose the enumeration in increasing order of size, e.g. $P_0=x(xx)$, $Q_0=(xx)x$, $P_1=x(x(xx))$, $Q_1=(xx)(xx)$, $P_2=(xx)(xx)$, $Q_2=((xx)x)x$, etc. Write $i_n=\max{i:|P_i|\le n}$.Then $M_n$ is the quotient of $J_1$ by the magma equivalence relation (= quotient inherits magma stricture) generated by $P_i(z)=Q_i(z)$ for all $z\in J_1$.
In the comment, you should add "...for $i\leqslant i_n$", right?
@მამუკაჯიბლაძე Yes, sorry, of course this why I introduced $i_n$.
Also in the answer, where you write "...is not associative", you actually need stronger statement "...is not power-associative", right? I believe this was part of the objection of @user158834
@მამუკაჯიბლაძე no it's correct (and correcting as you're suggesting would be correct too): for a 1-generated magma, associative and power-associative are the same by definition. I understand user158834's objection as understanding that $M_n$ is defined as the magma with finite presentation: 1 generator $x$, and, as relators, all associative relations of length $\le n$ w.r.t. $x$ (instead of taking also substitutions using these relations).
PS: the answer by user158834 was downvoted because of its initial tone, but what remains of it now is very interesting information.
It seems like you proved that there is no finite set of relations in one variable which characterize the power-associative magmas (because all such relations are necessarily reassociations, which are contained among the $\mathcal V_n$ relations), but there is still some work to be done to show that using more variables doesn't help. For example $((xx)x)y=(x(xx))y$ is also true in all power-associative magmas so it could conceivably be part of a basis.
@MarioCarneiro I don't follow your point. The notion of variety encompasses equalities with an arbitrary number of variables, and in particular $((xx)x)y=(x(xx))y$ is an equality of $\mathcal{V}_n$ for all $n\ge 3$.
Obviously my example relation is derivable for $n\ge 3$ (and I would be hard pressed to come up with an example that is not like this since I believe the theorem to be true), but as you defined it that relation is not literally contained among the specified finite set of relations used to define $\mathcal V_n$ for any $n$, and I don't see how you exclude the possibility that all the infinitely many basis relations follow from some relation on 2 or 3 variables, plus some 1 variable associativity assumptions from some $\mathcal V_n$.
Suppose I have a finite set of relations $R$ involving possibly multiple variables, and I take $n$ to be bigger than everything about $R$ (sum of number of symbols or something). It seems to be an implicit part of your argument that $\mathcal V_n\subseteq\mathcal V_R$, where $\mathcal V_R$ is the variety of magmas satisfying the relations $R$. To prove this we would need to show that each relation in $R$ follows from power-associations of $x^i$ for $i\le n$. In the example you can factor out the $(xx)x=x(xx)$ part but I don't see how to conclude this is always possible.
@MarioCarneiro If there's an implicit part, let me write it in the following more precise way. If you have an arbitrary sequence of magma varieties $V_n$ with $V_n\supset V_{n+1}$, define $V_\infty=\bigcap V_n$ and if $F$ is any finite set of identities of $V_\infty$, then there exists $n$ such that $F$ is a set of identities in $V_n$. (Indeed if $V_*(m)\subset F_m\times F_m$ is the set of relations in $m$ variables, $F_m$ free magma, then $V_\infty(m)=\bigcup_n V_n(m)$, ascending union.)
I see it now. The "$n$ large enough" here is based on the least $n$ which proves the relations, which we know must exist by compactness (applied to provability of each relation in $V_\infty$). Thank you.
The question has been answered, but I will add some remarks
about magma/groupoid/binar. This is in response to some
of the comments on this page:
What you can currently read on the English Wikipedia:
The term groupoid was introduced in 1927 by Heinrich Brandt
describing his Brandt groupoid (translated from the German Gruppoid).
The term was then appropriated by B. A. Hausmann and Øystein Ore (1937)
in the sense (of a set with a binary operation) used in this article.
…
The term magma was used by Serre [Lie Algebras and Lie Groups, 1965].
It also appears in Bourbaki's Éléments de mathématique, Algèbre,
chapitres 1 à 3, 1970.
What you can currently read on the French Wikipedia:
L'ancienne appellation « groupoïde de Ore », introduite par Bernard Hausmann et Øystein Ore en 1937 et parfois utilisée jusque dans les années 1960, est aujourd'hui à éviter, l'usage du terme groupoïde étant aujourd'hui réservé à la théorie des catégories, où il signifie autre chose.
[The old name groupoïde de Ore, introduced by Bernard Hausmann and Øystein Ore in 1937 and sometimes used until the 1960s, is to be avoided today, the use of the term groupoid being today reserved for category theory, where it means something else.]
What is currently not on either Wikipedia:
A. (magma)
Peter Shor of MIT has speculated that magma
might have been introduced
as a pun on Ore's name.
To distinguish a groupoid in the sense of Ore
from a groupoid in the sense of
Brandt, the phrase groupoid of Ore was used first,
then shortened to
magma, which in geology means a pile of molten ore.
B. (binar)
In June 1993, there was a conference at MSRI on
Universal Algebra and Category Theory
organized by Ralph McKenzie and Saunders Mac Lane.
At this conference, there was a discussion session where
several topics were discussed, including whether there
could be general agreement on the future
use of the word groupoid.
At this discussion session, George Bergman of
Berkeley proposed using binar to mean an algebraic structure
with a single binary operation. (He also proposed
unar for an algebraic structure with a
single unary operation.) It seemed to me that
Bergman thought this up on the spot, so it is plausible
that he is the source of binar.
I recall remarking to George Bergman around that time that there was a previous usage, from Star Trek TNG. I think he had the idea before the conference. Gerhard "Decided Not To Use Klingon" Paseman, 2020.06.07.
I've asked J-P. Serre about the invention of the term magma: his answer is "Il me semble que c'est Bourbaki qui a inventé ça. Autant que je souvienne, il voulait un mot qui n'ait pas été déjà utilisé en algèbre, et qui suggère quelque chose n'ayant aucune structure intéressante. Je crois que nous avions pensé à "fourbi", mais nous l'avons rejeté. En tout cas ça n'a rien à voir avec Ore." English rough translation: "I think Bourbaki invented this. As far as I know, he wanted a word not yet used in algebra, suggesting something that has no interesting structure. (...)
(...) I think we thought of "fourbi" but we rejected it. In any case it's unrelated to Ore"
@YCor Thanks for thi comment. Since I learned the word "magma" I intuitively thought of it as a kind of primordial chaos. I was appalled by the explanation in this answer, and I'm glad to know that it's not true.
The m-word sounds like an awful mockery of a name whether or not it is related to Ore. Also, “something that has no interesting structure” (primordial chaos) completely misses the point: Bourbaki, in their infinite wisdom, were apparently unaware that algebraists are not so much interested in the class of all groupoids (there is not much to say about it in this generality, and what is there to say also applies to algebras in other signatures), but they usual study varieties of groupoids satisfying this or that additional identity, precisely because they do have interesting structure.
@EmilJeřábek "magma" is a respectable geologic term which is not considered there as offensive or not pejorative whatsoever, so your superlatives are somewhat exaggerated. If Bourbaki gave them a name, this means they thought it was worth a name. As regards groupoids, I remember spending hours (I don't exaggerate) trying to figure out what a groupoid on 1 generator is, before I learnt that it was just used to mean a magma (I understood it in the meaning of small category with inverses).
@YCor I can't say I understand the point of your anecdote, but note that it's fairly common that the same word is used in different meanings in different fields, and if you want to read papers in some field, it sure helps to first make oneself familiar with standard terminology in that field. This is not going to be helped by Bourbaki inventing a new word and trying to impose it from outside on practitioners in the field.
@EmilJeřábek What makes you believe anybody wants to impose a new word? It was created and "binar" also was, as far as I understand. I don't believe anyway that some subcommunity is owner of the terminology. For sure the fact that "groupoid" was mainly used for another meaning was a good reason to create a new one. Searching MO questions with "groupoids" yields 370 answers and if I'm correct, the totality of the 50 first answers points to the "small category" meaning. The point of my anecdote is that using without notice "groupoid" in a (now) rare meaning can lead to confusion.
|
2025-03-21T14:48:31.201774
| 2020-06-06T21:40:32 |
362373
|
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|
Stack Exchange
|
Central simple algebras via cohomology
I am following the book Central Simple Algebras and Galois Cohomology, by Gille and Szamuely (I am using the second edition). In section 2.4, the authors remark that the tensor product induces a natural map $\mathrm{CSA}_K(m)\times \mathrm{CSA}_K(n)\to \mathrm{CSA}_K(mn)$. Since we have previously identified degree $n$ central simple algebras split by $K$ with cohomology classes in $\mathrm{H}^1(\mathrm{Gal}(K/k),\mathrm{PGL}_n(K))$, there is a corresponding map
$$\mathrm{H}^1(\mathrm{Gal}(K/k),\mathrm{PGL}_m(K))\times \mathrm{H}^1(\mathrm{Gal}(K/k),\mathrm{PGL}_n(K))\to \mathrm{H}^1(\mathrm{Gal}(K/k),\mathrm{PGL}_{mn}(K))$$
The authors claim that this corresponding map is the one induced by the natural map $\mathrm{GL}_m(K)\times \mathrm{GL}_n(K)\to\mathrm{GL}_{mn}(K)$. I don't see why this is true. Any help?
To see the claim, you need to think through how the identification $\mathrm{CSA}_K(m) = \mathrm{H}^1(\mathrm{Gal}(K/k), \mathrm{PGL}_n(K))$ works. So it is hard to answer the question without knowing how well you understand that identification.
As @TheoJohnson-Freyd comments, the answer to your question is entirely dependent on your picture/explication of the first "identification" map. That is, not all of this is purely dictated by some universal functorial stuff...
|
2025-03-21T14:48:31.201885
| 2020-06-07T00:32:49 |
362381
|
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|
Stack Exchange
|
Examples of non-zero negative Steenrod operations
In JP May's paper A general algebraic approach to Steenrod operations, Steenrod operations are constructed in wide generality. In this context, it is not necessarily true that negative Steenrod operations are zero. However, I could not find any examples (not in May's paper, not even elsewhere), where one actually proves in some specific situation that some negative Steenrod operation is non-zero. It appears that sheaves of differential graded algebras $A$ (not concentrated in degree $0$) on a topological space $X$ should provide a counterexample, but I would like to see a specific example of $X$ and $A$, possibly with a proof.
Notice the switch in grading from homology to cohomology on May's page 182: $P^s(x) = P_{-s}(x)$. Operations that raise degree when graded homologically lower degree when graded cohomologically. A large part of the motivation for the paper was to give a common framework for the Dyer-Lashof operations in the homology of iterated loop spaces and the Steenrod operations in the cohomology of spaces. But if one grades these the same way, either homologically or cohomologically, one has positive operations and the other has negative operations. The grading May found most sensible for the Steenrod operations on the cohomology of cocommutative Hopf algebras (p. 226) also gives examples. There are a number of more recent papers that make serious use of the large Steenrod algebra (sometimes called the Kudo-Araki-May algebra) constructed using all operations.
Thank you very much for your reply. I am still curious about the last sentence of my question, i.e. the case of hypercohomology of sheaves of dgas (using cohomological grading), so let me elaborate a bit. I read here https://mysite.science.uottawa.ca/pparent/May.pdf (Remark 7 page 12, 'Not true for hypercohomology') that there should be (positive and) negative Steenrod operations, contrary to the case of sheaf of algebras (as studied by Epstein in the Inventiones paper). Do you have a specific example in mind?
What a terrible talk, beginning with a terrible typo. That talk was when I was trying to work out an alternative construction of Voevodsky's Steenrod operations in motivic cohomology. The attempt failed (see the last few pages). I'd still like to see an operadic construction of those operations, but this is not the place to explain the difficulties I was seeing then. There are surely examples that others will know off the top of their head, but they are not on the top of mine.
|
2025-03-21T14:48:31.202066
| 2020-06-07T00:40:33 |
362382
|
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"Igor Belegradek",
"Mustang GT",
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"url": "https://mathoverflow.net/questions/362382"
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|
Stack Exchange
|
Is there a classification of immersions of an n-sphere into a compact n-manifold without boundary?
Is there a classification of smooth immersions of an n-sphere into a compact n-manifold without boundary? I'm interested in the case where the dimensions of the two manifolds are the same. Any help would be appreciated.
Any such immersion is a covering map onto a connected component. The point is an immersion of closed connected $n$-manifolds is a covering map.
@IgorBelegradek Thank you!
|
2025-03-21T14:48:31.202135
| 2020-06-07T00:54:17 |
362383
|
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|
Stack Exchange
|
Inverse of a larger matrix where the inverse of the submatrix is known
Let $A, A^{-1} \in \mathbb{R}^{n \times n}$ be known matrices. Suppose we have an invertible matrix $B \in \mathbb{R}^{(n+1) \times (n+1)}$ of the following form:
$$B = \begin{bmatrix}
A & b\\
b^T & 1
\end{bmatrix}$$
where $b$ is a column vector and $c$ is a row vector. How can I calculate matrix $B^{-1}$ from known matrices $A$ and $A^{-1}$? Can the Sherman–Morrison formula be applied here? If so, how?
As far as I understand, it can be applied if some perturbation is made to $A$. However, the problem here is that $B$ has a different shape than $A$. Appending $A$ with zero entries in the beginning will not work either because the same matrix with zeros added in the bottom row and the right column is not necessarily nonsingular.
https://en.wikipedia.org/wiki/Schur_complement
You know $\begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & 1 \end{pmatrix}$, and from there you can make two successive rank-$1$ modifications, first adding $b$ along the last column, then $c$ along the last row. So, using Sherman–Morrison twice should work.
Presumably it is the same thing, but starting with $\pmatrix{A&b\0&1\}^{-1}=\pmatrix{A^{-1}&-A^{-1}b\0&1\}$ is good.
I actually have found a ready to use formula derived from Sherman-Morrison. I put the answer here: https://github.com/Bigpig4396/Incremental-Gaussian-Process-Regression-IGPR/blob/master/Inverse%20Matrix%20Update%20for%20Kernel%20Matrix.pdf
|
2025-03-21T14:48:31.202266
| 2020-06-07T01:15:56 |
362385
|
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|
Stack Exchange
|
Parseval-Plancherel identity involving absolute value
Let $\hat{f}$ be the fourier transform of $f$.
By Parseval-Plancherel identity, for suitable $f,g$, we have
$$\left\|\hat{f}*\hat{h}\right\|_{L^2_{\xi}}^2=\left\|f\cdot h\right\|_{L^2_{x}}^2.$$
Let $f,g$ be good enough functions. I want to know if for some universal constant $C>0$,
$$\left\||\hat{f}|*|\hat{h}| \right\|_{L^2_{\xi}}^2\leq C\left\||f|\cdot|h|\right\|_{L^2_{x}}^2.$$
Or is there a counterexample?
Thanks a lot!
Take $f, h$ as the characteristic functions of two disjoint intervals (or smoothed variants of them). Then the right-hand side is zero while the left-hand side is not.
@AlekseiKulikov, thanks a lot!
|
2025-03-21T14:48:31.202350
| 2020-06-07T05:47:21 |
362392
|
{
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"authors": [
"Gerhard Paseman",
"JoshuaZ",
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"Stanley Yao Xiao",
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"mathlove"
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|
Stack Exchange
|
Is every integer $\ge 312$ the sum of two integers with triangular divisors?
We say that a natural number $n$ has triangular divisors if it has at least one triplet of divisors $n = d_1d_2d_3$, $1 \le d_1 \le d_2 \le d_3$, such that $d_1,d_2$ and $d_3$ form the sides of a triangle (non degenerate)
E.g.: $60$ has triangular divisors because $60 = 3.4.5$ and $3,4,5$ form a triangle. Note that another triplet of divisors of $60 = 1.4.15$ does not form a triangle but because of the triplet $3,4,5$ the number $60$ qualifies a number with triangular divisors. On the other the number $10$ does not have any triplet of triangular divisors.
I found the following conjectures experimentally. Can they be proved or disproved?
Weak conjecture: Every integer $\ge 8$ which has triangular divisors can be written as the sum of two integers both of which has triangular divisors.
Strong conjecture: Every integer $\ge 8$ except $11, 14, 15,23, 38, 47, 55, 71, 103, 113$ and $311$ can be written as the sum of two integers both of which has triangular divisors.
Note: This question was posted in MSE 3 month ago year. It got some upvotes but to answers. Hence posting in MO.
Related question: How many numbers $\le x$ can be factorized into three numbers which form the sides of a triangle?
I think this question is quite hard. If one writes $N$ as the set of numbers with triangular divisors, then you are asking whether $N$ is an asymptotic additive basis of order 2. One knows that the analogous problem is unknown for primes (where necessarily only even numbers can be represented) which have density $X/\log X$ and it is known for sums-of-two-squares which have density $X/(\log X)^{1/2}$. $N$ is even thicker, but it has far less structure than either of these sets. In particular, it does not seem to be multiplicatively closed (continued...)
Therefore it is not clear to me what techniques would work. Problems relating additive and multiplicative properties of integers is notoriously difficult, as the Goldbach problem shows. In general we don't have a good handle on how to handle sums of multiplicative functions, which are probably the nicest class of arithmetical functions. Thus the answer to the question is "your conjecture is probably true, but we don't know how to prove it".
The second conjecture is false since $7$ cannot be written as the sum of two integers both of which has triangular divisors. It seems that you meant something like "Every integer $\ge 8$ except ...".
@mathlove Yes, I had missed the $\ge 8$. Added it. Thanks!
One can use standard results in number theory to show every positive number is the sum of four triangular numbers. What do you know about sums of three triangular numbers? Gerhard "Is This An Eureka Moment?" Paseman, 2020.06.07.
@GerhardPaseman Has triangular divisors is not the same as is a triangular number. The OP does not mean a number of the form n(n+1)/2.
Note that every perfect cube has triangular divisors. So one at least has that positive integer is expressible as the sum of at most 9 numbers with triangular divisor since every positive integer is expressible as a sum of at most 9 perfect cubes.
Indeed, poor wording on my part. And it has probably occurred to you by now that every square has triangular divisors. I think you can reword my comment now in your head. Gerhard "Maybe We Should Fix Notation" Paseman, 2020.06.07.
@GerhardPaseman, Ah yes. Since we can just for m=n^2 take (1,n,n).
@GerhardPaseman, Hmm, it is easy to show that for any n, the equation x^2 + y^2 +z^3 =n is solvable. But that requires that z be allowed to be negative. If this were to have solutions for sufficiently large n, that would then bump the required number of numbers with triangular divisors down by 1.
In a similar vein, one should be able to show every number (or almost every number) is the difference of two positive numbers, each with triangular divisors. Gerhard "What Else Can We Show" Paseman, 2020.06.07.
@GerhardPaseman . More is true; every powerful number (in the sense every prime factor is raised to at least the 2nd power) has triangular divisors. Heath-Brown proved that every sufficiently large number is the sum of 3 powerful numbers. So that gets us down to 3 if we allow sufficiently large. I don't know if anyone has made sufficiently large explicit in Heath-Brown's work.
Not a complete answer, but it made sense I think to summarize the comment thread above:
Theorem: Every sufficiently large positive integer is expressible as a sum of three numbers which have triangular divisors.
To see this, note that every perfect square, $n^2$, has triangular divisors $(1,n,n)$.
Also, if $m$ has triangular divisors, so does $mk^3$ for any $k$. This is because we can just take our three divisors $d_1$, $d_2$, and $d_3$ and scale them up to be $kd_1$, $kd_2$, $kd_3$.
So any number of the form $n^2k^3$ has triangular divisors. But these are precisely the powerful numbers, numbers in which all prime factors are raised to at least the second power. (Note these are sometimes called squarefull numbers.)
To prove the theorem we then use the theorem of Heath-Brown that every sufficiently large positive integer is expressible as the sum of three powerful numbers.
Another elementary observation: k*4^n has triangular divisors where k is less than 4^n (oops, less than 2^(n+1). This means every number m is within about sqrt(m) (again oops, more like m to the two thirds) of such a number, and gives a weak logarithmic bound to the number of terms such a sum. One might be able to tweak this to get better bounds. Gerhard "Solve Sum Bit By Bit" Paseman, 2020.06.09.
|
2025-03-21T14:48:31.202702
| 2020-06-07T06:36:41 |
362394
|
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|
Stack Exchange
|
How to do a multinomial theorem sum faster
For example we have this question :
Find the coefficient of $x^6$ in the following
$\frac{\left(x^{2}+x+2\right)^{9}}{20}$
So using multinomial Theorem which is this :
$\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}=\sum_{b_{1}+b_{2}+\cdots+b_{k}=n}\left(\begin{array}{c} n \\ b_{1}, b_{2}, b_{3}, \dots, b_{k} \end{array}\right) \prod_{j=1}^{k} x_{j}^{b_{j}}$
I have to first find out the combinations of power for which I am going to get $x^6$
then i have to calculate the multinomial coefficient multiplied by the $2$ raised to some power for all of those combinations.
Is there an easier way to somehow change this question into some other question then use Binomial or some other theorem to solve it quickly ?
If you don't insist on using the multinomial theorem, repeated squaring modulo $x^7$ seems more straightforward.
Extracting a сomplete square can save some time. For example, from $x^2+x+2=\frac{(2x+1)^2+7}{4}$, we get
$$[x^6]\, (x^2+x+2)^9 = \frac{1}{4^9} \sum_{k=0}^9 \binom{9}{k} \binom{2k}{6} 2^6 7^{9-k}.$$
Alternatively, we can employ the factorization $x^2+x+2 = (x-\alpha_1)(x-\alpha_2)$, where $\alpha_{1,2}=\frac{-1\pm I\sqrt{7}}2$, to get
$$[x^6]\, (x^2+x+2)^9 = \sum_{i=0}^6 \binom9i\binom9{6-i}(-\alpha_1)^{9-i} (-\alpha_2)^{3+i}.$$
Yet another approach is to represent $x^2+x+2=(x^2+x)+2$ and notice that $[x^6]\,(x^2+x)^k = \binom{k}{6-k}$. Then
$$[x^6]\, (x^2+x+2)^9 = \sum_{k=0}^9 \binom9k \binom{k}{6-k}2^{9-k}.$$
|
2025-03-21T14:48:31.202812
| 2020-06-07T06:44:02 |
362395
|
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|
Stack Exchange
|
Do height $h$ prime ideals in regular local rings contain regular sequences of length $h$?
Let $R$ be a regular local ring and let $P$ be a prime ideal of height $h$ in $R$.
Is it always the case that $P$ contains a regular sequence of lenght $h$?
This is clear if $h$ is $0,1$ or $\dim R$.
Since $R$ is regular, the localization $R_P$ is regular local of dimension $h$, and thus contains an $R_P$-regular sequence $f_1,\dots,f_h$, which can be chosen to live in $R$. However, it is a priori not clear that $f_1,\dots,f_h$ is a regular sequence in $R$.
The answer is yes. In fact, more generally, if $R$ is a Cohen-Macaulay ring, then the height of any prime ideal in it is equal to its depth, which is just the length of a maximal regular sequence contained in P. See for example Theorem 2.1.2 of the book
Cohen-Macaulay rings by Bruns and Herzog.
Thank you very much! Did you mean to write "grade on $R$" instead of "depth"?
The grade of an ideal is often called the depth of an ideal.
|
2025-03-21T14:48:31.202912
| 2020-06-07T06:59:14 |
362396
|
{
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|
Stack Exchange
|
A monotone countably unbounded function from $\omega^\omega$ to $\omega^{\omega_1}$
For a set $X$ we endow the set $\omega^X$ of all functions from $X$ to $\omega$ with the natural partial order $\le$ defined by $f\le g$ iff $f(x)\le g(x)$ for all $x\in X$. A function $\mu:\omega^\omega\to \omega^X$ is called monotone if for any $f\le g$ in $\omega^\omega$ we have $\mu(f)\le\mu(g)$.
Question. Is there a monotone function $\mu:\omega^\omega\to\omega^{\omega_1}$ which is countably unbounded in the sense that for every countable infinite set $A\subset\omega_1$ and function $f\in\omega^A$ there exist a function $g\in\omega^\omega$ and an infinite set $B\subseteq A$ such that $f{\restriction}_B\le\mu(g){\restriction}_B$?
Remark. By the answer of Johannes Schürz to this question, for every monotone function $\mu:\omega^\omega\to\omega^{\omega_1}$, there exists a countable set $A\subset\omega_1$ and a function $f\in\omega^A$ such that $f\not\le \mu(g){\restriction}A$ for every $g\in\omega^\omega$.
Is there some (diagonalization?) argument showing that given monotone $\mu:\omega^\omega \to \omega^{\omega_1}$ there is always $A\subseteq \omega_1$ countable and $f\in\omega^{\omega_1}$ such that for all $g\in\omega^\omega$ we have $f{\restriction}_A \not\leq (\mu(g)){\restriction}_A$?
@DominicvanderZypen Yes this is exactly what was proved by Johannes Schurz at https://mathoverflow.net/a/362314/61536
Oh ok, thanks Taras!
The answer to this problem is negative and follows from
Theorem. For any monotone function $\mu:\omega^\omega\to\omega^{\omega_1}$ there exists a countable infinite set $A\subset\omega_1$ such that for every $f\in\omega^\omega$ the function $\mu(f){\restriction}A$ is bounded.
Proof. For every $\alpha\in\omega_1$ consider the monotone function $\mu_\alpha:\omega^\omega\to\omega$, $\mu_\alpha:f\mapsto\mu(f)(\alpha)$. By Lemma~2.3.5 in this paper, for every $f\in\omega^\omega$ there exists $n\in\omega$ such that $\mu_\alpha[\omega^\omega_{f{\restriction}n}]$ is finite. Here $\omega^\omega_t=\{g\in\omega^\omega:t\subset g\}$.
Let $T_\alpha$ be the set of all (finite) functions $t\in\omega^{<\omega}$ such that $\mu_\alpha[\omega^\omega_t]$ is finite but for any $\tau\in\omega^{<\omega}$ with $\tau\subsetneq t$ the set $\mu_\alpha[\omega^\omega_\tau]$ is infinite. It follows (from the mentioend Lemma 2.3.5) that for every $f\in\omega^\omega$ there exists a unique $t_f\in T$ such that $t_f\subset f$.
Let $\delta_\alpha(f)=\max\mu_\alpha[\omega^\omega_{t_f}]\ge \mu_\alpha(f)$. It is clear that the function $\delta_\alpha:\omega^\omega\to\omega$ is continuous.
Consider the function $\delta:\omega_1\to C_k(\omega^\omega,\omega)$, $\delta:\alpha\mapsto\delta_\alpha$ and observe that $\delta(\alpha)(f)\ge \mu(f)(\alpha)$ for any $\alpha\in\omega_1$ and $f\in\omega^\omega$.
By Michael's Theorem 11.5 (in Gruenhage's survey), the function space $C_k(\omega^\omega,\omega)$ is an $\aleph_0$-space. In particular, it has a countable network. Using this fact, we can find a sequence of pairwise distinct ordinals $\{\alpha_n\}_{n\in\omega}\subset\omega_1$ such that the sequence $(\delta_{\alpha_n})_{n\in\omega}$ converges to $\delta_{\alpha_0}$ in the function space $C_k(\omega^\omega,\omega)$.
Consequently, for every $f\in\omega^\omega$ the sequence $(\delta_{\alpha_n}(f))_{n\in\omega}$ converges to $\delta_{\alpha_0}(f)$ and hence is upper bounded by some number $M_f$. Let $A=\{\alpha_n\}_{n\in\omega}$ and observe that for every $n\in\omega$ we have $\mu(f)(\alpha_n)\le\delta_{\alpha_n}(f)\le M_f$, which means that the function $\mu(f){\restriction}A$ is bounded.
|
2025-03-21T14:48:31.203243
| 2020-06-07T08:33:23 |
362400
|
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"András Bátkai",
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"url": "https://mathoverflow.net/questions/362400"
}
|
Stack Exchange
|
Existence of function $g$ such that $f(x,y)\le g(x)+g(y)$
Prove for each function $f:\mathbb Q\times\mathbb Q\to\mathbb R$ there exists a function $g:\mathbb Q\to\mathbb R$ such that $f(x,y)\le g(x)+g(y)\,\forall x,y\in\mathbb Q$.
Find a function $f:\mathbb R\times\mathbb R\to\mathbb R$ for which there is no function $g:\mathbb R\to\mathbb R$ such that $f (x,y)\le g (x)+g (y)\,\forall x,y\in\mathbb R $.
One should note that the problems are of purely settheoretic nature. In the first question $\mathbb{Q}$ can be replaced by $\mathbb{N}0$ and then $g(x) := \max{y,z \in \mathbb{N_0} :y+z \leq 2x} |f(x,y)|$ (of course very crude) does the job. For the second question I think $\mathbb{R}$ should be well ordered and then ?
I am probably missing something, but why was the question closed? Is there an obvious answer to the second question?
Such questions are closed, though I was not involved, because there is no question. It sounds like a (hard) exercise for extra credits.
The second "question" was answered in a more general setting in the comments of this subsequent question where the exercise-style was not followed and that question was well-received. This makes it a duplicate, but rather than closing on this grounds, one could remove the second question and focus on the 1st one. But at the same time the first question is possibly too easy to deserve a separate question (it's claimed as easy in the linked question).
|
2025-03-21T14:48:31.203372
| 2020-06-07T09:16:00 |
362402
|
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"Alexandre Eremenko",
"Dima Pasechnik",
"Gordon Royle",
"Timothy Chow",
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"url": "https://mathoverflow.net/questions/362402"
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|
Stack Exchange
|
My research paper involves computing additional terms of an existing OEIS sequence. Should I first amend the sequence or publish the results?
In the course of my research I computed terms of an existing OEIS sequence that are currently unknown. Having prepared my paper for publication, I am now faced with a (small) dilemma:
Do I first contribute my findings to the OEIS database, then submit my paper for publication, or
Go through the whole (lengthy) peer review process, then once the paper is published add the new terms to OEIS?
Both of these options have their advantages/disadvantages. On the one hand, the very purpose of the OEIS is to accommodate new computational results and there is no better place to publish them than there. However, given that these new terms required months to compute, they should probably go through some review process first.
On the other had, some results in my paper explicitly depend on these newly computed terms. Contributing them to the OEIS before the lengthy publication process bears its fruit might prompt others to use these new terms to find and publish the same (or similar) results.
Surely, I am not the first one to ask this question. What is the correct course of action in this case?
Why not post the paper on arXiv and then reference the preprint in OEIS entry? Once you are published in a journal you can update the reference.
I would even say: publish on arxiv and update the sequence on oeis at the same time. This is sufficient to stake a claim on a new result.
Dima is obviously correct. In addition, I think the number of people scanning newly-extended OEIS sequences in the hope of finding new data that will enable them to write papers gazumping the sequence extender may be lower than you think.
You should publish your result first. Because when amending OEIS you will need to insert a reference on your published paper.
@AlexandreEremenko : There are many OEIS entries with no published reference. If and when a relevant paper is formally published, that information can be added to the OEIS entry.
@Timothy Chow: I never published anything in that cite but if what you say is correct, this is regrettable. On my opinion, there must be published references:-) For a site to be trustworthy there should be a way to check everything which is there. Hope it is at least moderated.
@AlexandreEremenko : As someone who owes more than one published paper to the existence of the OEIS, including specifically an entry that had no published reference, I would say that the philosophy of the OEIS is that it is primarily a tool for discovery, and not primarily a compilation of absolute truths. I would much rather suffer ten false positives and score a magnificent hit than have zero false positives and miss the hit. Having said that, I will point out that the OEIS is indeed moderated.
@Timothy Chow: thanks for your explanation. Now I will be more careful when using OEIS:-) Of course it is OK, once people are warned, they can check the origin of any information it contains.
|
2025-03-21T14:48:31.203632
| 2020-06-07T09:48:23 |
362405
|
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"Liviu Nicolaescu",
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|
Stack Exchange
|
A completion equality in product of Sobolev spaces
Let $\Omega$ be a nice bounded domain of $\mathbb{R}^n$. For $s\ge 0$ we define
$$H_{0}^{s}(\Omega):=\overline{C_{c}^{\infty}(\Omega)}^{H^{s}(\Omega)}.$$
My question: is the following equality true?
$$\overline{\{ (f,f|_{\partial \Omega}), f\in C^{\infty}(\overline{\Omega}) \}}^{H^{s}(\Omega)\times {H^{s}(\partial\Omega)}}=H^{s}_0(\Omega)\times {H^{s}(\partial\Omega)}.$$
I expect that the result is true, but I am still stuck in giving a rigorous proof. Any hint would be useful.
The trace map $$H^s(\Omega)\ni f\mapsto f|_{\partial \Omega}\in H^{s-1/2}(\partial \Omega)$$ is a continuous surjection with kernel $H_0^s(\Omega)$ if $s>1/2$.
This is certainly not true, at least for two reasons:
since you are considering all smooth functions $f\in\mathcal C^\infty(\overline\Omega)$ without restrictions on the behaviour at the boundary, there is no resaon why in the first factor you should expect to retrieve the "zero-boundary space" $H^s_0$ instead of $H^s$
since the approximating pairs $(f,f|_{\partial\Omega})$ are really special structurally -- namely the second factor $f|_{\partial\Omega}$ must be the trace of the first factor in your product space before completeion -- you cannot expect to retrieve the whole product space $H^s_0(\Omega)\times H^s(\partial\Omega)$ upon completion. For example, how on earth would you approximate the constant (pair of) functions $(0,1)\in H^s_0(\Omega)\times H^s(\partial\Omega)$ by pairs of the form $(f_n,f_n|_{\partial\Omega})$? By continuity of the trace operator $\gamma:H^s(\Omega)\rightarrow H^{s-1/2}(\partial\Omega)$ (see @Liviu Nicolaescu's comment) for any such limit pair $(f,g)=\lim (f_n,f_n|_{\partial\Omega})$ you should get that $g=\lim f_{n}|_{\partial\Omega}=\lim \gamma( f_n) = \gamma(\lim f_n)=\gamma(f)=0$ (the limits hold in $H^{s-1/2}(\partial\Omega)$) so clearly the second factor $g$ should vanish.
Thank you! Do you have any idea on how to change the set ${ (f,f|_{\partial \Omega}), f\in C^{\infty}(\overline{\Omega}) }$ to get an affirmative result?
Sure: take independent pairs $(f,g)\in C^\infty_c(\Omega)\times C^\infty(\partial\Omega)$
|
2025-03-21T14:48:31.203784
| 2020-06-07T10:01:50 |
362407
|
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|
Stack Exchange
|
Maximum number of four cycles with no intersecting three vertex paths
What is the size of the largest set (asymptotic) of four vertex cycles in an $n$-vertex simple, undirected graph such that no two cycles from this set has a common three vertex path?
If we choose four cycles such that no two of them intersect at more than two vertices, then we get $\Theta(n^2)$ four cycles. On the other hand, it is easy to see that the upper bound is $O(n^3)$ because there are only $O(n^3)$ distinct three vertex paths in an $n$-vertex graph.
EDIT: This answer pertains to a misinterpretation of the question before it was later clarified, so does not answer the question as currently stated:
Two distinct 4-cycles share a 3-vertex path if and only if the union of those two 4-cycles is isomorphic to the complete bipartite graph $K_{2,3}$.
So, you are effectively asking for the maximum number $M_n$ of 4-cycles in a $K_{2,3}$-free graph on $n$ vertices.
Now, a graph is $K_{2,3}$-free if and only if every pair $x, y$ of vertices have at most two common neighbours -- and, if there are exactly two neighbours, $x$ and $y$ are diametrically opposite points on a 4-cycle. An upper bound on the number of 4-cycles in such a graph is therefore:
$$ M_n \leq \frac{1}{2} \binom{n}{2} $$
since there are $\binom{n}{2}$ pairs of distinct vertices (each of which can be the diagonal of at most one 4-cycle) and every 4-cycle has two diagonals.
A good lower bound can be extracted from https://arxiv.org/pdf/1306.5167.pdf by taking their Construction 3.15. In particular, let $p$ be an odd prime and take the graph $G_p$ whose vertex set is $(\mathbb{F}_p^2 \setminus \{(0, 0)\}) / \{ \pm 1 \}$ and contains an edge between $(a, b)$ and $(x, y)$ if and only if $ax + by \in \{ \pm 1 \}$.
$G_p$ has $\frac{1}{2}(p^2 - 1)$ vertices. For every pair of vertices $(a, b)$ and $(a', b')$, the common neighbours are the solutions to:
$$ ax + by, a'x + b'y \in \{ \pm 1 \} $$
Now, since the vertices are distinct there are four solutions, which we'll call $(c, d)$, $(-c, -d)$, $(c', d')$, and $(-c', -d')$. After taking the quotient, these four solutions correspond to two vertices. As $p \rightarrow 0$, the fraction of original pairs of vertices for which $\{ (a, b), (a', b'), (c, d), (c', d') \}$ are all distinct approaches 1. This implies that in these cases, the number of 4-cycles is asymptotically optimal: $\left(\frac{1}{2} - o(1)\right) \binom{n}{2}$.
For arbitrary $n$, we can let $p$ be the largest prime such that $\frac{1}{2}(p^2 - 1) \leq n$, and take the $n$-vertex graph to be the disjoint union of $G_p$ with $n - \frac{1}{2}(p^2 - 1)$ extra isolated vertices. Results on prime gaps (such as the theorem that there is a prime between every pair of sufficiently large cubes) show that the number of extra isolated vertices is $o(n)$.
Putting this all together, we get the following asymptotically tight bounds:
$$ \left(\frac{1}{2} - o(1)\right) \binom{n}{2} \leq M_n \leq \frac{1}{2} \binom{n}{2} $$
A union of two $C_4$ without a common $P_3$ can have a $K_{2,3}$. For example, consider the $K_{2,4}$ where one partition has vertices ${x, y}$ and the other partition has ${1,2,3,4}$. This graph is the union of two cycles $1,x,2,y,1$ and $3,x,4,y,3$. These cycles do not have a common $P_3$.
But your graph still has two 4-cycles -- such as $1,x,2,y,1$ and $3,x,2,y,1$ -- which share a common 3-vertex path ($x,2,y$).
If you didn't mean that (and you're only interested in finding a large collection of 4-cycles on a common vertex set which pairwise share no common 3-vertex path), then your mention of 'an $n$-vertex simple undirected graph' is a red herring.
Yes. I was only asking for a large collection with no common $P_3$. I can see how my question is confusing. I will edit it.
For the corrected question, a similar argument used for your lower bound works to obtain an $\Omega(n^3)$ bound.
|
2025-03-21T14:48:31.204038
| 2020-06-07T10:03:57 |
362408
|
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|
Stack Exchange
|
Chromatic self-maps for almost disjoint families
Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say that ${\cal A}\subseteq [\omega]^\omega$ is an almost disjoint family if $A \neq B \in {\cal A}$ implies $|A\cap B|< \aleph_0$.
Let $X\neq\varnothing$ be a set and let ${\cal E}\subseteq {\cal P}(X)\setminus\{\varnothing\}$ be a collection of non-empty subsets. We say that a map $f: {\cal E}\to X$ is a chromatic self-map if
$f(e) \in e$ for all $e\in {\cal E}$, and
if $e_1\neq e_2 \in {\cal E}$ and $e_1\cap e_2 \neq \varnothing$, then $f(e_1)\neq f(e_2)$.
Question. Does every almost disjoint family ${\cal A}\subseteq [\omega]^\omega$ have a chromatic self-map?
Remark. It suffices to answer the question for maximum almost disjoint families ("MAD families").
I find your use of the term "self-map" a bit confusing. I guess you mean functions with the property $f(e)\in e$? I'm used to seeing those called choice functions, or selectors. And a "self-map" of a set $S$ is a map $f:S\to S$.
The map would be injective: if $f(e_1)=f(e_2)=n$ then $n\in e_1\cap e_2$, so by condition 2 you'd get $e_1=e_2$.
If such an $f$ exists then $\mathcal A_n=\{e\in\mathcal A:f(e)=n\}$ is a collection of pairwise disjoint subsets of $\omega$, so $\mathcal A_n$ is countable, so $\mathcal A=\bigcup_n\mathcal A_n$ is countable. So the answer is "no" if $\mathcal A$ is uncountable. Of course it is "yes" if $\mathcal A$ is countable.
|
2025-03-21T14:48:31.204150
| 2020-06-07T10:48:46 |
362411
|
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|
Stack Exchange
|
Amenable groups with all subgroups finitely generated
Does anyone know an example of an amenable group with all subgroups finitely generated that is not elementary amenable?
No it's certainly unknown. (For elementary amenable groups, noetherian means virtually polycyclic, but you probably know this.) Very few noetherian groups are known: virtually polycyclic, quasi-finite f.g. groups, quasi-cyclic ones (all constructed by Olshanskii), and some immediate variants or combinations.
Such groups "probably" exist though?
|
2025-03-21T14:48:31.204219
| 2020-06-07T11:04:37 |
362412
|
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"Adrien",
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"Konstantinos Kanakoglou",
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"url": "https://mathoverflow.net/questions/362412"
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|
Stack Exchange
|
Further developments of Cartier–Gabriel–Kostant–Milnor–Moore Structure Theorem for cocommutative Hopf algebras
A very well-known theorem in Hopf algebra theory (see, for example, Lorenz - A tour of representation theory or the EGNO book (Etingof, Gelaki, Nikshych, and Ostrik - Tensor categories)) states that if $H$ is a cocommutative Hopf algebra over an algebraically closed field $\Bbbk$ of characteristic zero then $$H \cong U(P(H)) \mathbin\# \Bbbk G(H)$$ as Hopf algebras, where $P(H)$ is the space of primitive elements of $H$ and $G(H)$ the group of group-like elements. As far as I know, this theorem is attributed differently to many people, but I believe the list in the title should be comprehensive of all of them. For example, I know that the Milnor–Moore part of the theorem is the one stating that an irreducible cocommutative Hopf algebra $H'$ over a field of characteristic zero satisfies $H' \cong U(P(H'))$.
I know that Nichols proved, in The Kostant structure theorem for $K/k$ Hopf algebras, an analogue of the full CGKMM theorem for Hopf algebroids of the form $(K,H)$, where $K$ is a field extension of $\Bbbk$, that Moerdijk and Mrčun proved, in On the universal enveloping algebra of a Lie algebroid, an analogue of Milnor–Moore for bialgebroids and that, later, Kališnik and Mrčun proved, in A Cartier–Gabriel–Kostant structure theorem for Hopf algebroids, an analogue of the full Cartier–Gabriel–Kostant–Milnor–Moore for Hopf algebroids over the algebra $\mathcal{C}^{\infty}_c(\mathcal{M})$ of smooth functions with compact support over a smooth real manifold $\mathcal{M}$.
Is anybody aware of some further developments/extensions of this result to other classes of Hopf algebras (for example, weak Hopf algebras, general Hopf algebroids, (co)quasi-Hopf algebras, etc.)?
Otherwise, is there anybody aware of counter-examples that suggest this could not be generalized further in some direction?
Edit: Just to add some further motivation to this quest. Let $G$ be a linear algebraic group over an algebraically closed field $\Bbbk$ of characteristic zero. The coordinate algebra $\mathscr{O}(G)$ is a commutative Hopf algebra. If we consider the subspace $\mathscr{O}(G)^\circ$ of $\mathsf{Hom}_\Bbbk\left(\mathscr{O}(G),\Bbbk\right)$ formed by all those linear functionals which vanish on a finite-codimensional ideal, this is a cocommutative Hopf algebra. By the CGKMM Theorem we know that
$$\mathscr{O}(G)^\circ \cong U\left(P\left(\mathscr{O}(G)^\circ\right)\right) ~\#~ \Bbbk G\left(\mathscr{O}(G)^\circ\right).$$
It happens that $P\left(\mathscr{O}(G)^\circ\right) \cong \mathfrak{g}$, the tangent Lie algebra to $G$ at the neutral element, and $G\left(\mathscr{O}(G)^\circ\right) \cong \mathsf{Alg}_\Bbbk\left(\mathscr{O}(G),\Bbbk\right) \cong G$, so that
$$\mathscr{O}(G)^\circ \cong U\left(\mathfrak{g}\right) ~\#~ \Bbbk G,$$
where on the right-hand side $G$ is considered as a discrete group and $U(\mathfrak{g}) \cong \mathsf{Dist}_G$, the hyperalgebra of distributions on $G$.
Is there anybody aware of some further results like this one for groupoids?
There is an operadic version of Milnor-Moore's theorem which characterizes enveloping algebras in arbitrary symmetric monoidal categories (whenever this notion makes sense). This is theorem 6.1 in https://arxiv.org/abs/math/0306212.
The super version of the theorem, refers to "hopf superalgebras", or "$\mathbb{Z}_2$-graded hopf algebras" or "hopf algebras in the braided monoidal category of $\mathbb{CZ}_2$-modules":
Let $\mathcal{H}$ be a super-cocommutative hopf superalgebra over an algebraically closed field $k$ of char zero. Then we have the hopf superalgebra isomorphism:
$$
\mathcal{H}\cong k[G(\mathcal{H})]\ltimes_{\pi} U\big(P(\mathcal{H})\big)
$$
where, $k[G(\mathcal{H})]$ is the group algebra of the the group $G(\mathcal{H})$ of the grouplikes of $\mathcal{H}$, $U\big(P(\mathcal{H})\big)$ is the universal enveloping algebra of the lie superalgebra $P(\mathcal{H})$ of the primitive elements of $\mathcal{H}$ and the smash product $\ltimes_{\pi}$ is with respect to the representation of $G(\mathcal{H})$ on $P(\mathcal{H})$ determined by: $\pi:G\to Aut(P)$, $\pi(g)x=gxg^{-1}$, for all $g\in G$, $x\in P$.
This has been an old result, first shown by Kostant, at:
B.Kostant, "Graded manifolds, graded Lie theory and prequantization", Differential Geometrical Methods in Mathematical Physics. Lecture Notes in Mathematics, vol 570, p. 177-306, (1977))
P.S.: What i do not know, is whether there is a corresponding version of the theorem for Hopf algebras in braided monoidal categories. I am not aware of some reference in that direction (although, i think the generalization should not be that difficult to prove). The only related reference i know of is Braided bialgebras of Hecke-type, A. Ardizzoni, C.Menini, D. Stefan, Journal of Algebra 321 (2009) 847–865, where a result -analogous to the Milnor-Moore part of the theorem- is proved for connected, braided bialgebras which are infinitesimally cocommutative (see theorem 5.5 and the resulting discussion).
Edit: another possible line of generalization of the theorem , has to do with the class of quasitriangular hopf algebras. These, generalise the cocommutative hopf algebras, so it is natural to consider the possibility of a "quasitriangular version" of the theorem. See for example: Classification of quasitriangular Hopf algebras
You have my sincere gratitude for all your inputs. Alas I cannot +1 you more than once. I am leaving the question unanswered to see if somebody else shows up with other references
Thank you. I have just now realized that you have been a coauthor of A. Ardizzoni and D. Stefan, so maybe, you were already aware of their work mentioned in my PS above :)
|
2025-03-21T14:48:31.204592
| 2020-06-07T12:46:52 |
362415
|
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|
Stack Exchange
|
simple formula for a finite sum of multinomial numbers
Let $k,m$ and $r$ be positive integers.
Define
$$\Omega(k,m,r) = \binom k {m-2r}\binom {k-m+2r} r$$
and
$$\Omega(k,m) = \sum_{r=\max\{0,m-k\}}^{[\frac{m}{2}]}\Omega(k,m,r).$$
Question.
1. Is $\Omega(k,m)$ has a simple formula?
2. Is $\frac{\sum_{r=\max\{0,m-k\}}^{[\frac{m}{2}]}r\Omega(k,m,r)}{(k+m)\Omega(k,m)}$ has a simple formula?
Background:
If we want to calculate the numbers of $m$ white balls insert in $k-1$ black balls with the constraint that continuum three white balls $\circ\circ\circ$ is forbidden, then we can derive the formula above.
First notice that
$$\Omega(k,m,r) = \binom{k}{m-2r,r,k-m+r}.$$
It follows that $\Omega(k,m)$ equals the coefficient of $x^{m-k}$ in $(1+x+x^{-1})^k$, which is the same as the coefficient of $x^{m}$ in $(1+x+x^2)^k$, also known as the trinomial coefficient $\binom{k}{m-k}_2$. That is, $\Omega(k,m) = \binom{k}{m-k}_2$.
The sum $\sum_r r\Omega(k,m,r)$ can be obtained by differentiating $(1+xy+x^{-1})^k$ at $y=1$, which yields $k\binom{k-1}{m-k-1}_2$.
If we let $m=\rho k$,$k$ goes to infinity, can we obtain the limit $\lim_{k\rightarrow\infty}\Omega(k+1,m)/\Omega(k,m)$?
@YuhuanLei: Check out this paper: Asymptotic Estimate for the Multinomial Coefficients.
thank you very much! The result in that paper considers the case that $q>3$ and $c$ is a positive integer. Is there any result about the case that $q=3$ and $c\in[-1,1]$?
I have asked Jiyou Li,he said the case I concerned is also valid.
|
2025-03-21T14:48:31.204720
| 2020-06-07T13:04:43 |
362416
|
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|
Stack Exchange
|
Roadmap to Floer homotopy theory?
I am a young postdoc working in symplectic topology.
Recently I became intrigued by Floer homotopy, especially after seeing it had been applied to classical questions in symplectic topology. (e.g. Abouzaid and Kragh).
This revelation made me excited about the new possibilities that this approach opens up, and I want to try and find other applications.
I am very comfortable with classical Floer theory, both Hamiltonian and Lagrangian, and with constructions such as symplectic homology, TQFT operations, etc. But when it comes to Floer homotopy, my knowledge is pretty much restricted to the slogan, that Floer homotopy cooks up a spectrum from the moduli spaces of Floer solutions, and that this spectrum should see "more information" than mere homology. I am mostly curios about understanding this extra information and how it is utilized to solve problems in Hamiltonian dynamics or topology of Lagrangians.
Which reading (and / or video talks) would you recommend to get a working knowledge of the tools and theory?
In a way, I think I am asking the following: Imagine you had a fresh student who wanted to get into the field. What would the reading list you would compile for them contain?
You may be familiar with these older papers of Ralph Cohen http://math.stanford.edu/~ralph/morse.ps http://math.stanford.edu/~ralph/floer.pdf The germs of the ideas are there. Manolescu's work in Seiberg-Witten theory may be closer to what you want. In the symplectic case there are some subtleties that escape me.
Cohen wrote a more up to date review of his take, which also appeared in the handbook of homotopy theory.
|
2025-03-21T14:48:31.204863
| 2020-06-07T14:21:18 |
362420
|
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|
Stack Exchange
|
Does the following operator have a unique fixed function? Do iterations of the operator converge?
The functions considered are positive real for positive real $x$, and monotonically increasing to infinity. Given such a function $f(x)$, define the (nonlinear) operator $f\mapsto f^*$ by
$$
f^*(x) := 1 + \frac {x}{f(1)}
+ \frac{x^2}{ f(1)f(2)}
+ \frac{x^3}{ f(1)f(2)f(3)} +\dots
= 1 + \sum\limits_{n=1}^{\infty}
\frac{x^n}{\prod\limits_{k=1}^n f(k)}.
$$
The new function $f^*$ is again positive real for positive real $x$ and monotonically increasing to infinity. (In fact $f^*$ is an entire function, not that we need that.) So this process can be iterated. It seems that, no matter what function we start with, the result converges pointwise.
soooo, the question is what, exactly? contained in the title?
Yes. it seems the title is enough, no?
|
2025-03-21T14:48:31.204948
| 2020-06-07T14:39:59 |
362422
|
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|
Stack Exchange
|
Extension of the definition of entropy to $\mathbb{Z}^d$ and $\mathbb{N}^d$
I read the paper Entropie d'un groupe abélien de transformation by Conze and the part of the book Dynamical systems of Algebraic Origin by Schmidt about the entropy for $\mathbb{Z}^d$ actions. I was wondering if it is possible to generalize the definition of topological and measure-theoretic entropy to $\mathbb{N}^d$ actions using the same idea. If so, what happens if we compute the entropy of a $\mathbb{Z}^d$ action using the definition for $\mathbb{N}^d$? Do we get the same result?
The extension of both topological and measure entropy to $\mathbb N^d$-actions is straightforward (in the case of measure entropy you just need to remember that you are taking inverse images of sets, and so the maps do not need to be invertible). Another, equivalent, approach is to take the "natural extension" of the action, using inverse limits, to obtain a $\mathbb Z^d$-action with the same entropy. If you start with a $\mathbb Z^d$-action and think of it as an $\mathbb N^d$-action, the natural extension is just the $\mathbb Z^d$-action, so has the same entropy.
More interesting is the problem of isomorphism, which can lead to quite different answers in the two cases. If a $\mathbb Z^d$-action is considered as an $\mathbb N^d$-action, then an isomorphism can only depend on coordinates in $\mathbb N^d$, which is a severe restriction and in general much harder to obtain.
|
2025-03-21T14:48:31.205091
| 2020-06-07T15:42:49 |
362426
|
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|
Stack Exchange
|
Distributivity of ! over?
Has anyone studied a variant of linear logic, or of its semantic counterpart (exponential modalities on linearly distributive categories / $\ast$-autonomous categories / polycategories) for which there is a distributive law
$$!? \to ?!$$
Presumably it would need to interact sensibly with the (co)monoidal structure of ! and ?. I haven't thought about whether there is a nice sequent calculus or resource interpretation. I ask for two reasons:
Such a distributive law should imply that there is a "mixed Kleisli category" whose morphisms from $A$ to $B$ are the morphisms $!A \to ?B$. This would be more symmetrical and pleasing than the usual translation of classical logic involving morphisms $!?A \to ?B$.
In $\rm Chu(Cat,Set)$ I believe there is such a distributive law, or at least a pseudo-distributive law, and a morphism $!A\to ?B$ should be precisely a profunctor from $A^+$ to $B^-$. So this would give a nice way of recovering $\rm Prof$ from $\rm Chu(Cat,Set)$.
The closest thing I can think of is differential linear logic, in which $A\multimap,!A$ (and therefore $?A\multimap A$) is provable, so the distributive law you mention becomes provable, but the structure is too symmetric (the converse distributivity holds in the same way). In his Ph.D. thesis, Stéphane Gimenez considered differential linear logic with a "super-promotion" rule (which is asymmetric), but he didn't have distributive laws in mind, I'm not sure how relevant his work is for what you have in mind.
Their is the paper On the Unity of Logic: a Sequential, Unpolarized Approach by Norihiro Yamada that adds the distributive law to a substystem of linear logic.
@Potato44 Thanks! That certainly answers the question ("Has anyone studied..."), so if you post it as an answer I'll accept it.
I would tend to say "no".
However, besides my comment above, which is not very pertinent, let me mention the paper Combining effects and coeffects via grading, by Marco Gaboardi, Shin-ya Katsumata, Dominic Orchard, Flavien Breuvart and Tarmo Uustalu. They consider "graded" monads and comonads, which include usual monads and comonads as the special case in which the grading is trivial (I don't know who introduced these first, I learned of graded comonads from this note by Paul-André Melliès). The programming language underlying their work is linear and their comonads are graded in a semiring (rather than just a monoid) so that they are a generalization of the exponential modality $!(-)$ of linear logic (the additive structure of the semiring grades weakening/contraction, i.e. the monoidal structure of the comonad, while the multiplicative structure of the semiring grades the actual comonad structure, i.e., the counit and the comultiplication). They then study graded versions of the usual distributive law between monads and comonads in order to account for the simultaneous presence of "quantitative" effects and coeffects in programming languages (e.g., not just tell whether a program may raise an exception but tell, if possible, how many exceptions it will raise, or whatever. This is what the increased expressiveness given by the grading is meant to be used for).
So, forgetting the grading, i.e., if we grade everything with the trivial semiring and the trivial monoid, we are close to what you describe, but not quite: while the trivially graded version of their comonad is, indeed, the $!(-)$ modality of linear logic, the $?(-)$ modality does not fit into their axiomatization, because the trivially graded version of their monad is necessarily strong, and $?(-)$ is not a strong monad. So, strictly speaking, this work says nothing about the distributive law you are looking at.
I skimmed through the references given by Gaboardi et al. in relation with distributive laws and none of them seems to mention linear logic. This supports my belief that no one has ever introduced/studied the variant of linear logic you mention... but of course I can't be sure!
|
2025-03-21T14:48:31.205494
| 2020-06-07T15:46:12 |
362427
|
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|
Stack Exchange
|
Small flag triangulations
In geometric group theory and low-dimensional topology, asking for a triangulation of a specific space to be flag can often be somewhat more cumbersome than just turning a CW-complex into a simplicial complex. Cheap ways to achieve this include taking successive barycentric subdivisions, which in practice makes the number of vertices and edges involved quite bloated.
For which properties of spaces (such as the properties of their fundamental group) are the smallest flag triangulations known?
Here, 'smallest' means one of 'vertex-minimal' or 'edge-minimal' (or 'triangle-minimal').
It would be useful to have a zoo of examples, for testing conjectures and quick computations. I expect that more may be known about this than appears at a first glance; since there could be various spaces or properties that this has been answered for, which only appear as small lemmas in literature focused on the applications, rather than the complexes themselves. On the other hand, there may be triangulations that we intuitively understand to be minimal, but rigorous proof may require heavy computation.
For example, in this paper, vertex-minimal flag triangulations of several common spaces are found.
More specifically, I am wondering what the smallest flag complex with torsion-free perfect fundamental group is.
I don't know about smallest but one can easily construct flag triangulations of the presentation 2-complex of Higman's group. Also presentation 2-complexes of perfect small cancellation groups are fairly easy to construct and would guarantee non-trivial fundamental group.
I completely agree that a zoo of small examples would be very useful (though I don't see why the triangulations need to be provably minimal, as long as they are small enough to be tractable). One nice such "zoo" has been developed recently by Caprace--Conder--Kaluba--Witzel. The paper doesn't seem to be on the arXiv yet, but you can see a video of Caprace giving a talk about them here: https://www.newton.ac.uk/seminar/20200220160017002 .
In the direction of a zoo: it's sporadically updated, but have you seen http://www.eg-models.de/models_noapplet.html ? It has many small triangulations of various nice spaces, with pictures + data.
|
2025-03-21T14:48:31.205671
| 2020-06-07T16:15:30 |
362430
|
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|
Stack Exchange
|
Stochastic processes with convergent fi-di distributions but no tightness
I am looking for natural and interesting examples of stochastic processes with convergent finite-dimensional distributions but no tightness. As for the space, it is either $C[0,1]$ equipped with the uniform norm or $D[0,1]$ equipped with the Skorokhod topology $M_2$. Here I've chosen the topology $M_2$ since it is weaker than the Skorokhod topologies $J_1$ and $M_1$, so tightness in $M_2$ is the least restrictive.
What I am up to are examples of interest by their own and not specifically crafted to answer the question. Say, some type of stochastic processes that was the subject of a reasonable paper, where one of the main results proves only convergence of fi-di distributions, at least in some cases.
UPD: My best example is on partial sums of moving averages of heavy-tailed i.i.d. random variables, where the coefficients have different signs. The main point of this example is that it combines naturally few modes of tightness and non-tightness in various Skorokhod topologies.
Let $(X_n)_{n \ge 0}$ be i.i.d. r.v.'s from the domain of normal (for simplicity) attraction of an $\alpha$-stable law with $\alpha \in (0,2)$. Consider the sequence of moving averages $Y_n:=a_0 X_n + a_1 X_{n-1}$ and the processes of their partial sums $S_n(t):= \sum_{k=1}^{[nt]} Y_k$, where $t \in [0,1]$. Put $a:=a_0 + a_1$ and assume that $a \neq 0$.
Then $S_n(t) / n^{1/\alpha}$ converges weakly to $a S(t)$ for every $t$, where $S$ is the limiting $\alpha$-stable process. The same clearly holds for all fi-di distributions. Then:
-If $a_0 \neq 0, a_1= 0$, then $S_n / n^{1/\alpha}$ converges weakly to $aS$ in $(D, J_1)$.
-If $a_0 \neq 0, a_1 \neq 0$, then the distributions of $S_n / n^{1/\alpha}$ are not $J_1$-tight.
-However, if $a_0 a_1 \ge 0$, then $S_n / n^{1/\alpha}$ converge weakly to $aS$ in $(D, M_1)$ (Avram and Taqqu, 1992).
-If $a_0 a_1 <0 $, then the distributions of $S_n / n^{1/\alpha}$ are not even $M_2$-tight.
There are examples of moving averages of higher orders where there is $M_2$-convergence (with proofs due to Basrak and Krizmanic, 2014).
Here's a couple of "naturally occurring" situations when no tightness is known, but alas in a more complicated setting. One is loop ensemble in the double dimer model, see arXiv:1809.00690; the tightness is expected to hold but the proof is missing. Another is the energy field in the critical 2d Ising model https://archive-ouverte.unige.ch/unige:18163; the scaling limits of correlations blow up so badly on the diagonal that no distribution-valued continuous field with such correlations exists.
How about $\min\{n\cdot \mathrm{dist}(t,X),1\}$, where $X$ is your favorite point process, say, the Poisson process?
For an example with Markov property, consider a continuous time Markov chain with two states $0,1$ and transition intensities $1$ and $n$.
In a somewhat different flavor, take the process that is given by independent Bernoulli at the points of $n^{-1}\mathbb{Z}$ and extended to be constant on each $\left[\frac{k-\frac{1}{2}}{n};\frac{k+\frac{1}{2}}{n}\right)$.
Did you mean something like $\max{ n - n \cdot dist(t, X), 0}$? Otherwise your sequence, divided by $n$ (which is a natural thing to do) converges to $dist(t,X)$. Am I wrong?
oh yes, I meant min not max.
Thanks! Still, this does not feel to me as a natural example: it seems that it is neither related to something substantial nor it is a particular case of something more general. Here it is clear the limit process should be described in a different way.
Btw, I removed my original example since it was not that good. Yours is better, in a sense.
@Vysotsky, I added two more examples, maybe they would feel more natural... As for the general principle, all examples have a high-frequency component that is either not seen by point evaluations or prevents the limiting process from existing in the first place.
Thanks for the further ideas! They are natural as sort of counter-examples, I agree. Apparently, I was not good in explaining what I actually wanted. So I updated the question, and rewritten my original example. Should I explain more/better?
|
2025-03-21T14:48:31.205959
| 2020-06-07T16:19:45 |
362431
|
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|
Stack Exchange
|
$x^3+x^2y^2+y^3=7$, and solvable families of Diophantine equations
(a) Do there exist integers $x$ and $y$ such that $x^3+x^2y^2+y^3=7$ ?
(b) Is this equation belongs to some family $F$ of equations for which there is a known algorithms for testing if they have an integer solution? For example, such algorithms are known for quadratic equations (in $n$ variables) and for cubic equations in 2 variables, but this equation has degree 4.
(c) Reference request: can you recommend a good book/survey/website which would help to determine if a given (reasonably simple) Diophantine equation belong to some solved case, or is new?
(a) No. There are no integer solutions. The curve $C$ you give has genus $3$ and it has an obvious automorphism $\phi(x,y) = (y,x)$. The quotient curve is an elliptic curve. In particular, if you let $X = -(x+y)$ and $Y = xy$, then the equation becomes $E : Y^{2} + 3XY = X^{3} + 7$. So any integer point on your curve gives rise to an integer point on $E$. As you note, cubic equations in two variables have algorithms for finding their integer points. This elliptic curve has rank $1$ and its integral points are $(-3,4)$, $(-3,5)$, $(186,-2831)$ and $(186,2273)$. None of these lead to integer points on $C$.
(b) There are not many general families of such curves. For curves of genus $0$, there are algorithms for finding their integer points. For curves of genus $1$, the existing procedures rely on being able to compute the Mordell-Weil group and it is currently open whether there is an algorithm to do so. One other family that might be worth mentioned is the family of Thue equations, those of the form $F(x,y) = k$ where $k$ is a constant and $F$ is a homogeneous polynomial in two variables.
(c) For a survey of what number theorists can and cannot handle algorithmically, you might consult Henri Cohen's book "Number Theory, Volume I: Tools and Diophantine Equations." Chapter 6 in that book gives a nice survey of simple Diophantine equations together with some techniques that might suffice to handle those. The class of equations that can be definitely be handled in a systematic or algorithmic way is quite small. (If I had been unable to compute the Mordell-Weil group of $E$ above, the software would not have been able to provably find all the integral points.)
Thank you, Jeremy!
Let me post an elementary answer to this question, which I have found recently.
(a) Assume that $(x,y)$ is an integer solution. From symmetry, we may assume that $|y|\geq |x|$. If fact, cases $|y|=|x|$, $|y|=|x|+1$, and $|x|\leq 3$ can be checked directly, hence we may assume that $|y|\geq |x|+2>5$.
Now rewrite the equation as
$$
(x^2+y)(y^2+x)=xy+7,
$$
from which it follows that
$$
k = \frac{xy+7}{y^2+x}
$$
is an integer. However,
$$
y^2+x \geq |y|(|x|+2)-|x|=|xy|+2|y|-(|y|-2)=|xy|+|y|+2>|xy|+7\geq |xy+7|
$$
hence $|k|<1$, and it can be an integer only if $k=0$, or $xy+7=0$, which is impossible for $|y|\geq |x|+2>5$.
(b) This equation belongs to family of equations satisfying Runge's condition, for which there exists a general algorithms, see, for example
Walsh, P. A quantitative version of Runge's theorem on Diophantine equations, Acta Arithmetica, 62, 157--172, 1992.
|
2025-03-21T14:48:31.206174
| 2020-06-07T16:43:30 |
362433
|
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|
Stack Exchange
|
Research Problem combining Algebraic Geometry and QFT
A student who specialises in algebraic geometry has contacted me to ask if they could collaborate with me on some problem which relates mathematical physics and algebraic geometry. I think his idea is that he knows more about algebraic geometry and I know more about physics and QFT, so he is asking if there is some problem which combines both where we would both contribute decisively.
I would be keen to stay away from string theory for now if possible, so I was wondering if there is some interesting research problem at the moment which combined both QFT and algebraic geometry (I can be more specific about the type of algebraic geometry which he does if necessary if the problem needs to be narrowed down further).
The problem of categorification can be seen as in the intersection of TQFT and algebraic geometry, see here.
This older paper of a classmate of mine was in this intersection of areas: https://arxiv.org/pdf/1112.6208.pdf and may be of interest.
There has been a lot of work on the algebraic geometry of Feynman integrals (just search for "periods, motives, Feynman integrals"). You might also consider the direction started by Arkani-Hamed et al. in https://arxiv.org/abs/1212.5605
If you count matrix models or more general integrable systems as QFT, these have a lot to do with the algebraic geometry of the spectral curve. See the review article https://arxiv.org/abs/1510.04430 .
|
2025-03-21T14:48:31.206310
| 2020-06-07T17:01:09 |
362436
|
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|
Stack Exchange
|
Can we make Pres *-autonomous?
The category $\mathbf{Sup}$ of sup-lattices (posets admitting all supremum and supremum preserving map between them) is a well known example of a $*$-autonomous category:
The internal Hom is simply the set of supremum preserving map with the pointwise ordering between them, map $X \otimes Y \to Z$ corresponds to order preserving map $X \times Y \to Z$ that preserves supremum in each variable, the $*$-involution takes a poset $P$ to its opposite ordering and map $f:P \to Q$ to its right adjoint, seen as a supremum preserving map $Q^\text{op} \to P^\text{op}$. The dualizing object is $\Omega^\text{op}$, with $\Omega$ the subobject classifier (so $\{0,1\}$).
In particular, the tensor product can be described by the usual:
$$ X \otimes Y = \text{Hom}(X,Y^\text{op})^\text{op} $$
i.e. as the set of infimum preserving maps $X^\text{op} \to Y$.
The category $\mathbf{Sup}$ is the category of "locally presentable posets, i.e. locally presentable $(0,1)$-categories". But as soon as we move one step up the categorical ladder, to the (bi)category $\mathbf{Pres}$ of locally presentable categories and colimit preserving (equivalently, left adjoint) functors between them, we no longer have a $*$-autonomous category.
Still much of this structure is still present on $\mathbf{Pres}$: It is a monoidal closed category, where the internal hom is given by the category of all left adjoint functor $C \to D$. The tensor product was constructed by Bird (see section 5 of his thesis) following previous work of Kelly, and is such that colimit preserving functors $C \otimes D \to E$ are the same as functors $C \times D \to E$ preserving colimits in each separate variables.
And in fact, $\mathbf{Pres}$ still feel very much like a $*$-autonomous category, again with opposite category and taking right adjoint as the $*$-involution, in the sense that we still have formulas like:
$$ \text{Ladj}(C,\text{Set}^\text{op}) \simeq C^\text{op} $$
$$ C \otimes D \simeq \text{Ladj}(C,D^\text{op})^\text{op}$$
But it is not a $*$-autonomous category for the good reason that the opposite a locally presentable category is almost never locally presentable (it is if and only if it is an object of $\mathbf{Sup} actually).
My question is, can we modify $\mathbf{Pres}$ (by adding new objects) to make it a $*$-autonomous category in a nice way ?
I am relatively open about the sort of category (let's call it $\mathbf{A}$) I want, here is a list of the type of properties I would like to have.
$\mathbf{A}$ should be a $*$-autonomous category.
$\mathbf{A}$ should contains $\mathbf{Pres}$ as a monoidal full subcategory, closed under exponential.
But I still would like objects of $\mathbf{A}$ to be category in some sense:
, and the formula above fits into a $*$-autonomous structure of $\mathbf{A}$:
$\mathbf{A}$ should have a forgetfull functor $U$ to the category of categories and left adjoint functor between them.
$U$ restricted to $\mathbf{Pres}$ should be the obvious functor.
$U(C^*)$ should be $C^\text{op}$ and if $F$ is a functor $U(F^*)$ should be a right adjoint of $U(F)$.
I don't think it is reasonable to expect that $U$ could be fully faithful, so I'm fine with $\mathbf{A}$ being a category of "category with structure", as long as $U$ is fully faithful on $\mathbf{Pres}$.
$\mathbf{Pres}$ is also known to have all small limits and colimits, I would like these to be preserved by the inclusion to $\mathbf{A}$. I don't think it is reasonable to expect that $\mathbf{A}$ will have all small limits and colimits as well, but the more it has the better.
Any negative results showing that this whish list, or some other reasonable wish list is too strong would also be greatly appreciated.
Precision: I'm aware of Mike Shulman recent preprint showing that any monoidal closed category can be fully faithfuly embeded in a $*$-autonomous category, with an embedding preserving tensor product and exponential (and even some colimits). In fact I had this question in mind for a long time, but I've thinking about it again recently because of this preprint.
So, assuming Mike's construction apply to $\mathbf{Pres}$, despite it not being a $1$-category nor locally small, that is definitely a good place to start. But I havn't really been able to parse what this construction would give in this case, nor if it comes with a forgetful functor to Cat. Also, I got the impression that this construction was maybe to general for this example ? Any comment on this would also be very welcome.
I disagree with you, Simon. The analogue of Sup is the 2-category of cocomplete categories. It happens that a cocomplete poset is complete, thus the opposite of a suplattice is a suplattice. For categories, you could maintain the analogy by looking at complete and cocomplete categories. Notice that even in Cat a cocomplete category is always very close to be complete (Adamek et al: Cocompleteness almost implies completeness).
The fact that loc. pres. categories are complete is just a phenomenon of the more general pattern that every total category is complete. The generator forces the limits to exist computing them as "sups of lower bounds".
Ok, let say that what I really mean is that the $0$-categorical analogue of $\mathbf{Pres}$ is $\mathbf{Sup}$, which I think you will agree with. I could argue that the $0$-categorical analogue of the category $\mathbf{CoComp}$ of co-complete categories is really the category $\mathbf{SUP}$ of possibly large posets having supremum of small families, which would contradict what you say. Though these matters are somehow subjective, so I guess we can agree to disagree ^^
Ahah, you are somehow missing my point. The 2-category of complete and cocomplete categories (and cocontinuous functors) is $*$-autonomous replacing adjoints with (co)continuous functors. The existence of a generator in a loc. pres category just makes the whole situation tamer. In a way, I am claiming that you already have the answer, you just care too much about (co)generating sets and smallness assumptions. From my point of view, we agree to agree.
Let me rephrase my previous comment. There exists a very natural fully faithful embedding of $\textbf{Pres}$ in a $*$-autonomous category preserving the monoidal closed structure. That is the inclusion of $\textbf{Pres}$ in the (2-)category of complete and cocomplete categories and cocontinuous functors.
Is it ? I don't think the tensor product of complete & co-complete categories is defined. Also if you want 'op' to be a contravariant functor you need morphisms to have right adjoint.
You are just right, I'll be thinking about this. One could try to fix what I said in a technical way (choosing only left adjoints, choosing total categories instead of complete and cocomplete ones), but I will try to come back with a more mature observation before shooting in the sky.
|
2025-03-21T14:48:31.206739
| 2020-06-07T17:29:59 |
362439
|
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|
Stack Exchange
|
A question on the use of fractional derivatives in Riemann Hypothesis
We already know that Riemann-zeta function on the critical band is defined as follows:
$$(1-2^{1-\alpha})\zeta(\alpha) = \sum_{k=1}^{\infty} (-1)^{k+1}k^{-\alpha},\quad \Re(\alpha) \in ]0, 1[ $$
Is it possible to say that
$$ (1-2^{1-\alpha})\zeta(\alpha) = \left[\frac{d^{-\alpha}}{dw^{-\alpha}} \sum_{k=1}^{\infty}(-1)^{k+1}e^{iwk}\right]_{w=0} = \left.
\frac{d^{-\alpha}}{dw^{-\alpha}}\frac{e^{iw}}{1+e^{iw}}\right|_{w=0} \;?
$$
how can this fractional derivative
$$\left.
\frac{d^{-\alpha}}{dw^{-\alpha}}\frac{e^{iw}}{1+e^{iw}}\right|_{w=0} = \left.
\frac{d^{\alpha-1}}{dw^{\alpha-1}}\frac{e^{iw}}{1+e^{iw}}\right|_{w=0} \ $$
be evaluated or numerically estimated using the fractional derivative or antiderivative definition that has exponentials as eigenfunctions?
Thanks
Note. I know this method does not fully comply with absolute convergence and I am not sure if using such fractional derivatives makes sense.
(Disclaimer I need to verify if your integral of $\frac{1}{1+e^{iw}}$ trick works but here's an alternative framework which should meet your needs)
So we wish to evaluate
$$(1-2^{1-a} ) \zeta(a) = \sum_{k=1}^{\infty} (-1)^{k+1} k^{-a} $$
Our first trick is to consider the function
$$ f = \frac{e^x}{1+e^x} = e^x - e^{2x} + e^{3x} - ... $$
So if we take integrals of this we end up with your function namely:
$$ \frac{d^{-a}}{dx^{-a}} \left[ f(x) \right] |_{x = 0} = (1-2^{1-a}) \zeta(a)$$
Now if you pick your integration bounds carefully there probably is way to make that work. If we want to use the fractional "Derivative" as opposed to the integral our strategy changes a bit.
We want to fractionally differentiate the following function
$$ g= e^{x} - e^{\frac{1}{2}x} + e^{\frac{1}{3}x} - ... $$
But this series does not converge the way its given, so we can expand all the exponentials to yield
$$ g = \left( 1 - 1 + 1 - ...\right) + \left( 1 - \frac{1}{2} + \frac{1}{3} - ... \right) x + \left( 1 - \frac{1}{4} + \frac{1}{9} - .... \right) $$
We can sum these
$$ g = \frac{1}{2} + \ln(2)x + \frac{\pi^2}{12}x^2 ... $$
So now we have that for this magic $g$
$$ \frac{d^{a}}{dx^a} [g] |_{x=0} = (1 - 2^{1-a}) \zeta(a) $$
But this is not as profound as it may seem initially...
Really given any function $f(a)$ if we consider the taylor series
$$ h(x) = \frac{1}{0!}f(0) + \frac{1}{1!}f(1)x + \frac{1}{2!}f(2)x^2 + ... $$
Then:
$$ \frac{d^{a}}{dx^{a}}[h(x)]|_{x=0} = f(a) $$
And what we have done above is a restatement of that... (Quite literally this is how you get the definition of the gamma function).
Now I'm actually pretty poor with complex integration so I need to review how to evaluate your integral for the first definition (although I suspect i'll end up refinding the functions you found after using a suitable contour).
Thank you, you helped me spot a missing $e^{iw}$
Btw if you liked this answer please accept :)
|
2025-03-21T14:48:31.206919
| 2020-06-07T17:44:25 |
362442
|
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|
Stack Exchange
|
Finding automorphisms and cyclic modules via QPA
Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.
Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)
Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.
Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?
Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?
Question 4: Assume we know that $A$ as a bimodule admits a projective resolution (not necessarily minimal) of the form:
$... \rightarrow A^e \rightarrow ... A^e \rightarrow A^e \rightarrow A \rightarrow 0$,so that every term can be choosen to be the regular module of the enveloping algebra. This means that $\Omega_{A^e}^i(A)=x_i A^e$ for some elements $x_i \in A^e$. Is there a canonical choice of the $x_i$ or some nice behavior? Can one obtain the $x_i$ in a nice form via QPA?
More generally, can we obtain a minimal free resolution of a module in QPA instead of a minimal projective resolution?
This is possible by filtering through all elements but in practise this takes too long.
Assume that we start with an admissible quotient $A = kQ/I$ for a path algebra over a field $k$ and a finite quiver $Q$. In addition assume that $\Omega^n_A(A/rad) \simeq A/rad$ as right $A$-modules, and if necessary $\Omega^n_A(S)$ is simple for all simple right $A$-modules $S$. This is all you need of assumptions. It will follow that the algebra is selfinjective.
Question 1:
Find $n$ such that $\Omega^n_A(S)$ is simple for all simple right $A$-modules $S$.
Find $B = \Omega^n_{A^e}(A)$.
Construct the natural algebra homomorphism $A \to A^e$ via the command g := TensorAlgebrasInclusion( A^e, 2 ).
Find $B_A$ via RestrictionViaAlgebraHomomorphism( g, B ).
Find isomorphism $\varphi\colon A_A\to B_A$ using IsomorphismOfModule( A, B ).
Find $\varphi( 1 )$. Call it $b$.
Define $f\colon A\to A$ by letting $f(a) = \varphi^{-1}(ab)$.
Then the automorphism $f$ is found.
Question 2: Don't know any better than the above. All taken from Theorem 1.4 in
Green, Edward L.; Snashall, Nicole; Solberg, Øyvind, The Hochschild cohomology ring of a selfinjective algebra of finite representation type., Proc. Am. Math. Soc. 131, No. 11, 3387-3393 (2003). ZBL1061.16017.
Question 3: The element $x$ can be taken as the element $\varphi(1) = b$ given in Question 1.
Question 4: Here already for $n = 2$ the assumptions needed for the above should be satisfied just using dimension arguments. Using the above, or maybe just taking the minimal generator of $\Omega^i_{A^e}(A)$ would give you the $x_i$ that you want.
|
2025-03-21T14:48:31.207231
| 2020-06-07T18:01:59 |
362443
|
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|
Stack Exchange
|
An alternative proof of Künneth spectral sequence, independent of Künneth formula for homology
I am currently reading Künneth spectral sequence, which is given below.
Let $R$ be a ring and A$=\big\{A_n,d_n:A_n\longrightarrow A_{n-1}\big|d_{n-1}\circ d_n=0\big\}_{n\in \Bbb Z}$ be a chain complex of right $R$-modules with each $A_n$ is flat and $A_n=0$ when $n<0$. Let C$=\big\{C_n,\partial_n:C_n\longrightarrow C_{n-1}\big|\partial_{n-1}\circ \partial_n=0\big\}_{n\in \Bbb Z}$ be a chain complex of left $R$-modules with $C_n=0$ if $n<0$. Then, We have two spectral sequences $^\text{I}E,\ ^\text{II}E$ both converging to $H_n(\textbf{A}\otimes_R\textbf{C})$ such that $$^\text{II}E^2_{p,q}\simeq\bigoplus_{l+k=q}\text{Tor}^R_p\bigg(H_k(\textbf{A}), H_l(\textbf{C})\bigg)\text{ and }^\text{I}E^2_{p,q}=\begin{cases} H_p(\textbf{ A}\otimes_R\textbf{ C}) & \text{ if }q=0,\\0 & \text{ if }q\not=0.\end{cases}$$
Now, the proof based on Künneth formula for homology. Also, these spectral sequences can give an alternative proof of Künneth formula for homology, when chain complexes are positive. So, the argument is somewhat cyclic. I am looking for an alternative proof of Künneth spectral sequence. Is it possible? Any help will be appreciated. Thanks in advance.
I know a simple proof of the Künneth SS that does not use the Künneth formula. However, it uses the derived tensor product and the derived categories. Would that be satisfactory?
Could you tell me some more, like, where it can be found?
It's essentially the proof of theorem 8.1 in Elmendorf, Kriz, May rephrased in the language of complexes instead of spectra (or, if you want, just by taking modules over the Eilenberg-MacLane spectrum $HR$)
|
2025-03-21T14:48:31.207622
| 2020-06-07T18:44:28 |
362444
|
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|
Stack Exchange
|
Two definitions of $L^p$ spaces that are not always equivalent
There are two definitions of $L^p(S, \Sigma,\mu)$ in the literature. (Here $S$ is a set, $\Sigma$ is a $\sigma$-algebra of subsets of $S$ and $\mu$ is a positive measure.) The two definitions are the same in some (most?) cases (such as if $\mu$ is $\sigma$-finite) but are not always the same. One seems to be standard in most graduate textbooks and the other is from Dunford and Schwartz's Linear Operators, Part I: General Theory (abbreviated DSI). Can anyone recount the history and present day
usage of the two definitions? For easier reference the two definitions are recalled here.
The first definition is taken from: Michael E Taylor, Measure Theory and Integration, Graduate Studies in Mathematics, volume 76, Amer Math Soc. 2006, page 43. It is:
Definition 1.
$L^p(S, \Sigma,\mu)$ is the set of all (equivalence classes) of $\Sigma$-measurable functions $f$ so that
$$
\|f\|_{p}^p:= \int_S \lvert f\rvert^p d \mu < \infty \text{ if $1\le p <\infty$}
$$
or the essential supremum of $\lvert f\rvert$ is finite if $p=\infty$.
(The equivalence classes are for the relation $f \equiv g$ if
$f=g$ $\mu$-almost everywhere.)
Here $\Sigma$-measurable is (according to DSI page 240) defined as: A function $f:S \to \mathbb{C}$ is $\Sigma$-measurable if $f^{-1}(B) \in \Sigma$ for all Borel subsets $B$ of the complex plane $\mathbb{C}$.
The second definition is from DSI, page 119:
Definition 2.
It is exactly the same as Definition 1 except that “$\Sigma$--measurable” is replaced by “$\mu$-measurable”.
Let us recall the definition of $\mu$-measurable from DSI.
First they introduce (DSI page 101) the topology associated with convergence in $\mu$ measure.
A function $f:S \to \mathbb{C}$ is said (page 106) to be totally $\mu$-measurable if it belongs to the closure of the set of all $\Sigma$-measurable simple (complex valued) functions.
A function $f:S \to \mathbb{C}$ is said to be $\mu$-measurable if $\chi_E f$ is totally $\mu$-measurable whenever $E \in \Sigma$ has finite $\mu$ measure. (Here $\chi_E$ denotes the characteristic function of $E$.)
In general $\Sigma$-measurable implies $\mu$-measurable but not conversely. Hence Definition 1 defines a smaller set $L^p$ than Definition 2. Is it strictly smaller? Not if $\mu$ is $\sigma$-finite. When then is there a difference?
Following DSI, page 296 we define $\Sigma_1$ as follows. We suppose (without loss of generality)
that $(S,\Sigma,\mu)$ is complete (so that $\Sigma$ contains all subsets of sets in $\Sigma$ of measure zero).
Define $\Sigma_1$ to be the family of all subsets $E$ of $S$ so that $A \cap E \in \Sigma$ whenever $A \in \Sigma$
has finite measure. Certainly $\Sigma_1 \supseteq \Sigma$ and if the containment is proper and if $E \in \Sigma_1\setminus \Sigma$ then $\chi_E$ belongs to $L^\infty(S,\Sigma, \mu)$ as defined in DSI (second definition), but is not in $L^\infty(S,\Sigma, \mu)$ as defined in the first definition. Constructing an example where $\Sigma_1 \neq \Sigma$ seems to be straightforward.
Surely this is known and noted somewhere in the literature. Can anyone direct me?
When you quote the definition of $L^p$ from Taylor, you still refer to DSI for the definition of the subterm "$\Sigma$-measurable". Is their definition the same as Taylor's?
Taylor calls $\Sigma$--measurable, simply measurable like many authors.
Re: Whatever Taylor calls the notion, is the definition the same as the one you cite from DSI? (I don't have access to either book right now, so I can't check.)
Yes. (I had to choose one author's terminology) Denis
The property Σ=Σ_1 amounts to (X,Ε,μ) being locally determined.
A measure space (X,Σ,μ) is locally determined if μ is semifinite
and A∈Σ if and only if A∩F∈Σ for all F∈Σ such that μ(F) is finite.
See Fremlin, Measure Theory, Definition 211H.
Almost all measurable spaces that arise in practice (e.g., from Radon measures)
are strictly localizable and therefore locally determined by Theorem 211L(d) in op. cit.
A measure space (X,Σ,μ) is strictly localizable if it can be partitioned
into a disjoint family of measurable subsets of finite measure such that
(X,Σ,μ) is the disjoint union (coproduct) of the resulting measure spaces.
The closely related (but slightly weaker) property of being localizable
amounts to saying that μ is semifinite and the Boolean algebra Σ/N
is complete, where N is the σ-ideal of sets of μ-measure 0.
Localizability is equivalent to the following results in measure theory:
the Hahn–Jordan decomposition theorem for signed measures that are absolutely continuous with respect to μ, the Riesz representation theorem,
the Radon–Nikodym theorem, and the fact that bounded measurable
functions form a von Neumann algebra.
Thus, once you move outside of the domain of localizable measure spaces,
all of measure theory falls apart and you have much bigger problems
than L^p-spaces.
As for the remaining case of localizable measure spaces that
are not strictly localizable, the known examples are quite pathological,
but in any case, they can be improved to a complete locally
determined measure space
(see Proposition 213D in op. cit.)
and by Proposition 213H(d) there is no practical difference
between the two spaces.
In fact, for any localizable measure space
we can find a strictly localizable measure space
with an isomorphic Boolean algebra Σ/N,
which for all practical purposes can be used instead of the original space.
The Hahn-Jordan decomposition theorem is stated on Wikipedia without any preconditions
@ogogmad: It's a different Hahn–Jordan decomposition theorem (one measure instead of two). I added a clarification.
Sounds like the Lebesgue decomposition theorem, no?
@ogogmad: No. I am not even sure what you are trying to say here, since the Lebesgue decomposition theorem is trivial in the case of absolute continuity, which is a precondition for the Hahn–Jordan theorem.
Wikipedia says that Lebesgue decomposition requires that the measures $\mu$ and $\nu$ be $\sigma$-finite. This can surely be relaxed, but it doesn't hold unconditionally
@ogogmad: First of all, the Lebesgue decomposition theorem does not require σ-finiteness or any other conditions (see the nLab). Secondly, as I already explained, it has nothing to do with the Hahn–Jordan decomposition, which decomposes a signed measure into its positive and negative components.
Can you link to a reference to your Hahn-Jordan decomposition theorem?
@ogogmad: This is Theorem 2 in Kelley's paper that is linked in the answer.
The name of the linked paper by Kelley: Decomposition and representation theorems in measure theory.
|
2025-03-21T14:48:31.208065
| 2020-06-07T19:09:11 |
362448
|
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|
Stack Exchange
|
Hypercylic operators with sets of hypercyclic vectors almost covering the space
Let $\{T_i\}_{i \in I}$ be a family of hypercylic operators on a separable Banach space $X$. From the transitivity theorem, we know that $HC(T_i)$, the set of vectors $x \in X$ with $\{T_i^n(x):n \in \mathbb{N}\}$ dense in $X$, is itself a dense $G_{\delta}$ subset of $X$.
It is known, or when is it the case that:
There exists a finite-dimensional subspace $Y$ of $X$ such that
$$
X-\bigcup_{i \in I} HC(T_i)\subset Y$$
There is no finite-dimensional subspace $Z\subseteq X$ such that
$$
X-\bigcup_{j=1}^n \, HC(T_{i_j})\subseteq Z
$$ for any finite sequence $i_1,\dots,i_n$ in $I$
|
2025-03-21T14:48:31.208154
| 2020-06-07T19:10:07 |
362449
|
{
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|
Stack Exchange
|
Categorification of probability theory: what does a "probability sheaf" tell us (if anything) about probability theory?
Disclaimer: I only have a superficial knowledge of what category theory and related subjects are concerned with.
So, my understanding is that category theory and related fields of higher mathematics are meant to (A) better organise knowledge within a fields of mathematics (B) build powerful bridges between otherwise dislocated branches of mathematics (C) elevate concepts to a higher level of generality / abstraction so as to expose them to a more unified treatment (which can give one a better understanding of problems / questions in these fields, etc.).
I found Simpson - Probability sheaves and the Giry monad, in which a sheaf for probability theory was constructed. In my layman's quick reading, the manuscript doesn't seem to (1) tell anything new / non-trivial about probability theory (2) help in anyway organize concepts already present within probability theory. So my question is
Question
How does the construction of a sheaf for probability theory (see paper cited above) help probability theory?
Has any such attempts (to "categorify" probability theory) been made before?
Thanks in advance for any enlightenment!
I think of sheafification and categorification as different things, but you seem to be conflating them. Is that so, or are you just referring to the fact that the cited work seems to do both?
@LSpice I've heard that a topos is a perculiar kind of category, and topos theory is just a a sytnhesis of the different kinds of sheaves (P. Cartier). So, I wouldn't consider my wording (i.e referring to construction of sheaves as "categorification") too conflating. No ?
@dohmatob: both “sheafification” and “categorification” have acquired rather specific meanings that are a bit different from just “applying the methods of sheaves/categories”. Calling this “sheafification” would sound entirely wrong to me (that means the construction turning a presheaf into a sheaf). Calling this “categorification” sounds much less wrong, but still a bit misleading (that can sometimes mean “reorganising with categorical methods”, but more usually means “replacing set-based structures with analogous category-based structures”).
@PeterLeFanuLumsdaine OK, thanks for the explanation. Indeed, "categorification" sounds "less wrong" here.
Probably related? It is a question of mine: https://mathoverflow.net/questions/357184/are-étalé-spaces-a-thing-for-probability-spaces
I'm not an expert on the sheaf-theoretic approach to probability theory, but a quick look at the paper you're asking about shows that it's a 6 page conference proceeding from 2017 that defines a new notion and reads like a set of lecture notes. I think it's a bit early to be asking for big applications to probability theory. That said, if you use Google Scholar to search for who cites Alex Simpson's paper, you find two. The first is about constructive measure theory, so should be of interest to logicians and others wanting a firmer foundation for probability theory. The second is a PhD thesis that goes much more in depth than Simpson's paper about the properties of "probability sheaves" and investigates connections to topos theory.
From the abstract: "In this dissertation, we emphasize how sheaves and monads are important tools for thinking about modern statistical computing." The abstract goes on to advertise applications to hypothesis testing and the analysis of data sets with missing data, pretty important topics. Chapter 3 contains lots of history of previous attempts to bring probability theory under the umbrella of category theory, and also includes applications to probabilistic programming (whatever that is). If you're interested in this field, I think you will want to read these references (plus the blog post by Tao that Simpson cites), and you may need to give it time before super compelling applications arrive.
Thanks for the detailed input and refs.
That dissertation looks interesting. A potential application might be that a more categorical approach to probability theory could lead to approaches to statistical computing which are more natural for functional programming, especially with languages like Haskell that are designed with category theory in mind.
Coincidentally, a day after you asked this question, Alex Simpson gave a nice talk (video, slides) where he gave a synthetic formulation of probability theory. In this formulation, random variables are a primitive notion, not maps from a sample space to a measurable space. Hence there's no need to keep track (or even mention) sample spaces at all. That's basically how probability theory was done informally, long before it was encoded in set theory. Several prominent mathematicians (Rota, Tao, Mumford) had suggested that such a reformulation of probability theory would be desirable.
I'd consider this an application of "categorical" probability theory, since I suspect that Simpson arrived at this axiomatisation via the categorical sheaf model he had constructed earlier.
In the sprit of being "constructively curmudgeonly": does the talk explain whether the hope/claim is to refute the quote from Williams's book, or merely to say that Williams's admittedly sweeping claim should not be taken as gospel?
FWIW I don't think that a synthetic approach to RVs is impossible, but I would need more convincing that the subtleties of stochastic processes can also be handled this way. My inexpert intuition is that continuity/measurability issues for stochastic processes are akin to size issues with constructions in category theory, as discussed at https://mathoverflow.net/questions/365947/when-size-matters-in-category-theory-for-the-working-mathematician/ -- the technical issues can be dealt with, but they are real, and work has to be done to handle them, just as the traditional approaches involve work
|
2025-03-21T14:48:31.208576
| 2020-06-07T19:10:11 |
362450
|
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|
Stack Exchange
|
Decomposing a compact connected Lie group
I want to prove the following. Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak g$ and center $Z_G.$ It is not hard to prove that $\mathfrak g$ is reductive. Therefore, we can decompose $\mathfrak g=Z_{\mathfrak g}\oplus[\mathfrak g,\mathfrak g]$ as direct sum of ideals where $[\mathfrak g,\mathfrak g]$ is semisimple. Let $G_{\mathrm{ss}}$ be the analytic subgroup of $G$ with Lie algebra $[\mathfrak g,\mathfrak g].$ Then we have $G=(Z_G)_0G_{\mathrm{ss}}$ where $(Z_G)_0$ is the identity component of $Z_G.$
I am reading "Lie Groups Beyond Introduction" by Knapp where the following argument is given. Choose covering groups $\widetilde{G_{\mathrm{ss}}}$ of $G_{\mathrm{ss}}$ and $\widetilde{(Z_G)_0}$ of $(Z_G)_0$ respectively. Then $\widetilde{G_{\mathrm{ss}}}\times\widetilde{(Z_G)_0}$ is a covering group of $G_{\mathrm{ss}}\times Z_G$ with Lie algebra $Z_{\mathfrak g}\times [\mathfrak g,\mathfrak g]$ which is isomorphic to $\mathfrak g.$ Therefore, we must have $\widetilde{G_{\mathrm{ss}}}\times\widetilde{(Z_G)_0}$ is a covering group of $G.$ I understand till this point. But it seems that from this one can prove at once what I have proposed. How is it so?
You might argue along the lines that $\widetilde{G_{ss}}\times \widetilde{(Z_G)0}$ is a covering group of $(Z_G)0 G{ss}$, that $(Z_G)0 G{ss}\subset G$, and $G$ cannot be any bigger than $(Z_G)0 G{ss}$ and still have $\widetilde{G{ss}}\times \widetilde{(Z_G)_0}$ as a covering group
@fierydemon. I think much simpler argument avoiding covering groups exists. Consider the map $(g,h)\mapsto gh$ from $(Z_G)0\times G{ss}$ to $(Z_G)0G{ss}.$ The derivative of this map is clearly onto and hence an isomorphism as the corresponding Lie algebras have the same dimension. Thus the map is local diffeomorphism. As the exponential map is a local diffeomorphism and $G$ is connected. The map $(g,h)\mapsto gh$ is surjective.
|
2025-03-21T14:48:31.208718
| 2020-06-07T19:12:16 |
362451
|
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|
Stack Exchange
|
Partitioning bases of vector spaces
Let $V$ be a $\mathbb{F}_p$-vector space of dimension $d$. Set $W=\bigoplus_{1\leq i\leq n} V$ and let
$$S=\{w_i=(v_{i1},\dots,v_{in}): 1\leq i\leq nd\},$$
be a basis for $W$. I am wondering if the following statement holds: $S$ can be partitioned into $n$ sets, $B_1,\dots, B_n$, each of size $d$, such that for any $1\leq \ell\leq n$ the following set
$$
\{\pi_\ell(w): w\in B_\ell\}
$$
is a basis for $V$, where $\pi_{\ell}(x_1,\dots,x_n)=x_\ell$.
I don't know if the statement holds but I couldn't find a simple counterexample.
Use induction on $n$. For the step of induction, apply the following lemme to $V$ and $\bigoplus_{2\leq i\leq n} V$.
Lemma. If $w_i=(u_i,v_i)$ constitute a basis of $U\oplus V$, then the $w_i$ can be split into two groups such that the $U$-components of the first group and the $V$-components of the second group form bases in the corresponding spaces.
To prove the lemma, express the $w_i$ via some basis in $U$ and some in $V$. In the obtained $(a+b)\times (a+b)$ non-degenerate matrix, you need to find complementary minors of orders $a$ (in the first $a$ rows) and $b$ (in the last $b$ rows) both of which are non-degenerate. The existence of such minors is guaranteed by the general Laplace expansion of the determinant.
Dear IIya: Many thanks for your answer. Of course we all work with the Laplace expansion but actually I didn't know about the general Laplace expansion and it is indeed beautiful. Thanks again for your answer.
|
2025-03-21T14:48:31.208844
| 2020-06-07T19:49:59 |
362453
|
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|
Stack Exchange
|
Does sum of i.i.d. Bernoulli random variables with parameter $\lambda/n$ asymptotically converge to Gamma distribution?
Actually the question has more details than what it says in the title. Sorry about that I may described the question wrongly.
Let $X_1^n, X_2^n,\dots$ be i.i.d. Bernoulli random variables with parameter $\lambda/n$, i.e. $X_1^n \overset{d}{=}\operatorname{Be}(\lambda/n)$ with fixed $\lambda > 0$. Consider $$
T_i^n := \inf\{k : X_1^n + \cdots + X_k^n = i\}.$$
And I want to show that $$
\frac{T_i^n}{n}\xrightarrow[n\to\infty]{d}\text{Gamma}(i,\lambda).$$
This confuse me since we know that the sum of Bernoulli random variables asymptotically converges to Poisson distribution and I don't see any relationship between Poisson and Gamma distribution.
Can anyone help me out?
$\newcommand\la\lambda$ $\newcommand\nt{\lfloor nt\rfloor}$
For any natural $i,n,k$, let $S^n_k:=X^n_1+\dots+X^n_k$, with $S^n_0:=0$. Then for any real $t>0$
\begin{align}
P(T^n_i/n>t)&=P(T^n_i>\nt) \\
&=P(S^n_{\nt}<i)\to P(S_{\la t} <i) \\
&=\frac{\la^i}{\Gamma(i)}\int_t^\infty u^{i-1} e^{-\la u}\,du
\end{align}
(as $n\to\infty$), where $S_{\la t}\sim Poisson(\la t)$;
the convergence holds by the Poisson limit theorem;
the last displayed equality can be obtained by integrating by parts $i-1$ times.
Thus indeed, the distribution of $T^n_i/n$ converges to the gamma distribution with the shape paratemer $i$ and the rate $\la$.
How do you know that convergence relationship, i.e. $P(S^n_{\lfloor nt\rfloor}<i)\to P(S_{\lambda t} <i)$?
@MathislikeFriday : I have added this detail.
Thank you sir :)
|
2025-03-21T14:48:31.208975
| 2020-06-07T20:12:27 |
362458
|
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|
Stack Exchange
|
Random walk with cyclic randomness
Let $S=\{0,1\}^n$ be a binary string of length $n$. Suppose you pick a number $r$ at random from any distribution on $\{1,\ldots,R\}$ of your choice and randomly generate $r$ boolean hash functions $h_0, h_1,\ldots, h_{r-1}: \{0,1\}\to\{0,1\}$. Using these hash functions, you take a one-directional random walk on $S$, starting at time $t=1$ and index $i=1$ so that if $h_{t \bmod{r}}(S[i])=1$, then you move from index $i$ to $i+1$. In other words, if the value of the hash function, indexed by your current time, of the character of your current position is $1$, then you move forwards in $S$. The random walk ends when you've processed the entire string $S$, i.e., when you've hit index $n+1$.
Conditioned on the random walk ending, do there exist $R=\text{polylog}(n)$ and constant $c>0$ for which we can upper bound the probability that the random walk ends at time $t\equiv 0 \bmod{r}$ by at most $O\left(\frac{1}{R^c}\right)$? Note that for $R=O(1)$, the probability is bounded by $\frac{1}{2}$ depending on the realization of $h_{r-1}$ and for $R=\Omega(n)$ with sufficiently large constant, the probability is bounded by $O\left(\frac{1}{\sqrt{n}}\right)$ from anti-concentration.
Should the hash domain be larger than ${0,1}$?
No, the input to the hash functions is the symbol of the binary string $S$ at the current time/position, so I think ${0,1}$ is fine.
ok, fine, the term "hash" is misleading though, it is usually associated with a length compressing string function. You may have said that you added a random noise bit $\pmod 2$ to the $S[i].$
I think you misstated something. What's to stop us from taking large $R$, but always taking $r=1$ (since we are allowed to have it be taken from any probability distribution on ${1,\ldots,R}$)?
Yeah you're right, for $r=1$, it may be possible that $h_0(0)=h_1(0)=0$, so that the random walk never progresses (and in general), so I should have asked whether we can upper bound the probability, conditioned on the random walk terminating. Thanks for bringing it up, I've corrected the question.
Note that for $h_0(0)=h_0(1)=1$, the random walk always terminates at some time congruent to $0\pmod{1}=0\pmod{r}$, so we cannot upper bound the probability by $\frac{1}{R^c}$.
Yea, you can’t upperbound by any function of $R$ alone (e.g., the support could be much smaller than $R$). You should maybe say something about the entropy of the distribution...? I can imagine a bound like that coming in.
Not sure I understand; we get to pick the distribution on $R$, so if we get a distribution for which we cannot upper bound the probability by any function of $R$ alone, doesn't that just mean we picked a bad distribution?
|
2025-03-21T14:48:31.209175
| 2020-06-07T20:31:39 |
362459
|
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|
Stack Exchange
|
Meromorphic functions on a modular curves of genus $0$ that take each value exactly once
Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb Z)$, and let $\mathfrak H$ be the upper half-plane. Let $X(\Gamma)$ be the compactification of $\Gamma\backslash\mathfrak H$. Then $X(\Gamma)$ is a compact Riemann surface.
Suppose that $X(\Gamma)$ has genus $0$. Then also according to a general theorem there exists an analytic isomorphism $X(\Gamma)\longrightarrow \mathbb P_\mathbb C^1$. In the case of $\Gamma=\operatorname{SL}_2(\mathbb Z)$ this isomorphism is given by the $j$-invariant.
What is the simplest way to prove that there exists an analytic isomorphism $X(\Gamma)\longrightarrow \mathbb P_\mathbb C^1$ in this special case? That is, how much advantage can our knowledge that $X(\Gamma)$ is a modular curve give us?
Is there an algorithm to determine such isomorphism in terms of known functions (such as Dedekind $\eta$-function or $j$-invariant)?
For example, consider
$$\Gamma=\bigg\lbrace\gamma\in \operatorname{SL}_2(\mathbb Z)\colon \gamma\equiv \begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix},\begin{pmatrix}0 & 1 \\ 1 &0\end{pmatrix}\text{mod }2\bigg\rbrace.$$
Then an isomorphism $X(\Gamma)\longrightarrow \mathbb P_\mathbb C^1$ is determined by the function
$$j_\Gamma(\tau)=-\frac{\eta\left(\frac{\tau+1}{2}\right)^{24}}{\eta(\tau)^{24}}=q^{-1/2}\prod_{n=1}^{\infty}(1+q^{n-1/2}).$$
Can't you just take the $j$ invariant to answer both questions?
@WillSawin Thank you, I will modify my question.
probably what you are looking for is the notion of Hauptmodul.
|
2025-03-21T14:48:31.209307
| 2020-06-07T20:32:08 |
362460
|
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|
Stack Exchange
|
Jordan algebra identities
A Jordan algebra is a vector space with a commutative bilinear operation $\circ$ obeying an identity that's often written as
$$ (x \circ y) \circ (x \circ x) = x \circ (y \circ (x \circ x)) . $$
I find this identity rather obscure. If we write $x^2 = x \circ x$ and use $L_a$ to stand for left multiplication by $a$, we can rewrite it in a more appealing form:
$$ L_{x^2} L_x = L_x L_{x^2} .$$
However, I'd be even happier if this were a special case of a more general identity
$$ L_{x^m} L_{x^n} = L_{x^n} L_{x^m} \qquad (\ast) $$
holding for all $n, m \in \mathbb{N}$.
This more general identity parses in any Jordan algebra, because any Jordan algebra is power-associative: expressions like $x \circ \cdots \circ x$ are independent of how you parenthesize them, so $x^n$ is well-defined. But is this more general identity $(\ast)$ true in every Jordan algebra?
Seems not what you want but multilinear form of the Jordan identity does not look so obscure. It is$$[L_{ab},L_c]+[L_{bc},L_a]+[L_{ca},L_b]=0$$(where $[-,-]$ is usual commutator of operators). Another equivalent form:$$\operatorname{as}(ab,d,c)+\operatorname{as}(bc,d,a)+\operatorname{as}(ca,d,b)=0$$where $\operatorname{as}(x,y,z)=(xy)z-x(yz)$ is the associator.
I'm sorry to be foolish, but aren't some of these right multiplications? For example, $L_{x^2}L_x y$ should be $x^2 \circ (x \circ y)$, but we want $(x \circ y) \circ x^2$.
A Jordan algebra is commutative so left multiplication is the same as right multiplication.
This identity (*) is, indeed, true, and is, in fact, a step in one of the standard ways to prove that Jordan algebras are power-associative: see McCrimmon's 2004 book A Taste of Jordan Algebras, exercise 5.2.2A (question (2)) on page 201.
Edit: another reference, which has the better taste of not being an exercise and of being earlier in a book: Jacobson, Structure and Representations of Jordan Algebras (1968), page 35, just above formula (56). (I also just realized that the fact is mentioned in the Wikipedia article on Jordan algebras, with that reference.)
For completeness of MathOverflow, I might as well copy the essence of the argument: first prove the identity
$$
L_{(b\circ d)\circ c} = L_{b\circ d}L_c + L_{c\circ d}L_b + L_{b\circ c}L_d - L_b L_c L_d - L_d L_c L_b
$$
by linearizing the Jordan identity (here we use the fact that the characteristic is not two). Apply this to $b=a^k$ and $c=d=a$, giving
$$
L_{a^{k+2}} = 2L_{a^{k+1}}L_a + L_{a^2}L_{a^k} - L_{a^k} L_a^2 - L_a^2 L_{a^k}
$$
From this it follows by induction that all $L_{a^k}$ belong to the algebra generated by $L_a$ and $L_{a^2}$ and, since these commute, they all commute.
Wow, thanks! I don't know how I overlooked that. I now claim this is the true definition of Jordan algebra.
@JohnBaez assuming there should be a "true definition"... one advantage of the usual definition is its simplicity, and the fact that it is given by a finite number of identities. Certainly a student whose exercise is to check that some given Jordan is indeed Jordan will be grateful that the definition is the usual one. Also the usual one makes it easy to implement an algorithm checking whether a (non-associative) algebra given by structure constants (in a computable field) is Jordan.
@YCor why not have a conceptually clear definition, like John would like, and then a theorem that gives a finite (short!) list of identities to check that suffices to imply the full definition?
@DavidRoberts why not, I have no problem with it, but don't see why it should be called the only "true" definition. From another intuition one might find the original definition simpler. Also, what do you think of associativity? $(ab)c=a(bc)$ may sound nice because we're used to it, but what's important is that all bracketings of products $x_1\dots x_n$ lead to the same element. In spite of this I'm happier with the usual short definition.
[This addresses the edited post, not the current stream of comments] Actually I once checked that the naive definition of Jordan algebra in char 2 (commutative + Jordan identity) is so bad that it does not pass to extension of scalars (for extensions of fields of char 2, probably $F_2\subset F_4$). The correct definition of Jordan ring should be that a ring satisfying all identities with coefficients in $\mathbf{Z}$ satisfied by, say, all complex Jordan algebras. When 2 is not invertible it implies we should add the polarizations among the axioms.
@YCor: From what I understand of McCrimmon's book (but he may be biased in that he invented the concept), the “right” version of Jordan algebras in characteristic $2$ requires a change in perspective and to use the operator $U_x$ which is defined as $2L_x^2 - L_{x^2}$ in characteristic $\neq 2$, or rather the map $x \mapsto U_x$, giving rise to the concept of quadratic Jordan algebra. But this gets us again back to the question of what the “right definition” even means.
No, because the previous "true definition" comments were about two ways to define the same object. Here it might be two non-equivalent definitions. The definition I'm suggesting is not specific to Jordan, and makes it meaningful to any ring (not just algebras over a field). For instance in this way (taking $\mathbf{Z}$-identities from char zero and claiming them as axioms) one gets the right (universally accepted) notion of Lie ring. That for Jordan in char 2 this results in something which is a questionable definition is possible, but I'd need to have a more careful look.
|
2025-03-21T14:48:31.209799
| 2020-06-07T22:33:18 |
362468
|
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|
Stack Exchange
|
Reference for Rellich Kondrachov theorem on bounded domains and spaces with finite measure
I read at the top of the page $580$ of this paper that the imbedding $W^{1,2}(\Gamma,d\mu) \hookrightarrow L^2(\Gamma,d\mu)$ is compact for a bounded domain $\Gamma \subset \mathbb{R}^n$ and a measure $\mu$ as defined in the second paragraph of the section $2$. I would like a reference for this result.
Thanks in advance!
I think the author assumes that a cone property holds at the beginning of Section 2. The corresponding result can then be found in the book of Adams. (Reference [2] in the paper, which is also mentioned there in relation to the cone property.) The mentioned embedding will not in general not be compact for an arbitrary domain, but the conditions are rather mild I would say. (For example it would be enough to have an embedding $W^{1,2}(\Gamma) \hookrightarrow L^r(\Gamma)$ with an $r>2$, and this allows for outward cusps in $\Gamma$.)
@Hannes: Why would it be enough to have an embedding $W^{1,2}(\Gamma) \hookrightarrow L^r(\Gamma)$ for some $r > 2$?
@JochenGlueck By considering $W^{1,2}_0$ in the interior and using the gap $r>2$ for a small "boundary layer": let $\tau$ be smooth cutoff function associated to $\Gamma_0$ with $|\Gamma\setminus\Gamma_0| = \varepsilon$ and with support strictly contained in $\Gamma$. If $u_n \to 0$ weakly in $W^{1,2}(\Gamma)$, then $(\tau u_n) \subset W^{1,2}_0(\Gamma)$ has a subsequence going to zero strongly in $L^2(\Gamma)$. For $((1-\tau)u_n)$, we use the Hölder inequality: $|(1-\tau)u_n|2 \leq |1-\tau|{2r/r-2} |u_n|_r \lesssim \varepsilon^{(r-2)/2r}$. I learned this from a paper of Daners :-)
@Hannes: That's a great argument, thank you! Maybe I should spend more time reading Daniel Daners' papers instead of writing new ones with him ;-). ("Everybody writes, nobody reads"...)
@Hannes, can you cite the corresponding theorem in Adams' book that you commented please? If you are talk about the Rellich-Kondrachov theorem in Adams' book, then I can not see how the theorem in the book can be used to prove the remark that I want a proof. Other thing, the author of the paper comments that $\Gamma \subset \mathbb{R}^n$ is a unbounded domain with the cone and segment property at the beginning of the section $2$, but he did not comment this in the remark for bounded domains. Can I assume these properties hold for bounded domains?
@George I did mean the R-K Theorem in Adams' book, why can it not be used? (Is it the measure? That should be equivalent to the Lebesgue measure here and then the involved spaces are the same, no?) Regarding the domain assumption, personally I personally would understand that the assumption on $\Gamma$ at the beginning of Section 2 holds for the rest of the paper and the remark in question only considers the additional property that $\Gamma$ is in fact bounded.
The immersion compact is guaranteed by the theorem is $W^{m,p}(\Omega) \longrightarrow W^{m,p}(\Omega_0^k)$ , where $\Omega_0$ is a subdomain of $\Omega$ and $\Omega_0^k$ is the intersection of a $k$-plane in $\mathbb{R}^n$ with $\Omega_0$. The immersion holds for $0 < n - mp < k \leq n$ and $1 \leq q < \frac{kp}{n - mp}$ or $n = mp$, $1 \leq k \leq n$ and $1 \leq q < \infty$, but think better I think is not a problem because you can take $\Omega_0 = \Omega$ in the case of $\Omega$ be bounded and $k = n$ to obtain $\Omega_0^k = \Omega$.
Adams considers the Lebesgue measure, but I have a doubt now: in what sense and how the measure defined in the paper is equivalent to the Lebesgue measure? Do you have a reference for this statement?
@George $\mu$ is a weighted Lebesgue measure in the paper, with the weight $\rho$ being $L^\infty$ and locally bounded away from zero on $\overline\Gamma$, so for compact $\overline\Gamma$, bounded away from zero. Thus $\mu$ is equivalent to the Lebesgue measure in the sense of $\rho_\bullet |A| \leq \mu(A) \leq \rho^\bullet |A|$ for every measureable set $A$, where $0<\rho_\bullet\leq\rho^\bullet$ are constants. (This is also mentioned and used e.g. in the proof of Prop. 2.2 in your paper.) The definition of the $\rho$-weighted Sobolev spaces then shows that these are just the usual ones.
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