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2025-03-21T14:48:31.131958
| 2020-06-03T13:52:20 |
362101
|
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|
Stack Exchange
|
Upper semi-continuity of intersection numbers
Consider a smooth projective morphism of schemes $X \rightarrow S$ with relative dimension $n$ (the application I have in mind is with $S$ = an open subset of $\text{Spec } \mathbb{Z}$) and assume that $V$ and $W$ are two closed subscheme of $X$, flat over $S$, such that $\text{codim}(V) + \text{codim}(W) = n$.
Denote the fibers of $X,V$ and $W$ over $s \in S$ by $X_s,V_s$ and $W_s$ and write $A^{\bullet}(X_s)$ for the Chow ring of $X_s$. The proper pushforward of the structure morphism of $X_s$ induces a morphism $\text{deg} : A^{n}(X_s) \rightarrow \mathbb{Z}$.
For every $s \in S$, the intersection of $V_s$ and $W_s$ in $X_s$ gives an intersection number $\text{deg}([V_s].[W_s]) \in \mathbb{Z}$. Is there an upper semi-continuity result on the function $s \mapsto \text{deg}([V_s].[W_s]) \in \mathbb{R}$ ?
My question is related to this post : Semi-continuity of intersection numbers but the example given in the comments does not satisfy the flatness assumption.
In fact the intersection numbers are constant: by [Fulton, Cor. 20.3], the specialisation maps $\sigma \colon A^*(X_\eta) \to A^*(X_s)$ are ring homomorphisms if $S$ is a Dedekind scheme with generic point $\eta$ and closed point $s$. By [Fulton, §20.2] the intersection product for flat cycles is just the fibre product:
$$[V] \underset S\times [W] = \left[ V \underset S\times W \right].$$
Finally, degree of $0$-cycles is preserved under specialisation (use that a finite flat $S$-scheme is locally free). Thus, we can compare $\deg([V_s] \cdot [W_s])$ to $\deg([V_{s'}] \cdot [W_{s'}])$ by identifying both with $\deg([V_\eta] \cdot [W_\eta])$. $\square$
References.
[Fulton] W. Fulton, Intersection theory. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 2. Berlin: Springer (1998). ZBL0885.14002.
I understand your answer in the case where $S$ is the spectrum of a discrete valuation ring. For a Dedekind scheme, the specialization map in Fulton is a map from the Chow ring of the fiber of $X$ over $S-s$ and not over the generic point. How do you get your result from this ? Thank you
Apply the above to each DVR $\mathcal O_{S,s} \subseteq \mathcal O_{S,\eta}$.
|
2025-03-21T14:48:31.132146
| 2020-06-03T14:40:06 |
362103
|
{
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"Derek Holt",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
What is the Schur multiplier of the Mathieu group $M_{10}$
It is well known that the automorphism group of the alternating group $A_6$ is $P\Gamma L_2(9)$. There are three different index $2$ subgroups of $P\Gamma L_2(9)$, namely the symmetric group $S_6$, the projective general linear group $PGL_2(9)$, and the Mathieu group $M_{10}$. By checking the ATLAS (http://brauer.maths.qmul.ac.uk/Atlas/v3/), the Schur multipliers of those groups $A_6,S_6, PGL_2(9)$ are the cyclic group $Z_6$. What about the Mathieu group $M_{10}$?
Also, I don't know why, in the following book, Page 302, Table 4.1, the Schur multiplier of the group ${\sf C}_2(2)$ is listed to be the cyclic group $Z_2$?
Gorenstein, Daniel, Finite simple groups. An introduction to their classification, Moskva: Mir. 352 p. R. 2.50 (1985). ZBL0672.20010.
According to SAGE it is cyclic group of order 3: at Sage Cell Server I entered MathieuGroup(10).homology(2) and obtained Multiplicative Abelian group isomorphic to C3
$H_2(M_{10},\mathbb Z)\cong H^2(M_{10},\mathbb C^\times)\cong H^3(M_{10},\mathbb Z) = \oplus_{ p | 720} H^3(M_{10},\mathbb Z)_{(p)}$ with $p\in\lbrace 2,3,5\rbrace$. A $p$-primary component is isomorphic to the set of $M_{10}$-invariant elements of $H^3(\text{Syl}_p(M_{10}))$. We can check that $M_{10}$ has semi-dihedral Sylow 2-subgroups and cyclic Sylow 5-subgroups (Wikipedia), both of whose Schur multiplier is trivial. So we're left with the 3-primary component. We can check that $M_{10}$ has elementary abelian Sylow 3-subgroups (Wikipedia), all isomorphic to $\mathbb Z_3^2$, and thus $H^3(M_{10})_{(3)}\cong H^3(\mathbb Z_3^2)^{H/\mathbb Z^2_3}$ by Swan's theorem, where $H$ is the normalizer of $\mathbb Z_3^2\subset M_{10}$. This invariant subgroup is the full group $H^3(\mathbb Z_3^2)\cong\mathbb Z_3$ (see below), so $H_2(M_{10},\mathbb Z)\cong\mathbb Z_3$.
To see that $H^3(\mathbb Z_3^2)^{H/\mathbb Z_3^2}\cong H^3(\mathbb Z_3^2)$, we first note that our $H$ is a maximal subgroup of order 72 so we may as well take it to be the Mathieu group $M_9=\mathbb Z_3^2\rtimes Q_8$ sitting naturally inside $M_{10}$, where $Q_8$ acts as the faithful two-dimensional irreducible representation over $\mathbb Z_3$. So we need to check that $H^3(\mathbb Z_3^2)^{Q_8}\cong H^3(\mathbb Z_3^2)$. For the standard generators $\{I,J\}$ of $Q_8$ and basis $\{a,b\}$ of $\mathbb Z_3^2$ we have $I(a)=a+b$, $I(b)= a-b$, $J(a)=-a+b$, and $J(b)=a+b$. Noting that $H^\ast(\mathbb Z_3^2,\mathbb Z_3) = \mathbb Z_3[x,y]\otimes\Lambda(u,v)$ with $|x|=|y|=2$ and $|u|=|v|=1$, the element $uv$ is $Q_8$-invariant. The induced map $\delta:H^2(\mathbb Z_3^2,\mathbb Z_3)\to H^3(\mathbb Z_3^2,\mathbb Z)$ under the short exact sequence $\mathbb Z\hookrightarrow\mathbb Z\twoheadrightarrow\mathbb Z_3$ is surjective and $Q_8$-equivariant, with image $\langle\delta(uv)\rangle$, and so $H^3(\mathbb Z_3^2)^{Q_8}\cong H^3(\mathbb Z_3^2)$.
We've essentially shown that this agrees with the Schur multiplier of $M_9$ by the same technique (note that the Schur multiplier of $Q_8$ is trivial), $H_2(M_9)\cong H^3(\mathbb Z_3^2)^{Q_8}\cong\mathbb Z_3$.
Yes that's right. Note that the $3$-part of the Schur Multipliers of the two other extensions $S_6$ and ${\rm PGL}(2,9)$ of $A_6$ by groups of order $2$ are both trivial, because the outer involutions in those two groups invert the central $C_3$ subgroup. So it is centralized by their product, which is the outer element of $M_{10}$.
|
2025-03-21T14:48:31.132362
| 2020-06-03T15:27:23 |
362106
|
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"Federico Barbacovi",
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|
Stack Exchange
|
Admissibility of intersection of subcategories
Let $\mathscr{T}$ be a triangulated category, and $\mathscr{A}$ be a right admissible subcategory, which means that $i_{\mathscr{A}} : \mathscr{A} \rightarrow \mathscr{T}$ has a right adjoint $i_{\mathscr{A}}^R$. Let $\mathscr{B}$ another subcategory of $\mathscr{T}$ (not necessarily right/left admissible). Consider the subcategory $\mathscr{A} \cap \mathscr{B} \subset \mathscr{B}$. It's natural to ask whether this subcategory is still right admissible. The first thing one would try is to consider the right adjoint $i_{\mathscr{A}}^R$ and check whether it sends $\mathscr{A} \cap \mathscr{B}$ to $\mathscr{B}$. If this is the case, we get right admissibility. Let's assume however that this is not the case, i.e. there exists an object $E \in \mathscr{A} \cap \mathscr{B}$ such that $i_{\mathscr{A}}^R(E) \notin \mathscr{A} \cap \mathscr{B}$. Is it possible the there exists another functor which is right adjoint to $i_{\mathscr{A}} \vert_{\mathscr{A} \cap \mathscr{B}} : \mathscr{A} \cap \mathscr{B} \rightarrow \mathscr{B}$ and which is not the restriction of $i_{\mathscr{A}}^R$? I tried to apply the Yoneda lemma to prove that this is not possible, but the only case in which this works is the degenerate case in which $\mathscr{A} \cap \mathscr{B}$ weakly generates $\mathscr{B}$, i.e. if $\left( \mathscr{A} \cap \mathscr{B} \right)^{\perp} = 0$ in $\mathscr{B}$, which makes me think that maybe it is possible. Thank you in advance.
It is possible. Given $\mathscr{T}$, $\mathscr{A}$ and $i_\mathscr{A}^R$, it is often possible to find a subcategory $\mathscr{B}$ of $\mathscr{T}$ so that $\mathscr{A}\cap\mathscr{B}=0$ (and so certainly $\mathscr{A}\cap\mathscr{B}$ is a right admissible subcategory of $\mathscr{B}$), but $i_\mathscr{A}^R(\mathscr{B})\neq0$.
For example, let $\mathscr{T}=K^-(\text{Mod-}\mathbb{Z})$ be the homotopy category of bounded above complexes of abelian groups, and let $\mathscr{A}=K^-(\text{Proj-}\mathbb{Z})$ be the homotopy category of bounded above complexes of free abelian groups, so $i_\mathscr{A}^R$ takes a complex to its projective resolution. And finally let $\mathscr{B}$ be the homotopy category of bounded above complexes of torsion abelian groups.
Then $\mathscr{A}\cap\mathscr{B}=0$ (there are no nonzero maps from $\mathscr{B}$ to $\mathscr{A}$), but if $B\in\mathscr{B}$ is not acyclic, then $i_\mathscr{A}^R(B)\neq0$.
Ok, that’s clear, thanks!
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2025-03-21T14:48:31.132779
| 2020-06-03T15:36:58 |
362108
|
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"Ben McKay",
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|
Stack Exchange
|
how to normalize curvature to be between -1 and -1/4
On paper "Harmonic maps into singular spaces and p-adic superrigidity for lattices in groups of rank one", lemma 8.1, when the author tried to construct finite energy map from Quaternion hyperbolic space or Cayley plane, they normalize the curvature to be between -1 and -1/4. My question is under what condition can we do this and how to realize this? Any reference will be much appreciated.
There is more than one standard convention for the usual metrics on the quaternion hyperbolic plane, and on the Cayley hyperbolic plane, but differentt conventions agree up to positive constant rescaling. In any such rescaling, the ratio of maximum to minimum of absolute sectional curvature is 4, so we can pick the constant rescaling to get the required curvature range. See Besse, Einstein Manifolds, I think, for a more detailed discussion.
Thank you for the reference, Ben.
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2025-03-21T14:48:31.132882
| 2020-06-03T15:42:44 |
362109
|
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"Henri",
"Jean",
"Luka Thaler",
"Wojowu",
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|
Stack Exchange
|
injective holomorphic mapping between unit disk and unit polydisk
In $\mathbb{C}^n,\ n\geq 2$, there is no bijection between unit disk $B^n(0,1)$ and unit polydisk $P^n(0,1)$. But if we wish to find injective holomorphic mapping from unit disk to polydisk(whose image contains origin), inclusion is the obvious mapping or suitable automorphisms of unit disk(which is sort of inclusion, after applying automorphism). But can we go beyond unit disk in polydisk, by means of injective holomorphic mapping. Are there results in this direction available?
I'm not exactly sure what your question is - to find an injective map from the unit ball to the polydisk with the image not in the unit ball? You can map $B^n(0,1)$ to some small in the "corner" of the polydisk.
I want to see, if we can find a map, whose image contains origin, but also consists points out of $B^n(0,1)$. Edited my question a bit.
How about $z\mapsto 2z$?
@Henri Its image not be within polydisk $P^n(0,1)$
Ok, I hadn't seen that you wanted the image to be contained in the polydisk.
So if I understand correctly, given a point in $x\in P^n(0,1)- B^n(0,1)$ you are asking if there is an injective map $f:B^n(0,1)\rightarrow P^n(0,1)$ such that $f(B^n(0,1))$ contains $0$ and $x$? If so then (think for simplicity $n=2$) take $\epsilon>0$ very small and $r=|x|+\epsilon$ such that the image of the unit ball under the map $(z,w)\rightarrow (r z,\epsilon w)$ is almost a flat disk then use a unitary tranformation $A\in U(2)$ to rotate this thickened disk in direction of $x$.
Take $f(z_1,z_2):=(z_1,sz_2+(1-s)z_1)$, $0<s<1$. For example, $s=1/2$. Then $f$ is an injective holomorphic mapping from $B(0,1)$ to $P(0,1)$, ($n=2$), sending $0$ to $0$. Moreover, $\|f(s,0)\|>1$ for $s<1$, near $1$, i.e. $f(s,0)\notin B(0,1)$..
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2025-03-21T14:48:31.133018
| 2020-06-03T15:54:59 |
362111
|
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"Dirk Werner",
"Piotr Hajlasz",
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"url": "https://mathoverflow.net/questions/362111"
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|
Stack Exchange
|
A good reference for Bochner spaces
I am looking for some references on Bochner spaces containing basic stuff such as measurability, convergence and $L^p$ theory. I already have the Analysis in Banach Spaces: Volume I book which covers quite well the theory but I would like some other reference, one possibly containing more applications on evolution problems and convergence in $L^p(S,X)$. Any suggestions?
What about the classics Dunford/Schwartz or Diestel/Uhl?
Diestel, J.; Uhl, J. J., Jr. Vector measures. Mathematical Surveys, No. 15. American Mathematical Society, Providence, R.I., 1977.
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2025-03-21T14:48:31.133091
| 2020-06-03T16:05:25 |
362114
|
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"Evgeny Shinder",
"R. van Dobben de Bruyn",
"THC",
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|
Stack Exchange
|
$0$-Dimensional $k$-varieties in the Grothendieck ring $K_0(V_k)$
Let $k$ be a field, and let $K_0(V_k)$ be the Grothendieck ring of $k$-varieties. Let $K_0(V_k^0)$ be the Grothendieck ring of $0$-dimensional $k$-varieties. I also assume that $k$ is perfect.
(1) I have read that $K_0(V_k^0)$ is generated by the classes $[\mathrm{Spec}(K)]$ with $K$ a finite field extension of $k$.
(2) I have also read that $K_0(V_k^0)$ is the free $\mathbb{Z}$-module generated by the isomorphism classes of transitive permutation representations of $\mathrm{Gal}(\overline{k}/k)$.
As I am come from far outside Galois theory or cohomology theories, I wondered why both statements are true, and if someone could provide some clear references. What happens if $k$ is not perfect ?
(1) is easy: a $0$-dimensional scheme is a finite union of points, and $[Z^{\operatorname{red}}] = [Z] \in K_0(\mathbf{Var}_k)$ for any $k$-scheme $Z$. Since isomorphism classes of finite transitive $\operatorname{Gal}(\bar k/k)$-sets is the same thing as finite separable field extensions, at least this shows it's generated by such. I'm not sure why there are no relations.
@R.vanDobbendeBruyn : "Since isomorphism classes of finite transitive Gal(¯/)-sets is the same thing as finite separable field extensions" -- could you explain this in more detail ? (Thanks !!) -- THC
This is Galois theory. To a separable field extension $k \to \ell$, associate the set $\operatorname{Hom}_k(\ell, \bar k)$, which has a natural Galois action by postcomposition. Conversely, a finite transitive $\operatorname{Gal}(\bar k/k)$-set is of the form $G/H$ for some closed subgroup $H \subseteq G = \operatorname{Gal}(\bar k/k)$, so take $\ell = (\bar k)^H$. Galois theory says that these constructions are inverses.
@R.vanDobbendeBruyn : does $(\overline{k})^H$ denote the fixed elements subfield in $\overline{k}$, of $H$ ?
In characteristic zero, classes of fields are linearly independent by the Larsen-Lunts theorem, as in the answer here: https://mathoverflow.net/questions/354718/field-extensions-in-grothendieck-rings/354805#354805
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2025-03-21T14:48:31.133249
| 2020-06-03T16:12:19 |
362115
|
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"Dominic van der Zypen",
"Gerhard Paseman",
"YCor",
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|
Stack Exchange
|
Are complete regular linear hypergraphs on $\omega$ isomorphic?
If $H_i = (V_i, E_i)$ are hypergraphs for $i=1,2$ then we say they are isomorphic if there is a bijection $f: V_1 \to V_2$ such that for $A \subseteq V_1$ we have $$A\in E_1 \text{ if and only if } f(A) \in E_2.$$
We say that $H=(\omega, E)$ is a complete regular linear hypergraph on $\omega$ if
$e_1\neq e_2\in E \implies |e_1\cap e_2| = 1$ and
for all $n\in \omega$ we have $|\{e\in E: n \in e\}| = \aleph_0$.
Question. Assuming ${\sf ZFC}$, if $H_i = (\omega, E_i)$ are complete regular linear hypergraphs for $i = 1,2$, are $H_1$ and $H_2$ necessarily isomorphic?
Since we won't always know if there is a bijection between aleph nought and its square, I suspect the answer is very dependent on the underlying set theory. My guess is ZFC shows there is an isomorphism. Gerhard "See What Projective Geometry Says" Paseman, 2020.06.03.
Thanks for your comment @GerhardPaseman - my framework for this question is ZFC, but I forgot to add this to the question - will do!
In the absence of set-theoretic specification I guess ZFC is assumed.
Since the underlying set is well ordered, this induces a well ordering on the edge set. However, choice may still be needed to establish an isomorphism. If both edge sets are recursively computable, then it may be possible to construct a recursive isomorphism without choice. However this assumes a lot. Gerhard "Not Sure Which To Choose" Paseman, 2020.06.03.
If $K$ is a field of cardinality $\aleph_0$, then the points and lines of the projective plane over $F$ constitute a complete regular linear hypergraph. The field $K$ can be recovered (up to isomorphism) from the hypergraph, so this produces lots of non-isomorphic such hypergraphs.
No. Given a hypergraph $H$, the $2$-shadow of $H$ is the $2$-uniform graph where we take each pair which is contained in an edge of $H$. This is isomorphism invariant.
Now consider the following graph on $\mathbb{Q}^2\cup (\mathbb{Q}\cup\{\infty\})$, where we think of $\mathbb{Q}\cup\infty$ as a slope. For each line in $\mathbb{Q}^2$, we draw an edge containing the points of the line and its slope; this is $H_1$. This is a complete regular linear hypergraph: obviously there are infinitely many lines through any given point, or of any given slope, and any two lines either have the same slope and no points in common, or otherwise different slopes and one point in common. The 2-shadow of this is a complete graph with all the edges between different slopes removed. In other words, it has one maximal independent set and every other pair is an edge.
Now we draw another graph $H_2$, whose vertices will be all rational lines in $\mathbb{Q}^3$ together with all rational normals in $\mathbb{Q}^3$. For each plane in $\mathbb{Q}^3$, we put an edge containing all the lines in that plane, together with the normal to the plane. Again, this is a complete regular linear hypergraph; for any line or normal, there are infinitely many planes through the line or with the normal, but any two planes either have the same normal and no line in common, or different normals and exactly one line in common.
But the 2-shadow of $H_2$ is more complicated than that of $H_1$. The set of normals is still an independent set, and it is a maximal independent set, because any line is contained in a plane which has a normal. But there are pairs of lines which are not in any plane (skew ones), so there are non-edges outside this maximal independent set.
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2025-03-21T14:48:31.133609
| 2020-06-03T16:33:55 |
362117
|
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|
Stack Exchange
|
Galois theory of ramified coverings vs classical Galois theory
That's an exact copy of my former MSE question I asked a couple of weeks ago and unfortunately not got the answer I was looking for.
The question adresses reuns' answer in this thread: Algebraic closure of $k((t))$
In the answer, reuns used a theory connecting classical Galois theory with Galois theory of ramified coverings. Unfortunately, he uses also some details which aren't clear to me, see below. Firstly, what I understoond:
In the setting of the linked thread we consider a finite Galois extension $L/\Bbb{C}(z)$. The theory says that one can associate to the field $L$ a unique connected Riemann surface $Y_L$ and a map $f_L: Y_L \to \Bbb{PC^1}$ which is a ramified cover over $\Bbb{PC^1}$ of degree $n=[L:\Bbb{C}(z)]$.
Ramified means that that there exist a finite subset $S \subset \Bbb{PC^1}$ (= the "branch points") such that the restricted map $Y_L \backslash f_L^{-1}(S) \to \Bbb{PC^1} \backslash S$ is a $n$-cover known from basic topology.
We recover $L$ as the field of meromorphic functions on $Y_L$ (and $\Bbb{C}(z)$ " $\Bbb{PC^1}$).
The deepth of the theory gives a beautiful identification for Galois group $$Gal(L/\Bbb{C}(z))= \pi_1(\Bbb{PC^1} \backslash S,z_0)/Q$$
where $\pi_1(\Bbb{PC^1} \backslash S,z_0)$ is the fundamental group and the subgroup $Q \subset \pi_1(\Bbb{PC^1}\backslash S,z_0)$ corresponds via classical covering theory to the cover $p:U_{\Bbb{PC^1}\backslash S} \to Y_L \backslash f_L^{-1}(S)$. Here $U_{\Bbb{PC^1}\backslash S}$ is the universal cover of $\Bbb{PC^1}\backslash S$. That's the background.
NOW THE TWO QUESTIONS:
reuns wrote:
For $L/\Bbb{C}(z)$ a finite Galois extension then its elements are locally meromorphic on $\Bbb{C}$ minus a few branch points (the zeros of the discriminant of the minimal polynomials) and $Gal(L/\Bbb{C}(z))$ consists of the analytic continuations along closed loops enclosing some of those branch points.
Proof: with $\gamma_1(a),\ldots,\gamma_m(a)$ the analytic continuations of $a$ then the coefficients of $h(X)=\prod_{l=1}^m(X-\gamma_m(a))$ stay the same under analytic continuation, thus they are meromorphic on the Riemann sphere, ie. they are in $\Bbb{C}(z)$ so $h(X)$ is the minimal polynomial of $a$.
QUESTION #1: I learned this theory from Szamuely's "Galois Groups and Fundamental Groups". In the construction there unfortunately wasn't explicitly explaned why the branch points of $f_L$ are the common zeros of the discriminants of minimal polynomials $F_i(z,T)$ of the generators $g_i \in L$ (in case $L= \Bbb{C}(z)(g)$ of $L$, while outside of the zeros/poles of the discriminants there are no branch points.
In spirit of this problem I'm looking forliterature explaning this construction of $Y_L$ from $L$ in that way there is explicitly pointed out why the branch points correspond to the common zeros of the discriminants $Disc(F_i(z,T))$ of minimal polynomials of the generators of $L$.
QUESTION #2:
How a $l \in \pi_1(\Bbb{PC^1}\backslash S,z_0)$ represents
an element of the Galois group $Gal(L/\Bbb{C}(z))$ explicitely?
Szmuely's book proves the isomorphism
$\pi_1(\Bbb{PC^1}\backslash S,z_0)/Q= Gal(L/\Bbb{C}(z))$
using category equivalences and unfortunately this proof not gives a good geometric
inside look what is going on there. In other words I "understand"
the proof step by step but have no visual intuition on this identification.
What we need is that such $l$ act as automorphism on
the meromorphic function from $L$ and fixes $\Bbb{C}(z)$.
Reuns explaned this identification as
" analytic continuations along closed loops enclosing some of
those branch points."
Unfortunately I can't imagine what he means by this analytic continuation and I would like to understand what he means here. ie wlog say our
$l$ is a class of such closed loop around a pranch point.
What/ which object is here "analytically continued along $l$"?
How does this construction work?. The object of our interest
is a meromorphic function $m \in L$ of $Y_L$.
How it can be "continued" by $l$? Isn't it already determined
on a dense open subset of $Y_L$ exept at it's poles.
Thus why and how it is continued along $l$? I not
understand it.
|
2025-03-21T14:48:31.133887
| 2020-06-03T16:51:50 |
362119
|
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"Fedor Petrov",
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|
Stack Exchange
|
Inf sup condition for discrete Stokes problem
I am considering the discrete inf-sup condition for the discrete Stokes problem
$ \inf_{q \in Q_h} \sup_{v \in V_h} \frac{(q, \nabla \cdot v)}{\| q \| \| \nabla v\|} \geq \beta > 0$
This means that
$ \inf_{q \in \mathbb{R}^p} \sup_{v \in \mathbb{R}^n} \frac{v^T B q}{\| q \| \| v\|} \geq \beta > 0$
Can a do the following argumentation:
$ \inf_{q \in \mathbb{R}^p} \sup_{v \in \mathbb{R}^n} \frac{v^T B q}{\| q \| \| v\|} = \inf_{q \in \mathbb{R}^p, \|q \| = 1 } \sup_{v \in \mathbb{R}^n, \| v \| = 1} v^T B q = \inf_{q \in \mathbb{R}^p, \|q \| = 1 } |B q|$
Is this correct? Or am I missing something?
I would be very grateful about any help.
Best,
frieder
I do not know what are $B_h$ and $Q_h$, but after that everything is ok.
ok, I should have mentioned more details. $Q_h$ is a finite element pressure space and $B$ is then the matrix related to $(q,\nabla \cdot v)$
|
2025-03-21T14:48:31.133990
| 2020-06-03T18:17:59 |
362121
|
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"url": "https://mathoverflow.net/questions/362121"
}
|
Stack Exchange
|
A question about Schwartz-type functions used in analytic number theory
In analytic number theory we like to weigh our counting functions with a smooth function $f$, so that we may apply Poisson's summation formula and take advantage of Fourier transforms. Typically the weight function $f$ will be a Schwartz type function with the following properties:
1) $f(x) \geq 0 $ for all $x \in \mathbb{R}$;
2) $f(x) = 1$ for $x \in [-X,X]$ say;
3) $f(x) = 0$ for $|x| > X + Y$; and
4) $f^{(j)}(x) \ll_j Y^{-j}$ for $j \geq 0$.
In most applications the dependence on $j$ in the last condition does not matter, since $j$ would be bounded. However, in a problem I am considering it might be worthwhile to make $j$ a (slow growing) function of $X$ so it then becomes relevant to know how the bound depends on $j$. Is it possible to give an explicit example of a function $f$ satisfying the above properties for which the dependence can be made explicit?
In https://arxiv.org/abs/1707.01576 (on page 9 just below (3.8)) a very nice and very explicit function $\omega$ is given which easily allows to build a function doing almost what you want. The caveat is that it is not smooth but instead features a parameter $s$ ensuring $\omega\in C^s$. In many application this might still be good enough.
The answer to your question is yes, and it is a pretty well-understood topic.
First of all $X$ is more or less irrelevant for the bounds in 4) so let us take $X = Y$, say, for convenience.
Second, we can always scale everything down by $Y$. So without loss of generality $Y = 1$.
Put $g = f'$. Let us also for simplicity assume that $f$ is symmetric so it is enough to study $g$ on $[1, 2]$. The problem therefore is reduced to the following: for which sequences $t_j$ we can find a function $g$ such that $g = 0, x\notin [1, 2]$, $\int g = 1$ and $|g^{j}(x)|\le C t_j$ for some constant $C$.
The answer to this question in more or less full generality is given by the Denjoy-Carleman Theorem: if the sequence $M_j = \frac{t_j}{j!}$ is logarithmically convex (i.e. $\frac{M_{j+1}}{M_j}$ is increasing in $j$) then such a function exists if and only if $\sum_j \frac{1}{jM_j^{1/j}} < \infty$. For example there exists a function $f$ such that
\begin{equation}\label{bound}
|f^{(j)}(x)| \le CY^{-j}j^{(1+\varepsilon)j}
\end{equation}
for any fixed $\varepsilon > 0$ (this is related to the so-called Gevrey classes).
Actually, since you mentioned Fourier transform, let me write about a different result which is more directly applicable to this type of problems: Beurling-Malliavin multiplier theorem. It reads as follows:
Let $w:\mathbb{R}\to \mathbb{R}$ be a nonnegative Lipshitz function (this is a small technical condition). Then there exists a nonzero compactly supported function $f$ with $|\hat{f}(\xi)| \le e^{-w(x)}$ if and only if integral
$$\int_\mathbb{R} \frac{w(x)}{x^2 + 1}dx$$
is convergent. Moreover, support of the function can be arbitrary small.
Lastly, if you want an explicit function $g$ (and therefore $f$), satisfying the above bound, you can take
$$g(x) = e^{-(1-x)^{-m}}e^{-(x+1)^{-m}}\chi_{(-1, 1)},$$
see e.g. this paper, section 3.1.
Interesting paper you cited. This kind of info about eigenvalues was a key ingredient of the work by Connes on the trace formula and Riemann Hypothesis https://link.springer.com/article/10.1007/s000290050042
Very nice! Thank you!
Not sure what $X$ and $Y$ are. Positive constants? Also, the meaning of the symbol $\ll_j$ should be explained. I may have misunderstood condition 4) but it might be too strong and imply analyticity of $f$ which contradicts 2) and 3).
It seems what the OP needs is some explicit bumpfunctionology. Start with $f(x)=0$ for $x\le 0$ and $f(x)=\exp\left(-\frac{1}{x}\right)$ for $x>0$. Then take shifts, reflections, and products.
Integrate and also shift, reflect, and take products of the antiderivative. One typically obtains in this way functions that are Gevrey with explicit bounds on derivatives as a function of $j$. These bounds can be proved via Cauchy's Theorem with complex analysis or combinatorially (my favorite approach). For a fully worked out example of how to do the bounds combinatorially, see my answer to this MO question:
Gevrey estimate of derivatives
|
2025-03-21T14:48:31.134273
| 2020-06-03T19:16:37 |
362126
|
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|
Stack Exchange
|
Explicit solution of a free boundary problem for heat equation
Consider the free boundary problem
$$
\min\{u_t - u_{xx} -1, u \} = 0 \qquad \text{ in } (0,T)\times (-1,1) \\
u(0,\cdot) = 0 \qquad \text{ in } (-1,1)\\
u(\cdot, -1) = u(\cdot, 1) = 0 \qquad \text{ in } (0,T)
$$
Can we obtain an explicit solution for it?
From the maximum principle, the solution $v$ of
$$v_t=v_{xx}+1,$$
together with the data $v=0$ at the boundary and initial time, is positive. Therefore your $u$ is nothing but $v$. It turns out that it can be expressed explicitly, using the heat kernel.
|
2025-03-21T14:48:31.134347
| 2020-06-03T20:03:35 |
362132
|
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"ofer zeitouni"
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|
Stack Exchange
|
Reversing the order of conditioning in a sum to compare conditional variances
Suppose $Z=X+Y$ where $X$ is independent of $Y$ and $Y\sim N(0,1)$. I would like to compare $\text{var}(E(X|Z))$ to $\text{var}(E(Z|X))$. Obviously, $\text{var}(E(Z|X))=\text{var}(X)$.
In particular, my guess is that $\text{var}(E(Z|X)) > \text{var}(E(X|Z))$. If $Z$ is normal, then this is easy to prove directly from known formulas for conditional distributions of a multivariate Gaussian. Furthermore, if $Y$ is non-normal, then this appears to be false as well. The difficulty, as I see it, is in getting a handle on $E(X|Z)$ while accounting for the fact that $Y$ is Gaussian. Another approach is to compute (or control somehow) the marginal $P(Z)$, which would help understand $P(X|Z)=P(X,Z)/P(Z)$.
@NateEldredge Essentially that is what I use in my answer. Maybe you can help me make the second part completely bulletproof.
@S.Surace: It's definitely true that if $\operatorname{Var}[X \mid Z]=0$ then $X$ is $Z$-measurable, i.e. $X =f(Z)$. The last part I have to think about.
This is a nice question. The issue can be reformulated as finding a lower bound on the estimation error of $X$ given $Z$. Now, if the law of $X$ has finite fisher information, then the Gaussianity of $Y$ together with Van-Tree's version of Cramer-Rao would give a strictly positive lower bound on the variance. This however does not prove the general case.
@oferzeitouni Can you elaborate on this part: "...would give a strictly positive lower bound on the variance. This however does not prove the general case."? Which variance and what kind of lower bound? I think my confusion is purely semantic, i.e. there are several "variances" in the question. Apologies if this is not clear!
@PeteJorgensen Yes. The question you are asking can be rephrased as a question about the mean square error of the optimal estimator of $X$ given $Z$. Namely, your conjecture is equivalent to the claim that $\sigma^2:=\inf_{W} E(X-W)^2>0$, where the infimum is over all $Z$-measurable, $L^2$ random variables (the infimum is achieved by $W=E(X|Z)$, and this is the link to your question). $\sigma^2$ is the variance I refer to in my answer.
@PeteJorgensen Now, there is a lower bound on $\sigma^2$, known as the van-Trees version of the Cramer Rao lower bound from statistics, which gives a lower bound on $\sigma^2$ in terms of the inverse of a sum of two Fisher informations: one related to $p(Z|X)$ (which involves the Gaussianity of $Y$, and hence is bounded), and the other is the Fisher information of the law of $X$. If the latter is bounded (e.g., if $X$ has a nice smooth density that does not vanish too fast at a point) then the latter is bounded and you get a lower bound on $\sigma^2.
@PeteJorgensen The argument for the general case (ie when the density of $X$ does not have a bounded Fisher information) escapes me.
EDIT: It seems that one can remove the integrability condition.
Suppose that $\text{Var}[X]<\infty$ and $Z=X+Y$, where $Y\sim\mathcal{N}(0,1)$ and $X$ and $Y$ are independent.
Because of the law of total variance,
$$\text{Var}\{E[Z|X]\}=\text{Var}\{X\}\\=E[\text{Var}\{X|Z\}]+ \text{Var}\{E[X|Z]\}\geq \text{Var}\{E[X|Z]\},$$
with equality iff $ \text{Var}\{X|Z\}=0$. So to prove your conjecture we only need to show $ \text{Var}\{X|Z\}>0$.
Theorem $\text{Var}\{X|Z\}=0$ if and only if $X$ is constant.
Proof: If $X$ is constant then $\text{Var}[X]=0$ and therefore also $\text{Var}\{X|Z\}=0$. For the converse, suppose that $\text{Var}\{X|Z\}=0$. Let $\mathbb{P}$ be the measure under which $Z=X+Y$, where $X$ has law $P_X$, $Y$ is standard Gaussian and independent of $X$, and $\mathbb{\tilde P}$ be the measure under which $Z$ is standard Gaussian and independent of $X$, which has the same law as under $\mathbb{P}$. Let
$$ L := \exp\left[X Z-\frac12 X^2\right].$$
Since $L$ is nonnegative, we have by Tonelli's theorem and the MGF of $\mathcal{N}(0,1)$ that
$$\mathbb{\tilde{E}}[L]=\int_{\mathbb{R}^2}e^{xz-\frac12 x^2}\mathbb{\tilde P}_X(dx)\otimes \mathbb{\tilde P}_Z(dz)=\int_{\mathbb{R}}e^{-\frac12 x^2}\left(\int_{\mathbb{R}}e^{xz}\mathbb{\tilde P}_Z(dz)\right)\mathbb{\tilde P}_X(dx)\\=\int_{\mathbb{R}}e^{-\frac12 x^2}e^{\frac12 x^2}\mathbb{\tilde P}_X(dx) = 1.$$
Thus $\mathbb{P}\ll\mathbb{\tilde P}$, $ \frac{d\mathbb{P}}{d\mathbb{\tilde P}} = L$, and we have an abstract Bayes formula
$$\mathbb{E}[\varphi(X)|Z]=\frac{\mathbb{\tilde E}[\varphi(X)L|Z]}{\mathbb{\tilde E}[L|Z]}.$$
Let $M(Z):=\mathbb{\tilde E}[L|Z]$, which is finite by assumption. Then the conditional variance of $X$ given $Z$ can be written as
$$\text{Var}\{X|Z\}=\frac{M''(Z)}{M(Z)}-\left(\frac{M'(Z)}{M(Z)}\right)^2 = \frac{d}{dz}\frac{M'(z)}{M(z)} \Bigg|_{z=Z}.$$
Since $\text{Var}\{X|Z\}=0$, we have
$$\frac{M'(Z)}{M(Z)} = \frac{M'(0)}{M(0)} = \frac{\mathbb{E}[Xe^{-\frac12 X^2}]}{\mathbb{E}[e^{-\frac12 X^2}]} . $$
But we also have
$$ X = \mathbb{E}\left[ X|Z\right] = \frac{M'(Z)}{M(Z)}.$$
It therefore follows that $X$ is constant.
I think you must use normalcy of Y somewhere because of this: if X takes only 1/4, 3/4 and Y only 0,1, then when you see X+Y you know both X and Y so E(X|Z) = X and you don't get strict inequality.
Yes, @mike, I noticed that this does not get us much further. At present I do not know how to complete the argument.
@S.Surace Nice! just a quick remark: since $zX-X^2$ is bounded above, its expectation is also bounded above, so the condition you wrote holds trivially. On the other hand, you need a bit more I think, namely (as a minimum) you need the expectation of $dP/d\tilde P$ to be bounded, and for that you need $E(e^X^2/2)$ finite.
@oferzeitouni, thanks! I made a correction including the Novikov-like condition.
|
2025-03-21T14:48:31.134720
| 2020-06-03T20:05:36 |
362133
|
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|
Stack Exchange
|
When is a formal group smooth?
This is a question that I suspect is simply a matter of technical issues written down or clarified somewhere in the literature, but which I can't find.
Suppose we're working over an arbitrary base scheme $S$, maybe with some unspecified basic niceness assumptions. Following, e.g., the terminology used on p. 493 of Hazewinkel's tome Formal Groups and Applications, we can define the category of formal groups over $S$ simply as the group objects in formal schemes. Some of these formal groups' underlying spaces can be written as a relative formal spectrum, for example in the one-dimensional case $\underline{\operatorname{Spf}}_S\mathcal{O}_S[[t]]$, in which case they are called smooth by Hazewinkel. I've also variously seen the smooth ones called formal Lie groups, or even the term "formal groups" used exclusively to refer to these special ones, as in the notes of Weinstein on The geometry of Lubin–Tate spaces.
In any case, these formal groups that possess such a framing give rise to formal group laws, which is great. My question is, what are the most natural/general conditions on a formal group that imply that it has a framing? For example, I guess certainly at minimum it should be connected and formally smooth, and I believe this is enough if $S$ is the spectrum of a DVR, or even a local ring.
But in general it seems like the sheaf on $S$ given by the relative tangent space also needs to be free. For example, if we take a non-stacky open modular curve over $\mathbb{Z}[1/N]$ or even $\mathbb{Q}$, and look at the formal completion of the universal elliptic curve over it, I believe it can't have a framing since the corresponding vector bundle is in general nontrivial. (This point is what motivated the question, because on the bottom of p. 19 of Artin and Mazur - Formal groups arising from algebraic varieties, it seems to be claimed that a pretty general class formal groups of including the universal elliptic curve example is of Lie type, which seems false.)
Is this triviality of the relative tangent bundle sufficient? It seems like you could probably prove something like this using formal smoothness by lifting jets over increasing thickenings, but I'd love a place this is all written down.
I don’t know a better reference than this, and surely there is one, but the argument you describe is recorded as Proposition 7 in Lecture 11 of Lurie’s course notes on chromatic homotopy theory: http://people.math.harvard.edu/~lurie/252xnotes/Lecture11.pdf .
interesting, thank you! but actually it seems he doesn't do what i suggested and instead requires by fiat that there's a zariski-local trivialization; the rest is then a quick descent argument. it remains to figure out exactly what you need to coordinatize a formal group over a local ring
Your Artin–Mazur link pointed to "http://Is%20this%20sufficient?", which is (probably?) not what you intended. I changed it to Artin and Mazur - Formal groups arising from algebraic varieties, which I hope is correct.
oops, thank you!
I think an adaptation of Schlessinger's argument from Functors of Artin rings should work. Let $\Lambda$ be a Noetherian ring, and suppose we have a connected formal group $\mathcal{G}$ formally smooth over $\Lambda$ with relative tangent space $\mathcal{G}(\Lambda[t]/t^2)\cong \Lambda^n$. We also assume that at any closed point $\Lambda \to k$, $\mathcal{G}$ has a formal group law, hence consists only a single point (the reduction of the identity). In practice, this seems like very little to ask, though it'd be nice to figure out a more natural way to state this hypothesis in terms of $\mathcal{G}$ over $\Lambda$. (Edit: oh I guess it's probably more natural to just require $\mathcal{G}(\Lambda)$ to be a singleton; I thought there was a problem with this, but it's actually fine.)
We claim $\mathcal{G}$ can be pro-represented by some topological Noetherian $\Lambda$-algebra $R$, which is furthermore complete with respect to the ideal $\mathfrak{a}$ giving the identity section $R/\mathfrak{a} \cong \Lambda$. Indeed, we can always replace any representing object $O$ by the completion of its local ring at the identity point $\widehat{\mathcal{O}(O)_{id}}$, since for an arbitrary local Artin algebra $A$ over $R$ and an $A$-point of $\mathcal{G}$, the image of this ideal under any induced map of rings $\mathcal{O}(O)_{\text{id}}\to A$ must be nilpotent - which is just to say that the unique closed point of $\text{Spec }A$ must map to a reduction of the identity, as mentioned earlier.
Using the map $\mathcal{G}(\Lambda[t]/t^2)\cong \Lambda^n$, iteratively lift the $n$ basis vectors on the RHS by formal smoothness to elements of $\mathcal{G}(\Lambda[t])/t^k)$ for increasing $k$; in the limit, we obtain a continuous map $R\to \Lambda[[t_1,t_2,\ldots, t_n]]$ which induces an isomorphism on tangent spaces. We claim this is an isomorphism. The key lemma is an adaptation of Lemma 1.1 from Schlessinger:
Lemma: Let $S$ and $T$ be local topological Noetherian $\Lambda$-algebras, each with a distinguished $\Lambda$-point given by the respective ideals $\mathfrak{m}$ and $\mathfrak{n}$, and complete with respect to these ideals. If $\phi: S\to T$ is an adic map (sending $\mathfrak{m}$ into $\mathfrak{n}$) respecting these $R$-points, and it induces a surjection on relative tangent spaces over $\Lambda$, then it is a surjection.
Proof: The problem is to lift the surjection $\mathfrak{m}/\mathfrak{m}^2\twoheadrightarrow \mathfrak{n}/\mathfrak{n}^2$ to a surjection $\mathfrak{m}\twoheadrightarrow \mathfrak{n}$, as then we would be done since $\phi$ respects the $R$-point. We can further reduce this into showing $\mathfrak{m}^k/\mathfrak{m}^{k+1}\twoheadrightarrow \mathfrak{n}^k/\mathfrak{n}^{k+1}$ via a similar dévissage argument. This last assertion follows by pushing forward the $k=1$ case along the surjections $\text{Sym}^k(\mathfrak{m}/\mathfrak{m}^2)\twoheadrightarrow \mathfrak{m}^k/\mathfrak{m}^{k+1}$ and similarly for $\mathfrak{n}$. $\square$
In the end, the crucial thing was just that formal groups' coordinate rings can be taken to be complete with respect to the ideal at a distinguished point over an arbitrary base.
|
2025-03-21T14:48:31.135142
| 2020-06-03T20:32:06 |
362135
|
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|
Stack Exchange
|
Does the plane clustered to minimize sum distances^2 to clusters centers ( inertia / "K-means") produce hexagonal clusters / hexagonal lattice?
"K-means" is the most simple and famous clustering algorithm, which has numerous applications.
For a given as an input number of clusters it segments set of points in R^n to that given number of clusters.
It minimizes the so-called "inertia" i.e. sum distances^2 to clusters centers = $\sum_{i ~ - ~ cluster~ number} \sum_{X - points~ in ~i-th ~ cluster} |X_{in ~ i-th ~ cluster} - center_{i-th~ cluster} |^2 $
By some reasons let me ask, what happens for the plane i.e. there is no any natural clusters, but still we can pose minimization task and it will produce something. Let us look on the example:
So, most clusters look like hexagons. Especially the most central one which is colored in red.
Well, boundary spoils things, also may be not enough sample size/iteration number - simulation is not a perfect thing - but I made many and pictures are similar...
Hexagonal lattice appears in many somewhat related topics, so it might be that some reason exists.
Question 0 What is known on "inertia" minimization on the plane/torus ? (torus - to avoid boundary effects.) (Any references/ideas are welcome). Do hexagons arise as generic clusters ?
Question 1 Consider a torus of sizes R1,R2 , consider the number of clusters to be mn ,
is it true that hexagonal lattice will provide the global minima for "inertia" ? (At least for consistent values of R1, R2, m,n (R1=am, R2 = a*n) ).
(Instead of finite number of points we can consider the continuous case and substitute summation over points by the integral. Or we can sample large enough uniform datacloud - as done in simulation).
Let me mention beautiful survey by Henry Cohn at ICM2010, where lots of optimization problems of somewhat related spirit are discussed and which sound simple, but remain unsolved for years (see also MO78900). That question is not discussed there unfortunately.
The Python code for the simulation above. One can use colab.research.google.com - to run it - no need to install anything - can use google's powers for free.
from sklearn.cluster import KMeans
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial.distance import cdist
import time
#import datetime
t0 = time.time()
N = 10**5 # Number of uniformly scattered point
d = 2 # dimension of space
X = np.random.rand(N,d) # Generate random uniform N poins on [0,1]^d
n_clusters = 225 # Number of clusters for Kmeans
clustering = KMeans(n_clusters=n_clusters,
init='k-means++', n_init=10, max_iter=600, # max_iter increased twice from default
tol=0.0001, random_state=None, algorithm= 'full' ).fit(X) # Run K-means with default params
# https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html#sklearn.cluster.KMeans
print(time.time() - t0, ' secs passed' ) # Print time passed
cluster_centers_ = clustering.cluster_centers_ #
predicted_clusters = clustering.labels_ #
####################################################################
# Choose the most central classter - hope boundary effect on it would be negligble
central_point = 0.5 * np.ones(d) # Choose central pint
idx_most_central_cluster = np.argmin( cdist( central_point.reshape(1,-1), cluster_centers_ ) ) # Find cluster most close to central point
coords_most_central_cluster_center = cluster_centers_[idx_most_central_cluster,: ]
mask = idx_most_central_cluster == predicted_clusters
Xm = X[mask,: ] # Select only points from the most central cluster
#######################################################################
# Plotting
fig = plt.figure( figsize= (20,10 ) ) # 20 - horizontal size, 6 - vertical size
plt.scatter( X[:,0], X[:,1], c = predicted_clusters ) # scatter plot all the points colored according to different clusters
plt.scatter( cluster_centers_[:,0], cluster_centers_[:,1], c = 'blue' ) # Mark centers of the clusters
plt.scatter( Xm[:,0], Xm[:,1], c = 'red' ) # Color by red most central cluster
plt.title('n_sample = '+str(N) + ' n_cluster = ' + str(n_clusters))
plt.show()
Since a point p is assigned to center c iff c is the closest center to p otherwise the cost decreases, cluster centers partition the space into voronoi diagrams (https://en.wikipedia.org/wiki/Voronoi_diagram). So your question is more about how voronoi diagrams of random points look like
Not quite. Of course, points are clustered as most close to cluster centers, so it is, of course, Voronoi diagram for cluster centers. But, the main point of the question, is that clusters centers behave not randomly , but seems to tend to hexagonal lattice. Please pay attention, that we are optimizing clusters centers positions, they are not fixed.
The answer is yes, at least in the limiting case where the number of points tends to infinity.
Specifically, this is known as the quantizer problem (see Chapter 2 of Sphere Packings, Lattices and Groups by Conway and Sloane). The two-dimensional version of the problem was solved by Fejes Tóth, who showed that the hexagonal lattice is optimal.
László Fejes Tóth, 1959: Sur la représentation d'une population infinie par un nombre fini d'éléments
The way that the quantizer problem is formalised in Sphere Packings, Lattices and Groups is to take a large compact ball $B \subsetneq \mathbb{R}^n$ and ask for the limit (as $M \rightarrow \infty$) of the infimum (over all arrangements of $M$ points in the ball) of the normalised mean squared error from a uniform random point in the ball to the closest of the $M$ points:
$$ \dfrac{1}{n} \dfrac{\frac{1}{M} \sum\limits_{i=1}^{M} \int\limits_{V(P_i)} \lVert x - P_i \rVert^2 \; dx}{\left( \frac{1}{M} \sum\limits_{i=1}^{M} \textrm{Vol}(V(P_i)) \right)^{1 + \frac{2}{n}}} $$
Here, $V(P_i) \subseteq B$ is the Voronoi cell of $P_i$. The connection with $k$-means (where $k = M$ and the ambient dimension is $n$) is that the minimiser of this expression must have each $P_i$ be the centroid of its Voronoi cell $V(P_i)$, and therefore the optimal solution is a fixed point of the $k$-means iteration. The complicated normalisation is to ensure that the limit is sensible (e.g. not $0$ or $\infty$).
For $n = 2$, the limit as $M \rightarrow \infty$ of the infimum of the above expression is $\frac{5}{36 \sqrt{3}} \approxeq 0.0801875$, and is the same as the limit as $M \rightarrow \infty$ of the expression where the points are centred at the vertices of a hexagonal lattice (scaled to have exactly $M$ points inside $B$).
For $n = 3$, the best lattice is the body-centred cubic lattice, but there are more efficient nonlattice arrangements and the quantizer problem is unsolved.
In higher dimensions the problem is unsolved.
Thank you very much for your excellent answer ! If it would be possible to suggest references on the dimension 3 and higher cases - that would be very kind of you . The paper by Toth, unfortunatelt is behind the pay-wall, if it possible to suggest some later references, may be just surveys - that would be great...
Let me clarify one point about dimension 2, does the hexagonal lattice is THE ONLY solution, or the there can be some other ?
@AlexanderChervov In terms of surveys, you'll want to get hold of a copy of the book Sphere Packings, Lattices and Groups by Conway and Sloane; my entire answer is basically just paraphrasing relevant parts of Chapter 2.
@AlexanderChervov Yes, the hexagonal lattice is the unique optimal solution. This is implied by the proof given here, together with the observation that equality holds in (3) if and only if the polygon in question is regular. https://dmg.tuwien.ac.at/gruber/gruber_arbeiten/short1.pdf
Thank you very much for your comments !
|
2025-03-21T14:48:31.135722
| 2020-06-03T20:55:02 |
362137
|
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"Pruthviraj",
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}
|
Stack Exchange
|
Problem related to transforming polynomial
Given $a_1,a_2,...,a_m$ positive integer. Denominator $d$ is smallest positive integer for $b_l$ integer coefficient.
$$\sum_{k=1}^m\binom{n}k a_k= \frac{1}d\sum_{l=1}^mb_ln^l$$
Now consider $n=dt+r$ where $d>r\ge 0$.
can it be shown that, above equation transform as
$$\frac{1}d\sum_{l=1}^mb_l(dt+r)^l=\sum_{u=0}^{m-1}(x_ut+y_u)(dt+r)^u$$
With $x_u$ and $y_u$ integers.
Example
Let $a_1=1,a_2=1,a_3=3$
then $\sum_{k=1}^3\binom{n}k a_k=(n^3-2n^2+3n)/2$ here $d=2$
Case$(1)$ $n=2t$,
$\frac{n^3-2n^2+3n}2=(t-1)n^2+n+t$
Case$(2)$ $n=2t+1$,
$\frac{n^3-2n^2+3n}2=(t-1)n^2+(t+2)n$
Thank you.
Denote $\sum b_l(x+r)^l=\sum c_l x^l$. The numbers $c_l$ are still integers, $c_0=\sum b_l r^l$ is divisible by $d$, and we have, denoting $dt+r=z$,
$$
\frac1d\sum b_l(dt+r)^l=\frac1d\sum c_l (dt)^l=\frac{c_0}d+\sum_{l>0} c_l t\cdot (dt)^{l-1}=\\
\frac{c_0}d+\sum_{l>0} c_l t\cdot (z-r)^{l-1}=
\frac{c_0}d+\sum_{l>0,0\leqslant j\leqslant l-1} c_l {l-1\choose j}(-r)^{l-1-j}z^jt,
$$
that has desired form if you group all terms with $z^j$ together.
Thanks for your solution. It's helpful to progress on my other problems. In given claim, $d\mid c_0=0$.
|
2025-03-21T14:48:31.135818
| 2020-06-04T02:21:14 |
362145
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362145"
}
|
Stack Exchange
|
GCD of polynomial values over a box
Let $f,g$ be two polynomials with integer coefficients and equal degree $r \geq 3$, and co-prime as polynomials. I am interested in the following question: what is the distribution of $\gcd(f(m), g(n))$ as $m,n$ vary over a box?
In particular, I want to count the number of pairs of integers $(m,n)$ with $|m|, |n| \leq X$ and such that $\gcd(f(m), g(n)) > X^2$.
In the special case I am interested in, $f,g$ are given by $f(x) = F(x,\alpha), g(x) = F(x,\beta)$ for a binary form $F$ with integer coefficients and degree $r \geq 3$, with $\alpha, \beta$ distinct integers having size at most $X$.
|
2025-03-21T14:48:31.135887
| 2020-06-04T02:38:23 |
362148
|
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"Alapan Das",
"VS.",
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"https://mathoverflow.net/users/156029"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362148"
}
|
Stack Exchange
|
Minimum size integer accommodating some divisors within some prescribed gaps
Assume we pick $t$ uniformly random integers $l_1$ to $l_t$ independently from $1$ to $2^v$.
Assume $k_1$ through $k_t$ are similarly picked from $1$ to $2^r$.
What is the minimum size of non-negative integer $a$ such that at every $i$ from $1$ to $t$ we have $k_i|(a-l_i)$?
I think by CRT it is $\prod_{i=1}^t k_i$ size at most.
I just was going to write that. But the upper limit is true when $k_i$ are mutually co-prime. If $(k_i,k_j)=x$ and $x \nmid (l_i-l_j)$ for any two $i,j$, then there won't exist any such $a$.
|
2025-03-21T14:48:31.135958
| 2020-06-04T03:36:11 |
362150
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362150"
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|
Stack Exchange
|
Euclidean volume of symmetric matrices in operator norm
This is a nearly identical question to Euclidean volume of the unit ball of matrices under the matrix norm except in the symmetric case.
Let $\mathrm{Sym}_{n \times n}(\mathbb{R})$ be the space of real-valued $n \times n$ symmetric matrices. Let $\phi : \mathbb{R}^{n(n+1)/2} \mapsto \mathrm{Sym}_{n \times n}(\mathbb{R})$ embed $\mathbb{R}^{n(n+1)/2}$ into $\mathrm{Sym}_{n \times n}(\mathbb{R})$.
Consider the set $H_n = \{ v \in \mathbb{R}^{n(n+1)/2} : \| \phi(v) \| \leq 1 \}$, where $\|M\| = \max_{\|x\|=1} \|Mx\|$ is the $\ell_2 \mapsto \ell_2$ operator norm.
What is the formula for $\mathrm{Vol}(H_n)$, where $\mathrm{Vol}(\cdot)$ is the Lebesgue measure on $\mathbb{R}^{n(n+1)/2}$?
@RodrigodeAzeved that was a typo, thanks for pointing it out
|
2025-03-21T14:48:31.136042
| 2020-06-04T03:50:18 |
362151
|
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"authors": [
"Fedor Petrov",
"Pedja",
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"joro"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/362151"
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|
Stack Exchange
|
Primality test for specific class of $N=8k \cdot 3^n-1$
This question is related to my previous question.
Can you prove or disprove the following claim:
Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$
Let $N=8k \cdot 3^n-1$ such that $n>2$ , $k>0$ , $8k <3^n$ and
$\begin{cases} k \equiv 1 \pmod{5} \text{ with } n \equiv 0,1 \pmod{4} \\ k \equiv 2 \pmod{5} \text{ with } n \equiv 1,2 \pmod{4}
\\ k \equiv 3 \pmod{5} \text{ with } n \equiv 0,3 \pmod{4}
\\ k \equiv 4 \pmod{5} \text{ with } n \equiv 2,3 \pmod{4} \end{cases}$
Let $S_i=S_{i-1}^3-3S_{i-1}$ with $S_0=P_{18k}(3)$ , then $N$ is prime iff $S_{n-2} \equiv 0 \pmod N$ .
You can run this test here. I have verified this claim for $k \in [1,300]$ with $n \in [3,1000]$ .
Is double $i-1$ typo?
@joro No, it isn't.
Where does this question come from and why do you expect this to be true?
@FedorPetrov I made it by myself. I was inspired by Lucas-Lehmer-Riesel primality test.
Assume that $N$ is prime. Then we prove $S_{n-2}\equiv 0\pmod N$, the assumption that $8k<3^n$ is not used. I do not know how to prove it in the opposite direction.
We have $P_m(2\cos t)=2\cos mt$, so they are Chebyshev polynomials and satisfy $P_{mn}=P_n\circ P_m$. Note that $x^3-3x=P_3$, thus $S_{i}=P_{18k\cdot 3^i}(3)$, $S_{n-2}=P_{2k\cdot 3^n}(3)=P_{(N+1)/4}(3)=P_{(N+1)/2}(\sqrt{5})$.
We prove that $2^{3(N+1)/2}P_{(N+1)/2}(\sqrt{5})$ is divisible by $N$ in the ring $\mathbb{Z}[\sqrt{5}]$. This would imply $$\frac{2^{3(N+1)/2}P_{(N+1)/2}(\sqrt{5})}N\in \mathbb{Z}[\sqrt{5}]\cap \mathbb{Q}=\mathbb{Z},$$
thus $N$ indeed divides $P_{(N+1)/2}(\sqrt{5})$. Further we write congruences modulo $N$ in the ring $\mathbb{Z}[\sqrt{5}]$.
Using quadratic reciprocity and the explicit calculations of powers of 3 modulo 4, we see that your additional condition means that 5 is a quadratic non-residue modulo $N$.
This means that $5^{(N-1)/2}\equiv -1$, or $5^{N/2}\equiv-\sqrt{5}$. We have
$$
2^{3(N+1)/2}P_{(N+1)/2}(\sqrt{5})=2^{N+1}\left(\left(\sqrt{5}+1\right)^{(N+1)/2}+\left(\sqrt{5}-1\right)^{(N+1)/2}\right)\\ =\left(\sqrt{5}-1\right)^{(N+1)/2}\cdot\left(
\left(\sqrt{5}+1\right)^{N+1}+2^{N+1}\right)\\
\equiv\left(\sqrt{5}-1\right)^{(N+1)/2}\cdot\left(
(\sqrt{5}+1)(5^{N/2}+1)+2^{N+1}\right)\\\equiv
\left(\sqrt{5}-1\right)^{(N+1)/2}\cdot\left(
(\sqrt{5}+1)(1-\sqrt{5})+2^{N+1}\right)\equiv 0.
$$
I guess the same reasoning can be applied here.
Yes, it looks so. But again only in one direction.
|
2025-03-21T14:48:31.136200
| 2020-06-04T04:33:08 |
362152
|
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"Yemon Choi",
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"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/763",
"leo monsaingeon",
"maria_c"
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|
Stack Exchange
|
Convergence as measure vs in $H^{-2}$
Let the domain be the two dimensional torus $\mathbb{T}$, and let $f_n $ be a sequence bounded in $H^1$, such that $\sup_n |f_n|\le 1$, and $f_n \to f$ weakly in $H^1$. Let $u_n = f_n \frac{\partial f_n}{\partial x} \frac{\partial f_n}{\partial y}$.
My Question: do I have the lower semicontinuity result
$$\liminf_n \|\nabla f_n\|_{L^2 }^2 +\epsilon \int_{\mathbb{T}} u_n dx\ge \|\nabla f\|_{L^2 }^2 +\epsilon \int_{\mathbb{T}} f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}dx $$
at least for sufficiently small $\epsilon$?
My Attempt: clearly the $L^2$ norm of the gradient is lower semicontinuous.
To get the full semicontinuity, I try to show
$$\int_{\mathbb{T}} u_n dx\to
\int_{\mathbb{T}} f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}dx . $$
Since $f_n$ is uniformly bounded in $H^1$, we have $f_n \to f$ strongly in $L^2$, and
$\frac{\partial f_n}{\partial x} \to \frac{\partial f}{\partial x} $ and $\frac{\partial f_n}{\partial y} \to \frac{\partial f}{\partial y} $ weakly in $L^2$. Thus $u_n\to f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$ in some weaker topology, e.g. $H^{-2}$. Moreover, since $\sup_n|f_n|,|f|\le 1$, and $f_n\to f$ weakly in $H^1$, we have $u_n$ is uniformly bounded in $L^1$, hence it converges weak-* to some Radon measure $\mu$.
So (up to subsequence), $u_n$ converges to $\mu$ in the weak-* topology, and to $f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$ in $H^{-2}$.
Main issue: can I say that $\mu=f \frac{\partial f}{\partial x} \frac{\partial f}{\partial y}$?
My main worry is that $H^2$ functions are not dense in $L^\infty$...
Any help is greatly appreciated, thanks!
How on earth do you get that $f_n \partial_xf_n\partial_yf_n\to f\partial_xf\partial_y f$ "in a weak topology" (presumably $H^{-2}$, as you seem to believe)? you cannot simply take the product of weakly converging sequences and conclude that it is weakly converging to the product of the limits... I recommend cross-posting your question to Math.StackExchange, people there will certainly explain why your naive hope cannot hold
For what it's worth I think that the question may as well stay here, even though there is a problem with the reasoning as Leo Monsaingeon has pointed out. For instance, it is not immediately clear to me that the original question has a negative answer, merely that the OP's reasoning along the way is incorrect
@leomonsaingeon The reason I "hope" this holds is gut feeling: what I hope is actually the lsc of the elastic part of the Landau de Gennes energy $|\nabla Q|{L^2}^2 + \epsilon\sum{i,j,k,l} Q^{kl} \partial_{x_k}Q^{ij} \partial_{x_l}Q^{ij}$, where $Q$ is a 2x2 Q-tensor. I just wrote one of the terms in the cubic term here, since I don't see lots of simplifications coming from the cubic term. With an added bulk term in $Q$, it was shown for the gradient flow to exist, under small initial data (Xu, Iyer, Zarnescu 2015). So for a gradient flow to exist, I'd guess its leading part to be lsc...
well I'm no expert in Ladau de Gennes theory, but I can tell you right ahead that something really intricate is going on here. In your simplified setting, and if all you know is that $f\in H^1$, the term $\int f \partial_x f\partial_y f$ may not even be integrable. Indeed $2^*=\frac{2d}{d-2}=+\infty$ at least formally, but this is a borderline case for the Sobolev embedding and $H^1$ functions fail to be $L^\infty$ in general in dimension $d=2$. Since the best one can hope for the crossed term is $\partial_x f\partial_yf\in L^2L^2=L^1$ clearly your $u$ term may very well fail to be integrable.
I would bet that the lower semi-continuity cannot really work, at least not for the weak $H^1$ convergence: because of my previous comment, and because you cannot take products of weak limits, there is no hope for a full "weak continuity" argument to work (as you were trying to use). So really the only hope is for true lower semicontinuity (by that I mean a real inequality between $\int\leq \liminf\int $, not an actual equality in general). But note that changing $f$ to $-f$ should give the reversed inequality, if any, so I really don't think that such a "true" lower sc can work here.
@Yemon Choi: fair enough!
@leomonsaingeon Ok got it, it seems one really need to use the particular structure of the flow, instead of something general. Thanks!
|
2025-03-21T14:48:31.136486
| 2020-06-04T05:27:16 |
362153
|
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"Tom Copeland",
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|
Stack Exchange
|
Derivatives of Riemann $\xi$ and traces of zeros
Looking for references essentially corroborating (to authoritatively satisfy some editors) the sketch below of the relationship between even power (2,4,...) sums (traces) of the imaginary part of the complex zeros above the real axis of the Riemann zeta function $\zeta(s)$ and the derivatives evaluated at $t =0$ of $\Omega(t) = \xi(1/2+it)/\xi(1/2)$ where $\xi(s)$ is the Landau Riemann xi function. Please inform me if you feel there are unleapable gaps or flaws in the analysis below.
The Landau Riemann function $\xi(s)$ investigated in "Relations and positivity results for the derivatives of the Riemann ξ function" by Coffey can be used to define the real-valued, entire, even (recall $\xi(s)=\xi(1-s)$) function
$$\Omega(t) = \xi(1/2 + it)/\xi(1/2)$$
which can be expanded in the Taylor series
$$\Omega(t) = \sum_{n \geq 0} (-1)^n \frac{\xi^{(2n)}(1/2)}{\xi(1/2)} \frac{t^{2n}}{(2n)!}.$$
The numerical values for the first few derivatives are given in Coffey (as well as some ways to compute them and others).
The log of the Hadamard product (Weierstrass factorization) for $\xi$ allows the derivatives of $\Omega(t)$ to be expressed in terms of the even power sums (traces $Tr_{2n}$) of the reciprocals of the imaginary parts of the zeros of the Riemann zeta above the real axis, and conversely, the power sums can be calculated in terms of the derivatives, all through the Newton identities.
For example, for a polynomial
$$p(x) = \sum_m c_m x^m = \prod_m (1-x/x_m),$$
$$=\exp[\log(p(x))] = \exp[\sum_m\log(1-x/x_m)]$$
$$=\exp[\sum_{k \geq1} (-\sum_m \frac{1}{x_m^k})\frac{x^k}{k}] =\exp[\log(1-Tr.x)]$$
$$= \sum_n ST1_n(Tr_1,Tr_2,..,Tr_n) \frac{x^n}{n!},$$
and this is precisely the exponential generating function for the signed Stirling partition polynomials of the first kind, a.k.a., the cycle index partition polynomials for the symmetric groups, given in A036039, known more commonly as the Newton identity expressing the elementary symmetric polynomials in terms of the power sums. The Faber polynomials give the Newton identity expressing the power sums in terms of the elementary symmetric polynomials.
Now for some sanity checks:
$Tr_{2n} = \sum_m 1/z_m^{2n}$ for $n=1,2,3,4$ for the first several thousands of the zeros were calculated and presented by Gottfried Helms in an MSE question. $m$ indexes the zeros above the real axis and $z_m$ denotes the imaginary part.
Using the Newton identities with all odd-indexed indeterminates nulled, I have the numerical consistency checks between Coffey's and Helms' values:
A) From Coffey's derivative estimates,
$$Tr_2 = \xi^{(2)}(1/2) / (2\xi(1/2)) \simeq .022972/(2 \cdot .497) \simeq .02311,$$
B) and Helms' sum is
$$Tr_2 = .02307.$$
C) From Coffey's estimates,
$$\xi^{(4)}(1/2)/\xi(1/2) \simeq .0.002963/.497 \simeq .005962,$$
D) and from Helms,
$$3(-2Tr_2)^2 + 6(-2Tr_4) = 12(Tr_2^2-Tr_4) \simeq 12((.02311)^2-.0000372) \simeq.005962.$$
E) From Coffey's estimates,
$$2 Tr_4 = -4 \Omega^{(4)}(0)/4! + 2 (\Omega^{(2)}(0)/2)^2$$
$$\simeq -4 \cdot .005962/4! + 2 (-.04622/2)^2 \simeq .00007448,$$
F) and from Helms
$$2Tr_4 \simeq 2 \cdot .000037173 = .00007435.$$
If there is doubt about the order of $\xi$ and whether $Tr_2$ converges, the analysis here, since the sums are truncated in the computation by Helms, can be regarded as for a corresponding truncation of the Hadamard product for $\xi$. Also, the power sum of order two over the full complex zeros is convergent and can be checked with estimates of derivatives of $\xi(s)$ at $s=0$ by using precisely the Newton identities as above.
Edit (June 5, 2020):
Affirming the convergence, see the comments in the MSE question, extracted from the Titchmarsh references.
From the example, $$-\log(g(x)) = -\log[1+a_1x+a_3x^2+ \cdots] = \sum_{n > 0} F_n(a_1,..,a_n) \frac{x^n}{n}= \sum_{n>0} Tr_n \frac{x^n}{n}$$ where $F_n$ are the Faber polynomials. Using the Appell raising op, a simple recursion relation can be developed for the $a_n$ in terms of a convolution of lower order $a_k$ and $Tr_k$, another form of the Newton identities.
https://mathoverflow.net/questions/111165/riemann-zeta-function-at-positive-integers-and-an-appell-sequence-of-polynomials is an example of this type of analysis, for the entire reciprocal gamma function.
See also e.g.f. versions--the logarithmic polynomials of https://oeis.org/A263634 and the cumulant expansion polynomials of https://oeis.org/A127671 --of the Faber polynomials https://oeis.org/A263916.
A note by Terry Tao on the sector through which the RH has been numerically checked: https://www.google.com/amp/s/terrytao.wordpress.com/2018/09/06/polymath15-tenth-thread-numerics-update/amp/
It is not a good idea to compute
$$\sum_{n=1}^\infty \gamma_n^{-k}$$ by computing a partial sum of several thousands of terms. The series converges but too slowly for this.
(see "Computation of the secondary zeta function" is on arXiv now: https://arxiv.org/abs/2006.04869).
For simplicity assume that the Riemann Hypothesis is true, then $\gamma_n$ are the ordinates of non-trivial zeros $\rho$ of $\zeta(s)$ with $\textrm{Im}(\rho)>0$.
Here I can not give a complete account of how to compute these sums efficiently. But I implemented this in mpmath some years ago. You can use this in Sage. For example writing
from mpmath import *
mp.dps=50
for n in range(2,6):
$\quad$ print secondzeta(n)
You get the approximate values with 50 digits (it is true the last four of five will not be good, as can be seen computing to more precision)
$$\begin{array}{c}
0.023104993115418970788933810430339014003381760397422\\
0.00072954827270970421587551856909397050335150570355395\\
0.000037172599285269686164866262471740578453650889730014\\
0.0000022311886995021033286406286918371933760764310879243
\end{array}
$$
I will try to revise my old paper about this, and put it on arXiv in a few days.
In mpmath the function
secondzeta(s)$=\sum_n \gamma_n^{-s}$. This extends to a meromorphic function on the complex plane. So secondzeta(1) is the value of this extended function. The series do not converge. For comparison we should give the values for second zeta(2n) these are:
$$\begin{array}{c}
0.000037172599285269686164866262471740578453650889730014\\
0.00000014417393140097327969538155609482090703688300853254\\
0.00000000066303168025299086987327208196135724847369284211165\\
0.0000000000032136641506166012161021165998346551415628219091519
\end{array}$$
@Tom Copeland Sage is free. from mpmath import *, load mpmath to Sage. mp.dps=50, put your computation to 50 digits. secondzeta(s) compute the value of this function at the point s. Even in cases the series is divergent, because it computes the only meromorphic extension.
@ Tom Copeland No, my numbers is those in which you are interested. If $\rho = \frac12+i\gamma$ is a zero of zeta then I compute $\sum_{n=1}^\infty \gamma_n^{-s}$.
@ Tom Copeland In fact my numbers coincide with those of Helm, as you can check from my values.
Thanks for revising and clarifying that for me. I see you use "ordinate" in the revised post rather than "coordinate" and your results are consistent with Helms. Thanks for corroborating that.
Let us continue this discussion in chat.
|
2025-03-21T14:48:31.136939
| 2020-06-04T05:56:44 |
362154
|
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|
Stack Exchange
|
When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable?
I was reading David Carchedi's answer for a question on Grothendieck topology for a non-small category. It "reads" like people "choose" if they allow manifolds to be Hausdorff and/or second countable. When I came across the notion of smooth manifolds for the first time, by definition, smooth manifolds are Hausdorff and second countable.
When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable?
I have observed this when I am reading about stacks.
Is it because of the constructions when we do in the set up of manifolds (stacks) which resulted in spaces that are "same as" manifolds but are not Hausdorff or not second countable?
Is it because of the influence of algebraic geometry?
Is it because of the influence of algebraic geometry (where the very first topological space (Zariski topology) we come across is non-Hausdorff)?
The Zariski topology on the maximal spectrum of a C^∞-ring is Hausdorff, so this example is not relevant.
@DmitriPavlov oh, I did not knew (do not know yet) about C^\infty rings..
@DmitriPavlov It is easy to blame anything on Algebraic geometry :P It has all kind of surprising properties/set up/constructions.. :D
One elementary reason for contemplating non-second countable manifolds is that it allows you to view any set, not just countable ones, as a discrete manifold. If one replaces second countable with paracompact, any connected component of a manifold is still second countable.
@PraphullaKoushik: I think you obviously misread David Carchedi's answer, which says “Whether one takes manifolds to mean 2nd countable + Hausdorff, or whether one removes these conditions and considers all topological spaces with a smooth atlas…”. As you can see, he uses the term “topological space with a smooth atlas” and does not mention the term “manifold” at all when he talks about non-Hausdorff or nonparacompact objects.
@DmitriPavlov In the comment below his answer he says "I just meant to say, any smooth manifold, without the requirement that it is 2nd countable, Hausdorff, paracompact... just a bare atlas."... Did I misunderstand something?
@PraphullaKoushik: Yes. He was merely responding to an inquiry about the meaning of the term “topological space with a smooth atlas” that confused it with something else. You cannot make inferences about acceptable terminology from such a clarifying remark. See also https://ncatlab.org/nlab/show/red+herring+principle about this.
@DmitriPavlov I understand.. This is the second reference to same "red herring" :D
The étale space construction produces non-Hausdorff and nonparacompact spaces (e.g., smooth manifolds)
in many practical examples that have nothing to do with algebraic geometry.
The étale space is often non-Hausdorff because two germs can coincide
on some nontrivial part of the domain without being equal.
For example, germs of continuous real-valued functions around 0 can be equal on (−ε,0),
but different on (0,ε) for arbitrarily small ε>0.
Nonparacompactness arises for the same reason.
For connected manifolds, paracompactness coincides with second countability.
The étale space, even if connected, is often not second countable because there are uncountably
many disjoint open subsets, e.g., germs of constant real-valued functions.
By etale space construction do you mean the one from sheaves on a topological space $X$? This is the only etale space construction I know.
"non-Hausdorff and nonparacompact spaces (e.g., smooth manifolds)" so, for you smooth manifolds are not necessarily Hausdorff and paracompact??
@PraphullaKoushik: The étale space construction works for sheaves of sets on any site. In particular, for sheaves of sets on manifolds it produces a (typically) non-Hausdorff non-paracompact manifold.
@PraphullaKoushik: No, and you cannot make such inferences from such a use of language. Just like when somebody says “nonunital ring”, it does not mean that rings are not assumed to be unital. See https://ncatlab.org/nlab/show/red+herring+principle.
:) I understand what you mean.. Yes, Yes, sheaves of sets on a site.. can you suggest me to look at some paper which (one of which) started using this possibility of non-Hausdorff non-paracompact manifolds..
@PraphullaKoushik: Moerdijk and Mrčun: Introduction to foliations and Lie groupoids.
Oh.. I will check that :)
Hmm, I don't know anything about stacks and all that, but below is a situation where one cares about non-Hausdorff 1-manifolds.
If one considers a codimension-1 foliation of a manifold, a very useful thing to do in dimension 3, it is often fruitful to study the leaf space of the universal cover of this manifold. That is to lift the foliation to the universal cover and examine the quotient space you get by collapsing every leaf to a point.
The generic case of this space is a non-Hausdorff, second countable, locally Euclidean topological space. The leaf space is homeomorphic to $\mathbb{R}$ if and only if it is Hausdorff. When the leaf space is $\mathbb{R}$ the foliation is called $\mathbb{R}$-covered.
The leaf space contains a remarkable amount of information about the manifold, the original foliation, and its fundamental group. This comes by way of studying how the fundamental group naturally acts on the leaf space, which is by deck transformations before quotienting. Non-Hausdorff 1-manifolds are in the literature often referred to as $\mathbb{R}$-trees, and perhaps less commonly as order trees.
Thank you.. I heard many times that one would come across non-Hausdorff spaces when studying foliation groupoids (manifolds).. I just need some time to respond..
@PraphullaKoushik Not only the leaf space but some times the groupoid associated to foliation is sometimes non Haussdorff. But this can not occured in the real analyytic case.
Search for example Ann. Maria Torpe: K theory for foliation with Reeb components
Right, I meant to say something about $C^*$ algebras, K-theory, and foliations, but I think it's better that I forgot, as I'm not really qualified to address that aspect of the theory.
@guest but non Haussdorfness of leaf space is not a surprising fact however the graph of a foliation and non Haussdorfness of it is more interesting. In 2 master thesis I tried to understand these objects with collaboration of my students. The graph of a foliation is also an interesting paper.
https://link.springer.com/article/10.1007/BF02329732
In fact in most onteresyong cases, the leaf space does not have remarkable information about foliation Kronecker foliations are obvious examples.
Non-Hausdorffness shows up in several contexts when dealing with Lie groupoids: the integration (Lie's 3rd Theorem) for Lie algebroids to Lie groupoids will typically produce a non-Hausdorff one, if it works at all. So there seems to be a large part of oid-geometry involved with non-Hausdorff manifolds.
However, to abandon second countability is a more serious step, at least in my eyes. It is not so much the existence of partition of unities (which requires paracompactness) but the second countability itself which is extremely useful. Consider the following statement:
A bijective immersion is a diffeomorphism
This is a theorem in differential geometry which you definitely want to be true and it relies directly (well, a bit hidden) on second countability. To see that it fails right on the nose if you drop second countability, consider a manifold $M$ in the usual sense of positive dimension and $M_{\mathrm{discrete}}$ as a zero-dimensional manifold with uncountably many connected components, each of which is paracompact (it's a point...) Then $\mathrm{id}\colon M_{\mathrm{discrete}} \longrightarrow M$...
Now why is this theorem important: Lie theory depends strongly on it. Any group would be a zero dimensional Lie group otherwise. In particular, the manifold structure of a Lie group would not be determined by the group structure. This would also imply that a transitive smooth group action of a Lie group on a manifold is not the same thing as a homogeneous space $G/H$ and many more problems...
So before asking why one should abandon a property, it might be good to understand what it is good for. For Hausdorffness the situation seems to be very different to second countability.
Thanks for the answer.. I wanted to read about integration of Lie algebroids.. "before asking why one should abandon a property, it might be good to understand what it is good for" +1
"In particular, the manifold structure of a Lie group would not be determined by the group structure.": The manifold structure is never determined by the group structure, unless the group is countable. For instance, the discrete group R of real numbers can be equipped with uncountably many manifold structures that turn it into a Lie group because there are uncountably many automorphisms of R.
If you're asking a historical question, it is probably because of Whitney, who set out a clear account of differentiable manifolds and proved the embedding theorem, that the Hausdorff and second countability assumptions have been regarded as "standard." Nevertheless, violations of these assumptions had been contemplated prior to Whitney's work, notably the non-second-countable Prüfer surface, which dates back to the 1920s.
In addition to the examples mentioned by others, note that people studying general relativity have sometimes considered non-Hausdorff spacetime manifolds (see here for a non-paywalled version). The consensus, however, seems to be that these are mathematical curiosities that are not physically significant. Luc and Placek amusingly quote Penrose as saying, "I must … return firmly to sanity by repeating three times: spacetime is a Hausdorff differentiable manifold; spacetime is a Hausdorff …"
Names of references, in case of linkrot: Prüfer surface -> Gabard - A separable manifold failing to have the homotopy type of a CW-complex; non-Hausdorff spacetime manifolds -> Luc and Placek - Interpreting non-Hausdorff (generalized) manifolds in general relativity.
I once posted this question (but not on MathOverflow). It was pointed out to me that non-Hausdorff manifolds arise in the study of dynamical systems. Unfortunately, I know nothing beyond that.
I am not sure if I am reading it correctly.. You are saying because of Whitehead’s theorem (which required space to be second countable and Hausdorff), people start taking Hausdorff and second countable as “standard” assumptions in definition of a manifold.. Before that, manifold was not assumed to be Hausdorff and second countable?
@PraphullaKoushik: It's Whitney's theorem, not Whitehead's. And there is no “before”: the 1931 paper by Veblen and Whitehead defined abstract smooth manifolds for the first time, and Whitney's paper followed shortly. Veblen and Whitehead assume smooth manifolds to be Hausdorff, but not necessarily paracompact.
@DmitriPavlov Yes, a mistake.. I mean Whitney's theorem :)
@PraphullaKoushik : Often, mathematicians study important examples, then gradually sense that it would be useful to make an abstract definition. There may be several "false starts" as people fumble around for the "right" definition. The "right" definition is often the one that yields a good theorem. People certainly had some vague idea of what a manifold was, or ought to be, before the 1930s. But then Whitney came along and laid out everything cleanly and proved a wonderful theorem. So it was natural to think of an abstract manifold as "the thing that Whitney's theorem was about."
@TimothyChow "[--]So it was natural to think of an abstract manifold as "the thing that Whitney's theorem was about."[---]".. Yes... :) absolutely reasonable thing to do..
|
2025-03-21T14:48:31.137758
| 2020-06-04T06:46:38 |
362158
|
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|
Stack Exchange
|
Involutive automorphism of simple Lie algebra
I am sorry if this question is too elementary to be posted here, but no experts answer this question when I post it on Math Stackexchange.
Let $\mathfrak{g}=\mathfrak{k}+\mathfrak{p}$ be a Cartan decomposition for a noncompact real simple Lie algebra $\mathfrak{g}$ corresponding to a Cartan involution $\theta$, where $\mathfrak{k}$ is the maximal compact subalgebra of $\mathfrak{g}$. Suppose that $\sigma$ is another involutive automorphism of $\mathfrak{g}$ such that $\sigma\theta=\theta\sigma$. Then $\sigma$ preserves the Cartan decomposition, and $\sigma|_\mathfrak{k}:\mathfrak{k}\rightarrow\mathfrak{k}$. By the classification for symmetric pairs, it seems true that $\sigma|_\mathfrak{k}$ is never the identity map, but how to prove this fact theoretically (instead of case by case)?
In other words, there is no (non-Cartan) involutive automorphism $\sigma$ of a noncompact real simple Lie algebra $\mathfrak{g}$ such that the subalgebra $\mathfrak{g}^\sigma$ of the fixed points under the action of $\sigma$ on $\mathfrak{g}$ contains a maximal compact subalgebra. How to prove it? I shall be grateful if experts here may offer any hint.
Did you delete the question on M.SE? I tried to find it, to link between the two, but I couldn't.
@DavidRoberts Yes, I deleted the question on M.SE after I posted the question here. Do I need to undelete it? The link is https://math.stackexchange.com/questions/3703866/involutive-automorphism-of-simple-lie-algebra, but I am not sure whether it appears.
No, it's ok. If it got no traction over there (like: no answers at all, no substantial comments), then it's better to not have a duplicate and zombie question floating around. The situation would be different if someone had put up a helpful answer but which still wasn't a full answer, or answered a slightly different question.
I guess $\sigma=id$ doesn't count?
@TorstenSchoeneberg I think that in general people do not regard the identity map as an involutive automrphism, but I am not sure. Anyway, here I just suppose that $\sigma$ is not identity on $\mathfrak{g}$.
Let $\mathfrak g$ be a noncompact simple Lie algebra and let $\mathfrak g=\mathfrak k+\mathfrak p$ be a Cartan decomposition. The simplicity of $\mathfrak g$ implies that
the adjoint representation of $\mathfrak k$ on $\mathfrak p$ is irreducible (indeed,
if $\mathfrak p_1$ is an $\mathrm{ad}_\mathfrak k$-invariant subspace of
$\mathfrak p$, one can show$^\dagger$ that $\mathfrak g_1:=[\mathfrak p_1,\mathfrak p_1]+\mathfrak p_1$ is an ideal of $\mathfrak g$). In particular, $\mathfrak k$ is a maximal (not only maximal compact) subalgebra of $\mathfrak g$. Indeed, if $\mathfrak h$ were a subalgebra containing $\mathfrak k$, then
$\mathfrak h =\mathfrak k +\mathfrak h\cap\mathfrak p$ and $\mathfrak h\cap\mathfrak p$ would be an $\mathrm{ad}_{\mathfrak k}$-invariant subspace of $\mathfrak p$.
$\dagger$ Write $\mathfrak p=\mathfrak p_1+\mathfrak p_2$ $\mathrm{ad}_{\mathfrak k}$-invariant decomposition. Let $X_i\in\mathfrak p_i$ and $Y=[X_1,X_2]\in\mathfrak k$. Denote the Cartan-Killing form of $\mathfrak g$ by $B$; it is negative definite on $\mathfrak k$ and positive definite on $\mathfrak p$; in particular, we may assume $B(\mathfrak p_1,\mathfrak p_2)=0$. Now $B(Y,Y)=B(Y,[X_1,X_2])=B([Y,X_1],X_2)\in B(\mathfrak p_1,\mathfrak p_2)=0$, so $Y=0$. We have shown that $[\mathfrak p_1,\mathfrak p_2]=0$. Using Jacobi, one completes the check that $\mathfrak g_1$ is an ideal of $\mathfrak g$.
Thank you for the answer. So you just show that a maximal compact subalgebra in a simple Lie algebra is automatically a maximal subalgebra. Interesting.
Do you know whether the statement that $\frak{l}$ is maximal in $\frak{g}$ also hold at the group level? Namely, given $K$ the compact subgroup with Lie algebra $\frak{l}$ and $G$ the (connected) Lie group with Lie algebra $\frak{g}$, is $K$ maximal in $G$? Does this only hold only for certain Lie groups with Lie algebra $\frak{g}$? (i.e. with additional properties such as simple-connectedness).
@AnotherUser Yes, it is true. I do not have time now to think of an argument, but look at the book by Helgason, exercise A.3 in ch. VI.
Thanks for the reference!
|
2025-03-21T14:48:31.138293
| 2020-06-04T07:46:57 |
362161
|
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|
Stack Exchange
|
Signs of difference matrices (sum of submatrices)
Given matrix $A \in \mathbb{R}^{m \times n}$, are there any results related to its difference array
$$A^* \triangleq \left[sign(a_{i,j} + a_{r, s} - a_{r, j} - a_{i, s})\right]_{i<r, j<s}?$$
Or equivalently, given matrix $A \in \mathbb{R}^{m \times n}$, are there any results related to
$$\hat{A} \triangleq \left[sign\left(\sum_{x=i}^r \sum_{y=j}^s a_{x, y} \right)\right]_{i\le r, j\le s}?$$
Here, $sign: \mathbb{R} \mapsto \{+, -, 0\}$ is the sign function.
What kind(s) of results are you hoping for?
Anything is useful to me since I don't even know a name for this. I am trying to characterize this 4-dimensional pattern.
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2025-03-21T14:48:31.138366
| 2020-06-04T09:17:01 |
362164
|
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|
Stack Exchange
|
Almost quadratic difference sets
Does there exist a characterization of sets $S$ such that $|S-S|$ is "almost quadratic" in $|S|$? For instance, what are some examples of sets such that $|S-S|$ is on the order of $\frac{{|S|}^2}{\log |S|}$?
As an example of what I mean, we can consider $S = A + R$ where $A$ is an arithmetic progression of size $\log n$ and $R$ is a noise, random subset of size $\frac{n}{\log n}$; $S$ consists of random translations of an arithmetic progression. We expect $|S|$ to be on the order of $n$, and $|S-S|$ to be on the order of $\frac{n^2}{\log n}$.
I was wondering, besides examples resembling the one I gave above, whether there are "constructive" sets with this difference set property. For instance, are there any such results regarding the difference sets of squares or Fibonacci numbers?
|
2025-03-21T14:48:31.138458
| 2020-06-04T09:33:41 |
362168
|
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|
Stack Exchange
|
Existence of connected component with large boundary?
Question 1. Let $\Gamma=(V,E)$ be a connected
graph with $n$ vertices, all of degree $d\geq 4$. Assume every vertex has $d$ distinct neighbors. (We can think of $d$ as being much smaller than $n$, but not necessarily bounded.)
As is customary, for a set of vertices $W\subset V$, we define the boundary $\partial W$ to be the set of vertices not in $W$ that have at least one neighbor in $W$. Call a set $W\subset V$ connected if the corresponding subgraph $\Gamma|_{W}$ of $\Gamma$ is connected.
Write $|S|$ for the number of elements of a set $S$.
What sort of lower bound can we give on $\max_{\text{$W\subset V$ connected}} |\partial W|$?
Question 2. What happens if you remove the assumption that all vertices have the same degree, and just require them to have degree between $3$ and $d$, say?
Are you at least assuming that $\Gamma$ is connected ? Otherwise the optimal lower bound seems to be $d$ (consider a disjoint union of complete graphs of degree $d$)...
Yes, I was assuming that - thanks. I've added the assumption.
If all degrees are at least 3, there exists a spanning tree with at least $n/4+2$ leaves (D. J. Kleitman and D. B. West, Spanning trees with many leaves, SIAM J. Disc. Math. 4(1991), 99-106), the сomplement of these leaves gives you a connected set with boundary of size at least $n/4+2$.
Nice! As the paper says, the result is a little bit older.
For better visibility: Kleitman and West prove that if the minimum degree of a graph is $d\geq 4$, then there is a spanning tree with at least $2n/5 + 8/5$ leaves.
The first question asked is the just the maximum leaf number of the graph. The problem of finding it is in general NP-Hard. For references, I think a good one is this, which is algorithmic. A recent paper is here. Note that the maximum leaf number is $n-d(G)$ where $d(G)$ is the connected domination number of the graph $G$.
By the way, your notation seems confusing. Not all vertices can have $d$ distinct neighbors if the graph is $d$-regular. The adjacent vertices always have one common neighbor, isnt it?
Can you send me your actual name, so that I can add you to the acknowledgements?
@HAHelfgott sorry, I didnt get you. Are you trying to acknowledge me in some paper or conference or something similar?
Paper, acknowledgements section.
F. Petrov is already there :).
@HAHelfgott I dont think I did something new in this answer. Anyways, you can cite me as is, by my username. You can give me an upvote for this answer as the acknowledgement !! When I would try to write a number theory paper, I would try to collaborate with you. Then, I would let you know my full name!! I think you would be open to email discussions?
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2025-03-21T14:48:31.138695
| 2020-06-04T09:47:19 |
362170
|
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Stack Exchange
|
Smooth Morse function from Forman's discrete Morse function
Let $M$ be a smooth manifold and $K$ a triangulation of $M$, so $K$ is a regular CW-complex and in particular a simplicial complex. Assume that $M$ is compact so $K$ is finite. Let $f\colon K \to \mathbb{R}$ be a discrete Morse function (in the sense of Forman). Is is possible to define a smooth Morse function $f'\colon M \to \mathbb{R}$ with the same critical points as $f$ (and satisfying a correspondence between the indexes of the critical points)? Is it possible to do it in "an algorithmic way" (I mean that the proof is constructive)?
As far as I know, the converse was addressed by Gallais and Benedetti, am I right?
I apologize in advance if the questions are to vague or the answers are well-known. Thanks in advance for your time.
Doesn't $f$ have critical simplices (not just points of $M$)? Because then I don't know what you mean.
@Chris A critical k-dimensional simplex should correspond to (part of) the stable manifold of an index k critical point by analogy to handlebody decompositions.
@MikeMiller That is exactly what it is shown in the reference I mentioned in my answer.
You can do the next best think. To a Forman-Morse function $f$ one can associate a flow on the manifold whose stationary points are precisely the barycenters of the faces of your simplicial decomposition. The Conley index of the barycenter of a critical face has the homotopy type of a sphere of the dimension of that face. The Conley index of the barycenter of a non-critical face is homotopically trivial.
Additionally, one can construct a continuous function $\tilde{f}$ on the manifold that decreases along the trajectories of this flow and whose value at a barycenter is equal to the value of $f$ on the corresponding face. As Mike Miller correctly pointed out, a critical face is filled out by the trajectories exiting the barycenter.
For details see Chapter 11 of this paper. The faces of the barycentric subdivision of your simplicial complex are invariant sets of this flow, and on such a face the flow is depicted in Figure 2, p.16 of the above paper.
It took me a while to realize that in Morse theory the gradient flow associated to a Morse function is more important than the function itself. The function plays a sort of accounting role and the Morse condition restricts the nature of the stationary points of the gradient flows.
Remark A while ago I asked this question on MathOverflow that is related to the abundance of discrete Morse functions. They are extremely rare as opposed to the usual smooth Morse functions that are generic.
Do you know that this is impossible for smooth Morse functions, let's say for the trivial discrete Morse function $\mu(\sigma) = \dim \sigma$? That is, is there some smooth triangulation of a smooth manifold $M$ so that there is not a smooth function $f$ on $M$ so that the barycenters of the simplices are the critical points of $f$, and the unstable manifold at a critical point is the interior of the corresponding simplex? This may be a naive question, I don't know.
@MikeMiller Given a triagulation the function that associates to each face its dimension is a discrete Morse function. However, if you randomly assign numbers to faces, then the probability that you get a discrete Morse function is small exponentially so as the number of faces increases.
|
2025-03-21T14:48:31.138965
| 2020-06-04T10:04:18 |
362171
|
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|
Stack Exchange
|
Can this problem be rephrased as optimization on a manifold?
I have question. I have a Riemmanian manifold $\mathcal{M}$, like an $n$-dimensional regular surface in $\mathbb{R}^n$. And I have a smooth scalar field defined on this manifold $f:\mathcal{M} \to \mathbb{R}$ (positive function).
My actual problem is discretized but I'm trying to generalize it, I don't have much experience with optimization on manifold but I do know about smooth manifolds and riemannian manifolds.
I was wondering if the problem was more suitable for a standard optimization problem (if you have like a single chart maybe you take $f \circ \phi_{\alpha} : \mathbb{R}^n \to \mathbb{R}$ than this would cast in a standard optimization problem), or if problems like this (where you know in advance you have a manifold with some metric) are actually better tackled in this form. The idea would be to work out an algorithm at the last.
I have few references about optimization on manifolds and I'm slowly reading through them, it seems most of the time you have a closed form description of your manifold which actually seems to enable closed form expressions of gradients, hessians and retractions etc. In my case instead I don't really have that description (because I have a discretized mesh).
Is there any benefit you can highlight maybe? (Please let me know if I can give more details).
What are you trying to optimize? What is the role of the Riemann metric? The Fermat principle works on smooth manifold. If $x_0$ is a local minimum of $f$ then $x_0$ is a critical point of $f$, i.e., $df(x_0)=0$. Also at a critical point there is a well defined concept of Hessian, and if this Hessian is positive definite then that critical point is a local min.
@LiviuNicolaescu As an example I'm trying to optimize curvature, I have this manifold and I want to find the point whose curvature is maximum (or a local optimizer).
Here again, there are many flavors of curvature: the curvature tensor, the sectional curvatures, the Ricci curvature and the scalar curvature. Only the last is a scalar quantity and, usually, only scalar quantities can be optimized. Without any assumption on the manifold or the quantity you cannot expect a precise answer.
And again, given my assumption (i.e. $f : \mathcal{M} \to \mathbb{R}$) in your list you can pick the scalar curvature. I think I've given already a stricter assumption already, i.e. regular submanifolds in $\mathbb{R}^n$, or isn't this enough?
Regular submanifold is no assumption at all since by Whitney's embedding says that manifold can be properly embedded as a regular submanifold. Is this manifold compact? If it is not compact, then maybe there is no absolute maximum or there could be many local maxima and no absolute maxiumum, or there could be an absolute maximum. A special case of your question is: given a function $f:\mathbb{R}\to\mathbb{R}$ find its maximum. possible answers to this question are all possible in the more general case, and a bit more.
@LiviuNicolaescu I'll make my question as more specific as possible then.
That will help. The more specific, the more helpful the answer.
@LiviuNicolaescu I had a thought about this problem I don't think apart from compactness I cannot assume more.
In this case minima/maxima exist. Such a point is a critical point, i.e., the differential of $f$ at such a point is zero. Find all the critical points (not easy in general) and then pick the critical point where $f$ has the smallest value (not easy). For a Random function will have many critical points and very few will be local minima.
|
2025-03-21T14:48:31.139222
| 2020-06-04T10:19:04 |
362173
|
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|
Stack Exchange
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Centralizer of elements in the upper-triangular matrices
Let $p$ be a prime number and $G=\operatorname{GL}_n ( \mathbb{Z} / p \mathbb{Z}
)$ 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 $G$.
Question 1: Let $M\in U$. Is there a special case for the group $C_{U}(M)$ to be Abelian?
Question 2: What is the order of $C_{U}(M)$?
Any help would be appreciated so much. Thank you all.
I think that in the extreme case ${\rm rank}(M-I) = n-1$, we do have $C_{U}(M)$ Abelian.
Yes, since then $M$ is a cyclic matrix (one Jordan block) and so its centralizer in the larger group $\mathrm{GL}_n(\mathbb{Z}/p\mathbb{Z})$ is the invertible matrices in the algebra in generates, and so is commutative.
I'm not sure there is a nice answer here. Certainly it depends on more than just the Jordan type of $M$. For example, in the $3 \times 3$ matrices, the matrix $Z = I + E_{13}$ is in the centre of the Sylow subgroup $U$, so $C_U(Z) = U$ is non-commutative. But $M = I + E_{12}$, which has the same Jordan type, has centralizer $C_U(M) = \langle M, Z \rangle$ which is commutative.
@MarkWildon: Yes, the second question is complicated, and even the first is more complicated than I first thought ( though I knew the rank(M-I) = n-1 case was clear. More generally, as you say, just knowing the JNF is not enough (though the centralizer (in GL_n) is hard enough to calculate, given the JNF- that question is considered (and worked out) in the famous Hall-Higman paper of 1956).
That $n\le p$ allows to treat $G$ as a Lie algebra over $\mathbf{Z}/p\mathbf{Z}$, which can ease computations. That is, passing to the $\log$ reduces to the same question in the Lie algebra of upper triangular nilpotent matrices.
Computer calculations show that if $n \le 7$ and $p=2$ then there are matrices of every Jordan type in $U$ whose centralizer in $U$ is abelian, with the obvious exception of $(1^n)$ and, less obviously, $(2,1,1,1)$ when $n=5$, $(2,1,1,1,1)$, $(3,1,1,1)$, $(4,1,1)$ when $n=6$ and $(2,1^5)$, $(2,2,1^3)$, $(3,1^4)$, $(3,3,1)$, $(4,1,1,1)$, $(4,3)$ when $n=7$.
I think there is an interesting question here but probably it is hard. As far as I know there is no good description of the conjugacy classes in $U$ for general $n$, even when $n < p$. The number $k(U)$ is the subject of a long-standing 'PORC' conjecture of Higman: see https://www.math.ucla.edu/~pak/papers/pattern_groups9.pdf
I can’t find a good answer, but Lemma 4.1 in {}[https://sci-hub.tw/10.1007/s10801-011-0288-2] may be interesting to answer to the second question.
The lemma in the Azad et al paper is essentially the quantitative version of Geoff Robinson's observation about the centralizers of cyclic matrices.
I think an answer to this question may exist in the literature but, because of lockdown I can't access all the relevant bits so I'll mention what I've found. First, refer to Lawther's "maximal abelian sets of roots" which addresses a related question in great generality: https://www.repository.cam.ac.uk/bitstream/handle/1810/254694/Lawther-2016-Memoirs_of_the_American_Mathematical_Society-AM.pdf?sequence=1&isAllowed=y ...
Lawther notes in Section 1.1 that in Section 3.3 of Gorenstein-Lyons-Solomon (vol 3) they show that "in an untwisted finite group of Lie type in characteristic $p$, abelian $p$-subgroups correspond to abelian sets of roots" and hence Lawther's memoir yields a solution to this question. Actually for $GL_n$, the associated root system is just $A_{n-1}$ and calculating the maximal abelian sets of roots should be easy enough directly -- but, anyway, Lawther's work covers it. The bit I can't check is what, exactly, GLS3 asserts because I don't have access from home. Perhaps someone else can check.
The final piece of the jigsaw, once one knows what the maximal abelian subgroups in the upper-triangular matrices are, is to check which of them are centralizers. I'm not quite sure how to do this but am hoping that it can be "read off" from the root group information.
|
2025-03-21T14:48:31.139561
| 2020-06-04T11:54:32 |
362178
|
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Stack Exchange
|
Prove that a given distribution is tempered
Suppose I have a distribution $E$ such that $\phi \ast E$ is square-integrable for all $\phi \in C_c^\infty \left( \mathbb{R}^d \right)$. Is it possible to prove that $E$ is tempered? It seems plausible to me, but I only get so far:
For brevity, define
\begin{equation}
G_\phi = \phi \ast E
\end{equation}
for any $\phi \in C_c^\infty$.
Now, because convolution is commutative, for all $\phi, \psi \in C_c^\infty$ we have
\begin{equation}
\mathcal{F} \left( \phi \ast \psi \ast E \right)
= \tilde{G}_{\phi \ast \psi}
= \left( 2 \pi \right)^{d/2} \tilde{\phi} \cdot \tilde{G_\psi}
= \left( 2 \pi \right)^{d/2} \tilde{\psi} \cdot \tilde{G_\phi}
\end{equation}
where the tildes denote Fourier transformed quantities.
Then, we get
\begin{equation}
\frac{\tilde{G}_\phi}{\tilde{\phi}} = \frac{\tilde{G}_\psi}{\tilde{\psi}} =: F
\end{equation}
which is of course what we expect, since we would like to interpret $F$ as the Fourier transform of our distribution $E$.
We can now deduce that $F \in L^1_{\mathrm{loc}} \left( \mathbb{R}^d \right) \cap L^2_{\mathrm{loc}} \left( \mathbb{R}^d \right)$ by
\begin{equation}
\left \Vert F \right \Vert_{L^1 \left( K \right)}
\le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert_{L^1 \left( K \right)}
\le \left \Vert \frac{1}{\tilde{\phi}} \right \Vert_{L^2 \left( K \right)} \left \Vert \tilde{\phi} F \right \Vert_{L^2 \left( K \right)}
< \infty
\end{equation}
and
\begin{equation}
\left \Vert F \right \Vert_{L^2 \left( K \right)}
\le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert_{L^1 \left( K \right)}
\le \sqrt{\left \Vert \frac{1}{\left \vert \tilde{\phi} \right \vert^2} \right \Vert_{L^\infty \left( K \right)}} \left \Vert \tilde{\phi} F \right \Vert_{L^2 \left( K \right)}
< \infty
\end{equation}
for any compact $K$ by using some $\phi \in C_c^\infty \left( \mathbb{R}^d \right)$ with strictly positive Fourier transform (These do exist).
I seem to be unable to get any further though.
It is now evident, that $F$ is in fact a distribution (as it is locally integrable) and it remains to show that we can approximate any Schwartz function under the integral in a (Schwartz)-continuous way.
Thank you! I removed that statement.
Got it, thanks.
In Laurent Schwartz's Théorie des Distributions (page 245, chap. VII, §5) you can find something similar: A distribution $T\in \mathscr D'(\mathbb R^d)$ is tempered if and only if all regularizations $T \ast \varphi\in \mathscr O_M$ for
$\varphi\in\mathscr D(\mathbb R^d)$, where $\mathscr O_M$ is the space of slowly increasing $C^\infty$-functions $f$, i.e., for every $\alpha\in\mathbb N_0^d$ there is $k\in\mathbb N$ such that $\partial^\alpha f(x)/(1+|x|)^k$ is bounded.
Sweet! I will check that reference tomorrow!
As you accepted this hint to the certainly highly relevant work of Schwartz as an answer: Is it clear that this solves your problem as stated? I guess that the answer is negative but I don't have an example.
Hm, I may be wrong (it is getting late again), but since $\phi \ast E \in L^2$ and all its derivatives are as well, we should be able to invoke a Sobolev statement? Thus $\phi \ast E$ as well as its derivatives are in $L^\infty$. Then they are also trivially in $\mathscr O_M$.
You are right. Schwartz's result indeed implies your statement.
@JochenWengenroth the result from Schwartz's book you quoted is a remark ("Remarque importante" in page 244-245) which seems to be only stated, not really proved. Is the proof elsewhere in the book (I just didn't see it), or is there another reference with the proof? Thanks in advance.
Edit: even if another answer has been accepted, I edited mine in order to correct (hopefully) the issues raised in the comments, and
possibly list more easily readable references to Łojasiewicz's solution of the division problem, and
prove that the tempered distribution $S$, found by using Łojasiewicz's division theorem and such that $\phi\ast S=\phi\ast E$, is equal to $E$.
prove a stronger property that the one required by the asker: namely, if $\phi\ast E\in\mathscr{S}^\prime$ for a single function $\phi\in C^\infty_c(\Bbb R^n)$, then $E\in\mathscr{S}^\prime$.
The result can be proved by using Stanisław Łojasiewicz's solution of the division problem in $\mathscr{S}^\prime(\Bbb R^n)$ (see [2] and [3] or [4] pp. 99-101 or [6] chapter VI, §VI.1): the equation
$$
\Phi S=T\label{div}\tag{DIV}
$$
has a tempered distribution solution $S\in\mathscr{S}^\prime(\Bbb R^n)$ for every non-null real analytic function $\Phi\in \mathscr{A}(\Bbb R^n)$ and every datum $T\in\mathscr{S}^\prime(\Bbb R^n)$. Indeed, since
$$
G_\phi=\phi \ast E\in L^2(\Bbb R^n)\qquad \forall \phi\in C_c^\infty(\Bbb R^n),
\label{1}\tag{1}
$$ we have also that $G_\phi\in\mathscr{S}^\prime(\Bbb R^n)$ as a distribution, and thus $\hat{G}_\phi\in\mathscr{S}^\prime(\Bbb R^n)$ by the isomorphism theorem for the Fourier transform in $\mathscr{S}^\prime$ (see [1], chapter VII, §7.1, theorem 7.1.10, p. 164). Then we can choose a test function $\phi\not\equiv 0$ and, by using the division theorem, find a tempered distribution $S$ such that
$$
\hat{\phi}\hat{S}=\hat{G}_\phi\label{2}\tag{2}
$$
since
$$
\phi\in C_c^\infty(\Bbb R^n) \implies\phi\in \mathscr{E}^\prime(\Bbb R^n)
$$ as a distribution, and therefore $\hat{\phi}\in \mathscr{A}(\Bbb R^n)$ i.e. $\hat{\phi}$ is a complex valued real analytic function (see, for example, [1], chapter VII, §7.1, theorem 7.1.14 pp. 165-166). Now applying the inverse Fourier transform to both sides of equation \eqref{2} and considering equation \eqref{1} we have
$$
\phi\ast S=\phi\ast E\iff \phi\ast(S-E)=(S-E)\ast\phi =0 \label{3}\tag{3}
$$
Lemma. Equation \eqref{3} implies $S=E$.
Proof for a given $\phi\in C^\infty_c(\Bbb R^n)$, consider the following convolution equation: $\DeclareMathOperator{\invs}{\small{inv}}$
$$
\phi\ast\psi(x)=-\varphi(-x)=-\varphi\circ\invs(x) \quad \forall \varphi(x)\in C^\infty_c\Bbb R^N \label{4}\tag{4}
$$
where $\Bbb R^n \ni x\mapsto \invs(x)=-x$ is the point reflection map. Again by the division theorem, this equation is solvable and its solution, i.e.
$$
\psi(x)= -\mathscr{F}^{-1} \left[\hat{\phi}^{-1}\right] \ast \varphi\circ\invs(x)
$$
apart from being tempered as a distribution, is a $C^\infty$ function since it is equal to the convolution of a tempered distribution whit a compactly supported and $C^\infty$-smooth function. Now define $\eta_r\in C_c^\infty(\Bbb R^n)$, $r>0$, as
$$
\eta_r(x) =
\begin{cases}
1 & |x|<r\\
0\le\text{ and }< 1 & r\le |x|\le r+1\\
0 & |x|>r+1,
\end{cases}
$$
Then the family
$$
\{\psi_r(x)\}_{r>0}=\{\eta_r(x)\cdot\psi(x)\}_{r>0}
$$ is a family of compactly supported $C^\infty$ functions converging to $\psi$.
Now consider the structure of the convolution on the left side of \eqref{3}: we have that
$$
(S-E)\ast \phi = \big\langle S-E ,\phi(x-\cdot)\big\rangle
$$ and thus
$$
\begin{split}
\langle (S-E)\ast \phi, \varphi\rangle & = \int_{\Bbb R^n}\langle S-E ,\phi(x-\cdot)\rangle\varphi(x)\mathrm{d} x\\
& = \left\langle S-E,\int_{\Bbb R^n}\phi(x-y)\varphi(x)\mathrm{d} x\right\rangle\\
& = \big\langle E- S,\phi\ast\varphi\circ\invs\big\rangle
\end{split}\label{5}\tag{5}
$$
so, again considering the relation \eqref{3} and the definition of the family of compactly supported functions $\{\psi_r(y)\}_{r>0}$ we have
$$
\begin{split}
\lim_{r\to+\infty}\big\langle (S-E)\ast \phi,\psi_r\big\rangle=\langle S-E,\varphi\rangle=0\quad \forall \varphi\in C_c^\infty(\Bbb R^n)
\end{split}
$$
and thus $E-S=0\iff E=S\;\blacksquare$.
Finally, the above lemma implies $E=S\iff E\in\mathscr{S}^\prime$.
Notes.
The expositions for Łojasiewicz's solution of the division problem, apart from the original works [2] and [3] are the books by Bernhard Malgrange [4] and Jean-Claude Tougeron [5] (the latter two deal with the work of Malgrange which generalizes Łojasiewicz's solution to systems and even to $C^\infty$ division in some special cases): however, neither of them is particularly readable to the accustomed to the "ordinary" theory of distributions and its application, since the techniques used are more from the theory of analytic sets (varieties) and from the (however related) theory of ideals of smooth functions than from functional analysis. Nevertheless I like the work of Malgrange [4], partly because of its generality and partly because of its newly improved digital version produced by the Tata Institute of Fundalmental research. However, as stated above, they are not easy read.
Jochen Bruning and iolo point out that the solution of \eqref{4} is unique if we chose $\phi$ in such a way that $\hat{\phi}(\xi)>0$ for all $\xi\in\Bbb R^n$: this is alway possible by the Paley-Wiener theorem.
The relation \eqref{5} can also be proved directly by using the standard definition of convolution of distributions: however, using the fact that $\varphi\in C^\infty_c$ simplifies bit the formal development.
References
[1] Lars Hörmander (1990), The analysis of linear partial differential operators I, Grundlehren der Mathematischen Wissenschaft, 256 (2nd ed.), Berlin-Heidelberg-New York: Springer-Verlag, ISBN 0-387-52343-X/ 3-540-52343-X, MR1065136, Zbl 0712.35001.
[2] Stanisław Łojasiewicz (1959), "Sur le problème de la division" (French),
Studia Mathematica 18, 87-136, DOI: 10.4064/sm-18-1-87-136, MR0107168, Zbl 0115.10203.
[3] Stanisław Łojasiewicz, Sur le problème de la division, (French), Rozprawy Matematyczne 22, pp. 57 (1961), MR0126072, Zbl 0096.32102.
[4] Bernard Malgrange, Ideals of differentiable functions, (English) Studies in Mathematics. Tata Institute of Fundamental Research 3. London: Oxford University Press, pp. 106 (1966), MR0212575, Zbl 0177.17902.
[5] Jean-Claude Tougeron, Ideaux de fonctions différentiables (French) Ergebnisse der Mathematik und ihrer Grenzgebiete. Band 71. Berlin-Heidelberg-New York: Springer-Verlag. pp. VII+219 (1972), MR0440598, Zbl 0251.58001.
I do not agree with your first point since $\hat{E}$ is not à priori defined because we do not know from the start whether $E$ is tempered. But since we can perform the Fourier transform (in L^2) and the result is in fact a tempered distribution, this is not an issue.
I admit I have trouble reading your refs: First, i fail to see that Łojasiewicz actually writes about tempered distributions, but that might be to a language barrier. Second, it seems we need in fact to use a $\hat{\phi} > 0$ to obtain a unique solution - they exists, so this is again a non-issue.
@iolo, of course you do not know a priori if $E$ is tempered or not. However, you know that its convolution with any $\phi$ is tempered, and thus you can solve a division problem like \eqref{2} and determine a tempered solution distribution which is unique except for a solution of the homogeneous equation, i.e. $$ \phi\ast E_o=0$$ Now, are solutions of this equation tempered? I’ll have a look this evening to this problem.
I don't see how your edit solves the problem that, a priori, $E$ cannot be Fourier transformed. Anyway, Lojasiewicz's theorem is a tremendous sledge hammer but I am somehow missing the point why it is relevant, here: For all test functions $\varphi$ there is a a tempered distribution $S$ with $\hat S \hat \varphi = \widehat{E\ast \varphi}$ and thus $S\ast \varphi=E\ast\varphi$. Why does this imply $E\in\mathscr S'$?
Since $\phi \ast E \in L^2$ it is a tempered distribution and its Fourier transform $T$ is a tempered distribution as well. Now we can pose the division problem $\tilde{\phi} S = T$ for a $\phi \in \mathcal{D}$ with a strictly positive Fourier transform $\tilde{\phi}$, which incidentally is also analytic by Paley-Wiener. Thus the above theorem applies and there exists a (unique, because $\tilde{\phi} > 0$) solution $S$ to the above equation. However, I do not see where Lojasiewicz's theorem states that this solution is tempered. If so, however, it must be the Fourier transform of $E$.
Apparently, I am too stupid, today, but I still don't get the point: That $S$ is tempered is part of Lojasiewicz's theorem, and if $\hat\varphi$ is strictly positive one can conclude $\widehat{E\ast\varphi}/\hat\varphi \in \mathscr S'$. But I don't understand how to argue with THE Fourier transform of $E$. Arbitrary distributions don't have a Fourier transform.
@iolo the division theorem is proved for the quotient spaces $\mathscr{P}^\prime(Y;X) = \mathscr{D}^\prime(X)/\mathscr{D}^\prime(Y)$, and $\mathscr{S}^\prime(\Bbb R^n)\simeq\mathscr{P}^\prime({\infty},\Bbb S^n)$ ($\Bbb S^n$ is the one point compactification of $\Bbb R^n$) thus the division holds for tempered distributions (I was surprised to see that it also holds in $\mathscr{D}^\prime$, but this is not useful for our purposes: have a look at chapter VII in Malgrange's book [4] particularly page 102 for the identification of $\mathscr{S}^\prime$).
@JochenWengenroth, apart from what you and iolo noticed, things are also trickier for the fact that the $S$ such that $\phi\ast S=\phi\ast E$ seems to depend on the chosen $\phi\in C_c^\infty(\Bbb R^n)$, i.e. $S=S_\phi(\varphi)$. I'll settle this and the other problem later on: as now I have to work on something different (as a matter of fact, I am commenting during my working pauses).
Okay, eventually I agree: We know $S\ast\varphi=E\ast\varphi$ for some $\varphi\in\mathscr D$ with $\hat\varphi>0$. To conclude $S=E$ is is enough to prove $S\ast\psi=E\ast\psi$ for all test functions $\psi$ (because $E(\psi)=(E\ast \check{\psi})(0)$) and this follows from $\widehat{S\ast\psi}=\widehat{S\ast\varphi\ast\psi}/\hat\varphi=\widehat{E\ast\varphi\ast\psi}/\hat\varphi=\widehat{E\ast\psi}$.
|
2025-03-21T14:48:31.140430
| 2020-06-04T12:17:34 |
362181
|
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|
Stack Exchange
|
Can we construct a sequence of trigonometric polynomials that converges pointwise to a given continuous function on the torus?
Consider any continuous function $f$ on an $m$-dimensional Torus $\mathbb{T}^m$. Can we construct a sequence of band limited functions (trigonometric polynomials), with the band width (degree of the trigonometric polynomial) along any direction, being non decreasing, in such a way that the sequence converges pointwise to the function $f$?
Refined and posted another question here, which adds computability condition, after reading the answer by Yuval. https://mathoverflow.net/q/362189/14414
The multidimensional Fejer series, i.e the Cesaro averages of the Fourier series of f, will converge uniformly to f. See https://arxiv.org/pdf/1206.1789.pdf or https://www.sciencedirect.com/science/article/pii/S0022247X12000546 for a lot more detailed information.
That would be a different question. The paper you cite involves computability from finite samples. Rather than editing your question when it is answered, it is better to post a new question.
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2025-03-21T14:48:31.140534
| 2020-06-04T12:45:05 |
362183
|
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|
Stack Exchange
|
Existence of plurisubharmonic functions on complex manifolds
Let $X$ be a noncompact complex manifold which contains no positive dimensional compact analytic sets.
Conjecture: There must be strictly plurisubharmonic functions on $X$ .
Is it true?
Is it true for $X$ being the blow-up of $\mathbb{C}^2$ at the origin minus a point in the fiber $\mathbb{C}P^1$ over the origin? (Just asking, I know nothing about strictly psh functions, but the above is a basic example of a non-quasi-affine variety with no positive dimensional compact subvarieties.)
I don't think it's true. Take for instance $Y$ to be a non-Kahler compact complex surface which has no compact complex curves, for instance an Inoue surface. If $p\in Y$ is some arbitrary point, set $X=Y\setminus\{p\}$. Then obviously $X$ has no compact curve, and if $\varphi$ is some strictly plurisubharmonic function on $Y$, then $\omega:=i\partial\bar\partial\varphi$ is a positive $(1,1)$-current, which, since $p$ has codimension $2$ in $X$, can be extended to $Y$ by Shiffman's extension theorem. So you have $\tilde\omega$ a a closed positive current on $Y$ which is smooth and $>0$ on $X$, but then there is a (more or less) standard procedure which allows you to remove the singularity at $p$, and to obtain a smooth $(1,1)$-form, strictly positive on $Y$. This procedure is due to Miyaoka. But this is impossible since the surface $Y$ is non-Kahler.
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2025-03-21T14:48:31.140648
| 2020-06-04T13:09:23 |
362187
|
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|
Stack Exchange
|
Which prime numbers fit into my equation similiar to Fermat's little theorem?
Fermat's little theorem says:
$a^{p-1} \equiv 1 \pmod p$
I have a equation which is similar to this but it's not covered by Fermat's theorem nor by Euler's theorem.
My equation is:
$2^{(p-1)/3} \equiv 1 \pmod p$
Of course this has to be true:
$1\equiv p\pmod 3 $
So is it possible to say when my equation is true? Which primes apply to my equation?
For the prime numbers 31, 43, ... the equation is correct.
For the prime numbers 7, 13, 19, 37, ... the equation is incorrect.
It looks randomly for me, which prime numbers apply.
$2^{(p-1)/3}\equiv 1\pmod p$ iff $2$ is a cubic residue, which happens iff $p$ is of the form $x^2+27y^2$. Look up cubic reciprocity.
That's great! Maybe you can also help me with n > 4 here
|
2025-03-21T14:48:31.140734
| 2020-06-04T13:20:08 |
362189
|
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"Fedor Petrov",
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|
Stack Exchange
|
Can we construct a computable sequence of trigonometric polynomials that converges pointwise to a given continuous function defined on the torus?
Consider any continuous function $f$ on an $m$-dimensional torus $\mathbb{T}^m$. Can we construct a sequence of band limited functions (trigonometric plynomials), with the band width (degree of the trigonometric polynomial) along any direction, being non decreasing, and additionally each element of the sequence being computable from a finite set of samples of $f$, in such a way that the sequence converges pointwise to the function $f$? (clarity: The finite set of samples of $f$ for computability of each element of the sequence could be different).
Without this additional computability from finite samples condition, an example given by Yuval Peres here, the multi dimensional Fejer series would be an example. But the computability from finite samples condition would exclude this, due to this non-computability theorem.
This question was refined from here, after the answer from Yuval there.
Is it correct that for approximation of a continuous function on $[0, 1]$ by polynomials Bernstein polynomials do this work?
@FedorPetrov : Although, they do approximate, Bernstein polynomials aren't band limited, I looked up this page https://en.wikipedia.org/wiki/Bernstein_polynomial#:~:text=In%20the%20mathematical%20field%20of,form%20is%20de%20Casteljau's%20algorithm.
Can you approximate your continuous function by a piecewise linear function (this requires a finite number of samples), then approximate that with a sufficiently far out term of the Fejer series?
Its been done here with trigonometric polynomials like Bernstein polynomials : https://www.hindawi.com/journals/tswj/2014/174716/
|
2025-03-21T14:48:31.140842
| 2020-06-04T14:07:36 |
362190
|
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"authors": [
"Max Lonysa Muller",
"Robert Israel",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/93724"
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"url": "https://mathoverflow.net/questions/362190"
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|
Stack Exchange
|
What is known about products of zeta values?
A couple of years ago, I asked this MSE question on the evaluation of the product of even zeta values: $$ \prod_{n=1}^\infty \zeta(2n) \approx 1.82 \quad .$$ While it can be shown that the product converges, the exact value is -- as far as I know -- unknown (please correct me if this isn't the case).
I wonder to what extent more general products of (Riemann) zeta values, e.g.
$$\prod_{n=1}^{\infty} a_{n} \zeta(c_{n} \cdot n), $$ for some sequences $(a_{n})$ and $(c_{n})$ have been studied, and what kind of results have been obtained. Even though a thing or two is known about rational zeta series, information on their product counterparts seems to be sparse. Any pointers to relevant literature are much appreciated.
See OEIS sequence A080729 and Kellner's paper referenced there.
@RobertIsrael Thank you.
|
2025-03-21T14:48:31.140927
| 2020-06-04T15:11:51 |
362193
|
{
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"Henri Cohen",
"LSpice",
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|
Stack Exchange
|
Reciprocity theorem with $n \ge 5$
If $p \equiv 1 \pmod n$, what additional conditions are needed to ensure that
$$2^{(p-1)/n} \equiv 1 \pmod p?$$
I know:
For $n=3$ (cubic reciprocity) the form is $p=x^2+27y^2$.
For $n=4$ (biquadratic reciprocity) the form is $p=x^2+64y^2$.
The solution has to deal with Gaussian integers and Artin's reciprocity theorem.
Can someone please show me how exactly to do this, maybe with the example $n=5$.
I found it very hard to understand the question, so I re-phrased it, hopefully without changing the meaning. Feel free to revert or re-edit if I did not succeed. I think you also mean to assume that $p$ is prime. I also mention your earlier question https://mathoverflow.net/questions/362187/which-prime-numbers-fit-into-my-equation-similiar-to-fermats-little-theorem , which was the case $n = 3$.
If $n$ is a prime and $\mathbb Q(\zeta_n)$ has class number 1, then you can use Eisenstein reciprocity to give a similar criterion.
I strongly suggest that you read the excellent and very complete book of Franz Lemmermeyer (who is on MO) on the subject.
|
2025-03-21T14:48:31.141038
| 2020-06-04T15:45:39 |
362194
|
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|
Stack Exchange
|
Enlarging a compact set in order to improve its shape
In my previous question it was established that if $X$ is a metrizable, connected, locally path connected space and $K\subset X$ is compact, then there is a Peano continuum $L\subset X$ such that $K\subset L$. This motivated the following questions.
Let $X$ be as above and $\pi_k(X)$ is trivial, for $k=0,...,n$ (resp $X$ is contractible). If $K\subset X$ is a Peano continuum, can we find a compact $L\subset X$ such that $K\subset L$ and $\pi_k(L)$ is trivial, for $k=0,...,n$ (resp $L$ is contractible)?
Regarding the "contractible version", here is an idea that does not work: let $F:X\times[0,1] \to X$ be a homotopy from the identity to a constant. It is tempting to try to show that $F(K\times [0,1])$ is the set that we are looking for. However, if we started with $K$ a singleton, the obtained set can be any Peano continuum, and so not necessarily contractible.
Edit: There is a good candidate for a counterexample suggested by HJRW. Let $X$ be the Whitehead manifold, which is a contractible $3$-dimensional manifold, not homeomorphic to $\mathbb{R}^3$. It is proven in the paper Brick & Mihalik - The QSF property for groups and spaces, that $X$ fails the so called QSF property. This amounts to existence of subcomplex $A$ of $X$ such that there is no celluar map $f:K\to X$ of a finite simply connected complex such that $f|_{f^{-1}(A)}\to A$ is a homeomorphism. In particular, there is no finite simply connected subcomplex $L\supset A$.
Hence, we are almost done: there is a compact set (in a contractible manifold) that cannot be included into a simply-connected subcomplex. The missing ingredient therefore is:
Can every simply connected $M\subset X$ be included in a simply connected finite subcomplex of $X$?
Interestingly, in Definition 1.5 this article, the QSF property is defined with no reference to complexes, and in fact is exactly what i need. However, that's the only place i saw this definition, so it's not 100% clear it is equivalent to the more common one.
Maybe the technique of "engulfing" is helpful, e.g. Chapter 3 of Daverman-Venema, Embeddings in manifolds.
Regarding the contractible version, you might like to think a bit about the Whitehead manifold: https://en.wikipedia.org/wiki/Whitehead_manifold . It's a contractible 3-manifold, but I believe that it cannot be filtered by compact, simply connected subsets.
@HJRW what does "filtered by compact, simply connected subsets" mean? sorry if this is a stupid question
I’d call a manifold $M$ “filtered by compact, simply connected subsets” if it can be written as an ascending union $\bigcup_{n\geq 0} K_n$, where each $K_n$ is compact and simply connected.
@HJRW that would completely answer my question (even more). Do you know a reference for that claim?
You could start with: Brick & Mihalik, The QSF property for groups and spaces, Math. Z. (1995) . The QSF property is a bit more complicated than the definition I gave above, but I think it's in the same spirit.
@HJRW please tell me if i understand correctly. It says after Theorem 1.1 that the Whitehead manifold $X$ does not have QSF. Since it is simply connected, it is its own uiversal cover. Thus, there is a finite subcomplex $A$ of $X$ such that there is no celluar map of a finite simply connected complex $K$ which homeomorphically "covers" $A$. In particular, there is no finite simply connected $L\supset A$. The missing ingredient however is: can every simply connected $M\subset X$ be included in a simply connected complex?
Interestingly, in Definition 1.5 by the link below, everything is stated in "general-topological language", with no mentioning of complexes. However, that's the only place i saw this definition, so it's not 100% clear they are equivalent. https://arxiv.org/pdf/0709.1576.pdf
@erz: I think pointing you in the direction of this literature is pretty much the limit of my expertise! You could also check out Whitehead's original paper. My one other suggestion is to look at Part V of Ross Geoghegan's book Topological methods in group theory, which I think covers some of these kind of notions.
@HJRW thank you for your help in any case! Your suggestion of QSF was very helpful! I'll check the book.
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2025-03-21T14:48:31.141320
| 2020-06-04T15:49:19 |
362195
|
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"Aurelio",
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|
Stack Exchange
|
Why are quotient sets (types) called quotients -- are they the inverse of some product?
There seems to be a beautiful relation between natural numbers and sets (and types),
as in the size of a discriminate union, cartesian product, and function type,
is described by the sum, product, exponential of the sizes of the components. (As I learned from type theory). This also makes it easy to see why the symbols + and x are used for discriminate union and cartesian product (sum type and product type).
$$
\forall A, B, C : \text{sets} \\
A + B = C ~ \implies ~ |A| + |B| = |C|\\
A \times B = C ~ \implies ~ |A| \times |B| = |C|\\
A → B = C ~ \implies ~~~~~~~~ |B|^{|A|} = |C|
$$
However, why are quotient sets (and quotient types) called quotients and use the symbol $/$?
That does not seem to make sense to me. At the very least, to deserve the name quotient, I would expect them to somehow be the inverse of some product. I first thought they should be the inverse of the cartesian product, I tried to google this, but I cannot find anything. Is there some relation between quotient (sets) and (cartesian) products, that I am missing?
If you consider quotients of groups, there is a convincing argument showing why the notation makes sense.
In the spirit of @Aurelio's answer, perhaps the next best thing to a product is a bundle; given groups $H \le G$, we have the bundle $G \to G/H$ with fibres $H$. (Although there are set maps $G \cong H \times G/H$, they respect no interesting structure in general, even if $H$ is normal.) A general quotient of set $X$ by equivalence relation $\sim$ doesn't admit $X \cong X/{\sim} \times \text?$ in any reasonable sense, even as sets, but we still have the projection $X \to X/{\sim}$.
Really this is the wrong forum for your question. Anyway, here is an analogy.
Suppose you have many cookies, say 24 of them, and you want to share them with two of your friends. You may believe in equity, and you want some of the cookies, and to give one person more than another may be socially awkward, so you want to arrange the cookies into piles or boxes so that you and your friends can check that they each have the same number. If the cookies are of different sorts, they might check that each person got the right quantity of each sort (one may prefer oatmeal to ginger and the other may have the opposite preference, say).
You arrange the cookies into three boxes, and (if I did the arithmetic right) each box got eight cookies. (This assumes none ate cookies during the process.) In arithmetic, the three is the dividend, and the eight is the quotient.
If you have one or three or five friends, you can do this process again with similar outcomes, and everyone gets an equal number of cookies in their box. If you have four friends though, there is a problem: five does not divide 24 evenly. You have to put four in each box ,and you have a remainder of four left over. For equal division, you may need another method, like breaking cookies. Or you can put the remainder in its own box for later processing.
The idea is to split up some space into pieces so that you can deal with the collections in some way. Oddly, the three boxes are consider the quotient space, and one thinks about how the groups relate to one another. It may be the remainder box is important to the quotient space. Just as important are the properties that are preserved as well as those that are dropped in passing to the quotient. The idea is not that the quotient is an inverse, but that it is a result of something that is complementary to something like Cartesian product.
Gerhard "Can I Have The Leftovers?" Paseman, 2020.06.04.
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2025-03-21T14:48:31.141724
| 2020-06-04T18:53:20 |
362200
|
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|
Stack Exchange
|
Show identity for a norm on Fréchet differentiable functions on a Banach space
Let $E$ be a $\mathbb R$-Banach space, $v:E\to(0,\infty)$ be continuous with $$\inf_{x\in E}v(x)>0\tag1,$$ $r\in(0,1]$ and$^1$ $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v^r\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$ Note that $\rho$ is a well-defined metric on $E$. Let $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$ and $$\operatorname{Lip}(\rho):=\left\{f:E\to\mathbb R\mid|f|_{\operatorname{Lip}(\rho)}<\infty\right\}.$$ $|\;\cdot\;|_{\operatorname{Lip}(\rho)}$ is a semi-norm on $\operatorname{Lip}(\rho)$. Let $\mu$ be a probability measure on $(E,\mathcal B(E))$ with $$\int\rho(\;\cdot\;,0)\:{\rm d}\mu<\infty\tag2$$ By $(2)$, $$\operatorname{Lip}(\rho)\subseteq\mathcal L^1(\mu)$$ and $$\left\|f\right\|_{\operatorname{Lip}(\rho)}:=\left|\int f\:{\rm d}\mu\right|+|f|_{\operatorname{Lip}(\rho)}\;\;\;\text{for }f\in\operatorname{Lip}(\rho)$$ is a norm.
Let $f\in\operatorname{Lip}(\rho)$ be Fréchet differentiable. How can we show that $$\left\|f\right\|_{\operatorname{Lip}(\rho)}=\sup_{x\in E}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}+\int f\:{\rm d}\mu?\tag3$$ In particular, how can we show that $$\sup_{\substack{y\:\in\:E\\0\:<\left\|x-y\right\|_E\:<\:\varepsilon}}\frac{|f(x)-f(y)|}{\rho(x,y)}\xrightarrow{\varepsilon\to0}\frac{\left\|{\rm D}f(x)\right\|_{E'}}{v^r(x)}\tag4$$ for all $x\in E$?
EDIT: Let $x\in E$. It's clear to me that $$\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}\xrightarrow{\delta\to0+}\left\|{\rm D}f(x)\right\|_{E'}\tag5.$$ So, writing $$\frac{|f(x)-f(y)|}{\rho(x,y)}=\frac{|f(x)-f(y)}{\left\|x-y\right\|_E}\frac{\left\|x-y\right\|_E}{\rho(x,y)}\;\;\;\text{for all }y\in E\tag6,$$ it only remains to show $$\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{\rho(x,y)}{\left\|x-y\right\|_E}\xrightarrow{\delta\to0+}v^r(x)\tag7.$$ Now let $$c(t,y):=(1-t)x+ty\;\;\;\text{for }(t,y)\in[0,1]\times E$$ is clearly continuous and $$\sup_{t\in[0,\:1]}\left\|c(t,y)-x\right\|_E=\left\|x-y\right\|_E\xrightarrow{y\to x}0\tag8.$$ Moreover, \begin{equation}\begin{split}&\sup_{y\in B_\delta(x)}\left|\frac1{\left\|x-y\right\|_E}\int_0^1(v^r\circ c)(t,y)\left\|\frac{{\rm d}c}{{\rm d}t}(t,y)\right\|_E\:{\rm d}t-v^r(x)\right|\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\sup_{y\in B_\delta(x)}\int_0^1\left|(v^r\circ c)(t,y)-(v^r\circ c)(t,x)\right|\:{\rm d}t\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\sup_{(t,\:y)\:\in\:[0,\:1]\times B_\delta(x)}\left|(v^r\circ c)(t,y)-(v^r\circ c)(t,x)\right|\xrightarrow{\delta\to0+}0\end{split}\tag9\end{equation} for all $y\in E$. So, this yields at least \begin{equation}\begin{split}&\sup_{y\in B_\delta(x)\setminus\{x\}}\frac{\rho(x,y)}{\left\|x-y\right\|_E}-v^r(x)\\&\;\;\;\;\;\;\;\;\;\;\;\;\le\sup_{y\in B_\delta(x)\setminus\{x\}}\int_0^1|(v^r\circ c)(t,y)-v^r(x)|\:{\rm d}t\xrightarrow{\delta\to0+}0.\end{split}\tag{10}\end{equation} How can we show the other inequality?
To get started: If $c:[0,1]\to E$ is a $C^1$-curve from $x$ to $y$ you have
$$ |f(x)-f(y)|=\left|\int_0^1(f\circ c)'(t)dt\right|\le \int_0^1 \|Df(c(t))\|_{E'} \|c'(t)\|dt $$ $$=\int_0^1 \|Df(c(t))\|_{E'} \frac{v^r(c(t))}{v^r(c(t))} \|c'(t)\|dt$$ $$ \le \sup\{\|Df(z)\|_{E'}/v^r(z):z\in E\} \varrho(x,y)(1+\varepsilon)$$ for a suitable curve.
This gives you an inequality for (4) (I guess that $x$ is fixed, there). For the other inequality I would try to take a direction $r\in E$ which (almost) maximizes $|Df(x)(r)|$ and take $y=x+\varepsilon r$. However, I did not try to work this out.
Thank you for your answer. Please take note of my edit.
I guess you are missing a differential on the right-hand side of your first equality and your $\varrho$ should be a $\rho$. Where does the division by $v^r(z)$ come from in your supremum?
Right, the derivative was missing. I have expanded the calculation.
Thanks for the edit. What do you say to my edit? I've got the feeling we only need to show that $$\frac{\rho(x,y)}{\left|x-y\right|_E}\to v(x)$$ and for this it should be enough to show that there is a curve $c$ connecting $x,y$ with $$\left|\frac1{\left|x-y\right|_E}\int_0^1(v\circ c)(t)\left|c'(t)\right|_E:{\rm d}t-v(x)\right|=\frac1{\left|x-y\right|_E}\left|\int_0^1(v\circ c)(t)\left|c'(t)\right|_E:{\rm d}t-\left|x-y\right|_Ev(x)\right|\to0.$$
I think your last inequality is wrong. It seems like you've thought that $\rho$ is the supremum (not the infimum) over the curves. However, we should still be able to obtain it by letting $\varepsilon>0$ be arbitrary and choosing $c$ with $\int_0^1v^r(c(t))\left|c'(t)\right|_E:{\rm d}t<\rho(x,y)+\varepsilon$.
Of Course, that was meant.
Maybe you've misinterpreted $(4)$ (and since I've mistakenly taken the supremum over $x$ and $y$ before, that would be my fault), but I don't understand how your upper bound in terms of $\sup{|Df(z)|_{E'}/v^r(z):z\in E}$ could be useful. I still think the easiest way to show $(4)$ is showing $(7)$, since we already know $(5)$. (Edited the question once more btw.)
You can take the supremum over a small ball containing the path instead of all E.
But that still doesn't help. We need to bound the left-hand side by $\frac{\left|{\rm D}f(x)\right|_{E'}}{v^r(x)}\rho(x,y)$; not a supremum.
|
2025-03-21T14:48:31.142012
| 2020-06-04T19:41:09 |
362204
|
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|
Stack Exchange
|
Simulating measurement data from a dynamical system and its derivative
I am implementing the SINDy algorithm in Python.
Discovering governing equations from data by sparse identification of nonlinear dynamical systems
I have a question concerning the simulation of dynamical systems.
Lets say we have measurement data of the following form and that the way features $x, y$ and $z$ change in time gets governed by a dynamical system with ODE's that we do not know:
$$
\begin{array}{ccc}
x(t_1) & x(t_2) & \ldots & x(t_n) \\
y(t_1) & y(t_2) & \ldots & y(t_n) \\
z(t_1) & z(t_2) & \ldots & z(t_n) \\
\end{array}
$$
In this example we have 3 features that got measured at $n$ different points in time.
I implemented an algorithm in Python that takes as an input this measurement data and its derivatives:
$$
\begin{array}{ccc}
x'(t_1) & x'(t_2) & \ldots & x'(t_n)\\
y'(t_1) & y'(t_2) & \ldots & y'(t_n)\\
z'(t_1) & z'(t_2) & \ldots & z'(t_n)\\
\end{array}
$$
The algorithm serves to identify the underlying dynamical system. In general these derivatives can be computed by the total variation regularized derivative algorithm, or TVregdiff.
My problem is that in order to test my algorithm and see if it identifies the correct dynamical system, I want to simulate measurement data from a given dynamical system but I am not sure if I am doing it correctly.
Lets say I want to have measurement data from the Lorenz Attractor, so we have the following dynamical system($\rho=54, \sigma=10, \beta=8/3 $):
$$
\begin{cases}
x' = -\sigma\cdot(x_0 - y_0) \\
y' = \rho\cdot x_0 - y_0 - x_0\cdot z_0 \\
z' = -\beta\cdot z_0 + x_0\cdot y_0
\end{cases}
$$
Now in order to find the data for $x, y$ and $z$, I integrate over n timesteps (using scipy's integrate.odeint) in order to solve the differential equations.
To find the derivatives I took the dynamical system and just inserted the $x(t)$, $y(t)$ and $z(t)$. So in order to find $x'(t_3), y'(t_3)$ and $z'(t_3)$ I would do this:
$$
\begin{cases}
x'(t_3) = -\sigma\cdot(x(t_3) - y(t_3)\\
y'(t_3) = \rho\cdot x(t_3) - y(t_3) - x(t_3)\cdot z(t_3) \\
z'(t_3) = -\beta\cdot z(t_3) + x(t_3)\cdot y(t_3)
\end{cases}
$$
I do this for all time steps and from that I find the derivatives of $x,y,z$ at all timesteps.
Could you please tell me if there are any errors in my thoughts about this? Thanks a lot in advance
Yes it is correct.
@RodrigodeAzevedo The ODE's are unknown, just measurement data is known. The algorithm used is SINDy.
@RodrigodeAzevedo Thanks to point that out, I will edit the question.
X-posted: https://math.stackexchange.com/q/3705900/339790
|
2025-03-21T14:48:31.142188
| 2020-06-04T19:49:19 |
362205
|
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|
Stack Exchange
|
Question on models for $EG$ for a $G$-CW complex
I am having trouble finding information on a definition in P. Hanham's PhD thesis paper. recall that given a discrete group $G$ a $G$-CW-complex $X$ is a CW-complex equipped with a topological $G$ action such that if $G$ acts on a cell in $X$ trivially, then $G$ acts on each point in said cell trivially. Hanham says a free $G$-CW-complex ($G$-CW-complex with trivial stabilizers) is said to be a model for $EG$ if $X/G$ is a $K(G,1)$ Eilenberg-Maclane space.
My question is, why $EG$? Is this just a definition or is there more that i can learn about this '$EG$'. Hanham says nothing more on the topic. I'm pretty sure there is something more to this but i am having difficulty finding this online. If someone could point me in the right direction, I would greatly appreciate it.
Thanks in advance!
$EG$ is standard notation for the total space of a principal $G$-bundle.
@MarkGrant Thank you!
|
2025-03-21T14:48:31.142275
| 2020-06-04T20:17:57 |
362207
|
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|
Stack Exchange
|
Basic questions and reference on Grothendieck ring of varieties
$\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\Spec{Spec}$I am looking for a comprehensive reference on the theory of the Grothendieck ring of varieties over a field $K$, denoted $K_0(K)$ here, that includes answers to such basic questions as:
(1) If L/K is a Galois extension of fields, then $\Gal(L/K)$ acts on $K_0(L)$. Is "$\Spec(K_0(L))//\Gal(L/K) \simeq \Spec(K_0(L))$" as stacks?
(2) What are the prime ideals of $K_0(K)$?
(3) If $R$ is an integral domain, $K$ its field of fractions, and $k$ the reduction of $R$ modulo a prime ideal, what is the relation between $K_0(K)$ and $K_0(k)$? Is there a nice fibration?
(2) is presumably very hard, because it is equivalent to "what are all additive invariants valued in a domain". For (3), have a look at this paper (which answers the question in characteristic $0$ for a certain widely studied quotient of $K_0$).
(1) is a bit strange: RHS is a scheme, so why do you ask for isomorphism of stacks? The natural question is whether $K_0(L)^G$ is isomorphic to $K_0(K)$.
|
2025-03-21T14:48:31.142393
| 2020-06-04T22:00:12 |
362210
|
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"Gabe K",
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"Pierre PC",
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|
Stack Exchange
|
Is there a Riemannian metric on the configuration space of $n$ distinct points with "nice" geodesics?
Let $C_n = C_n(\mathbb{R}^3)$ denote the configuration space of $n$ distinct points in $\mathbb{R}^3$. Is there a Riemannian metric $g$ on $C_n$ such that given any two configurations in $C_n$, there is a unique geodesic joining them?
In addition, it would be nice if $g$ was also geodesically complete, and if $g$ came from natural considerations in Physics (for instance if it is the kinetic term of some naturally occuring Lagrangian etc.).
Edit 1: I accepted Andy Putman's answer below, because it does answer negatively my question (thank you!). However, could someone please indicate whether or not there exists a complete Riemannian metric $g$ on $C_n$? Is it more appropriate to create another post perhaps? I just found out that Nomizu and Ozeki proved that any connected smooth (second countable) manifold admits a complete Riemannian metric. This is nice. However, is there a known explicit such complete Riemannian metric $g$ on $C_n$? If two of the points say are going towards each other and seem about to collide, there has to be a repulsive force that forbids collision (in physical terms).
What about bijectively mapping the configuration of $n$ points to a sum of Dirac measures and then using the optimal transport (Wasserstein-2) metric? Can this yield a Riemannian metric on $C_n(\mathbb{R}^3)$ by pullback?
See the related question https://mathoverflow.net/questions/301908/tangent-space-and-gradient-on-subspace-of-wasserstein-space-given-by-finitely-su
@S.Surace, thank you. I have not studied optimal transport yet. Do you have an introductory reference to the optimal transport metric? I will look it up.
Due to Mather's avoidance principle, the Wasserstein-2 metric will indeed induce a metric on $C_n(\mathbb{R}^3)$ (this is the repulsive force you want) . The geodesic between two configurations will not necessarily be unique, but the metric will be complete. Also, you can compute the distances reasonably quickly using linear programming and it should be feasible to determine when two configurations are in the cut locus.
@GabeK, thank you so much! I also thank @S.Surace!
I would suggest to take a look at Villani's books. More details can be found in Ambrosio, Gigli and Savaré's book from 2008.
Ah. I realized I made a mistake last night, which I should correct. The metric is not complete, but it is geodesically complete in that the shortest path between any two configurations remains a configuration. If you follow a geodesic too long though, points can collide.
The answer is no. This relies on two things:
A uniquely geodesic proper metric space is contractible; see here for a proof.
$C_n$ is not contractible. Indeed, it has many nontrivial homology groups (there is a huge literature on this).
This answers the questions under the assumption that the metric is complete, correct? Using the hypothesis that “it would be nice if $g$ was also geodesically complete”?
This is just a long comment, and a pretty speculative one at that. However it might perhaps be of interest to you since:
there is a natural connection to physics,
the construction only works in three dimensions,
the construction is equivariant with respect to the action of the permutations.
I have in mind the (conjectured) map described by Atiyah in [1] which maps configurations of points to the complex flag manifold:
$$
C_n(\mathbb{R}^3) \to U(n) / T^n.
$$
Since the flag manifold is homogeneous, this map would provide a metric on $C_n(\mathbb{R}^3)$ if we could spot a natural metric on the fibres. I don't know if this is possible but for $n=2$, the fibres of the map are pairs of distinct points defining the same direction (first point looking at second point) and so are naturally parameterised by their midpoint $m$ and distance apart $t$. It's a bit of a stretch but if we give this fibre the metric of $dm^2 + (dt/t)^2$ then we get something you might regard as "nice".
[1] Atiyah, M., "Configurations of Points", R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 359 (2001), no. 1784, 1375-1387.
Thank you for your comments. I am actually motivated by this specific problem you have mentioned. A small correction. For $n=2$, the map factors through the sphere, which is the relative direction from $x_1$ to $x_2$, and the factor map from the sphere to $U(2)/T^2$ is actually a diffeomorphism. Permuting $x_1$ and $x_2$ corresponds to the antipodal map on the sphere, which corresponds to interchanging the two columns of the corresponding two by two matrix.
Please note though, that the fibers of those maps (assuming linear independence) are not so easy to describe (in part because these maps involve the polar decomposition, at the end, which is a Weyl equivariant orthogonalization process).
I knew I had heard of your name before. I think I have read part of your thesis a while back. I think you had worked on the problem: "what is the natural geometry on the moduli space of hyperbolic monopoles?" and related problems.
You have a good memory! I did indeed write my thesis on that problem, though it has been many years since I've thought about it. I have just looked at your personal website and I see that you have already published results on Atiyah's configurations conjecture so I now realise that you were aware of all that I said. (Nice work btw!) I agree with your remarks except I am confused by the correction: note that I claimed that the (conjectured) map is equivariant, not invariant. Anyway, very good luck with your interesting research.
Thank you. Oh, it is just a minor point. I just meant that the fiber over a point in the image of the map from the sphere of relative directions to $U(2)/T^2$, for $n=2$, consists of just a single point on the sphere, rather than a pair of antipodal points. Maybe I misunderstood what you had written. It is just a minor point. Thank you for your nice comments.
|
2025-03-21T14:48:31.142795
| 2020-06-04T22:51:56 |
362215
|
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|
Stack Exchange
|
Discriminants of Gleason's period-$n$ polynomials for the Mandelbrot set
Gleason's polynomials are the sequence of monic integer polynomials defined recursively by
$$
\prod_{d \mid n} G_d(c) = (((c^2+c)^2+c)^2+\cdots+c)^2+c \quad \quad \quad [\textrm{$n$ iterates}],
$$
for $n=1,2, \ldots$. Thus they start out like:
$$
\begin{align}
G_1(c) &= c, \\
G_2(c) &= c+1, \\
G_3(c) &= c^3 + 2c^2 + c + 1, \\
G_4(c) &= c^6 + 3c^5 + 3c^4 + 3c^3 + 2c^2 + 1
\end{align}
$$
They give the period-$n$ centers for the hyperbolic components of the Mandelbrot set in complex dynamics. In many ways they resemble the cyclotomic polynomials, which would result if we had $c^n-1$ on the right-hand side of the recursive definition; or the dynatomic polynomials in the dynamical plane (this is their version in the parameter plane parametrizing the quadratic iterations $z^2+c$).
For example, $\mathrm{Res}(G_n,G_m) = \pm 1$ for any pair $n \neq m$, just as for the cyclotomic polynomials. This is proved for instance in Corollary 4.8 of this paper by Hutz and Towsley. (Courtesy of Matt Baker for this reference. An aside: What are the exact signs?)
My question is about the discriminants of these polynomials:
What are the lower and upper growth rates of $\delta_n := \log{|\mathrm{Disc}(G_n)|}$? Does $\frac{1}{n}\sum_{d \mid n} \delta_d \asymp \log{n}$?
(The latter $\sim \log{n}$ for the cyclotomics, and the upper and lower growth rates $\sim \phi(n)\log{n}$ for the individual cyclotomic discriminants are $n\log{n}$ and $e^{-\gamma}n\frac{\log{n}}{\log{\log{n}}}$, by Mertens's theorem.)
I was wondering in what ways would the Gleason discriminants behave similarly to the cyclotomic discriminants (size-wise?), and in what ways they are markedly different. One marked difference is that the prime factors of the $\mathrm{Disc}(G_n)$ are quite unpredictable; a casual look at the first few prime factorizations of the discriminants of $G_3, G_4, G_5$ and $G_6$ reveals
$$
23 \times 2551, \quad 13 \times<PHONE_NUMBER>1639909, \quad 13^2 \times 949818439 \times<PHONE_NUMBER>068386528993226361, \quad 8291 × 9137 × 420221 ×<PHONE_NUMBER>89 × 4813<PHONE_NUMBER>32 730513 × 2<PHONE_NUMBER>11<PHONE_NUMBER>29 × 1<PHONE_NUMBER>52<PHONE_NUMBER>13<PHONE_NUMBER>14 699601
$$
The one thing that is easy and very useful to see is that these discriminants are odd: this is how Gleason established that the complex roots of the polynomials are all distinct (no multiplicities), which is non-obvious from the definition. We certainly have the trivial lower $\delta_n \gg n$ (from Minkowski) and upper $\delta_n = o(n^2)$ (from noting that the Mandelbrot set has logarithmic capacity $1$) estimates, but both of these are on purely general grounds, and they leave out a large margin.
$\sum_{d|n} \delta_d$ is the logarithm of the discriminant of this iteration, right?
@FedorPetrov: Yes, exactly. Just because it should presumably be a smoother function of $n$ (as for the cyclotomic case). By the way, the irreducibility of these polynomials $G_n$ is conjectured but not proved: they should be the irreducible factorization of the iteration.
Roughly speaking, the primes $\mathfrak p$ dividing Disc$(G_n)$ are the primes where two $c$-values merge modulo $\mathfrak p$, or more-or-less equivalently, the primes that ramify in $\mathbb Q(c)$. Since $c$ lives in a moduli space, it seems that a more natural analogy would be to look at the ramification in the fields $\mathbb Q(j(\mathfrak a))$, where $\mathfrak a$ is an ideal in a quadratic imaginary field, i.e., the fields generated by CM $j$-invariants of elliptic curves. The CM discriminants have a lot of nice structure, again in contrast to your Gleason discriminants.
|
2025-03-21T14:48:31.143107
| 2020-06-05T02:10:46 |
362223
|
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|
Stack Exchange
|
On the $\mathsf{LCM}$ of a set of integers defined by moduli of powers
For integers $a,b,t$ define $$\mathcal R_t(a,b)=\{q\in\mathbb Z\cap[1,\min(a^t,b^t)]: a^t\equiv b^t\bmod q\}$$ and $\mathsf{LCM}(\mathcal R_t(a,b))$ to be $\mathsf{LCM}$ of all entries in $\mathcal R_t(a,b)$.
Similar reasoning to On $\mathsf{LCM}$ of a set of integers gives $$\mathsf{LCM}(\mathcal R_t(a,b))\leq\mathsf{LCM}(T_t(a,b))$$ where $T_t(a,b)$ is defined as $$T_t(a,b)=\Big\{q\in\mathbb Z\cap[1,\infty]:q|(a^t-b^t)\Big\}.$$
If $a,b\in\big[\frac r2,r\big]$ hold and are coprime then what is the probability $\mathsf{LCM}(\mathcal R_t(a,b))<\beta r^{t-\alpha}$ at some $\alpha\in(0,t)$ and $\beta>0$?
Do you mean $a,b$ are chosen randomly?
Yeah but on set of coprimes but I think without coprime condition and coprime condition should be same.
We have $\mathrm{LCM}(T_t(a,b))=|a^t-b^t|$. On the other hand, the equality $\mathrm{LCM}(R_t(a,b))=\mathrm{LCM}(T_t(a,b))$ does not always hold.
$q|(a^t-b^t)\iff a^t\equiv b^t\bmod q$ and thus we can take $LCM$ of all $q$ with $q|(a^t-b^t)$ which is exactly all $q$ with $a^t\equiv b^t\bmod q$?
I see $\infty$ and $\min(a^t,b^t)$?
Corrected definition of $\mathcal R_t(a,b)$
I corrected defintion now (I did not necessitate $b+1<a<2b$).
$9^7\equiv 7^7\bmod 1979713\implies 1979713\in\mathcal R_7(9,7)$ and so $\mathsf{LCM}(\mathcal R_7(9,7))>1979713\not<2$ no (and in fact $2\in\mathcal R_7(9,7)\wedge\mathsf{GCD}(2,1979713)=1\implies\mathsf{LCM}(\mathcal R_7(9,7))=3959426=\mathsf{LCM}(T_7(9,7))$)?
Ok,,,,,,,,,,,,,,,,
With the modified definition, $T_t(a,b)=R_t(a,b)$ (if $a^t\ne b^t$) and the lcm of either is $|a^t-b^t|$, so what is the point of all these definitions?
I wrote $\min(a^t,b^t)$ before instead of $|a^t-b^t|$ in definition of $R$. I made a mistake in making equality of $R$ and $T$. With $\min(a^t,b^t)$ the probability question made sense. Corrected now. I was thinking about coprimes and did not think of scenario when $b\ll a$ held.
@VS.: Why do you need to define $T_t(a,b)$ while we simply have $\mathrm{LCM}(T_t(a,b))=|a^t-b^t|$ ?
@MaxAlekseyev Similarity of definitions contrast to difference in content.
|
2025-03-21T14:48:31.143241
| 2020-06-05T07:51:32 |
362226
|
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|
Stack Exchange
|
Intersecting subsets of $\{1,\ldots,n\}$
Is there $n\in\mathbb{N}$ and a collection ${\cal C}$ of subsets of $\{1,\ldots,n\}$ with the following properties?
$|{\cal C}| = n$,
$|c| > 1$ for all $c\in {\cal C}$,
$c\neq d \in {\cal C} \implies |c\cap d|=1$, and
$\big|\{|c|: c\in {\cal C}\}\big| > 2$.
No, there isn't. This is essentially the dual version of the De Bruijn-Erdos theorem if the elements of $\mathcal C$ are the points, and the elements from $\{1,\ldots,n\}$ are the lines. The original proof is here.
|
2025-03-21T14:48:31.143308
| 2020-06-05T08:09:15 |
362228
|
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|
Stack Exchange
|
A variant of the capset problem
Let $p > 2$ be a prime of bounded size.
Suppose $A$ is a subset of $G = \mathbf{F}_p^n$ with only degenerate solutions to
$$x + y = 2a,\\
x+z = 2b,\\
y + z = 2c,$$
where a solution is considered degenerate if any of $x,y,z$ are equal.
Is there some $\theta = \theta(p) < p$ such that $|A| \leq \theta^p$?
My question (which arose in discussions with Freddie Manners) is motivated by trying to understand the limitations of the slice rank method.
The above system has complexity 1 in the sense of additive combinatorics,
so it is "controlled by Fourier analysis", and considered not much more difficult than the 3AP system,
but it does not seem to yield to an obvious adaptation of the solution for ordinary capsets, as far as I can see.
In applications of the slice rank method, there are normally two steps: bound the slice rank above and bound it below. For this system, neither part is clear.
To bound the slice rank above, one expresses the configuration tensor $T$ in a clever way that exhibits low slice rank.
In this case, it looks natural to write
$$
\sum_{\text{solutions}} e_x \otimes \cdots \otimes e_c
= \sum_{x,y,z,a,b,c\in G} \prod_{i=1}^n (1 - (x_i+y_i-2a_i)^{p-1}) (1 - (x_i+z_i-2b_i)^{p-1}) (1 - (y_i+z_i-2c_i)^{p-1}).
$$
But as a polynomial this has total degree $3(p-1)n$, which is exactly typical for a polynomial in 6 variables,
so no obvious bound for the slice rank comes out of this.
To bound the slice rank below, one restricts $T$ to $A^6$ and uses the diagonal-like structure.
This part is not clear either, because the presence of partially degenerate solutions muddy the water.
If $A$ has only degenerate solutions then $T$ is a sum of terms like
$$ e_x \otimes e_x \otimes e_z \otimes e_x \otimes e_{(x+z)/2} \otimes e_{(x+z)/2},$$
and it's not clear why such a tensor should have large slice rank.
Eric Naslund (https://arxiv.org/pdf/1701.04475.pdf) came up with a generalisation of slice rank to 'partition rank' to deal with a similar problem where it was morally of complexity 1, but there are partially degenerate solutions. Have you tried running this problem through the more general partition rank machinery?
Oh I see, he builds the distinctness into the tensor rather than deal with the partial degeneracy latter. Interesting, will investigate.
To clarify, I think that probably solves my point 2, in some sense, but only at the cost of pushing the degree of configuration tensor up even higher, so my point 1 is as problematic as ever.
@SeanEberhard: The Partition Rank solves point 2, and reduces it to point 1 without causing degree issues. However, point 1 is a problem.
It suffices to observe that if two of your equations are trivially solved, then so is the third. Using the delta notation in my paper, consider your tensor multiplied by $$(1-\delta(x,y,a)-\delta(x,z,b)-\delta(y,z,c)).$$ The term involving $\delta(x,y,a)$ must have $b=c$, and so it is equivalent to multplying by $\delta(x,y,a)\delta(b,c)$ and the Partition Rank Method provides a bound. The problem is the $1$ term - i.e. the original tensor
|
2025-03-21T14:48:31.143901
| 2020-06-05T08:41:36 |
362230
|
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"Gerhard Paseman",
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|
Stack Exchange
|
Solving a recurrence relation involving binomial coefficients
This question originates from a graph neural network architecture (see 1) in which an edge-labelled graph $G=(V,E,\eta)$ of size $|V|=n$ with $\eta:E\to \mathbb{R}^{s_0}$ is represented as a "tensor'' $\mathbf{A}_G^{(0)}\in \mathbb{R}^{n\times n\times s_0}$, and in which for round $t>0$, a new tensor $\mathbf{A}_G^{(t)}\in\mathbb{R}^{n\times n\times s_t}$ is defined in terms of $\mathbf{A}_G^{(t-1)}\in\mathbb{R}^{n\times n\times s_{t-1}}$. The dimension $s_t$ is defined as ${n + s_{t-1}-1 \choose s_{t-1}-1}$ for $t>0$.
Question: Given $n$ and $a_0$ in $\mathbb{N}$, define $a_t:={n+a_{t-1}\choose a_{t-1}}$ for $t>0$. Is there an explicit formulation of $a_t$ in terms of $n$, $a_0$ and $t$? In other words, how to "solve" such a recurrence relation? What can one say about $a_t$ when $t$ is large?
A quick remark rather than answer is that
$$
a_t={\rm dim}_{\mathbb{C}}\{\ S^n(S^n(\cdots S^n(\mathbb{C}^{a_0})\cdots))\ \}
$$
where you iterate taking the $n$-th symmetric power $t$ times.
Thanks, this is indeed an expression for $a_t$ in terms of $n$, $a_0$ and $t$ but I was aiming for a more explicit function. Nice connection though.
Rewrite the recursion by using n rather than a_{t-1} for the bottom coefficient. Now you have an explicit formula as an iterated polynomial (as in the other answer) and a_t grows like a_0^(tn).
Gerhard "But Do Mind The Coefficients" Paseman, 2020.06.05.
Indeed, the nth root of the coefficient will come from Stirling's approximation for factorial, and so a_t will look like ((ea_0)/n)^(tn). Gerhard "Don't Need No Lower Powers" Paseman, 2020.06.05.
Thanks, I see that $a_t$ could be alternatively defined as ${n+a_{t-1} \choose n}$ but when writing this as a polynomial, the coefficients also seem to depend on $n$?
Yes. The constant coefficient is 1, and the rest depend on n, including the first which is 1/n! . Gerhard "Leaves The Rest To You" Paseman, 2020.06.05.
|
2025-03-21T14:48:31.144073
| 2020-06-05T09:03:53 |
362231
|
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|
Stack Exchange
|
propagation of a invariance along some PDE
Consider the following non linear PDE on $\mathbb{R}^n$
$$ \partial_t u_t(x) \,=\, F\big(x, u_t(x), D u_t(x)\big)$$
with given initial condition $u_0(x)$.
Assume that:
$u_0$ is rotation invariant, namely $u_0(Ox)=u_0(x)$ for all rotation matrix $O$;
if $u$ is a rotation invariant function, then $F(x,u(x),Du(x))$ is rotation invariant
Can I conclude that the solution of the PDE $u_t$ is rotation invariant for any $t\geq0$? In discrete time this would be a very intuitive result...
We need to know more about the functions involved, I think. But if $F$ is real analytic, and we demand that $u_t$ be real analytic, then uniqueness of solution follows from the Cauchy--Kovalevskaya theorem, and then invariance is clear, since any noninvariant solution is taken by a rotation to another solution with the same initial conditions.
If initial data give rise to unique solutions of your equation, then it is sufficient to prove that there exists a rotation invariant solution. Try to write $u_t(x) = v_t(r)$, where $r$ is the radius variable. Reduce the original PDE to a PDE for $v_t(r)$. If the reduced equation has a solution, it is guaranteed to be rotation invariant.
|
2025-03-21T14:48:31.144201
| 2020-05-31T20:18:45 |
361835
|
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|
Stack Exchange
|
Do Sierpiński numbers of Izotov type have a covering set?
Sierpiński number is an odd integer $k$ such that $2^nk+1$ is composite for all $n\in{\mathbb N}$. In the paper Sur un probleme concernant les nombres $k\cdot2^n+1$, zbl:0093.04602 (1960), Sierpiński proved that there are infinitely many $k$ with this property. All integers $2^nk+1$ constructed by Sierpiński are divisible by at least one of the primes in an explicit small set (a covering set).
Then, in A note on Sierpinski numbers, zbl:0849.11016 (1995), Izotov has given infinitely many Sierpiński numbers of a new type $(*)$:
for $n=4m+2$, the number $2^nk+1$ has an easy algebraic factorization;
for all other $n$, the number $2^nk+1$ is divisible by at least one of the primes $\{3,17,257,641,65537,6700417\}$ (a covering set).
(The author notes that the above covering set does not work when $n=4m+2$.)
Two decades later, in the paper Sierpiński and Carmichael numbers, zbl:1325.11010 (2015),
Banks et al. write:
Every currently known Sierpiński number $k$ possesses at least one covering set $\cal P$,
which is a finite set of prime numbers with the property that $2^nk+1$ is divisible
by some prime in $\cal P$ for every $n\in{\mathbb N}$.
So, for Sierpiński numbers of the Izotov type $(*)$, was a bigger covering set found between 1995 and 2015?
Related: Primes in a shifted geometric sequence.
Full reference: W.Banks, C.Finch, F.Luca, C.Pomerance, P.Stănică, Sierpiński and Carmichael numbers. Trans. Am. Math. Soc. 367, No. 1, 355-376 (2015).
See https://math.stackexchange.com/questions/1683082/does-every-sierpinski-number-have-a-finite-congruence-covering where the question of Sierpinski without covering was discussed (but not settled).
So, for Sierpiński numbers of the Izotov type $(*)$, was
a bigger covering set found between 1995 and 2015?
No, it was not. And it is conjectured that none exists. In the Math Stack Exchange thread linked by Gerry Myerson in a comment to the question, I give other examples of Sierpiński numbers for which it is not likely that they should possess such a covering set $\cal P$.
(I do not know why Banks et al. repeated the obsolete claim that all known Sierpiński numbers possess a covering.)
|
2025-03-21T14:48:31.144368
| 2020-05-31T20:22:44 |
361836
|
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|
Stack Exchange
|
Problem regarding existence of a divisor representing line bundle
We consider a normal irreducible variety $X$ and a line bundle $L$. The question is when $L$ is induced by a Cartier divisor $D$. We know that if $s$ is a rational section of $O_X(D)$, where $D$ is a Cartier divisor of $X$, then $(s)\sim D$, so the idea is to build a rational section $s$ of the line bundle $L$. In that case the divisor $D:=(s)$ will induce the line bundle $O_X(D)\cong L$. Thus the problem is equivalent to prove the existence of a rational section of the line bundle $L$.
In a coomology point of view, considering the short exact sequence
$0\to O_X^\times\to M^\times_X\to M^\times_X/O^\times_X\to 0$
we get the long exact sequence
$0\to H^0(X,O_X^\times)\to H^0(X,M_X^\times)\to H^0(X,M^\times_X/O_X^\times)\to^\alpha H^1(X,O_X^\times)\to H^1(X,M_X^\times)\to H^1(X,M_X^\times/O_X^\times)\to\cdots$
It is clear that $PDiv(X)=H^0(X,M_X^\times)$, $CDiv(X)=H^0(X,M_X^\times/O_X^\times)$ and $Pic(X)=H^1(X,O_X^\times) $ corresponds to the group of the line bundles on $X$. Moreover the connection morphism $\alpha$ corresponds exactly to the map associating to each Cartier divisor, its line bundle, in the usual sense. Our statement corresponds to require the morphism $\beta: H^1(X,O_X^\times)\to H^1(X,M_X^\times)$ is trivial. In fact we would have $\alpha$ surjective and so
$Pic(X)\cong CDiv(X)/PDiv(X)$
What does it mean $\beta$ trivial?
That for each line bundle $L=\{(\underline{U}, g_{ji})\}$, $\beta(L)=0$ in $H^1(X,M_X^\times)=\ker(\delta_1)/\Im(\delta_0)$, i.e. there exists some $s\in C^0(\underline{U},M_X^\times)$ such that
$g_{ji}=\delta_0(s)_{ji}=\frac{s_j}{s_i}$ in $U_i\cap U_j$
In other words $s\in C^0(\underline{U},M_X^\times)$ is the rational section that induces the line bundle $L$. Is it correct this reasoning?
In the Griffiths-Harris book, Principle of Algebraic geometry, it is proved the existence of a rational section in the case in which $X$ is an irreducible projective variety:
Let $H$ be an ample divisor of $X$. By Serre-Grothendieck theorem we know $L\otimes O_X(mH)$ is g.g. for $m$ sufficiently large, i.e there exists a global regular section $t\in H^0(X, L\otimes O_X(mH))$. But $H$ is ample, so there exists a global regular section $v$ of $O_X(H)$. Thus $\frac{t}{v^m}$ will be the rational section of $L$.
However I think there is in general (not only for a projective variety) a simple way to build it:
If $\psi_i: \pi^{-1}(U_i)\to U_i\times \mathbb{C}$ is a local trivialization of the line bundle $L$, then one can define the rational section
$s_i: U_i\to \pi^{-1}(U_i)\subseteq L$, $s_i(p):=\psi_i^{-1}(p,1)$
What am I missing? It is not possible it is so simple to build a rational section in general.
Hartshorne seems to give a counterexample Ex 1.3, on p.9 of the book Ample subvarieties of algebraic varieties, due to Kleiman. (which it seems you must log in to amazon to see) https://www.amazon.com/reader/3540051848?_encoding=UTF8&page=22
Related: S. Kleiman, https://www.e-periodica.ch/digbib/view?pid=ens-001:1979:25::101#372
|
2025-03-21T14:48:31.144578
| 2020-05-31T20:55:35 |
361839
|
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"ABIM",
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|
Stack Exchange
|
Test for OU-Process
Suppose that I'm given a sample from time-series $(x_n)_{n=1}^N$ and want to decide if it comes from an OU process or not. Is there a (rigorous) test I can use?
So far, everything I've seen is hand-wavy...
$\newcommand\th{\theta}$ $\newcommand\Si{\Sigma}$
The Ornstein--Uhlenbeck (OU) process is a Gaussian process, with the mean and covariance functions being known functions depending on $s\le4$ unknown real-valued parameters of the OU process. So, for the sample $X:=(x_1,\dots,x_n)$, we know the functions $\mu$ and $\Si$ given by $\mu(\th):=E_\th X$ and $\Si(\th):=Cov_\th X$ for all $s$-tuples $\th$ of the real-valued parameters. Then
$$Z(\th):=\Si(\th)^{-1/2}(X-\mu(\th))$$
has the standard normal distribution in $\mathbb R^n$.
Now using appropriate estimates of $\th$ and partitioning $\mathbb R^n$ into some number $k>s+1$ of disjoint Lebesgue-measurable sets of nonzero (say the same) Gaussian measure, we can use e.g. a chi-square goodness-of-fit test, with a test statistic distributed approximately as $\chi^2$ with $k-s-1$ degrees of freedom, as described e.g. by Watson.
Another goodness-of-fit test, based on a strict negative definiteness of Euclidean distance, is described in Section 3 of the paper by Székely and Rizzo.
What is the advantages of the latter method?
In Sections 5 and 6 of their paper, Székely and Rizzo provide comparisons with a few other tests.
Thanks, Iosif. This is very interesting and useful actually.
I am glad this helped.
|
2025-03-21T14:48:31.144723
| 2020-05-31T20:57:27 |
361841
|
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|
Stack Exchange
|
Solutions to the Diophantine equation $a^xy+x=c$
Fix positive integers $a,c$ with $a>2$. Is it possible that the Diophantine equation
$$a^xy+x=c$$
has infinitely many solutions (in positive integers $x$ and $y$)?
Do you have any experimental material? Mostly I find either no or one solution. For $a=3$ there are at least two solutions when $c=85, 166, 247, 328, 409, 490, 571$; for $a=4$ there are at least two solutions with $c=1029$. Do you know any other cases? In particular, do you know any cases when there are more than two solutions?
I haven't numerically investigated the equation. I assumed that this was easy or intractable, and I am not an expert in these things. But I appreciate your effort and enthusiasm!
I don't know about the number of solutions. But that number is definitely always finite. x must be at most c since $a^x y$ is positive, and $y$ must be at most $c$ by similar logic. So unless I'm missing something here, for fixed c, there are at most $c^2$ solutions even if a is allowed to vary over positive integers (not even $a>2$). It isn't hard to improve this to $\frac{c^2}{2}$ given your restrictions.
The smallest $c$ such that the Diophantine equation $3^xy+x=c$ has three positive solutions is $c=38180350917190281105854137945200663220016794$. The solutions correspond to $x=1, 4, 85$.
As was observed by @JoshuaZ, for given $a, c\in\mathbb{N}$, the equation $(*)\; a^xy+x=c$ has only finitely many solutions. On the other hand, I show that for any $a$ and any $N\in\mathbb{N}$ one can find $c=c(a, N)$ such that the equation $(*)$ has at least $N$ solutions. To see this, let us observe that $(*)$ has a solution if and only if
$$
c\equiv x\pmod{a^x}.
$$
Thus, to show that $(*)$ it is enough to find posittive integers $x_{1}<x_{2}<\ldots<x_{N}$ such that the system of congruences
$$
(**)\quad c\equiv x_{1}\pmod{a^{x_{1}}},\; c\equiv x_{2}\pmod{a^{x_{2}}},\;\ldots, \;c\equiv x_{N}\pmod{a^{x_{N}}}
$$
has a solution.
Applying general form of chinese remainder theorem we need to find $x_{1}, x\ldots, x_{N}$ such that for each $i, j\in\{1,\ldots, N\}, i\neq j$ we have $\gcd(a^{x_{i}},a^{x_{j}})=a^{x_{i}}|x_{j}-x_{i}$. To construct suitable sequence it is is enough to define
$$
x_{1}=1, x_{n}=a^{x_{n-1}}+x_{n-1}
$$
and observe (using simple induction on $k$) that
$$
x_{n+k}=\sum_{i=0}^{k-1}a^{x_{n+i}}+x_{n}.
$$
This clearly implies that $a^{x_{i}}|x_{j}-x_{i}$ for $j>i$. Computing now the solution $c=c(a, N)$ of the system $(**)$ we get that there are positive integers $y_{1}, \ldots, y_{N}$ such that the equation $(*)$ has solutions $(x_{1}, y_{1}), \ldots, (x_{N}, y_{N})$.
|
2025-03-21T14:48:31.144923
| 2020-05-31T21:39:31 |
361845
|
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|
Stack Exchange
|
Embedding of spheres which satisfies intersection rules
Let $S = \{S_1, \dots ,S_n\}$ be a finite set of $d$-dimensional spheres with the same radius, and let $E$ be a combination of intersections between them, where an intersection is a rule of the form $S_i \cap S_j \subset S_k$. Given any number of spheres and any combination of intersections, is it always possible to find a configuration of spheres embedded in $\mathbb R^d$ which satisfies all and only the intersections in $E$? Hence, this configuration must not contain any intersection that is not present in $E$.
Side question:
If the answer is negative, but dependent on the dimension, is the following true: Given any number of spheres and any combination of intersections, there exist a finite dimension d such that is it always possible to find a configuration of d-dimensional spheres embedded in $\mathbb R^d$ which satisfies all and only the intersections in $E$?
The problem is similar to that of specifying for each $i,j$ whether or not $S_i$ is inside $S_j$, where now the spheres need not have the same radius. Felsner, Trotter, and Fishburn showed that some specifications could not be achieved in any dimension. See http://page.math.tu-berlin.de/~felsner/Paper/sphere.pdf.
If I am not misunderstanding,
it could be that this $\mathbb{R}^2$ example shows that $S_5$
cannot simultaneously satisfy these relationships:
\begin{eqnarray}
S_1 \cap S_2 & \subset & S_5 \\
S_3 \cap S_4 & \subset & S_5 \\
S_2 \cap S_3 & \not\subset & S_5 \\
S_1 \cap S_4 & \not\subset & S_5 \\
\end{eqnarray}
Here four circles have same radii, but in general the radii may differ.
This is related to OEIS A250001.
I recall that Jonathan Wild proved the impossibility of the above configuration,
but I have no reference.
This example may not resolve your question, because
I am uncertain if Wild's conditions on drawing circles
are exactly
equivalent to your intersection relations.
It actually looks relevant, thanks!
Are you aware of any multi-dimensional extension of this?
@Alfred: I am unaware of extensions to $\mathbb{R}^d$ for $d \ge 3$. It is already quite difficult for $\mathbb{R}^2$: $16951$ ways to draw $5$ circles!
|
2025-03-21T14:48:31.145090
| 2020-05-31T22:12:51 |
361849
|
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|
Stack Exchange
|
Reference request: sufficiently smooth functions on the plane belong to the projective tensor square of $L^2$ of the line
Let $\newcommand{\ptp}{\widehat{\otimes}}\ptp$ denote the projective tensor product of Banach spaces. Some back of the envelope calcuations, using the Fourier transform and Plancherel/Parseval, suggest to me that the following should be true:
$\newcommand{\Real}{{\bf R}}$
for some $k > 1$, every compactly supported $C^k$-function $\Real^2\to\Real$ belongs to $L^2(\Real)\ptp L^2(\Real)$.
(Recall that in contrast, there are continuous functions on $[0,1]^2$ that do not belong to $C[0,1]\ptp C[0,1]$.)
If the claim above is true, I would like to know if there are standard references, perhaps from the world of integral kernel operators or Sobolev spaces, which I could cite, rather than reinventing the wheel (and probably getting suboptimal values of $k$).
In a slightly different direction, I would also be interested to know of references which prove analogous resuts for $C^k$ functions (suitably interpreted) on compact connected Lie groups.
I do not know of a good reference, but the Fourier transform argument works very well for every $k>1$: by the compact support assumption, you can as well periodize the problem and work with a $C^k$ function $(\mathbf{R}/{\mathbf{Z}})^2 \to \mathbf{R}$. Then the Fourier expansion gives that $|f|{L^2 \hat \otimes L^2} \leq \sum{i,j} |\hat f(i,j)| \leq C(\varepsilon) |f|{H^{1+\varepsilon}}$ for every $\varepsilon >0$ (here $H^s$ is the usual Sobolev of functions such that $((1+|i|+|j|)^s\hat f(i,j)){i,j}$ belongs to $\ell^2$).
For your second question, the fact that you have a Lie group is not relevant: in any compact manifold (or any compactly supported function on a manifold), working in local coordinates with a partition of unity reduces the problem to $(\mathbf{R}/\mathbf{Z})^d$, in which case the above Fourier transform argument shows that $C^k$ functions work whenever $k>d/2$.
Hi @MikaeldelaSalle - yes, your first comment was more or less the argument I had in mind, but it is good to get confirmation (I was taking $k\geq2$ just to play safe but as you observe this is unnecessary). I had overlooked your second point that the group structure is unnecessary and that one can localize to $L^2$ of a $d$-cube (hence a $d$-torus)
Another way to see this (for $k=1$) is by rephrasing your question in terms of integral operators; what you are asking is whether an integral operator on $L_2$ with a $C^1$-kernel is nuclear. By compactness of the support one can assume that $k\in C^1([0,1]^2)$ vanishes at the boundary; let $T_k$ be the associated integral operator. Let $k_2(s,t) = \frac{\partial}{\partial t} k(s,t)$ and $V(f)(t)=\int_0^t f(s),ds$. Then $T_k = T_{k_2}V$ by integration by parts; hence $T_k$ factors as a product of two Hilbert-Schmidt operators and is therefore nuclear.
@DirkWerner thanks! I like this argument, since in one version of the intended application, integral operators occur quite naturally, and this factorization trick might be useful for independent reasons
@DirkWerner : thanks, I also like this argument.
|
2025-03-21T14:48:31.145334
| 2020-05-31T22:47:31 |
361851
|
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|
Stack Exchange
|
Upper bounds for difference of entries between matrices and their inverses in $\mathsf{GL}_k(\mathbb Z)$
Let $a(M)$ be the maximum absolute value of entries of matrix $M\in\mathsf{GL}_k(\mathbb Z)$.
$M^{-1}\in\mathsf{GL}_k(\mathbb Z)$ holds.
What is a good upper bound for $|a(M)-a(M^{-1})|$?
I am thinking whether the dependence could be a little smaller than fully exponential in $k$ for $a(M)\cdot a(M^{-1})$ which will reflect upper bound for $|a(M)-a(M^{-1})|$.
A good upper bound for $|a(M)-a(M^{-1})|$ is $|a(M)-a(M^{-1})|$ itself :) you probably want an upper bound not depending too much on $M$, for instance, $a_{k,n}=\sup_{|a(M)|\le n}|a(M)-a(M^{-1})|$: estimate $a_{k,n}$? (for $k$ fixed)
I think $a(M)\cdot a(M^{-1})\leq(\max(a(M),a(M)^{-1}))^k$ might be possible and can that $k$ be $o(k)$ or just even $k-\alpha$ with $\alpha\gg1$ and not just $k$?
A trivial upper bound is $a(M^{-1})\le (k-1)!a(M)^{k-1}$. For a Jordan matrix $M_n=I+nJ$ in size $n$ one has $a(M_n)=n$ and $a(M_n^{-1})=n^{k-1}$, so in this case $a(M_n)a(M_n^{-1})=n^k$.
[You wrote $a(M)a(M^{-1}\le \max(a(M),a(M^{-1}))^k$, but this is trivial with $k=2$: $ab\le \max(a,b)^2$... this might be a typo]
For $k=2$, the upper bound is zero.
For $k>2$, there is no upper bound. E.g., let $$M=\pmatrix{1&1&1\cr9&10&11\cr-n&n&3n-1\cr}$$ Then $$M^{-1}=\pmatrix{10-19n&2n-1&-1\cr38n-9&1-4n&2\cr-19n&2n&-1\cr}$$ and $a(M)-a(M^{-1})=(3n-1)-(38n-9)=-(35n-8)$ is unbounded.
actually the matrix $M_n=\begin{pmatrix}1 & n & 0\ 0 & 1 & n\0 & 0 & 1\end{pmatrix}$ with $a(M_n)=n$ even has $a(M_n^{-1})=n^2$, since $M_n^{-1}=\begin{pmatrix}1 & -n & n^2\ 0 & 1 & -n\0 & 0 & 1\end{pmatrix}$.
So the gap is $(a(m))^{k-1}$?
If the elements of $M$ are bounded in absolute value by $b$, then the elements of $M^{-1}$ can't be any bigger than $c_kb^{k-1}$ for some constant $c_k$ depending only on $k$.
|
2025-03-21T14:48:31.145474
| 2020-05-31T22:48:57 |
361852
|
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|
Stack Exchange
|
(Dis)continuity of periodic functions with non-summable Fourier series
Let $f : [0,2 \pi)^d \rightarrow \mathbb{R}$ be a square-integrable periodic function in $L^2( [0,2 \pi)^d )$ with $d \geq 1$.
We assume moreover that the square-summable Fourier coefficients of $f$, denoted by $c_n(f)$ for $n \in \mathbb{Z}^d$, are such that:
$c_n(f)$ is real, strictly positive, with $c_{n}(f) = c_m(f)$ for any $n,m \in \mathbb{Z}^d$ with equal Euclidian norms $\lVert n \rVert = \lVert m \rVert$,
the sum $\sum_{n \in\mathbb{Z}^d} c_n(f) = \infty$, and
the $c_n(f)$ are decreasing in the sense that $c_n(f) \leq c_m(f)$ as soon as $\lVert n \rVert \geq \lVert m \rVert$.
Can we prove that $f$ is necessarily discontinuous and even unbounded at $0$ with these assumptions?
Context: I am especially interested by the situation where $c_n(f)$ behaves asymptotically like (or even is equal to for $n \neq 0$) $\lVert n \rVert^{- \alpha}$ for some $\alpha \in (d/2, d]$.
The case $d = 1$ with $2\pi$ periodic functions is already interesting and may be quite simpler.
At least in the context situation, this sounds right, since at least in the continuous case $|t|^{-\alpha}$ has FT $c|x|^{\alpha -d}$, and the perturbation has a smooth FT, so won't affect the behavior at $x=0$.
I am not sure to understand: what is the perturbation you are talking about? Do you know if there is anything like the result $\mathcal{F} { \lvert t \rvert^{-\alpha} }(x) = c \lVert x \rvert^{\alpha - d}$ for the periodic setting? (I realized there was a mistake in the context for the parameter $\alpha$, I corrected it.)
The perturbation in my comment refers to $d_n$, if you write $c_n=|n|^{-\alpha}+d_n$.
I see, the comparison with the continuous case is effectively interesting.
|
2025-03-21T14:48:31.145614
| 2020-06-01T01:00:38 |
361856
|
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|
Stack Exchange
|
Analytic and algebraic definitions of intersection multiplicity of two complex algebraic curves coincide
There are two definitions of intersection multiplicity of two complex algebraic plane curves. One of them is given in An introduction to algebraic curves by Griffiths. Let $t \mapsto (t^k, y(t) )$ be the local normalization of $V_f$ then $(V_g \cdot V_f)_P = ord_P(g(t^k, y(t)))$. The other one is given in Algebraic Geometry by Hartshorne. $(V_g \cdot V_f)_P = \mathrm{length}(O_p/(f,g))$ as $O_P$ module.
Do these definitions coincide?
There may be some approaches. Showing that completion preserves length may help cause you can thereby catch the normalization in complex analysis to algebraic one. Showing that $\mathrm{length}(O_P/(g, f_1 f_2))= \mathrm{length}(O_P/(g, f_1))+\mathrm{length}(O_P/(g, f_2))$ is also important.
If your algebraic surface is smooth, a good (conceptual) way to compare the two is by comparing both to the cup product in cohomology.
A proof (for the $n$-dimensional version) is given in Chapter 4 of https://arxiv.org/abs/1806.05346. Sorry for the self-promotion, but one reason for including this element in the (draft of the) book was that I could not locate a direct proof in the literature.
|
2025-03-21T14:48:31.145732
| 2020-06-01T02:00:43 |
361860
|
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|
Stack Exchange
|
How many trees have $n$ nodes with fewer than three neighbors?
We want to know how many trees have $n$ nodes with fewer than three neighbors. For $n=1$, the only possibility is a single node. For $n=2$, the only possibility is two connected nodes. For $n=3$, the only possibilities are a node adjacent to three nodes or a node adjacent to two nodes. A set of drawings produces the following (possibly incorrect) sequence: $1$, $1$, $2$, $4$, $9$, $25$, $70$. My question is whether this enumeration is a new problem or one that has been dealt with. See OEIS A335342.
https://oeis.org/search?q=1%2C+1%2C+2%2C+4%2C+9%2C+19 ?
Questions should be self-contained. While a descriptive title is good, you should not remove the question from the body of the post.
Are you thinking of isomorphism classes? If instead we count labeled trees on $N$ vertices, the number of such trees with given degree sequence $d_1,\dots, d_N$ is the multinomial coefficient $\binom{N-2}{d_1-1,\dots, d_N-1}$, from which it ought to be possible to derive a reasonably closed expression (summing over the possible values for $N$).
These are not labeled trees.
It might be easier to first try to figure out the number of trees having $n_1$ nodes with one neighbor and $n_2$ nodes with two neighbors.
The revised sequence is not in OEIS (https://oeis.org/search?q=1%2C+1%2C+2%2C+4%2C+9%2C+25%2C+70 ), so, if it's correct, you should submit it.
What kind of answer would you like? An asymptotic formula?
I need to check it for correctness. I do a lot of OEIS submissions.
Re answer: I think I can work out a method along the lines of free trees, but I don't want to do the work if someone already has.
1, 1, 2, 4, 9, 25, 70 are correct and it continues 226, 753, 2675, 9785, 37087, 143487, 566952 (the last one is for 14). For parameter 14, the number of vertices is 14-26, with counts like this: 1, 127, 2091, 13744, 48236, 102116, 140588, 130263, 82616, 35419, 9961, 1655, 135.
2274967 for parameter 15. Getting up to about 20 is plausible.
I have just submitted A335342 to the OEIS. It includes a Mathematica program and a B-file for n<=60 (about 7 minutes on my 2010 iMac). The results match those in the latest comments. Thanks for all the help; I guess we are done here.
Here are some observations which may inform a partial enumeration.
Such a tree has $t+n$ vertices, where the $l$ leaves are part of the $n$ nodes and the rest of the $n$ nodes look like they are interior vertices that are part of a path. The $t$ vertices each have three or more branches, so we must have $3t \lt n+t+t-1$, where the $t-1$ is a correction for over counting edges between two branch vertices. This means $t \leq n-2$.
Now if $t$ is $0$, the only tree is a path, while if $t=1$ there is a bijection between these trees and partitions of $n$ into at least three parts. For larger $t$, we need to decide on how to arrange and color the edges between the $t$ nodes. For $t=2$ we need to decide how many of the n nodes go between, and then make sure there are at least four leftover to distribute among the remaining branches. For $t=3$, there are two internal edges to color with a total of $0$ up to $n-5$ nodes, and then distribute the remainder on the leafy portions, being sure to save at least one for the middle $t$ node.
We now see that this becomes a problem of counting branchy trees on $t$ nodes, and then partitioning $n$ nodes on the edges, and then worrying about overcounting. At this point I leave the problem to the professionals.
Gerhard "Neither Professional Accountant Nor Gardener" Paseman, 2020.06.04.
|
2025-03-21T14:48:31.146177
| 2020-06-01T03:12:08 |
361862
|
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|
Stack Exchange
|
What are some examples of theorem requiring highly subtle hypothesis?
I would like you to expose and explain briefly some examples of theorems having some hypothesis that are (as far as we know) actually necessary in their proofs but whose uses in the arguments are extremely subtle and difficult to note at a first sight. I am looking for hypothesis or conditions that appear to be almost absent from the proof but which are actually hidden behind some really abstract or technical argument. It would be even more interesting if this unnoticed hypothesis was not noted at first but later it had to be added in another paper or publication not because the proof of the theorem were wrong but because the author did not notice that this or that condition was actually playing a role behind the scene and needed to be added. And, finally, an extra point if this hidden hypothesis led to some important development or advance in the area around the theorem in the sense that it opened new questions or new paths of research. This question might be related with this other but notice that it is not the same as I am speaking about subtleties in proof that were not exactly incorrect but incomplete in the sense of not mentioning that some object or result had to be use maybe in a highly tangential way.
In order to put some order in the possible answers and make this post useful for other people I would like you to give references and to at least explain the subtleties that helps the hypothesis to hide at a first sight, expose how they relate with the actual proof or method of proving, and tell the main steps that were made by the community until this hidden condition was found, i.e., you can write in fact a short history about the evolution of our understanding of the subtleties and nuances surrounding the result you want to mention.
A very well known and classic example of this phenomenon is the full theory of classical greek geometry that although correctly developed in the famous work of Euclides was later found to be incompletely axiomatized as there were some axioms that Euclides uses but he did not mention as such mainly because these manipulations are so highly intuitive that were not easy to recognize that they were being used in an argument. Happily, a better understanding of these axioms and their internal respective logical relations through a period of long study and research lasting for millennia led to the realization that these axioms were not explicitly mention but necessary and to the development of new kinds of geometry and different geometrical worlds.
Maybe this one is (because of being the most classic and expanded through so many centuries and pages of research) the most known, important and famous example of the phenomena I am looking for. However, I am also interested in other small and more humble examples of this phenomena appearing and happening in some more recent papers, theorems, lemmas and results in general.
Note: I vote for doing this community wiki as it seems that this is the best way of dealing with this kind of questions.
I have the sense that the spectral theorem for unbounded operators is perhaps a good example, with the "subtle hypothesis" being self-adjointness (not merely symmetric). But I don't know the history well enough to write a proper answer.
Probably a lot of the basic results in analysis require somewhat "subtle" hypotheses, as evinced by the famous (mythical?- see https://mathoverflow.net/questions/176693/did-cauchy-think-that-uniform-and-pointwise-convergence-were-equivalent) story of how Cauchy got tripped up by pointwise vs. uniform convergence
Just as an observation, this is essentially asking for a particular kind of badly written mathematics, which one would hope a referee picks up. I've seen examples of this; maybe the worst being a statement 'proved by induction' followed by a calculation. In this example the n=-1 and 0 cases of the induction were trivial, and the n\ge2 cases were proved by induction with the given calculation. And the calculation did not work in the n=1 case; here one had to notice that this was a hypothesis of the lemma. The final version of this paper made this point explicit.
The Axiom of Choice was (unconsciously) used in tons of proofs before it was realized to be an axiom.
This question is about results which were accepted but incorrect, so I think many examples there apply here as well.
I won't make this an answer since I don't know much about the topic myself, but I suppose the topic of constructive mathematics could be of relevance here - seeing how the use of the law of excluded middle impacts mathematics.
When first learning about universal covers and Galois correspondence of topological spaces, I remember feeling that conditions like "path-connected and semi-locally simply connected" were rather subtle.
The convergence conditions for the Fourier series of a function $f:S^1 \to \mathbb{R}$ are a good example. The investigation of convergence conditions for Fourier series was a major motivation for Cantor's set theory and Lebesgue's measure theory. Depending on what kind of convergence you want, the conditions can be very subtle. For example, if you want the Fourier series of a continuous function to converge pointwise everywhere, then I don't think that there is any nice set of necessary and sufficient conditions known. Various sufficient conditions are known, e.g., the Dirichlet conditions, which are fairly subtle.
Nowadays, I think it is generally considered that asking for convergence everywhere is the "wrong question"; one should ask for convergence almost everywhere. Then the most famous theorem is Carleson's theorem that the Fourier series of a function in $L^2$ converges almost everywhere. The hypothesis here is easy to state, but the way that the hypothesis is used is subtle. There are various proofs known now but none of them is easy. Note for example that Kolmogorov's first paper gave an example of a function in $L^1$ whose Fourier series diverges almost everywhere.
Do you have a reference for this investigation being a motivation for Cantor's set theory?
@Wojowu : It's sort of clear if you just look at Cantor's papers, where he started off studying Fourier series, and naturally segued into set theory. But if you want a historical account that spells this out, see for example this and this. That's what I found after a quick search; there may be better references.
Interesting, I had no idea this is what sparked Cantor's work!
@Wojowu: See The trigonometric background to Georg Cantor's theory of sets by Joseph Warren Dauben. Among other things, Cantor proved the uniqueness theorem for trigonometric series, which says that if two trigonometric series agree for all $x \in [-\pi,,\pi],$ then their corresponding coefficients are equal (i.e. the two trigonometric series are the same). (continued)
Cantor then managed to prove this in which a finite exceptional set is allowed (i.e. change "for all" to "for all but possibly finitely many"). After this, Cantor managed to allow certain types of infinite exceptional sets, first to sets each of whose set of limit points is finite (I'm not sure if this was separately published), and eventually to sets each of which after a finite application of the operation of taking limit points leaves a finite set (equivalently, leaves an empty set, by taking one more limit point operation). From this it's a small step to ordinal numbers (I'm out of space).
@DaveLRenfro Thank you for this summary! I have since yesterday read a bit about the topic (I've liked this article). It is quite an interesting story.
There is Euler's formula $$V - E + F = 2.$$ Today, we might not think of the hypotheses as being especially tricky. But Lakatos's classic Proofs and Refutations makes an entertaining case for its subtlety.
If Lakatos does not convince you, consider Euler's Theorem for Tilings. Suppose we have a tiling of the plane; take a finite portion of it, apply the standard Euler formula, and divide by $F$. Intuitively, as we take larger and larger portions, $V/F$ and $E/F$ approach limiting values $v$ and $e$ respectively, and we obtain Euler's Theorem for Tilings:
$$v - e + 1 = 0.$$
However, even if the limits $v$ and $e$ exist, they do not necessarily satisfy Euler's Theorem for Tilings unless the tiling satisfies certain subtle hypotheses. For example, in the heptagonal tiling below (taken from Grünbaum and Shephard's book Tilings and Patterns), the heptagons get skinnier and skinnier as one moves out from the center, creating a "singularity at infinity." It is not hard to see that $v=7/3$ and $e=7/2$, so $v-e+1 = -1/6$ and not zero.
In the notes to Chapter 3, Grünbaum and Shephard write:
Euler's Theorem for Tilings and its various corollaries are often quoted and used—usually without any indication of restrictions that must be imposed on a tiling to give meaning and validity to this procedure. In contrast to many other cases—in which a cavalier attitude towards mathematical rigor is an aesthetic shortcoming that does not affect the outcome—here many authors have claimed to have proved statements that are actually false. As recent examples we may mention Walsh (Geometriae Dedicata 1 (1971), 117–124) and Loeb (Space Structures: Their Harmony and Counterpoint, especially Chapter 9).
I tried a [5,3] version with pentagonal faces and order-3 vertices, to force $v-e+1>0$. I start with the central pentagon and can generate two surrounding layers of five pentagons each ... but then I cannot generate any more edges while keeping the [5,3] connectivity. It's effectively the opposite singularity from the one described here.
This example has been mentioned elsewhere on MO but seems worth reproducing here. The abstract of Amnon Neeman's paper A counterexample to a 1961 “theorem” in homological algebra says:
In 1961, Jan-Erik Roos published a “theorem”, which says that in an [AB4∗] abelian category, lim1 vanishes on Mittag–Leffler sequences. … This is a “theorem” that many people since have known and used. In this article, we outline a counterexample. We construct some strange abelian categories, which are perhaps of some independent interest.
It turns out that the theorem can be repaired by adding some relatively weak hypotheses that are usually satisfied in practice. That the need for such hypotheses apparently went unnoticed for so long is perhaps evidence that they are "highly subtle."
Interesting! Does someone know if the hypotheses needed for this result to be true were actually used unnoticeably in the 'proof' of Jan-Erik Roos?
I asked Prof. Neeman, and he responded, "Yes. As long as one can guarantee that the map from the inverse limit to the $n,$th term in the sequence has the same
image as the map from the $N,$th term (for all $N\gg n$), then Roos' argument works. So the counterexample amounts to
constructing an abelian category, and inside it an inverse sequence of epimorphisms whose inverse limit vanishes.
It's counter-intuitive, but turns out to exist."
This is one which I've seen trip up a number of students when first learning the material: the hypothesis of admissibility (or acceptability - I learned the latter, but the former seems more common) in the context of numberings of unary partial computable functions (or equivalent objects like c.e. sets).
Results like Rice's Theorem and the Recursion Theorem are generally presented for a specific numbering whose details are quickly forgotten; the motto "all reasonable numberings work the same" is introduced somewhere around this point, and is mostly true. However, the right notion of "reasonability" is not usually obvious, since presentations tend to focus on the following two features of the canonical numbering $\Phi:=(\varphi_e)_{e\in\mathbb{N}}$:
The numbering construed as a partial binary function $\langle e,x\rangle\mapsto\varphi_e(x)$ should itself be computable.
For every unary partial computable $f$ there should be some $e$ with $f\simeq \varphi_e$.
By themselves these properties are not enough to get the standard results to apply: the usual extreme counterexample is a Friedberg numbering, which is a numbering satisfying the two properties above such that every partial computable $f$ has exactly one index (so Rice's Theorem and the Recursion Theorem each fail basically trivially).
Instead, we need to strengthen the second bulletpoint above as follows:
(Admissibility/acceptability): For every binary partial computable $f$ there is some total computable unary $g$ such that for each $e$ we have $$f(e,-)\simeq \varphi_{g(e)}.$$
This amounts to a kind of "universality" of the numbering in question; roughly speaking, every other numbering needs to be translatable into it. This turns out to be exactly what we need to deduce all the basic results about the usual numbering, and indeed so far as I'm aware there really are no essential differences between admissible numberings. Moreover, once this sort of universality occurs to us as something important we're led to consider general comparisons between numberings of various systems, and this leads to several interesting topics (see especially Rogers semilattices).
I was unfamiliar with most of this — this is a wonderful example, thank you!
In fact, I would expect most uses of the term "admissibility" in math describe conditions that are at least a little bit subtle.
This reminds me that in computational complexity theory, hierarchy theorems need a hypothesis that the functions in question be constructible or proper. Otherwise weird things can happen, e.g., the gap theorem.
Theorem. Assuming the axiom of choice, the countable union of countable sets is countable.
Proof. Let $\{A_n\mid n\in\Bbb N\}$ be a family of countable sets, and so we can write $A_n$ as $\{a_{n,m}\mid m\in\Bbb N\}$.
Let $A$ be the union, and define $f(a) = 2^n3^m$ such that $n$ is the least such that $a\in A_n$, and $a=a_{n,m}$. Easily, this is an injection so the union is countable.
The trained eye, of course will notice the use of the axiom of choice immediately. We choose an enumeration of each $A_n$. But this is very subtle and usually people will not notice that at first.
And of course, this use of choice is necessary. Indeed, it is consistent that the real numbers are a countable union of countable sets! (Still uncountable, though.)
"Usually people will not notice that": As far as I know, "the countable union of countable sets is countable" was proved by Cantor, and his proof was accepted by the community, long before Zermelo pointed out the axiom of choice.
Well, hardly gets any subtler than that.
A related example: The result that $\varepsilon$-$\delta$-continuity implies sequential continuity in metric spaces has some use of choice.
@QiZhu: Indeed. Most indeedity.
This is not a perfect example because the subtle hypotheses in question were not "unnoticed"; nevertheless I think it fulfills several of your other criteria. Let us define the "Strong Fubini theorem" to be the following statement:
If $f:\mathbb{R}^2 \to \mathbb{R}$ is nonnegative and the iterated integrals $\iint f\,dx\,dy$ and $\iint f\,dy\,dx$ exist, then they are equal.
The Strong Fubini theorem looks innocent enough, but without any measurability hypotheses, it is independent of ZFC. For example, Sierpinski showed that Strong Fubini is false if the continuum hypothesis holds.
In the other direction, a paper by Joe Shipman investigates a variety of interesting hypotheses that imply Strong Fubini, e.g., RVM ("the continuum is real-valued measurable"), which is equiconsistent with the existence of a measurable cardinal.
Here's another one: Let $\kappa$ denote the minimum cardinality of a nonmeasurable set, and let $\lambda$ denote the cardinality of the smallest union of measure-zero sets which covers $\mathbb{R}$. Then the assertion that $\kappa < \lambda$ implies Strong Fubini.
Title of the paper: Shipman - Cardinal conditions for strong Fubini theorems.
Some of Euclid's theorems rely on axioms of betweenness that he was not aware of.
Hilbert's axioms: https://www.math.ust.hk/~mabfchen/Math4221/Hilbert%20Axioms.pdf
This is already mentioned in the question itself.
Gödel's ontological proof requires a subtle assumption that if $\varphi$ is an essential property of $x$ then $x$ must possess $\varphi$. The first time that Gödel showed his proof to anyone was in 1970, when he provided Dana Scott with a terse writeup. In this writeup, Gödel's definition of the essence of $x$ omitted the conjunct $\varphi(x)$, thereby omitting the subtle assumption.
Many years later, Christoph Benzmüller and Bruno Woltzenlogel Paleo discovered, using a computer proof assistant, that omitting the subtle assumption not only creates a gap in the proof, but leads to a contradiction. See their 2016 paper, The Inconsistency in Gödel’s Ontological Argument: A Success Story for AI in Metaphysics.
To be fair to Gödel, it seems likely that he knew the assumption was needed, since older private notes of his include the assumption. Also, when Dana Scott presented the proof to others, he included the assumption without indicating that he was doing anything more than reporting what Gödel had told him. (For further information, see page 392 of Volume III of Gödel's Collected Works.) Nevertheless, it was subtle enough that Gödel did omit it on at least once crucial occasion (even if by accident), and the fact that its omission leads to a contradiction seems to have gone unnoticed for over 40 years.
How about the classic equality $0.999...=1$?
Proofs of this fact either assume the series defined by this infinite decimal representation converges, or initially prove convergence using the fact that the common ratio of the geometric series has norm less than unity. But the convergence works only because we use a number system where that norm is in fact less than unity. To illustrate how this cannot be taken for granted, render the decimal expansion as the series
$(9/10)+(9/100)+(9/1000)+...$
and render each term into $3$-adics. Since the initial term $9/10$ and the common ratio $1/10$ can each be expressed as a $3$-adic integer, so can every term in the series. But they all end with $...00$ and thus the sum can never converge to $...01$. Nor can the sum converge to anything else since the $3^2$ place is $1$ in every term. The latter gives a hint of where the breakdown occurs: the terms of the geometric series do not tend to zero because in $3$-adics the common ratio actually gives $|1/10|=1$, not $|1/10|<1$.
|
2025-03-21T14:48:31.147523
| 2020-06-01T03:31:51 |
361863
|
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|
Stack Exchange
|
The normalizer of block diagonal matrices
Let $G=\mathrm U_n$ or $\mathrm{GL}_n(\mathbf C)$ and $H$ the subgroup of block diagonal matrices respecting a partition $n=n_1+\dots+n_k$. Is the normalizer $N=N_G(H)$ computed anywhere in the literature?
I guess, but haven’t proved, that it is generated by $H$ and the permutations (“transpositions”) exchanging the partition’s same-length segments ($\smash{n_i=n_j}$, if any). I also suspect this may be discussed in these papers, to which I don’t have access:
Koĭbaev, V. A., On subgroups of the full linear groups containing a group of elementary block diagonal matrices. ZBL0521.20027. (1982, translation 1983; other translation?)
Borevich, Z. I.; Vavilov, N. A., Ordering of subgroups, containing a group of block-diagonal matrices, in the general linear group over a ring. ZBL0512.20031. (1982; translation?)
Vavilov, N. A., Subgroups of the general linear group over a semi-local ring containing the group of block-diagonal matrices. ZBL0509.20035. (1983; ever translated?)
Your guess is correct. Because there are clearly elements in the normaliser arbitrarily permuting same-sized blocks, if you've got an element of the normaliser, you may assume that it fixes each block. But then it must induce an inner isomorphism on each block, and so you can further assume it fixes each block pointwise. But this forces it to be block scalar.
@LSpice Fantastic, thank you. I am still mildly interested in whether this is printed anywhere, but meanwhile will gladly accept your comment as an answer.
By request, from my comment: Your guess is correct. Because there are clearly elements in the normaliser arbitrarily permuting same-sized blocks, if you've got an element of the normaliser, you may assume that it fixes each block. But then it must induce an inner isomorphism on each block, and so you can further assume it fixes each block pointwise. But this forces it to be block scalar.
I'm not sure where this would be written down, but someone must have worked it out as part of an example in some notes. If you're willing to combine some theorems, then it may help to think of your block diagonal matrices are examples of Levi subgroups. Let $G$ be an appropriate general linear or unitary group, $M$ a block diagonal subgroup of the sort you describe, and $A$ the subgroup of diagonal matrices. If $g$ belongs to $\operatorname N_G(M)$, then $g A g^{-1}$ is (the group of rational points of) a split maximal torus in $M$, hence is conjugate in $M$ to $A$ (that's one of the theorems; p. 387 of Murnaghan); so, modulo $M$, it suffices to ask which elements of $\operatorname N_G(A)$ preserve $M$. This is an easy computation, since $\operatorname N_G(A)/A$ is a symmetric group (that's another theorem; Example 8.1 of Murnaghan). If this is an argument that it would be satisfactory to cite, I could easily dig up some references; here I've indicated where you can look in the notes of Murnaghan on linear algebraic groups in CMP 4 (MSN), which just state the results but were the first place to look that occurred to me.
Thanks again! Ironically the argument for $\operatorname N_G(M)\subset\operatorname N_G(A)M$ is one I forgot essentially making here. AFAICT the “easy computation” to then exclude Weyl group elements that don’t permute same-size segments still requires unpleasant though manageable bookkeeping.
I now see that Corollary 12.11 (= Exercise 20.4) of Malle-Testerman (2011) also has that argument.
|
2025-03-21T14:48:31.147787
| 2020-06-01T04:30:08 |
361866
|
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|
Stack Exchange
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Why do people study Weyl asymptotics and partial-spectral-projections?
The major focus of the research that my advisor has me doing centers around the idea of asymptotic behavior of partial-spectral-projections on compact manifolds. In a few sentences, here is the context for the research:
$(M,g)$ is a compact Riemannian manifold without boundary, and $-\Delta_g$ is the (positive) Laplace-Beltrami operator of the metric $g$.
The operator $\sqrt{-\Delta_g}$ is defined in the usual way its collection of $L^2$-normalized eigenfunctions is denoted by $\{e_j(x)\}_{j=0}^{\infty}$, with eigenvalues $0 = \lambda_0 < \lambda_1 \leq \lambda_2 \leq \cdots \to \infty$.
For a fixed $\lambda > 0$, we then define the partial-projection operator $$ \sum_{j=0}^{\infty}\langle f,e_j\rangle e_j(x) = f(x) \mapsto \sum_{\{j \,:\, \lambda_j \in [\lambda, \lambda+1)\}} \langle f,e_j \rangle e_j(x) $$ as the projection of $f(x)$ onto the direct sum of eigenspaces which have eigenvalues in the unit-interval $[\lambda, \lambda+1)$.
We then denote the Schwartz kernel of the corresponding integral operator as $K(x,y;\lambda)$, where $$ f(x) \mapsto \int_{M} K(x,y;\lambda)f(y) \,dV_g(y) $$ agrees with the partial-sum definition above.
The goal of our research is to then analyze the big-oh behavior of this Schwartz-kernel as $\lambda \to \infty$. Usually this is formulated as $$ \sup_{x,y \in M}\big| K(x,y;\lambda) - F(x,y;\lambda) \big| = O(\lambda^{n-1}), $$ where the term $F(x,y;\lambda)$ comes from some parametrix approximation or something.
At this point I'm a little embarrassed to say admit that while I can do the mathematical research needed, I am unsure as to why people actually care about such a specific kind of linear operator?
I understand that the Weyl law is an old result in functional and harmonic analysis, but sadly I'm not sure why this specific problem is useful in the larger field of research. I've tried asking this of my advisor before, but he has not offered me all that much in the way of an answer. Also, while reading through the literature of similar problems to my own, I find many references to the myriad of results and slightly-different hypotheses, but still an answer of WHY? eludes me.
Specifically, why does everyone also study these partial-projections onto unit-length interval? What would be different if we projected only an interval of length 2? Or length $L$? Or onto a compact set of some fixed, finite measure?
Any insight into these kinds of problems, and their important to the mathematical body at-large would be much appreciated. Thank you in advanced, as always.
If you dilate $M$ by a factor $c$, don't the eigenvalues of $\sqrt{-\Delta}$ scale by a factor of $c^{-1}$, or something like that? That could explain why you're specifically looking at unit intervals, because going to arbitrary finite lengths really isn't any more general.
But to the broader point, I don't think you should expect a problem given to you by your advisor to have great, fundamental importance. Problems of great importance that can be solved by a typical grad student are quite rare. Some people explode out of the gates, but for most of us it takes many years of experience to get to the point where we can come up with important problems whose solutions are within reach.
The unit length hypothesis at here is not important, and very crude estimates are available using Sobolev embedding only. The main issue is that studying the spectrum on the manifold itself is not enough for recovering underlying topological/geometric information of the manifold. This is a subtle topic even for 2 dimensional surfaces, where a lot of work has been done.
For very recent work, check some papers by Sogge and Xi:
https://arxiv.org/abs/1711.04707
I would suggest that instead of working through the detailed estimates (on a sphere, on a torus, on negatively curved manifold, etc), think about some other ways to understand the spectrum of the Laplacian on the manifold. For example, a compact Riemann surface of genus $g\ge 2$ can be realized as the quotient of the upper half plane $\mathcal{H}/\Gamma$. There is a lot of interesting work that can be done to understand the relationship between the group action and the spectrum. The interplay between the algebraic nature of the surfaces and flexibility of the analysis tools made the subject really interesting.
A survey paper by Sanark may be a good start:
http://web.math.princeton.edu/facultypapers/sarnak/baltimore.pdf
For 3-manifolds this becomes deep and is related to heat kernels in geometric analysis. The subject is related to Ricci flow and there are plenty written up online already.
Thank you for taking the time to post these insights; they are rather interesting. However I'm not sure if I understand some of the bigger implications. Are you saying that studying the spectrum of the manifold (as a whole) is insufficient? But that by looking at the spectrum's distribution amongst higher frequencies (spread out among compact intervals) we can glean more important topological/geometric information about the manifold?
And if this indeed the case, then can you explain briefly/broadly why know this kind of information about the spectrum tells you that kind of information?
@Patch: I did not say "by looking at the spectrum's distribution amongst higher frequencies (spread out among compact intervals) we can glean more important topological/geometric information about the manifold? " anywhere in the answer. You put word into my mouth. I suggest you talk to your advisor. I would be direct: I do not think this is an interesting thesis topic, and I suggest you work on other things. Hope this is clear enough.
Sorry, I never mean to put words in your mouth. Since I was asking why people would be interesting in these kinds of estimates, I was thinking you were implying something more. It's one thing to say knowing the spectrum alone isn't enough, but that didn't tell me why projecting onto these restricted Eigenspaces would be of interest. Thank you, again, for your time.
|
2025-03-21T14:48:31.148344
| 2020-06-01T04:37:21 |
361867
|
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|
Stack Exchange
|
Cauchy's Integral with quadratic exponential term
As I was studying the Cauchy's integral formula, I tried to do the integral:
\begin{equation}
I = \int\limits_{-\infty}^{\infty} \frac{1}{x - a} e^{(i A x^2 + i B x)} dx
\end{equation}
with $A>0, B>0$ and $a > 0$.
Consider an integral on a complex plan:
\begin{equation}
J = \int\limits_{C + C_R} \frac{1}{z - a} e^{(i A z^2 + i B z)} dz
\end{equation}
where $C$ is along the real axis $-\infty \rightarrow +\infty$ and $C_R$ is the upper half circle $z = Re^{i\theta}$ with $R \rightarrow \infty$ and $\theta \in [0, \pi]$.
Naively, I would expect $C_R$ part of the integral gives zero and $C$ part of the integral gives $I$, then the $I$ can be derived by Cauchy's integral formula.
However, as I tried to check the $C_R$ part of the integral, I found that ($z = Re^{i\theta}$):
$$
\begin{split}
I_R &= \int\limits_0^{\pi} d\theta \frac{iRe^{i\theta}}{Re^{i\theta} - a} \exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big) \\
|I_R| &\leq \int\limits_0^{\pi} d\theta\left |\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \Big|\exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big)\Big|
\end{split}
$$
where the first term
\begin{equation}
\left|\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \leq \frac{R}{R-a} \rightarrow 1 \ as\ R \rightarrow \infty
\end{equation}
and the second term
\begin{equation}
\left|\exp(iAR^2e^{2i\theta}+iBRe^{i\theta})\right| \leq e^{-AR^2\sin(2\theta) - BR\sin(\theta)}
\end{equation}
will not approach to zero because of $e^{-AR^2\sin(2\theta)}$.
Is there anything wrong in my approach? And is there any other way I can perform this integral $I$?
Thanks a million for advises!
Have you tried a rectangle instead of circle?
@DuFong. Thanks for your suggestion. Yes, I tried a contour $C_1:-R \rightarrow R$, $C_2: R \rightarrow R+i\epsilon$, $C_3: R+i\epsilon \rightarrow -R+i\epsilon$ and $C_4: -R+i\epsilon \rightarrow -R$. The integrals of $C_2$ and $C_4$ approach to zero as $R\rightarrow \infty$. But the one of $C_3$ faces the same problem : $\int_{-\infty}^{\infty} dx e^{-2A\ x\ \epsilon}$ will not. Any other suggestions, please?
Before anything else, you should define what you mean by integrating over the point $x=a$. Is this meant to be a principal value?
This integral is apparently related to the time-dependent linear Schrodinger equation with step function initial data, have you check the StackExchange: this. And to study this, inverse scattering is possible the most commonly used method in mathematical physics. By the way, in order to make sense, $\frac{1}{x-a}$ should be regarded as a distribution. As it is not easy to find the exact solution, one may consider finding asymptotics as $A\rightarrow \infty$
@MichaelEngelhardt. Yes, it's a principal value.
@DuFong. Thanks for your advice. I will check it out!
Let me first remove the $Bx$ term by completing the square,
$$I=\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2+iBx}}{x - a}\,dx=e^{-iB^2/4A}\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a-B/2A}\,dx.$$
Mathematica evaluates the Cauchy principal value of the integral in terms of Meijer G-functions,
$$I=-\tfrac{1}{8} \pi ^{-5/2} e^{-iB^2/4A}\biggl\{G_{3,5}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
0,\frac{1}{4},\frac{3}{4} \\
0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\
\end{array}
\right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
\end{array}
\right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
\frac{1}{4},\frac{1}{2},\frac{3}{4} \\
0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\
\end{array}
\right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
\end{array}
\right)\biggr\},$$
with
$$\alpha=\left(a+\frac{B}{2A}\right)^4\frac{A^2}{4}.$$
|
2025-03-21T14:48:31.148582
| 2020-06-01T05:47:13 |
361868
|
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|
Stack Exchange
|
Isomorphism between two families of curves over the Teichmueller space
In his construction of the Teichmueller space of curves of genus $\geq 2$ Grothendieck states in Corollaire 2.4 that the map $$\underline{Isom}_S(X,Y) \xrightarrow{} S$$ is finite. The map represents the functor of isomorphisms between the two families of curves over S.
I understand that the fibres are finite. Hence one is left to prove properness. I know the valuative criterion of properness Lemme 2.5.
But I do not understand: Why does Grothendieck invoke Lemme 2.5 to prove Cor. 2.4? Why does Cor. 2.4 satisfy the assumption of Lemme 2.5, i.e. why
$$u_K:X_K \xrightarrow{\simeq} Y_K \implies u:X \xrightarrow{\simeq} Y$$
i.e. why does isomorphy of the two families over a Zariski open subset of $S$ imply isomorphy over all of $S$?
Could you explain what is it you don't understand? Lemma 2.5 is exactly the expression of the valuative criterion of properness for $\underline{Isom}_S(X,Y)$, and Lemma 2.6 contains the proof.
@abx I made an edit to my question.
Sorry I still don't understand (and I will stop trying). "why does isomorphy..." is what you have to prove (for the valuative criterion), and this is what Grothendieck proves in the Lemma.
@abx Sorry for the inconvenience concerning my question. So let me just ask: How do you prove Corollary 2.4? According to which logical steps? Thank you.
As you say, we want to prove properness applying the valuative criterion. In our case, this amounts exactly to prove Lemma 2.5. Then we (I mean, Grothendieck) reduce this Lemma to Lemma 2.6, which we prove.
@abx OK. Now I understand the logic of the proof: The proof of Lemme 2.6 and a posteriori Cor. 2.4 builds on the fact that all morphisms from P^1 to a curve of genus >= 2 are constant (Riemann-Hurwitz), analytically speaking. Thank you for helping :-)
|
2025-03-21T14:48:31.148741
| 2020-06-01T06:28:17 |
361869
|
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Stack Exchange
|
Counting number of special subset of vertices in a tree
As defined in this article, an ordered pair $ (X,Y) $ of disjoint subsets of the vertices of a graph $ G $ with $ \vert X \vert = \vert Y \vert =2 $, is called an odd pair if the number of edges with one endpoint in $ X $ and another in $ Y $ is odd. Denote the number of odd pairs in $ G $ by $ s(G) $ (Note that if $ X \neq Y$, then $(X, Y) \neq (Y, X)$).
If $ G $ is the star graph $ S_{n} $, it is easily seen that $s(G) = 0$.
If $ G $ is the graph of order $n$ with the sequence of degrees $ \lbrace n-2, 2, 1, 1, \ldots ,1 \rbrace $, I have proved that $s(G) = 4(n-3)^{2}$. (Here, $G$ is very similar to star graph.)
Now, I guess that for every tree $T$ of order $n$ such that $T$ is not star graph, $s(T) \geq 4(n-3)^{2}$. I have tested this by computer search, but I couldn't prove that. Can anyone help? I think using induction can lead to the proof.
Thanks in advance.
I'm really sorry. You are right. I made a mistake. Now, the question is modified.
What's the point of counting ordered pairs instead of unordered pairs?
There's no particular point. Only following the convention in the article.
Let us prove the desired bound $2(n-3)^2$ for the number of unordered odd pairs by induction on $n$, base case being $n=4$.
Suppose that $n\geq 5$. Take a leaf $a$ with a unique neighbour $b$; let $\deg b=k+1\leq n-2$. We assume that $a$ is chosen so as to maximize $k$. In particular, this ensures that $G-a$ is not a star, so it contains at least $2(n-4)^2$ odd pairs. Moreover, if $k=n-3$, we get the extremal graph which has been already investigated. So we assume $k\leq n-4$.
Now it suffices to find $2(n-3)^2-2(n-4)^2=2(2n-7)$ additional odd pairs containing $a$. These are almost provided by the following collections.
$(1)$ Pairs of the form $(\{a.b\},\{c,x\})$ where $c\in N(b)$ and $x\notin \{a,b\}\cup N(b)$. There are $k(n-2-k)$ such pairs.
$(2)$ Let $xy$ be an edge in $G-\{a,b\}$ (there are $n-k-2$ such). Any vertex $d\notin\{a,b,x,y\}$ is not adjacent to one of $x$ and $y$ --- say to $x$. Then $(\{a,x\},\{d,y\})$ is a desired odd pair; there are at least $(n-4)(n-k-2)$ such pairs.
$(3)$ Pairs of the form $(\{a,c\},\{b,d\}$, where $abcd$ is a path of length 3 --- there is at least one such.
All in all, this provides
$$
(n-k-2)(k+n-4)+1=(n-3)^2-(k-1)^2+1\geq (n-3)^2-(n-5)^2+1=2(2n-7)-1
$$
additional odd pairs. We need thus to find one extra additional pair in the case $k=n-4$ (otherwise the above inequality is strict).
If $n\geq 6$ (so $k\geq 2$), then the estimate in $(2)$ is not sharp: for any edge $xy$ in $G-\{a,b\}$ there exists a neighbor $c\neq a$ of $b$ which is adjacent to neither $x$ nor $y$, so both pairs $(\{a,x\},\{c,y\})$ and $(\{a,y\},\{c,x\})$ work. This already increases the bound by $2$.
If $n=5$, then $k=1$ only when $G$ is a path. In this case we find $8$ odd pairs by hands.
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2025-03-21T14:48:31.148968
| 2020-06-01T07:27:28 |
361871
|
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|
Stack Exchange
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The number of unitary circulant matrices over a finite field $\mathbb{F}_{q^2}$
I asked this question in MSE few days ago but there was no response.
Suppose $\mathbb{F}=\mathbb{F}_{q^2}$, where $q$ is a prime power. The conjugate of elements in $\mathbb{F}$ is defined by $\overline{x}=x^q$. I need to find the number of $n\times n$ unitary circulant matrices over $\mathbb{F}$.
The number of invertible circulant matrices over a finite field can be seen elsewhere, such as when $n,q$ coprime and my question when $n=\operatorname{char} q$.
Is there any better method to calculate this number other than considering each entry?
This is equivalent to the order of the centraliser of the permutation matrix of $(1,2,\dots,n)$ in $\operatorname{GU}_n(q)$.
Added on 30 May 2020 MSE:
Let $C$ be the subgroup of $\operatorname{GL}_n(q^2)$ of all circulant matrices. Is $C\operatorname{GU}_n(q)$ a subgroup of $\operatorname{GL}_n(q^2)$? That is, is $C\operatorname{GU}_n(q)=\operatorname{GU}_n(q)C$? If that is correct then $C\operatorname{GU}_n(q)=\operatorname{GL}_n(q^2)$ and so $|C\cap\operatorname{GU}_n(q)|$ follows. Here we denote by $\operatorname{GU}_n(q)$ the general unitary group over $\mathbb{F}_{q^2}$.
Let $\tau$ denote the permutation matrix corresponding to $(1,2,\ldots,n)$. Consider it first as an element of $\mathrm{M}_n(q^2)$.
This matrix has minimal polynomial equal to $X^n-1$, which is equal to its characteristic polynomial. It is therefore cyclic, and its centralizer is isomorphic to $\mathbb{F}_{q^2}$-algebra $\mathbb{F}_{q^2}[X]/(X^n-1)$. For simplicity, I'll assume $\mathbb{F}_{q^2}$ has no $N$-th root of unity, so that this algebra is isomorphic to $\mathbb{F}_{q^{2n}}$. Mapping $X\to \bar{X}^t$ defines a field automorphism of order $2$ of $\mathbb{F}_{q^{2n}}$ which, by Hilbert 90 (which is an overkill, but does the job), restricts to a surjective map $\mathbb{F}_{q^{2n}}^\times\to \mathbb{F}_{q^n}^\times$. The centralizer you're seeking is precisely the kernel of this map, and is of carinality
$$ \frac{q^{2n}-1}{q^n-1}=q^n+1.$$
Does this make sense?
Now, if $X^N-1$ splits in $\mathbb{F}_{q^2}$, then $\mathbb{F}_{q^2}[X]/(X^N-1)$ is a product of finite fields, and the same type of argument works per coordinate, which should result in a formula of the form $\prod_{i=1}^r q^{d_i}+1$ for suitable $d_i$'s such that $\sum d_i=n$.
Thank you for your answer. But what if $(n,q)\ne 1$ or in particular $n=\operatorname{char}q$?
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2025-03-21T14:48:31.149145
| 2020-06-01T08:04:56 |
361874
|
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|
Stack Exchange
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Are measures better thought of as densities than differentials?
The standard notation for integrating with respect to a measure $\mu$ is:
$$\int f(x)\,d\mu(x).$$
But I've wondered if it could be better written as:
$$\int f(x)\mu(x)\,dx$$
where $\mu(x)$ is now thought of as a density. Then applying the measure $\mu$ to a set $A$ can be expressed as $\int \mathbf 1_A(x) \mu(x)\,dx$ or $\langle \mathbf 1_A, \mu\rangle$.
In particular, if $\mu$ is the Dirac $\delta$ measure, then integrating with respect to $\delta$ can be written $\int f(x)\delta(x)\,dx$ instead of the more awkward $\int f(x)\,d\delta(x)$. If $\mu$ is the Lebesgue measure, then we can denote it as $1$, and write $\int f(x)\cdot 1\,dx$ for the integral of $f$ with respect to $1$. If $\mu$ is a probability measure $p$, then we can write $\int f(x)p(x)\,dx$ for $\mathbb E_p[f(X)]$.
Regarding Stieltjes measures, the appropriate notation for the Stieltjes measure of a monotonic right-continuous function $g$ is $g'$, not $dg$.
I got this idea from reading the nLab entry on the Radon-Nikodym Theorem. There, it's pointed out that the Radon-Nikodym "derivative" of $\nu$ with respect to $\mu$ can be written as $\frac\nu\mu$. Notice that when written this way, it doesn't actually look like a derivative. So perhaps, the terminology "Radon-Nikodym derivative" is misleading. Besides, if it were really useful to see it as a derivative, then surely it would satisfy the product rule, but it doesn't.
Are there any disadvantages of this notation?
What would be the advantage, really? Although I'm an analyst I personally also like the probabilistic notation $\int f(x)\mu(dx)$
@leomonsaingeon To me that reads "Sum of $f(x)$ times $\mu(dx)$" where $\mu(dx)$ doesn't make much sense on its own, but maybe I'm being too literal. I like to think of $\int f(x),dx$ as literally being the sum of $f(x)$ times $dx$
yes, precisely! the only difference between a "sum" and the more rigorous integral formalism is that $\mu(dx)$ is the ininitesimal mass contained in the infinitesimal set $dx$ (in fact as far as I can tell probabilists often write $dx$ for measurable sets, so non infinitesimal at all) After all, an integral is nothing but a fancier weighted sum, is it not? (the $\int$ sign is even reminiscent from that idea)
@leomonsaingeon I guess that makes sense. But now you can't write $\int f(x) , dx$ because $dx$ is some "infinitesimal" set. You have to write $\int f(x) \mu(dx)$. But I guess that apart from that awkwardness, it could work
Well, this is some kind of residual notational ambiguity due to the fact that, when one learns integration, usually the Lebesgue integral is given a priviledged role (for obvious historical and convenience reasons). For example (some) probabilists would never actually write $\int f(x) dx$, but rather $\int f(x)\lambda(dx)$ or $\int f(x) \mathcal L^d(dx)$, where $\lambda$ or $\mathcal L^d$ stand for the $d$-dimensional Lebesgue measure. Actually the second notation is somehow also standard in geometric measure theory and calculus of variations, especially in the Italian school.
For me its a strange discussion. The proposal seems only to make sense if we have probability measures on $\mathbb{R}$ or some locally compact group. But what if we have a measure on an abstract measure space $(\Omega,\cal{A})$? Then $dx$ with respect to what?
The Radon-Nikodym derivative has its name obviously from the fundamental theorem of calculus. I have never seen someone getting confused by the name.
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2025-03-21T14:48:31.149715
| 2020-06-01T09:50:09 |
361876
|
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|
Stack Exchange
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On grades of torsion modules in noetherian rings
Let $A$ be a (not necessarily commutative) two-sided noetherian ring with minimal injective coresolution $(I_i)$ of the regular module $A$ as a right module.
Say that $A$ has dominant dimension $n$ in case $n$ is the smallest integer such that $I_n$ is not flat (or say that the dominant dimension is infinite in case all $I_i$ are flat).
For a finitely generated module $M$ let $ev_M:M \rightarrow M^{**}$ be the evaluation map, where $M^{*}:=Hom_A(M,A)$ and let $t(M):=ker(ev_M)$ be the torsion module of $M$.
Recall that the grade $gr(M)$ of a module $M$ is defined as $gr(M):= inf \{ i \geq 0 | Ext_A^i(M,A) \neq 0 \}$.
In the article "Homological theory of noetherian rings" Reiten proved in theorem 5.2. that the following are equivalent for a noetherian ring of global dimension 2:
a) $gr(t(M)) \geq 2$ for all finitely generated $M$.
b) $A$ has dominant dimension at least 2.
Question: Is the following more general statement true?
Let $A$ be a noetherian ring (so we have no restriction on the global dimension now) and $n \geq 2$. Then $A$ has dominant dimension at least $n$ iff $gr(t(M)) \geq n$ for all finitely generated $M$.
One can show that dominant dimension at least $n$ implies $gr(t(M)) \geq n$ for all finitely generated $M$, but I am not sure about the other direction.
Computer experiments suggest that it might be true.
|
2025-03-21T14:48:31.149859
| 2020-06-01T10:32:08 |
361878
|
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|
Stack Exchange
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Relations between quantum groups at roots of unity, modular representation theory, and physics
I understand that quantum groups at roots of unity are related to physics because they are used in the construction of Reshetikhin-Turaev invariants, conjectured by Witten. Are there other relations of quantum groups at roots of unity to physics? Also, modular representation theory of Lie algebras is related to quantum groups at roots of unity via Andersen-Jantzen-Soergel. Modular representation theory is a very active area of research (cf. work of Lusztig, Bezrukavnikov, Williamson, and others), and I am wondering if there are relations between results/questions in this area and physics.
Witten conjectured that every unitary modular tensor category would come from a Chern-Simons-Witten TQFT. Quantum groups at roots of unity provide an algebraic setting for studying anyons and topological phases of matter. Such phenomena are (expected to be) relevant to various physical systems; most notably the fractional quantum Hall effect. A further application of these ideas is in the construction of a 'topological quantum computer' or memory. There's a lot of literature on this topic and several related questions on MO.
@SValera Thank you very much! 1) Would you be able to give some links (to preprints, papers, books, questions on MO) about these applications (to anyons, topological phases of matter, fractional quantum Hall effect, topological quantum computers) of quantum groups at roots of unity that you mention? 2) Would you be able to tell if/how the existing research/questions on modular representation theory of Lie algebras have a bearing on these physical applications via the equivalence of certain categories of representations of Lie algebras in pos. char. and quantum groups at roots of unity?
Modular representations (representations in spaces over a field of nonzero characteristics) have been used in physics by Felix Lev to construct a quantum theory that is based on a finite number field (rather than on $\mathbb{C}$).
F.M. Lev,
Finiteness
of physics and its possible consequences.
F.M. Lev,
Modular
representations as a possible basis of finite physics.
F.M. Lev,
Why
is quantum physics based on complex numbers?
Thank you very much! I am sorry, my education is in mathematics rather than physics, but may I ask if there is much work in physics which uses these ideas of F. Lev?
Also, sorry, I saw that some articles on this topic by F. Lev are posted on viXra, which seems to suggest that they may not reflect a very mainstream point of view in physics. Again, my understanding of physics is quite limited, so it would be great to receive a clarification from an expert.
The articles which Carlo mentions are all published in journals to be fair, I think the majority of prolific academics have some preprints which never see publication for whatever reason (usually just because of not being able to find the right journal for them).
While true, the first two of those papers have about 30 citations, and only two of them are by authors other than Lev. I don't think there's much evidence that other authors have adopted these ideas. Chasing through the citations of the final paper seems more promising.
The area seems to be very broad, let me just give some remarks, which are somewhat close to me.
Many interesting conformal field theories are "rational", in some simple cases it means that some parameter like a central charge is rational/integer $k$.
By some reasons people consider expressions like $(P)\exp( 2\pi i/k J(x) )$, where $J(x)$ are some generators of symmetries of field theories—currents.
And they appeared to be related to quantum groups. So if $k$ is an integer, then $\exp(2\pi i /k)$ is a root of unity. So the point is that roots of unity are related to "rationality" of some field theories.
The oversimplified example is just to consider canonical commutation relations: $[X,Y] = 2\pi i/k$, which after exponentiation gives a quantum-group-style relation: $\exp(X)\exp(Y) = q\exp(Y)\exp(X)$, for $q=\exp(2\pi i/k)$—so $q$ will be a root of unity for $k$ integral.
(Although this is oversimplified, from some very high-level point of view the whole story is about it.)
Probably the most famous example is the Kazhdan–Lusztig equivalence between the category of certain
integrable representations of the Kac–Moody algebra at a negative level
and the category of (algebraic) representations of the "big" (a.k.a. Lusztig's) quantum group.
So here you see that we need "negative level", i.e. central charge $k$ is a negative integer, and what you get is a quantum group at $q = \exp(2\pi i/k)$—a root of unity. (I might forget to shift $k$ by dual Coxeter number.)
I guess mathematical proof does not use explicitly calculations like I mentioned above—just take "currents" $J(x)$—and show that $P\exp( J(x)/k)$ generate quantum group, but that's what some physicists were doing.
To put that idea in the right framework - let us think of the famous Drinfeld–Kohno theorem which says that the monodromy of the representation of the Knizhnik–Zamolodchikov equation is given by the corresponding quantum group.
Again you can see that integer values of $k$ would correspond to root of unity for $q$, by the trivial reason that monodromy locally is given by exponent.
In some sense that statement is closely related to Kazhdan–Lusztig theorem—the KZ-equation is given by "currents" $J(x)$ in tensor product of evaluation modules,
one considers its Pexp (i.e. monodromy) and gets the quantum group.
Another example: in
Integrable Structure of Conformal Field Theory III. The Yang-Baxter Relation,
V. V. Bazhanov, S. L. Lukyanov, A. B. Zamolodchikov explicitly construct quantum-group-like relations from the conformal field theory operators.
For certain special values of parameters one can get quantum groups at roots of unity.
If I remember correctly they exploit that in some papers.
Would it be all right with you if I TeXed this?
@LSpice you are welcome
Dear Alexander, do you have some reference about this sentence ? "Although this is oversimplified, from some very high-level point of view the whole story is about it." Thank you !
|
2025-03-21T14:48:31.150275
| 2020-06-01T10:59:29 |
361881
|
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|
Stack Exchange
|
Modified straightline complexity of almost square of sums
Assume every linear operation (such as inner product with constant vector) can be performed in one step and multiplication by variables (quadratic operation) can be performed in one step.
We know the straight line complexity of $$(\sum_{i=1}^kx_i)^2$$ is linear in $k$ basic model and $2$ in this new modified model (adding one step and squaring one).
What is the modified complexity of $$(\sum_{i=1}^k\alpha_ix_i^2)+\sum_{i=1}^k\beta_ix_i(\sum_{\substack{{j=1}\\ i\neq j}}^kx_j)$$ where $\alpha_i>\frac{\beta_i^2}2+\beta_i$ holds at every $i\in\{1,\dots,k\}$?
I am just wondering if it is $o(k)$ at some special $\alpha_i$,$\beta_i$ with $\alpha_i>\frac{\beta_i^2}2+\beta_i$?
I am NOT reasonably sure the complexity is similar to $\alpha_i<\frac{\beta_i^2}2+\beta_i$.
|
2025-03-21T14:48:31.150361
| 2020-06-01T11:03:09 |
361882
|
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"Henry Adams",
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|
Stack Exchange
|
How much smaller is the Čech complex than the Vietoris-Rips complex?
The Čech complex
is a subcomplex of the
Vietoris-Rips complex.
The V-R complex
includes as a simplex a set of points with pairwise
distances at most $\epsilon$,
whereas the Č complex
includes as a simplex a set of points
with non-empty intersection of diameter $\epsilon$-balls
centered on the points.
One advantage of the Č complex is it can be
(and generally is) smaller than
the V-R complex. My question is essentially: How much smaller?
Q. What results are known for the relative sizes of
the two complexes for random point clouds?
By size I mean some measure of combinatorial complexity, such as the total number of simplicies.
I am open to any definition of what constitutes
a "random point cloud":
uniformly distributed within a sphere,
multidimensional Gaussian distribution, benchmark
data sets, ...
I'm primarily interested in points in $\mathbb{R}^3$ but
higher-dimensional results would be equally welcomed.
Hi Joseph, I do not know any results along these lines, but would be very interested to learn more! One minor comment on your post is that if you want the Cech complex to be a subset of the Vietoris-Rips complex, then you should use balls of radius $\epsilon/2$ (unless by $\epsilon$-balls you mean balls of diameter $\epsilon$?).
@HenryAdams: Thanks for the diameter-$\epsilon$ correction, now incorporated.
The expected size of the Cech complex is discussed in: https://mathoverflow.net/questions/235288/throwing-darts-at-a-barn-and-putting-a-bullseye-around-them-in-higher-dimensions?noredirect=1&lq=1. For Rips, it is discussed in https://mathoverflow.net/questions/133750/what-fraction-of-n-point-sets-in-the-unit-ball-have-diameter-smaller-than-1
@alesia: Thanks. Interesting that neither post mentions the name of the complex. I'm not finding it easy to extract a comparison from the two posts. I'll work on it.
I'm going to offer an answer mainly to get an idea off my brain and maybe someone will point out why this is incorrect. However, in my view, a lot of discussions about Čech Vs Vietoris-Rips seem to overlook a key connection: The Vietoris-Rips Complex is a particular type of Čech complex.
Given a topological space $X$ and a cover $\mathcal{U} = \{U_i\}$ of $X$, the Čech complex is like a free vector space generated by higher intersections of elements of the cover.
If the $U_i$'s are $\epsilon$-balls centered at finitely many chosen points then this is still a Čech complex and it is commonly referred to as "the Čech complex" when the finitely many chosen points are exactly the point cloud in question.
If the $U_i$'s are $\epsilon'$-cubes centered at finitely many chosen points then this is still a Čech complex and it is commonly referred to as "the Vietoris-Rips complex" when the finitely many chosen points are exactly the point cloud in question.
So most of the relationships and discussions around Čech vs V-R complexes to me are really overly-complicated conversations about cubes and balls. Once we let $\epsilon$ vary, the discussions are not really important anymore since we are supposedly imagining our data points are sitting on some unknown manifold and so the nerve theorem says that cubes and balls will eventually compute (co)homology if epsilon is small enough.
Looking forward to seeing if this is a valuable thought to others. (I didn't realize any of this until I was working with some undergrads on this stuff.)
|
2025-03-21T14:48:31.150730
| 2020-06-01T12:15:46 |
361887
|
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|
Stack Exchange
|
Completeness of certain formal deduction system
Consider a certain formal system with only axiom Excluded Middle -$EM$
and 18 inference rules:
9 implicative ruules (clearly not independent)
and 9 tautological rules:
If we have substitution at hands as well but we are restricted no to use conditional proof.
Is this particular system complete?
I always believed that the system is complete.
But once I decided to prove or disprove completeness I stuck.
I neither can find any reference neither nor can proof completeness.
If someone is familiar with this system I would be grateful to have some reference or proof.
Where does the system come from?
This is a system we usually offer to students in first common logic course for students of non-mathematical profile, in order they understand what is concept of proof.
What exactly do you mean by “tautological rules”? I assumed, as the notation suggests, that these are just inference rules that can be applied in both directions (on whole lines in the proof). However, do you by any chance mean that they can be actually applied as replacement rules anywhere deep inside the formulas? If the latter, then the calculus is easily seen to be complete.
@Emil Jeřábek Yes, you see it correctly, tautological rules work both direction. How do we see completeness of the calculus?
That does not answer what I asked. Can the tautological rules be applied only to whole formulas, or also to subformulas?
You are right, I want to figure out if it is possible to prove completeness without replacement of a sub-formulas but with substitution if needed.
Ok. Then I do not see how to either prove or disprove completeness.
|
2025-03-21T14:48:31.150873
| 2020-06-01T12:20:02 |
361888
|
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|
Stack Exchange
|
Symmetric functions for multidimensional variables
I have $N$ variables (let's call them $X$) of dimensionality $D$, that I want to make permutation invariant.
For $D=1$ I know I can use e.g. the elementary symmetric polynomials to accomplish this. What if $D>1$ (only integer dimensions, not fractals :) and $\bar{X}$ are vectors? Is there a way to systematically generate functions which are symmetric under permutations of their variables ( i.e. $f_i(\bar{X}_i, \bar{X}_j)$ = $f_i(\bar{X}_j, \bar{X}_i)$), $f_1, f_2, ..., f_N$ so that $f_i(\bar{X}_1, ..., \bar{X}_N)$ generates a vector of dimension $D$?
If this is not possible in general, can we accomplish it for the limited case when $N=2$ and $D>1$?
As far as I can see, you're not really telling us what operations you want your object to be symmetric under. Once you're clear about this, you can always pick an arbitrary representative and sum over the equivalence class. That's by construction symmetric.
right, symmetric under permutations of $\bar{X}_i$. Thanks for pointing this out. I've updated the question
Basically you are asking about the coordinate ring of $((\Bbbk^D)^{\otimes N})/S_N$, I think?
If it is Zach's interpretation which is meant in the post then the relevant search term is the theory of "multisymmetric functions". It is a very difficult subject. Good luck.
One obvious construction is $f_i (\bar{X}{1},\ldots,\bar{X}{N} ) = \sum_{P} g_i (\bar{X}{P(1)},\ldots,\bar{X}{P(N)} )$ where $P$ denote all the permutations of $N$ objects, and $g_i $ can be anything you like. If $g_i $ is a bona fide tensor, $f_i $ will inherit that property - that aspect is independent of the symmetrization issue.
Okay, thanks! I appreciate the solution may be very complicated or computationally expensive now.
|
2025-03-21T14:48:31.151025
| 2020-06-01T13:26:55 |
361891
|
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|
Stack Exchange
|
Existence of Liouville vector fields on symplectic manifolds
Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X}
\omega=\omega$. The existence of a Liouville vector field implies that $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.
My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$?
Thanks!
Given any 1-form $\alpha$, there is a vector field $V$ so that $i_V \omega = \alpha$. This is non-degeneracy. So if $\omega = d\lambda$ then the unique $V$ so that $i_V \omega = \lambda$ is Liouville. The existence of a Liouville field is equivalent to exactness. For surfaces this is equivalent to nonempty boundary.
@MikeMiller Thank you!
If the symplectic form integrates to a nonzero quantity on a compact surface in your manifold, it is not exact. For example, on $M=S^2\times S^1\times [0,1]$ with symplectic form $dA_{S^2} + d\vartheta \wedge dt$.
This is in fact an equivalence, since every 2-dimensional homology class is represented by a closed surface.
Thank you for the clarification!
|
2025-03-21T14:48:31.151134
| 2020-06-01T13:29:39 |
361892
|
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|
Stack Exchange
|
R-matrices and symmetric fusion categories
Given a $\mathbb{C}$-linear braided fusion category $\mathcal{C}$ containing a fusion rule of the form e.g.
\begin{equation}X\otimes Y\cong A\oplus B \oplus C\end{equation}
(where $A,B, C, X$ and $Y$ are all simple objects with $A, B, C$ non-isomorphic), we can write the $R$-matrix $R^{XY}=\text{diag}(R^{XY}_{A}, R^{XY}_{B}, R^{XY}_{C})$. My intuition has always been that these scalars for two distinct objects cannot be the same (much like eigenvalues for distinct eigenspaces should not be the same). Is this misguided? That is,
(Q) Can we have scalars $R^{XY}_{A}=R^{XY}_{B}$ for $A\not\cong B$?
If $\mathcal{C}$ is symmetric, then I believe that diagonal matrix $R^{XY}$ can only have $\pm1$s along its diagonal. However, if the answer to (Q) is no, then this would mean that $\mathcal{C}$ can only contain fusion rules of the form $X\otimes Y\cong pA\oplus qB$ and $X\otimes Y\cong pA$ (where $p$ and $q$ are positive integers). This seems too strong.
Thanks!
Basically any representations of any groups will give you counterexamples in the symmetric case. The simplest case, where X and Y are the defining representation of SU(2), already works. For non-symmetric categories you'd expect that "generically" they'll look different, e.g. if you look at the category of representations of a quantum group and vary q, then for all but finitely many q the eigenvalues will be different. But there's plenty of special q where you'll get repeats, and not just symmetric cases. For example, take SU(2) at a level which has a $D_{2n}$ de-equivariantization, then you have a transparent invertible object and if you take the tensor square of the "middle" object it will have both the trivial and the transparent as summands with the same eigenvalue.
Thanks very much, this is exactly what I was looking for. A follow-up: if e.g. $R^{XY}_{A}$ is now a $k$ by $k$ matrix with $k>1$ (i.e. $A$ appears in the decomposition of $X\otimes Y$ with multiplicity $k$) then suppose we diagonalise this $R$-matrix. Is it also possible for these diagonals to be distinct?
A partial answer to this for $\mathcal{C}$ ribbon: we know $\left[R^{XY}{A}R^{YX}{A}\right]{\mu\nu}=\frac{\vartheta{A}}{\vartheta_{X}\vartheta_{Y}}\delta_{\mu\nu}$. In some choice of gauge, we should be able to get $R^{XY}{A}=R^{YX}{A}$. For such a choice of gauge, the diagonalisation of $R^{XY}{A}$ will have scalars $\pm\sqrt{\frac{\vartheta{A}}{\vartheta_{X}\vartheta_{Y}}}$ along the diagonal. The 'eigenvalues' thus differ by a minus sign at most in this choice of gauge, and if $X=Y$ then the argument is gauge-invariant.
|
2025-03-21T14:48:31.151335
| 2020-06-01T15:39:27 |
361898
|
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|
Stack Exchange
|
Is a polytope that has in-spheres for faces of all dimensions already regular?
Let $P\subset\Bbb R^d$ be a convex polytope (convex hull of finitely many points).
A $k$-in-sphere of $P$ is a sphere centered at the origin to which each $k$-face of $P$ is tangent. So a 0-in-sphere contains all the vertices and is actually a circumsphere, and a $(d-1)$-in-sphere is completely contained in $P$.
$\qquad\qquad\qquad\qquad\qquad$
Question: If $P$ has $k$-in-spheres for all $k\in\{0,...,d-1\}$, is $P$ a regular polytope?
By definition, all these spheres are centered at the origin, hence are concentric.
The answer to the question is Yes for polygons. For $d\ge 3$ note that this property of $P$ is inherited by its faces, and it follows that all 2-faces of $P$ are regular polygons and all edges are of the same length.
Do you assume that the spheres are concentric, or do you say that they must be concentric?
@WlodekKuperberg I assume this. Otherwise e.g. any triangle (not necessarily regular) would be an example as well.
This is true in all dimensions, and can be proved by induction (on $d$) applied to the following (slightly stronger) hypothesis:
Theorem: If $P$ is a convex $d$-polytope with $k$-in-spheres for all $k \in [0, d-1]$, then:
$P$ is regular.
$P$ is determined (up to an element of the orthogonal group $O(d)$) by that $d$-tuple $(r_0, r_1, \dots, r_{d-1})$ of $k$-in-radii.
$P$ is determined completely if additionally a facet (codimension-1 face) $Q$ of $P$ is specified.
Proof: If the polytope $P$ has squared $k$-in-radii $(r_0^2, r_1^2, \dots, r_{d-1}^2)$, then every facet of $P$ has squared $k$-in-radii $(r_0^2 - r_{d-1}^2, r_1^2 - r_{d-1}^2, \dots, r_{d-2}^2 - r_{d-1}^2)$. By the first two parts of the inductive hypothesis, all facets of $P$ are therefore regular and congruent to each other (being determined by these $k$-in-radii).
Now, given a facet $Q$ of $P$ and a facet $R$ of $Q$, let $\Pi$ be the hyperplane through the origin which contains $R$. Let $Q'$ be the other facet of $P$ which contains $R$. Because the $k$-in-spheres of $Q'$ are the reflections (in $\Pi$) of the $k$-in-spheres of $Q$, and they share a common facet $R$, it follows (from the third part of the inductive hypothesis) that $Q'$ is the reflection of $Q$ through the hyperplane $\Pi$.
As the boundary $\partial P$ (i.e. the union of all facets) is homeomorphic to $S^{d-1}$, we can reach any facet $Q_1$ from any facet $Q_0$ by a 'path' of 'adjacent' (i.e. sharing a common subfacet) facets. Consequently, we can transform any facet into any other facet by a sequence of reflections in hyperplanes through the origin. As each facet is flag-transitive, it therefore follows that $P$ is flag-transitive (i.e. regular) as desired.
Moreover, this reflection procedure of building $P$ from a single facet $Q$ establishes the third part of the theorem.
This leaves the second part of the theorem. Suppose $P$ and $P'$ are two polytopes sharing the same set of $k$-in-spheres. Let $Q$ be an arbitrary facet of $P$, and $Q'$ be an arbitrary facet of $P'$. By the inductive hypothesis, $Q$ and $Q'$ are congruent; let $f$ be an isometry of the ambient space which maps $Q$ to $Q'$. The origin is either mapped to itself or (if we chose the 'wrong' isometry) to $2v$, where $v$ is the centroid of $Q$; we can if necessary reflect again in the hyperplane containing $Q$ to ensure the origin is preserved by $f$. Consequently, $f$ is an element of the orthogonal group $O(d)$ which maps $Q$ to $Q'$. By the third part of the theorem (which we've already proved), $f$ must map $P$ to $P'$, establishing the second part of the theorem.
In $R^3$, since the spheres are concentric, not only all faces are regular, but also all edges are of the same length, and all faces are inscribed in circles of the same radius, hence are congruent. Also, all dihedral angles between faces with a common edge are equal, which implies that all vertices are of the same valence. This makes the polytope regular. It seems that this reasoning can be generalized to all dimensions.
How exactly you show that "all faces are inscribed in circles of the same radius" (I know that they are inscribed, but why same radius)? Also I would be interested in how exactly this generalizes.
Each face lies on a plane which is tangent to the insphere, and intersects the circumsphere in a circle (namely the circumcircle of that face). The squared radius of this circumcircle is the difference between the squared radius of the circumsphere and the squared radius of the insphere (by Pythagoras).
@M.Winter: All centers of these circles are at the same distance from the origin, and all circles lie on one sphere, the one containing the vertices.
I apologise for my wrong edit. I actually read it carefully and thought I knew what you meant, but I should have asked rather than changing it when I realised I wasn't sure.
@LSpice -- Thanks, no harm done.
|
2025-03-21T14:48:31.151739
| 2020-06-01T15:39:33 |
361899
|
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|
Stack Exchange
|
Path-covering for vertex-transitive graphs
I have the following dummy problem:
Claim - There exists $N$ such that for $n > N$, if $G_n$ be a connected directed vertex-transitive graph with $n$ vertices, then there exists a set $S$ of paths on $G_n$, such that $$\bigcup_{P \in S} P = V(G_n),$$ $$|S| < \frac{n}{\log(n)^4},$$ $$\sum_{P \in S} |P| < \left( 1+
\frac{1}{\log(n)^4}\right)n.$$
I am interested as my research involves trying to prove a similar claim with additional restrictions on $P$, but I am not sure how to begin attacking such a problem.
One idea I’ve thought might be useful is a Theorem of Gallai and Milgram:
If a digraph $G$ has independence number $\alpha$, then $G$ can be partitioned into $\alpha$ paths.
|
2025-03-21T14:48:31.151858
| 2020-06-01T15:47:53 |
361901
|
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|
Stack Exchange
|
extending functor from a dense subcategory
Let $A$ and $B$ be two cocomplete categories (i.e. closed under small colimits) and $A'$ be a dense subcategory of $A$ i.e. any object in $A$ is a colimit of objects in $A'$. Given a functor $F': A' \to B$, does there always exist an extension of functor $F :A \to B$ preserving all colimits?
I think you mean "cocomplete", rather than "complete"? And $A'$ is a dense subcategory of $A$, rather than $B$?
Do you want some condition on $F'$? Otherwise you could just take $A'=A$, with $F'$ some functor that doesn't preserve colimits.
@JeremyRickard Uhhh, sorry for the typos.. right, I meant cocomplete and $A'$ a dense subcategory of $A$. Maybe I need to say $F'$ preserves colimits.
If $A'$ is full in $A$ and skeletally small, then yes: even more is true, there is also an adjunction $[A',Set]\leftrightarrows B$; replace $A'$ with its small skeleton, and use Yoneda lemma. Otherwise, subtle set theory comes in, and I guess you might want to say a bit more on the context of the question :-)
Hi @Fosco . One example in my mind: let $C$ be a category (say category of schemes over a fixed base), $A$ be its category of presheaves of sets and $A'$ the subcategory of representable presheaves. Does $F'$ extend to colimit preserving $F$ in this case?
@Fosco What's your definition of $[A', Set]$ here? Btw do you have a reference for such related results?
I actually meant $[(A')^{op},Set]$, sorry for the typo! The category $[(A')^{op},Set]$ has objects the contravariant functors from $A'$ to the category of sets. The fact that every functor $F : X \to B$ with cocomplete domain has an extension to a cocontinuous functor $\bar F : [X^{op},Set]\to B$ "is the Yoneda lemma" in a suitable sense.
@Fosco I see. Do you have a reference for the detailed treatment of what you said? Anyways what I am really interested in is that $A'$ is the fppf site over a fixed base scheme which I think is not small in any sense and $A$ the category of sheaves $A'$? can we still say something similar? thanks btw!
https://arxiv.org/abs/1501.02503 see 3.1.1 here; not because there are no other reference, just because that's the most convenient source to quote for me :) in case $A'$ is not small, see here: https://ncatlab.org/nlab/show/small+presheaf
No. This is true only when $A$ is a free cocompletion of $A'$, i.e., the category of small contravariant functors from $A$ to $Set$.
I think I have problem with the condition of the functors being small here. If A′ is small, such condition should be empty. However, in practice (in algebraic geometry), say if we consider A′ be the fppf site over a fixed base scheme S which I think is not really small, there is no category of presheaves because of the size issue. But does the category of sheaves make sense here? if so call it $A$, can we say anything about extending a functor from $A'$ to $A$?
There is no size problem if you take small presheaves, i.e., small colimits of representables.
Sorry for being vague. I mean I eventually want to deal with arbitrary SHEAF. My thought was that all sheaves come from presheaves (via sheafification) and hence it'd be nice if we can talk about arbitrary presheaf, small or not. Now my problem is (sorry for asking another question here), is it true that any fppf sheaf is the sheafification of some small presheaf ? If true, that will save my day. Thanks very much!
I would expect that sheafifications of small presheaves are small sheaves. All what I know about this subject is contained in my two joint papers with Boris Chorny (arXiv:1110.0605 and arXiv:1110.4252).
Thanks for the answer! I don't really know the precise definition but I guess small sheaves are by definition the sheafification of small presheaves. In some cases these small sheaves seems to be all sheaves. For example, if we consider small etale site over a scheme (which is an essentially large category), all etale sheaves here would be small because the category of sheaves on the small etale site coincides with the category of sheaves on small affine etale site (which turns out to be essentially small). But this does not seem to work for the big sites e.g. big fppf site.
|
2025-03-21T14:48:31.152194
| 2020-06-01T17:19:33 |
361902
|
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|
Stack Exchange
|
Complexity of edge coloring of class 1 graphs
We know that the decision problem of classifying the graphs as class $1$ or class $2$ (with respect to edge coloring) is NP-complete. But, suppose we have to prove a graph to be in class $1$. Does it always imply that we have to find a polynomial time algorithm for the edge coloring of that graph? Like, is it possible to prove a graph to be in class $1$ without describing an explicit algorithm to do so, like, say by computing the Lovasz number of the line graph of the graph?
Most of the problems I have seen invariably lead me to think that one could prove the graph is in class $1$ iff there is a polynomial time algorithm to color the edges of the graph. Is this right? Thanks beforehand.
I‘m not sure I understand your question. If you want to prove that a fixed graph is class 1 (which I assume means that the number of colors equals the max degree), it is enough to give a coloring with that many colors and checking this. Checking can be done in polynomial time, implying that the problem is in NP. On the other hand, if you can find a poly time algorithm which for every graph in class 1 finds a coloring with the right number of colors, you can separate the two classes in polynomial time, thus proving P=NP (again, assuming my interpretation of „class 1“ and „class 2“ is correct).
@PatrickSchnider you interpreted the class1 and class 2 concepts right, but I think you misunderstood the question. When you say that giving a coloring with colors equal to maximum degree and then checking it, the very process of assigning that many colors to the graph satisfying the adjacency condition is NP-hard in general. What I am asking is that to prove a graph is class 1, is it necessary to give a polynomial time algorithm for such a coloring, or is it possible by not providing an algorithm as such
As written the question is ill-posed: class 1 or class 2 is a property of a particular graph, whereas polynomial time is a property of an algorithm applied to an infinite family of graphs.
Patrick Schnider points out in the comments that the first fix to the question doesn't work: if we have a polynomial time algorithm for finding a colouring when it exists, we certainly have a polynomial time algorithm for testing whether a colouring exists.
The deeper question is perhaps this. Suppose we have an efficient algorithm for determining whether a particular structure (such as a colouring of a graph) exists. Can we use the algorithm to find it?
Sometimes the answer is yes. Suppose we have an efficient algorithm for deciding whether a graph is $k$-colourable. Then we can use it to find a $k$-colouring (when it exists) of a graph $G$ as follows. Test each non-edge $e$ of $G$ to see whether $G+e$ is also $k$-colourable. If it is, switch attention to $k$-colouring $G+e$. If it isn't, move on to the next edge. Note that we never have to test an edge more than once, as if $G+e$ is not $k$-colourable then $G+H+e$ is never $k$-colourable either, for any $H$. The end-point of this process is a complete $k$-partite graph, which is easily $k$-colourable.
But sometimes the answer is no, or at least "we don't know yet". Primality testing is in P, so we have an efficient algorithm to detect whether an integer can be factorised. But we don't have an efficient (classical) algorithm to factor integers, and it's not clear that we ever will. If there was a mechanical process to go from existence of objects to providing examples then this would not be such a notoriously difficult problem.
so the problem i somewhat a rephrasing of whether $P=NP$, right?
It is not completely clear to me how you can use a polynomial time algorithm that finds colourings if they exist to actually determine whether or not they exist. Suppose I give you the algorithm and a graph to be tested... what do you do next?
@GordonRoyle, suppose I can find colourings when they exist in time at most $f(n)$. I input the graph I'm wondering about into the algorithm and try to run it for $f(n)$ steps. The possible results are that it finds a colouring, or that it fails to find a colouring, possibly because it encounters a state it doesn't know what to do with (the promise of existence having been broken). Which result I get tells me whether a colouring exists.
@vidyarthi, It's not unrelated, but it's not exactly the same question. I'd say it's more to do with the difference between decision problems (which just have yes/no answers) and more complicated types of question which are asking for something close to what we'd usually think of as a "computation".
@BenBarber But most polynomial-time algorithms only come with an $O(f(n))$ time-bound, not a precise number-of-steps bound.
@GordonRoyle I suppose that this could cause a practical difficulty if you happened not to know the implicit constant, or how large $n$ needed to be for some known constant to be valid. But then running for say $nf(n)$ steps would work eventually; and if only $O(\cdot)$ level information is acceptable to call a decision process efficient, we should extend the same courtesy to the run length of the algorithm for producing witnesses.
How we should treat algorithms which are known to terminate but whose run length exceeds any computable function of the input size is on a level of subtlety on which I am not competent to comment.
|
2025-03-21T14:48:31.152795
| 2020-06-01T19:21:32 |
361913
|
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|
Stack Exchange
|
Morphism of sketches inducing a (co)reflective subcategory
Certain morphisms of theories gives rise to (co)reflective subcategories of their models. For the sake of simplicity, I'm going to fix my definition of sketch to be a category $\mathbb{C}$ with a chosen class of limits $L$, $(\mathbb{C},L)$. If $L' \subseteq L$, then we certainly have that $Mod(\mathbb{C},L')$ is a reflective subcategory of $Mod(\mathbb{C},L)$. For a coreflective category, we can take the inclusion of vector bundles into the category of double vector bundles, where $A \xrightarrow{q} M$ becomes the double vector bundle $ M \xleftarrow{q} A \xrightarrow{q} M $ (the base is just identity $M = M$) and the core construction gives the coreflector (see more details on the core here).
I can't seem to find any reference on essentially algebraic theories or limit sketches that gives any clear conditions for when a morphism of sketches induces a (co)reflective morphism of theories.
|
2025-03-21T14:48:31.152889
| 2020-06-01T19:21:49 |
361914
|
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"sort": "votes",
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|
Stack Exchange
|
Decomposition of Gaussian spaces with respect to covariance function
Let $K(t,s):T^2 \to \mathbb{R}$ be a kernel symmetric and type positive (for every $n$ $\sum^n_{i,j}u_iu_jK(t_i,t_j) \geq 0$ and $(u_1,\dots,u_n) \in \mathbb{R}^n$) where $T$ is any set. Thus, it is possible get a suitable probability space ($\Omega, \mathcal{F},P$) and a Gaussian process $(X_t)_{t \in D}:\Omega \to \mathbb{R}$ with values in $\mathcal{H}$ (a closed subspace of $L^2(\Omega, \mathcal{F},P$) of Gaussian process with mean 0) and $K$ is the covariance function of $X_t$. I know that if I can write $X_t = Z^{1}_t + Z^2_t$ where $Z^{1}_t$ and $ Z^2_t$ are independent centered Gaussian processes then
$$K(t,s) = K_1(t,s) + K_2(t,s),$$
where $K_1$ and $K_2$ are the covariance function of $Z^{1}_t$ and $Z^2_t$ respectively. I'm asking for a converse of it: If I have $K(t,s) = K_1(t,s) + K_2(t,s)$, then I can get $Z^{1}_t$ and $ Z^2_t$ independent centered Gaussian processes. Thanks for any help.
|
2025-03-21T14:48:31.153000
| 2020-06-01T20:26:43 |
361920
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Furdzik Zbignew",
"Gaspar",
"Pierre PC",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361920"
}
|
Stack Exchange
|
Intuition behind Gubinelli derivative
I apologise for the confusion of the following sentences. I'm lazy to give more information about Rough path theory as Is a fairly broad subject.
On page 14 of "A Course on Rough Paths
With an Introduction to Regularity Structures" by Peter K. Friz & Martin Hairer has written:
For $\alpha \in (1/ 3; 1 /2]$, define the space of $\alpha$-Hölder rough paths (over V ),
in symbols $\mathcal C^{\alpha} ([0,T]; V )$, as those pairs $(X; \mathbb X) =: \mathbf{X}$ such that
$$
\|X\|_{\alpha}:= \sup_{
s\neq t \in [0;T ]}
\frac{|X_{s,t}|}{|t-s|^{\alpha}}
< \infty , \quad \|X\|_{2\alpha}:=\sup_{
s\neq t \in [0;T ]}
\frac{|\mathbb X_{s,t}|}{|t-s|^{\alpha}}
< \infty ,
$$
and such that the algebraic Chen relation ( is satisfed.
And on page 56 it hase written:
Given a path $X \in \mathcal C^{\alpha}([0, T ]; V )$, we say that $Y \in \mathcal C^{\alpha}([0, T ]; \hat{W} )$, is controlled by $X$ if there exists $Y' \in \mathcal C^{\alpha}([0, T ]; \mathcal L(V , \hat{W}))$, so that the remainder term $R^Y$ given implicitly through the relation $$ Y_{s,t }= Y_{s0} X_{s,t} + R_{s,t}^Y , $$ satisfes $\|R^Y\|_{ 2 \alpha}< 1$. This defines the space of controlled rough paths, $(Y, Y') \in \mathcal D_X^{2α}([0, T ]; \hat{W })$: Although $Y'$ is not, in general, uniquely determined from Y. We call any such $Y'$ the Gubinelli derivative of $Y$ (with respect to $X$). Here, $R_{s,t}^Y$ takes values in $\hat{W}$, and the norm $\| \cdot\|$.
Question : What is the intuition behind this idea of the Gubinelli derivative?
Any help is appreciated with thanks in advance.
In a way it is very much like a usual derivative. Recall first that for a regular function $Y$, its derivative $Y'_s$ at a point $s$ is the (unique) number such that
$$
Y_{t,s}=Y'_s(t-s)+ R_{s,t},
$$
where $R_{s,t}\to0$ faster than linearly. If $Y$ is twice differentiable, then $R_{s,t}\lesssim |t-s|^2$. That is, as a function of $t$, $Y_t$ "looks like" the linear function $Y_s+Y'_s(t-s)$, in the neighborhood of $s$.
Now simply replace the linear function by $X$. So we impose
$$
Y_{t,s}=Y'_sX_{t,s}+R_{s,t}
$$
with the remainder $R_{s,t}\to0$ faster than the first term, that is, faster than $|t-s|^\alpha$ (The condition $R_{s,t}\lesssim|t-s|^{2\alpha}$ from Friz-Hairer corresponds to the twice differentiable scenario in the previous case). Then as a function of $t$, $Y_t$ "looks like" the path $Y_s+Y'_sX_{s,t}$. This is great news for integration: we can of course integrate $Y_s$ against $dX_t$ (since as a function of $t$ it is just constant), and we can also integrate $X_{s,t}$ against $dX_t$ (by the definition of a rough path).
Actually, I wouldn't focus so much on assigning a meaning to $Y'$ itself, but rather focus on what the existence of a $Y'$ means for $Y$.
I like this answer very much. With this in mind, in many situations the approximation $Y_{t,s}\simeq Y'sX{t,s}$ gives a good idea of what $Y'$ should be: intuitively, if $Y=f(X)$, then $Y_{t,s}\simeq f'(X_s)X_{t,s}$; if $Y=\int A\mathrm dX$ for some process $A$, then $Y_{t,s}\simeq A_s\cdot X_{t,s}$; etc.
Morality: this derivative is none other than the usual derivative modulo the regularities as it has just been pointed out in the commentary of Pierre PC.
Cool answer! One side question, though... Does the space $\mathcal{D}_X^{2\alpha}([0,T],\mathcal{L}(V,W))$ depend on the rough path $\mathbf{X} = (X,\mathbb{X})$ or solely on the (pre-lift) path $X$? I ask this because in the book Remark 4.8 says that the Banach space of controlled rough paths depends on $\mathbf{X}$. However, nothing in its definition makes use of $\mathbf{X}$, solely $X$.
@Gaspar Indeed it only depends on $X$.
We want to define $\int_0^T f(X_s) dX_s$ for smooth bounded $f$ with bounded derivatives of all orders. Using linearity and a partition ${t_k}$ of $[0,T]$, we have
\begin{align*}\int_0^T f(X_s) dX_s&=\sum_k\int_{t_k}^{t_{k+1}}f(X_s) dX_s\\&=\sum_k\int_{t_k}^{t_{k+1}}f(X_{t_k})+f'(X_{t_k})(X_s-X_{t_k})+O(|s-t_k|^{2\alpha})dX_s\\&=\sum_k f(X_{t_k})(X_{t_{k+1}}-X_{t_k})+f'(X_{t_k})\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s+O(|t_{k+1}-t_k|^{3\alpha})\end{align*}
As $3\alpha>1$ the third term goes to zero as the mesh size goes to $0$. The first term is just a Riemann integral. The second term is the "rough path" term. $f'(X_{t_k})$ is the Gubinelli derivative and $\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s$ is your area process.
That is for a smooth function $f$ where the Gubinelli derivative is intuitively seen as a velocity or tangent vector at a point $X_{t_k}$, but what about function with some regularity. Thanks for your answer.
|
2025-03-21T14:48:31.153342
| 2020-06-01T20:38:40 |
361921
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Mizar",
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|
Stack Exchange
|
Does this formula correspond to a series representation of the Dirac delta function $\delta(x)$?
Consider the following formula which defines a piece-wise function which I believe corresponds to a series representation for the Dirac delta function $\delta(x)$. The parameter $f$ is the evaluation frequency and assumed to be a positive integer, and the evaluation limit $N$ must be selected such that $M(N)=0$ where $M(x)=\sum\limits_{n\le x}\mu(n)$ is the Mertens function.
(1) $\quad\delta(x)=\underset{N,f\to\infty}{\text{lim}}\ 2\left.\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\ n}\ \left(\left\{
\begin{array}{cc}
\begin{array}{cc}
\cos \left(\frac{2 k \pi (x+1)}{n}\right) & x\geq 0 \\
\cos \left(\frac{2 k \pi (x-1)}{n}\right) & x<0 \\
\end{array}
\\
\end{array}
\right.\right.\right),\quad M(N)=0$
The following figure illustrates formula (1) above evaluated at $N=39$ and $f=4$. The red discrete dots in figure (1) below illustrate the evaluation of formula (1) at integer values of $x$. I believe formula (1) always evaluates to exactly $2\ f$ at $x=0$ and exactly to zero at other integer values of $x$.
Figure (1): Illustration of formula (1) for $\delta(x)$
Now consider formula (2) below derived from the integral $f(0)=\int_{-\infty}^{\infty}\delta(x)\ f(x)\, dx$ where $f(x)=e^{-\left| x\right|}$ and formula (1) above for $\delta(x)$ was used to evaluate the integral. Formula (2) below can also be evaluated as illustrated in formula (3) below.
(2) $\quad e^{-\left| 0\right|}=1=\underset{N,f\to\infty}{\text{lim}}\ 4\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\ n}\frac{n\ \cos\left(\frac{2\ \pi\ k}{n}\right)-2\ \pi\ k\ \sin\left(\frac{2\ \pi\ k}{n}\right)}{4\ \pi^2\ k^2+n^2}\,,\quad M(N)=0$
(3) $\quad e^{-\left| 0\right|}=1=\underset{N\to\infty}{\text{lim}}\ \mu(1)\left(\coth\left(\frac{1}{2}\right)-2\right)+4\sum\limits_{n=2}^N\frac{\mu(n)}{4 e \left(e^n-1\right) n}\\\\$ $\left(-2 e^{n+1}+e^n n+e^2 n-e \left(e^n-1\right) \left(e^{-\frac{2 i \pi }{n}}\right)^{\frac{i n}{2 \pi }} B_{e^{-\frac{2 i \pi }{n}}}\left(1-\frac{i n}{2 \pi },-1\right)+e \left(e^n-1\right) \left(e^{-\frac{2 i \pi }{n}}\right)^{-\frac{i n}{2 \pi }} B_{e^{-\frac{2 i \pi }{n}}}\left(\frac{i n}{2 \pi }+1,-1\right)+\left(e^n-1\right) \left(B_{e^{\frac{2 i \pi }{n}}}\left(1-\frac{i n}{2 \pi },-1\right)-e^2 B_{e^{\frac{2 i \pi }{n}}}\left(\frac{i n}{2 \pi }+1,-1\right)\right)+2 e\right),\quad M(N)=0$
The following table illustrates formula (3) above evaluated for several values of $N$ corresponding to zeros of the Mertens function $M(x)$. Note formula (3) above seems to converge to $e^{-\left| 0\right|}=1$ as the magnitude of the evaluation limit $N$ increases.
$$\begin{array}{ccc}
n & \text{N=$n^{th}$ zero of $M(x)$} & \text{Evaluation of formula (3) for $e^{-\left| 0\right|}$} \\
10 & 150 & 0.973479\, +\ i\ \text{5.498812269991985$\grave{ }$*${}^{\wedge}$-17} \\
20 & 236 & 0.982236\, -\ i\ \text{5.786047752866836$\grave{ }$*${}^{\wedge}$-17} \\
30 & 358 & 0.988729\, -\ i\ \text{6.577233629689039$\grave{ }$*${}^{\wedge}$-17} \\
40 & 407 & 0.989363\, +\ i\ \text{2.6889189402888207$\grave{ }$*${}^{\wedge}$-17} \\
50 & 427 & 0.989387\, +\ i\ \text{4.472005325912989$\grave{ }$*${}^{\wedge}$-17} \\
60 & 785 & 0.995546\, +\ i\ \text{6.227857765313369$\grave{ }$*${}^{\wedge}$-18} \\
70 & 825 & 0.995466\, -\ i\ \text{1.6606923419056456$\grave{ }$*${}^{\wedge}$-17} \\
80 & 893 & 0.995653\, -\ i\ \text{1.1882293286557667$\grave{ }$*${}^{\wedge}$-17} \\
90 & 916 & 0.995653\, -\ i\ \text{3.521050901644269$\grave{ }$*${}^{\wedge}$-17} \\
100 & 1220 & 0.997431\, -\ i\ \text{1.2549006768893629$\grave{ }$*${}^{\wedge}$-16} \\
\end{array}$$
Finally consider the following three formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ where all three convolutions were evaluated using formula (1) above for $\delta(x)$.
(4) $\quad e^{-\left|y\right|}=\underset{N,f\to\infty}{\text{lim}}\ 4\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\ n}\frac{1}{4\ \pi^2\ k^2+n^2}\ \left(\left\{
\begin{array}{cc}
\begin{array}{cc}
n \cos\left(\frac{2\ k\ \pi\ (y+1)}{n}\right)-2\ k\ \pi\ e^{-y} \sin\left(\frac{2\ k\ \pi}{n}\right) & y\geq 0 \\
n \cos\left(\frac{2\ k\ \pi\ (y-1)}{n}\right)-2\ k\ \pi\ e^y \sin\left(\frac{2\ k\ \pi}{n}\right) & y<0 \\
\end{array}
\\
\end{array}\right.\right),\ M(N)=0$
(5) $\quad e^{-y^2}=\underset{N,f\to\infty}{\text{lim}}\ \sqrt{\pi}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\\\\$ $\ \sum\limits_{k=1}^{f\ n}e^{-\frac{\pi\ k\ (\pi\ k+2\ i\ n\ y)}{n^2}}\ \left(\left(1+e^{\frac{4\ i\ \pi\ k\ y}{n}}\right) \cos\left(\frac{2\ \pi\ k}{n}\right)-\sin\left(\frac{2\ \pi\ k}{n}\right) \left(\text{erfi}\left(\frac{\pi\ k}{n}+i\ y\right)+e^{\frac{4\ i\ \pi\ k\ y}{n}} \text{erfi}\left(\frac{\pi\ k}{n}-i\ y\right)\right)\right),\ M(N)=0$
(6) $\quad\sin(y)\ e^{-y^2}=\underset{N,f\to\infty}{\text{lim}}\ \frac{1}{2} \left(i \sqrt{\pi }\right)\sum\limits _{n=1}^{\text{nMax}} \frac{\mu(n)}{n}\sum\limits_{k=1}^{f n} e^{-\frac{(2 \pi k+n)^2+8 i \pi k n y}{4 n^2}} \left(-\left(e^{\frac{2 \pi k}{n}}-1\right) \left(-1+e^{\frac{4 i \pi k y}{n}}\right) \cos\left(\frac{2 \pi k}{n}\right)+\right.\\\\$ $\left.\sin\left(\frac{2 \pi k}{n}\right) \left(\text{erfi}\left(\frac{\pi k}{n}+i y+\frac{1}{2}\right)-e^{\frac{4 i \pi k y}{n}} \left(e^{\frac{2 \pi k}{n}} \text{erfi}\left(-\frac{\pi k}{n}+i y+\frac{1}{2}\right)+\text{erfi}\left(\frac{\pi k}{n}-i y+\frac{1}{2}\right)\right)+e^{\frac{2 \pi k}{n}} \text{erfi}\left(-\frac{\pi k}{n}-i y+\frac{1}{2}\right)\right)\right),\qquad M(N)=0$
Formulas (4), (5), and (6) defined above are illustrated in the following three figures where the blue curves are the reference functions, the orange curves represent formulas (4), (5), and (6) above evaluated at $f=4$ and $N=39$, and the green curves represent formulas (4), (5), and (6) above evaluated at $f=4$ and $N=101$. The three figures below illustrate formulas (4), (5), and (6) above seem to converge to the corresponding reference function for $x\in\mathbb{R}$ as the evaluation limit $N$ is increased. Note formula (6) above for $\sin(y)\ e^{-y^2}$ illustrated in Figure (4) below seems to converge much faster than formulas (4) and (5) above perhaps because formula (6) represents an odd function whereas formulas (4) and (5) both represent even functions.
Figure (2): Illustration of formula (4) for $e^{-\left|y\right|}$ evaluated at $N=39$ (orange curve) and $N=101$ (green curve) overlaid on the reference function in blue
Figure (3): Illustration of formula (5) for $e^{-y^2}$ evaluated at $N=39$ (orange curve) and $N=101$ (green curve) overlaid on the reference function in blue
Figure (4): Illustration of formula (6) for $\sin(y)\ e^{-y^2}$ evaluated at $N=39$ (orange curve) and $N=101$ (green curve) overlaid on the reference function in blue
Question (1): Is it true formula (1) above is an example of a series representation of the Dirac delta function $\delta(x)$?
Question (2): What is the class or space of functions $f(x)$ for which the integral $f(0)=\int\limits_{-\infty}^\infty\delta(x)\ f(x)\ dx$ and Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ are both valid when using formula (1) above for $\delta(x)$ to evaluate the integral and Fourier convolution?
Question (3): Is formula (1) above for $\delta(x)$ an example of what is referred to as a tempered distribution, or is formula (1) for $\delta(x)$ more general than a tempered distribution?
Formula (1) for $\delta(x)$ above is based on the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ defined in formula (7) below. Whereas the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ evaluated using formula (1) above seems to converge for $y\in\mathbb{R}$, Mellin convolutions such as $f(y)=\int\limits_0^\infty\delta(x-1)\ f\left(\frac{y}{x}\right)\ \frac{dx}{x}$ and $f(y)=\int\limits_0^\infty\delta(x-1)\ f(y\ x)\ dx$ evaluated using formula (7) below typically seem to converge on the half-plane $\Re(y)>0$. I'll note that in general formulas derived from Fourier convolutions evaluated using formula (1) above seem to be more complicated than formulas derived from Mellin convolutions evaluated using formula (7) below which I suspect is at least partially related to the extra complexity of the piece-wise nature of formula (1) above.
(7) $\quad\delta(x+1)+\delta(x-1)=\underset{N,f\to\infty}{\text{lim}}\ 2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\ n}\cos\left(\frac{2 k \pi x}{n}\right),\quad M(N)=0$
The conditional convergence requirement $M(N)=0$ stated for formulas (1) to (7) above is because the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ defined in formula (7) above only evaluates to zero at $x=0$ when $M(N)=0$. The condition $M(N)=0$ is required when evaluating formula (7) above and formulas derived from the two Mellin convolutions defined in the preceding paragraph using formula (7) above, but I'm not sure it's really necessary when evaluating formula (1) above or formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ using formula (1) above (e.g. formulas (4), (5), and (6) above). Formula (1) above is based on the evaluation of formula (7) above at $|x|\ge 1$, so perhaps formula (1) above is not as sensitive to the evaluation of formula (7) above at $x=0$. Formula (1) above can be seen as taking formula (7) above, cutting out the strip $-1\le x<1$, and then gluing the two remaining halves together at the origin. Nevertheless I usually evaluate formula (1) above and formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ using formula (1) above at $M(N)=0$ since it doesn't hurt anything to restrict the selection of $N$ to this condition and I suspect this restriction may perhaps lead to faster and/or more consistent convergence.
See this answer I posted to one of my own questions on Math StackExchange for more information on the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ and examples of formulas derived from Mellin convolutions using this representation. See my Math StackExchange question related to nested Fourier series representation of $h(s)=\frac{i s}{s^2-1}$ for information on the more general topic of nested Fourier series representations of other non-periodic functions.
Since this question mentions the Mertens function, rather than several analysis tags I would tag it as (analytic) number theory
@Mizar Thanks for your suggestion. I changed the fourier-transform tag to analytic-number-therory.
$$\int_{-\infty}^\infty \delta(x)f(y-x),dx$$ makes no sense in traditional math (e.g. see https://encyclopediaofmath.org/wiki/Generalized_function).
@user64494 $g(x)\to\delta(x)$ if $\forall,f(x)\in C^\infty_c(\Bbb{R}), \int_{-\infty}^\infty g(x)f(x)dx\to f(0)$. Most representations of $\delta(x)$ are limit representations (e.g. see formulas 34-40 at https://mathworld.wolfram.com/DeltaFunction.html and https://functions.wolfram.com/GeneralizedFunctions/DiracDelta/09/). Formula (1) above is of interest to me because it is a series representation.
My question above, original answer, and this new answer are all based on analytic formulas for
$$u(x)=-1+\theta(x+1)+\theta(x-1)\tag{1}\,.$$
and
$$u'(x)=\delta(x+1)+\delta(x-1)\tag{2}\,.$$
My question and original answer are both based on the analytic formula
$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(n)}{n}\,\left(1+2\sum\limits_{k=1}^{f\,n}\cos\left(\frac{2\,\pi\,k\,x}{n}\right)\right)\right)\tag{3}$$
where the evaluation frequency $f$ is assumed to be a positive integer and
$$M(N)=\sum\limits_{n=1}^N \mu(n)\tag{4}$$
is the Mertens function.
Formula (3) above simplifies to
$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\left(2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\ n}\cos\left(\frac{2 k \pi x}{n}\right)\right)\tag{5}$$
since
$$\sum\limits_{n=1}^\infty\frac{\mu(n)}{n}=\frac{1}{\zeta(1)}=0\,.\tag{6}$$
I originally defined formulas (3) and (5) above in this answer I posted to one of my own questions on Math StackExchange.
The formula in my question above wasn't quite right as it wasn't a smooth function at $x=0$ (there's a discontinuity in the first-order derivative corresponding to $\delta'(x)$). My original answer fixed this problem but still required the upper evaluation limit $N$ be selected such that $M(N)=0$.
This new answer is based on the analytic formula
$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \mu(n) \left(-2 f \text{sinc}(2 \pi f x)+\frac{1}{n}\sum\limits_{k=1}^{f\,n} \left(\cos\left(\frac{2 \pi (k-1) x}{n}\right)+\cos\left(\frac{2 \pi k x}{n}\right)\right)\right)\right)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \mu(n) \left(-2 f\,\text{sinc}(2 \pi f x)+\frac{\sin(2 \pi f x) \cot\left(\frac{\pi x}{n}\right)}{n}\right)\right)\tag{7}$$
which no longer requires $N$ to be selected such that $M(N)=0$.
Formula (7) above is a result related to this answer I posted to another one of my questions on Math Overflow and this answer I posted to a related question on Math StackExchange.
I believe formula (7) above is exactly equivalent to
$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{f\to\infty}{\text{lim}}\left(2 f\ \text{sinc}(2 \pi f (x+1))+2 f\ \text{sinc}(2 \pi f (x-1))\right)\tag{8}$$
in that formulas (7) and (8) above both have the same Maclaurin series.
My original answer and this new answer are based on the relationship
$$\delta(x)=\frac{1}{2}\left(u'(x+1)+u'(x-1)-\frac{1}{2} u'\left(\frac{x}{2}\right)\right)\tag{9}$$
which using formula (7) above for $u'(x)$ leads to
$\delta(x)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\mu(n) \Bigg(f (-\text{sinc}(2 \pi f (x+1))-\text{sinc}(2 \pi f (x-1))+\text{sinc}(2 \pi f x))+\right.$ $\left.\frac{1}{2 n}\left(\sum\limits_{k=1}^{f\,n}\left(\cos\left(\frac{2 \pi (k-1) (x+1)}{n}\right)+\cos\left(\frac{2 \pi k (x+1)}{n}\right)+\cos\left(\frac{2 \pi (k-1) (x-1)}{n}\right)+\cos\left(\frac{2 \pi k (x-1)}{n}\right)\right)-\frac{1}{2} \sum\limits_{k=1}^{2 f\,n}\left(\cos\left(\frac{\pi (k-1) x}{n}\right)+\cos\left(\frac{\pi k x}{n}\right)\right)\right)\Bigg)\right)$
$$=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\mu(n)\left(f (-\text{sinc}(2 \pi f (x+1))-\text{sinc}(2 \pi f (x-1))+\text{sinc}(2 \pi f x))+\frac{\sin(2 \pi f (x+1)) \cot\left(\frac{\pi (x+1)}{n}\right)+\sin(2 \pi f (x-1)) \cot\left(\frac{\pi (x-1)}{n}\right)-\frac{1}{2} \sin(2 \pi f x) \cot\left(\frac{\pi x}{2 n}\right)}{2 n}\right)\right)\tag{10}$$
I believe the formula (10) above is exactly equivalent to the integral representation
$$\delta(x)=\underset{f\to\infty}{\text{lim}}\left(\int\limits_{-f}^f e^{2 i \pi t x}\,dt\right)=\underset{f\to\infty}{\text{lim}}\left(2 f\ \text{sinc}(2 \pi f x)\right)\tag{11}$$
in that formulas (10) and (11) above both have the same Maclaurin series.
Now consider the slightly simpler analytic formula
$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1}\left(\frac{1}{2}+\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k \cos\left(\frac{\pi k x}{2 n-1}\right)\right)\right)=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1} \sec\left(\frac{\pi x}{4 n-2}\right) \cos\left(\pi x \left(2 f+\frac{1}{4 n-2}\right)\right)\right)\tag{12}$$
which also no longer requires $N$ to be selected such that $M(N)=0$.
I believe formula (12) above is exactly equivalent to formulas (7) and (8) above in that all three formulas have the same Maclaurin series.
The Maclaurin series terms for formula (12) above can be derived based on the relationship
$$\sum\limits_{n=1}^\infty\frac{\mu(2 n-1)}{(2 n-1)^s}=\frac{1}{\lambda(s)}\,,\quad\Re(s)\ge 1\tag{13}$$
where $\lambda(s)=\left(1-2^{-s}\right)\,\zeta(s)$ is the Dirichlet lambda function. I believe formula (13) above is valid for $\Re(s)>\frac{1}{2}$ assuming the Riemann hypothesis.
The relationship in formula (9) above and formula (12) for $u'(x)$ above leads to
$\delta(x)=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1}\left(\frac{3}{4}+\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k \left(\cos\left(\frac{\pi k (x+1)}{2 n-1}\right)+\cos\left(\frac{\pi k (x-1)}{2 n-1}\right)\right)\right.\right.$ $\left.\left.-\frac{1}{2}\sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k \cos\left(\frac{\pi k x}{2 (2 n-1)}\right)\right)\right)$
$=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{4}\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1} \left(\sec\left(\frac{\pi (x+1)}{4 n-2}\right) \cos\left(\pi (x+1) \left(2 f+\frac{1}{4 n-2}\right)\right)+\sec\left(\frac{\pi (x-1)}{4 n-2}\right) \cos\left(\pi (x-1) \left(2 f+\frac{1}{4 n-2}\right)\right)-\frac{1}{2} \sec\left(\frac{\pi x}{2 (4 n-2)}\right) \cos\left(\frac{1}{2} \pi x \left(4 f+\frac{1}{4 n-2}\right)\right)\right)\right)\tag{14}$
which I believe is exactly equivalent to formula (10) above and the integral representation in formula (11) above in that all three formulas have the same Maclaurin series.
$$\sum_k e^{2i\pi kx} = \sum_m \delta(x-m)$$
Convergence in the sense of distributions
$$\lim_{N\to \infty,M(N)=0}\sum_{n=1}^N \frac{\mu(n)}{n} \sum_k e^{2i\pi kx/n} =\lim_{N\to \infty,M(N)=0}\sum_{n=1}^N \mu(n) \sum_n\delta(x-mn)$$ $$=\lim_{N\to \infty,M(N)=0}\sum_{l\ge 1}(\delta(x+l)+\delta(x-l))\sum_{d| l,d\le N} \mu(d)
=\delta(x+1)+\delta(x-1)$$
Can you ground your "Convergence in the sense of distributions". TIA.
I suspect the original formula for $\delta(x)$ defined in my question above is not quite correct as the associated derived formula for $\delta'(x)$ has a discontinuity at $x=0$. The definition of $\delta(x)$ in formula (1) below eliminates the piecewise nature of my original formula which resolves this problem and also seems to provide simpler results for formulas derived via the Fourier convolution defined in formula (2) below. The formula for $\delta(x)$ defined in formula (1) below also seems to provide the ability to derive formulas for a wider range of functions via the Fourier convolution defined in formula (2) below. The evaluation limit $f$ in formula (1) below is the evaluation frequency and assumed to be a positive integer. When evaluating formula (1) below (and all formulas derived from it) the evaluation limit $N$ must be selected such that $M(N)=0$ where $M(x)$ is the Mertens function. Formula (1) is illustrated in Figure (1) further below. I believe the series representation of $\delta(x)$ defined in formula (1) below converges in a distributional sense.
(1) $\quad\delta(x)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\sum\limits_{n=1}^N\frac{\mu(n)}{n}\left(\sum\limits_{k=1}^{f\ n}\left(\cos\left(\frac{2 \pi k (x-1)}{n}\right)+\cos\left(\frac{2 \pi k (x+1)}{n}\right)\right)-\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n}\cos\left(\frac{\pi k x}{n}\right)\right)$
(2) $\quad g(y)=\int\limits_{-\infty}^\infty\delta(x)\,g(y-x)\,dx$
Formula (1) for $\delta(x)$ above leads to formulas (3a) and (3b) for $\theta(x)$ below (illustrated in Figures (2) and (3) further below) and formula (4) for $\delta'(x)$ below (illustrated in Figure (4) further below). Note formula (3b) for $\theta(x)$ below contains a closed form representation of the two nested sums over $k$ in formula (3a) for $\theta(x)$ below.
(3a) $\quad\theta(x)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\frac{1}{2}+\frac{1}{\pi}\sum\limits_{n=1}^N\mu(n)\left(\sum\limits_{k=1}^{f\ n}\frac{\cos\left(\frac{2 \pi k}{n}\right) \sin\left(\frac{2 \pi k x}{n}\right)}{k}-\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n} \frac{\sin\left(\frac{\pi k x}{n}\right)}{k}\right)$
(3b) $\quad\theta(x)=\underset{\underset{M(N)=0}{N\to\infty}}{\text{lim}}\quad\frac{1}{2}+\frac{i}{4 \pi}\sum\limits_{n=1}^N\mu(n) \left(\log\left(1-e^{\frac{2 i \pi (x-1)}{n}}\right)-\log\left(1-e^{\frac{i \pi x}{n}}\right)+\log\left(1-e^{\frac{2 i \pi (x+1)}{n}}\right)-\log\left(1-e^{-\frac{2 i \pi (x-1)}{n}}\right)+\log\left(1-e^{-\frac{i \pi x}{n}}\right)-\log\left(1-e^{-\frac{2 i \pi (x+1)}{n}}\right)\right)$
(4) $\quad\delta'(x)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\pi\sum\limits_{n=1}^N\frac{\mu(n)}{n^2}\left(\sum\limits_{k=1}^{f\ n} -2 k \left(\sin \left(\frac{2 \pi k (x-1)}{n}\right)+\sin \left(\frac{2 \pi k (x+1)}{n}\right)\right)+\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n} k\ \sin\left(\frac{\pi k x}{n}\right)\right)$
The following formulas are derived from the Fourier convolution defined in formula (2) above using the series representation of $\delta(x)$ defined in formula (1) above. All of the formulas defined below seem to converge for $x\in\mathbb{R}$. Note one of the two nested sums over $k$ in formula (6) below for $e^{-y^2}$ has a closed form representation. Both of the nested sums over $k$ in formulas (5), (8), and (9) below have closed form representations which were not included below because they're fairly long and complex.
(5) $\quad e^{-|y|}=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\sum\limits_{n=1}^N\mu(n)\ n\left(\sum\limits_{k=1}^{f\ n}\frac{2 \left(\cos\left(\frac{2 \pi k (y-1)}{n}\right)+\cos\left(\frac{2 \pi k (y+1)}{n}\right)\right)}{4 \pi^2 k^2+n^2}-\sum\limits_{k=1}^{2\ f\ n}\frac{\cos\left(\frac{\pi k y}{n}\right)}{\pi^2 k^2+n^2}\right)$
(6) $\quad e^{-y^2}=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\sqrt{\pi}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\left(\sum\limits_{k=1}^{f\ n} e^{-\frac{\pi^2 k^2}{n^2}} \left(\cos\left(\frac{2 \pi k (y-1)}{n}\right)+\cos\left(\frac{2 \pi k (y+1)}{n}\right)\right)-\frac{1}{4}\sum\limits_{k=1}^{2\ f\ n} \left(e^{-\frac{\pi k (\pi k+4 i n y)}{4 n^2}}+e^{-\frac{\pi k (\pi k-4 i n y)}{4 n^2}}\right)\right)$
$\qquad\quad=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\sqrt{\pi}\sum\limits_{n=1}^N\frac{\mu (n)}{n}\left(\frac{1}{2} \left(\vartheta_3\left(\frac{\pi (y-1)}{n},e^{-\frac{\pi^2}{n^2}}\right)+\vartheta_3\left(\frac{\pi (y+1)}{n},e^{-\frac{\pi^2}{n^2}}\right)-2\right)-\frac{1}{4} \sum\limits_{k=1}^{2\ f\ n} \left(e^{-\frac{\pi k (\pi k+4 i n y)}{4 n^2}}+e^{-\frac{\pi k (\pi k-4 i n y)}{4 n^2}}\right)\right)$
(7) $\quad\sin(y)\ e^{-y^2}=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\sqrt{\pi } \sum\limits_{n=1}^N\frac{\mu (n)}{n}\left(2 \sum\limits_{k=1}^{f\ n} e^{-\frac{\pi^2 k^2}{n^2}-\frac{1}{4}} \cos\left(\frac{2 \pi k}{n}\right) \sinh\left(\frac{\pi k}{n}\right) \sin\left(\frac{2 \pi k y}{n}\right)-\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n} e^{-\frac{\pi^2 k^2}{4 n^2}-\frac{1}{4}} \sinh\left(\frac{\pi k}{2 n}\right) \sin\left(\frac{\pi k y}{n}\right)\right)$
(8) $\quad\frac{1}{y^2+1}=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\pi\sum\limits_{n=1}^N\frac{\mu (n)}{n}\left(2 \sum\limits_{k=1}^{f\ n} e^{-\frac{2 \pi k}{n}} \cos\left(\frac{2 \pi k}{n}\right) \cos\left(\frac{2 \pi k y}{n}\right)-\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n} e^{-\frac{\pi k}{n}} \cos\left(\frac{\pi k y}{n}\right)\right)$
(9) $\quad\frac{y}{y^2+1}=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\quad\pi\sum\limits_{n=1}^N\frac{\mu(n)}{n}\left(2\sum\limits_{k=1}^{f\ n} e^{-\frac{2 \pi k}{n}} \cos\left(\frac{2 \pi k}{n}\right) \sin\left(\frac{2 \pi k y}{n}\right)-\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n} e^{-\frac{\pi k}{n}} \sin\left(\frac{\pi k y}{n}\right)\right)$
The remainder of this answer illustrates formula (1) for $\delta(x)$ above and some of the other formulas defined above all of which were derived from formula (1). The observational convergence of these derived formulas provides evidence of the validity of formula (1) above.
Figure (1) below illustrates formula (1) for $\delta(x)$ evaluated at $f=4$ and $N=39$. The discrete portion of the plot illustrates formula (1) for $\delta(x)$ evaluates exactly to $2 f$ times the step size of $\theta(x)$ at integer values of $x$ when $|x|<N$.
Figure (1): Illustration of formula (1) for $\delta(x)$
Figure (2) below illustrates the reference function $\theta(x)$ in blue and formulas (3a) and (3b) for $\theta(x)$ in orange and green respectively where formula (3a) is evaluated at $f=4$ and formulas (3a) and (3b) are both evaluated at $N=39$.
Figure (2): Illustration of formulas (3a) and (3b) for $\theta(x)$ (orange and green)
Figure (3) below illustrates the reference function $\theta(x)$ in blue and formula (3b) for $\theta(x)$ evaluated at $N=39$ and $N=101$ in orange and green respectively.
Figure (3): Illustration of formula (3b) for $\theta(x)$ evaluated at $N=39$ and $N=101$ (orange and green)
Figures (2) and (3) above illustrate formulas (3a) and (3b) above evaluate at a slope compared to the reference function $\theta(x)$, and Figure (3) above illustrates the magnitude of this slope decreases as the magnitude of the evaluation limit $N$ increases. This slope is given by $-\frac{3}{4}\sum\limits_{n=1}^N\frac{\mu(n)}{n}$ which corresponds to $-0.0378622$ at $N=39$ and $-0.0159229$ at $N=101$. Since $-\frac{3}{4}\sum\limits_{n=1}^\infty\frac{\mu(n)}{n}=0$, formulas (3a) and (3b) above converge to the reference function $\theta(x)$ as $N\to\infty$ (and as $f\to\infty$ for formula (3a)).
Figure (4) below illustrates formula (4) for $\delta'(x)$ above evaluated at $f=4$ and $N=39$. The red discrete portion of the plot illustrates the evaluation of formula (4) for $\delta'(x)$ at integer values of $x$.
Figure (4): Illustration of formula (4) for $\delta'(x)$
Figure (5) below illustrates the reference function $\frac{y}{y^2+1}$ in blue and formula (9) for $\frac{y}{y^2+1}$ above evaluated at $f=4$ and $N=101$.
Figure (5): Illustration of formula (9) for $\frac{y}{y^2+1}$
|
2025-03-21T14:48:31.154423
| 2020-06-01T20:52:29 |
361923
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361923"
}
|
Stack Exchange
|
expectation of the function of Wishart matrix eigenvalues
For Given a $N×M$ random complex gaussian matrix $X$ where $M=XX^H$, let $\lambda_1>\lambda_2>\cdots>\lambda_N$ be the ordered eigenvalues of $M$ my objective is to get an estimation of
$$
f = \left(\sum_{i = 1}^{\min(N,M)} \frac{1}{\sqrt {\lambda_i} } \right)^2
$$
Let me take $N< M$. For $N\gg 1$ the fluctuations in the sum $\sum_i \lambda_i^{-1/2}$ are smaller than the expectation value by a factor $1/N$, so we may estimate
$$\mathbb{E}[f]\approx\left(\mathbb{E}\left[\sum_i\lambda_i^{-1/2}\right]\right)^2=N^2\left(\int \rho_{\rm MP}(\lambda)\lambda^{-1/2}d\lambda\right)^2,$$
with $\rho_{\rm MP}(\lambda)$ the Marchenko-Pastur distribution.
For $N=M$ the expectation value of $f$ is divergent.
|
2025-03-21T14:48:31.154523
| 2020-06-01T21:10:37 |
361924
|
{
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"authors": [
"Nate Eldredge",
"Serge Resnick",
"https://mathoverflow.net/users/158955",
"https://mathoverflow.net/users/4832"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361924"
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Stack Exchange
|
Can one change the dimension of a Bessel process by a Girsanov change of measure?
Recall that a (squared) Bessel process $X_t$ with the dimension $\delta_0>0$ is the solution of the SDE
$$d X_t = 2\,\sqrt{X_t}\,d W_t+\delta_0\,d t.$$
A naive application of the Girsanov Theorem seems to imply that, in some equivalent new measure (i.e., absolutely continuous w.r.t. the old one), the same process may have a different value $\delta_1$ of the dimension parameter.
I believe that this is not really possible to do rigorously, at least not for an arbitrary pair of values $\delta_0,\delta_1\geq 0$. What is the exact statement of result here --- is it ever possible to change the dimension of a Bessel process by an absolutely continuous change of measure, and if so, under what conditions on $\delta_0$ and $\delta_1$?
I think this is Proposition 2.2 of Lawler's notes.
Thank you, but up to the hitting time of 0 it seems much easier: the process can be bounded from below, so the drift correction remains bounded from above and Girsanov Theorem applies. So it probably implies that if both old and new dimensions are at least 3, then the answer to my question is "yes" because the hitting time is then infinite a.e. And in the other cases?
|
2025-03-21T14:48:31.154626
| 2020-06-01T22:22:05 |
361926
|
{
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"Danny Ruberman",
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"url": "https://mathoverflow.net/questions/361926"
}
|
Stack Exchange
|
(symplectic) $h$-cobordism from $S^1\times S^2$ to itself
I ran into an oriented smooth $h$-cobordism from $S^1\times S^2$ to itself in my project. I wish to argue that it is diffeomorphic/homeomorphic to the product.
From this question 4-dimensional h-cobordisms, it seems that in dimension $4$, $h$-cobordism is automatically $s$-cobordism. But $s$-cobordism is not necessarily trivial in general.
I wonder whether one can tackle the question, with the further constraint that the boundaries are both $S^1\times S^2$.
My cobordism is also symplectic, where the boundaries are the contact boundary of the subflexible domain $T^*S^1\times \mathbb{C}$. It seems from the mentioned post symplectic property also plays a role (at least for elliptic boundaries).
You don't need any special argument to conclude that your h-cobordism is an s-cobordism: the Whitehead group of the integers is trivial. (This is an old theorem of Higman.)
|
2025-03-21T14:48:31.154703
| 2020-06-01T22:50:31 |
361928
|
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"url": "https://mathoverflow.net/questions/361928"
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Stack Exchange
|
Limit of an integral / Boundary behaviour of a Gaussian convolution / single layer potential
Let $k(t,x)$ be the transition density of Brownian motion $$ k(t,x) := \frac{1}{\sqrt{2 \pi t}} \exp \left\{ \frac{-x^2}{2t} \right\} , \quad t \geq 0, x \in {\mathbb R.}$$
Question
Let $0 < x < a$. Show that
$$ \lim_{x \nearrow a}\int_0^t \frac{a-x}{s} k(s,x-a)k(t-s,a)ds = k(t,a).$$
Can someone offer some intuition as to why this is true?
First note that
$$ \partial_x k(s, x) = -\frac{x}{s}k(s,x),
\quad s > 0, x \in {\mathbb R},$$
so your integral becomes
$$
q_a(t,x) := \int^t_0 \partial_x k(s,x-a)k(t-s,a)ds.
$$
Now suddenly your integral $q_a(t,x)$ becomes a representation of the unique classical solution to the boundary value problem for the heat equation (see e.g Cannon's book One-Dimensional Heat Equation, Ch 4):
\begin{align}
\partial_x q_a(t,x) &= \frac{1}{2} q_a(t,x), \\
q_a(0,x) &= 0, \quad x < a, \\
q_a(t,a) &= k(t,a) \quad t > 0.
\end{align}
The constant boundary $x = a$ is regular for this problem, and the solution achieves the boundary values continuously, i.e.,
$$
\lim_{x \nearrow a} q_a(t,x) = k(t,a).
$$
|
2025-03-21T14:48:31.154813
| 2020-06-01T23:15:10 |
361929
|
{
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"David Corwin",
"Laurent Moret-Bailly",
"Will Sawin",
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Stack Exchange
|
Non-representability of Weil Restriction
Let $f \colon R \to S$ be a ring homomorphism and $X$ an $S$-scheme. We define the functor $R_{S/R}(X)$ on $R$-algebras $T$ by
$$
R_{S/R}(X)(T) = X(T \otimes_R S).
$$
If $S/R$ is a field extension (more generally, finite and locally free), and $X$ is quasi-projective, then this functor is known to be representable.
Is there an example where this is not representable? Can one find an example that is smooth and finite type over $S$? What about if $X$ is a simple non-separated scheme, such as the affine line with two origins?
I should note that I found a related discussion for the Hilbert functor in https://arxiv.org/pdf/math/0603473.pdf
I think for a counterexample we can take $R$ a DVR, $S$ the field of fractions, and $X = \mathbb A^1$. Or are we still assuming finite locally free?
Let $R =\mathbb R$, $S = \mathbb C$ (any quadratic field extension works) and let $X$ be the doubled line.
We have a map of functions from the Weil restriction of $X$ to the Weil restriction of $\mathbb A^1$, which is $\mathbb A^2$. So if the representing scheme exists then it must map to $\mathbb A^2$.
Now take $T = \mathbb C[x,y]$ so $T \otimes_R S = \mathbb C[x,y] \times \mathbb C[x,y]$. Consider the map from $T \otimes_R S$ to $X$ where the coordinate on the affine line is $ (x+iy, x-iy)$ but on the first component we map to the affine line with the first origin and on the second we map to the affine line with the second origin.
This defines a map from $T$ to the Weil restriction. If the Weil restriction is a scheme then there must be an open affine $\operatorname{Spec} T'$ in $\operatorname{Spec} T$, that maps to an open affine $\operatorname{Spec} U$ in the Weil restriction, which maps to the Weil restriction, which maps to $\mathbb A^2$. Dualizing we have maps of rings $\mathbb R[x,y] \to \mathbb C[x,y] = T \to T'$ and $\mathbb R[x,y] \to U \to T'$ forming a commutative diagram.
The image of $U$ inside $T'$ either is contained in $\mathbb R(x,y)$ or not. In either case we will derive a contradiction.
If the image is contained in $\mathbb R[x,y]$, then because $\operatorname{Spec}
T'$ contains the origin, the image must be a subring of $\mathbb R(x,y)$ that the map $\mathbb R[x,y] \to \mathbb C$ sending $x$,$y$ to $0$ factors through. Thus it must be $\mathbb R[x,y]$ adjoining some rational functions that are well-defined at $0$. So in fact we have a factorization $\mathbb R[x,y] \to U \to \mathbb R \to \mathbb C$ sending $x,y$ to $0$. But this implies that the $T$-point of the Weil restriction we wrote down earlier, restricted to the origin, descends to $\mathbb R$, which would make it Galois-invariant. But it is not Galois-invariant, because the two components, exchanged by Galois, map to the two different origins.
If the image is not contained in $\mathbb R$, then it contains some rational function $f$ in $\mathbb C(x,y)$. Choose some $u,v \in \mathbb R$, not both $0$, such that the point $(u,v) \in \mathbb A^2$ lies in $\operatorname{Spec} T'$, and for which $f(u,v)$ is well-defined but not an element of $\mathbb R$. This is easy because these are all generic conditions. For such an $x,y$, the induced map $\mathbb R[x,y] \to U \to T' \to \mathbb C$ sending $x$ to $u$ and $y$ to $v$, sends $f$ to $f(x,y)\notin \mathbb R$ and thus has image $\mathbb C$.
Now if the map from the ring of functions of any affine neighborhood of a $\mathbb C$-point of a scheme to $\mathbb C$ has image $\mathbb C$, that point does not descend to $\mathbb R$ (because that can be checked affine-locally). However, our chosen point $(u,v)$ does descend to $\mathbb R$, because $u, v \in \mathbb R$ and $u,v$ are not both $0$ so $u+iv, u-iv \neq 0$, and thus we lie in the affine line with doubled origin. So this contradicts our conclusion that the image of $U$ in $\mathbb C$ is $\mathbb C$.
After writing this I came up with a probably faster proof which is to compute the base change of the Weil restriction from $\mathbb R$ to $\mathbb C$ - it's the product of the affine line with two origins with itself - and check that one of the points lying over the origin doesn't have any Galois-invariant affine-neighborhood, which it would if the scheme descended from $\mathbb C$ to $\mathbb R$.
For a separated example, see my answer here: https://mathoverflow.net/q/194326
|
2025-03-21T14:48:31.155091
| 2020-06-02T00:41:43 |
361932
|
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"François Brunault",
"https://mathoverflow.net/users/120548",
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"xir"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361932"
}
|
Stack Exchange
|
Jacobi forms and Kato's modular units
this is pretty much just a silly literature question; apologies in advance. Kato uses the following theta function (or slight variants thereof) in his construction of his Euler system:
$$\Theta(\tau, z) = q^{1/12}(e^{\pi iz} - e^{-\pi iz}) \prod_{n\ge 1} (1-q^n e^{2\pi iz})(1-q^ne^{-2\pi i z})$$
where $q=e^{2\pi i \tau}$. This is the theta function I like as someone who leans heavily towards the algebraic side of things; it's essentially the de Rham realization of a polylogarithm class in the $H^{1,1}$ motivic cohomology of some universal elliptic curve. However, I can't quite figure out the relation to some classical formulas for the Jacobi theta function. It's been stated in several places (for example, Scholl's notes) that the above theta function is essentially the same as the classical "half-integral weight Jacobi form," which I guess is the series
$$ \vartheta(\tau, z)=\sum_{n\in \mathbb{Z}} e^{\pi i n^2 \tau + 2\pi i n z}= \sum_{n\in \mathbb{Z}} q^{n^2/2} e^{2\pi i n z}$$
which has the similar-looking product expansion
$$\vartheta(\tau, z)= \prod_{n=1}^\infty (1-q^n)(1-e^{2\pi i z}q^n)(1-e^{-2\pi iz}q^n)$$
(In particular, Scholl says it essentially coincides with the Jacobi theta series $\vartheta_1$, which didn't prove so helpful since I've found that naming conventions for these several classical Jacobi forms are really confusing and inconsistent in the literature. But all of them have very similar forms and are related by a few simple transformations, from what I can tell. It seems they're basically the orbit of the Jacobi form I wrote down under translation by the half-integral lattice, I gather this has something to do with transformations under the action of the metaplectic cover but don't fully understand the significance.)
These almost look same, but I can't quite figure how to get from one to the other. The denominator in the leading power of $q$ bothers me since it implies a monodromy obstruction in Kato's case. Most notably, the latter function is a triple product whereas the former isn't; the Jacobi triple product identity is my understanding of how to expand the classical Jacobi-type products, so the fact that Kato's theta function is missing one of the "triple product" ingredients throws me off, and I can't work out how to relate them.
Is this discussed somewhere in the literature? I'd really love to be able to clarify the relationship here.
Wikipedia does not quite have the same expression for Jacobi's $\vartheta$, see https://en.wikipedia.org/wiki/Jacobi_triple_product
ah sorry my fault, i was silly and mixed up the formulas. i'll fix it
The triple product in the Jacobi theta function $\vartheta(\tau,z)$ can be rewritten
\begin{equation*}
\prod_{n \geq 1} (1-q^n) \prod_{n \geq 0} (1-q^n e^{2\pi i(z+\frac{\tau}{2}+\frac12)}) \prod_{n \geq 1} (1-q^n e^{-2\pi i(z+\frac{\tau}{2}+\frac12)})
\end{equation*}
($2\pi i n$ should be replaced by $2\pi iz$ in your equation).
On the other hand, the theta function used by Kato is essentially the Weierstrass $\sigma$ function (see for example Silverman, Advanced topics in the arithmetic of elliptic curves, I.6.4). This is not surprising as the $\sigma$ function can be used to construct elliptic functions with prescribed divisors (op. cit., I.5.5).
If you work things out, you will see that the discrepancy is essentially the cube of the infinite product $\prod_{n \geq 1} (1-q^n)$, which brings into play $\Delta^{1/8}$, explaining the factor $q^{1/8}$.
Note that Kato really uses the function $\Theta(\tau,z)^{c^2} \Theta(\tau,cz)^{-1}$, which is associated to the divisor $c^2(0) - \sum_{x \in E[c]} (x)$ on the universal elliptic curve. Since this divisor has degree 0, the term $\prod (1-q^n)$ will cancel out.
oh this fully explains it, thank you! it should've been plain to me if i hadn't gotten all muddled by an irrational fear of big series
|
2025-03-21T14:48:31.155355
| 2020-06-02T00:46:12 |
361933
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361933"
}
|
Stack Exchange
|
Closed curves with minimal total curvature in the unit circle
Chakerian proved in this paper that a closed curve of length L in the unit ball in $\mathbb{R}^n$ has total curvature at least L.
In this later paper Chakerian gave a simpler proof and noted that equality holds iff the curve is of length $2\pi n$ and winds round the unit circle $n$ times.
Assume that the closed curve is smooth (no corners) and we are working in $\mathbb{R}^2$. My question is:
For closed curves with length $4\pi> L>2\pi$ within the unit circle in $\mathbb{R}^2$ (and hence non-convex) what is the minimal total curvature and for which curves is this attained?
|
2025-03-21T14:48:31.155691
| 2020-06-02T01:48:24 |
361935
|
{
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"Steven Stadnicki",
"Timothy Chow",
"Ville Salo",
"https://mathoverflow.net/users/123634",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361935"
}
|
Stack Exchange
|
Does the Angel have to be really smart?
My question is about the computational complexity of the Angel's strategy in the Angels and Devils game, tl;dr does the Angel have a polynomial time strategy.
I'm a big Conway fan, so as you can guess I found myself looking through his works in April. One thing that occurred to me was it might be fun to implement the Angels and Devils game, as I have never actually played it / seen it played, and I figured it's not the kind of game you want to play with a human, since it seems likely that the interesting stuff happens only in very large-scale games.
Let me recall the $k$-angel game for completeness. Two players, Angel and Devil, playing alternately. The set of game positions is
$$ X = \{(x, p) \in \{0,1\}^{\mathbb{Z}^2} \times \mathbb{Z^2} \;|\; x_p = 0 \}, $$
and the initial position is $(0^{\mathbb{Z}^2},(0,0))$. A move of the Angel consist in moving $p$ by $k$ chess king moves so that after the move the new pair $(x,p')$ is in $X$ (so the constraint on $p'$ is $|p - p'|_\infty \leq k$). A move of the Devil consists in turning a coordinate of $x$ to $1$, so that after the move the new pair $(x',p)$ is in $X$. The Devil wins if the Angel does not have a valid move after finitely many steps, and otherwise the Angel wins.
This is an open game so one of the players have a winning strategy for each $k$. Indeed, it was proved in 2007 that the $2$-angel wins in Máthé, András, The angel of power 2 wins, Comb. Probab. Comput. 16, No. 3, 363-374 (2007). ZBL1222.91012. and also in Kloster, Oddvar, A solution to the angel problem, Theor. Comput. Sci. 389, No. 1-2, 152-161 (2007). ZBL1210.91023. . (Two more papers from the same year proved this for larger $k$.)
Let's concentrate on the $2$-angel situation, I think it's the most interesting one, especially in light of how the solutions work. (Also, it turns out that there is a way to generalize this to real speeds so that any "$(2-\epsilon)$-angel" loses, so $2$ is the most interesting speed in that sense too.)
Quick descriptions of the $2$-angel solutions. The paper of Máthé is very short and easy: the strategy is "follow the left wall", up to a reduction to another game (where there are some additional requirements for the players). The reduction is also not much different than one already explained by Conway's 1996 advertisement paper for this problem. So obviously my first thought was to implement that. Unfortunately it seems that Máthé's proof does not actually easily give a computable strategy; you get an "explicit" strategy, but you can't actually compute moves from it algorithmically, or at least I don't see how to.
Kloster's paper has a similar strategy, but a longer proof. You walk along the left border of the set of positions you have deemed dangerous (I don't recall the exact terminology they use). Kloster's strategy is explicitly computable, as they themselves note, but you have to quantify over all / an exponential number of paths so it's not so great from a computational complexity perspective.
So let me state my questions now. Observe first that we may clearly base our strategies on only the current state of the board matters, so a strategy for either player is just a function that produces a new state from old, according to the rules. I am interested in computational complexity of strategies, so we have to be a bit more careful with encodings than for just computability considerations. I give two possible formulations that I think are in the correct spirit.
Define the list encoding of $(x,p) \in X$ as the word $w_0 3 w_1 3 w_2 3 ... 3 w_n$ where $w_i \in \{0,1\}^* 2 \{0,1\}^*$, and $w_0$ gives the coordinates of $p$ in binary, while other $w_i$ list the support of $x$.
Is there a polynomial time winning strategy for the Angel in the $2$-angel game? I.e., is there a function in FP that gives a winning strategy for the $2$-angel, when the current position is given with the list encoding?
More generally, "what is the complexity"? (One could obviously also ask things like, if the Devil is playing with a polynomial strategy, can the Angel win with a polynomial strategy, and so one.)
Another possibility is to define the $n$-finitary $2$-angel game as the same game played with $\mathbb{Z}$ replaced by the box $\{-n, -n+1, ..., n\}^2$, and the Angel wins by exiting the box. (There's a relevant compactness phenomenon so this captures the same game as the original in a sense.) Now we can give the complete board state explicitly to each player.
Is there a polynomial time winning strategy for the Angel in the $n$-finitary $2$-angel game?
More generally, "what is the complexity"? In this sense, Kloster's strategy is explicitly in EXPTIME if I'm not mistaken. You quantify over paths that minimize a certain cost function that I won't reproduce here. I don't know if this strategy can already be computed in polynomial time. I would not be surprised if it is possible to encode some NP-hard (or more) problem to it at the very least. (Also I think just from general considerations there is a PSPACE strategy.)
As for implementing the game, since this turned out to be a can of worms I ended up implementing Conway's Soldiers instead, as I hadn't played that either. It turned out to be the easiest game ever. If this question is not answered, I may later try implementing the obvious heuristic strategy based on Kloster's proof, i.e. pick random paths (with some reasonable distribution), optimize locally, take the path that minimizes the cost. If you have more intelligent practical suggestions you can share those too.
In the $n$-finitary case, I assume you're giving the board in unary (i.e., as a 0-1 sequence with a 1 denoting a destroyed square)? Then it seems that your list encoding could be exponentially more succinct than a unary encoding, and that could affect the definition of "polynomial time." Maybe this doesn't matter because things only get interesting when a large fraction of squares in the vicinity of the Angel have been destroyed, but at first glance the succinctness of the list representation seems to make the problem harder.
You are correct about how I meant the board to be encoded in the $n$-finitary case. You are also correct about the exponential succinctness, and things could absolutely depend on these encoding issues. If someone can give an answer with some other encoding, or explain why some encodings give a silly or non-silly hardness result, that is already something.
The situation is actually a bit more subtle than just one encoding being more succinct than the other. By basic CS nonsense (it's a polylength game), for the $n$-finitary case there is an optimal "PSPACE-Angel", i.e. better than what I got from Kloster's paper. The Angel wins the infinitary game, so this PSPACE strategy wins all $n$-finitary games. Yet the CS doesn't imply there is any computable strategy as far as I can tell. It could be the for the first gazillion $n$ the Angel's winning strategy can start by walking upwards for two steps, but after that it has to start by moving diagonally.
(My guess is for practical purposes, the PSPACE-Angel provided by theory is worse than the EXPTIME algorithm I extracted from Kloster's paper. Maybe it's possible to do Kloster's algorithm in PSPACE by being a bit smarter. Of course we don't even know $\mathrm{PSPACE} \neq \mathrm{EXPTIME}$. And for practical purposes I guess PSPACE isn't much better than EXPTIME.)
Another question related to the issues of complexity: since the Angel wins it's hard to speak of an 'optimal' strategy for the Devil, but let's assume that D's goal is to (asymptotically) minimize $\max(|p|)$. What's the diameter of the Devil's 'working space'; i.e., given that $n$ moves have happened, what's the max of $|x|$ that a Devil needs to consider for good moves under some metric of good?
|
2025-03-21T14:48:31.156205
| 2020-06-02T02:46:22 |
361936
|
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"Angelo Lucia",
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|
Stack Exchange
|
Mackey theory for semidirect products: equivalence between constructions for modules
I am learning about the so-called "Mackey Machine" for unitary irreps of semidirect products of locally compact groups. Let $G = N \rtimes H$ where $N$ is a closed normal abelian subgroup and $H$ is a closed subgroup which acts on $N$ by $\phi: H \to Aut(N)$. Given an unitary irrep $\sigma$ of $H$ and a irreducible representation $p$ of $N$, I have found two (I believe equivalent) ways in which an irrep of $G$ is constructed by different authors.
If $H_p$ is the semigroup of $H$ which stabilizes $p$, and $G_p = N \rtimes H_p$, then we have the following commutative diagram of inclusions.
$\require{AMScd}$
\begin{CD}
H_p @>>> H\\
@VVV @VVV\\
G_p @>>> G
\end{CD}
If I am understanding it correctly, Folland, in A Course in Abstract Harmonic Analysis (see the discussion leading to Theorem 6.43, on pp. 199-201 of the second edition) goes "first down then right": defines a representation of $G_p$ as $p\otimes \sigma$, then takes the representation of $G$ induced by it (i.e. the same way it is presented in this question). On the other hand, Etingof et al. - Introduction to representation theory (section 4.26 on pp. 76) takes the route "right then down": first they consider the representation of $H$ induced by $\sigma$, then they extend it to $G$ with a skew-product.
Question 1: How to express these two constructions in terms of modules?
Question 2: Since the two constructions are equivalent, the modules resulting from the first and the second path should be isomorphic. Can we see that isomorphism explicitly?
Let me pretend that the groups are finite, since I believe it should work equally well without any topological complicacies.
What I tried
Let $\mathcal{H}_\sigma$ the $\mathbb{C}(H_p)$-module given by $\sigma$. Since $p$ is a 1-dimensional representation of $N$, we can give to $\mathbb{C}$ the structure of a $\mathbb{C}(N)$ module, which I will denote with $\mathbb{C}_p$. Then Folland's construction is (I am keeping the structure of the diagram above, to match each module to the corresponding group, but the arrow have no further meaning):
\begin{CD}
\mathcal{H}_\sigma \\
@VVV \\
\mathbb{C}_p \otimes_{\mathbb{C}(H_p)} \mathcal{H}_\sigma @>>> \mathbb{C}(G) \otimes_{\mathbb{C}(G_p)} \mathbb{C}_p \otimes_{\mathbb{C}(H_p)} \mathcal{H}_\sigma
\end{CD}
Let me explain what I think should be going on: $\mathbb{C}_p$ can be made into a $(\mathbb{C}(G_p), \mathbb{C}(H_p))$-bimodule, thanks to the action $\phi$ (which stabilizes $p$).
But I am not sure what module is given by the other route
\begin{CD}
\mathcal{H}_\sigma @>>> \mathbb{C}(H) \otimes_{\mathbb{C}(H_p)} \mathcal{H}_\sigma \\
@. @VVV\\
@. ?
\end{CD}
Although the conceptual meaning is pretty clear from the context, it may be worth clarifying that your arrows between Hilbert spaces are "construct from" arrows, not Hilbert-space morphisms.
Also, (1) where do these two constructions occur? (The closest I can find is Folland, Theorem 6.38; Etingof et al., Section 4.26.) (2) Where are the parentheses in $\mathbb C_p \rtimes \mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$? (3) In the notation $a\phi_h(b)$, I think that $b$ is a complex number, but $\phi_h$ is an automorphism of $N$; so what does $\phi_h(b)$ mean?
If I'm right to look at Folland, Theorem 6.38, and Etingof et al., Section 4.26, then the two constructions they describe, in 'module language', are $\mathbb C(G) \otimes_{\mathbb C(G_p)} (\mathbb C_p \otimes_{\mathbb C} \mathcal H_\sigma)$ (where the $N$ piece of $G_p = N \rtimes H_p$ acts only on $\mathbb C_p$, and the $H_p$ piece acts only on $\mathcal H_\sigma$); and $\mathbb C_p \otimes_{\mathbb C} (\mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma)$ (similar action convention). It is clear that these are the same construction.
($(n \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v) \mapsto p(n)z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v)$ and $(1 \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v) \leftarrow z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v)$ (stuck with $\leftarrow$ because we don't seem to have \mapsfrom).)
@LSpice You are right, I made a mistake: the skew-product I wrote did not make any sense. I have added some more specific references, and clarified the meaning of the arrows as you suggested. Since my construction was flawed, I have modified my answer so that you can post your both comments as an answer.
By request, I post my comments (1 2). I think that they answer the revised question, but let me know if not.
Question 1. Folland's construction corresponds to extending a $\mathbb C(H_p)$-module $\mathcal H_\sigma$ of $H_p$ across $N$ by $p$ to $G_p = N \rtimes H_p$, giving $\mathbb C_p \otimes_{\mathbb C} \mathcal H_\sigma$; and then inducing up to $G$, giving $\mathbb C(G) \otimes_{\mathbb C(G_p)} (\mathbb C_p \otimes_{\mathbb C} \mathcal H_\sigma)$. (I'm not sure how to make sense of your proposed $\mathbb C_p \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$, since $\mathbb C_p$ is not a $\mathbb C(H_p)$-module in any obvious-to-me way; notice that, since $\mathbb C_p$ is $1$-dimensional, specifying the module structure would be equivalent to specifying a homomorphism $H_p \to \mathbb C^\times$. The tensor product over $\mathbb C(H_p)$ will collapse to a $1$-dimensional representation, on which $H_p$ acts trivially (if $\mathcal H_\sigma$ is not isomorphic to $\mathbb C_p$ as $\mathbb C(H_p)$-modules) or by the relevant character (if they are isomorphic).)
Etingof et al.'s construction corresponds to inducing $\mathcal H_\sigma$ up to $H$, giving $\mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$; and then extending across $N$ by $p$ to $G = N \rtimes H$, giving $\mathbb C_p \otimes_{\mathbb C} (\mathbb C(H) \otimes_{\mathbb C(H_p)} \mathcal H_\sigma)$.
Question 2. The isomorphism from Folland's to Etingof's construction is given by $(n \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v) \mapsto p(n)z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v)$. The inverse isomorphism is given by $z \otimes_{\mathbb C} (h \otimes_{\mathbb C(H_p)} v) \mapsto (1 \rtimes h) \otimes_{\mathbb C(G_p)} (z \otimes_{\mathbb C} v)$.
Regarding your question about the $\mathbb{C}(H_p)$ structure of $\mathbb{C}_p$: I think that every character $p:N \to \mathbb{C}^{\times}$ (I am abusing notation and calling it $p$ again) could extended to a character of $G_p$ (and thus by restriction of $H_p$), as $\tilde{p}(nh) = p(n)$ for $n\in N$ and $h\in H_p$. This is an homomorphism due to the fact that $G_p$ leaves $p$ invariant, so if $n,m\in N$ and $h,k \in H_p$ then $\tilde{p}(nh)\tilde{p}(mk) = \tilde{p}(n) \tilde{p}(m)$ is the same as $\tilde{p}(nhmk) = \tilde{p}(n\phi_h(m)hk)=\tilde{p}(n)\tilde{\phi_h(m)}$.
@AngeloLucia, you are right that it is a homomorphism; it corresponds to choosing the trivial action of $H_p$ on $\mathbb C_p$. Then we always have $\mathbb C_p \cong \mathbb C_p \otimes_{\mathbb C(H_p)} \mathcal H_\sigma$ as $\mathbb C(G_p)$-modules via any map $z \mapsto z \otimes_{\mathbb C(H_p)} v$ with $v \ne 0$, which doesn't seem to be what you want.
|
2025-03-21T14:48:31.156756
| 2020-06-02T02:56:08 |
361937
|
{
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"Chris",
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|
Stack Exchange
|
Complex Monge-Ampere equation with degenerate right hand side
Given a Kahler manifold $(X, \omega_0)$, and a smooth function $f$, suppose that I have a smooth solution to the following complex Monge-Ampere equation:
$(\omega_0 +i \partial \bar \partial \varphi)^n = e^{f} \omega_0^n$
By the classical result of pluripotential theory, $\sup_{X}|\varphi|$ can be estimated by $\|e^f\|_{L_p( \frac{\omega_0^n}{n!})}$, for $p > 1$. Now suppose instead I only have control for $\|e^{f+g}\|_{L_p( \frac{\omega_0^n}{n!})}$, where $g$ is $-\infty$ on some measure zero set and smooth outside this set. I wonder whether there has been any results on bounding $sup_{X} |\varphi|$ in terms of $\|e^{f+g}\|_{L_p( \frac{\omega_0^n}{n!})}$.
By degenerate I mean that $e^{pg} \frac{\omega_0^n}{n!}$ is a positive measure but not a volume form. Not sure whether this is the right terminology.
The same paper that proves the classical result that you state also proves what you want: https://projecteuclid.org/euclid.acta/1485891130 Theorem 2.5.2 and Example 2 after that
I see, and thank you. Never really actually read the original paper.
|
2025-03-21T14:48:31.156870
| 2020-06-02T03:30:57 |
361940
|
{
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"authors": [
"Carlo Beenakker",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361940"
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|
Stack Exchange
|
What is the current state of research in Chern-Simons theory?
I'm a PhD student in mathematical physics looking for some orientation. As asked in the title, I would like to know the current state of research in Chern-Simons theory. More specifically, what are some of the directions that people are currently pursuing in this field.
I did not ask for more general topological field theories, but this is largely due to personal interests. Of course, answers related to other TFTs are more than welcome.
I feel this question has been asked before. But since this is a question about the current state of research, I think it deserves an update.
Thank you very much.
this book from 2009 gives an overview of open problems, and for a mature field such as this it should be recent enough: https://books.google.nl/books?id=Qex4AAAAQBAJ
|
2025-03-21T14:48:31.156967
| 2020-06-02T04:56:49 |
361942
|
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"Groups",
"John Shareshian",
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|
Stack Exchange
|
Existence (or the number) of generating triple of involutions of $\operatorname{PGL}_2(p)$ with some conditions
Let $G=\operatorname{PGL}_2(p)$, where $p\ge 5$ is a prime. Is there a generating triple of involutions $(x,y,z)$ of $G$ such that $|xy|=p$, $|xz|=p+1$ and $|yz|=p-1$? That means, $\langle x,y,z\rangle=G$ with
$x^2=y^2=z^2=1$;
$\langle x,y\rangle \cong D_{2p}$;
$\langle x,z\rangle\cong D_{2(p+1)}$;
$\langle y,z\rangle\cong D_{2(p-1)}$.
If there exists, then what is the number of such triples?
We know that when $p\equiv 1\pmod 4$ we have $x,y\in\operatorname{PSL}_2(p)$, while when $p\equiv 3\pmod 4$ there is exactly one of them in $\operatorname{PSL}_2(p)$ (but we do not know which one is).
Is there any result on this?
Such triples exist, I think.
First, embed $PGL_2(p)$ in $S_{p+1}$ through its action on $1$-spaces from ${\mathbf F}_p^2$. This maps elements of order $p+1$ in $PGL_2(p)$ to $(p+1)$-cycles, and elements of order $p-1$ to $(p-1)$-cycles. In particular, every element of order $p+1$ or $p-1$ gets mapped to an odd permutation.
There are two conjugacy classes of elements of order two in $PGL_2(p)$. Elements from one class fix two $1$-spaces and elements from the other class fix none. So, elements of one class map to even permutations and elements of the other map to odd permutations under the given embedding.
It follows now that if $a,b \in PGL_2(p)$ have order two with $|ab| \in \{p-1,p+1\}$ then one of $a,b$ fixes two points and the other fixes none.
The stabilizer $B$ of a $1$-space in $PGL_2(p)$ is an extension of a cyclic group of order $p$ by a cyclic group of order $p-1$ that acts faithfully. It follows that if $a,b \in B$ are distinct elements of order two, then $|ab|=p$.
Now we are in good shape. Pick $g \in PGL_2(p)$ with $|g|=p+1$. Let $z=g^{(p+1)/2}$. As $g$ acts as a $(p+1)$-cycle, $z$ is a fixed-point-free involution. Now find involutions $a,b$, each with two fixed points, such that $|az|=p+1$ and $|bz|=p-1$. (This is certainly possible by the arguments above.) Now pick any $1$-space $V$ from ${\mathbf F}_p^2$. As $\langle g \rangle$ acts transitively on $1$-spaces, there exist a $\langle g \rangle$-conjugate $x$ of $a$ and a $\langle g \rangle$-conjugate $y$ of $b$ such that both $x$ and $y$ fix $V$.
As $g$ centralizes $z$, we get that $|xz|=p+1$ and $|yz|=p-1$. So, $x \neq y$. Now, as $x$ and $y$ fix a common $1$-space, $|xy|=p$.
Many thanks for your answer but I have some questions on your solution. Firstly I think the involutions without any fixed points could also be mapped to an even permutation when $p\equiv 3\mod 4$. And could you give more details on the existence of involutions $a,b$ making $|az|=p+1$ and $|bz|=p-1$?
You are correct that fixed-point-free involutions will act as even permutations when $p \equiv 3 \bmod 4$. This does not matter. Elements of one of the two classes will act as even permutations and elements of the other will act as odd permutations. Which does which depends on $p \bmod 4$. The key point is that elements of order $p-1$ or $p+1$ will act as odd permutations and therefore, if one of these elements is the product of two involutions, then these two involutions are not in the same class. So, one of the two involutions is fixed point free and the other is not.
As you say, $PGL_2(p)$ contains dihedral subgroups of order $2(p+1)$ and $2(p-1)$. Therefore, there exist elements $r$ and $s$ of respective orders $p+1$ and $p-1$ and involutions $c,d,f,h$ such that $cd=r$ and $fh=s$. By comment (1) above, we may assume that $d$ is fixed point free and so is $h$. Now find $u,v$ such that $u^{-1}du=z$ and $v^{-1}hv=z$. Set $a=u^{-1}cu$ and $b=v^{-1}fv$.
You might find helpful the paper of Liebeck and Shalev "Classical Groups, Probabilistic Methods, and the (2,3)-Generation Problem". There, they show that there are three involutions that generate all but finitely many simple classical groups. Indeed, the same is true if you require the involutions to be conjugate.
Their technique is to choose elements at random, and show that with high probability they avoid being all contained in a maximal subgroup. Perhaps this technique can be adjusted for $PGL$, or perhaps their result for $PSL$ will suffice for your application.
Results of the type that you're looking for often go along with work on the (2,3)-generation problem, which asks whether a group is generated by an involution together with an element of order 3. (This question is interesting partly because such groups are exactly the quotients of the modular group $PSL_2(\mathbb{Z})$.)
Thank you for your answer. However, I think in this case the probabilistic method cannot apply because both $\langle x,z\rangle\cong D_{2(p+1)}$ and $\langle y,z\rangle\cong D_{2(p-1)}$ are maximal in $G$ and we have so many restrictions on the orders of products.
|
2025-03-21T14:48:31.157289
| 2020-06-02T05:19:30 |
361945
|
{
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"authors": [
"Ben McKay",
"Gerry Myerson",
"Lagrida Yassine",
"https://mathoverflow.net/users/13268",
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"https://mathoverflow.net/users/51189",
"zeraoulia rafik"
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"url": "https://mathoverflow.net/questions/361945"
}
|
Stack Exchange
|
For which values of $ x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges?
The copy of this question is posted here
I have tried to to determine values of real $x$ for which $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges but I can't , Presumably the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ then $\tan(n!)$ will be large arbitrarily infinitely often, and the difference $\tan(n!)-n!$ would be probably small. However I don't know anything about limit of the fraction $\frac{\tan(n!)}{n!}$ for large $n$, And I have used Stiriling formula but it weren't helpful, This really forbide me to determine how $x$ values should be to get the titled series converge, then my question is :
Question:
For which values of real number $x$ we have $\sum\limits_{n=0}^{\infty}x^{\tan(n!)-n!}$ converges ? We may avoid trivial case $x=0$ , In particular how we prove the set of values $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ ?
Note:I have used this trick for approximation : $n! = e(-1)^{n+1}/(n+1) + O(1/n^2) (\mod \mathbb Z)$ implies that $\tan(\pi n!)-n!=\pi e((-1)^{n+1}/(n+1) + O(1/n^2)) -\frac{1}{\pi})(\mod \mathbb Z)$ but this what could saying to me ? Probably conditional convergence could be happen here .
Addedendum
I'm interested to evaluate the above series because this need to prove $\{ n! \bmod 2\pi\mathbb Z\}$ is dense in $[0,2\pi]$ which is related to irrationality measure and Bounds approximation of $\pi$ which it is interesting in number theory and in the same time looking to the behavior of $\tan(n!)-n!$ for large $n$ which it help to get bounds for some arithmitic functions , Also to loook to the relationship of convergence of series to lacunar Fourier series which we have a lot of results on convergence for them
using mathematica code for $n=400$, $\tan(n!)-n!$ I have got the below plot and this is the code ,
Block[{$MaxExtraPrecision = 800},
lst = Table[Tan[n!] - n!, {n, 0, 400}];
ListPlot[N[Accumulate[lst], 200], PlotRange -> All]
]
Does this question fit into a larger picture? Why is this question important to you?
@BenMcKay, Sorry I missed to add motivation of that question, I'm already adding that in Addedendum
However , I don't know why downvote however that question is in the level of research and in the same time I have showed its motivation
Your question is intersting whene i plot the function $f(t)=\tan(t)-t$, we can see that almost values of $f(n)$ with $n$ positive integers are negative and diverge fast.
@Lagrida , probably you missed factorial in t , and in the same time the power of x is the approximation of n! by (2k+1) pi/2
12 versions in 12 hours.
|
2025-03-21T14:48:31.157497
| 2020-06-02T05:42:34 |
361947
|
{
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"authors": [
"Jochen Glueck",
"Ramiro Lafuente",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/14708",
"https://mathoverflow.net/users/33741",
"leo monsaingeon"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361947"
}
|
Stack Exchange
|
Principal eigenvalue of non self-adjoint elliptic operators on closed manifolds
Consider the elliptic operator $Lu = - \Delta u + \langle \nabla u , X \rangle + c \, u $ acting on functions on a closed Riemannian manifold $M$. Here $\Delta$ denotes the Laplace-Beltrami operator, $X$ is an arbitrary smooth vector field, and $c \geq 0$ is a smooth function on $M$ which does not vanish identically. Does $L$ have a so-called `principal eigenvalue' $\lambda_1 > 0$, whose corresponding (unique up to scaling) eigenfunction does not change sign?
A similar statement holds for smooth domains in $\mathbb{R}^n$, as shown for instance in Evans' PDE book, chapter 6. Moreover, in this paper it is sated that this fact is equivalent to the operator satisfying a maximum principle (which is indeed the case for the above $L$).
Yes, this is a consequence of the Krein-Milman theorem.
Do you mean the Krein-Rutman theorem perhaps? do you know a reference for the above result in the closed manifolds case?
Oooops, indeed I meant Krein-Rutman, sorry. I don't have any reference for manifolds, but I'm pretty sure the standard proof should carry through
Here's an outline of the general Krein-Rutman based strategy that was suggested by @leomonsaingeon: (i) Choose a space to work on - for instance, $L^2$ over the manifold. (ii) Show that all spectral value of $L$ have real part $\ge \varepsilon$ for some $\varepsilon > 0$. (iii) Show compactness of the resolvent (for instance by showing that the domain of $L$ embeds compactly into $L^2$) (iv) Use the maximum principle to show that $L^{-1}$ a positive operator. [to be continued]
[continuation] (v) Apply the Krein-Rutman theorem to $L^{-1}$ in order to see that its spectral radius is in the spectrum and that it has a positive eigenvector. (vi) Use the spectral mapping theorem for resolvents to get back to the operator $L$.
That’s very helpful, thanks!
|
2025-03-21T14:48:31.157652
| 2020-06-02T06:43:24 |
361950
|
{
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"authors": [
"Ben Wieland",
"Mikhail Borovoi",
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"https://mathoverflow.net/users/798",
"skupers"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361950"
}
|
Stack Exchange
|
Cohomology ring of special linear group over finite fields
I am trying to find about the cohomology ring $H^*(SL_n(\mathbb{F}_q),\mathbb{Z}/2\mathbb{Z})$ where $q$ is odd. For $n=2$, an explicit description is given. But for $n>2$, I didn't come across a thing.
If not the cohomology ring, then a description of some cohomology groups will also be helpful.
Do you mean group cohomology? Then $$H^1({\rm SL}_n(\mathbb{F}_q),\mathbb{Z}/2\mathbb{Z})={\rm Hom}({\rm SL}_n(\mathbb{F}_q),\mathbb{Z}/2\mathbb{Z}),$$ which is, I think, trivial except for some small values of $n$ and $q$.
For general linear groups instead of special linear groups, this is a famous computation of Quillen, "On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field."
Here is Quillen's 1970 ICM address, 5 pages. In part 1, he computes the mod p cohomology of every finite group modulo nilpotent elements. Another approx strategy is to pull back the Sylow subgroup, which is the normalizer of the torus. These probably combine to compute it exactly for $SL_n$. In part 2, he computes the coprime cohomology of the finite points of every algebraic group. For the multiplicative structure and $Z/2$ coeff, something goes wrong, but you can probably control it by comparing $SL_n$ with $GL_n$.
|
2025-03-21T14:48:31.157767
| 2020-06-02T06:59:51 |
361951
|
{
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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/361951"
}
|
Stack Exchange
|
Surjectivity of multiplicative map (in more specific case)
(I have asked the question Surjectivity of multiplicative map. I ask here the more specific case.)
Let $S$ be a smooth complex algebraic surface, and $D$ be a divisor on $S$ such that $D^2>0$ and $H^i(S, \mathcal{O}(nD)) = 0 $ for all $i > 0, n \in \mathbb{Z}_{\geq 1}$ ($D$ is not necessarily ample , but nef & big).
Write $W = S \times S$ and the coherent sheaf $\mathcal{G} = \mathcal{O}_{S}(nD) \boxtimes \mathcal{O}_{S}(nD)$ on $W$.
Suppose that we have an exact sequence
$$ 0 \rightarrow H^0(W, \mathcal{G} \otimes I_{\Delta}^{n}) \rightarrow H^0(W, \mathcal{G}) \xrightarrow{\varphi} H^0(W, \mathcal{G} \otimes \mathcal{O}_W/I_{\Delta}^{n}) \rightarrow H^1(W, \mathcal{G} \otimes I_{\Delta}^{n}) \rightarrow 0.$$
where $\Delta$ is a diagonal embedding $S \hookrightarrow S \times S = W$ and $I_{\Delta}$ is an idela sheaf of $\Delta$.
By the Kunneth formula, we have $H^0(W, \mathcal{G}) = H^0(S, \mathcal{O}_S(nD)) \otimes H^0(S, \mathcal{O}_S(nD)).$
So, $\varphi$ can be regarded as a multiplicative map followed by some inflation into infinitesimal neighborhood.
I want to show the surjectivity of the map $\varphi$, or equivalently, the vanishing of $H^1(W, \mathcal{G} \otimes I_{\Delta}^{n})$ (under the action of $\mathfrak{S}_2$ in practice) for sufficiently large $n$.
The sheaf $\mathcal{O}_W/I_{\Delta}^{n}$ has a filtration by $I^k_{\Delta}/I^{n}_{\Delta}$ with associated graded factors
$$ I^k_{\Delta}/I^{k+1}_{\Delta} \simeq Sym^k(\Omega_S), \ 0 \leq k \leq n-1. $$
So, I think we have to check the surjectivity on each $H^0(S, \mathcal{O}(2nD)\otimes Sym^{k}(\Omega_S))$ (is it right?).
I have shown $H^i(S, \mathcal{O}_S(2nD) \otimes Sym^k(\Omega_S)) = 0$ for all $i > 0$ and $0 \leq k \leq n-1$.
I found the paper of Mark L. Green Koszul cohomology and the geometry of projective varieties(I read part of it) which says that:
Corollary(4.e.4)(The Explicit $H^0$ Lemma). Let $C$ be a smooth curve of genus $g$, and $L \rightarrow C$ and $M \rightarrow C$ analytic line bundles.
Assume that $\deg L \leq \deg M$ and that $|L|$ is base-point free.
If either
$$\deg L + \deg M \geq 4g + 2$$
or
$$\deg M = 2g+1, \deg L = 2g$$
then the multiplication map
$$H^0(C, L) \otimes H^0(C, M) \rightarrow H^0(C, L \otimes M)$$
is surjectictive.
My question is that can we generalize this statement to the case of surfaces?
How can I prove the surjectivity of $\varphi$ ?
I appreciate any advice, answers, or references.
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