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2025-03-21T14:48:30.519215
| 2020-05-07T16:17:49 |
359647
|
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"Deane Yang",
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|
Stack Exchange
|
Variation of the (Chern) curvature with respect to the metric
Let $E\rightarrow X$ be a holomorphic vector bundle, for any Hermitian metric $h$ on $E$ we denote by $F_h$ the curvature of the Chern connection associated to $h$. Fix a metric $h_0$ and consider a one-parameter family of metrics $h_t=h_0(g_t\cdot ,g_t\cdot)$ with $g_t\in \Gamma(End(E))$.
I am interested in finding a formula for
$$\dfrac{\partial}{\partial t}F_{h_t}$$
If someone has a reference ill be grateful
You can put the formula for $h_t$ into the formula for the curvature of a metric, differentiate, and set $t=0$. Look at papers on curvature flows for guidance on how this goes, if not the exact formula itself.
|
2025-03-21T14:48:30.519298
| 2020-05-07T17:32:58 |
359650
|
{
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359650"
}
|
Stack Exchange
|
Sets of finite perimeter: intersection with an half space
I have a question regarding sets of finite perimeter. In particular I'm interested to find
$$\mu_{E \cap H_t}, \label{1}\tag{1}$$
where $E$ is a set of finite perimeter in a generic open set $\Omega \subset \mathbb{R}^n$ and $H_t$ is a half-space, i.e., $H_t =\left\{ x \in \mathbb{R}^n:x\cdot e<t \right\}$ for some $e \in S^{n-1}\mathrel{:=}\left\{ x \in \mathbb{R}^n:\lvert x\rvert=1 \right\}$ and $t \in \mathbb{R}$.
The notation $\mu_E$ means $\mu_E=\nu_E \mathcal{H}^{n-1} \mathop\lfloor \partial^* E$, where "$\lfloor$" denotes the restriction.
In the book Sets of finite perimeter and Geometric Variational problems by F. Maggi (MSN), I found a partial answer to my question:
Exercise 15.3 (Intersection with half spaces). If $E$ is a set
of finite perimeter in $\mathbb{R}^n$, then $E \cap H_t$ is a set of
finite perimeter in $\mathbb{R}^n$, and, for a.e. $t \in \mathbb{R}$,
$$ \mu_{E \cap H_t}=\mu_E \mathop\lfloor H_t +e \mathcal{H}^{n-1} \mathop\lfloor
\left( E \cap \partial H_t\right), \label{2}\tag{2}
$$
where "$\lfloor$" denotes the
restriction and $\mu_E=\nu_E \mathcal{H}^{n-1} \mathop\lfloor \partial^* E$.
Question I: does \eqref{2} hold true even if $E$ is a set of finite perimeter in a generic open set $\Omega$ and not in $\mathbb{R}^n$?
If not, how I can derive \eqref{1}?
In the same book, the author gives the formula for the general intersection between two sets of finite perimeter:
Theorem 16.3 (Set operations on Gauss–Green measures). If $E$ and $F$ are two sets of finite perimeter, then
$$\mu_{E \cap F}=\mu_E \mathop\lfloor F^{(1)} +\mu_F \mathop\lfloor E^{(1)}+\nu_E
\mathcal{H}^{n-1} \mathop\lfloor \left\{ \nu_E = \nu_F \right\},\label{3}\tag{3}$$
where
$$\left\{ \nu_E = \nu_F \right\}=\left\{ x \in \partial^*E \cap
\partial^*F : \nu_E (x)=\nu_F (x) \right\}.$$
Question II: How can one deduce \eqref{2} from \eqref{3} when $F=H_t$? And why $E^{(1)}$ in \eqref{3} does not compare in \eqref{2}?
Some help or advice?
In particular does the equation (2) hold true if $E$ is a set of locally finite perimeter, i.e.$E$ has finite perimeter for every compact set $K \in \mathbb{R}^n$? Thanks.
|
2025-03-21T14:48:30.519456
| 2020-05-07T17:47:14 |
359651
|
{
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"Andy Sanders",
"Igor Rivin",
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|
Stack Exchange
|
Mirzakhani's hyperbolic method generalized to moduli space of stable maps
I've been learning about Mirzakhani's use of hyperbolic geometry to compute Weil-Petersson volumes of moduli space of curves, and the application to proving Virasoro constraints for a point. Why have these methods not been directly extended to higher dimensional target spaces? I don't know much about the structure of the moduli space of stable maps into a projective variety. Is there expected to exist a reasonable definition of volume for these spaces? I'm further confused by the fact that "topological recursion" introduced by Eyndard-Orantin, and inspired by Mirzakhani's recursion relations, was used to compute Gromov-Witten invariants for toric Calabi-Yau 3-folds. I suppose my question is: to what extent might one expect that intersection theory on the moduli space of stable maps into a projective variety is related to the hyperbolic geometry of the domain Riemann surface?
I'm not expert enough to know if this is a "good" question, but it's definitely a very interesting question.
Mirzakhani's computation of volumes of deformation spaces is very heavily based on the work of Greg McShane (McShane's identity). McShane's theory has been extended to other deformation spaces, see, for example
Labourie, François; McShane, Gregory, Cross ratios and identities for higher Teichmüller-Thurston theory, Duke Math. J. 149, No. 2, 279-345 (2009). ZBL1182.30075.
(a search of Google Scholar for "McShane Labourie" reveals a fair bit of other work). So, for the last question, the answer is "not very", but feel free to read the papers...
Thanks for the answer, but I think I’m confused. How do the spaces in that paper relate to moduli spaces of stable maps into a projective variety (such that appear in Gromov-Witten theory?)
@JohnRached I did not say they did relate. Your question was whether the methods are intrinsically hyperbolic-geometric, and these references show that they are not.
|
2025-03-21T14:48:30.519617
| 2020-05-07T17:50:13 |
359653
|
{
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|
Stack Exchange
|
Invariance under derived equivalence of a Gorenstein projective bimodule
A module $M$ over an finite dimensional algebra $A$ is called Gorenstein projective in case there exists an exact complex $(P_i)$ of projective $A$-modules such that the complex stays exact after applying the functor $Hom_A(-,A)$ and $M$ is isomorphic to a kernel of some map in the complex $(P_i)$.
Equivalently $M$ is Gorenstein projective in case $Ext_A^i(M,A)=0=Ext_A^i(Tr(M),A)$ when $M$ is the Auslander-Bridger transpose of $M$.
Question: In case two algebras $A$ and $B$ are derived equivalent and $A$ as an $A$-bimodule is Gorenstein projective, is also $B$ as a $B$-bimodule Gorenstein projective?
|
2025-03-21T14:48:30.519698
| 2020-05-07T18:45:39 |
359657
|
{
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"authors": [
"Deane Yang",
"Guillaume Aubrun",
"Hermann",
"Igor Rivin",
"M. Winter",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/11142",
"https://mathoverflow.net/users/157696",
"https://mathoverflow.net/users/613",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359657"
}
|
Stack Exchange
|
Is Minkowski sum of boundary convex again?
Consider a closed, bounded and convex set $C \subset \mathbb{R}^{2}$ and denote its boundary with $\partial C$. It is very well-known that the Minkowski sum of two convex sets is convex again. What about the Minkowski sum of its boundary?
Is the Minkowski sum $\partial C + \partial C$ again a convex set and how can one prove that?
Does this property hold in other dimensions?
@IgorBelegradek No, $\partial C +\partial C = 2C$ in that case. I believe this equality is true in general (equivalently, any point in a convex body is the midpoint of boundary points) and should be provable by a topological argument.
Take $C$ to be the unit disk and $D$ to be the disk of radius, say, $1/100$. The sum of the convex sets is the disk of radius $11/10$ but the sum of the boundaries is the annulus with inner radius $9/10$ and outer radius $11/10$.
@DeaneYang The question is about the sum of $\partial C$ with itself.
@M.Winter, oy. Thanks.
Yes, $\partial C + \partial C$ is convex since it equals $2C$. Equivalently, every point in $z \in C$ is a midpoint of two boundary points. This is obvious if $z \in \partial C$. Otherwise, let $f :S^{n-1} \to \mathbf{R}$ be the continuous function which sends $u$ to the length of the segment going from $z$ to $\partial C$ in direction $u$. Since $n > 1$, this function takes equal values at a pair of antipodal points (a very simple corollary to Borsuk-Ulam, if you want), which gives the desired property.
If you consider segments going from $z$ to $\partial C$ in directions $u$ and $-u$, and you compare their lengths, then I think all you need here is the intermediate value theorem to find that they will be equal in length at some point (by continuously moving $u$ to $-u$).
@GuillaumeAubrun Although it seems intuitively correct, can you give an argument why f is continuous?
This follows e.g. from the fact that the gauge function of a convex body is continuous. The gauge function $g_K$ of a convex body $K$ is defined on $\mathbf{R}^n$ by $g_K(x) = \inf { t \geq 0 , : , x \in tK }$ ; it is convex hence continuous. When $z=0$ (which you can assume), our $f$ is the restriction of $1/g_C$ to $S^{n-1}$.
Convex hull of Minkowski sum is the Minkowski sum of convex hulls. The proof is Theorem 1.1.2 in
Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001.
...But the answer to the question is: NO. See the picture in https://www.geometrie.tuwien.ac.at/peternell/ms_paper_v1.pdf (Peternell, Minkowski sum of boundary surfaces...) Can't find a citation.
I don't see how this answers the question. Also, the desired property is false in dimension $1$, so this restriction should enter at some point.
@GuillaumeAubrun Fair point.
@GuillaumeAubrun but see the edit.
@GuillaumeAubrun And thanks for the downvote!
I did note downvote your answer and I'm sorry that you believe this! I think the picture in the paper you mention addresses a different question, namely whether $\partial C + \partial D$ is convex.
@GuillaumeAubrun OK, thanks for not downvoting! Otherwise, I thought that WAS the OP's question (whether the sum of the boundaries was convex).
|
2025-03-21T14:48:30.519954
| 2020-05-07T18:57:28 |
359659
|
{
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"authors": [
"Angelo",
"Hacon",
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|
Stack Exchange
|
Polarization of an abelian variety made by the sum of two divisors
Let $X$ be an abelian variety of dimension $n$, and let $L$ be a polarization, that is, an ample line bundle on $X$, with $\chi(L)=3$.
In my specific case, I have that $L=\mathcal{O}_X(\Theta + D)$, where $\Theta$ is an ample divisor with $\chi(\Theta)=1$ and $D$ is an effective Cartier divisor.
I want to show that $(D^2)=0$ (self-intersection of $D$), or equivalently that $(\Theta^{n-2}.D^2)=0$.
For $n=2$, $X$ is a surface, and using Riemann-Roch I have that $2\chi(L)=6=(\Theta^2)+2(\Theta.D)+(D^2)$, where the first two intersection numbers are strictly positive because of ampleness of $\Theta$ (in particular $(\Theta^2)=2$). If I suppose $(D^2)\ne 0$, then $(D^2)=2$ and so $(\Theta.D)$ must be 1. But this is impossible by the index theorem, because we have $4=(\Theta^2)(D^2)\le (\Theta.D)^2$.
But for dimension $n>2$, I don't know how to procede, because in Riemann-Roch formula $n!$ increases too fast, so it seems impossible to make the same argument.
Thanks for help!
Note: I have already posted this question on Math StackExchange, but maybe it is better to post it here.
Why do you expect to get 3 in general?
@Angelo, what do you mean? 3 is simply the degree of the polarization.
For any $P\in Pic^0(A)$ consider the map $|\Theta +P|\times |D-P|\to |\Theta +D|\cong \mathbb P ^2$. Since $D$ is effective, there is an abelian subvariety $T\subset Pic ^0(A)$ such that $|D-P'|\ne \emptyset$ for any $P'\in P+T$ and if $t=\dim T$ the $D^t\ne 0$ but $D^{t+1}=0$. If $t\geq 3$, then a general element in the image of the above map may be written as $\Theta _{P'}+D_{P'}$ in for infinitely many $P'\in P+T\subset Pic ^0(A)$. Thus $\Theta _{P'}+D_{P'}=\Theta _{P''}+D_{P''}$, but then $D_{P''}\geq \Theta _{P'}$ (as $\Theta _{P'}\in |\Theta +P'|$ is unique and different from $\Theta _{P''}$) and hence $\chi (\Theta +D)\geq \chi (2\Theta )>3$. Finally, if $t=2$, the above argument shows that any element $G\in |\Theta +D|$ can be written as a sum of elements $\Theta _{P'}\in |\Theta +P'|$ and $D_{P'}\in |D-P'|$ and $\dim |D-P'|=0$. Thus, the corresponding rational map $T\to \mathbb P ^2$ is generically finite, and of degree $>1$ (as $T$ is not rational). But then, for general $G\in |\Theta +D|$, we have $G=\Theta _{P'}+D_{P'}=\Theta _{P''}+D_{P''}$ which implies $\Theta _{P'}=D_{P''}$ and hence $\chi (L)=4$.
NB I originally misread the question so I have edited the answer appropriately
Thanks for the answer! A couple of things I don't understand: why there exists such an abelian subvariety T? And why its dimension determines self-intersections of D?
Let $D$ be an effective divisor on an abelian variety $A$. If $D$ is not ample, there exists a quotient abelian variety $A\to B$ such that $D$ is the pull-back of an ample divisor on $B$. Thus $D^t\ne 0$ and $D^{t+1}=0$ where $t=\dim B$. Also $|D+P|\ne 0$ for any $P\in Pic^0(B)\subset Pic ^0(A)$. If $D$ is nef, then there will be some $P\in Pic^0(A)$ such that $|D+P|\ne \emptyset$. Since you are actually assuming $D\geq 0$ then we can assume $P=0$.
Another question: why for $t \ge 3$ there are infinitely many ways for writing an element in the image, while for $t=2$ there is only one? I was thinking about $V^0(D)$, but I don't know how to see its dimension. Thanks for your help
If $t\geq 3$, then there is a $t+d-1\geq 3$-dimensional family of divisors in $|D-P|$ where $d=h^0(D)$ and $P\in Pic ^0(B)\subset Pic ^0(A)$; in fact these divisors are parametrized by a $\mathbb P ^{d-1}$ bundle over $T$ and they are all distinct. The point is that $D$ is the pullback of an ample divisor say $\bar D$ on $B$ and then $h^0(\bar D-P)=h^0(\bar D)\geq 1$ for any $P\in Pic^0(B)$. If $t=2$ and $h^0(D)=1$, then we only have a 2 dimensional family of divisors $D_P\in |D-P|$. But in this case the family is parametrized by the abelian surface $Pic^0(B)$.
|
2025-03-21T14:48:30.520356
| 2020-05-07T19:08:47 |
359660
|
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"authors": [
"Emil Jeřábek",
"YCor",
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|
Stack Exchange
|
Centralizer of a single element in the monoid of self-maps of a set
This is a follow-up to this question: For what sets $X$ do there exist a pair of functions from $X$ to $X$ with the identity being the only function that commutes with both?
Let $X$ be a set, and $X^X$ the monoid of self-maps $X\to X$ (also known as full transformation monoid of $X$). In my answer to the above question, I checked that there exists a pair in $X^X$ whose centralizer is reduced to $\{\mathrm{id}_X\}$. Of course we can't expect this with a single element since it commutes with its own powers. So the question (which I initially asked in a comment) is
Let $X$ be a set with $|X|>c\,(=2^{\aleph_0})$. Does there exist $f\in X^X$ whose centralizer is reduced to $\{f^n:n\ge 0\}$? Or at the opposite, is it true that every $f\in X^X$ has a centralizer cardinal $2^{|X|}$?
What does it mean that a centralizer is “reduced”?
@EmilJeřábek nothing, but I wrote "reduced to", which means "is only equal to", in which the mathematical contents is "is equal to", and the remainder is rhetorical, emphasizing that one inclusion is trivial.
Let $|X| > \mathbb{N}$ and $f \in X^X$, we construct a function $g$ that commutes with $f$ but is not a power of $f$. Let $G$ be the directed graph of $f$. Split $G$ into connected components. Call a connected component boring if it is of the form $a_1, a_2, \cdots$ or $\{a_i\}_{i \in \mathbb{Z}}$ with $f(a_i) = a_{i+1}$. If $G$ has no non-boring components, then for size reasons, two must be isomorphic so we can take $g$ to be an isomorphism between the two and the identity on the rest of $X$. So suppose $C$ is a non-boring component and consider three cases
1: There is an $x \in C$ with $f^{-1}(x) = \emptyset$. Then since $C$ is not boring, there is an $n \geq 1$ with $y \in f^{-1}f^n(x)$ and $y \notin\{x, f(x), f^2(x), \cdots\}$. Then we can take $g(x) = y$ and for $z \neq x, g(z) = f^{n-1}(z)$.
2: 1 is not the case and $C$ contains no cycles. Then let $\{a_i\}_{i \in \mathbb{Z}} \subset C$ be a sequence with $f(a_i) = a_{i+1}$. For every element $c \in C$, let $d(c)$ denote the minimal non-negative integer for which $f^{d(c)}(c) \in \{a_i\}_{i \in \mathbb{Z}}$ and let $r(c)$ be the minimal integer for for which $f^{d(c)}(c) = a_{r(c)}$. Set $g(c) = a_{r(c) - d(c)}$. Since there is an infinite chain disjoint from $\{a_i\}_{i \in \mathbb{Z}}$, for every $n \geq 0$, $f^n$ sends some element outside $\{a_i\}_{i \in \mathbb{Z}}$ so $g$ commutes with $f$ but is not a power of $f$.
3: 1 is not the case and $C$ contains a cycle. Then there exists $\{a_i\}_{i \in \mathbb{Z}} \subset C$ with $f(a_i) = a_{i+1}$, $a_1$ is in a cycle and $a_j$ is not in a cycle for $j < 1$. Let $k$ be the smallest positive integer greater than $1$ with $a_1 = a_k$. For every element $c \in C$, let $d(c)$ denote the minimal non-negative integer for which $f^{d(c)}(c) \in \{a_i\}_{i \in \mathbb{Z}}$ and let $r(c)$ be the minimal integer for for which $f^{d(c)}(c) = a_{r(c)}$. Set $g(c) = a_{r(c) - d(c)}$. For the reasons in 2, $g$ commutes with $f$ but is not a power of $f$.
Thank you. It would be nice to have a lower bound on the cardinal of the centralizer then.
|
2025-03-21T14:48:30.520585
| 2020-05-07T19:28:30 |
359662
|
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"authors": [
"Nate Eldredge",
"Nick S",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359662"
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|
Stack Exchange
|
Separating compact sets in locally compact spaces
It is known that not every locally compact Haussdorff space is normal, see for example
here
But it seems that the following is true, I just want to make sure I am not making any mistake:
Lemma Let $X$ be a locally compact Haussdorff space and let $K, W \subset X$ be compact with $K \cap W = \emptyset$. Then, there exists open sets $K \subset U, W \subset V$ such that $U \cap V =\emptyset$.
Proof:
We have $K \subset (X \backslash W)$. Since $K$ is compact, $X \backslash W$ is open and $X$ is locally compact Haussdorff space, by Theorem 2.7 in Rudin, "Real and Complex Analysis" we can find an open set $U$ (with compact closure) such that $K \subset U \subset \bar{U} \subset X \backslash W$.
Let $V:= X \backslash \bar{U}$. Then $V$ is open and satisfies the above conditions.
\qed
Is this correct? I am a bit uneasy, especially since the proof only uses the fact that $W$ is closed, compactness is only needed for $K$.
Yes, this is true in any Hausdorff space; you don't even need local compactness. The proof is a nice exercise. With local compactness you get regularity and then indeed it's enough to have one set closed and the other compact.
@NateEldredge Perfect, thank you.
Roughly speaking, anywhere in the separation axioms that you see "point", you can upgrade it to "compact set" for free.
@NateEldredge If I understand right, the reason for this is because for each of the points in the compact set we can find a pair of separating open sets. Then, we pick a finite subcover, and take the corresponding finite intersection of open sets, which is open since finite. The last step seems to also be the reason why these cannot be extended from compact to closed.
Yes, that's it exactly.
Yes, this is known and you don't need local compactness even as long as $X$ is completely regular (locally compact spaces are completely regular).
I think that the best way to prove is to go via the Čech–Stone functor, which is available precisely for completely regular spaces. Consider $\beta X$ and note that $K, W\subset X\subset \beta X$ remain compact in $\beta X$.
|
2025-03-21T14:48:30.520759
| 2020-05-07T20:15:03 |
359667
|
{
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"authors": [
"Laura",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359667"
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|
Stack Exchange
|
Effect of extending scalars on maps of modules
Let $k$ be a field and $R$ be a $k$-algebra. Let $M$ and $N$ be left $R$-modules. Finally, let $\ell$ be a field extension of $k$. We thus have an $\ell$-algebra $\ell \otimes R$, and both $\ell \otimes M$ and $\ell \otimes N$ are left modules over it.
Question: What is the relationship between the $k$-vector space $\text{Hom}_R(M,N)$ and the $\ell$-vector space $\text{Hom}_{\ell \otimes R}(\ell \otimes M,\ell \otimes N)$?
I am most interested in the case where $\text{char}(k)=0$ and $R = k[G]$ for a finite group $G$ and $\ell$ is a finite extension of $k$ and $M$ and $N$ are finitely generated.
We have that $\text{Hom}_{\ell \otimes R}(\ell \otimes M,\ell \otimes N)$ is isomorphic to $\ell \otimes \text{Hom}_R(M,N)$ as $\ell$-vector spaces for a general (possible infinite dimensional) $k$-algebra $R$ in case $M$ has finite $k$-dimension, see for example Lemma 7.4. in the book "A first course in noncommutative rings" by Lam.
That reference is exactly what I was looking for, thanks!
|
2025-03-21T14:48:30.520861
| 2020-05-07T20:26:40 |
359668
|
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|
Stack Exchange
|
Does $\mathbb{R}$ have a partite subbase?
If $X\neq \varnothing$ is a set we say that ${\frak P} \subseteq {\cal P}(X)$ is a partition of $X$ if
$\bigcup{\frak P} = X$, and
$P\neq Q \in {\frak P} \implies P\cap Q = \varnothing$.
Let $H = (V,E)$ be a hypergraph with $V \neq \varnothing$ and $\bigcup E = V$. A partition ${\frak P}$ of $V$ is said to be splitting if for all $e\in E$ and $P \in {\frak P}$ we have $|e \cap P| = 1$, and we call such a hypergraph admitting a splitting partition partite.
It is easy to see that every edge in a partite hypergraph $H$ has the same cardinality $\kappa$, and that $\kappa$ is also the cardinality of every splitting partition of $H$.
Let $\tau$ be the Euclidean topology on $\mathbb{R}$. It is easy to see that $(\mathbb{R}, \tau\setminus\{\varnothing\})$ cannot have a splitting partition since $\mathbb{R} \in \tau$.
Question. Is there a subbase ${\cal S}\subseteq (\tau\setminus\{\varnothing\})$ such that $(\mathbb{R}, {\cal S})$ is partite?
The restriction $X\neq\emptyset$ is non-standard, nevertheless it is usual to assume that components of a partition are nonempty. For instance for $X={1,2,3}$ you currently allow both ${{1},{2,3}}$ and ${\emptyset, {1},{2,3}}$ as distinct partitions. Probably you want to remove $X\neq\emptyset$, and add $P\in\mathfrak{P}$ $\Rightarrow$ $P\neq\emptyset$ in the axioms.
A near miss: $\mathcal S$ is all open intervals of length $1$, and $\mathfrak{P}$ is all shifts of $\mathbb Z$.
May I ask what the motivation for your question is? and why do you ask for a subbase and not for a base?
A base would contain (an infinite chain of) sets nested by inclusion, but a partite hypergraph has no pair of nested edges.
If it can help, it is not too difficult to see that any $P$ in a possible $\mathfrak P$ would be closed discrete, and there would be for all $[-M,M]$ an $\varepsilon>0$ (depending only on $M$) such that the distance between two points of $P\cap [-M,M]$ would be at least $\varepsilon$ (the distance between two consecutive points of $P$ is locally bounded below, uniformly in $P$).
@JanKyncl That's right - so first I was wondering whether ${\mathbb R}$ had a subbase without nested elements and then found out this to be true, so I could turn towards partiteness
@RamirodelaVega Thanks for your question about the motivation. My questions often arise out of some playfulness. In this case it happened this way. I was reading this article about Ryser's conjecture and wondered whether the concept of "partiteness" that appears on slide 7 can be applied to topological spaces. I quickly realized that nestedness of open sets posed a problem - therefore the question about a subbase
|
2025-03-21T14:48:30.521089
| 2020-05-07T21:10:02 |
359671
|
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|
Stack Exchange
|
Membership problem in general linear group
This is surely a very well known problem, but I could not find an answer on MO or on Google, so here I am.
Given some finitely generated free subgroup $H$ of $\operatorname{GL}_n(\mathbb{Z}[t,t^{-1}])$, and some arbitrary $\omega \in \operatorname{GL}_n(\mathbb{Z}[t,t^{-1}])$, is it decidable whether $\omega \in H$? If so, what is the algorithm for checking this?
No, it's already false in $\mathrm{GL}_2(\mathbf{Z})$. Indeed there are uncountably many subgroups, while membership to a subgroup is decidable only for countably many subgroups. In $\mathrm{GL}_4(\mathbf{Z})$ there are even finitely generated subgroups with non-decidable membership problem, this has certainly been mentioned here somewhere.
@YCor for which subgroups is it decidable? Would it be decidable if $H$ is free?
Maybe you mean "finitely generated free" otherwise my counting argument applies. It's certainly unknown in general for which subgroups it is decidable. For finitely generated free subgroups I don't know (this could make a reasonable question).
@YCor—Sorry for the confusion. I am only interested in finitely generated $H$.
I think this is the same as asking whether a f.g. free subgroup can be more than recursively distorted. I guess it's unknown (but don't know if anybody ever thought about it).
Actually, since $GL_2(\mathbb{Z})$ is virtually free there are algorithms due to Markus-Epstein and others for the membership problem for f.g. subgroups. For $n>2$ the membership problem is open for f.g. subgroups, I believe, and it's not clear how freeness helps. @YCor, how does distortion connect to the membership problem?
... ahh I got it. Enumerate elements of the free group, and look at their images as a subgroup sitting inside $GL_n(S)$. A bound on distortion will enables you to produce a have a complete list of all elements within a ball of radius $m$ inside $GL_n(S)$.
Actually, both in $\mathrm{GL}_2(\mathbf{Z}[t^{\pm 1}])$ and $\mathrm{GL}_3(\mathbf{Z})$ I don't know if the membership problem holds for f.g. subgroups, and even for f.g. free subgroups.
@NWMT exactly. In a f.g. group with solvable word problem, a f.g. subgroup has solvable membership problem iff its distortion function (relative to fixed finite generating subsets) is recursive, iff it's bounded above by a recursive function.
@YCor: Yes, see the question raised here. I am quite sure that the existence of free subgroups with non-recursive distortion is an open problem. But, in view of hydra examples, for all practical purposes, the membership problem in free subgroups is undecidable.
|
2025-03-21T14:48:30.521287
| 2020-05-07T21:28:53 |
359673
|
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|
Stack Exchange
|
Prime-like numbers that avoid Green-Tao?
I would like to understand the conditions that support
the Green-Tao Theorem, which established that
the primes contain arbitrarily long arithmetic progressions.
I am wondering:
Q. Is it difficult to define an infinite
set $S$ of natural numbers that mimics the
$\sim n / \log n$ distribution of
the primes (and perhaps other prime properties), but yet fails to contain arbitrarily long arithmetic progressions?
For example, $S$ might be the
lucky numbers
(OEIS A000959),
or
Hawkins' "random primes."
Do either of these avoid arbitrarily long arithmetic progressions?
My intuition is that it is easier to avoid such progressions
than it is to establish their existence.
MSE question unanswered:
Do lucky numbers contain arbitrarily long arithmetic progressions?
I believe a famous conjecture of Erdős says that it is not possible to define such a set: https://en.wikipedia.org/wiki/Erd%C5%91s_conjecture_on_arithmetic_progressions.
(I also remember hearing from experts that the density from Erdős's conjecture is not really believed to be the 'true' boundary for containing arbitrarily long arithmetic progressions, so in some sense the conjecture is 'misleading.')
See Section 2 of this survey of Gowers, which discusses this problem: https://arxiv.org/abs/1509.03421
I can define such a set greedily, but I do not know its density. Pick the first member to be 1. Now pick the next smallest unpicked and unmarked member: this will 2. Now mark all numbers which form a three term arithmetic progression with previously picked members, and disallow those from the sequence. I get 1,2,4,5,10,11,13,14,22 . If you disallow just four terms, the sequence becomes more dense (and more challenging to construct). Gerhard "This May Be In OEIS" Paseman, 2020.05.07.
@GerhardPaseman: see https://en.wikipedia.org/wiki/Stanley_sequence
I now see that you asked a very similar question here 5 years ago: https://mathoverflow.net/questions/198387/most-dense-subset-of-numbers-that-avoids-arbitrarily-long-arithmetic-progression?rq=1
@Sam: Now that's embarrassing! I guess I am still wondering 5 years later, despite the "likely impossible" answers.
@PeterLeFanuLumsdaine: Yes, see the remark of Sam Hopkins.
This is likely impossible.
Indeed the largest sets known to be free of arbitrarily long arithmetic progressions asymptotically satisfy $| [ A \cap [1, n] | \lesssim_{k} n / \log^{k} n$ for all $k>1$, and it is widely believed that these examples are near maximal.
The Erdos-Turan conjecture mentioned in the comments is almost equivalent to the claim that any set of relative density greater than $n/ \log n$ contains arbitrary long arithmetic progressions. As remarked above, this is likely true for even sparser sets.
Does this mean that proving that the number $T(x)$ of twin primes below $x$ fulfills $T(x)\asymp x/\log^{2}x$ would imply that the sequence of twin primes contains arbitrarily long arithmetic progressions?
@SylvainJULIEN Well, one would need to prove density of twin-primes and prove this generalization of Erdos-Turan, bit. :-)
|
2025-03-21T14:48:30.521522
| 2020-05-07T23:14:27 |
359678
|
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|
Stack Exchange
|
Expected number of directed cycle in a directed complete graph
Consider the randomized, directed complete graph G = (V, E) where for each pair of vertices u, v ∈ V, we add either the directed edge (u → v) or the directed edge (v → u) chosen uniformly at random. What is the expected number of directed cycles of 3 vertices, as a function of the number of vertices n?
A good question, but too simple for MO.
@JohnCartor, this kind of graph is called a random tournament. This keyword might help you find information elsewhere.
|
2025-03-21T14:48:30.521598
| 2020-05-07T23:53:44 |
359679
|
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|
Stack Exchange
|
Where is the flaw in this argument with $p$-adic extensions?
I cannot find what I am missing in the following computation. Let $K=\mathbb{Q}_p(p^{1/{(p-1)p^{\infty}}})$ and $L=\mathbb{Q}_p(\zeta_{p^{\infty}})$, where $\zeta_{p^n}$ is a primitive $p^n$-th root of unity. I think I have computed that $\mathcal{O}_K/p\cong \mathcal{O}_L/p\cong \mathbb{F}_p[t^{1/p^{\infty}}]/t^{p-1}:=R$, which is perfect (I can provide more details for this computation if needed). We should then have that for the $p$-adic completions $\widehat{\mathbb{Z}_p[p^{1/{(p-1)p^{\infty}}}]}\cong \widehat{\mathbb{Z}_p[\zeta_{p^{\infty}}]}\cong W(R)$ since there is a unique strict $p$-ring having a given perfect ring $R$ of characteristic $p$ as its residue ring (Theorem 5, pg.39 Local Fields Serre). However, I know this cannot be true. Maybe there is an easier way to see it but here is an argument: By Corollary 5.10 in Kuhlmann - The algebra and model theory of tame valued fields, we get that $L$ is relatively algebraically closed in its completion. In particular, if the above isomorphism is true, this would give us that $p^{1/p}\in L$. But since $L/\mathbb{Q}_p$ is abelian, this would imply that $\mathbb{Q}_p(p^{1/p})$ is Galois, which is not true.
p.s. This is related to a question that I posted on mathstack: https://math.stackexchange.com/questions/3662451/what-is-wrong-with-this-application-of-ax-kochen
For a ramified extension $\mathbf Z_p \subseteq \mathcal O_K$, the ring $\mathcal O_K/p$ is never perfect, because an element $x$ with valuation $0 < v(x) < v(p)$ satisfies $x^{p^r} = 0$ in $\mathcal O_K/p$ for $r \gg 0$, but $x$ is not zero in $\mathcal O_K/p$.
Perfect here means that the Frobenius is surjective, not bijective.
Edit: But you might be right; in Serre's book pg.39 he defines a ring to be perfect if the Frobenius is actually an automorphism. So probably the theorem that I used doesn't apply.
As far as I know, the usual definition of perfect ring is that the Frobenius is an isomorphism, but I agree that the wording in Serre is confusing.
A characteristic $p$ ring in which Frobenius is merely surjective is apparently sometimes called semiperfect. (also, I believe the word "perfect" also meant "bijective" in the lecture we were attending this week :) )
I think that both completions $\hat{L}$ and $\hat{K}$ are isomorphic, and are both perfectoid fields which are untilts of the same perfectoid field $\mathbb{F}_p((t))/t^{p-1}$. In particular, they are isomorphic. As for Kuhlmann, observe that $L^c/L$ is not algebraic, hence not separable. For tilts/unitlts/perfectoid, you can certainly look at Scholze's IHES paper, or at lectures 2 and 3 by Lurie here https://www.math.ias.edu/~lurie/205.html
@Wojowu Hi Wojtek, I think that in that context perfect really meant semiperfect; what we were calling perfect $\mathbb{F}_p$-algebras were things like $A/pA$ for strict $p$-rings $A$(see definition 3.3.1 https://people.maths.ox.ac.uk/gulotta/hodge_notes.pdf). These are never perfect in the bijective sense for ramified $A$ as R. van Dobben de Bruyn pointed out. Right?
@FilippoAlbertoEdoardo Hi Fillippo, thanks for your answer.The fact that they are untilts of the same perfectoid field does certainly not imply they are isomorphic; the completions of $\mathbb{Q}_p(p^{1/p^{\infty}})$ and $\mathbb{Q}p(\zeta{p^{\infty}})$ are standard examples of non-isomorphic fields having the same tilt up to isomorphism.
@KostasKartas To be honest, I am still not 100% sure about that. I know that Kedlaya in his article requires perfect algebras to have bijective Frobenius, and the frequent uses of $x^{p^{-n}}$ (which otherwise would not be well-defined) make me thing our lectures demands that too. I agree bijectivity fails for $A/pA$ for $A$ ramified, but it doesn't seem to me we ever need to use results about perfect algebras for them (e.g. we never form $W(A/pA)$, only $W(\mathcal O_K)$ when $K$ has characteristic $p$).
@Wojowu It seems you are right, sorry for misleading you with my comment in the chat of the lecture!
|
2025-03-21T14:48:30.522217
| 2020-05-08T00:29:33 |
359680
|
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|
Stack Exchange
|
Geometric/Algebraic intersection numbers of curves on surfaces
I have the following problem, and struggling to find some references.
Suppose I start with a homology class of a curve on a closed genus $g$ surface $$h=(a_{1},b_{1},\dots,a_{g},b_{g})\in H_{1}(\Sigma_{g};\mathbb{Z}).$$
If $h$ is primitive, I can find a simple closed curve $\gamma$ such that $[\gamma] = h$. Now I want to find another disjoint simple closed curve $\gamma'$ so that $\gamma$ and $\gamma'$ don't intersect. We have that $[\gamma'] = (a_{1}',b_{1}',\dots,a_{g}',b_{g}')\in H_{1}(\Sigma_{g};\mathbb{Z})$ for some $a_{i}',b_{i}'\in\mathbb{Z}$.
What restrictions can we place on the $a_{i}'$ and $b_{i}'$ to ensure that $\gamma'$ is disjoint from $\gamma$?
Clearly we would require that we have for the algebraic intersection number $\hat{i}([\gamma],[\gamma'])=0$, but it seems that the algebraic intersection number is not easily computable. I think I found the apocryphal formula
$$\hat{i}((a_{1},b_{1},\dots,a_{g},b_{g}),(a'_{1},b'_{1},\dots,a'_{g},b'_{g})) = \sum_{i=1}^{g} a_{i}b'_{i}-a'_{i}b_{i},$$
but can't find a proof.
Any references, advice, or solutions would be greatly appreciated!
What are $a_i$ and $b_i$ in terms of the geometry of the surface? The proof at the level of detail you're requesting seems to hinge entirely on the particular choice of basis of homology.
The formula you mention is correct. The algebraic intersection number is a symplectic form on $H_1(\Sigma_g)$ which takes on exactly the form you suggest. See Farb and Margalit's A Primer on Mapping Class Groups, chapter 6.
Of course, asking for algebraic intersection $0$ is not nearly enough to have geometric intersection $0$.
@EthanDlugie, is there an obvious counterexample I could think about of a homology class $[\gamma']$ where $\hat{i}([\gamma],[\gamma'])=0$, but $[\gamma']$ does not contain a representative that has zero geometric intersection number with $\gamma$?
Let $a_i,b_i,$ be the standard geometric basis for the first homology of $\Sigma_3$ (see F&M, p. 165). Let $\gamma$ be $b_2$ and let $\gamma'$ be the simple closed curve which wraps around the 1st and 3rd holes and "under" the second. As as a homology class, this is $a_1+a_3$. Then $\hat i(\gamma,\gamma')=0$ but $\gamma$ and $\gamma'$ intersect twice without any bigons, so these intersections are essential.
Awesome, thankyou for your help.
Collecting my previous comments here. The formula you give for the algebraic intersection number is correct, and points to the fact that the intersection number is a symplectic form on $H_1(\Sigma_g)$ (with $\mathbb{R}$-coefficients if you want to make it a vector space). A reference for this is Farb and Margalit's A Primer on Mapping Class Groups, Chapter 6.
You said that algebraic intersection number zero is necessary for geometric intersection number zero, which is certainly true. However it is not sufficient. Just take this example on the genus 3 surface (or any higher genus):
The trouble is that specifying a homology class fails to really specify a homotopy class of simple closed curve. For instance if you chose $\gamma'$ to go above rather than below the middle hole in this picture, you'd have a homologous curve which is now disjoint from $\gamma$.
However in the genus $\leq 2$ case, I believe $\hat i=0$ does imply $i=0$ for simple closed curves representing primitive homology classes. This is easy on the torus, since simple closed curves here in minimal position intersect with the same orientation at all points. I haven't been able to come up with a proof yet in the genus two case beyond lack of counterexample.
Edit: @AndyPutman observed that I was wrong in the genus 2 case. Taking sufficiently complicated pseudo-Anosov elements in the Torelli group lets you construct curves which intersect very highly but do so in an algebraically trivial manner.
I just came upon this post. Your final paragraph is false in genus 2. Here's a counterexample: https://imgur.com/a/BTWc0wB
@AndyPutman I spy a bigon between those two curves!
Whoops, that's embarrassing! Here's another one that (hopefully) works: https://imgur.com/a/I4igvY3
By the way, here's how I know without drawing a picture that this must be false. If $\gamma$ is a simple closed curve and $f$ is a pseudo-Anosov that lies in the Torelli group, then for large $n$ the curves $f^n(\gamma)$ and $\gamma$ will intersect many times but be homologous, and hence have algebraic intersection number $0$. And starting in genus $2$ the Torelli group will contain many pseudo-Anosov elements (just think of your favorite construction of pseudo-Anosovs using Dehn twists, and do it with separating twists).
And I've just realized from your example that you could do something even easier: take a separating (hence nullhomologous) curve and then some curve that intersects it nontrivially.
|
2025-03-21T14:48:30.522561
| 2020-05-08T00:31:34 |
359681
|
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|
Stack Exchange
|
Constructing Turing degrees near $\mathcal{O}$
I'm looking for some examples of constructions for Turing degrees close to, but not above, Kleene's $\mathcal{O}$. I know of one by Jockusch and Simpson using forcing with hyperarithmetic, uniformly pointed perfect trees, and this construction has the property that it lets you force the triple jump of the degree constructed. I'd like some other examples of constructions, particularly ones that let you force the double jump but not the triple jump. Does anyone know of any?
Sorry for the very late and non-productive comment but I stumbled upon this by browsing through posts and was wondering if you found any relevant answer in the meantime?
@H.CManu No, I didn't.
I see, I was hoping otherwise because I find the question very interesting. Can you provide me the reference for the one you know of by Jockush and Soare please?
@H.CManu It's this one: https://www.sciencedirect.com/science/article/pii/0003484376900231
Oh it's free too, thank you
|
2025-03-21T14:48:30.522799
| 2020-05-08T01:24:38 |
359684
|
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|
Stack Exchange
|
Why 'excedances' of permutations?
For a permutation $\pi=\pi_1\pi_2\cdots\pi_n$ written in one-line notation, an index $i$ for which $\pi_i > i$ is usually called an 'excedance.' To me, this seems like a mispelling of what should be 'exceedance': many dictionaries list 'exceedance' as a valid word, but none I can find consider 'excedance' a correct spelling of any word. Also, as far as I know, 'excedance' is pronounced like 'exceedance' (that is, ex-ceed-ance). But, while both 'excedance' and 'exceedance' are to some extent used for this permutation concept, it seems that 'excedance' is much more common.
Question: Does anyone know the origin of the spelling 'excedance' for this permutation concept? Is it an error which has become standard?
Sounds like a bad translation of either Latin or the French excédence.
You could ask linguists.
Although your reputation indicates you know the spirit of the site better than I do, it's hard for me to understand in what sense this is a question about research mathematics.
I've seen questions here before about terminology, but if this is deemed inappropriate for the site I would understand that as well.
https://en.wiktionary.org/wiki/excedance explicitly lists excedance as a misspelling of exceedance.
I think this is a reasonable deviation and diversion from the forum scope. Gerhard "Don't Do It Again Soon" Paseman, 2020.05.08.
@CarloBeenakker: interestingly that page has now been edited to list excedance as the correct spelling of the mathematical concept.
Mea culpa. Comtet used the term excédence. When writing EC1 I needed an English term for this concept. For some reason I didn't like the word exceedance. I thought it looked better without the double e, analogous to proceed and procedure. Thus I made up the word excedance.
Couldn't imagine a more authoritative response. Thanks!
It looks better to me too without the double e, but that might be because I've gotten used to it.
|
2025-03-21T14:48:30.523000
| 2020-05-08T02:27:49 |
359686
|
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|
Stack Exchange
|
Conjecture by Ekedahl on Weyl groups and Abelian varieties
A conjecture was made on p.14 in "Cycle Classes of the E-O Stratification on the Moduli of Abelian Varieties" by Torsten Ekedahl (late, excellent contributor to MO) and Gerard Van Der Geer concerning number of elements of the Weyl groups defined in the initial paragraph of the paper and supposing these can be identified with OEIS A000629.
Conjecture. Fix a positive integer $g$. Let $W_g$ be the subgroup
\begin{align}
\left\{\sigma \in S_{2g} \mid \sigma\left(i\right) + \sigma\left(2i+1-g\right) = 2g + 1 \text{ for all } g \right\}
\end{align}
of the symmetric group $S_{2g}$; this is a Coxeter group of type $C_{g}$. It is isomorphic to the semi-direct product $S_g \ltimes \left( \mathbb Z / 2 \mathbb Z \right)^g$, where $S_g$ acts on $\left( \mathbb Z / 2 \mathbb Z \right)^g$ by permuting the factors.
Let $w_\varnothing \in W_g$ be the permutation that sends $1, 2, \ldots, g, g+1, g+2, \ldots, 2g$ to $g+1, g+2, \ldots, 2g, 1, 2, \ldots, g$, respectively. Let $\leq$ denote the Bruhat order on $W_g$. Then, the number of all $w \in W_g$ that satisfy $w \leq w_\varnothing$ is
\begin{align}
\left. \left(x \dfrac{d}{dx}\right)^g \left(\dfrac{1}{1-x}\right) \right|_{x=1/2}
\end{align}
(OEIS sequence A000629).
What is the status of this conjecture? Confirmed or not?
Lemma 2.14 of the paper shows that an $w \in W_g$ satisfies $w \leq w_\varnothing$ if and only if all $i \in \left\{1,2,\ldots,g\right\}$ satisfy $w\left(i\right) \leq g+i$. Thus, it is not necessary to understand the Bruhat order to approach this conjecture.
Edit (May 14, 2020):
The sequence A000629 is of some general importance in algebra and combinatorics. It is the row sums of the unsigned partition polynomials A263634 and A127672, related to the logarithmic derivative of e.g.f.s, or formal Taylor series, and, consequently, to the raising op for Appell sequences and, thence, to Weyl and Heisenberg algebras and series and integral convolutions, to the cumulant expansion theorem, and to the Faber polynomials A263916 (and, therefore, the symmetric polynomials/functions). So, any further combinatorial proofs of the conjecture would perhaps inform these constructs.
In fact, Getzler alludes to necklaces in a natural generalization of the identity above to A263634 in "The semi-classical approximation for modular operads."
This looks like a rook theory problem.
I think it's true, assuming that the "a(n) = Sum_{k=0..n} Stirling2(n+1, k+1)*k!. - Paul Barry, Apr 20 2005" formula on the OEIS is true. Indeed, let me rename $g$ as $n$. How many ways are there to find a permutation $w \in W_n$ satisfying $w \leq w_\varnothing$ ? Clearly, it suffices to choose the $n$ values $w\left(1\right), w\left(2\right), \ldots, w\left(n\right)$, since the other $n$ values of $w$ will then be uniquely determined by the definition of $W_n$. In choosing these $n$ values $w\left(1\right), w\left(2\right), \ldots, w\left(n\right)$, we need to satisfy the relation ...
$w \left(i\right) \leq n+i$ for each $i \in \left{1,2,\ldots,n\right}$ (by Lemma 2.14 of the paper), and furthermore ensure that $w\left(i\right) \neq w\left(j\right)$ and $w\left(i\right) + w\left(j\right) \neq 2n+1$ for any distinct $i, j \in \left{1,2,\ldots,n\right}$. So we can proceed as follows: In Step 1, we decide how many elements (say, $k$ many) $i$ of $\left{1, 2, \ldots, n\right}$ will satisfy $w\left(i\right) \leq n$. (We don't decide which ones these will be.) In Step 2, we choose these $k$ many elements $i$, calling them homebound (since they are $\leq n$ and ...
... will also be sent to elements $\leq n$), and we also choose the values $w\left(i\right)$ for all non-homebound elements $i$ of $\left{1,2,\ldots,n\right}$. These need to be chosen subject to the requirements that $w\left(i\right) \leq n+i$ for each $i \in \left{1,2,\ldots,n\right}$. In Step 3, we finally choose the values $w\left(i\right)$ for all homebound elements $i$ of $\left{1,2,\ldots,n\right}$. These need to be chosen subject to the requirements that $w\left(i\right) \neq w\left(j\right)$ and $w\left(i\right) + w\left(j\right) \neq 2n+1$ for any ...
... distinct $i, j \in \left{1,2,\ldots,n\right}$. Note that the first of these requirements simply forces them to be distinct, while the latter forces them to avoid $n-k$ numbers corresponding to the values at the non-homebound elements chosen in Step 2; the condition $w\left(i\right) \leq n+i$ will be satisfied automatically. It is now easy to see that the number of options in Step 2 is the Stirling number of the second kind $S_2\left(n+1, k+1\right)$, while the number of options in Step 3 will be $k!$. So ...
... the total number of $w$'s is $\sum\limits_{k=0}^n k! S_2\left(n+1, k+1\right)$, which is OEIS sequence A000629 according to the above-quoted comment by Paul Barry. Please check.
By the way, I have no idea what a "necklace of partitions of n+1 labeled beads" is.
@darijgrinberg: maybe it means an ordered set partition of $[n+1]$, but considered up to cyclic rotation of the blocks?
In which case the formula makes sense: if we had $(k+1)!$ instead of $k!$ we would get ordered set partitions; but replacing $(k+1)!$ by $k!$ groups all rotation classes, which always have $k$ set partitions in them, together.
Nice! @SamHopkins
Nice. Certainly related to factorial rook polynomials since $(xD)^n(1/(1-x) ) = \sum_{k \geq 0} k^nx^k$ (see around p. 61 of "Rook polynomials" by Gessel). This sum is related to the Eulerian numbers and therefore to the combinatorics of the permutahedra and thence to the Bernoulli numbers and quite a lot of core integer sequences and functions, such as the logarithmic integrals (cf. https://oeis.org/A131758).
The expansion of the Euler / theta / state op $(xD)^n$ in the ops $x^kD^k$ gives the Stirling numbers of the second kind (unsigned), which again are related to the number of faces of the permutahedra (and therefore to complete graphs).
Gessel's slide presentation:http://people.brandeis.edu/~gessel/homepage/slides/wash-rook.pdf
Darij, I know you are perhaps answer averse until you are sure the answer is correct, but I'd encourage you to post these comments as an answer. (Think of it as a discourse rather than an Olympiad test question.)
The following is transcribed from @darijgrinberg's comments 1 2 3 4 5, made CW to avoid reputation, per @TomCopeland's request and absent any reply after a few days. If @darijgrinberg prefers to post it themselves, or to have it not posted at all, then I will be happy to delete it.
I think it's true, assuming that the "a(n) = Sum_{k=0..n} Stirling2(n+1, k+1)*k!. - Paul Barry, Apr 20 2005" formula on the OEIS is true. Indeed, let me rename $g$ as $n$. How many ways are there to find a permutation $w \in W_n$ satisfying $w \le w_\emptyset$? Clearly, it suffices to $w(1), \dotsc, w(n)$, since the other $n$ values of $w$ will then be uniquely determined by the definition of $W_n$. In choosing these $n$
values $w(1), w(2), \dotsc, w(n)$, we need to satisfy the relation
$w(i) \le n + i$ for each $i \in \{1, 2, \dotsc, n\}$ (by Lemma 2.14 of Ekedahl and van der Geer - Cycle classes of the E-O stratification on the moduli of Abelian varieties), and furthermore ensure that $w(i) \ne w(j)$ and $w(i) + w(j) \ne 2n + 1$ for any distinct $i, j \in \{1, 2, \dotsc, n\}$. So we can proceed as follows: In Step 1, we decide how many elements (say, $k$ many) $i$ of $\{1, 2, \dotsc, n\}$ will satisfy $w(i) \le n$. (We don't decide which ones these will be.) In Step 2, we choose these $k$ many elements $i$, calling them homebound (since they are $\le n$ and
will also be sent to elements $\le n$), and we also choose the values $w(i)$ for all non-homebound elements $i$ of $\{1, 2, \dotsc, n\}$. These need to be chosen subject to the requirements that $w(i) \le n + i$ for each $i \in \{1, 2, \dotsc, n\}$. In Step 3, we finally choose the values $w(i)$ for all homebound elements $i$ of $\{1, 2, \dotsc, n\}$. These need to be chosen subject to the requirements that $w(i) \ne w(j)$ and $w(i) + w(j) \ne 2n + 1$ for any
distinct $i, j \in \{1, 2, \dotsc, n\}$. Note that the first of these requirements simply forces them to be distinct, while the latter forces them to avoid $n - k$ numbers corresponding to the values at the non-homebound elements chosen in Step 2; the condition $w(i) \le n + i$ will be satisfied automatically. It is now easy to see that the number of options in Step 2 is the Stirling number of the second kind $S_2(n + 1, k + 1)$, while the number of options in Step 3 will be $k!$. So
the total number of $w$'s is $\sum_{k = 0}^n k! S_2(n + 1, k + 1)$, which is OEIS sequence A000629 according to the above-quoted comment by Paul Barry. Please check.
Ha, he's probably writing a short paper generalizing the results, from past experience.
@TomCopeland: No, not this time -- this is a combinatorial argument, not an algebraic one, so I doubt I'll find a good way of making it slick and readable. Also, Sam Hopkins has contributed to the argument (by justifying Paul Barry's formula).
@darijgrinberg, certainly not trying to pressure you but the refinements of this seq, as I allude to in comments, have Coxeter A written all over them and geometric alg if you take a look at the Getzler ref, so maybe on a slow day ... .
|
2025-03-21T14:48:30.523516
| 2020-05-08T02:33:22 |
359687
|
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|
Stack Exchange
|
Surjectivity of the Abel-Prym map
It is well known that the Abel-Jacobi map restricted to $\text{Eff}_g(C)$ surjects onto the Jacobian $\text{Jac}(C)$, since every divisor of degree $g$ is effective.
Is there an analogous statement for Prym varieties? That is, given an unramified double cover $\widetilde C\to C$ with involution $\tau$, consider the map $f:\text{Eff}_d(\widetilde{C})\to\text{Prym}(\widetilde{C}/C)$ given by $f(D)=D-\tau(D)$. Is $f$ surjective if, for instance, $d=g-1$?
First of all, note that your definition is not correct: when $d$ is odd, the image of your map does not land in the Prym variety -- you have to add a constant term. When this is done, the answer is yes, for the following reason. Let $X$ be the image of $\tilde{C} $ in $P:=\operatorname{Prym}(\tilde{C}/C ) $. Let me put $h:=g-1=\dim P$. What you want to prove is that the addition map $X^{h}\rightarrow P$ is surjective, that is, of degree $>0$. Now this degree is computed by the Pontryagin product $[X]^{*h}$, where $[X]$ is the class of $X$ in $H^{2h-2}(P,\mathbb{Z})$. We know that this class is $2\dfrac{\theta ^{h-1}}{(h-1)!} $, where $\theta $ is the class of the principal polarization.
So we just have to prove that $\theta ^{*h}\in H^{2h}(P,\mathbb{Z})$ is nonzero. This is true for any principally polarized abelian variety $(P,\theta )$ of dimension $h$: it suffices to prove it for a Jacobian $J\Gamma $, and this amounts to say that the Abel-Jacobi map $\Gamma ^h\rightarrow J\Gamma $ is surjective, as you recall in your post.
Thanks for your answer! You're right, I meant to say that either there is a parity condition on the degree, or that the image falls in a translation of the Prym. Why does it suffice to prove the claim for a Jacobian?
Because this is a question about cohomology classes, hence invariant by deformation. But you can also compute $\theta^{*h}=h!$ directly.
I thought the Prym is the kernel of the norm map, and since a divisor of the form $D - \tau(D)$ always maps to $0$ via the norm map it represents a point in the Prym by definition?
@zudumazics: no, the kernel of the norm map has two components, the Prym being the component of zero. So $D-\tau(D)$ is in the Prym variety iff $\deg(D)$ is even.
@abx thanks for the correction. Why in the above did you assume $\dim P = g-1?$, i.e. the genus of $C$ is 1? Also, can you elaborate a bit more about why $[X]^{\ast h}$ has that expression in terms of $\theta$? To me a priori $X$ seems quite arbitrary a subset of $P$.
@zudumazics: I am sorry but I don't think this is the right place to explain the theory of Prym varieties. Mumford's paper (Prym varieties I) is an excellent introduction.
@abx thanks for the pointer, I will look into it. May I ask again about the point above: even if $\deg(D)$ is odd, say $D = p$ for simplicity, can't we just send $p$ to a ramification point of $\tilde{C} \rightarrow C$ to get to $ p - \tau(p) = 0$?
@abx oh I see: you are working with an unramified covering, yes? Do you know if your claims hold for ramified covering then?
|
2025-03-21T14:48:30.523773
| 2020-05-08T03:34:50 |
359689
|
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"R. van Dobben de Bruyn",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359689"
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|
Stack Exchange
|
Extending vector bundles on divisors
Let $X$ be a smooth variety. Let $Y$ be the union of the four sub-varieties of $X\times \mathbb{P}^1 \times \mathbb{P}^1$ determined by $X \times \mathbb{P}^1 \times \{0\}$, $X \times \mathbb{P}^1 \times \{\infty\}$, $X \times \{0\} \times \mathbb{P}^1$ and $X \times \{\infty\} \times \mathbb{P}^1$. Can any vector bundle on $Y$ be extended to $X\times \mathbb{P}^1 \times \mathbb{P}^1$? (What are minimum requirements for this to be possible?)
2nd question: A similar question can be asked by replacing $\mathbb{P}^1 \times \mathbb{P}^1$ by $\mathbb{P}^2$ and instead of 4 lines we can consider three general projective lines in $\mathbb{P}^2$ and product of them with $X$. What is the answer for the extensions problem in this scenario? What are the obstructions for the extensions?
The answer to the 1st and 2nd question seems to be negative by the comment below.
3rd question: Now assume the same question but $\mathbb{P}^1$ replaced by $\mathbb{A}^1$. The obstructions for the extensions (This is probably true if we assume $X$ is affine and believe in Bass-Quillen conjecture and assume that going around the loop gives the identity as the automorphism)
What do you mean by "extended"? $Y$ is not contained in $X$.
This already fails for line bundles if $X$ is a point: you can glue any $\mathcal O_{\mathbf P^1}(d_i)$ to each other along a square of $\mathbf P^1$s, but to come from $\mathbf P^1 \times \mathbf P^1$ the restriction to the two vertical lines (or the two horizontal lines) has to be the same. Another thing you can do is create a loop of trivial line bundles where you glue each consecutive pair by a constant such that the product around the loop is not equal to $1$.
@ abx I fixed the problem.
For $\mathbf A^1$ the situation is no different: a loop (triangle, square, ...) of $\mathbf A^1$s has nontrivial line bundles obtained from trivial line bundles on each component by glueing them using a set of constants whose product does not equal $1$.
@R. van Dobben de Bruyn For the $\mathbb{A}^1$ case is it possible to give an obstruction for the extension? (When $X$ is affine I think probably the automorphism constructed by going around the loop needs to be identity in order to be extendable)
|
2025-03-21T14:48:30.523957
| 2020-05-08T05:57:15 |
359692
|
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"Dexter Chua",
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|
Stack Exchange
|
Model categories and chain complexes
I'm fairly new to thinking about homological algebra and chain complexes in their own right, i.e outside of isolated examples such as for constructing simplicial homology, or for computing $Ext$ groups for some Hopf algebroid.
Given an abelian category $\mathscr{A}$ with a category of chain complexes $Ch(\mathscr{A})$, the homotopy category
$K(\mathscr{A})$ is defined as the naive homotopy category (i.e. replace chain maps with chain homotopy classes of chain maps) but with quasi-isomorphisms inverted (these are the maps which induce isomorphisms on homology). This is obviously the result of placing some kind of model structure on $Ch(\mathscr{A})$. This leads me to consider a few obvious questions.
Are there alternate, interesting model structures for $Ch(\mathscr{A})$?
(How) has the advancement of model category theory impacted the study of things such as chain complexes, and perverse sheaves for example?
Are there any good treatments of homological algebra which makes use of the rich theory of model categories?
Note that there are in fact two model structures that are commonly used - the injective and projective model structures, which correspond to taking injective and projective resolutions respectively. (At least naively, each of these require different boundedness conditions on the chain complexes because we resolve in different directions. Putting a model structure on unbounded chain complexes is trickier. This should not be surprising because we cannot inductively construct injective/projective resolutions for unbounded chain complexes)
@DexterChua of course! When I say "alternate" model structures I certainly mean alternate to (at least) these two model structures. Could you say anything about the unbounded case? For example I'm guessing it still has weak equivalences the quasi-isomorphisms and I would certainly hope that it is cofibrantly generated
I'll take your question as license to advertise a relatively recent paper in a slightly more specialized but concretely calculational direction: http://nyjm.albany.edu/j/2014/20-53p.pdf. Its title is
Six model structures for DG-modules over DGAs: Model category theory in homological action. The theme is how different model structures can illuminate concrete calculations. I had computed the cohomology of various homogeneous spaces, way back in the 1960's, using some strange looking explicit cochain complexes. Tobi Barthel, Emily Riehl and I found that those turn out to be explicit examples of a variant kind of cofibrant approximation.
Yes, there are zillions of model structures on Ch(A), corresponding to whatever class of projectives you choose to use for your homological algebra. This is all spelled out in the paper "Quillen model structures for relative homological algebra" by Christensen and Hovey.
More generally, the theory of cotorsion pairs builds model structures in many algebraic settings (including quasi-coherent sheaves; I'm not sure about perverse sheaves, not even if they satisfy bicompleteness as a category). Hovey's seminal paper is here. A great survey by Gillespie is here. This material has also appeared in books. A recent one by Marco Perez is here.
|
2025-03-21T14:48:30.524216
| 2020-05-08T05:59:35 |
359693
|
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|
Stack Exchange
|
Expectation of the trace of inverse of a Gaussian random matrix
Given a $N×M$ random complex gaussian matrix $X$ and $N×K$ random complex gaussian matrix $Y$ I'm interested in approximating the expectation expressed as:
\begin{align}
E[trace({(aX{X^H} + I)^{ - 1}}Y{Y^H})]
\end{align}
a Is a positive given variable. I know that $XX^H$ and $YY^H$ have Wishard distribution, However. I can not approximate the problem value. Thanks!
For context, this problem relates to the quality of MIMO communication link
I assume the matrices $X$ and $Y$ are independent. Since the trace commutes with the expectation value, and since the expectation value of the product of independent random variables is the product of expectation values, we have
$$
F(a)=\mathbb{E}\bigl[{\rm tr}\,\bigl({(aX{X^H} + I)^{ - 1}}Y{Y^H}\bigr)\bigr]={\rm tr}\bigl(\,\mathbb{E}[(aX{X^H} + I)^{ - 1}]\mathbb{E}[Y{Y^H}]\bigr).$$
The second factor is simply $K$ times the unit matrix, so
$$F(a)=K\,{\rm tr}\,\mathbb{E}[(aX{X^H} + I)^{ - 1}].$$
We can now again exchange trace and expectation value, to rewrite this as an integral over the eigenvalues $\mu_k$ of $XX^H$, with density $\rho(\mu)$,
$$F(a)=K\int \rho(\mu)(a\mu+1)^{-1}\,d\mu.$$
The density $\rho(\mu)$ is known, for large matrix size it is the Marcenko-Pastur distribution.
For the Marcenko-Pastur distribution, so for $M\geq N\gg 1$, I find
$$F(a)=\frac{K}{2a} \left(\sqrt{a^2 (M-N)^2+2 a (M+N)+1}+a (N-M)-1\right).$$
please can I employing the same approach can I used Wishart distribution??
I am using the Wishart distribution: firstly when I use that $\mathbb{E}[YY^H]=K$ times the identity, and secondly the Marcenko-Pastur distribution is derived from the Wishart distribution.
Is the answering missing a factor of min(M, N) since the trace is the summation of all eigenvalues?
The following argument is quite similar to Carlo Beenakker's.
For simplicity, I only consider the real case. I'm also going to use different symbols for the sizes of the matrices. Let $X$ be an $n \times d$ random matrix with iid rows from $N(0,\Sigma)$ let $Y$ be a $m \times d$ random matrix independent of $X$, with iid rows from $N(0,S)$. Set $\widehat \Sigma := X^\top X/n$ and $\widehat S := Y^\top Y/m$. Let $F(a)$ be the expression you wish to evaluate and set $\lambda=1/(na)$. Note that
$$
F(a) = \frac{m}{na} \mathbb E\,[\operatorname{tr} \widehat S(\widehat\Sigma + \lambda I_d)^{-1} ]
$$
Then, in the limit $n,d \to \infty$ such that $d/n \to \phi \in (0,\infty)$, we have
$$
\mathbb E\,[\operatorname{tr} \widehat S (\widehat\Sigma + \lambda I_d)^{-1} \mid Y] \simeq \operatorname{tr} \widehat S(\Sigma + \kappa I_d)^{-1},
$$
where $\kappa = \kappa(\lambda,n)$ is the unique nonnegative solution to the equation $\kappa - \lambda = \kappa \operatorname{df}_1(\kappa)/n$, where
$$
\operatorname{df}_1(\kappa) := \operatorname{tr}\Sigma (\Sigma + \kappa I_d)^{-1}.
$$
We deduce that
$$
\mathbb E\,[\operatorname{tr} \widehat S(\widehat \Sigma + \lambda I_d)^{-1} ] \simeq \mathbb E\,[\operatorname{tr} \widehat S (\Sigma + \kappa I_d)^{-1}] = \mathbb E\,[\operatorname{tr}S (\Sigma + \kappa I_d)^{-1}].
$$
In your question, you have $\Sigma=S=I_d$, and so
$$
\mathbb E\,[\operatorname{tr}\widehat S (\widehat \Sigma + \lambda I_d)^{-1}] \simeq \frac{d}{1+\kappa}.
$$
Moreover, $\kappa$ is now given by the following well-known formula which is reminiscent of the Marchenko-Pastur distribution
$$
\kappa = \frac{\lambda + \overline \phi + \sqrt{(\lambda + \overline\phi)^2 + 4\lambda}}{2},
$$
where $\overline \phi := \phi-1$.
Simplify...
|
2025-03-21T14:48:30.524478
| 2020-05-08T06:15:37 |
359694
|
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"leo monsaingeon"
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|
Stack Exchange
|
An interpolation inequality
I am interested in the following statement. Let $q>p$. Then there are positive numbers $\alpha$ and $\beta$ so that for all $f\in C^1(\mathbb{R}^n)$, one has
$$ \left(\int|\nabla f|^p dx\right)^\frac{1}{p} \leq C \left(\int|f|^p dx\right)^\frac{\alpha}{p} \left(\int|\nabla f|^q dx\right)^\frac{\beta}{q}. $$
Does someone have an idea or a reference?
Thank you, I learned that it was a special case of Gagliardo-Nirenberg inequality. Will the proof simplify in my case?
The problem is that you're in the whole space, right? So my strategy would be trying to control the decay of $|\nabla f|$ by $|f|_{L^p}$, up to trading-off with a small fraction of the higher $|\nabla f|_q$, Have you tried for radial functions $f(x)=\frac{1}{|x|^\delta}$ to identify possible candidates for $\alpha,\beta$?
I deleted my answer, as it is NOT a special case of Gagliardo-Nirenberg (since you want identical orders of derivatives in both sides of the inequality, $j=m=1$ and therefore $\alpha=1$ in Wikipedia's notations, hence Gagliardo-Nirenberg is completely vacuous in this setting)
Thank you. $\alpha$ and $\beta$ will have to solve $\alpha+\beta=1$ and $-1+\frac{n}{p} = \alpha\frac{n}{p} + \beta(-1 + \frac{n}{q})$ because of scaling invariance. I cannot read what you mean with $\delta$?
well, since the problem is decay at infinity, the natural "test-case" for your inequality is my $\frac{1}{|x|^\delta}$ functions (away from zero, of course). So I suggest you test your inequality with your specific values $\alpha,\beta=[\dots]$ and see if it holds for this whole family of functions. If it does it's a good sign. If it doesn't then it's the end of the road.
Radial functions do not seem to provide a counterexample...
Just for clarification, $\alpha$, $\beta$ and $C$ are independent of $f$? And, why the term interpolation inequality?
I want to indicate that one of the norms on the right hand side is weaker than the one on the left hand side while the other is stronger. We have that $\alpha$ and $\beta$ depend on $p$, $q$ as explained above. Also $C$ is meant independent of $f$.
Such an inequality is false. Take $n=1$, $\phi$ a standard cut-off function supported in $[0,1]$ and $f=\sum_n c_n \phi (\frac{x-n}{r_n})$. Then $\|f\|_p \approx \sum_n |c_n|^p r_n$, $\|f'\|_p \approx \sum_n \frac{|c_n|^p}{r_n^{p-1}}$, $\|f'\|_q \approx \sum_n \frac{|c_n|^p}{r_n^{q-1}}$. If $p=1$, $q=2$, $c_n=1/n$ and $r_n=(\log n)^{-2}$ then $f\in L^1$, $f' \in L^2$, $f' \not \in L^1$.
True. Thank you.
|
2025-03-21T14:48:30.524684
| 2020-05-08T06:19:02 |
359695
|
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|
Stack Exchange
|
subanalytic realization of smooth abstract stratification
Consider an $C^\infty$ abstract stratification $A$ (in the Thom-Mather sense, see Mather's note).
Can we embed $A$ in some $\mathbb{R}^n$ (or in an analytic manifold) as a subanalytic set?
If not, what could be the problem?
This apparently was shown by Laurent Noirel in his thesis for the $C^k$ case, $1 \leq k < \infty$.
|
2025-03-21T14:48:30.524899
| 2020-05-08T10:17:02 |
359705
|
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|
Stack Exchange
|
The origin(s) of the word "elliptic"
The word elliptic appears quite often in mathematics; I will list a few occurrences below. For some of these, it is clear to me how they are related; for instance, elliptic functions (named after ellipses, see here) are the functions on elliptic curves over $\mathbb C$. For others, I do not know if there is a relationship at all.
Ellipses
Elliptic integrals
Elliptic functions
Elliptic curves
Elliptic genera (in the sense of Hirzebruch)
Elliptic (as opposed to parabolic or hyperbolic) isometries of the hyperbolic plane
Elliptic partial differential operators, elliptic PDEs
Elliptic cohomology
I am interested in the etymology of this word, in particular, the origins of the different usages listed above. More precisely, I was wondering whether there is, in a way, a single "strain" for all uses of elliptic in mathematics, going all the way back to ellipses in Euclidean geometry.
Might be more suitable for History of Science and Mathematics SE.
My rough understanding is that elliptic integrals arose when trying to find arc lengths of ellipses, and the these integrals can be viewed as integrals over curves of the form $y^2=$cubic or quartic, hence elliptic curves. From there you get elliptic cohomology and genera. On the flip side, elliptic operators have positive-definite associated quadratic forms, just like equations defining ellipses. It seems to be more of a tree of connections than a single strand.
The level curves of the principal symbol of an elliptic PDE are ellipses
Early calculations of the arc length of an ellipse by Euler led to elliptic integrals, then to elliptic functions (Legendre, Abel) and elliptic curves. That is one strain. The elliptic PDE's are a different strain, more closely related to the geometric shape of an ellipse.
The etymology for "elliptic genus" can be found on the nLab page. It is named after elliptic functions. Elliptic cohomology came later, of course. https://ncatlab.org/nlab/show/elliptic+genus
I think it's kind of neat to know that 'ellipses' itself is etymologically derived from 'ellipsis', an omission, and is part of a linguistic, not just mathematical, continuum from 'ellipse' to 'parabola' to 'hyperbola': an ellipse leaves out some eccentricity ($e < 1$); a parabola has just the right amount of eccentricity ($e = 1$); and a hyperbola has too much eccentricity ($e > 1$).
I answered a question like this in math.SE.
@Wojowu could you expand a little more the part of your comment about how elliptic cohomology appears from looking at elliptic integrals? Thank you
@DanielD. I only meant that etymologically. Elliptic cohomology is "elliptic" because it involves elliptic curves (in its definition).
@Wojowu I see my confusion, thank you again
I have always found it amusing that one can speak elliptically (omitting words), parabolically (in parables), or hyperbolically (exaggerating), all in contrast to speaking straight.
@LSpice that's interesting! Is the part about the eccentricity historically accurate?
@TimothyChow apparently the use of these word that we mathematicians find amusing is the original one. (Except for straight talk)
@MichaelBächtold, I don't have an authoritative reference that it refers to eccentricity, but at least they are named via the relevant ancient Greek (indirectly in the case of ellipses): https://en.wiktionary.org/wiki/ellipse#Etymology and https://en.wiktionary.org/wiki/ellipse#Etymology_2 ; https://en.wiktionary.org/wiki/parabola#Etymology ; and https://en.wiktionary.org/wiki/hyperbola#Etymology .
@Lspice I looked on Jeff Millers website and found a references. It's not exactly the eccentricity, but along those lines. See my answer.
@MichaelBächtold, thanks! It's nice to have a better reference than "it's gotta be". The answer.
Your saying "elliptic functions are the functions on elliptic curves over $\mathbb C$" is somewhat misleading, I think. First came elliptic integrals measuring arc-length on an ellipse. These are generalizations of the inverse trig functions (take the ellipse to be a circle). The inverse functions to the elliptic integrals are elliptic functions. It was noted that the integrand of an elliptic integral is (after a change of variables) of the form $dx/\sqrt{f(x)}$, where $f(x)$ is a cubic (or quartic, depending on your preference). This in turn leads to elliptic curves, which are curves of the form $y^2=f(x)$, since then the integrand is $dx/y$, and the integral is on the curve. At this point, one sees that the use of the word elliptic in elliptic curve is quite unfortunate, since the geometry of an elliptic curve is quite different from the geometry of the ellipse from which it derives its name. Further, there is the distinction between a (smooth algebraic) curve of genus $1$, and such a curve with a marked base point that serves as the identity element for its group law. This is especially important if one is working over a non-algebraically closed field, but even over $\mathbb C$, if elliptic curve includes the group law, then it presupposes the choice of a point.
It might be worth adding that the arc-length of a hyperbola is also given in terms of elliptic integrals, so that they just as well might have been named 'hyperbolic curves' (but that would probably have been an even worse name clash!)
The origin of all these uses is very different. Joe Silverman explained the genesis of the sequence ellipse $\rightarrow$ elliptic integral $\rightarrow$ elliptic function $\rightarrow$ elliptic curve.
Another large class of occurrences of the word "elliptic" is connected with
"trichotomies", that is classifications of some objects into three classes. Such classifications occur very frequently, and most of them can be traced
to the simplest trichotomy "positive, negative, zero". Historically, the oldest non-trivial triple classification is the classification of irreducible plane conics (ellipse, parabola, hyperbola). Later, there was a tendency to use the words elliptic, parabolic, hyperbolic for any such trichotomy (metrics of positive, negative and zero curvature, Riemann surfaces, PDE's, singular points of dynamical systems, linear-fractional transformations, etc.)
Then,
when various generalizations occur, mathematicians like to keep the term even in absence of a trichotomy. For example, according to Gromov, a Riemannian manifold is called elliptic if it receives a non-constant quasiregular map from $R^n$. Here the term elliptic comes from the theory of Riemann surfaces (which can be elliptic parabolic or hyperbolic), though in Gromov's situation there is no hyperbolic or parabolic case anymore. Similarly a dynamical system can be "hyperbolic", while the
terms "elliptic, parabolic" have no such standard meaning in this case.
Indeed it's quite disturbing that an elliptic curve is a parabolic Riemann surface...
Also one can have a 'hyperbolic' or a 'parabolic' group, whereas elliptic group would have no standard meaning.
@Francois Brnault: It is much more disturbing from my point of view that algebraic CURVE is a Riemann SURFACE:-)
@Alexandre Eremenko:
@HollisWilliams Is
elliptic subgroup not considered standard usage?
OK, my mistake. But I guess there is still a distinction of sorts, one can have a hyperbolic group and elliptic and parabolic subgroups.
@Alexandre Eremenko: just noticed that I posted an empty comment by mistake - that's weird. I just wanted to say: some - me included, to an extent - would find cool, rather than disturbing, that a Riemenn surface is called a complex curve :)
@Qfwfq: this "cool" feature causes great inconveniences when you write a complicated text involving curves $F(x,y)=0$ and Riemann surfaces, and curves $\gamma(t)$ on these surfaces.
@AE: Can't disagree on that!
As to why the conic section got called ellipse, the introductory chapter of Toomer, Diocles, On Burning Mirrors is interesting. He does not give a conclusive answer, but here's an excerpt, p. 7:
Apollonius found symptomata for all three curves, and defined them by the method of "application of areas", which was the standard Greek procedure for formulating geometrically problems which are, algebraically, equations of the second degree. In the parabola, if the ordinate is $y$ and the abscissa $x$, he represented the symptoma corresponding to equation $y^2 = px$,
by saying that the rectangle of side $x$ and area equal to $y^2$ is applied (παραβάλλɛται) to the line-length $p$. In the case of hyperbola [...]
he represents the relationship $y^2=x(p+\frac{p}{a}x)$ by saying (see Fig. I) that a rectangle of side $x$ and area equal to $y^2$ is applied to $p$ so that it exceeds it (ύπερβάλλɛι) by a rectangle similar to $\frac{p}{a}$. Similarly [for the ellipse] he represents $y^2=x(p-\frac{p}{a}x)$ by saying (see Fig. II) that a rectangle of side $x$ and area equal to $y^2$ is applied to $p$ so that it falls short of it (έλλείπɛι) by a rectangle similar to $\frac{p}{a}$. Hence he gives the curves the names "parabola", "hyperbola" and "ellipse" respectively.
Toomer then discusses at length "what features in this account appear difficult to sustain in the light of Diocles' treatise".
This may not be "etymological", but may perhaps shed some light on the relationship between E/P/H things in mathematics:
A. Rastegar, EPH-classifications in Geometry, Algebra, Analysis and Arithmetic (2015).
The origins of all of the companion terms parabola, hyperbola, and ellipse were coined by Apollonius of Perga, in his classic text "On Conic Sections." (He was born about 262 BC, approximately 25 years after the birth of Archimedes.)
The terms we use are direct descendants of the Greek words. Now, the reason they have the names they is that in the construction of conic sections produced by passing a plane through a cone, a parabola (something literally "thrown beside") differs from a hyperbola (something thrown over) and an ellipse (left out). These are in reference to a "parameter" (Apollonius's term) something "measuring alongside," a parameter exceeding, and a parameter being deficient by (leaving out), an amount.
We use those same Greek words in a variety of unrelated contexts, of course. Hyperbolae refers to statements that are over the top, excessive, and by extension unbelievable. Parabolic seems to be generally in reference to the conic section, but ellipse means something that has been left out, in the same way that Apollonius used it in reference to the construction of his conic section. Generally we think of an ellipse as a figure that can be described as an oval, with two foci. (A circle is a degenerate ellipse, because there is only one focus.) However the reason Apollonius called an ellipse "deficient" is because the same parameter was lacking, in the same way that it was excessive in the hyperbola.
The classic text "On Conic Sections" is part of the Great Books of the Western World Collection, which comprise the core of the Great Books programs taught at St. John's College (Santa Fe, NM, and Annapolis, MD), the University of Chicago, and a few other offshoots -- St. Mary's Moraga, CA for example.
Marklan
For a nonmathematical appearance of “parabola”. we have the word “parable”, and more: it would seem that there was a Latin verb “parabolare”, to parabolate, I suppose, which is the origin of the Italian verb “parlare” and the French “parler”. So, “parliament”, etc.
Excellent point, Lubin. I thought of parable this morning, and was going to edit my post accordingly but see that you mentioned it before I could do so. Thank you. Marklan
|
2025-03-21T14:48:30.525802
| 2020-05-08T10:21:09 |
359706
|
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|
Stack Exchange
|
On weaker forms of the abc conjecture from the theory of Hölder and logarithmic means
In this post (the content of this post is now cross-posted from Mathematics Stack Exchange see below) we denote the radical of an integer $n>1$ as the product of disctinct primes dividing it $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ with the definition $\operatorname{rad}(1)=1$. The abc conjecture is an important problem in mathematics as one can to see from the Wikipedia abc conjecture. In this post I mean the formulation ABC conjecture II stated in previous link.
I was inspired in the theory of generalized mean or Hölder mean (see [1]) to state the following claim (Mathematics Stack Exchange 3648776 with title A weak form of the abc conjecture involving the definition of Hölder mean asked Apr 28 '20).
Claim. On assumption of the abc conjecture $\forall \varepsilon>0$ there exists a constant $\mu(\epsilon)>0$ such that for triples of positive integers $a,b,c\geq 1$ satisfying $\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1$ and $a+b=c$ ones has for real numbers $q>0$ that the following inequality holds
$$c<\mu(\varepsilon)\left(\frac{\operatorname{rad}(a)^q+\operatorname{rad}(b)^q+\operatorname{rad}(c)^q}{3}\right)^{3(1+\varepsilon)/q}.\tag{1}$$
Remark 1. Thus as $q\to 0$ from the theory of Hölder mean we recover the abc conjecture.
In a similar way I was inspired in the definition of the logarithmic mean and its relationship to the arithmetic mean to pose the following conjecture (Mathematics Stack Exchange 3580506 with title Weaker than abc conjecture invoking the inequality between the arithmetic and logarithmic means asked Mar 14 '20).
Conjecture. For every real number $\varepsilon>0$, there exists a positive constant $\mu(\varepsilon)$ such that for all pairs $(a,b)$ of coprime positive integers $1\leq a<b$ the following inequality holds $$2\,\frac{b-a}{\log\left(\frac{b}{a}\right)}\leq \mu(\varepsilon)\operatorname{rad}(ab(a+b))^{1+\varepsilon}.\tag{2}$$
Remark 2. Thus I think that previous conjecture is weaker than the abc conjecture by virtue of the relation between the artihmetic and logarithmic means.
Question. I wondered what work can be done to prove/discuss unconditionally (I mean on assumption of the cited requirements/conditions, but without invoking any formulations of the abc conjecture) the veracity of previous Claim for the smallest $q>0$ close to* $0$ that you are able to prove. Similarly**, is it possible to prove Conjecture? Many thanks.
*I'm curious to know what is the smallest $q>0$ close to $0$ such that the inequality in Claim is true, I think that the right discussion is for $q>0$ but if you want to discuss $|q|$ very close to $0$ because you think that it makes sense, feel free to study our inequality for real numbers $|q|$ very close to $0$.
$^{**}$On the other hand I think that should be possible to prove the Conjecture, since I think that this statement is much weaker than the abc conjecture.
I was inspired in the Wikipedia articles for Generalized mean and Logarithmic mean. I add references to bilbiography. I know the statement of formulation ABC conjecture II for example from [3].
References:
[1] P. S. Bullen, Handbook of Means and Their Inequalities, Dordrecht, Netherlands: Kluwer (2003).
[2] B. C. Carlson, Some inequalities for hypergeometric functions, Proc. Amer. Math. Soc., 17: in page 36 (1966).
[3] Andrew Granville and Thomas J. Tucker, It’s As Easy As abc, Notices of the AMS, Volume 49, Number 10 (November 2002).
(1/2) I'm curious about if these questions in my Question have a good mathematical content. I bring here this post asking to know if the Question is interesting in the context of the abc conjecture. I've asked in this site MathOverflow On variants of the abc conjecture in terms of Lehmer means, with identificator 350998 (asked on Jan 23 '20). If from your experience and knowledges you can explain if this kind of inequalities can be potentially interesting, please feel free to add your comments or explain it in your answer for the Question in this post. It is appreciate.
(2/2) I have hope that my inequalities are interesting, feel free to refer these if you know that some colleague (a professor) have studied the abc conjecture. On the other hand I am trying to read other questions posted on this MathOverflow about weak forms of the abc conjecture, and I know that in the literature there are also articles that were written by professors about relaxations of the abc conjecture.
abc implies $b-a$ can't be much smaller that $a+b$ and we have $\log(b-a) \sim \log(a+b)$
Many thanks for your contribution, and many thanks for the upvoter. As aside/unrelated comment the logarithmic mean, as refers the Wikipedia Stolarsky mean, is a particular case of the Stolarsky mean, I say it if can be inspiring for some user. If I understand well Lehmer was a number theorist and Stolarsky studies in particular diophantine approximation. Hölder provided us, in particular, his important inequality as refers the Wikipedia Hölder's inequality (I refer these facts for all users, since I like to study certain mathematical details when I try to evoke certain relationships).
Many thanks to the upvoter and for all those persons that read the post for his/her attention.
abc implies your conjecture with $b-a$.
Case 1 Let $a,b,c=a+b$ be bad abc triple,i.e. $c < rad(ab(a+b))$.
We have $rad(ab(a+b)) > c > b - a$.
Case 2 Let $a,b,c=a+b$ be good abc triple,i.e. $c>rad(ab(a+b))$.
Then $T : (b-a)^2,4ab,(a+b)^2$ is good abc triple too.
The radical is divisor of $ab(a+b)(b-a)$ and we have $(a+b)^2 > (a+b)(b-a)$.
If $\log(b-a) < (1-C) \log(b+a)$ this will give infinitely many
good abc triples with quality $2/(2-C)$, which contradicts abc.
In summary, abc implies there are only finitely many good abc triples
satisfying $\log(b-a) < (1-\epsilon) \log(b+a)$
Many thanks for your excellent and interesting answer. I was disconeccted (yesterday) and I don't know if I loss the opportunity to award your answer with the bounty that was expiring/ending (now I don't know if my words make sense, since I don't know how works a bounty for the last day in which is offering: this bounty expires just 7 hours ago and you answer is edited 15 hours ago).
I've flagged my (question) post asking, if possible, to award your answer with the bounty, I was disconnected (yesterday) and I don't know if I loss the opportunity to award the available answer while the bounty was expiring (I don't know if now is possible). Isn't required a response and good day.
The moderators told me about my flag that they don't have a method to restore bounties. Thus I'm sorry if as a consequence of that I was disconnected I can not award the bounty.
|
2025-03-21T14:48:30.526295
| 2020-05-08T11:07:40 |
359710
|
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|
Stack Exchange
|
Two results about (shifted) symplectic structures
I am now interested in shifted symplectic structures.
I found Zhang's results about symplectic structures (2011, p.3-4, arXiv link,
Comm. Anal. Geom. 2017) and Pantev–Toen–Vaquié–Vezzosi's results on shifted symplectic structures (2011, see p.1-2, arXiv link, Publ. IHES 2013).
I thought Zhang's results were completely included in the results of Pantev–Toen–Vaquié–Vezzosi. Is this idea correct?
(About the reason of asking the question, see my comment below)
Any comment welcome! Thank you.
"I thought were completely included": you mean "Zhang's results were completely included"? Zhang's paper was posted on arXiv 2 weeks after the other one. Possibly the comparison is relevant but I'm not sure this is the right place to compare papers and discuss their merits.
That's right, I'm sorry.
The reason I posted was that I first found Zhang's results and thought that similar results would be worth doing for moduli stacks of other kinds of sheaves (torsion-free sheaves and more generally coherent sheaves). I found the result of Pantev–Toen–Vaquié–Vezzosi and thought it was obvious to them.
First of all, let me say that it is not always easy to recover non-derived results from derived ones.
Second of all, it seems to me that Zhang's definition of a symplectic stack may not be accurate, in the sense that it doesn't generalize the notion of a smooth symplectic algebraic variety (or, homomophic symplectic manifold). Indeed, their definition only requires to have a non necessarily closed non-degenerate 2-form. This is manifest in Example 6.7 of Zhang's paper.
Now, can one deduce the result of Zhang from the one of Pantev-Toën-Vaquié-Vezzosi (PTVV, for short)? I do think so.
Indeed, as a special case of PTVV's general existence result of shifted symplectic structures on derived mapping stacks, one gets a 0-shifted symplectic structure on the derived moduli stack of perfect sheaves on a K3 surface. PTVV explain in their paper how one can recover the usual symplectic structure on the coarse moduli of simple sheaves (this is in subsection 3.1). The comparison with Zhang's non-degenerate pairing on the tangent complex of semi-stable sheaves on a K3 shall be very similar.
More precisely, in PTVV, if one forgets about the closedness of the 2-form (which is really the hardest part of the story) and only cares about the induced pairing on the tangent complex, it is just the standard construction (also used in Zhang's paper) with the Atiyah class and Serre duality. Then for the comparison, one shall just compare the cotangent complexes of the derived stack from PTVV and of the underived stack from Zhang, over the semi-stable locus. They do not exactly match, but I suspect that the discrepancy between the two is something like $\mathcal O[-1]\oplus\mathcal O[1]$ (see PTVV for more details).
Thank you very much for your comment !
|
2025-03-21T14:48:30.526531
| 2020-05-08T11:15:10 |
359711
|
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|
Stack Exchange
|
Constructing real numbers using Bolzano-Weierstrass Theorem?
$\mathbb{R}$ is generally constructed as equivalence classes of Cauchy sequences. As Cauchy Completeness and Archimedean Property together are equivalent to the Bolzano-Weierstrass Theorem, there should be a method to construct $\mathbb{R}$ from $\mathbb{Q}$ using the Bolzano-Weierstrass Theorem.
So, is there such a method? If so, could someone point me towards some resources related to this?
A corresponding [math.se] post: Constructing Real Numbers using Bolzano-Weierstrass Theorem?
That's my post. I didn't get a response there, so I've asked here.
Yes, it is recommended to link the posts to each other when [meta-tag:cross-posting]. That's why I commented under both psots.
I think the difficulty is that, whereas a Cauchy sequence or a Dedekind cut determines a single real, a bounded sequence determines, by Bolzano-Weierstrass, a nonempty set of reals, namely the set of accumulation points. So to produce a definition of the reals, you'd need to pick out one real from that set. It can certainly be done; e.g., the set has a largest element. But it looks ugly to me.
"Didn't get an answer" in 13 hours? I recommend allowing 3 days before concluding you won't get an answer.
Thanks @MartinSleziak. I did not know of this policy.
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2025-03-21T14:48:30.526674
| 2020-05-08T11:25:09 |
359712
|
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|
Stack Exchange
|
Space of holomorphic functions multiplied by smooth functions taking real values
Suppose we have a fixed function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ (sufficiently regular, say $C^\infty$ ). The question is: for which $f$ there exists a scalar function $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that the product $fg = (f_1g, f_2g)$ is holomorphic (using the standard identificacion $\mathbb{C} \sim \mathbb{R}^2$)? Are there any sufficient or necessary conditions for $f$ in order to exist such a $g$?
In other words, is there any "nice" description of the product-set of the set of holomorphic functions on the plane with the set of real (sufficiently regular) functions on the plane?
(Of course, one can try to solve Cauchy-Riemann but I get a pair of transport PDEs $(f_1g)_x = (f_2g)_y$ and $(f_1g)_y = -(f_2g)_x$ and is not clear that there is a common solution to both of them). Any references would be very much appreciated
EDIT: As someone commented I am looking for $g\neq 0$, this value of $g$ gives a trivial solution
$g \equiv 0$ always works. Presumably you look for a nontrivial solution.
First, if $fg$ were holomorphic and nontrivial, then $\{g = 0\}$ has to be discrete. This implies that $g$ has to be either $\geq 0$ or $\leq 0$. So we can assume WLOG $g \geq 0$.
Restricting away from its zero set, we can study the function $v = \ln g$. The Cauchy-Riemann relations becomes
$$ \begin{cases} f_1 v_x - f_2 v_y = (f_2)_y - (f_1)_x \\
f_2 v_x + f_1 v_y = - (f_1)_y - (f_2)_x \end{cases} $$
Unless $f_1 = f_2 = 0$ (note also that the set on which this happens is also discrete), the matrix $A_f = \begin{pmatrix} f_1 & -f_2 \\ f_2 & f_1\end{pmatrix}$ is invertible, and hence you can explicitly solve the expression for $v_x, v_y$ in terms of $f$ and its first derivatives.
To solve the "local" problem, all you are missing is the integrability conditions, which is $(v_x)_y = (v_y)_x$. If you write the vector
$$ V = A_f^{-1} \begin{pmatrix} (f_2)_y - (f_1)_x \\ - (f_1)_y - (f_2)_x \end{pmatrix} \tag{Vdef}$$
the integrability condition is $$ (V_1)_y = (V_2)_x. \tag{INT}$$
Summary
Starting from $f$, construct $V$ by (Vdef), which is well-defined away from the set $\{f = (0,0)\}$.
If (INT) fails to hold, then there exists no solution to your problem.
If $U$ is a simply connected domain on which (INT) holds, then there exists a function $v:U\to \mathbb{R}$ such that $V = \nabla v$. Then on $U$ you have $g = e^v$ solves your problem.
Additional necessary conditions
By Picard's Theorem, necessarily that every direction must be represented in the image of $f$ for there to be a solution $g$.
|
2025-03-21T14:48:30.526876
| 2020-05-08T11:34:49 |
359713
|
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|
Stack Exchange
|
Lyapunov condition for CLT for asymptotically independent sequence
Suppose I have some triangular array $\{X_{n,j}\}$ of random variables, which need not be independent or identically distributed. Suppose I further know that
$$Var\left(\sum_{j=1}^n X_{n,j}\right)\to \sigma^2 \text{ and } \sum_{j=1}^n \mathbb{E}|X_{n,j}|^{2+\epsilon}\to 0,$$
for some $\epsilon>0$. Then this triangular array satisfies the Lyapunov condition. However, for Lyapunov's CLT we further require independence for each $n$. Suppose instead that we have asymptotic independence in the sense that for any $t\in\mathbb{R},$
$$\left|\mathbb{E}\left[\prod_{j=1}^n\exp\left(itX_{n,j}\right)\right] - \prod_{j=1}^n\mathbb{E}\left[\exp\left(itX_{n,j}\right)\right]\right|\to 0,$$
where $i$ denotes the imaginary unit. Can we then still conclude that
$$\sum_{j=1}^n X_{n,j}\to N(0, \sigma^2)?$$
It is easy to see (by a Levy continuity theorem argument) that this is true if the covariances also fade fast enough, so that
$$\sum_{j=1}^nVar\left( X_{n,j}\right)\to \sigma^2,$$
but do we need this or can we also conclude asymptotic normality without this additional condition?
The answer is no. E.g., suppose that
\begin{equation*}
(X_{n,j})=(X_{n,j})_{j=1}^n\sim(1-u^2/n)N_n(0,I_n/n)+(u^2/n) N_n(0,J_n/n),
\end{equation*}
where $u\in(0,\infty)$, $I_n$ is the $n\times n$ identity matrix, and $J_n$ is the $n\times n$ matrix with all entries equal $1$, so that $N_n(0,J_n/n)$ is the distribution of the random vector $(Y,\dots,Y)/\sqrt n$ with $Y\sim N(0,1)$.
Thus, the distribution of $(X_{n,j})$ is the mixture of the $n$-variate normal distributions $N_n(0,I_n/n)$ and $N_n(0,J_n/n)$ with respective weights $1-u^2/n$ and $u^2/n$.
The idea here is to make the variance of the sum $\sum_j X_{n,j}$ greater (than it would be with $u=0$) without significantly affecting the distribution of the sum, which will still be $\approx N(0,1)$ even if $u\ne0$.
Then
\begin{equation*}
Cov(X_{n,j},X_{n,k})=(1-u^2/n)1_{j=k}/n+(u^2/n)/n
\end{equation*}
and hence
\begin{equation*}
Var\sum_jX_{n,j}=\sum_{j,k}Cov(X_{n,j},X_{n,k}) \\
=(1-u^2/n)+n(n-1)(u^2/n)/n\to1+u^2=:\sigma^2. \tag{1}
\end{equation*}
Next, for any real $\epsilon>0$
\begin{equation*}
E|X_{n,j}|^{2+\epsilon}=(1-u^2/n)/n^{1+\epsilon/2}+(u^2/n)/n^{1+\epsilon/2}
=1/n^{1+\epsilon/2}
\end{equation*}
and hence
\begin{equation*}
\sum_jE|X_{n,j}|^{2+\epsilon}=n/n^{1+\epsilon/2}\to0.
\end{equation*}
Similarly, $Ee^{itX_{n,j}}=e^{-t^2/(2n)}$ and hence
\begin{equation*}
\prod_1^n Ee^{itX_{n,j}}=e^{-t^2/2}.
\end{equation*}
Further,
\begin{equation*}
E\prod_1^n e^{itX_{n,j}}=(1-u^2/n)e^{-t^2/2}+O(u^2/n)\to e^{-t^2/2},
\end{equation*}
so that the condition
\begin{equation*}
E\prod_1^n e^{itX_{n,j}}-\prod_1^n Ee^{itX_{n,j}}\to0
\end{equation*}
holds as well.
However,
\begin{equation*}
\sum_j X_{n,j}\sim(1-u^2/n)N(0,1)+(u^2/n) N(0,n)\to N(0,1)\ne N(0,\sigma^2),
\end{equation*}
in view of (1).
Very nice example, I very much liked the intuitive explanation you added. Thank you!
|
2025-03-21T14:48:30.527194
| 2020-05-08T12:14:50 |
359715
|
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|
Stack Exchange
|
complemented subspace of the direct sum of two Banach spaces
When I was reading a paper, I saw something like:
If $F$ and $E$ are Banach spaces with symmetric bases (precisely, they are symmetric sequence spaces), and $F$ is isomorphic to a complemented subspace of $l_2\oplus E$, then $F=l_2$ or $F$ is isomorphic to a complemented subspace of $E$.
The author claimed that the result follows by standard elementary arguments. We omit the details. I don't know what the argument is. Any clue?
I don't understand the claim. What about $F=l_2 \oplus E$?
@JochenWengenroth I guess if $l_2\oplus E$ has a symmetric basis, then $l_2\oplus E \hookrightarrow l_2$ or $E$. I am not sure. My question came from the second sentence in the proof of Corollary 3.3. https://www.sciencedirect.com/science/article/pii/0022123681900525
This is well known but not trivial. It follows primarily from Theorem 2.c.13 of [LT]. The Theorem says (applied to this situation) if every operator $T:E\to \ell_2$ is strictly singular, then every complemented subspace of $\ell_2\oplus E$ is of the form (up to isomorphism) $X'\oplus E'$ for some complemented subspace $E'$ of $E$, and $X'$ is either finite dimensional or isomorphic to $\ell_2$. So if $E'$ is finite dimensional then of course $F$ is isomorphic to $\ell_2$. In other cases, since $F$ has symmetric basis, $F$ isomorphic to $\ell_2\oplus E'$ implies, $F$ is either isomorphic to complemented subspace of $E'$ (and hence of $E$) or $\ell_2$. For the last statement see Prop 3.b.8 of [LT].
If there is an operator $T:E\to \ell_2$ which is not strictly singular, then, using the definition of not strictly singular and the fact that every subspace of $\ell_2$ is complemented, you get a complemented subspace $E'$ of $E$ isomorphic to $\ell_2$. But then $E\oplus \ell_2$ is isomorphic to $E$ by decomposition method (using $E$ has symmetric basis).
[LT] Lindenstaruss-Tzafriri, Classical Banach spaces vol 1.
Well, it is a nice proof and I would say that it is truly not trivial. Very appreciate!
|
2025-03-21T14:48:30.527351
| 2020-05-08T13:08:38 |
359721
|
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|
Stack Exchange
|
Results on: (path/initial condition)-dependent variant of exponential map generates compactly supported diffeomorphisms
Let $M$ be a connected and simply connected Riemannian manifold, non-compact, and suppose that $\{V_p\}_{p \in M}$ is a family of vector fields on $M$ indexed by $M$. Suppose moreover that the map
$$
p\to V_p(x),
$$
is smooth and define the differential equation
$$
f_t' = V_p(f_t) ,\, f_0=p. \qquad (*)
$$
So in $(*)$ there is an explicit dependence on $p$ both in the initial condition (IC) and in the "dynamics" of the ODE itself. Therefore, let's relabel this explicit dependence on $p$ of a solution to $(*)$ by $f_{t,p}$.
Let $X$ be the collection of all compactly-supported smooth vector-fields on $M$ such that
A solution $f_{t,p}$ exists up to time $1$ for each $p$,
The map taking $(V_p)_{p \in M}$ to the time $1$ solution of $(*)$ is a diffeomorphism,
$p\to V_p(x)$ is non-constant for at-least one value of $x\in M$.
Denote this latter map by $\widetilde{\operatorname{Exp}}:X\to \operatorname{Diff}_{c,0}(M)$. It is known, see this post for references, that if there is no dependence on $p$ then $\operatorname{Exp}$ generates $\operatorname{Diff}_{c,0}(M)$. Are such results known in this case, i.e.: where vector fields depend non-trivially on the ICs?
Just to be sure, what is $\mathrm{Diff}_{c,0}$?
It's the identity component of the space of $C^{\infty}$ diffeomorphisms with compact support (in the sense of A. Banyaga; namely that they are the identity outside some compact in M).
Thanks. Got it. Here is another (dumb) question. Is $X$ nonempty?
Yes, ofcourse, we assume that to be the case (it can be shown also).
|
2025-03-21T14:48:30.527514
| 2020-05-08T13:45:06 |
359722
|
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"Arno Fehm",
"Daniel Loughran",
"Frank",
"Olivier Benoist",
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|
Stack Exchange
|
A property of varieties between unirational and retract rational
EDIT: The vague question Q1 below is partially answered, while the concrete question Q2 seems to be still open.
Let $V$ be a geometrically integral variety over a field $K$.
I consider the following properties:
(1) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for $n={\rm dim}(V)$.
(2) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is unirational)
(3) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with geometrically
integral generic fiber, for some $n$.
(4) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with a right inverse $V\dashrightarrow\mathbb{P}_K^n$, for some $n$. ($V$ is retract rational)
(5) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V\times\mathbb{P}_K^m$ for some $m,n$. ($V$ is stably rational)
(6) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is rational)
We have that $(6)\Rightarrow(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(2)\Leftrightarrow(1)$. For curves all of these properties are equivalent, but they diverge in higher dimension. From browsing the literature I gather that it is known that $(2)\not\Rightarrow(4)$ and $(5)\not\Rightarrow(6)$, and it is expected that $(4)\not\Rightarrow(5)$.
However, I am interested in property (3), which I could not find anywhere in the literature.
Q1: Does property (3) occur in the literature? Does it have a name? Is it equivalent to either (2) or (4)?
EDIT: As the answer of Daniel Loughran shows, the Châtelet surfaces described below are examples for $(3)\not\Rightarrow(5)$ over $K=\mathbb{R}$.
Phrased more down to earth, here is a very concrete question I am interested in:
Q2: Is every intermediate field $F$ of $\mathbb{R}(X,Y,Z)/\mathbb{R}$ which is algebraically closed in $\mathbb{R}(X,Y,Z)$ purely transcendental over $\mathbb{R}$?
Of course this is clear for $F$ of transcendence degree $0$, $1$ or $3$ over $\mathbb{R}$, so it is really just a question about surfaces. The equivalent question over $\mathbb{C}$ has a positive answer, as every unirational complex surface is known to be rational. The closest to a counterexample I found in the literature is the surface over $\mathbb{R}$ given by $x^2+y^2=f(z)$ with $f$ of degree $3$ with three real roots, which I think satisfies (2) but not (5), but I don't know if it satisfies (3).
EDIT: As explained in the answer of Daniel Loughran, such a surface $V$ satisfies (3). But it seems not clear whether the $n$ in $\mathbb{P}_K^n\dashrightarrow V$ with geometrically integral generic fiber can be chosen to be 3, which would be needed to answer Q2 negatively.
Are your maps in (1), (2), etc, $K$-defined?
Yes, everything defined over K. I'll change the $\mathbb{P}^n$ to $\mathbb{P}^n_K$ to make this clear.
I do not know if (3) has been studied much, but I have seen a similar condition but for covering families of curves in Kollár's notes https://arxiv.org/abs/0710.5516 Section 8 - called the Lefschetz condition. You might find some more references there.
Let $k$ be a field of characteristic $0$, $a \in k$ and $f$ a separable polynomial of degree $3$.
The projective surface $X$, given as the minimal smooth compactification of the affine surface
$$X: \quad x^2 - ay^2 = f(z)$$
you have written down is an example of a Châtelet surface. (Note that $X(k) \neq \emptyset$ always as there is a rational point at infinity). These have been studied in great detail by Colliot-Thélène and his collaborators. The key paper relevant to your question is:
Arnaud Beauville, Jean-Louis Colliot-Thélène, Jean-Jacques Sansuc and Peter Swinnerton-Dyer - Variétés Stablement Rationnelles Non Rationnelles, Annals of Mathematics.
Such surfaces are non-rational provided $a$ is not a square in any of the residue fields of the irreducible factors of $f$. Moreover, in the above paper, it is shown that they are stably rational provided certain assumption hold (e.g. $f$ is irreducible with Galois group $S_3$). But as remarked in the comments, there are examples which are also not stably rational.
They prove this using universal torsors
$$T \to X.$$
For a good overview of the theory of universal torsors, I would recommend the book
Skorogobatov - Torsors and rational points
I will just remark that these are torsors under the Néron-Severi torus, in particular the generic fibre is geometrically integral.
A sufficient criterion for the existence of a universal torsor is $X(k) \neq \emptyset$; but as already explained we have this property so universal torsors exist. There may be many universal torsors in general; but the twists of a given torsor gives a parametristation of the rational points of $X$. So there is always some torsor with a rational point. But it turns out that such torsors $T$ are birational to a complete intersection of two quadrics in projective space, which is shown to be a rational variety (details in the above paper). So this shows that (3) holds.
Altogether, this shows that there are Châtelet surfaces which satisfy (3), (5), but not (6), and also those which satisfy (3) but not (5) nor (6). This seems to give a complete answer to your question.
Further details on these results and constructions can be found in the seminar Bourbaki report:
Laurent Moret-Bailly - Variétés stablement rationnelles non rationnelles
Thanks, I will have a closer look at that paper. I am a bit confused though: $X(\mathbb{R})$ is a real manifold with two connected components, hence so is $(X\times\mathbb{A}^n)(\mathbb{R})$. Shouldn't this imply that $X\times\mathbb{A}^n$ and $\mathbb{A}^m$ are not birationally equivalent (over $\mathbb{R}$)?
I think you are both right. The (smooth projective model of) $X$ does not satisfy (4) or (5) because it has non-connected real locus. It satisfies (3) because a universal torsor with a real point is stably rational (by Colliot-Thélène, Sansuc and Swinnerton-Dyer, Intersections of two quadrics and Châtelet surfaces, II, Theorem 8.1).
@OlivierBenoist: Thanks a lot. So if I understand correctly, the example would indeed answer the question. The only thing that is not clear to me yet is why a universal torsor with a real point necessarily exists. Is that a general fact? (A reference would be appreciated - I have not yet been able to access CT-S' La descente sur les variétés rationneles where apparently they were defined.)
Hi. Yes you are right - to prove stable rationality one needs some genericity assumptions, e.g. the polynomial $f$ is irreducible with Galois group $S_3$. Anyway, you seem happy with my answer so I will upgrade it and add more details.
As Daniel Loughran explained to me by email, the universal torsors of these Châtelet surfaces are in fact of dimension 8, so this example does show that $(3)\not\Rightarrow(5)$, but it does not answer question Q2.
|
2025-03-21T14:48:30.905651
| 2020-05-08T14:07:42 |
359725
|
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"Adittya Chaudhuri",
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|
Stack Exchange
|
What is the notion of a group object and its action in a 2-category?
It is well known that a group object in a category $C$ (with terminal object $1$ and such that any two objects of $C$ have a product) is defined as an object $G$ in $C$ with the following morphisms:
$m:G \times G \rightarrow G$, $e:1 \rightarrow G$, $\mathit{inv}: G \rightarrow G$
satisfying some conditions modelled on the group axioms such that $m$ behaves as the multiplication map, e behaves as the identity and $\mathit{inv}$ behaves as the inverse map.
The group action (right) of a group object $G$ on an object $X$ in $C$ can be defined as a morphism $\rho:X \times G \rightarrow X$ in $C$ such that the following two diagrams are commutative:
where $\mathit{id}_X$, $\mathit{id}_G$ are identity morphisms at $X$ and $G$ respectively and $\mathit{pr}_1$ is the first projection on $X$ from the product $X \times 1$.
My Question are the following:
(1) What is the analogue of the above notion in a $2$-category? I couldn't find any literature in this direction.
So I tried to guess it's definition roughly in the following way:
Let $C$ be a 2-category (with terminal object $1$ and such that any two objects of $C$ have a product in the context of 2- category as mentioned in https://ncatlab.org/nlab/show/2-limit#2limits_over_diagrams_of_special_shape ). I define a group object in the 2-category $C$ as an object $G$ with the following 1-morphisms:
$m \in C(G \times G ,G)$, $e \in C(1,G)$, $inv \in C(G,G)$ satisfying some conditions similar as above but in this case every equality in the conditions will be replaced by an invertible 2-morphism satisfying certain appropriate coherent conditions.
Correspondingly the action of $G$ on an object $X$ in $C$ will be defined exactly in the same way but the above two diagrams will be commutative only upto invertible 2-morphisms.
Is my guess correct?
Even if it is correct but writing all the details (taking care of all the invertible 2 morphisms) seems very complicated to me and seems an inappropriate definition to work with.(Both in the context of strict 2-category and bicategory)
So what should be an appropriate definition for a group object in a 2-category and its corresponding action on an object? (Both in the context of strict 2-category and bicategory)
Secondly,
It is well known that the Strict 2-Group is a group object in Cat(when Cat is considered as a 1 -category or a usual category ).
But then
(2) What is the group object in Cat (when Cat is considered as strict 2-category)?
I would be also very grateful if someone can refer some literatures in this direction.
Thank you.
4 upvotes, 4 bookmarks is a good number :D Hope some one answers your question..
@PraphullaKoushik I also hope that :D
I'll have a go at answering your question (although higher category-theory is absolutely not my area of expertise).
The conditions for $(G,m,e,inv)$ to be a group object is stipulated by the following relations
$m\circ (e\times inv)\circ\Delta=m\circ (inv\times e)\circ\Delta=id_G$
$m\circ (m\times id_G)=m \circ (id_G\times m)$
$m\circ(e\times id_G)=m\circ(id_G\times e)=id_G$
where $\Delta$ is the diagonal map and $(e,id_G)$ is the obvious map $G\to G\times G$.
Writing these out in full is a bit pedantic but it's important. The reason for this is that we can see that adding 2-morphisms to obtain a strict 2-category doesn't change anything about the structure of $G$.
In particular we don't gain any structure from the 2-category case. Thus the strict 2-groups are actually group objects in $\mathbf{Cat}$ viewed as a strict 2-category.
There are other ideas for "group-like" things in higher categories. You can ask for things like weak 2-groups where we consider weak 2-categories and thus only require the composition of morphisms to be associative and unital up to some 2-isomorphism. In particular a weak 2-group is a monoidal category with all morphisms invertible such that for any object $x$ we have $x\otimes x^{-1}$ and $x^{-1}\otimes x$ are only isomoprhic to the tensor unit 1. (This corresponds to (1).) We can also define the notion of a coherent 2-group where we make specific choices for $x^{-1}$ and specific isomorphisms $x\otimes x^{-1}\to 1$ and
$1\to x\otimes x^{-1}$ such that these form an adjoint pair. All of this is developed in this paper of Baez's & Lauda's (see definition 20).
In particular, a group object (or a strict 2-group) in a 2-category with finite products is a coherent 2-group $G$ except the natural isomorphisms defining the "coherent structure" of $G$ are all simply the identity (see Definition 29 of the above paper).
Another reference which spells out different group-like structures in 2-categories as well as how these things act on categories $X$ is given in this paper by Morten's & Picken's.
edit: edited for clarity (see comments below)
Why do you think " we don't gain any extra structure from the 2 category case" ? In fact when each of the equality of the conditions you mentioned is replaced by a specific choice of 2-isomorphisms and the notion of product and terminal object is replaced by 2 product and 2 terminal object(that is they come with an isomorphism) why do you think they will not add any extra structure? Also a strict 2 group can be seen as a 2 category with one object and invertible 1-morphisms and 2-morphisms... Then how will you describe this in this 2 category case?
@AdittyaChaudhuri I said in the case of a strict 2-category we don't gain any structure for the group object. Again this is because we simply have a category enriched in categories -- however those equalities still have to hold on the nose, not just up to 2-isomorphism! As for the equivalence of the two definitions this follows from the fact that internal categories in Grp are equivalent to group objects in Cat -- this is explained well in this short paper of Forrester-Barker's.
@AdittyaChaudhuri the point is that the structure of a group object is completely determined by conditions on the objects and 1-morphisms. When considering a strict 2-category we don't change these things at all! We merely say that we also have morphisms of 1-morphisms but this doesn't at all change the conditions for a group object.
@AdittyaChaudhuri sorry for the comment spam but you should be careful distinguishing between 2-morphisms in your strict 2-group and natural transformations of functors of categories! In particular the definition of one object with invertible 1-morphisms and 2-morphisms corresponds to a category object in Grp which corresponds to a group object in Cat (see my first comment)
Thanks for referring to https://arxiv.org/pdf/math/0307200.pdf. I have seen this paper previously also but did not notice that the answer to my question is hidden in the definition 29, in the section Internalisation. I got your point that for saying , some "entity" is a group object in "something" that entity need to have a representation as a category object in the category of groups. But there is no analogous description as " a 2 category object" in groups. So the closest one can get to "this sort of notion" is given in the reference I mentioned above in this comment.
For getting my answer +1 to your answer!
@AdittyaChaudhuri Note that in Definition 29 at the end they state that a strict 2-group in a 2-category is "...a group in the underlying category of this 2-category" which is what I was trying to communicate above. I'm glad you are satisfied! :)
Thanks for explaining.
For the sake of completeness of your answer can you please add the notion of definition 29 in https://arxiv.org/pdf/math/0307200.pdf along with the reference? Hope this will help the future readers. Thank you.
@AdittyaChaudhuri Let me know if these edits work and whether you'd like me to make any more changes. And thanks for the feedback!
It looks fine. :)
Hey @AdittyaChaudhuri! I see you unaccepted my answer. Was there a problem with it? :(
Actually I need some time to accept. As my real motive behind this question is to understand the action of "some notion of group object in a 2-category" on another 2- category. Now there exists a 2-category consists of 2-categories as objects , 2-functors as 1-morphisms and 2-nautural transformations as 2-morphisms. Now strict 2-group is a group object in Cat and we have a notion of action of strict 2-group on a category. So a proper analogue in 2-categories should have the same properties. But I am not able to figure out properly this notion in the 2-category of 2-categories .
Lastly even if the notion you mentioned makes sense , but incase of 2 category of 2 categories it looks very complicated to me and quite complex to work with. For example in Cat the group object is quite simple . May be I am missing something. Your answer is fine . May be the way I wrote the question was misleading. I may post a new question in this direction and mention the link here !
So once I am convinced about the doubt I mentioned I will accept your answer. Also it is highly appreciated if you can mention the case of 2 category of 2 categories with little explanation regarding it. Thanks!
! I am apologising for my behaviour (first accepting and then unaccepting) and the inconvenience caused due to it.. I am again accepting your answer as my doubt is not clearly expressed through the way I asked the question here. In fact according to the question I posted, your answer is fine. So I will clearly write my doubt in details in a separate question and will mention the link here. Thank you.
|
2025-03-21T14:48:30.907978
| 2020-05-08T14:37:23 |
359730
|
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|
Stack Exchange
|
A recurrence formula for the Legendre function $P_\mu^\nu(x)$
Im looking for a recurrence formula of type:
$$(\mu-\nu) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)=?, \quad \mu,\nu\in \mathbb R$$
where $P_\mu^\nu(x)$ is the Legendre function of the first kind (solution to the Legendre differential equation which is regular at the origin).
My goal is to rewrite the sum in one expression, i.e. $(\mu-\nu) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)= C P_\alpha^\beta(x) $
Any useful reference, I will be very grateful. Thank you in advance
why do you think there would exist such a recursion?
I don't know, I just asked the question to know if it exists or not
Try this. http://people.math.sfu.ca/~cbm/aands/intro.htm#006. If it's not there then you have a problem on your hands.
You can also check it in I.S. Gradshteyn and I.M. Ryzhik-Table of integrals, series, and products-Academic Press (2007) If it's not there, then it has little chances of being true.
It's not really a well-posed question. Of course one can just go to Abramowitz/Stegun or to the NIST handbook and pick out two recurrence relations that contain the two terms on the l.h.s., shift everything else over to the r.h.s. and one has an answer to the OP's question as stated. The r.h.s. will probably be ugly and the answer not particularly useful.
Let's take relation 14.10.3 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(\mu -\nu ) P_{\mu }^{\nu } (x) - (2\mu -1)x P_{\mu -1}^{\nu } (x)
+ (\mu+\nu -1) P_{\mu -2}^{\nu } (x) =0
$$
We can thus isolate the desired l.h.s.,
$$
(\mu -\nu ) xP_{\mu }^{\nu } (x) + P_{\mu -1}^{\nu } (x) =
((2\mu -1)x^2 +1) P_{\mu -1}^{\nu } (x) - (\mu+\nu -1)x P_{\mu -2}^{\nu } (x)
$$
The r.h.s. can be consolidated into an expression containing a single Legendre function by also invoking relation 14.10.4 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(1-x^2 ) \frac{d}{dx} P_{\mu -2}^{\nu } (x) =
(\nu-\mu +1) P_{\mu -1}^{\nu } (x) + (\mu -1)x P_{\mu -2}^{\nu } (x)
$$
Solving for $P_{\mu -1}^{\nu } (x)$ and inserting above, we end up with
$$
(\mu -\nu ) xP_{\mu }^{\nu } (x) + P_{\mu -1}^{\nu } (x) =
\left[ \frac{(2\mu -1)x^2 +1}{\nu -\mu +1} \left( (1-x^2 ) \frac{d}{dx}
- (\mu -1)x \right) -(\mu +\nu -1)x \right] P_{\mu -2}^{\nu } (x)
$$
Thank you very much M. Engelhardt
See equation 37 here. And now some more characters.
The formula 37 is for the associated Legendre polynomials $P^n_m$ (i.e, $n,m$ are positive integers). In our case: $\mu, \nu\in \mathbb R$.
This formula is valid for any real $n$, $m$.
Which is what i can take for $ \nu $ and $\mu$ to arrive at the formula $37$?
You can take two instances of this formula and construct your lhs, for example. The problem is that your question is not well formulated. Recursion in what? In $\mu $? In $\nu $? In both? Upwards? Downwards? Are derivatives allowed? What data can we assume to be known? Usually the point of a recursion is to be able to take a set of known instances and derive an unknown one. You haven't specified any of this, so the best we can do is point you to some general relations.
my goal is to rewrite the sum in one expression, i.e. $(\mu-\nu+1) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)= C P_\alpha^\beta(x) $
Can $C$ contain a derivative operator w.r.t $x$?
$C$ can depend on $x$ is not a problem
|
2025-03-21T14:48:30.908819
| 2020-05-08T15:24:27 |
359731
|
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|
Stack Exchange
|
Density property for Sobolev spaces
My question is as follows: is the space $ C_c^{\infty}(\mathbb{R}^3 \setminus \mathcal{C}) $ dense in $ H^1( \mathbb{R}^3) $ where $ \mathcal{C} $ is the circle $ \{(x,y,z) \in \mathbb{R}^3 \mid x^2 + y^2 = 1,\ z=0 \}$? In the radial case (i.e. $ H^1_{rad}( \mathbb{R}^3) $ ), does the proof become simpler and can we construct a sequence of approximations explicitely?
Thanks in advance.
Of course not, the result you get by completion is simply $H^1_0(\mathbb R^3\setminus C)$, the set of $H^1$ functions vanishing on the boundary.
This question would be better suited for Math.stackexchange, vote to close
I'm sorry for the mistake in writing. In fact $ \mathbb{R}^2 $ should be $ \mathbb{R}^3 $
what's $C$, then? you can edit your question after it's posted, please do so.
You edited the question so I edited my answer. The statement is the same and you still get $H^1_0$
|
2025-03-21T14:48:30.909184
| 2020-05-08T15:59:29 |
359733
|
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|
Stack Exchange
|
Rings whose finitely-generated modules are co-hopfian
Let $A$ be a unital, possibly noncommutative ring. Dischinger showed [1] that the following are equivalent:
For every $a \in A$, there exists $n \in \mathbb N$ such that $a^n A = a^{n+1} A$;
For every $a \in A$, there exists $n \in \mathbb N$ such that $A a^n = A a^{n+1}$;
Every cyclic right module $M \in Mod_A$ is co-hopfian (i.e. every injective endomorphism of $M$ is an isomorphism);
Every cyclic left module $M \in {}_A Mod$ is co-hopfian.
Such a ring $A$ is called strongly $\pi$-regular. Every strongly $\pi$-regular ring $A$ is $\pi$-regular in the sense that for every $a \in A$ there is some $n \geq 1$ such that $a^n$ is a von Neumann regular element of $A$ (i.e. $a^n = a^n b a^n$ for some $b \in A$). Every von Neumann regular ring is strongly $\pi$-regular, and if $A$ is commutative, then $A$ is strongly $\pi$-regular iff $A/nil(A)$ is von Neumann regular.
Dischinger also showed that the following are equivalent:
Every finitely-generated right module $M \in Mod_A$ is co-hopfian;
Every finitely-generated left module $M \in {}_A Mod$ is co-hopfian;
Every finite-rank matrix ring over $A$ is strongly $\pi$-regular;
Every ring right (or left) Morita equivalent to $A$ is strongly $\pi$-regular.
Let's call such a ring $A$ very strongly $\pi$-regular. Every von Neumann regular ring is very strongly $\pi$-regular.
Questions:
Is there a direct ring-theoretic, rather than module-theoretic, characterization of very strongly $\pi$-regular rings which doesn't explicitly mention matrix rings?
What is an example of a very strongly $\pi$-regular ring $A$ such that $A$ is not von Neumann regular but the center $Z(A)$ is von Neumann regular?
[1] Dischinger, Friedrich. “Sur Les Anneaux Fortement π-Réguliers.” CR Acad. Sci. Paris Sér. AB 283, no. 8 (1976): 571–573. link.
(2) Let $M$ be the monoid ${1,x,y}$ with $xy=xx=x$, $yx=yy=y$ and unit $1$. Let $K$ a field, and $A=KM$. Then the center of $A$ is reduced to $K$ (so is von Neumann regular). $A$ itself is not von Neumann regular because $x-y\notin (x-y)A(x-y)=0$. But since $A$ is a finite-dimensional $K$-algebra, clearly every finitely generated left $A$-module is cohopfian.
The algebra suggested by @YCor is isomorphic to 2x2 upper triangular matrices over K which obviously has the desired properties since it has a non-trivial Jacobson radical the strictly upper triangular matrices
In general the path algebra of a finite connected acyclic quiver with at least one edge over K has center K and is not von Neumann regular but is strongly $\pi$-regular.
Just would like to +1 Benjamin's comment: von Neumann regular does not imply strongly $\pi$-regular. A strongly (von Neumann) regular ring is strongly $\pi$-regular, and a von Neumann regular ring is $\pi$-regular. Here are a few other examples of VNR rings that aren't strongly $\pi$-regular.
Since "strong von Neumann regularity" isn't Morita invariant, it would not be possible to use it as a replacement as an example of "very strong $\pi$-regularity"
@rschwieb Wow, DaRT looks awesome! I wish it had "pure semisimple" and "finite representation type", but what it has so far is quite impressive!
@TimCampion I’m open to adding new conditions, especially when you can help provide interconnections between conditions and good papers for sources. Register and use a submission form and we can talk about it :)
|
2025-03-21T14:48:30.910436
| 2020-05-08T16:00:29 |
359734
|
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|
Stack Exchange
|
Free category with product and coproduct
Is there a known description of the free category with both product and coproduct?
That is, given a small category $C$, I want to consider a category $U C$ which has product and coproduct, a functor $C \to U C$ and such that $UC$ is universal for functor preserving both product and coproduct. The case $C = \emptyset$ is already interesting.
I'm also happy to focus on finite product and finite coproduct, especially if it avoids some size problems, though I don't think this is essential.
My guess is that this category should be a category of two-player games (player and opponent) with morphisms being simulation and where outcome of the game are marked by objects of $C$ (if $C = \emptyset$ we should just have a win/lose outcome):
The coproduct of a family of games is the game where the player first chooses which game he wants to play in the family, while their product is the game where the opponent chooses which game he wants to play. The initial object is the game where player loses at the start, and the final object is the one where opponent loses at the start.
But the details of this, and especially the proper definition of the morphisms are a bit involved, so I'm curious whether this has been worked out somewhere.
Of course, as soon as we assume compatibility between product and coproduct (for e.g. distributivity) there are simple description, but here I'm interested in the completely unconstrained situation.
A syntactic construction of the free finite product/coproduct bicompletion is given in Cockett–Seely's Finite sum–product logic. There are also the earlier, suggestively-named papers Free bicomplete categories and Free bicompletion of enriched categories of Joyal, though I couldn't find these online. Hu–Joyal's Coherence Completions of Categories and Their Enriched Softness describes the free completion under products, coproducts and a zero object.
Hughes' A canonical graphical syntax for non-empty finite products and sums treats nonempty finite products and coproducts on discrete categories, where they state that (at point of publication) a categorical formulation of the result for non-discrete categories and empty products/coproducts was an open question.
Thank you very much for the references, that sounds very interesting.
One last reference: the introduction to Cockett–Santocanale's On the word problem for ΣΠ-categories, and the properties of two-way communication is a nice survey, and also describes this category in terms of games. As far as I can tell, there hasn't been progress on this since, which would suggest the problem is still open.
@vakor : You should post these as an answer. I havn't finished studied all these paper, but they seem to answer my question. or at least give a pretty good account of what is known on the topic. especially the last one.
The general problem of giving a categorical construction of the free category with finite coproducts and products (or "free sum–product category") seems to still be open, though there are several works on special cases of the problem.
Cockett–Santocanale's On the word problem for ΣΠ-categories, and the properties of two-way communication gives a good introduction to the problem. They state:
There have been, directly or indirectly, a number of contributions towards
our goal in this paper. [...] These related results, however, work only for the fragment without units – or, more precisely, for the fragment with a common initial and final object. As far we know, there is no representation theorem for the full fragment with distinct units.
Special cases of interest include:
A syntactic construction of the free category with finite coproducts and products (and various characterisations): Finite sum–product logic, Cockett and Seely.
A construction of the free category with coproducts, products and a zero object: Coherence Completions of Categories and Their Enriched Softness, Hu and Joyal.
A construction of the free category with nonempty finite coproducts and products: A canonical graphical syntax for non-empty finite products and sums, Hughes.
Joyal has two related papers (at least for general colimits and limits), but unfortunately without explicit constructions or proofs:
Free bicomplete categories.
Free bicompletion of enriched categories.
Joyal's papers are available here and here.
|
2025-03-21T14:48:30.911504
| 2020-05-08T16:00:57 |
359735
|
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|
Stack Exchange
|
Area of an intersection of three ellipses
Let $\Delta := (A,B,C)$ be a triangle that is defined by three points in the Euclidean plane that are not collinear.
Let further $E_{(A,B),\,C},\,E_{(C,A),\,B},\,E_{(B,A),\,B}$ be the set of ellipses with two of the corners of $\Delta$ as their foci and the third on their boundary.
Question:
What is the area of the intersection of $E_{(A,B),\,C},\,E_{(C,A),\,B},\,E_{(B,A),\,B}$ expressed as a function of the triangle's sidelengths $a,\,b,\,c$?
You might look at Finding ellipse-ellipse intersections in $\mathbb{R}^2$. Calculating the intersection of two ellipses is already a complex calculation, involving finding roots of quartics.
|
2025-03-21T14:48:30.911765
| 2020-05-08T16:58:39 |
359738
|
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|
Stack Exchange
|
Vanishing of a Higher Brauer group of a field
Let $k$ be a field. I am interested in the notion of the higher Brauer group defined as follows: For X a smooth scheme over $k$, $Br^r(X):=H^{2r+1}_{et}(X, \mathbb{Z}(r))$, an etale motivic cohomology group. I would like to get a vanishing of this group for $X=Spec~ k$, in terms of the etale cohomological dimension of $k$.
My crude attempt at this is as follows: Fix a prime $l$. The $l$-primary torsion subgroup of $Br^r(k) $ is given by
$H^{2r+1}_{et}(Spec~k, \mathbb{Z}_{l}(r))$. Now $H^{2r+1}_{et}(Spec~k, \mathbb{Z}_{l}(r))\simeq H^{2r}_{et}(Spec~k,\mathbb{Q}_l/\mathbb{Z}_{l}(r))$.
Now depending on the prime $l$, for $l\neq char~k$,
$\mathbb{Q}_l/\mathbb{Z}_{l}(r)\simeq \lim_n\mu_{l^n}^{\otimes r}$.
For $l=char~k$, it is isomorphic to $\lim_n \nu_n(r)[-r]$. This is in B. Kahn's paper Definition 2.7.
Using these if $l\neq char~k$:
$H^{2r}_{et}(Spec~k,\mathbb{Q}_l/\mathbb{Z}_{l}(r))=0$ for $2r>cd(k)$,
and for $l= char~k$:
$H^{2r}_{et}(Spec~k,\mathbb{Q}_l/\mathbb{Z}_{l}(r))=H^{r}_{et}(Spec~k,\lim_n \nu_n(r))=0$
for $r>cd(k)$.
I am unsure about the last steps as the sheaves involved are non-torsion.
Any comments are welcome!
The sheaf $\mathbb Z(r)$ is $(-r)$-connective by definition, so $H^i(X,\mathbb Z(r))$ = 0 if $i > cd(X) + r$. In particular, $Br^r(X) = 0$ for $r > cd(X) - 1$.
|
2025-03-21T14:48:30.912161
| 2020-05-08T17:06:43 |
359739
|
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|
Stack Exchange
|
Exponential iterates of a complex number
Let $f:\mathbb C\to \mathbb C$ be defined by $f(z)=e^z-1$. Let $f^n$ denote the $n$-fold composition of $f$.
In my new paper Erdős space in Julia sets I show that $$Z:=\{z\in \mathbb C:\lvert\operatorname{Im}(f^n(z))\rvert\to\infty\}$$ contains a homeomorphic copy of the set of points in Hilbert space $\ell^2$ which have all rational coordinates. But I do not know of a specific complex number $z$ which belongs to this set. It is easy to find $z$'s for which the real part goes to infinity; $\operatorname{Re}(f^n(1+0i))\to\infty$ as $n\to\infty$, but the imaginary part of $f^n(1+0i)$ is always $0$. So the question is, can you give the precise coordinates of a point in the complex plane which belongs to $Z$?
How about an answer to this question for $f(z)=e^z$?
Would you be amenable to linking to the arXiv rather than to ResearchGate? (You've done so; thanks!) I always prefer interacting with the former. Anyway, what does it mean to say that a subset of $\mathbb C$ contains $\ell^2(\mathbb Q)$?
@LSpice Ok I changed the link to arXiv. I mean that $\ell^2(\mathbb Q)$ is homeomorphic to a subset of $Z$.
It depends on what you mean by precise coordinates. I am not sure that I would expect to find a number that has a specific closed form. But then, I do not know how to find a "precise" point where the real part tends to infinity under iteration either, had you not chosen a real parameter.
To find such a point up to arbitrary precision, on the other hand, is quite easy. For instance, let $N\geq 0$, define
$$ z_{N,N} := \log N + 2\pi i N. $$
and inductively let $z_{N,j}$ (for $j<N$) be the preimage of $z_{N,j+1}$ in the strip at imaginary parts between $(2j-1)\pi$ and $(2j+1)\pi$.
Then $z_{N,0}$ converges to a point $z_0$ such that $f^j(z_0)$ has imaginary parts between $(2j-1)\pi$ and $(2j+1)\pi$, and in particular the imaginary parts converge to infinity.
Here is some simple python code:
>>> n=100
>>> orbit = [0]*(n+1)
>>> orbit[n] = math.log(n) + 2*math.pi*1j*n
>>> for j in range(n):
orbit[n-j-1] = cmath.log(orbit[n-j]+1) + (n-j-1)*2*math.pi*1j
I get approximately $z_0 = 2.1302059107690132+1.1190548923421213j$. Of course, given the very strong expansion and instability, if you iterate forward this will only follow the desired orbit for a small number of iterations (for me, it stays close for 10 iterations), but Gottfried Helms has given a higher-precision estimate below.
EDIT. My notation was somewhat messed up above in the original post; it should be better now I hope. Note that the $\log N$ can be omitted in the definition of $z_{N,N}$, and you will converge to the same value.
I've tried my procedures adapted to the function $f:z \to \exp(z)-1$ and approximated your given value of $z_0$ further. Using z0= 2.130205910769013348261983648<PHONE_NUMBER>61596052319<PHONE_NUMBER>290821840652119357270 + 1.11905489234212128966<PHONE_NUMBER>034645067307<PHONE_NUMBER>01033623172<PHONE_NUMBER>337388*I one can iterate 24 times, and then it gets periodic (of course with increasing error, which can be reduced when internal precision for computations is for instance 800 dec digits). Your value agrees convincingly to the result given a $K$-vector in my notation as $K=[11,10,...1,0]$
Hi Lasse, I'm still interested in this approximation. Just wanted to see if there are any typos above because I don't quite understand the z's.
Also does this technique allow you to specify the external address beforehand?
@D.S.Lipham Yes, something was definitely wrong! Hopefully it makes more sense now. Yes, you can specify the external address, but you have to be a little bit careful about where you place the real part, and whether you want to get a point on the ray or the endpoint. Essentially you can use my model from my paper "Topological dynamics of exponential maps on their escaping sets" to tell where the n-th image of your desired point should roughly be, then pull back under the actual exponential map under consideration. If that makes any sense.
I have one idea, but am not sure whether this is really leading somewhere respective your question.
This is derived from a recent experimental work of mine, where I discuss a method to create $n$-periodic points for the iterated function $f: z \to \exp(z)$.
Giving a vector $K$ of $n$ branchindexes for the (iterated) logarithm as parameter, the implemented function gives back an initial value $z_0$ which can be iterated $n$ times with function $f$ and whose imaginary part increases (roughly by $2\pi î$) over that iterations.
For instance calling z0 = find([6,5,4,3,2,1,0]) find a value $z_0$ which is $7$-periodic. This does of course not solve your problem of an imaginary part growing to infinity, but might give an idea... :
The vector of branchindexes can be made arbitrarily long, for instance I've tried with length n=128, K=[127,126,...,0] and found the following result z0=find(K):
2.090728841+ 1.235766664*I <-- z0 (only 10 signif. digits displayed)
2.660235518+ 7.640965408*I ... list of iterates ....
3.023074791+ 13.97645147*I ...
3.289206714+ 20.28951149*I ...
3.499351193+ 26.59232200*I
3.672981867+ 32.88951626*I
3.820915018+ 39.18326906*I
3.949779448+ 45.47474630*I
4.063931345+ 51.76463340*I
4.166386797+ 58.05336160*I
4.259320451+ 64.34121681*I
...
6.676175218+786.9605354*I
6.682733533+793.2510338*I
0.8873292332+798.4983613*I
2.090728841+ 1.235766664*I (getting periodic after 128 iterations)
Comparing the results $z_{0,n}$ where $n$ is the length of the vector $K$ of branchindexes with $n \in \{16,32,64,128 \}$
display reduced to 60 dec digits, internal precision 800 dec digits
-----------------------------------------------------------------------
length(K) real part of z0
16 2.09072884145065670358930701024074056461462449074482469887391
32 2.09072884145065670358930571871821763780774559404297853095181
64 2.09072884145065670358930571871821763780774559404297853095179
128 2.09072884145065670358930571871821763780774559404297853095179
length(K) imag part of z0
16 + 1.23576666409482263688534788841502804976771359539736086858255*I
32 + 1.23576666409482263688534750071480578600396895244789044041851*I
64 + 1.23576666409482263688534750071480578600396895244789044041849*I
128 + 1.23576666409482263688534750071480578600396895244789044041849*I
... we get the impression, that there is some asymptotic for $z_0$ towards $n \to \infty$ which we can approximate arbitrarily near. (Note however, that for $n=128$ I used $800$ dec digits internal precision). (A visual example is in the paragraph about "aperiodic points" in the manuscript linked to below)
(If this seem to be useful for you at all, then a small manuscript at my webspace is more informative, where I explain this procedure a bit more. For readability I've introduced in the answer here the name find() for the main-function.
This is all written in Pari/GP and if you need this I can provide the scripts)
update By configuring the vector $K$ more exessive, for instance K=[2^31,2^30,2^29,...,2^0] we get initial values $z_0$ whose iterations gives exponentially growing imaginary values (until periodicity occurs).
Example:
2.665280329+7.622847729*I <-- z0
3.292353128+13.98978765*I ... iterates over f ...
3.951062797+26.61442216*I ...
4.627032158+51.78428650*I
5.311751681+102.0721534*I
6.000773471+202.6161507*I
6.691894308+403.6854928*I
7.384041486+805.8134993*I
8.076693109+1610.063510*I
8.769593941+3218.560203*I
9.462618394+6435.551762*I
10.15570435+12869.53388*I
10.84882095+25737.49759*I
11.54195286+51473.42471*I
12.23509241+102945.2788*I
12.92823577+205888.9869*I
13.62138104+411776.4031*I
14.31452727+823551.2354*I
15.00767397+1647100.900*I
15.70082092+3294200.229*I
16.39396798+6588398.887*I
17.08711510+13176796.20*I
17.78026225+26353590.84*I
18.47340941+52707180.10*I
19.16655659+105414358.6*I
19.85970376+210828715.7*I
20.55285094+421657429.8*I
21.24599812+843314858.1*I
21.93914530+1686629715.*I
22.63229248+3373259428.*I
23.32543966+6746518854.*I
2.088818122+1.349303771E10*I ... after this periodicity occurs ...
The initial value $z_0$ to $60$ dec digits is
z0 = 2.66528032862300130094954352169380883320313130819912077261863
+ 7.62284772864970968721488615058188954049634904915456329976660*I
It is perhaps of interest, that the values in the list above correspond to the $1$-periodic fixpoints, which can of course be adressed by the Lambert-W-function with its complex branches. See the following list
2.665280329+7.622847729*I -W(-1,-(2^1+1))= 2.653191974+13.94920833*I
3.292353128+13.98978765*I -W(-1,-(2^2+1))= 3.287768612+26.58047150*I
3.951062797+26.61442216*I -W(-1,-(2^3+1))= 3.949522742+51.76012200*I
.... ... ...
18.47340941+52707180.10*I -W(-1,-(2^24+1))= 18.47340941+105414358.6*I
19.16655659+105414358.6*I -W(-1,-(2^25+1))= 19.16655659+210828715.7*I
19.85970376+210828715.7*I -W(-1,-(2^26+1))= 19.85970376+421657429.8*I
20.55285094+421657429.8*I -W(-1,-(2^27+1))= 20.55285094+843314858.1*I
21.24599812+843314858.1*I -W(-1,-(2^28+1))= 21.24599812+1686629715.*I
21.93914530+1686629715.*I -W(-1,-(2^29+1))= 21.93914530+3373259428.*I
22.63229248+3373259428.*I -W(-1,-(2^30+1))= 22.63229248+6746518854.*I
23.32543966+6746518854.*I -W(-1,-(2^31+1))= 23.32543966+1.349303771E10*I
2.088818122+1.349303771E10*I -W(-1,-(2^32+1))= 24.01858684+2.698607541E10*I
It looks like when two consecutive W-values are taken as the edge-points of the rectangle in the complex plane, and they mark the anti diagonal, then the values of the list mark the top left of the main-diagonal, and approach the true main-diagonal edges more and more when the index of the entry in the list increases. Perhaps this gives a description for an asymptotic behaviour of the resulting coordinate if the index (and the complex component) approaches infinity.
The structure of periodic points of the exponential map (and cases such as $e^z-1$ is well-understood, this goes back to Devaney & Krych. There you can also find examples of points whose imaginary parts tend to infinity. I have posted an answer at https://math.stackexchange.com/a/3789416/16629 .
|
2025-03-21T14:48:30.914115
| 2020-05-08T17:18:43 |
359740
|
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"Gerhard Paseman",
"Sam Hopkins",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359740"
}
|
Stack Exchange
|
Value (not position)- based sorting; reference request
A recent answer of Ville Salo on the diameter of a Cayley graph induced by bubble sort generators (adjacent transpositions) has inspired this variation.
Many sort algorithms are position based: you go through a list of pairs of positions in the (integer) array, check the two values, and perform a swap or not. For many simple algorithms (especially in sorting networks) the list of position pairs is fixed, while for less simple algorithms the list may be created based on the input (viz. Quicksort).
The question above uses a limited list of pairs of positions (adjacent transpositions) to investigate. Suppose we use a limited list of values instead?
For the next bit of the post, assume we have scanned an array representing a permutation of the first n positive integers. We can peek all we want, but our swaps are limited to swapping adjacent values: if k and k+1 are both integers in the array, we are allowed to swap them, otherwise keep looking. Thus we never swap the two values 3 and 8 in this scheme, for example.
There may be a duality between indices (position labels) and values that could be exploited to answer questions like the diameter of the analogous graph, but I am not seeing it yet. The first question for me is: does such an exploitable duality exist?
The second and official question for this post is: has the idea of (restricted) value based sorting been explored in the literature? If so, what literature and by what name?
Gerhard "Sorting With A Different Priority" Paseman, 2020.05.08.
I didn't carefully read the whole post, but isn't the duality in permutations between indices and values just inversion?
(another way to say that would be: multiplying on the right vs. left.)
@SamHopkins You may be absolutely right. Unfortunately I am currently challenged in doing any categorical thinking or explaining. If you post a clear exposition on this as an answer, I will upvote it even if it turns out not to be the answer. Gerhard "Does Not Vote Very Often" Paseman, 2020.05.08.
An inversion of a permutation $\pi=\pi_1\pi_2\ldots\pi_n$ written in one-line notation is usually defined to be a pair of indices $(i,j)$ with $1 \leq i < j \leq n$ for which $\pi_i > \pi_j$. Inversions are "out-of-order positions." Traditional position-based sorting is about eliminating inversions. But note that the inversions of $\pi^{-1}$ are precisely the pairs $(\pi_{j},\pi_{i})$ for $(i,j)$ an inversion of $\pi$. In other words, these "inverse inversions" are "out-of-order values." So sorting based on values corresponds to eliminating inverse inversions, and has an exactly parallel theory to the usual kind of sorting.
One of these kind of inversions is called "left inversions" and the other "right inversions," but I always get mixed up about which is which. The point is it has to do with whether we are multiplying permutations on the left or right. Indeed, containment of inversion sets gives a partial order on the symmetric group called weak order; the Hasse diagram of weak order is the same as the Cayley graph of the symmetric group with the adjacent transpositions as generators, and the edges of this Hasse diagram correspond as one would expect to multiplying by an adjacent transposition (this graph is also the same as the 1-skeleton of the permutohedron). But there are two kinds of weak orders: left weak order and right weak order, and the distinction is whether we are using left or right inversion sets (equivalently, whether we are multiplying the adjacent transpositions on the left or the right).
I think this is what I seek; I'll mull over it. It definitely addresses the posted question and gives me the concept of weak order. You get an upvote from me. Gerhard "Encourages You To Add More" Paseman, 2020.05.08.
I posted the version involving a permutation on n integers to make the discussion easier. In applications, you may not know the value location relative to all the other elements (like you do with integers) but you still may be able to tell adjacency/nonadjacency between two objects. Assuming a zero cost adjacency test, how would you approach the analogous questions? Would it still be "multiplying on the right" even if you can't tell what the actual multiplicand is? Gerhard "Let's Fight For Human Lefts!" Paseman, 2020.05.08.
Actually, with cheap adjacency tests, you can do a quadratic pass to convert the arbitrary array into an integer array, so a better question is what to do when an adjacency test has a cost less than a swap. Gerhard "So Swap Those Notions Around" Paseman, 2020.05.08.
|
2025-03-21T14:48:30.914788
| 2020-05-08T18:02:28 |
359743
|
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|
Stack Exchange
|
Gromov-Hausdorff distance between graphs with edges as part of the space versus not part of the space
Let $G_1$ and $G_2$ be finite simple graphs viewed as metric spaces in the natural way where the edges are not part of the space. Let $G_1'$ and $G_2'$ be copies of $G_1$ and $G_2$ resp. but with the edges included in the space. Is it necessarily true that $$d_{GH}(G_1', G_2') \geq d_{GH}(G_1, G_2)$$
where $d_{GH}$ is the Gromov-Hausdorff Distance?
what happens for the example where $G_1$ is a circle with $n$ edges and $G_2$ is a circle with $n+1$ edges?
Just for formulation, you could define $G_i$ as the 1-skeleton and $G_i^0$ as its 0-skeleton; the question is then whether $G_i\mapsto G_i^0$ is 1-Lipschitz.
@HenrikRüping yes it works: let $\Delta_n$ be the circle with $n$ edges. Then clearly $d(\Delta_i^0,\Delta_j^0)\ge 1/2$ for $i<j$ (since otherwise two vertices of $\Delta_j$ would be at distance $<1/2$ of the same vertex of $\Delta_i$). But it is easy to see that $d(\Delta_3,\Delta_4)\le 1/4$.
@HenrikRüping I think you might be right that that is a counter-example. I thought of that but did the calculation and somehow thought that the obvious imbedding for the with edges case gave $\frac{1}{2}$.
|
2025-03-21T14:48:30.914973
| 2020-05-08T18:53:21 |
359751
|
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"Moishe Kohan",
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"url": "https://mathoverflow.net/questions/359751"
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|
Stack Exchange
|
Purely analytic proof of the Nielsen-Thurston classification theorem
I hope this question is appropriate for the site. I've been looking at the expositions of Bers' proof of the Nielsen-Thurston classification given in Hubbard's Teichmüller Theory
and Applications to Geometry, Topology, and Dynamics and in Farb-Margalit's A primer on mapping class groups. I was wondering if it's possible to circumvent the use of hyperbolic geometry in the proof.
More precisely, given $S = S_g$, elements are classified according to their translation distance when acting on Teichmueller space $(\mathcal{T}(S), d_{\mathrm{Teich}})$. Here the translation distance of an element is $D(\varphi) = \inf_{\mathcal{X} \in \mathcal{T}(S)} d_\mathrm{Teich}(\mathcal{X}, \varphi \cdot \mathcal{X}) $.
There are three cases:
If $D(\varphi) = 0$ and the infimum is realized, $\varphi$ is periodic
If $D(\varphi) >0 $ and the infimum is realized, $\varphi$ is pseudo Anosov.
If the infimum is not realized, $\varphi$ is reducible.
The first two cases can be handled using only the "analytic" definition of Teichmueller space, as the space of marked Riemann surface structures on $S$. In the first case, you can show that $\varphi$ is (or rather, has a representative that is) conjugate to an isometry of a Riemann surface structure $X$ on $S$, and since $\mathrm{Aut}(X)$ is finite, you are done. In the second, you show $\varphi$ is conjugate to a Teichmueller map and hence pseudo Anosov.
But in the third case, Mumford's compactness theorem is used in a crucial way to produce a multicurve that is preserved by $\varphi$. This involves looking at the lengths of closed geodesics with respect to the hyperbolic metric on a Riemann surface, and then controlling how much $\varphi$ can distort these lengths.
So, the questions that I have in mind are (the last two I'm interested in not only because of the proof of this particular theorem, but for their own sake):
Is there any other way of producing an invariant multicurve only using Riemann surface structures rather than hyperbolic structures on $S$?
Is there some translation for "hyperbolic length of the geodesic representative of the isotopy class of a simple closed curve" to Riemann surface language?
Is there an analog of Mumford's compactness theorem (describing some family of compact subsets of moduli space that also give an exhaustion of it) in "analytic" terms?
I'm not hoping for a positive answer to question 1, since I've never seen a complex analysis statement that says anything that would imply "this curve should go exactly here". It (and question 2) might also be a bit misguided, since hyperbolic and Riemann surface structures on a surface are so closely related that it wouldn't make sense to separate them. But maybe there is something that can be done. Thanks!
You can use the extremal length (which is defined complex-analytically) instead of the hyperbolic length.
|
2025-03-21T14:48:30.915321
| 2020-05-08T19:43:01 |
359754
|
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"Elwood",
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"url": "https://mathoverflow.net/questions/359754"
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|
Stack Exchange
|
Uniqueness condition for Hamilton-Jacobi equation?
Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that
$$
\partial_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}^d.
$$
As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance (say in $d = 1$) the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.
Assume that $f$ is a Lipschitz function that satisfies the equation above at every point of differentiability of $f$; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?
If we suppose that the function is smooth for some time, and the second derivative converges almost everywhere to something as we approach a possible blowup time, then I think we can use the assumptions to show that the second derivative converges in $L^1$; and this excludes the possibility of the most naive type of blowup (that is, a "reasonable" way for the second derivative to converge to something with a Dirac mass somewhere). But I'm not sure how to go further than this.
Proposition A.2 of https://arxiv.org/abs/2104.05360 shows that the answer is "yes". In fact, this is also valid for a general nonlinearity in the equation in place of the squared norm here.
|
2025-03-21T14:48:30.915790
| 2020-05-08T19:55:42 |
359756
|
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"sort": "votes",
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|
Stack Exchange
|
The bidual of the space of divergence-free vector fields
Consider the Banach space $L_1(\mathbb R^n, \mathbb R^n)$ of integrable vector fields $(n>1$) together with its subspace $N$ formed by those vectors fields whose divergence (computed in the distributional sense) is zero.
Is there any workable description of $N^{**}$?
Of course, even the description of $L_1^{**}$ is elusive, yet somewhat concrete (measures on the spectrum of $L_\infty$).
The annihilator of $N$ is called the space of closed currents.
$N$ is not complemented in $L_1(\mathbb R^n, \mathbb R^n)$ and $N^\perp$ is not complemented in $L_\infty(\mathbb R^n, \mathbb R^n)$.
$N$ is not an $L$-summand in $N^{**}$ (Godefroy--Lerner).
Is $N$ complemented in $N^{**}$?
|
2025-03-21T14:48:30.915940
| 2020-05-08T20:05:43 |
359757
|
{
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"DSM",
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|
Stack Exchange
|
The functional equation $f(xy) - 2 f(\frac{x+y}{2}) + f(x+y- x\cdot y) = 0$
Incidentally, I came across the following functional equation
$$f(xy) - 2 f(\frac{x+y}{2}) + f(x+y- x\cdot y) = 0$$
that is to hold for all $x,y\in \mathbb R$. Is there a neat way to find all solutions $f:\mathbb R\to \mathbb R$?
How did you come across it?
Taking $y = 0$ gives $f(x) = 2f(x/2)$, so that $f(0) = 0$ and $f : x \mapsto \lim_{n \to \infty} 2^n f(x/2^n)$ is entirely determined by its germ at $0$. Taking $y = -x$ gives $f(x^2) + f(-x^2) = 2f(0) = 0$, so $f$ is odd.
Taking $y=1$ gives $f(x)+f(1)=2f((x+1)/2)$, so combining with $f(x)=2f(x/2)$ gives $f(x'+1/2)=f(x')+f(1)/2$, where $x'=x/2$.
There is the trivial $f(z)=az+b$.
$f(xy) = axy+b$, $-2f((x+y)/2) = -2(a((x+y)/2)+b) = -(a(x+y)+2b)$, $f(x+y-xy) = a(x+y-xy)+b$. They add up to zero.
(I should have said that taking $y = 0$ gives $f(x) + f(0) = 2f(x/2)$, as @MichaelRenardy correctly points out in their answer below. Oops!)
Adding a constant to f does not change the property, so we may assume f(0)=0. Under this assumption, I claim that the property holds if and only if f is additive. The if part is obvious. For the only if part, we first set y=0 to obtain f(x)=2f(x/2)-f(0)=2f(x/2). Next, we set x+y=a, xy=b, and we obtain f(b)+f(a-b)=2f(a/2)=f(a), at least for b<0 (in which case there is always a corresponding x and y which is real). To deal with the opposite sign of b, simply reverse the roles of a and a-b. This proves f is additive. If f is also continuous, it must be linear. If you are pro-choice, you can characterize discontinuous solutions after you pick a basis for the real numbers as a vector space over the rationals.
|
2025-03-21T14:48:30.916170
| 2020-05-08T20:07:40 |
359758
|
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|
Stack Exchange
|
Origin of the term 'index of a subgroup'
The index of a subgroup $H$ in a group $G$ is the number of distinct cosets of $H$ in $G$.
Why did someone decide to call this an 'index'? What's the rationale for this?
The short answer is Cauchy, with only justification: “for short”. Burnside’s Theory of groups of finite order (1897) has a useful glossary stating, p. 382 (my bold):
The ratio of the order of a sub-group $H$, to the order of the group
$G$ containing it, is called by French writers the “indice,” by German
the “Index,” of $H$ in $G$. The phrase is most commonly used of a
substitution group in relation to the symmetric group of the same
degree.
The French writers in question are those mentioned in Burnside’s preface: Serret’s Cours d’Algèbre Supérieure (1866) (“the first connected exposition of the theory”) has, p. 287:
432. $\vphantom{\frac{\mathrm N}m}$ To abbreviate the discourse, I will call index of a system of conjugate substitutions formed with $n$ letters, the quotient obtained by dividing the product $\mathrm N=1.2.3\dots n$ by the number which expresses the system’s order. If $m$ denotes the index of a conjugate system of order $\mu$, one will have $m=\frac{\mathrm N}\mu$.
And already Cauchy’s memoir Sur le Nombre de Valeurs qu’une Fonction peut acquérir, lorsqu’on y permute de toutes les manières possibles les quantités qu’elle renferme, J. École Polytechnique 10 (1815) 1–28 (from which “the theory of groups of finite order may be said to date”) has, p. 6:
if one represents by $R$ the total number of essentially different values of the function $K$, $M$ being the number of values equivalent to $K_\alpha$ (...) one will have $RM=N$ (...) Hence $R$ (...) can only be a factor of $N$, that is, of the product $1.2.3\dots n$ (...) For short, I will henceforth call index of the function $K$, the number $R$ which indicates how many essentially different values this function can assume.
Note that the index is to “indicate” (but this could be said of any named count, especially ratio...), and that the paper’s title itself restates this very definition — it could be shortened “Sur l'Indice”.
Let me add that "indice" is used in French in the meaning "subscript".
|
2025-03-21T14:48:30.916500
| 2020-05-08T20:18:32 |
359760
|
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|
Stack Exchange
|
Is $\mathcal{S}(\mathbb{R}^n)$ a tame Fréchet space?
Hamilton's paper "The Inverse Function theorem of Nash and Moser" (1982, Bull. Amer. Math. Soc, vol. 7, n. 1, page $137$) proves that $C^{\infty}(M)$ is a tame Fréchet space when $M$ is a compact manifold. It was asked here on MO if this space is tame in the non-compact case, and in the second answer, there was an argument that for a space to be tame, every sufficiently large taming semi-norm would be a norm (which is false in $C^{\infty}(M)$ for the non-compact case), but this is true for $\mathcal{S}(\mathbb{R}^n)$ (the space of functions that are rapidly decreasing, along with its derivatives). So to disprove tameness, a different argument would be necessary.
So the question is, is $\mathcal{S}(\mathbb{R}^n)$ a tame Fréchet space (with the usual Fréchet topology)? Is there a reference?
Using Fourier series, $C^\infty(S^1)$ is ismorphic to the sequence space
$$s=\{(x_k)_{k\in \mathbb N}\in \mathbb C^{\mathbb N}: \sum_{k=1}^\infty k^{2n} |x_k|^2 <\infty \text{ for all $n\in\mathbb N$}\}$$
and this space is isomorphic to $\mathscr S(\mathbb R^d)$.
This is very classical, a modern treatment (with much information about structural properties of Fréchet spaces) is in the book Introduction to Functional Analysis of Meise and Vogt.
As a coda, there is a general rule of thumb: all Fréchet spaces of distribution theory are isomorphic. There are, of course, exceptions but the following holds—if $T$ is a self-adjoint operator on a Hilbert space, then the intersection of the domains of definition of ${T^n}$ is, in a natural way, a Fréchet space (Pietsch). If the spectrum of $T$ consists of a sequence of eigenvalues asymptotically like $(|n|^\alpha)$ for some positive $\alpha$, it is isomorphic to $s$. The classical Sturm-Liouville, Laplace and Schrödinger operators generate most spaces of test functions
I looked into the book by Meise and Vogt but cannot find any mention of tame Frechet spaces.. Moreover, all Frechet spaces are isomorphic as well-known. So, $\mathcal{S}(\mathbb{R}^n)$ is isomorphic to $C^\infty(M)$ for a non-compact (but locally compact) manifold. However, the OP says $C^\infty(M)$ is NOT tame...
So...I still find your answer incompete (or unconvincing to me at least..)
|
2025-03-21T14:48:30.916973
| 2020-05-08T21:04:55 |
359766
|
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|
Stack Exchange
|
How to make sense of recursively defined SPDE solutions, like in Hairer's "Solving the KPZ equation" paper?
In Martin Hairer's 2013 paper "Solving the KPZ equation", the process $X_\epsilon^\bullet$ is defined as the stationary solution to
$$
\partial_t X_\epsilon^{\bullet} = \partial_x^2 X_\epsilon^{\bullet} + \Pi_0^{\perp} \xi_\epsilon
$$
and then recursively, for each binary tree $\tau = [\tau_1, \tau_2]$ with $\bullet$ as the root,
$$
\partial_t X_\epsilon^{\tau} = \partial_x^2 X_\epsilon^\tau + \Pi_0^{\perp} (\partial_x X_\epsilon^{\tau_1} \partial_x X_\epsilon^{\tau_2})
$$
with $\Pi_0^{\perp} = 1 - \Pi_0$, $\Pi_0$ being the orthogonal projection onto constants in $L^2$.
How to make sense of these equations, which include the product of solution derivatives and noise mollification? Can it be seen from the semigroup framework, as in Da Prato and Zabczyk's book, or is there a more appropiate point of view (like Hida's white noise approach)? Any reference would be greatly appreciated.
There is some talks Hairer gave via zoom at a UCLA seminar, I haven't watch them but maybe you can find a hint there, another wild guess would be that it is kind of similar to a Zwanzig-Nakajima projection/equation
What I mean is that
$$
X_\epsilon^\tau(t) = \int_{-\infty}^t P_{t-s} \Pi_0^\perp (\partial_x X_\epsilon^{\tau_1}(s)\, \partial_x X_\epsilon^{\tau_2}(s))\,ds\;,
$$
where $P_t$ denotes convolution with the heat kernel. The product appearing on the right is the usual pointwise product of two random variables with values in the space of continuous functions. The projection $\Pi_0^\perp$ guarantees that the integral converges.
If you replace this product by the Wick product from white noise analysis you'll get a different process which will not lead to the physically relevant notion of solution.
|
2025-03-21T14:48:30.917205
| 2020-05-08T21:34:40 |
359769
|
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|
Stack Exchange
|
Projection formula for flat morphisms
Let $f\colon X\to Y$ be a flat morphism between two smooth projective varieties. Let $L$ be a locally free sheaf on $X$ and $\mathcal{F}$ a coherent sheaf on $Y$. How to prove $f_*(L\otimes f^*\mathcal{F})\cong f_*L\otimes\mathcal{F}$? I think it is a well-known result but I couldn't find a reference. You can assume $f$ is smooth and surjective if you want. Thank you.
Tag 0B54 gets pretty close, but not quite all the way there. Because $f$ is flat, you may write $f^$ for $Lf^$; because $Y$ is smooth quasi-projective, $\mathscr F[0]$ is perfect (you can find a finite locally free resolution). Because $L$ is locally free, you can write $\otimes$ on the RHS instead of $\otimes^{\mathbf L}$. This only leaves the derived tensor product on the LHS, but I don't see a reason why that should be usual tensor product in this situation. (This could be a way to look for a potential counterexample.)
By the way, this is true with no assumption on $f$ and $L$ (in fact, for any morphism of ringed spaces) if $\mathcal{F}$ is locally free.
Based on my comment, I constructed the following counterexample (which I believe is standard):
Example. Let $(E,O)$ be an elliptic curve, let $Y = E$ and $X = E \times E$, with $f \colon X \to Y$ the first coordinate projection. Let $\mathscr L = \mathcal O_{E \times E}(\Delta - E \times O)$, and let $\mathscr F = \mathcal O_O$.
Then $f_* \mathscr L = 0$, since $H^0(U \times E, \mathcal O_{U \times E}(\Delta|_U - U \times O)) = 0$ for every open $U \subseteq E$ as $\mathscr L|_U$ is a (fibrewise) degree $0$ line bundle that is not trivial.
On the other hand, $\mathscr L \otimes f^* \mathscr F = \mathcal O_{O \times E}$ since $\Delta|_{O \times E} = (O,O) = (E \times O)|_{O \times E}$. The short exact sequence
$$0 \to \mathcal O_X(-O \times E) \to \mathcal O_X \to \mathcal O_{O \times E} \to 0$$
gives a long exact sequence
$$0 \to \mathcal O_E(-O) \to \mathcal O_E \to f_*\mathcal O_{O \times E} \to \mathcal O_E(-O) \to \mathcal O_E \to R^1f_* \mathcal O_{O \times E} \to 0$$
since $R^if_* \mathcal O_X(-O \times E) = \mathcal O_E(-O)$ for $i \in \{0,1\}$ by the usual (derived) projection formula. Thus,
\begin{align*}
& & & & f_*\big(\mathscr L \otimes f^*\mathscr F\big) = \mathcal O_O \neq 0 = f_*\mathscr L \otimes \mathscr F. & & & & \square
\end{align*}
Remark. What's going on is that
$$Rf_* \mathscr L = \mathcal O_O[-1]$$
is not flat over $Y$ even though $\mathscr L$ is. There is a $\mathscr Tor_1$ term interfering in the derived projection formula of [Tag 0B54]. To see the above formula for $Rf_* \mathscr L$, use the short exact sequence
$$0 \to \mathcal O_X(-E \times O) \to \mathscr L \to \mathscr L|_{\Delta} \to 0.\tag{1}\label{1}$$
Since $\mathcal O_X(\Delta)|_\Delta = T_E = \mathcal O_E$, we get $\mathscr L|_\Delta = \mathcal O_E(-O)$. Note that $f$ induces an isomorphism $\Delta \to E$, so the long exact sequence of \eqref{1} reads
$$0 \to f_*\mathscr L \to \mathcal O_E(-O) \to \mathcal O_E \to R^1f_* \mathscr L \to 0.$$
Above we computed $f_* \mathscr L = 0$, so the map $\mathcal O_E(-O) \to \mathcal O_E$ is the natural inclusion, so $R^1f_* \mathscr L = \mathcal O_O$. $\square$
On the other hand, $Rf_* \mathcal O_X = \mathcal O_E \oplus \mathcal O_E[-1]$ is a complex of free modules, so the derived tensor product of the LHS of [Tag 0B54] is just a usual tensor product, as we saw implicitly in the computation of $Rf_* \mathcal O_{O \times E} = Rf_* f^* \mathcal O_O$ above.
Dear Remy, I think your counterexample is right and thank you. However, I am actually in a special situation that I demand the $L$ to be a torsion line bundle on an abelian variety. The $L$ in your example is not a torsion line bundle. In this case if $f_*L\neq0$, then the projection formula is true by base change theorem. If $f_*L=0$, then i don't know how to do it. I wish to prove that $h^0$ of $L$ restricted to each fibre would be 0 but I couldn't.
Even if $\mathscr L = \mathcal O_X$ I am not sure what to expect. For a smooth morphism in characteristic $0$ it is a highly nontrivial theorem that $R^if_* \mathcal O_X$ is locally free for all $i$ (uses Hodge theory), but in positive characteristic this is unknown as far as I'm aware. (Warning: in the case above the derived pushforward $Rf_* \mathcal O_X$ splits because this is true for $R\Gamma(E,\mathcal O_E)$, but in general it will be a little more complicated and you need to think about a spectral sequence for Tor as you filter $Rf_* \mathscr L$ by cohomological degree.)
That said, if your interest is specifically in characteristic $0$, then maybe the Hodge theory literature might be of help ― sometimes results for $\Omega^i_{X/Y}$ can be extended to torsion line bundles. I'm not super familiar with this, so I don't know any good references off hand.
Let's work with a fibration betweem abelian varieties over $\C$. It needs not to be complicated. When $L=\mathcal{O}X$, then $h^0$ is always 1. And you will have that the projection formula is true for any quasi-coherent $\mathcal{F}$ by base change. See Qing Liu's book algebraic geometry and arithmetic curves Remark 5.3.21(c) for example. When $L$ is only torsion and $f*L\neq0$, it's similar since in this case $L$ restricted onto fibre is trivial.
I think a more precise version well deserves a new question, so that other people can share their thoughts!
|
2025-03-21T14:48:30.917796
| 2020-05-08T22:03:33 |
359771
|
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|
Stack Exchange
|
Complete list of indecomposable representations of Temperley-Lieb algebras at roots of unity?
The Temperley Lieb algebra $TL_n$ at roots of unity is not semisimple. The standard representations $V_{n,p}$ are indecomposable but, in general, not irreducible. If $K_{n,p}$ is the sub-representation of $V_{n,p}$ given by the kernel of the usual bilinear form, the quotients $V_{n,p}/K_{n,p}$ are irreducible. For one dimensional representations, we have the trivial representation, and additionally, we sometimes have certain special one dimensional representations at roots of unity - for example at $\delta = 1$ ($q = \zeta_{12}$), we have two additional one dimensional representations - If we let $u_1,..,u_{n-1}$ be the generators of $TL_n$, we could send every $u_i$ to either $+1$ or $-1$, and this would define a representation. My question is: are there any others? Could there by some other indecomposable representation which is not listed above?
Thank you in advance.
David Ridout kindly informed me that the answer is in: https://arxiv.org/pdf/1605.05159.pdf. This is Theorem 3.10. The notation of that theorem is as follows: $S_k$ is the standard representation, $P_k$ is the projective cover of the irreducible obtained by quotienting by the radical (the kernel of the bilinear form). $B^l_k$ and $T^{l'}_k$ were new to me, and are defined in Section 3.1 (page 20).
|
2025-03-21T14:48:30.918001
| 2020-05-08T22:04:21 |
359772
|
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|
Stack Exchange
|
Ricci flow proof of isoperimetric inequality
It is well-known in geometric analysis that one can use curve-shortening flow to prove the isoperimetric inequality (where the general result requires curve-shortening flow for non-convex curves).
I was wondering if it might be possible to also prove such an inequality using Ricci flow (given hypotheses on convexity at the boundary and total curvature constraints if necessary).
Anthony Manning proved that the volume entropy decreases under volume normalized Ricci flow on surfaces of negative curvature. Question 4 at the end of his paper asks whether the Cheeger isoperimetric constant is a strictly increasing function of Ricci flow. So it looks like this is an open question.
There is a curious sort of topological isoperimetric inequality in three dimensions. Starting with a hyperbolic metric on an acylindrical 3-manifold with minimal boundary, one may run Ricci flow on the doubled manifold normalized by the minimum of the scalar curvature. Then this normalized volume is minimized by the hyperbolic metric with maximal area (totally geodesic) of the boundary. So the ratio of area to volume with sectional curvature at least $-1$ and minimal boundary is minimized in the totally geodesic case.
This is proved by monotonicity formulas of Hamilton and Perelman.
Let $\lambda(g)$ be the minimal eigenvalue of the operator
$-4\Delta_g + R$, where $R$ is the scalar curvature and $\Delta_g$ is the Laplacian. Define the quantity
$$V_{\lambda}(g) = Vol(M,g)(-\frac16 \min\{\lambda(g),0\})^{\frac32}$$
(there is a similar formula holding in any dimension).
Then $V_{\lambda}(g)$ is monotonically decreasing in dimension
3 for Ricci flow with surgery (the analogous quantity in dimension
2 will also be monotonic). One can actually see that
all of the normalized eigenvalues are monotonic.
In particular, if $R(g_0)$ is constant, then the first
eigenvalue $\lambda_1(g_0)$ will be determined by $R(g_0)$
and the second eigenvalue of $-4\Delta_g+R$. For a
surface of negative curvature, one also has monotonicity
of
$$V_R(g) = V(M,g) (-\frac16 \min\{R_{min}(g),0\})^{\frac32}.$$
As $t\to \infty$, $g_t$ will approach a constant curvature
metric, so I think this should give a relation between
the eigenvalue of the Laplacians for the initial and final
metrics. In turn, the first eigenvalue of the Laplacian is
related to the Cheeger constant by the Cheeger and Buser
inequalities. But this is probably not the sort of relation you're looking for.
|
2025-03-21T14:48:30.918323
| 2020-05-08T22:05:14 |
359773
|
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|
Stack Exchange
|
Postnikov invariants of the Brauer 3-group
Given a commutative ring $k$ there is a bicategory with
algebras over $k$ as objects,
bimodules as morphisms,
bimodule homomorphisms as 2-morphisms.
This is a monoidal bicategory, since we can take the tensor product of algebras, and everything else gets along nicely with that.
Given any monoidal bicategory we can take its core: that is, the sub-monoidal bicategory where we only keep invertible objects (invertible up to equivalence), invertible morphisms (invertible up to 2-isomorphism), and invertible 2-morphisms.
The core is a monoidal bicategory where everything is invertible in a suitably weakened sense so it's called a 3-group.
The particular 3-group we get from a commutative ring $k$ could be called its Brauer 3-group and denoted $\mathbf{Br}(k)$. It's discussed on the $n$Lab: there it's called the Picard 3-group of $k$ but denoted as $\mathbf{Br}(k)$.
Like any 3-group, $\mathbf{Br}(k)$ has homotopy groups which I will call $\pi_1, \pi_2, \pi_3$ (though there are choices of where we start numbering). These are well-known things:
$\pi_1$ is the Brauer group of $k$.
$\pi_2$ is the Picard group of $k$.
$\pi_3$ is the group of units of $k$.
My question is whether people have studied, or computed, the Postnikov invariants involving these things. The simplest is the map
$$ a : \pi_1^3 \to \pi_2$$
coming from the associator in the monoidal category of $k$-algebras (with isomorphism classes of bimodules as morphisms). Since the associator obeys the pentagon identity this is a 3-cocycle on $\pi_1$ with values in its module $\pi_2$, so it gives an element of $ H^3(\pi_1, \pi_2)$.
Is this element trivial? If not, what is it?
But in fact $\mathbf{Br}(k)$ is not just a 3-group but also a symmetric monoidal bicategory. So, it's what I call a symmetric 3-group, though some others call it a Picard 2-category. These have a number of other Postnikov invariants:
Nick Gurski, Niles Johnson, Angélica M. Osorno and Marc Stephan, Stable Postnikov data of Picard 2-categories.
Has anyone figured out any of these for $\mathbf{Br}(k)$?
As you point out, $\mathbf{Br}(k)$ is symmetric, and so the Postnikov invariants are (the destablizations of) stable cohomology operations. This highly constrains the question.
Br(k) is the 0th space of an HZ-module spectrum, so its Postnikov invariants are trivial (the Postnikov tower is noncanonically split).
Nice! Is this why you can compute the groups I'm calling $\pi_i$ using Galois cohomology as $H^{3-i}(\mathrm{Gal}(K|k), K^\star)$ where $K$ is the separable closure of $k$?
They have a common explanation. Assume k a field for simplicity, let K be a separable closure, and let G = Gal(K/k). Let Br(k) denote the classifying space for your 2-category (so it's the loop space of what you're denoting by Br(k) ). Then Br(K) carries a continuous action of G, and there's a natural map e from Br(k) to the (continuous) homotopy fixed points Br(K)^hG. The input you need is the following: 1) The map e is a homotopy equivalence (because the construction k -> Br(k) satisfies etale descent). (cont)
Br(K) is an Eilenberg-MacLane space K( K^, 2) (since the Picard and Brauer groups of a separably closed field vanish). Concretely this gives you a formula for pi_( Br(k) ) in terms of Galois cohomology. It also tells you that Br(k) admits the structure of a topologica/simplicial abelian group (since for an Eilenberg MacLane space Br(K), choosing such a structure is equivalent to choosing a base point, and the structure survives passage to homotopy fixed points when the base point is fixed by G).
Let me see if I understand what Jacob says in the comments. I think his argument can be summarized as: the Brauer 3-group is étale-locally an Eilenberg-MacLane spectrum, hence étale-locally an $\mathbb{Z}$-module spectrum, hence an $\mathbb{Z}$-module spectrum, hence the Postnikov tower splits. Do I have that right?
If so, I want to point out that the situation changes with one tweak, which is to allow superalgebras, super bimodules, etc. When $k = \mathbb{R}$ the super Brauer 3-group has a nontrivial homotopy operation $\pi_1 \to \pi_3$ given by taking the super dimension of the zeroth Hochschild homology of a superalgebra in the super Brauer group, and I computed an example of it taking a nontrivial value here. This implies that the Postnikov tower can't split but I don't know what the $k$-invariants are. I suppose following Jacob we could try to work things out as homotopy fixed points of the $\text{Gal}(\mathbb{C}/\mathbb{R})$-action on the super Brauer 3-group over $\mathbb{C}$ but this is beyond me.
The super Morita category is a great source of examples! For another example, the $k$-invariants for $\mathbb C$, see Dan Freed's Vienna notes around (1.42). Another proof, computed in a slightly different way, can be found here (see 4.3.5).
|
2025-03-21T14:48:30.919139
| 2020-05-08T22:21:02 |
359774
|
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"url": "https://mathoverflow.net/questions/359774"
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|
Stack Exchange
|
Residually amenable groups
I have two questions about residually amenable groups:
Is every finitely presented amenable group residually elementary amenable?
Given $n$, is the free Burnside group of exponent $n$ on two generators residually amenable?
Regarding 1., I know that Grigorchuk constructed an example of a finitely presented amenable group that is not elementary amenable, but I was unsure if it could be residually elementary amenable. Regarding 2., I believe it is an open question whether or not the free Burnside group on two generators is sofic, so if the answer to 2. is known, I'm guessing that it would be in the negative.
(1) No, there's a finitely presented ascending HNN extension of Grigorchuk's group, and it's an isolated group (it has a unique minimal nontrivial normal subgroup— more precisely every proper quotient is metabelian [Sapir-Wise]), so it's not residually elementary amenable. See details and references §5.7 in my 2007 paper with Guyot and Pitsch on isolated groups (arXiv link).
For (2) this group does not exist: it depends on an exponent and the answer depends on the exponent. However I think that the only cases for which the answer is known are those for which it's known to be finite (exponent $1,2,3,4,6$). That is, my guess is that even for large odd $n$ the failure of residual amenability is unknown, but this should be confirmed by specialists.
Thanks for the answer to (1). For (2), yes, I really did mean this family of groups.
Benjamin Weiss asked which free Burnside groups are sofic. https://www.jstor.org/stable/25051326
@YCor: it seems to me that, unlike either of the posted answers, your comment actually answers a part (i.e. part (1)) of the question. (It also seems that part (2) is open.) Perhaps you should post your comment as an answer?
@HJRW ok, I've just done so.
(From my initial comments)
No, there's a finitely presented ascending HNN extension of Grigorchuk's group, and it's an isolated group (it has a unique minimal nontrivial normal subgroup— more precisely every proper quotient is metabelian [Sapir-Wise]), so it's not residually elementary amenable. See details and references §5.7 in my 2007 J. Algebra paper with Guyot and Pitsch on isolated groups ArXiv link.
As confirmed by Mark Sapir for each given exponent and number $\ge 2$ of generators the question of residual amenability of the given Burnside group is open, except in the few cases where it's known to be locally finite (exponent $\le 4$ and $6$).
About 2), It is unknown if every bounded torsion amenable f.g. group is finite.
If that was true then for free Burnside groups residually amenable would be equivalent to residually finite. By Zelmanov's result, for large enough exponents, the free Burnside group with more than 1 generators is not residually finite.
|
2025-03-21T14:48:30.919911
| 2020-05-08T22:56:35 |
359776
|
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|
Stack Exchange
|
$\mathcal{M}_g$ and $\mathcal{A}_g$ have natural structures as quasi-projective varieties
Reading M. Hindry and J. H. Silverman (Diophantine Geometry-An Introduction), I find the claim that $\mathcal{M}_g$ and $\mathcal{A}_g$ have natural structures as quasi-projective varieties. Mumford and Fogarty's book (Geometric Invariant Theory) is indicated as a reference for this statement. However, it is an advanced book for me. I cannot identify where this is proven in the book of Mumford and Fogarty. Can anyone help me locate me ???
I think the answer might depend on why you want to know. (e.g. someone might instead recommend a later exposition.)
Can you recommend a reference, where it contains the proof that $\mathcal{M}_g$ and $\mathcal{A}_g$ are quasi-projective varieties, it would be good :)
In the 3rd edition of GIT, this is Theorem 7.10 for $\mathscr{A}_g$ and Corollary 7.14 for $\mathscr{M}_g$. The question would be more appropriate on MSE.
The GIT proof gives very nice compactifications of these spaces (and is the "right" way to do this), but they were known to be quasiprojective varieties long before GIT was developed.
The classical proofs depend on properties of theta functions. For $\mathcal{A}_g$, it should be attributed to some combination of Satake and Baily, and the appropriate references are
Satake, Ichiro
On the compactification of the Siegel space.
J. Indian Math. Soc. (N.S.) 20 (1956), 259–281.
and
Baily, Walter L., Jr.
Satake's compactification of Vn.
Amer. J. Math. 80 (1958), 348–364.
A textbook reference for this is
J. Igusa, Theta functions, Springer, New York, 1972.
For $\mathcal{M}_g$, the first person to show that it was a quasiprojective variety was Baily. In fact, what he did was show that the Schottky locus in $\mathcal{A}_g$ is an open dense subset of its closure in the Satake compactification. The reference is
Baily, Walter L., Jr.
On the moduli of Jacobian varieties.
Ann. of Math. (2) 71 (1960), 303–314.
Do any of these references work in characteristic p, or was the proof in GIT the first proof in general?
@Anonymous: These proofs only work over $\mathbb{C}$. I think the GIT proof was the first one that worked over any other fields.
Note that it is not at all clear that the Schottky locus is the same as $\mathcal{M}_g$ (the problem is with the hyperelliptic locus). This was proved much later by Oort and Steenbrink.
@abx: That's a good point, though if you just care about being quasiprojective it is not a huge deal since the map from $\mathcal{M}_g$ to the Schottky locus is bijective, and thus certainly quasi-finite.
|
2025-03-21T14:48:30.920297
| 2020-05-09T00:50:08 |
359778
|
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|
Stack Exchange
|
Can we say that : $ (A-B)\cap\overline{B}(0,r)\text{ is weakly compact, }\forall r>0 $
Let $X$ be a separable Banach space and $A,B$ are closed convex subsets of $X$ such that $B\subset A$ and
$$
A\cap\overline{B}(0,r) \text{ and } B\cap\overline{B}(0,r) \text{ are weakly compact, } \forall r>0.
$$
Can we say that :
$$
(A-B)\cap\overline{B}(0,r)\text{ is weakly compact, }\forall r>0
$$
with $A-B=\{a-b:a\in A,b\in B\}$.
No I made a mistake, I have to write $ B\subset A $.
I did not understand why $(A-B)\cap \overline{B}(0,r)$ probably is not closed?
Why $ (A−B)\cap\overline{B(0,r)}$ is relatively open in the convex?
Can you explain what you say in a few lines
@LSpice: $A-B$ is NOT set theoretic difference--look at the OP's definition. This is a reasonable question for MO.
@BillJohnson, thank you for correcting my misreading. @kakaHae, I apologise for misreading your question. I retracted my close vote, and will delete my inadvertently misleading comments.
Doesn't this already fail in finite dimensions? Take $X = \mathbb{R}^2$ and let $A = B =$
the closed region bounded by $y = \sqrt{x^2 + 1}$ and $y = x + 1$, both for $x \geq 0$. The intersection with any ball is closed and therefore compact, but $A - B$ is not closed: it contains points arbitrarily close to $(0, -1)$ but does not contain that point itself.
Oh, I was thinking the OP meant in the conclusion, in which case there is no counterexample in reflexive spaces. I now see a counterexample with "relatively weakly compact" but won't bother to add it if the OP did mean "weakly compact". @kakaHae: If you did mean "relatively weakly compact", please add this at the end of your question. Do not change the question as asked because that renders much of the discussion gibberish for anyone who does look at the edits to the question.
@BillJohnson I'm confused by your comment. Are you saying that in my example $A - B$ is not weakly compact but is relatively weakly compact? Or its intersection with every ball is weakly compact but not relatively weakly compact? It's not even closed, so I'm not sure what that could mean.
Not $A-B$, Nik, but $A-B$ intersected with every ball is what the OP is looking at. I thought the OP wanted relative weak compactness of $A-B$ intersected with balls, but I saw after your answer that he asked for weak compactness of $A-B$ intersected with closed balls. The version asking for relative weak compactness is a nice question, while, as you point out, what the OP asked is basically trivial.
Oh, I see. Well, that is an interesting misreading of the question ...
This is not an answer, but I put it here to stop close votes that are apparently based on a misreading of the question.
I do not know the answer to the OP's question, but I will point out that both hypotheses are necessary. For example, drop the hypothesis that $A$ is convex. Then there is a counterexample with $A=B$; namely, $A=\bigcup_n \{(n+1)e_n, ne_n\}$, where $(e_n)$ is the unit vector basis of $\ell_1$. Bounded subsets of $A$ are finite and $A-A$ contains $(e_n)$, which is not preweakly compact.
Secondly, drop the hypothesis that $B\subset A$ but keep the convexity hypotheses. Let $X= C[0,1]$ ($X$ could be any separable non reflexive space that contains an infinite dimensional reflexive subspace). It is known that there are quasi complementary subspaces $X_1$ and $X_2$ that are both isomorphic to $\ell_2$. This means that $X_1 \cap X_2 =\{0\}$ and $X_1 - X_2$ is norm dense in $X$, and hence $(X_1 - X_2) \cap B(0,1)$ is even norm dense in the non weakly compact unit ball of $X$.
@Bill_Johnson Thank you sir
Another way to achieve the conclusion as in the second half of Bill‘s answer: Take a reflexive $X$ and two closed subspaces $A$ and $B$ whose sum $Z=A+B=A-B$ is not closed and dense. Then the unit ball of $Z$ is not weakly compact, being a norm and weakly dense proper subset of the unit ball of $X$.
|
2025-03-21T14:48:30.920776
| 2020-05-09T01:00:48 |
359779
|
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|
Stack Exchange
|
Relationship between Quasicoherent sheaves and $\mathbb A^1$-fpqc modules over an fpqc stack
In what follows, assume several universes for simplicity.
Let $X$ be a stack in groupoids on the fpqc site of small affine schemes $\mathbf{Aff}_{\text{fpqc}}$. We can define $\mathbf{QCoh}(X)$ formally by considering the representable stack $$\mathbf{Hom}_{\operatorname{Stk}_{\text{fpqc}}(\mathbf{Aff})}(-,\mathbf{QCoh})$$
and restricting it to the full 2-subcategory $\operatorname{Stk}^{\text{Gpd}}_{\text{fpqc}}(\mathbf{Aff})^{\text{op}}$.
We can define the fpqc topos of $X$ to be the slice category $\operatorname{Shv}_{\text{fpqc}}(\mathbf{Aff})/X$, and we can make this topos locally ringed by pulling back $\mathbb{A}^1_X=X\times_{\operatorname{Spec}(\mathbb{Z})}\mathbb{A}^1.$
Since $\mathbb{A}^1_X$ is a local ring object in the topos, we can then define a category of $\mathbb{A}^1_X$-modules in $\operatorname{Shv}_{\text{fpqc}}(\mathbf{Aff})/X$. By the definition of $\mathbf{QCoh}(X)$, we can write down a forgetful functor $U$ to the category of $\mathbb{A}^1_X$-modules by sending a quasicoherent sheaf $F$ on $X$ to the evaluation of its pullback, that is, for $f:\operatorname{Spec}(R)\to X$, we have
$$U(F)(f)=\Gamma(\operatorname{Spec}(R),f^\ast F),$$
which is naturally an $\mathbb{A}^1_X$-module.
Question: Is the forgetful functor $U$ defined above fully faithful? If so, can we identify its essential image as the full subcategory of $\mathbb{A}^1_X$-modules admitting a presentation the way we can for schemes? Does either of these statements become true or stay true if we pass to derived everything? What if we restrict to the case where $X$ is an Artin stack?
Are both the words meant to be 'pseudofunctor' in "Kan extending the pseudofunctor … to a pseudofunctor"?
@LSpice Yeah, both are pseudofunctors. I guess one can do this more simply by considering the stack of categories $\mathbf{QCoh}$, taking the representable pseudofunctor $\operatorname{Hom}(-,\mathbf{QCoh})$ and restricting it to the full subcategory of groupoidal stacks, no Kan extension needed.
|
2025-03-21T14:48:30.921038
| 2020-05-09T03:23:37 |
359787
|
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|
Stack Exchange
|
Restriction of diffeomorphisms homotopic to identity to the boundary
Let $M$ be a smooth manifold with boundary $\partial M$. Let $Diff_0(M)$ be the group of all diffeomorphisms homotopic to identity. According to this article (Page 6, section "
Beyond mapping class group"), the restriction of a diffeomorphism to the boundary gives a well defined surjactive homomorphism $$\phi: Diff_0(M)\rightarrow Diff_0(\partial M).$$
I could not find a reference for the above two results. So I have two questions.
1) Why is $\phi$ well defined, i.e., why the restriction of a difeomorphism homotopic to identity in $M$ is homotopic to identity in $\partial M$.
2) Why every diffeomorphism homotopic to identity in $\partial M$ is a restriction of a diffeomorphism of $M$ which is homotopic to identity.
As I am not quite familiar with diffeomorphism groups, any suggestion/reference/comment will be extremely helpful. Also I would request you to improve the tags if possible.
Thanks in advance.
Both claims hold (with easy proofds) if you replace homotopy with isotopy; are you sure they use homotopy in the paper? The restriction map does not (in general, send diffeomorphisms homotopic to the identity to diffeomorphisms homotopic to the identity.
Good point. The paper states simply the "path component of identity". Also even for isotopy the first one is straightforward but how to prove the second one? Any reference for that.
Path component of the identity means isotopy. To get 1, just restrict the isotopy to the boundary, it remains isotopy. (The difference with homotopy is that boundary maps to boundary.) To get 2, use the existence of a collar neighborhood of the boundary which is the product $\partial M\times [0,1]$.
As discussed in comments, $Diff_0$ stands for the subgroup of the diffeomorphism group, consisting of diffeomorphisms isotopic (rather than homotopic) to the identity. With this in mind, the fact that the restriction map $\phi: Diff(M)\to Diff(\partial M)$ sends $Diff_0(M)$ to $Diff_0(\partial M)$ is clear. Let's prove surjectivity. First of all, $\partial M$ admits a "collar" $C$ in $M$, a closed neighborhood of $\partial M$ in $M$, $C$ is diffeomorphic to $\partial M\times [0,1]$. Now, given $h\in Diff_0(\partial M)$, let $H(x, t), t\in [0,1]$, denote the isotopy of $h= H(\cdot, 0)$ to $id_{\partial M}= H(\cdot, 1)$. I leave it to you to prove that $H$ can be chosen so that $H(x,t)=x$ for all $t\in [1/4, 1]$. Then, using the diffeomorphism $C\cong \partial M\times [0,1]$, extend $h$ first to $C$ and then, by identity, to the rest of $M$. Call the extension $\hat{h}$. Clearly, $\phi(\hat{h})=h$. It remains to prove that $\hat{h}\in Diff_0(M)$. To prove this, play the same game as before: Given an isotopy $H(x,t)$ from $h$ to $id_{\partial M}$, extend it to $C\cong \partial M\times [0,1]$ by
$$
(x,t,s)\mapsto H(x, t+s),
$$
and then by identity to the rest of $M$. This will be an isotopy $\hat{H}$ from $\hat{h}$ to $id_M$.
Here is an example to ponder: Let $M$ be the annulus $S^1\times [0,1]$. Consider the diffeomorphism $f(s,t)=(s, 1-t)$; $f: M\to M$ is homotopic to the identity, but its restriction to $\partial M$ is not.
|
2025-03-21T14:48:30.921427
| 2020-05-09T03:47:26 |
359788
|
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|
Stack Exchange
|
How to solve the integral equation $f(x)=\frac{\lambda\int_{x}^1 \sqrt{1+(f(t))^2}dt+c_1}{\int_{x}^1 \sqrt{1+(f(t))^2}dt+c_2}$?
Recently, I have asked a question about variational analysis (First moments of uniform distribution on a curve from (0,0) to (1,1) in two-space). Such the question can be addressed in some cases by solving the following integral function:
$f(x)=\frac{\lambda\int_{x}^1 \sqrt{1+(f(t))^2}dt+c_1}{\int_{x}^1 \sqrt{1+(f(t))^2}dt+c_2}$,
where $f(x)\in\mathbf{C}^1[0,1]$, and $\lambda$, $c_1$ and $c_2$ are constant numbers.
How to solve the above nonlinear equation?
Thanks for your answer, but I think that something is wrong because $f(0)=(-1)^{-0.5}$ is not allowed.
Yes. $q>1$ (the integral constant) .
Let us put $x=0$ in the equation under consideration. This implies $c=0$ and then $f(x)=\lambda$.
Thanks for your nice comment. The original posted question is indeed wrong, and now I have edited it.
Both original and edited questions are MSE topics. The integral equation under consideration can be reduced to a certain ODE of the first order by its differentiating and taking into account $$ \int_{x}^1 \sqrt{1+(f(t))^2}dt=-{\frac {f \left( x \right) c_{2}-c_{1}}{f \left( x \right) -\lambda}},,f(1)= \frac {c_1} {c_2}.$$ Maple 2019.1 solves it in terms of roots of a certain transcendental equation. That huge solution is useless.
Thanks for your answer. Is there any numericial method to solve this equation?
Precisely you get an ODE with separable variables, whence by integration you get a not too complicated expression for the inverse of $f$ in terms of elementary functions (the trascendental equation for $f$) combining logarithms, rational terms, and square roots). An explicit expression for $f$ seems out of reach though
$f(x).\int_{0}^{x} \sqrt{1+{f(t)}^2} dt =c+\lambda =a$
Say, $f(x)=y$
So, $\frac{d (\int_{0}^{x} \sqrt{1+{y(t)}^2} dt)}{dx} =\frac{d\frac{a}{y}}{dx}$
From Newton- Leibniz's formula
$\sqrt{1+y^2}=a\frac{d}{dx}(\frac{1}{y})$
Taking $z=(\frac{1}{y})^2$ , we get
$\sqrt{1+z}=a\frac{dz}{dx}$
or, $\frac{dz}{\sqrt{1+z}}= \frac{dx}{a}$
Or, $y=\frac{1}{\sqrt{(\frac{x}{a}+q)^2-1}}$ , ($q>1$ for $a>1$, $q<1$ for $a<0$)
Putting the result $y=\frac{1}{\sqrt{(\frac{x}{a}+q)^2-1}}$ we can easily see that the integral equation is satisfied.
Thanks. I have noticed that $f(x)\cdot\int_0^x\sqrt{1+f^2}dt=c+\lambda$ is used in your answer, which is not implied by the integral equation.
|
2025-03-21T14:48:30.921741
| 2020-05-09T03:49:51 |
359789
|
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|
Stack Exchange
|
Hopf fibration extended to bundle over $\mathbb{C}^2$
Consider the Hopf bundle $h:\mathbb{S}^3\rightarrow\mathbb{S}^2$. There is a connection $1$-form $\omega$ oh $h$ which is left $SU(2)$ invariant. In terms of the Euler angles $(\theta,\,\phi,\,\psi)$ this is given by $\omega=d\psi+\cos(\theta)d\phi$. Using $\omega$, we can form a covariant derivative on the associated bundle $\phi:\mathbb{S}^3\times_{U(1)}\mathfrak{u}(1)\rightarrow\mathbb{S}^2$. In the paper https://arxiv.org/abs/1705.02666, the author treats this as a covariant derivative on a bundle over a manifold with topology $\mathbb{C}^2$, such that the curvature is given by $d\omega$.
The only way I can think to extend $\phi$ to a bundle over $\mathbb{C}^2$ is pulling it back via $h$ to a bundle over $\mathbb{S}^3$, then extending radially over $\mathbb{C}^2\cong\mathbb{S}^3\times\mathbb{R}_{\geq 0}$. However, I am not sure how this would be well defined as the $\mathbb{R}_{\geq 0}$ component goes to $0$.
I would greatly appreciate if anyone could reconcile how to obtain such a bundle over $\mathbb{C}P^2$.
The bundle in question is the tautological line bundle. Its total space is $\mathbb{C}^2\setminus 0$ and the projection is the natural projection $\mathbb{C^2}\setminus {0}\to\mathbb{CP}^1$ that associates to a nonzero vector the line it determines
If I understand correctly what you are asking, one thing to point out is that the bundle over $S^2$ cannot extend to a bundle over $\mathbb{C}^2$, since then the Hopf bundle would be trivial—if by 'extend' you mean along some inclusion $S^2 \hookrightarrow \mathbb{C}^2$.
@DavidRoberts , this makes sense, thanks you. In essence when I say `extend' what I want is the following; I have a connection from $\mathbb{S}^2$ to some some line bundle, this connection has a curvature form $F$ associated to it, I want to somehow get a bundle over $\mathbb{C}^2$ with the same curvature form (which does seem like an inclusion actually).
If this is in fact what the author does, then it is indeed incorrect. Unless the author is using some kind of generalized covariant derivative, extending it implies extending the bundle, which is not possible here. As you say, it is indeed possible to extend it to the punctured plane but no further.
@DeaneYang, I may be wrong about the extension. What I am certain of is that he considers the curvature form $F=d\omega$ which should be over $\mathbb{S}^2$, however, he treats it as if it were over $\mathbb{C}^2$ (or $\mathbb{C}P^2\cong\mathbb{C}^2\cup\mathbb{S}^2$). The context is that he is considering a twisted Dirac operator, where the twisting is coming from a scaled version of the connection one form $\omega$.
Are you sure that form extends continuously to the origin? I'm not.
I think I may have (somewhat) sorted this out. Although hard to find in the paper I believe the curvature form I mentioned above is scaled by a function depending on $r$ (radial coordinate). The bundle I described above ends up coming through in the limit $r\to\infty$, where $\mathbb{C}^2$ approaches its asymptotic boundary $\mathbb{S}^2$. Although I am still not sure explicitly what the bundle on the bulk $\mathbb{C}^2$ is, I think it is well defined and a proposed "extension" in the way I was describing about is a bit naive. Thanks for t comments, I welcome any more thoughts on the question.
|
2025-03-21T14:48:30.922212
| 2020-05-09T04:12:07 |
359791
|
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|
Stack Exchange
|
Reduce ergodicity to the ergodicity of the coordinate process
Let $(E,\mathcal E,\lambda)$ be a probability space and $\lambda$ be a measurable map on $(E,\mathcal E)$ with $\lambda\circ\tau^{-1}=\lambda$.
I would like to show that $\tau$ is $\lambda$-ergodic by reducing this claim to the following well-known result: If $(E,\mathcal E)$ is a measurable space and $(Y_i)_{i\in\mathbb N}$ is an independent stationary $(E,\mathcal E)$-valued process on any probability space (we denote the law of $Y_i$ by $\mathcal L(Y_i)$ for $i\in\mathbb N$), then the coordinate process $(X_i)_{i\in\mathbb N_0}$ on $(\Omega,\mathcal A,\operatorname P):=(E^{\mathbb N},\mathcal B(E)^{\otimes\mathbb N},\bigotimes_{i\in\mathbb N}\mathcal L(Y_i))$ is independent and stationary. Thus, the shift $$\theta:\Omega\to\Omega\;,\;\;\;(x_i)_{i\in\mathbb N}\mapsto(x_{i+1})_{i\in\mathbb N}$$ is $\operatorname P$-ergodic.
Now assume $$\tau^i=\varphi\circ\theta^iY\;\;\;\text{for all }i\in\mathbb N.\tag1$$ for some $(\mathcal A,\mathcal E)$-measurable $\varphi:\Omega\to E$ and an $(E,\mathcal E)$-valued independent stationary process $(Y_i)_{i\in\mathbb N}$ on $(E,\mathcal E,\lambda)$.
Can we somehow use the identity $(1)$ to derive that $\tau$ is $\lambda$-ergodic from knowing that $\theta$ is $\operatorname P$-ergodic? I guess we only need to show that $\varphi$ is sufficiently nice for this conclusion to hold.
Feel free to assume that $(Y_i)_{i\in\mathbb N}$ is identically distributed, if necessary.
Remark: I've asked for the special case, where $(Y_i)_{i\in\mathbb N}$ is the dynamical system generated by the Bernoulli shift, back over on MSE: https://math.stackexchange.com/q/3657914/47771.
I think you’re asking whether a factor a Bernoulli process is ergodic. In fact it’s a simple fact that any factor of any ergodic process is ergodic.
You should think of $\phi$ as mapping the sequence of $Y$’s to the sequence of $\tau$’s. Now take any measurable shift-invariant subset of $\tau$ sequences. Its preimage is a measurable shift-invariant subset of $Y$-sequences. This therefore has measure 0 or 1. It follows that the subset of $\tau$ sequences has measure 0 or 1 as required.
I'm not sure whether I can follow. Let $\mathcal I_\tau:={B\in\mathcal E:\tau^{-1}(B)=B}$ and $\mathcal I_\theta:={A\in\Omega:\theta^{-1}(A)=A}$. Are you claiming that if $B\in\mathcal I_\tau$, then $\varphi^{-1}(B)\in\mathcal I_\theta$? This would require $\varphi$ to be $(\mathcal I_\theta,\mathcal I_\tau)$-measurable. It is clearly $(\mathcal A,\mathcal E)$-measurable and since $\mathcal I_\theta\subseteq\mathcal A$, we get $\varphi^{-1}(B)\in\mathcal A$. What am I missing?
I've asked for that separately here: https://math.stackexchange.com/q/3667845/47771. Maybe I'm missing something.
I think the additional ingredient is the fact that $\theta\circ\phi=\phi\circ\theta$. Here the first $\theta$ is the shift on $\tau$ sequences; and the second $\theta$ is the shift on $Y$ sequences. The fact that the inverse image of an invariant set is invariant follows from this.
I'm sorry, we've obviously got different backgrounds, which is why I really struggle to even understand your terminology (and maybe you've misunderstood mine at some point).
Could you please write down explicitly what your left and right $\theta$'s are (and please use different symbols for them). Moreover, what do you mean by a "$\tau$-sequence" or a "$Y$-sequence"? I've no idea what you could mean with these terms, since $\tau$ is a measurable map on the space $(E,\mathcal E)$ which is not a product space. Having said that, $\varphi$ maps into this space $(E,\mathcal E)$ and hence I've no idea what could be "shifted" (which is why $\theta\circ\varphi$ doesn't make sense to me).
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2025-03-21T14:48:30.922636
| 2020-05-09T04:15:39 |
359792
|
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|
Stack Exchange
|
Is there a fusion category not Grothendieck equivalent to a unitary one?
We refer to the book Tensor categories by Etingof-Gelaki-Nikshych-Ostrik (MR3242743) for the notion of (unitary) fusion category. Two fusion categories are Grothendieck equivalent if they have the same fusion ring.
Question: Is there a fusion category not Grothendieck equivalent to a unitary one?
The following citations (coming from above book) are almost on-topic, but not exactly, because they are about the existence of a certain structure on certain fusion categories, whereas above question relaxes up to Grothendieck equivalence, which is much weaker.
On page 284:
We note that we do not know an example of a fusion category over
$\mathbb{C}$ which does not admit a Hermitian structure, or a
pseudo-unitary fusion category which does not admit a unitary
structure.
On page 76:
Does every semisimple tensor category admit a pivotal structure? A
spherical structure? This is the case for all known examples. The
general answer is unknown to us at the moment of writing (even for
fusion categories over ground fields of characteristic zero).
Yes, according to Andrew Schopieray. He just provided a categorifiable fusion ring, of rank 6 and multiplicity 2, without pseudounitary categorification (so without unitary categorification), in the following preprint called Non-pseudounitary fusion.
https://arxiv.org/abs/2010.02958
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2025-03-21T14:48:30.923018
| 2020-05-09T05:21:31 |
359798
|
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"DuFong",
"Mateusz Kwaśnicki",
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|
Stack Exchange
|
Separating a Riemann-Hilbert problem
Consider a RHP on the real line a jump is piece-wise H\"older continuous(or $L^2$), say for example the jump is
$$g(x)=g_1(x)\chi_1+g_2(x)\chi_2,$$
where $g_j(x)$ are Holder continuous functions and $supp\chi_1\cap supp\chi_2=\emptyset$.
Do we have decomposition $m(z)=m_1(z)\chi_1(\Re(z))+m_2\chi_2(\Re(z))$, where $m_{j+}=m_{j-}g_j$ on the support of $\chi_j,j=1,2$.
My question: is my question even well-posed? Any existing theories in literature?
Maybe I misunderstood your question completely, but it makes no sense to me: $m(z)$ is supposed to be holomorphic in the upper and lower complex half-planes, while $m_1(z) \chi_1(\Re z) + m_2(z) \chi_2(\Re z)$, if non-zero, will necessarily be discontinuous almost everywhere on $x + i \mathbb{R}$ for every $x$ on the boundary of the support of $\chi_1$ or $\chi_2$.
That's right. And there are two ways. One is put new jumps one those boundaries (which may be designed my our own will, hopefully). Second way is to find another possible decomposition. The decomposition i provided is just a native guess though. @MateuszKwaśnicki
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2025-03-21T14:48:30.923217
| 2020-05-09T08:24:54 |
359804
|
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"Taras Banakh",
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|
Stack Exchange
|
Retracting off a compact set
Let $K$ be a compact set in $\mathbb{R}^n$ and let $U$ be a bounded open set that contains $K$. You may assume both are connected.
Can we always find an open $V$ such that $K\subset V\subset\overline{V}\subset U$ such that $U\backslash K$ retracts on $U\backslash V$?
For example, if we somehow find $V$ such that $\overline{V}$ is homeomorphic to the closed ball, then choosing any point of $x\in K$ to be its center, we can radially repel all the points of $V\backslash\{x\}\supset V\backslash K$ onto $\partial V$ and leave every other point fixed.
I tried to cover $K$ with cubic neighborhoods and run an induction over the number of cubes, but it is getting too messy. Perhaps there is a more clever way.
I hope that simplicial partitions of $\mathbb R^n$ could be of help. In other words the methods of PL-topology should work.
@TarasBanakh I am not familiar with the PL-topology. Perhaps you could illustrate your idea in dimension $2$, with a triangulation of the plane?
Divide the plane into squares, then each square into two triangles. Take the partition into squares so small that a single square does not intersect simultaneously $K$ and the complement to $U$. Divide each triangle by barycentric subdivision into 6 smaller triangles with a vertex in a center of the triangle. Take the union $T$ of all triangles of the barycentric subdivision that intersect $K$. Take the union of all triangles of the barycyntric subdivision that intersect $T$ and put $V$ be its interior in the plane. It seems that $V$ is the neighborhood of $K$ you are looking for.
For basic definitions of PL-topology look the first pages of this paper https://www.maths.ed.ac.uk/~v1ranick/papers/pltop.pdf (this was the first what google suggested for the search "PL-topology").
@TarasBanakh I thought for a while about separating the plane into squares or triangles, but still not sure how they help. What is the significance of the specific triangulation that you suggested?
The significance is that you can do everything locally (i.e., inside of triangles).
I wrote down the detail construction of the retraction for dimension 2 in my answer.
Let us show how to find such a retraction for $n=2$ (I do not know if this method generalizes to higher dimensions).
Given a compact set $C\subset\mathbb R^2$ and an open neighborhood $U\subseteq\mathbb R^2$ of $C$, choose a triangulation on $\mathbb R^2$ so fine that no triangle of the triangulation meets $C$ and $\mathbb R^2\setminus U$ simultaneously.
Replacing the triangulation by a finer triangulation, we can assume that for each triangle $T$ with $T\not\subseteq C$, one vertex of $T$ does not belong to $C$.
How to find such a triangulations? Assuming that $T\not\subseteq C$, we can find an interior point $v$ of $T$ that does not belong to $C$ and replace the triangle $T$ by 3 subtriangles having $v$ as a vertex.
Also we can assume that either $T\subseteq\mathbb R^2\setminus C$ or $T$ has a vertex in $C$. Assuming that $T$ has no vertices in $C$ but $T\cap C\ne\emptyset$, we can choose a point $c\in T\cap C$ and replace the triange $T$ by two or three triangles having $c$ as a vartex.
Therefore, we lose no generality assuming that each triangle $T$ of the triangulation has one of the following properties:
1) $T\subseteq C$;
2) $T\cap C=\emptyset$;
3) $T$ has one vertex in $C$ and one vertex outside of $C$;
4) If two vertices $u,v$ of $T$ do not belong to $C$, then the side $[u,v]$ does not intersect $C$.
A triangle $T$ will be called difficult if it has one vertex say $u$ outside $C$, two vertices $v,w$ in $C$ and the side $[v,w]$ is not a subset of $C$.
In this case choose any point $c[v,w]\in [v,w]\setminus C$. The points $c[v,w]$ can be chosen so that for two difficult triangle sharing the common side $[v,w]$ the point $c[v,w]$ is the same.
Now for every triangle $T$ of the triangulation we define a function $r_T\setminus C:T\to T\setminus C$ such that $r_T\circ r_T=r_T$ as follows. In case (1), let $r_T$ be the empty map and in case (2) $r_T$ be the idenity map of $T$.
In the remaining cases, the triangle $T$ has one vertex in $C$ and one vertex outside of $C$. If the triangle $T$ is not difficult, then it has two vertices $u,v$ such that the side $[u,v]$ either is contained in $C$ or is disjoint with $C$. If $[u,v]$ is contained in $C$, then let $r_T:T\setminus C\to\{w\}$ be the constant map into the unique vertex $w\notin C$ of $T$.
If $[u,v]\cap C=\emptyset$, then the third vertex $w$ of $T$ belongs to $C$ and we can apply the Urysohn lemma to find a function $r_T:T\setminus C\to[u,v]$ such that $r_T[[w,u]\setminus C]=\{u\}$, $r_T[[w,v]\setminus C]=\{v\}$, and $r_T(x)=x$ for every $x\in [u,v]$.
It remains to consider the case of a difficult triangle $T$. Such a triangle has one vertex $u$ outside of $C$, two vertices $v,w$ in $C$ and the point $c[v,w]\in [v,w]\setminus C$. Two cases are possible.
1) There exists a path $\gamma:[0,1]\to T\setminus C$ such that $\gamma(0)=u$ and $\gamma(1)=c[v,w]$. We can assume that $\gamma$ is injective and hence its image $A_T=\gamma[0,1]$ is an arc with endpoints $u$ and $c[v,w]$. Using the Urysohn Lemma, we can find a continuous function $r_T:T\setminus C\to A_T$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq \{u\}$, $r_T[[v,w]\setminus C]\subseteq\{c[v,w]\}$ and $r_T(a)=a$ for every $a\in A_T$.
2) No such a path $\gamma$ exists. Then the points $u$ and $c[v,w]$ belong to distinct connected components of $T\setminus C$. In this case we can choose a continuous map $r_T:T\setminus C\to\{u,c[v,w]\}$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq\{u\}$ and $r_T[[v,w]\setminus C]\subset\{c[v,w]\}$.
The definitions of the maps $r_T$ ensure that they agree on the intersections of their domains. Consequently, the union $r=\bigcup_T r_T$ of these maps is a continuous function $r:\mathbb R^2\setminus C\to\mathbb R^2\setminus C$ such that $r\circ r=r$. So, $r$ is a retraction onto the closed subset $F$ which can be written as the union of the triangles of the triangulation that do not intersect $C$, some vertices of the triangles that intersect $C$ and the arcs $A_T$ of difficult triangles (of the first type).
The choice of the triangulation $T$ (as sufficiently fine) implies that $V=\mathbb R^2\setminus F$ is a neighborhood of $C$ with $\bar V\subset U$. Then $r{\restriction}U\setminus C$ is the required retraction of $U\setminus C$ onto $U\setminus V$.
this is not much different from what i came up on my own, and here is where i was stuck all along: how to retract $\bar{\sigma}\backslash C$ onto $\partial\sigma\backslash C$? (by the way, i assume you meant $\sigma\in K^{(3)}$, right?) First, the latter set may be empty, even if the former is not (this can be solved by further dividing the cell). But what if the latter set is disconnected, while the former is connected? For example, imagine a triangle for which $C$ is an arc between two points in the triangle, but ventures into a neighboring triange.
@erz I corrected the definition of $K^{(n)}$ so $K^{(2)}$ is now defined correctly as the set of 2-dimensional simplexes. Concerning the retraction, then indeed, there is a problem in my argument. I will think how to correct it.
@erz I have totally rewritten the solution but only for dimension 2. I am not sure if this approach will work for higher dimensions. One can try to understand what is going on for wild compact sets: horned Alexander sphere, wild Cantor set, Antoine necklage, etc.
Thank you for your answer! Yeah, I also don't see how to apply this method to $n=3$: when you try to purify an edge, you spoil a face.
|
2025-03-21T14:48:30.924023
| 2020-05-09T13:59:36 |
359815
|
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|
Stack Exchange
|
The group of global sections of the automorphism bundle of the tangent bundle on a Grassmannian
Let $X={\rm Gr}(k,n)$ denote the Grassmannian of $k$-dimensional subspaces in ${\Bbb C}^n$.
We regard $X$ as an algebraic variety over $\Bbb C$.
Let ${T_X} \to X$ denote the tangent bundle on $X$. For an explicit description of ${T_X}$ see e.g.
here.
Consider the induced bundle ${\rm GL}({T_X})\to X$ whose fiber at $x\in X$ is the automorphism group ${\rm GL}(T_x)$ of the vector space $T_x$.
Question 1. What is the group $A={\rm Aut}_X\,{T_X}$ of regular global sections of ${\rm GL}({T_X})$ over $X$?
For any $\lambda \in{\Bbb C}^\times$ we have a global section of ${\rm GL}({T_X})$ taking value $\lambda I_x$ at $x$, where $I_x$ is the identity automorphism of $T_x$.
Thus we obtain a canonical embedding ${\Bbb C}^\times\hookrightarrow A$.
Question 2. Is it true that $A={\Bbb C}^\times\,$?
Question 3. In particular, is the answer to Question 2 "Yes" for $X={\rm Gr}(1,n+1)={\Bbb P}^n\,$?
I know that the answer to Question 2 is "Yes" for $X={\rm Gr}(1,2)={\Bbb P}^1$.
In this case ${\rm GL}({T_X})={\Bbb C}^\times\times {\Bbb P}^1\to {\Bbb P}^1$.
The situation is completely uniform, and what happens for $\mathbb{P}^1$ happens for all $G/P$ (at least in type ADE), see Boralevi: On simplicity and stability of the tangent bundle of rational homogeneous varieties (https://mathscinet.ams.org/mathscinet-getitem?mr=3202710). So $A=\mathbb{C}^\times$ for all Grassmannians.
@pbelmans: Excellent! Many thanks! Later I was going to ask a question about the tangent bundle of $G/P$ ...
I state here the relevant result of Ada Boralevi. Let $G$ be a simple algebraic group over $\Bbb C$ of one of the types ADE, and let $P\subset G$ be a parabolic subgroup. Set $X=G/P$. Then the tangent bundle $T_X$ of $X$ is simple, that is, ${\rm End}_X,T_X={\Bbb C}$. This answers in the positive my Questions 2 and 3. See Ada Boralevi, On simplicity and stability of the tangent bundle of rational homogeneous varieties, Theorem 4.4.
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2025-03-21T14:48:30.924280
| 2020-05-09T14:21:19 |
359817
|
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|
Stack Exchange
|
maximum modulus of polynomials
Suppose $p(z)$ is a polynomial of degree $n$ having no zeros in $|z|<1.$ Then for any $R\geq1$
we have a result $
\max_{|z|=R}|p(z)|\leq \frac{1+R^n}{2}max_{|z|=1}|p(z)|,\;\;R\geq 1.$
I could not find any literature on the similar $ (\leq)$ inequality when $R<1. $ Can I expect some input?
Trivially for $R<1$, $\max_{|z|=R}|p(z)|\leq\max_{|z|=1}|p(z)|$. This is unimprovable (think of contant)
Are you trying to reinvent Schwarz lemma? If $a_0=\dots=a_{k-1}=0$, then $\max_{|z|=R}|p(z)|\leq R^k\max_{|z|=1}|p(z)|$
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2025-03-21T14:48:30.924394
| 2020-05-09T14:23:26 |
359818
|
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|
Stack Exchange
|
What is a fast way to multiply a Schubert polynomial by an elementary symmetric polynomial (specifically $x_1\cdots x_k$)?
What is a computationally fast way to get the coefficients of Schubert polynomials in the expansion of the product of a Schubert polynomial and an elementary symmetric polynomial? I know "fast" is relative because this is a #P-complete problem, but I'm sure there are faster ways than what I am doing. I know the result is governed by Sottile's Pieri formula, but how to find the elements of the formula beyond brute force searching is not obvious.
I really only care about the product with polynomials of the form
$$x_1x_2\cdots x_k$$
In that case I tried using Monk's formula and the fact that
$$x_i=\mathfrak S_{s_i} - \mathfrak S_{s_{i-1}}$$
However this results in massive cancellation and hence it uses a lot of memory and is not very fast.
Edit: After posting this I thought of a much faster way. Some of the cancellation in multiplying by $\mathfrak S_{s_i}-\mathfrak S_{s_{i-1}}$ is predictable, so there's no need to directly use Monk's formula to multiply by $x_i$, you only need to compute the part that doesn't immediately cancel. I just tested this on a large example, the new method took $29$ seconds and the old method took $9$ minutes and $42$ seconds, so definitely a massive improvement. I am still looking for better ways though if anybody knows any.
By Sottile's formula I'm assuming you mean the Pieri rule in https://arxiv.org/abs/alg-geom/9505001? It seems like that would be a natural candidate at any rate.
@Sam Yes, that's the one I mean.
@Sam Usually positive, combinatorial formulas are not fast.
Ah, I see. Yeah I don't know much about the computational aspects.
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2025-03-21T14:48:30.924653
| 2020-05-09T14:38:50 |
359820
|
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|
Stack Exchange
|
Can we extract an injective envelope from a monomorphism?
Let $A$ be an artinian ring and $f : X \rightarrow \bigoplus_{j=1}^{n}I_{j}$ be a morphism of $A$-modules, where each $I_{j}$ is injective and indecomposable. If $f$ is a monomorphism, then can we conclude that there is an injective envelope $g : X \rightarrow \bigoplus_{t=1}^{m}I_{j_{t}}$? (Here $\{ j_1, \ldots, j_m \} \subset \{1, \ldots, n \}$)
For $n=1$, this would mean that any monomorphism into an injective module is an injective envelope...
@abx you are right... I changed the question now in a clever way. Thank you.
@abx it is not true that the injective envelope is indecomposable
You are right, sorry. I delete my comment.
Yes: given a monomorphism $f\colon X\to I$ with $I$ injective, as in the question, you can find a decomposition $I=I_0\oplus I_1$ such that $f=\begin{pmatrix}f_0&0\end{pmatrix}$ and $f_0\colon X\to I_0$ is left minimal, so it is an injective envelope.
Getting this decomposition does not require any assumption on the ring $R$, and only much weaker assumptions on $I$, for example that it is pure injective. A good reference is On minimal approximations of modules by Krause and Saorín.
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2025-03-21T14:48:30.924847
| 2020-05-09T14:47:13 |
359821
|
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|
Stack Exchange
|
Number of elements in $\mathrm{GL}(n,p)$ with maximal order
I learned reading this question that $\mathrm{GL}(n,p)$ elements have at most a multiplicative order of $p^n -1$.
I would like to know how many matrices have an order of exactly $p^n -1$. Do they represent a majority of matrices in $\mathrm{GL}(n,p)$?
The answer suggested to build matrices of maximal order this way.
Consider a degree n monic polynomial $P_n$ whose root is a generator $ξ$ of $F^∗_{p^n}$. Then a matrix with $P_n$ as its characteristic polynomial has order at least $p^n−1$ since $ξ$ is its eigenvalue.
I don't know how to create a such matrix, yet enumerate them...
Thank you so much for any help.
Rational canonical form is very relevant here. This is really a question about linear algebra and possible minimum polynomials.
To be precise, I think the number of such matrices is $\frac{\phi(p^{n}-1)|{\rm GL}(n,p)|}{n(p^{n}-1)}$.
The way to see this is to see that there are $\frac{\phi(p^{n}-1)}{n}$ conjugacy classes of elements of order $p^{n}-1$, and then note that if $A$ is a matrix of multiplicative order $p^{n}-1$ in $G = {\rm GL}(n,p)$, then $|C_{G}(A)| = p^{n}-1.$
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2025-03-21T14:48:30.925038
| 2020-05-09T15:21:31 |
359823
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Donu Arapura",
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/82179"
],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359823"
}
|
Stack Exchange
|
Two definitions of formal schemes
I have read two seemingly completely different definitions of formal schemes.
Hartshorne defined a formal scheme as the formal completion of a noetherian scheme $X$ along a closed subscheme $Y$ of $X$.
In Demazure's Lectures on p-Divisible Groups, a $k$-formal scheme is defined to be a left exact covariant functor from the category of finite dimensional $k$-algebras to the category of sets, where $k$ is a field.
Are these two definitions of formal schemes the same (or related to each other)?
(update) Now I have read about the definition in EGA I, which defines an affine formal scheme as the formal spectrum $\mathop{\mathrm{Spf}}(A)$ of an admissible ring $A$. As Murray answered, Demazure defines an affine formal scheme over $k$ as $\mathop{\mathrm{Spf}}(A)$ for some profinite $k$-algebra. It a profinite $k$-algebra always admissible? (It seems to me that this is not true, for example consider the the product of infinitely many $k$. So the definition in EGA is not the most general case?)
What you quote from Hartshorne is an example, not the definition, of a formal scheme. Demazure's definition isn't very general either. You can go to the original source EGA I section 10 for the actual definition. Probably the stacks project also treats this, but I haven't checked.
@DonuArapura: see Tag 0AHY.
Disclaimer: I am far from an expert. Feel free to ask for more details!
I was reading those lectures only recently. It is very important to acknowledge that Demazure uses the functor-of-points perspective for discussing schemes (see Chapter I.4 of those notes to see a discussion of the equivalence of this approach with the more standard one). Demazure gives 4 different (but equivalent) definitions of a formal scheme, one of those being the definition you stated. Another definition is stated in terms of profinite $k$-algebras. These are topological $k$-algebras $A$ such that (for some inverse system of open ideals $\{I_n\}$)
$A\cong \varprojlim_n A/I_n$
$A/I_n$ is finitely generated as a $k$-algebra.
Then the formal spectrum $\mathrm{Spf}(A)$ of $A$ is defined for a $k$-algebra $R$
as the set of continuous maps $\mathrm{Spf}(A)(R)$ (where $R$ is equipped with the discrete topology). Even better it is
$$
\mathrm{Spf}(A)(R)\cong \varinjlim_n \mathrm{Spf}(A/I_n)(R)
$$
Back to considering Hartshorne: A closed subscheme $Y\subset X$ gives rise to an ideal sheaf $\mathcal{I}\subset \mathcal{O}_X$. If we consider the case that $X=\mathrm{Spec}A$ is affine then $\mathcal{I}$ is simply an ideal of $A$ and $Y\cong \mathrm{Spec}{(A/\mathcal{I})}$.
We then place the $\mathcal{I}$-adic topology on $A$, check that it satisfies 1 and 2 above, and then the formal completion as given in Hartshorne exactly becomes
$$
\mathrm{Spf}(A)(R)\cong \varinjlim_n \mathrm{Spf}(A/\mathcal{I}^n)(R)
$$
As @DonuArapura commented the case presented in Hartshorne is an example of a more general definition where rather than considering adic topological rings, we consider admissible topological rings.
|
2025-03-21T14:48:30.925403
| 2020-05-09T15:24:36 |
359824
|
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"Dirk Werner",
"https://mathoverflow.net/users/127871",
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|
Stack Exchange
|
Determining the behavior of a contraction mapping with undefined points
Label $X$ as the real interval $[0, a]$ where $a \in \mathbb{R}^+$, so that $\text{int}(X) = (0, a)$ labels the interior of $X$ and $\partial X$ labels the boundary of $X$. I have a function $f:\text{int}(X) \rightarrow \text{int}(X)$ that is well defined on the interior of $X$, but ambiguous on the boundary of $X$; that is $f(\partial X)$ is not a defined mapping action.
Further, the system has the following properties: $f$ is a contraction mapping over $\text{int}(X)$, and if $a$ is not a positive integer, then sequential application of $f$ over any point $x \in \text{int}(X)$ will converge to a fixed point in $\text{int}(X)$. However if $a \in \mathbb{N}^+$, then application of $f$ over any point $x \in \text{int}(X)$ will converge to a boundary value in $\partial X$.
As such, I have a fixed map $f$ defined over a fixed interval $(0, a)$ where the limiting behavior of $f$ over initial point $x \in (0, a)$ is determined by parameter $a$. If $a$ is a positive integer, iteration of $f$ will converge to either 0 or $a$, where $f$ is not defined.
I want to prove that sequential application of $f$ over $x \in \text{int}(X)$ will converge to a fixed point in $\text{int}(X)$ if $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$. I am wondering if I can use the contraction mapping theorem to achieve this aim. After all, the contraction mapping theorem requires that the space is complete, however if action of $f$ is only defined over $(0, a)$ rather than $[0, a]$ then the space is not complete and the theorem cannot be invoked.
Does anyone know how I can then adequately set up this problem to eventually prove that iteration of $f$ over $\text{int}(X)$ will converge to a fixed point in $\text{int}(X)$ if $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$ using the contraction mapping theorem or something else?
Edit: Consider the following Theorem (Topology, Gamelin, Green) which will allow me to reword the above in a different way.
Theorem: Let $X$ be a complete metric space with metric $d$, and let $S$ be a metric space. Let $c$ be a fixed constant with $0 < c < 1$. Suppose that $(s, x) \rightarrow \Phi_s(x)$ is a continuous function from $S\times X$ to $X$ such that
$$
d(\Phi_s(x), \Phi_s(y)) \leq cd(x, y)
$$
for $x, y \in X, s \in S$. Then for each $s \in S$, there is a unique point $x_s^* \in X$ such that $\Phi_s(x_s^*) = x_s^*$. Furthermore, $x_s^*$ depends continuously on $S$.
Now, given the above theorem, we can let $a$ be the continuous parameter $s$ as stated in the above theorem, and I am now studying a function $\Phi_a: (0, a) \rightarrow (0, a)$ where the fixed point $x_a^*$ does not depend continuously on $a$, but rather depends piecewise discontinuously on $a$, and clearly $(0, a)$ is not a complete metric space as required.
As such, If I was to first establish that iteration of $\Phi_a$ is undefined when $a \in \mathbb{N}^+$, could I then apply the above theorem for all $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$ and put $\Phi_a: [0, a] \rightarrow [0, a]$ so that we have a complete metric space and the behavior of the map is well defined for all such values of $a$?
Maybe I don‘t understand the problem. You have a contraction $f:(0,a) \to (0,a)$. You seem to assume that iteration of any $x\in (0,a)$ leads to a fixed point of $f$. But you say you want to prove exactly this. Plus I don‘t understand why you are making a difference between integer and non-integer $a$.
@DirkWerner Yes, I would like to prove that statement. The reason I am distinguishing integer and non-integer values of $a$ is because the dynamics of the map over the interval are uniquely determined by the value of $a$, and it just so happens that when $a$ is an integer, this edge case occurs where the limiting behavior of the map is no longer well defined. As such, if I am seeking to prove the case when the behavior of the map is always well defined, I am not sure how to do this because it is only well defined over an open interval $(0, a)$ which is not complete.
@DirkWerner I edited the question which may help elucidate my problem.
I‘m sorry, but I still don‘t get it. The way you phrase it you have one interval and one function; i.e., $f:(0,a) \to (0,a)$. — Anyway, a remark worth keeping in mind is that a contraction on $(0,a)$ extends to a contraction on $[0,a]$.
In your new rendering of the question you seem to have a family of mappings $f_a: (0,a) \to (0,a)$. What is the relation between two such maps?
|
2025-03-21T14:48:30.925835
| 2020-05-09T15:39:49 |
359825
|
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"Dmitri Pavlov",
"Robert Furber",
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}
|
Stack Exchange
|
Image of probability measures under measurable mappings
Given two probability measures on two probability spaces, ($\mu, X$) and ($\gamma, Y$), what's the sufficient and necessary condition such that there is a measurable mapping $f:X\rightarrow Y$, such that $f^*\mu = \gamma$?
There is a complete classification of probability spaces up to a measure-preserving
isomorphism.
Specifically, consider a category whose objects are triples
(X,Σ,μ), where X is a set, Σ is a σ-algebra
of measurable subsets on Σ,
and μ is a probability measure on (X,Σ).
If we want to have a nice description of morphisms
in terms of equivalence classes of point-set maps, we must
also require that (X,Σ,μ) is a compact probability space
in the sense of Marczewski's 1953 paper
“On compact measures”, but without countability assumptions.
We also assume completeness, since this does not change
isomorphism classes, but makes it easier to define morphisms.
Morphisms (X,Σ,μ)→(X',Σ',μ') are equivalence classes of maps
of sets f:X→X' such that f*Σ'⊂Σ and μf*=μ'.
The equivalence relation identifies f and g
if for all σ∈Σ' the symmetric difference f*σ⊕g*σ
has measure 0.
The better-known equivalence relation of equivalence almost
everywhere reduces to the above one if the involved spaces are countably
separated (like the real line), but in general one must use the above definition.
The isomorphism classes of objects in the resulting category
admit a complete classification.
First, any object canonically decomposes as a coproduct (disjoint union)
of ergodic spaces, i.e., spaces for which the automorphism
group has no nontrivial invariant subsets.
Secondly, the ergodic spaces admit a complete classification as follows.
First, the discrete spaces are ergodic.
Secondly, the nondiscrete ergodic spaces are isomorphic
to the infinite product of I copies of {0,1} (a discrete space),
where I is an infinite set, whose cardinality is the only invariant
of the resulting space.
The classification of isomorphism
classes (first stated for Boolean algebras Σ/N) is known as the von Neumann–Maharam theorem.
A 1965 theorem by Cassius Ionescu Tulcea (with subsequent
improvements by Vesterstrøm–Wils, Edgar, Graf, Fremlin, Rinkewitz) shows that isomorphisms classes of such
Boolean algebras coincide with appropriately defined
isomorphism classes of probability spaces.
(This claim also holds if we remove the requirement that (X,Σ,μ) is compact,
in which case we must first localize the category at all morphisms f
such that the induced map of Boolean algebras Σ'/N'→Σ/N is
an isomorphism. Completeness can also be removed if the notion
of measurability is adjusted accordingly: elements in f*Σ' must be symmetric differences of elements in Σ and a subset of a measure 0 set.)
Thus, the original question can be answered as follows:
there is a measurable mapping f:X→Y such that f*μ=γ
if and only if the decompositions of X and Y into their ergodic
components have the same measures for each type of ergodic summand,
and the discrete parts are isomorphic.
It is not that $\Sigma$ is a compact class, but rather that there exists a compact class $\mathcal{K} \subseteq \Sigma$ such that $\mu$ is inner regular with respect to $\mathcal{K}$ (i.e. that $(X,\Sigma,\mu)$ is compact in the terminology used in Fremlin's Measure Theory volume 3). This cannot be replaced in the general case by any property of $\Sigma$, because there are $\sigma$-algebras that admit both compact and non-compact measures (e.g. the countable-cocountable $\sigma$-algebra), though of course all probability measures on a standard Borel space are compact.
You have also omitted the requirement that $(X,\Sigma,\mu)$ have a lifting. Shelah proved that it is relatively consistent to ZFC that $([0,1],\mathrm{Bo}([0,1]), \lambda)$, $\lambda$ being the Lebesgue measure restricted to Borel sets, have no lifting. Since complete probability spaces always have a lifting, it follows that the Borel restriction of Lebesgue measure is not isomorphic to its completion in this model. (Von Neumann proved, on the other hand, that this space does have a lifting under the continuum hypothesis).
And you may call it the "von Neumann-Maharam theorem", but from my perspective that only refers to the existence of an isomorphism of measure algebras as complete Boolean algebras, not of measure spaces. The existence of an isomorphism of measure spaces requires substantial extra work due to Fremlin.
@RobertFurber: I certainly meant a compact measure, not just a set-theoretic notion. I also added the completeness condition that I left out.
@RobertFurber: I must also say that the existence of point-set maps of measurable spaces is not due to Fremlin, but rather is a theorem by Cassius Ionescu Tulcea (see his 1965 paper in Comptes Rendus). There were several subsequent improvements, see my paper https://arxiv.org/abs/2005.05284, where I try to list (hopefully) all contributors.
That sounds interesting, but a search of the 1965 Comptes Rendus yields no papers by either Ionescu Tulcea.
@RobertFurber: Volume 261, pages 4961--4963. Link: https://gallica.bnf.fr/ark:/12148/bpt6k4027h/
Thanks. That was the fault of the crappiness of the French National Library's website.
Unfortunately Lemma 5.10 in your arxiv paper is not correct, unless you assume the nonexistence of real-valued measurable cardinals. The problem is that Lemma 4.37 can only be used to prove that you have a morphism preserving arbitrary suprema for $\sigma$-finite enhanced measurable spaces, not the general localizable case you intend to apply it to there. Counterexample in next comment.
As a counterexample, assume that $X$ is a set such that there exists a countably additive probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons (so $|X|$ exceeds the first real-valued measurable cardinal), and $\nu : \mathcal{P}(X) \rightarrow [0,\infty]$ is the counting measure. Let $f : (X, \mathcal{P}(X), \mu) \rightarrow (X, \mathcal{P}(X), \nu)$ be the identity mapping. The only nullset of $\nu$ is $\emptyset$, so this is a premap of enhanced measurable spaces. The join of all singletons in $X$ is 1, but each singleton has $\mu$-measure zero ...
So the join in $\mathcal{P}(X)$ modulo the $\mu$-nullsets is $0$. The good news is: 1. This is essentially the only possible counterexample. 2. If $X$ is compact and $Y$ is strictly localizable, this situation cannot occur and $\mathcal{ML}$ only produces complete Boolean algebra homomorphisms. If you like, I can write a letter to you explaining the proof of this, which you can incorporate into your arxiv paper.
@RobertFurber: Thanks a lot for pointing this out! Fortunately, the wrong half of Lemma 5.10 was not used anywhere, so I threw it away and added the missing reduction to the σ-finite case. I posted a new version here: https://dmitripavlov.org/equiv.pdf, which will be uploaded to arXiv sometime soon. Let me know if you see anything suspicious left or have any questions. I also added your example (with an acknowledgment in the introduction) as Remark 5.11.
The new version looks fine to me.
Most likely just standard probability spaces are sufficient for your purposes. All non-atomic such spaces are pairwise isomorphic.
One sufficient condition is that the source space is nonatomic and the target space has the Borel sets of a Polish space as the underlying $\sigma$-algebra. See here for pointers on how one can prove this.
|
2025-03-21T14:48:30.926480
| 2020-05-09T16:07:05 |
359827
|
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"C.F.G",
"Carlo Beenakker",
"Charles",
"Dmitri Pavlov",
"Francois Ziegler",
"Gabe Conant",
"Gro-Tsen",
"Joe Silverman",
"Najib Idrissi",
"R W",
"Sam Hopkins",
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|
Stack Exchange
|
How to cite a MathSciNet review
I want to cite a MathSciNet review. Is there a standard format for that? (I couldn't find anything)
It's a journal like any other, I would cite it as: A. Uthor (of the review). "Title of the article." Mathematical Reviews (20XX). https://link-to-the-review. I don't think they publish volumes or that reviews have page numbers anymore so it's a bit bare bones, of course.
At this point, I believe that every entry in MathSciNet has a unique MR identifier for the form MR#######, so that should certainly be included. But it's not clear if the reference should be sorted by the auhor of the article or the author of the review. I would tend to say the latter, since that's who wrote the material, but I'm sure others would disagree. In any case, both names should be listed, making it clear who wrote what.
@Joe Silverman Sure - of course I want to include the MR identifier, the names, etc. My question is rather about a template for the whole reference.
You can actually search Mathematical Reviews for itself (the phrase) in References and find hundreds of hits — e.g. one of the earliest being ref.3 of this paper.
Wait, if Math Reviews is a journal like any other, why aren't Math Reviews reviewed by Math Reviews?
@Gro-Tsen Perhaps a really extensive one would be!
Mathematical Reviews can be cited like any other journal, with a review number instead of volume and page numbers:
J. Smith. Review of the article “Regular doodads are widgits” by J. Doe. Mathematical Reviews 123456 (2009).
@article {review,
AUTHOR = {Smith, J.},
TITLE = {Review of the article ``{R}egular doodads are widgits'' by {J}. {D}oe},
JOURNAL = {Mathematical Reviews},
VOLUME = {123456},
YEAR = {2009},
ISSN = {0025-5629},
URL = {https://mathscinet.ams.org/mathscinet-getitem?mr=123456},
}
Yes - of course I understand that. I should have been more precise in my question - I am looking for a BibTeX template to do that.
@RW: I added some BibTeX code, but it is quite standard.
Thank you - this is probably the closest.
@DmitriPavlov: sorry, too late but is that possible to add the MathSciNet review link and not the paper link? i.e. the above link inside the BibTex (https://mathscinet.ams.org/mathscinet-getitem?mr=123456) is a link of paper. (I don't have access to MathSciNet, so I don't know its link format)
@C.F.G: I am not sure I understand your question. A hyperlink of the form https://mathscinet.ams.org/mathscinet-getitem?mr=123456 does indeed point to the MathSciNet review with the indicated number 123456. Where do you want to point it instead?
@DmitriPavlov: The downloadable PDF of review or mathscinet review page that contains the review. Perhaps the above link is as I wanted but for MathSciNet subscribers?
@C.F.G: The current hyperlink already points to the text of the review. The reviews are only accessible to subscribers, you cannot see them without a subscription.
I checked with the head of copy-editing at Mathematical Reviews and the Mathematical Reviews librarian. Their joint answer is (using the paper mentioned by Carlo Beenakker):
If you want to add something to a reference list, the form would be:
Udrişte, Constantin, review of ``Optimal approximations by piecewise smooth functions and associated variational problems,'' (in Comm. Pure Appl. Math. 42 (1989), no. 5, 577–685, by David Mumford and Jayant Shah), Mathematical Reviews/MathSciNet MR0997568 https://mathscinet.ams.org/mathscinet-getitem?mr=997568.
If you want to mention something in text,
Constantin Udrişte (see his review published in Mathematical Reviews/MathSciNet [MR0997568]) related the Mumford-Shah results to Griffiths' law of cracks in solid mechanics.
I hope this helps.
I once emailed Andres Caicedo (Associate Editor at Math Reviews) about this, and his response was essentially identical to this answer.
@Gabe Conant Good to know - thank you!
too long for a comment; here is a BibTeX template:
@article {MR997568,
AUTHOR = {Mumford, David and Shah, Jayant},
TITLE = {Optimal approximations by piecewise smooth functions and
associated variational problems},
JOURNAL = {Comm. Pure Appl. Math.},
FJOURNAL = {Communications on Pure and Applied Mathematics},
VOLUME = {42},
YEAR = {1989},
NUMBER = {5},
PAGES = {577--685},
ISSN = {0010-3640},
CODEN = {CPAMA},
MRCLASS = {49F20 (41A30 92A27)},
MRNUMBER = {997568 (90g:49033)},
MRREVIEWER = {Constantin Udri{\c{s}}te},
DOI = {10.1002/cpa.3160420503},
URL = {http://dx.doi.org/10.1002/cpa.3160420503},
}
This looks like a citation to an article, not a review.
That's fine for citing an article and giving a pointer to the Math Review, but I think that the OP actually wants to cite material that appears in the review, not the article, and thus giving appropriate credit to the person who wrote the review.
@Joe Silverman - yes, precisely.
OK, but if the OP asks for a BiBTeX template, this is it: a BiBTeX template should contain the full bibliographic information, what is actually displayed in the publication is then selected by a style file, not by removing material from the BiBTeX template.
|
2025-03-21T14:48:30.926864
| 2020-05-09T16:43:33 |
359831
|
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"GabS",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359831"
}
|
Stack Exchange
|
inequality involving the fractional Sobolev space
Let $X_{0}$ be the Sobolev space defined on $(1, 2)$ by $X_{0}(1,2)= \{u\in H^s(\mathbb R): u=0 \text{ in } \mathbb R-(1, 2)\}.$ Is it possible to determine the constant $C$ of the inequality
$$|u(x)| \leq C \|u\|_{X_{0}} $$
where $u\in X_{0}$, $H^s(\mathbb R)= W^{s, 2}(\mathbb R)$ and $$ \|u\|_{X_{0}}^2= \int_{\mathbb R} |(-\Delta)^{s/2} u|^2dx.$$
Does a weaker version of the inequality exists when $s=1/2.$
In the first place, you must have $s>1/2$. Next you write
$$
u(x)=\int_{\mathbb R} e^{2πix\xi}\underbrace{\hat u(\xi)(1+\xi^2)^{s/2}}_{\in L^2(\mathbb R)}\underbrace{(1+\xi^2)^{-s/2}}_{\in L^2(\mathbb R)}d\xi,
$$
entailing
$
\vert u(x)\vert\le \Vert u\Vert_{H^s(\mathbb R)}\left(\int_{\mathbb R}(1+\xi^2)^{-s} d\xi\right)^{1/2}.
$
Moreover, you have with $D_x=-i\partial_x$
$$
2\Re\langle D_x u, i x u\rangle_{L^2(\mathbb R)}=\Vert u\Vert^2_{L^2(\mathbb R)},
$$
so that if $u$ is supported in $(-1/2, 1/2)$, we have
$
\Vert u\Vert^2_{L^2(\mathbb R)}\le \Vert u\Vert_{L^2(\mathbb R)}\Vert D_xu\Vert_{L^2(\mathbb R)}
$
and thus $\Vert u\Vert_{L^2(\mathbb R)}\le \Vert D_xu\Vert_{L^2(\mathbb R)}$
proving that for functions whose support has a diameter $\le 1$ the $H^1$ norm is equivalent to the $L^2$ norm of the derivative. You get an answer for $s=1$, which can probably be extended to any $s>1/2$ (with different constants).
The interesting part the compact support of the function $u$ does not play a role.
For $u \in H^2(I)\cap H^1_0 (I)$, $I=(1,2)$, the inequality is equivalent to $$\|(-\Delta)^{-s/2}v\|_\infty \le C \|v\|_2, \quad v \in L^2(I).$$ Let us use
$$
(-\Delta)^{-s/2}v=\frac{1}{\Gamma (s/2)}\int_0^\infty t^{s/2 -1}T(t)v\, dt$$
where $T(t)$ is the semigroup generated by the Dirichlet Laplacian in $I$. If $\lambda_1$ is its first eigenvalue then $\|T(t)\|_{2 \to 2} \le e^{-\lambda_1 t}$ and $\|T(t)\|_{ 2 \to \infty} \le C_1t^{-1/4}$, by Gaussian estimates. The semigroup law yields $\|T(t)\|_{2 \to \infty} \le C_2 t^{-1/4}e^{-\lambda_1 t/2}$ and finally
$$\|(-\Delta)^{-s/2} v\|_\infty \le \frac{C_2}{\Gamma (s/2)}\|v\|_2 \int_0^\infty t^{s/2-5/4}e^{-\lambda_1 t/2}\, dt \le C_3 \|v\|_2$$
if $1/2<s<2$. Finally, note that all constants $C_i$ can be computed since $\lambda_1$ is known and the Gaussian estimates follow from domination with the heat semigroup in the whole space.
|
2025-03-21T14:48:30.927018
| 2020-05-09T17:02:44 |
359834
|
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}
|
Stack Exchange
|
How can I find minimum and maximum eigenvalue of non-positive define matrix
There is a power iteration method, but it only returns the greatest(in absolute value) eigenvalue of matrix. So when we have negative eigenvalues it'll give wrong results.
Is there any method, which work also for non-positive define matrices.
Find your greatest in absolute value eigenvalue, call it $E.$ If it is positive, the matrix $M + 2 EI$ is positive definite, so do whatever you do for positive definite matrices. If it is negative, $M - 2 E I$ is positive definite.
|
2025-03-21T14:48:30.927085
| 2020-05-09T17:29:32 |
359838
|
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"Harry Richman",
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|
Stack Exchange
|
Polytopes with large dihedral angles
The regular $d$-simplex has dihedral angle $\arccos(1/d)<90^\circ$, and the $d$-cube has dihedral angle exactly $90^\circ$.
The maximal dihedral angle of a prism over a $(d-1)$-simplex is also $90^\circ$.
For each $d\ge 2$, these are the polytopes with the smallest maximal dihedral angle among all convex $d$-dimensional polytopes.
Question: What are the $d$-polytopes with the next smallest maximal dihedral angle?
I am more interested in the value of the angle than the polyope, and I am okay with bounds rather than exact values.
More precisely, I wonder for which $d\ge 2$ there is a $d$-polytope $P\subset\Bbb R^d$ (not the simplex, the cube or a prism) with all dihedral angles smaller than
$\arccos(-\frac1d)>90^\circ$, or even
$\frac13(\pi + \arccos(-\frac1d))>90^\circ$.
For reference when $d=3$: https://en.wikipedia.org/wiki/Table_of_polyhedron_dihedral_angles. Is the answer known for small $d$?
As stated I think the answer is yes for all $d$: just take the cube and "squash" it slightly so some dihedral angles increase and some decrease.
@HarryRichman I see, that was unexpected. In my application I have additional assumptions on the polytope, namely, that its $(d-2)$-faces are simplices, so I suppose this is an essential assumption. Anyway, I will need to think about this and maybe edit the question.
|
2025-03-21T14:48:30.927208
| 2020-05-09T17:45:10 |
359841
|
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|
Stack Exchange
|
How many Shapes are possible to create using Voxels?
Let's suppose I have Big Cube of x cm by y cm by z cm, simmilar to this one:
This big cube is made of tiny little cubes of t cm.
All of this little cubes are transparent, but some of them are red
How many arrangements of little red cubes is it possible to do?
I think this is a permutations with repetition problem, and so far, this is the equation I came with:
$$ a= ( \sum_{n=0}^{A} P_{A}^{A-n,n} )^Z $$
Where:
$ a= $ Number of possible arrangements
$ Z= $ Value of the number of layers equal to $z/t$
$ A= $ Area of each layer equal to $(x/t)(y/t)$
$ P_n^{n_1,...,n_r}=$ Notation for Permutations with Repetition
So, for example, if it is used the values of the example, it is obtained this:
$$ a = ( \sum_{n=1}^{64} P_{64}^{64-n,n} )^8 = (18,446,744,073,709,551,616)^8 $$
But my problem is: How many arragenments of little cubes can I do, so that the all the red cubes touch each other, so that there is no isolated geometries?
For example, this one is valid, it is only one geometry:
This one is not valid, there are two shapes separated:
I suppose that the orginal formula, must have a term that gets rid of the cases with Isolated Shapes
$$ ( \sum_{n=0}^{A} P_{A}^{A-n,n} ) ^Z - (IsolatedShapes) $$
What's the value of that "IsolatedShapes" Term?, Could you give some ideas of how to tackle this problem?
Exact enumeration of connected shapes is a challenge. You might liken it to graph enumeration of graphs of bounded degree, with additional properties like being bipartite and connected (let vertices be cube centers, with an edge between adjacent cubes). There should be tables of earlier attempts on the web. Gerhard "Is Confronting Other Challenges Presently" Paseman, 2020.05.09.
But at least, Do you know if my formula for the General Case is correct?
I am unsure about your formula, as I do not understand it. One has an upper bound of 2^(Z^3) by considering subsets of cubes, but this assumes no symmetry collapse (e.g. there are Z^3 many figures, each with exactly one cube, in a different positions) and counts disconnected arrangements. You might find counting connected arrangements in a layer just as challenging, but then you could use that for tighter approximations. Gerhard "Go Ahead: Try A Slice" Paseman, 2020.05.09.
This reminds me a lot of the enumeration of LEGO structures, a famously difficult problem.
|
2025-03-21T14:48:30.927398
| 2020-05-09T17:50:18 |
359842
|
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|
Stack Exchange
|
Convergence of Fuchsian groups and existence of suitable homeomorphisms
Let $(\Gamma_n)_n$ ($\subset PSL(2,\mathbb{R})$) be a sequence of discrete groups, if we say that $(\Gamma_n)_n$ converges to a group $\Gamma$ this means that there exist isomorphisms $\tau_n:\Gamma\rightarrow\Gamma_n$ such that for all $\sigma\in\Gamma,\tau_n(\sigma)$ converge to $\sigma$.
Do someone have an idea or a reference about the proof of the following Lemma:
Lemma:
Let $(\Gamma_n)_n$ be a sequence of Fuchsian groups converging to a group $\Gamma$ and $\tau_n$ the isomorphisms given in the definition above then:
There exist homeomorphisms $\phi_n:\mathbb{H}^2/\Gamma\longrightarrow\mathbb{H}^2/\Gamma_n$ whose lifts $\tilde{\phi}_n$ satisfies for any $\sigma\in\Gamma$
$$\tilde{\phi}_n\circ\sigma=\tau_n(\sigma)\circ\tilde{\phi}_n$$
and such that $(\tilde{\phi}_n)_n$ converges to the identity map uniformly on compact sets i.e that $(\tilde{\phi}_n)$ satisfies,
$$\forall x\in\mathbb{H}^2, \tilde{\phi}_n(x)\underset{n\rightarrow\infty}{\longrightarrow}x$$
($\mathbb{H}^2$ is the hyperbolic plane.)
Thanks for any answer
I think, this was asked earlier at MSE. The answer, as before, that without further assumptions, your lemma is false.
@MoisheKohan why it is false? For me the Lemma is true like that
You can find a sequence of free groups of infinite rank $\Gamma_n$ converging to a free group of infinite rank $\Gamma$ such that $H^2/Gamma$ is a planar surface, while each $H^2/\Gamma_n$ has infinitely many handles. Hence, homeomorphisms $\phi_n$ do not exist.
@MoisheKohan Do you know if there’s a counterexample with finitely generated groups? Surfaces of finite type feel a bit more tame. Although if you use free groups you might be able to get a planar/handle mismatch...
@SantanaAfton: Lemma holds for finitely generated groups, but I am not sure the result is stated anywhere in the literature.
In fairness to the OP, the phrase "Fuchsian groups" is used inconsistently in the literature. Often, it includes the hypothesis of finite generation.
|
2025-03-21T14:48:30.927564
| 2020-05-09T17:52:29 |
359843
|
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|
Stack Exchange
|
Approximating expectation of the trace of inverse of a Gaussian random matrix combination
In order to characterize the performances of MIMO systems that depend directly on the distribution of the eigenvalues of random Hermitian matrix so I would like to feature the quality of some particular case that is defined as. For a given random complex gaussian matrix A $(N, M)$ and $B(N, K)$ Gaussian matrix and constant parameter $\alpha$, where ${\tilde A}$ indicate the Moore–Penrose inverse of $A$. The problem consists to find the expectations defined as:
\begin{align} E[\operatorname{trace}((aI + \frac{{\tilde A\,B{B^H}{{\tilde A}^H}}}{{\operatorname{trace}(\tilde A\,B{B^H}{{\tilde A}^H})}})({B^H}B))]\end{align}
@Carlo Beenakker yes its error I will correct it, thanks Sir
A few remarks: $AA^H$ is invertible, the MP inverse $\tilde{A}=A^H(AA^H)^{-1}$; so the trace in the denominator is
$${\rm tr}\,\tilde{A}BB^H\tilde{A}^H={\rm tr}\,A^H(AA^H)^{-1}BB^H(AA^H)^{-1}A={\rm tr}\,BB^H(AA^H)^{-1}.$$
The expectation value of this trace follows from your previous question and answer.
Now the full expectation value in this new question is unlikely to have a closed-form answer for any $N,M,K$, but for large values you can decouple the averages in numerator and denominator, because the trace is "self-averaging", meaning that you can replace it by its expectation value.
If this large $N,M,K$ limit is of interest, I may try to develop this a bit further.
sir For large N, M, K how can I separate the expectation because they are related? thanks
I try to solve the problem I find the result using Wishard distribution the problem is simplified to \begin{align} trace (((BB^H)^-1+\frac{1}{a}I)^{-1})\end{align}
|
2025-03-21T14:48:30.927717
| 2020-05-09T18:03:05 |
359844
|
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"Christian Remling",
"Stef1611",
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|
Stack Exchange
|
Eigenvalues of symmetric tridiagonal matrices with identical off diagonal elements
Is there a simple analytical solution to obtain eigenvalues (and eigenvectors) for this type of tridiagonal matrices ? ( Off diagonal elements are identical and the matrix is symmetric)
$$
\begin{pmatrix}
a_{1}-k & k & 0 & 0 \\\
k & a_{2}-2\cdot k & k & 0 \\\
0 & k & -a_{2}-2\cdot k & k \\\
0 & 0 & k & -a_{1}-k
\end{pmatrix}
$$
To obtain eigenvalues of symmetric tridiagonal matrices, I used the formula
$$
P_{k} (\lambda)=(\alpha_{k}−\lambda)P_{k-1}(\lambda)−\beta^2_{k-1}P_{k-2}(\lambda),
$$
but I could not find any simplification.
Thanks for answer.
No (I'm assuming you mean in general, not just for $n=4$). We're essentially dealing with the discrete Schrodinger equation $y_{n+1}+y_{n-1}+b_n y_n = \lambda y_n$, which can't be solved explicitly in general.
@Christian. n>4 would be better. But if a solution exists for n=4, I will be satisfied. There is also another constraint that I did not mention. a1=3*a2. I do not know if it could help. And yes, it is related to the discrete Schrodinger equation.
@Stef1611: For $n=4$ the characteristic polynomial has degree $4$, so there's a formula for the zeros (though it probably won't be pretty).
@Christian. I agree, they are not pretty at all. I used sympy to calculate them.
|
2025-03-21T14:48:30.927836
| 2020-05-09T18:07:56 |
359845
|
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"authors": [
"Alister Trabattoni",
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"R.P.",
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|
Stack Exchange
|
Fourier series of $\log(a +b\cos(x))$?
By numerical computation it seems like, if $a_0 < a_1$:
$$
\begin{multline}
\log({a_0}^2 + {a_1}^2 + 2 a_0 a_1 \cos(\omega t)) = \log({a_0}^2 + {a_1}^2) \\
+ \frac{a_0}{a_1}\cos(\omega t)
- \frac{1}{2}\frac{{a_0}^2}{{a_1}^2}\cos(2\omega t)
+ \frac{1}{3}\frac{{a_0}^3}{{a_1}^3}\cos(3\omega t)
- \frac{1}{4}\frac{{a_0}^4}{{a_1}^4}\cos(4\omega t) \ldots
\end{multline}
$$
If $a_0 > a_1$, the two term must be exchanged.
I'm quite confident in this solution but I cannot find a way to prove it mathematically...
Would be nice to know the reason of this result!
Looks like the real part of $\log(1+e^{i\omega t})$, give or take some constants.
You are totally right! This is the good approach.
Let's consider
$$
f(x) = \log (1+q^2+2q\cos x) = \log |1+qe^{ix}|^2 ,
$$
which differs from your function only by the additive constant $2\log a_1$ if we take $q=a_0/a_1$. Since $|q|<1$, we can use the Taylor series of $\log(1+z)$ to write
$$
\begin{align}
f(x) = 2\,\textrm{Re}\; \log (1+qe^{ix}) = 2\,\textrm{Re}\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} e^{inx} \\
= 2\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} \cos nx .
\end{align}
$$
Just the answer I was looking for. Perfect! Good idea to use the complex notation.
Sorry to be a stickler, but I think you need $(-1)^{n-1}$ in the summand of the Maclaurin series expansion above, unless I missed something?
@Procore: Yes, thanks, corrected now.
@ChristianRemling Do you thing it is possible to find a solution for the more general case proposed in the title? I'm trying to solve the case $\log({a_0}^2 + {a_1}^2 + 2 a_0 a_1 \cos(\omega t) + {a_2}^2)$ and just adding this little ${a_2}^2$ prevent me to find a nice solution.
@AlisterTrabattoni: I don't have any ideas for this off the top of my head, and my guess would be that these Fourier coefficients can't be found explicitly, but I could be wrong about this of course.
@ChristianRemling I found the way to apply your solution to the more general case. See the edit on my original post.
You could also add it as an answer, that would probably be clearer. @AlisterTrabattoni
To apply the solution proposed by @ChristianRemling to the more general case stated in the title you need to do the following:
The formula implies $a > 0$ and $|b| < a$. To solve the problem, we reformulate the formula as:
$$
\begin{align}
r|1 + qe^{ix}|^2
&= r(1 + qe^{ix})(1 + qe^{-ix}) \\
&= r(1 + 2q\cos x + q^2) \\
& = a + b\cos x \\
\end{align}
$$
This requires to solve the following system:
$$
\begin{align}
a &= r(1 + q^2) \\
b &= 2rq
\end{align}
$$
From the two possible solutions we take the one where $|q|<1$ which is mandatory for the Taylor expansion.
$$
\begin{align}
q = \frac{a - \sqrt{a^{2} - b^{2}}}{b} \\
r = \frac{a + \sqrt{a^{2} - b^{2}}}{2}
\end{align}
$$
Finally, Taylor series expansion solve the problem:
$$
\begin{align}
\log (a + b\cos x)
&= \log r
+ 2\log | 1 + qe^{ix} | \\
&= \log r
+ 2 \sum {(-1)}^{n-1} \frac{q^n}{n} e^{inx} \\
&= \log r
+ 2 \sum {(-1)}^{n-1} \frac{q^n}{n} \cos nx
\end{align}
$$
|
2025-03-21T14:48:30.928299
| 2020-05-09T18:17:03 |
359847
|
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|
Stack Exchange
|
Differential geometry of Donaldson-Thomas invariants
The Donaldson-Thomas invariants are defined by Thomas in the paper A holomorphic Casson invariant for Calabi-Yau 3-folds, and bundles on K3 fibrations, following the proposal in Gauge theory in higher dimensions by Donaldson and Thomas. Although the initial proposal was motivated by differential geometry, the rigorous definition uses tools from algebraic geometry, including moduli spaces of (semi-)stable sheaves and perfect obstruction theory.
The most successful application of DT theory so far seems to be the enumeration of curves, i.e. considering the DT invariants of ideal sheaves, see e.g. 13/2 ways of counting curves. So actually the question is two-fold:
Is there a way of counting solutions to some version of (perturbed) Hermitian-Yang-Mills equations which could presumably recover DT invariants of ideal sheaves?
As Gromov-Witten theory also makes sense for symplectic manifolds, is there a symplectic (actually, almost complex) counterpart of DT theory?
Theorem 1.4 of Bridgeland-Smith (https://arxiv.org/abs/1302.7030) implies that the DT invariants of certain quasi-projective CY 3-folds, defined with respect to the stability condition associated to the quadratic differential, are given by counting certain special Lagrangian submanifolds. See allso the work of Joyce: https://arxiv.org/abs/hep-th/9907013.
A symplectic theory should exist but is still some way off, even for curve counting invariants. See arXiv/1712.08383 by Doan-Walpuski for some progress for stable pairs.
|
2025-03-21T14:48:30.928561
| 2020-05-09T19:08:04 |
359851
|
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"url": "https://mathoverflow.net/questions/359851"
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|
Stack Exchange
|
difference between: Measurable multifunction integrably bounded and Measurable multifunction integrable
I read the article "Komlós Theorem for Unbounded Random Sets" by G. KRUPA (MSN), but I did not understand the difference between:
Measurable multifunction integrably bounded,
Measurable multifunction integrable
(see p. 239).
Let $F:\Omega\to 2^X$ be a measurable multifunction with closed values in the separable Banach space $X$. It seems to be implicitly assumed that $F$ jas nonempty values.
$F$ is integrably bounded if the function $\omega\mapsto \sup_{x\in F(\omega)}\|x\|$ is integrable.
$F$ is integrable if the function $\omega\mapsto \inf_{x\in F(\omega)}\|x\|=d(0,F(\omega))$ is integrable.
The point of these definitions might be better understood in terms of measurable selections. By the Kuratowski-Ryll-Nardzewski measurable selection theorem, $F$ admits measurable selections. If $F$ is integrably bounded, all selections will be integrable and dominated by a single integrable function. If $F$ is integrable, at least one selection will be integrable.
Let $\mathcal{P}{c}(X)$ be the family of all closed subsets of $X$. We know that the space of all measurable multifunctions $F:\Omega\to \mathcal{P}{c}(X)$ integrably bounded noted by $\mathcal{L}^1_{\mathcal{P}_{c}(X)}$. But what is the notation of the space of all the measurable multifunctions integrable?
I don't think the author introduces special notation for integrable multifunctions.
What are the advantages of the integrable multifunction? And why did the authors present this notion?
can you give me an example of multifunctions integrable but not integrably bounded ?
The multifunction with constant value $X$.
can you give me a book about multivalued integration
I think "Infinite Dimensional Analysis" by Aliprantis and Border is a great source for material on multifunctions (correspondences in the book).
Thank you very much, I still have a question that I did not understand : when we talk about the measurable function $f:\Omega\to X $, we assume that the $X$ is equipped with the Borelian $\sigma$-algebra $\mathcal{B}(X)$. but, to say that a multifunction $ F:\Omega\to 2^X $ is measurable. Which $\sigma$-algebra $\mathcal{A}$ should be equipped in $2^X$?
Measurability of a multifunction is generally not defined in terms of a $\sigma$-algebra on $2^X$. In special cases, something like that is possible though. If $X$ is a separable metrizable space and $\mathcal{K}$ the family of nonempty compact subsets of $X$ and if $f:\Omega\to 2^X$ has values in $\mathcal{K}$, then measurbility of $f$ corresponds to measurability with respect to the Hausdorff metric topology on $\mathcal{K}$. This is Theorem 18.10 in the mentioned book.
|
2025-03-21T14:48:30.928744
| 2020-05-09T19:53:53 |
359854
|
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"Christian Remling",
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|
Stack Exchange
|
Approximate identities: a converse question
Let $K(x)$ be a positive-valued function such that
$$\int_{-\infty}^{\infty} K(x) \ dx = 1$$
and let
$$K_{\lambda}(x) = \lambda K(\lambda x), \ \ \ (\lambda > 0);$$
that is to say, the family of functions $$\{K_{\lambda}(x)\}, \ \ \ \lambda \uparrow \infty $$
is an approximate identity generated by dilation.
Suppose $f(x)$ is a measurable function, but we do not know in advance whether or not $f(x) \in L^2$. Assume that for every $\lambda > 0$, the function
$$K_{\lambda}(x)*f(x)$$
is finite valued at every $x$.
Further, we know that there exists a function $g(x) \in L^2$ such that
$$ \lim_{\lambda \uparrow \infty} \int_{-\infty}^{\infty} [K_{\lambda}(x)*f(x) - g(x)]^{2} \ dx = 0.$$
Does it follow that $$g(x) = f(x) $$
almost everywhere?
If it makes the question easier to answer, one could add the assumption that $f(x)$ is locally integrable.
How do we define $K_{\lambda}*f$ if we only know that $f$ is measurable?
If $f$ is measurable, then the convolution function $K_{\lambda}*f$ is well-defined. I have added the condition that it is finite valued at every $x$.
If $f$ corresponds to, say, a bounded distribution (a continuous linear functional on the class of test functions $\phi$ such that all derivatives of $\phi$ are absolutely integrable), then $K_\lambda * f$ converges to $f$ in the space of bounded distributions, and to $g$ in $L^2$. Thus, $f = g$ almost everywhere. That said, I do not know if it is sufficient to assume that the integrals in the definition of $K_\lambda * f$ are absolutely convergent.
|
2025-03-21T14:48:30.928881
| 2020-05-09T19:56:30 |
359856
|
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"url": "https://mathoverflow.net/questions/359856"
}
|
Stack Exchange
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Any exact faithful functor is represented by a unique projective generator
In the book 'Tensor Categories' by Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik on page 10 it says:
'Conversely, it is well known (and easy to show) that any exact faithful functor $F : \mathcal{C} \rightarrow \text{Vec}$ is represented by a unique (up to a unique isomorphism) projective generator $P$.'
But I could not find any proof of that fact. Can someone tell me how to prove it or where I can find a proof?
Note: Here $\mathcal{C}$ is a finite k-linear abelian category for some field k.
Could you edit your question to include the hypotheses on $C$?
What do you mean exactly by a finite abelian category?
https://ncatlab.org/nlab/show/finite+abelian+category
Let $\mathcal C$ be the category of finite dimensional left modules over a finite dimensional ring $R$. Let $G: \mathcal C \to \mathrm{Vec}$ be an exact and faithful functor to finite dimensional vector spaces. We use $V^*$ to denote the dual vector space. For motivation, notice that if we had a representing object $M$, we would have $$G(R^*) = Hom_R(M,R^*) = Hom_R(R, M^*) = M^*.$$
Now $G(R^*)$ is a right $R$ module via the action of $R$ by left multiplication. So we define $P:= G(R^*)^*$ to be the dual left module and consider the functor $Hom_R(P,-)$.
This functor is tautologically left exact, and it takes the injective left module $R^*$ to $Hom_R(P,R^*) = Hom_R(R, G(R^*)) = G(R^*)$. Any other finite left module $M$ admits an injective presentation $$0 \to M \to R^{* \oplus a} \to R^{* \oplus b} \to $$
dual to the presentation of $M^*$ as a right $R$ module. So by left exactness, we see that there is a natural isomorphism $Hom_R(P,-) \simeq G(-)$.
Thus $G$ is represented by $P$. Since $G$ is right exact $P$ is projective, and since it is faithful $P$ is a generator.
Assuming that by $G(R)^$ you mean the vector space dual, then $P=G(R)^$ is not usually projective. Even when $G=\text{Hom}_R(R,-)$ is the forgetful functor, $G(R)^\cong R^$ which is not projective unless $R$ is self-injective.
@JeremyRickard Thanks, I was missing a dual. I edited the post to fix the mistake-- I think it is correct now.
Why is the Hom-functor tautologically right exact? I thought since Hom is right adjoint to the tensor functor, it usually just is left exact.
@S.Farr Sorry, I wrote it backwards-- I am using left exactness in the argument.
@PhilTosteson, good, then it makes sense. I think you also need to change the 'left exact' to 'right exact' in the last line.
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2025-03-21T14:48:30.929074
| 2020-05-09T20:28:02 |
359858
|
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"url": "https://mathoverflow.net/questions/359858"
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|
Stack Exchange
|
Question on Cuntz' proof of Bott periodicity
I am reading the presentation of Cuntz' proof of Bott periodicity for $C^*$-algebras in Wegge-Olsen (Thm. 11.2.1). Here one considers the short exact sequence of $C^*$-algebras
$$0 \longrightarrow \mathcal{T}_0 \longrightarrow \mathcal{T} \stackrel{q}{\longrightarrow} \mathbb{C} \longrightarrow 0,$$
where $\mathcal{T}$ is the Toeplitz algebra and $q$ is the homomorphism determined by sending the shift operator $S \in \mathcal{T}$ to $1$. Clearly, we also have the map $j: \mathbb{C} \rightarrow \mathcal{T}$, and $q \circ j = \mathrm{id}_{\mathbb{C}}$.
Now, the main step in the proof is to show that the induced map in $K$-theory $q_*: K_0(\mathcal{T}) \rightarrow K_0(\mathbb{C})$ is an isomorphism.
But then, it is claimed that it was also clear that also for any $C^*$-algebra $A$, one has that also $$(\mathrm{id}_A \otimes q)_*: K_0(A \otimes \mathcal{T}) \rightarrow K_0(A)$$ is an isomorphism.
I do understand that since all $C^*$-algebras in the short exact sequence are nuclear, the sequence
$$0 \longrightarrow A \otimes \mathcal{T}_0 \longrightarrow A \otimes \mathcal{T} \stackrel{\mathrm{id}_A \otimes q}{\longrightarrow} A \longrightarrow 0$$
is exact for any $C^*$-algebra. But I do not see why knowing that $q_*$ is an isomorphism implies the same thing for $(\mathrm{id}_A \otimes q)_*$. Any suggestions?
The conclusion drawn in the book of Wegge-Olsen is wrong (explained below), but can, however, easily be tweaked to a correct proof. What is shown is that $j\circ q$ is homotopic to the identity on $\mathcal T$ and hence the same is true after tensoring with $A$. It follows that $K_\ast(\mathcal T_0 \otimes A) = 0$ for every $C^\ast$-algebra $A$.
A quotient map $q$ of $C^\ast$-algebras is an isomorphism in $K$-theory if and only if $K_\ast(\ker q) =0$ (by six-term exactness). So the statement that a quotient map $q$ (with splitting $j$) induces an isomorphism in $K$-theory implies that $(q\otimes_\alpha id_A)_\ast$ is an isomorphism is equivalent to the statement that
$K_\ast(B)=0$ implies $K_\ast(B \otimes_\alpha A) = 0$ (for $\otimes_\alpha$ being either the maximal or minimal tensor product). And this is in general wrong. A $C^\ast$-algebra $A$ satisfies the maximal (respectively minimal) Künneth theorem if and only if $K_\ast(B \otimes_{\max} A) = 0$ (resp. $K_\ast(B \otimes_{\min{}} A) = 0$) for every $C^\ast$-algebra $B$ with $K_\ast(B) = 0$, see https://arxiv.org/abs/1111.7228 Theorems 3.1 and 4.1. This paper also contains examples of $C^\ast$-algebras that don't satisfy the maximal/minimal Künneth theorem. These examples are modelled after Skandalis' counterexamples to the Universal Coefficient Theorem.
Thanks, I thought so.
|
2025-03-21T14:48:30.929259
| 2020-05-09T22:46:29 |
359869
|
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|
Stack Exchange
|
Categories of modules generated under coproducts by a small set?
Question 1: For which rings $R$ does there exist a small set $S \subseteq Mod_R$ such that every module $M \in Mod_R$ is a direct sum of modules in $S$?
Equivalenty, for which rings $R$ does there exist a cardinal $\kappa$ such that every module $M \in Mod_R$ is a direct sum of modules generated by $\leq \kappa$-many elements?
Faith [1] says that "$Mod_R$ has a basis" if it answers to Question 1.
Background:
Kothe showed that if $R$ is an artinian principal ideal ring, then every $R$-module is a direct sum of cyclic modules (i.e. $R$ answers to my question with $\kappa = 1$).
The general question with $\kappa=1$ seems to be generally known as "Kothe's problem".
Cohen and Kaplansky showed that the converse holds if $R$ is commutative. Warfield extended this to show that if $R$ is commutative, then $R$ answers to my question if and only if $R$ is an artinian principal ideal ring.
So the question is only interesting when $R$ is noncommutative.
Nakayama constructed a (noncommutative) ring $R$ such that every module is a direct sum of cyclic modules, and yet $R$ is not an artinian principal ideal ring.
So Warfield's theorem apparently doesn't extend in the most straightforward way to the noncommutative setting.
Faith and Walker showed that there exists a cardinal $\kappa$ such that every injective right $R$-module is a direct sum of $\leq \kappa$-generated modules if and only if $R$ is right Noetherian. Faith ([1] cf. also Griffiths Thm 2.2) then showed that if $R$ answers to Question 1, then $R$ is right Artinian.
So any ring answering to Question 1 is right Artinian, though not necessarily a principal ideal ring by Nakayama's result.
I happen to be interested in a special case:
Question 2: Same as Question 1, but assuming that $R$ is hereditary.
[1] Faith, "Big Decompositions of Modules" AMS Notices 18, Feb 1971 p. 400
For some reasons a sort of dual question is also interesting: For which rings $R$ does there exist a small set of modules (equivalently a single module) $S$ such that every module is direct summand (=direct factor=retract) of a product of modules in $S$? (see https://mathoverflow.net/questions/68352/when-are-all-modules-direct-factors-of-a-direct-product-of-a-fixed-one )
The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book
Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.
or you might find it easier to access the older paper
Prest, Mike, Rings of finite representation type and modules of finite Morley rank, J. Algebra 88, 502-533 (1984). ZBL0538.16025.
As you say, such a ring must be right artinian.
It is known that a ring is both left and right pure semisimple if and only if it has finite representation type (i.e., every module is a direct sum of indecomposable modules, and there are finitely many isomorphism types of indecomposable module), which is a left/right symmetric condition. And there is a longstanding conjecture about a strengthening of this.
Pure Semisimplicity Conjecture: A right pure semisimple ring has finite representation type.
Or equivalently this says that pure semisimplicity should be a left/right symmetric condition. There are many positive results for particular classes of rings.
In Question $2$ you say that you are particularly interested in the hereditary case. This doesn't make the conjecture easier, as Herzog proved that if there is a counterexample then there is a hereditary counterexample. Combining this with a result of Simson, it turns out that to prove the conjecture it would be enough to prove that a right pure semisimple hereditary ring is left artinian.
There is a lot of work on rings of finite representation type, especially hereditary ones. The fundamental result is Gabriel's theorem classifying the finite dimensional hereditary algebras over an algebraically closed field with finite representation type as those Morita equivalent to path algebras of quivers whose underlying graph is a disjoint union of simply laced Dynkin diagrams. There are many generalizations; one for general hereditary artinian rings is
Dowbor, P.; Ringel, Claus Michael; Simson, D., Hereditary Artinian rings of finite representation type, Representation theory II, Proc. 2nd int. Conf., Ottawa 1979, Lect. Notes Math. 832, 232-241 (1980). ZBL0455.16013.
Thanks so much -- I hadn't dared hope for such a complete answer! It's interesting that the pure semisimplicity conjecture would imply that if $Mod_R$ has a basis, then it has a finite basis of indecomposable modules. This would be surprising, but definitely in the spirit of Warfield's theorem, which says that if $R$ is commutative and $Mod_R$ has a basis, then it has a basis of cyclic modules.
|
2025-03-21T14:48:30.929551
| 2020-05-09T23:12:32 |
359872
|
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|
Stack Exchange
|
Bimodule isomorphism for representation-finite blocks of the Schur algebra
Let $A$ be a representation-finite block of a schur algebra with $n \geq 2$ simple modules. Then the global and dominant dimension of $A$ are equal to $g=2n-2$. You can find quiver and relations for example in theorem 4.1. of https://www.sciencedirect.com/science/article/pii/S0022404918300380 when all lambdas are equal to one. For $n=2$ this is just the Nakayama algebra with Kupisch series [2,3].
Computer experiments suggest that we have $A \cong \Omega^g(D(A))$ as $A$-bimodules (while the projective dimension of $D(A)$ should be equal to $2g$).
Question 1: Is there an easy argument for this, avoiding heavy computation?
Question 2: Is there a deeper reason for such an isomorphism in case it is true, does it hold for a more general class of algebras?
My guess is that for certain algebras we have that $\Omega^i(A) \cong \Omega^j(D(A))$ for some i,j which would have interesting consequences for Hochschild homology and cohomology. But I was not able to think of any concrete conditions despite finding some examples.
Another class of examples seem to be the Auslander algebras of $K[x]/(x^n)$, where $A \cong \Omega^2(D(A))$.
|
2025-03-21T14:48:30.929669
| 2020-05-09T23:15:41 |
359873
|
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"Amir Sagiv",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359873"
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|
Stack Exchange
|
Average number of elements of a subset S of a matrix A after inducing the rows and columns of m randomly selected elements from subset S
Let $A_{N{\times}N}$ be an $N{\times}N$ matrix and $\mathcal{S_{k}}$ be a subset of elements in $A$ such that exactly $k$ elements from every row and column in $A$ are in $\mathcal{S_{k}}$. Thus, $\mathcal{S_k}$ has cardinality $N{\cdot}k$, with $k \in \{1,2,..,N\}$.
\begin{equation*}
A_{N,N} =
\begin{pmatrix}
a_{1,1} & a_{1,2} & \cdots & a_{1,N} \\
a_{2,1} & a_{2,2} & \cdots & a_{2,N} \\
\vdots & \vdots & \ddots & \vdots \\
a_{N,1} & a_{N,2} & \cdots & a_{N,N}
\end{pmatrix}
\end{equation*}
For instance, consider $A_{8{\times}8}$ as described below.
Let subset $S_{2}$ of the matrix $A_{8{\times}8}$ be given by the elements in bold in $A_{8{\times}8}$. Note that $S$ can be any subset having as its elements exactly $k=2$ elements per column and row of $A_{8{\times}8}$.
\begin{equation*}
A_{8,8} =
\begin{pmatrix}
\mathbf{a_{1,1}} & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & a_{1,6} & a_{1,7} & \mathbf{a_{1,8}} \\
\mathbf{a_{2,1}} & \mathbf{a_{2,2}} & a_{2,3} & a_{2,4} & a_{2,5} & a_{2,6} & a_{2,7} & a_{2,8} \\
a_{3,1} & \mathbf{a_{3,2}} & \mathbf{a_{3,3}} & a_{3,4} & a_{3,5} & a_{3,6} & a_{3,7} & a_{3,8} \\
a_{4,1} & a_{4,2} & \mathbf{a_{4,3}} & \mathbf{a_{4,4}} & a_{4,5} & a_{4,6} & a_{4,7} & a_{4,8} \\
a_{5,1} & a_{5,2} & a_{5,3} & \mathbf{a_{5,4}} & \mathbf{a_{5,5}} & a_{5,6} & a_{5,7} & a_{5,8} \\
a_{6,1} & a_{6,2} & a_{6,3} & a_{6,4} & \mathbf{a_{6,5}} & \mathbf{a_{6,6}} & a_{6,7} & a_{6,8} \\
a_{7,1} & a_{7,2} & a_{7,3} & a_{7,4} & a_{7,5} & \mathbf{a_{7,6}} & \mathbf{a_{7,7}} & a_{7,8} \\
a_{8,1} & a_{8,2} & a_{8,3} & a_{8,4} & a_{8,5} & a_{8,6} & \mathbf{a_{8,7}} & \mathbf{a_{8,8}}
\end{pmatrix}
\end{equation*}
Now, select randomly $m$ elements from $\mathcal{S_{k}}$ with replacement. Then, we create an induced matrix with just the rows and columns of $A_{N{\times}N}$ corresponding to the selected elements from $\mathcal{S_{k}}$.
For instance, if after selecting $m$ elements with replacement, the uniquely selected elements from $S_{2}$ are the following 5 elements: $a_{1,1},a_{3,2},a_{3,3},a_{6,6},a_{8,8}$, then the resultant matrix is
\begin{equation*}
A_{r} =
\begin{pmatrix}
\mathbf{a_{1,1}} & a_{1,2} & a_{1,3} & a_{1,6} & \mathbf{a_{1,8}} \\
a_{3,1} & \mathbf{a_{3,2}} & \mathbf{a_{3,3}} & a_{3,6} & a_{3,8} \\
a_{6,1} & a_{6,2} & a_{6,3} & \mathbf{a_{6,6}} & a_{6,8} \\
a_{8,1} & a_{8,2} & a_{8,3} & a_{8,6} & \mathbf{a_{8,8}}
\end{pmatrix}
\end{equation*}
Let $X$ be the number of elements in $\mathcal{S_{k}}$ of the resultant matrix ($A_{r}$). For the given example, $x=6$.
My Question: How we can calculate the average of $X$ for given $m,N,k$ ($E[X](N,k,m))$?
I have already calculated the average number of columns $E(C)$ when $m$ elements are randomly selected with replacement from $\mathcal{S_{k}}$. Note that for this case $E(C)$ is equal to the average number of rows $E(R)$ and can be calculated as:
$E[C](N,m)=N*P_{chosen}$, where $P_{chosen}=1-(1-(1/N))^m)$ is the probability that a column of the original matrix is selected at least once. Thus, $A_{r}$ is a $E[C] \times E[R]$ matrix.
Thank you for any help!
An alternative phrasing:
Write $I_N = \{1, 2, \cdots, N\}$. Let $S \in I_N \times I_N$ be a subset such that $|S \cap (\{i\} \times I_N)| = |S \cap (I_N \times \{i\})| = 2$. Note that $|S| = 2N$.
Then for given $0 \leq m \leq 2k$, what is the distribution of $|p_1(M)| |p_2(M_m)|$, where $M_m$ ranges uniformly over the set of $m$-element subsets of $S$, and $p_1, p_2$ are the projection functions?
I don't quite understand this question—for example, is it a question about a particular matrix $A$, or about any matrix?—but this seems to be an elementary probability question, not a research-level question, and so does not belong on MO. If you want to clarify it for MO, or ask it again on MSE, I encourage you to TeX it so that it will be easier to read.
Welcome to MO. It's good to have an example, but eventually the question is really unclear. It might be that the question would eventually fit MO, but I currently can't quite follow. Use crisper definitions and state a more precise question.
Thank you @LSpice for your comments and suggestions. I rephrased the problem, detailed it better, and used LaTex. I don't know if it fits better here or on MSE.
Thank you @AmirSagiv. I tried to explain better the problem. Please help me to define if it fits better here or on MSE.
It seems to me that it would be clearer to phrase this without the matrix.
I've added an alternative phrasing that I think captures the problem.
@user44191, thank you for rephrasing the question. Is the problem formulated in that way a known problem?
Suppose that we selected $m$ random elements of $S_k$. An element $s$ of $S_k$ appears in the induced matrix iff (i) there is a selected element in the row of $s$ in $A$; and (ii) there is a selected element in the column of $s$ in $A$. Call such an element $s$ lucky, and so $X$ is the number of lucky elements.
Under selection without replacement, the probability $P$ of a fixed element $s\in S_k$ to be lucky equals
$$P = 1 - \frac{2\binom{Nk-k}{m} - \binom{Nk-(2k-1)}{m}}{\binom{Nk}{m}},$$
where $\binom{Nk-k}{m}/\binom{Nk}{m}$ is the probability that nothing is selected from the row of $s$ in $A$, and $\binom{Nk-(2k-1)}{m}/\binom{Nk}{m}$ is the probability that nothing is selected from neither the row nor the column of $s$ in $A$.
Similarly, under selection with replacement, we have
$$P = 1 - \frac{2(Nk-k)^m - (Nk-(2k-1))^m}{(Nk)^m}.$$
Then
$$E[X](m,N,k) = Nk\cdot P.$$
Thank you @Max for your answer. It seems that in your solution $m$ is the number of unique selected elements, is it right?
@CarlosA.AstudilloTrujillo: Yes. Selection with replacement is similar - added it now
Another option to answer my initial question is just to consider $m$ as $m^{′}$ in your formula without replacement and calculate $m{′}=N{\cdot}k\left(1−B_{0}(n,p)\right)$, where $B_{r}(n,p)$ is the binomial distribution with parameters $n=m$ (as defined in the initial question), $p=\frac{1}{N{\cdot}k}$ and $r$ is the number of times an element in $S_k$ is selected. Thus, $1−B_{0}(n,p)$ is the probability that an element of $S_{k}$ is selected at least once in $n$ trials.
I tested your solution of selection with replacement and it is more accurate than my previous suggestion of considering $m^{'}$ and selection without replacement. Thank you again.
|
2025-03-21T14:48:30.930023
| 2020-05-09T23:18:21 |
359874
|
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|
Stack Exchange
|
Is this $1$-form harmonic?
Let $(M^3,g)$ be a compact, connected and oriented Riemannian $3$-manifold with boundary. For a harmonic map $u : M \to \mathbb{S}^1$ satisfying Neumann condition along $\partial M$, let $h = u^*(d \theta)$, so that $d h = 0$, $\delta h = 0$ and $h(\nu) = 0$, where $\nu$ is the unit exterior normal to $\partial M$. Let $i : \partial M \to M$ be the inclusion.
Denote by $J$ the complex structure of $\partial M$. Let $J(i^\ast h)$ denote the $1$-form dual to the vector field $JX$, where $X$ is the vector field dual in $\partial M$ dual to $i^\ast h$.
My question is: does it hold that $J(i^\ast h)$ is a closed $1$-form in $\partial M$? Is it harmonic?
Why is it relevant? If $S$ denotes a connected component of a regular level set of $u$, then the vector field $JX$ as above is tangent to $\partial S$ and nonzero along this set. So,
$$ \int_{\partial S} J(i^\ast h) \neq 0,$$
thus $\partial S$ represents a nontrivial class in $H_1(\partial M)$ provided that $J(i^\ast h)$ is a closed $1$-form.
I'm not sure but if I remember correctly, an isometric immersion pulls harmonic $1$-forms back to harmonic $1$-forms iff it is minimal. If that is the case for $\partial M$, and $J$ is compatible with $g$, then it seems that $J(i^*h)$ is harmonic, since $i^*h$ is harmonic and $J$ preserves the space of harmonic forms. (Also hello Eduardo, hope everything is fine at IME)
|
2025-03-21T14:48:30.930176
| 2020-05-10T03:36:54 |
359882
|
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Stack Exchange
|
Resources for divergent / asymptotic series
This series is divergent; therefore, we may be able to do something with it. -- Oliver Heaviside
[Edit (1/14/21) from the answer by Count Iblis to a recent MO-Q on math vids: An enthusiastic intro is that to the set of lectures by Carl Bender "Perturbation and Asymptotic Series." ]
Other than the usual references given in Wikipedia and Mathworld, which resources have you found helpful as intros to the topic and for advanced exploration?
I'll prime the pump with
"Divergent series:taming the tails" by M. V. Berry and C. J. Howls (cf. also refs in this MO-Q)
Sporadic examples in Heaviside's publications, see Heaviside's Operational Calculus, a post by Ron Doerfler.
A Singular Mathematical Promenade by Etienne Guys
Sum Divergent Series by the user mnoonan, a series of posts at The Everything Seminar
"Euler's constant:
Euler's work and modern developments" by Jeffrey Lagarias
"Uniform asymptotic methods for integrals" by Nico Temme
"On the Specialness of Special Functions (The Nonrandom Effusions of the Divine Mathematician)" by
Robert W. Batterman
For one example of the importance of such series, see the relation between the Harer-Zagier formula and the asymptotic expansion of the digamma function in Chapter 5 "The Euler characteristic of the moduli space of curves" of the course notes "Mathematical ideas and notions of quantum field theory" by Etingof.
Hardy's book Divergent Series? Or is that one of the usual references?
@Robert Furber, yep, under asymptotic series in Wiki. (Divergent Series has been google hijacked by Hollywood). Years ago when I had an excellent home library, I saved some early paper by Hardy in which he expressed what I have called The Hardy Heuristic. Goes something like: Apply two operations consecutively in one order then reverse the order. If one order is convergent and the other divergent, you have a summation method. If you can find that ref, would be a good one. Lost my library and have no access to a good University lib myself.
Écalle, Malgrange and Ramis work on Gevrey series may be also a good track.
I would add the tauberian theory bible of J Korevaar; there is also a survey of complex Tauberian theory by the author here (pdf linked) http://www.jointmathematicsmeetings.org/bull/2002-39-04/S0273-0979-02-00951-5/S0273-0979-02-00951-5.pdf
@Conrad, that's fitting. I found the Temme ref (or was reminded again of him) just last night in "Early work of N.G. (Dick) de Bruijn in analysis and
some of my own" by Korevaar.
@LoïcTeyssier, I found "Fonctions multisommables" by Malgrange and Ramis (in which Watson's "A theory of asymptotic series" is reffed). Which papers by E, M, and R would be a good intro to their work?
As was expressed the tauberian theorems are very important. On the side of divergent series I know from an informative point of view Kolmogorov studied divergent series (see History from the Wikipedia Carleson's theorem). Myself attempt of learning is to search and study concise articles that I can understand about divergent series, for example R. P. Agnew, A Slowly Divergent Series, The American Mathematical Monthly, Vol. 54, No. 5 (May, 1947) pp. 273-274, or T. S. Nanjundiah, Extensions of Olivier's Theorem, The American Mathematical Monthly, Vol. 76, No. 6 (Jul, 1969) pp. 666-667.
I can read these for free with my account of JSTOR.
Worth panning, but no time for me at the moment: "How Euler Did It" , an online MAA column, written by Ed Sandifer from 2003 to 2010. These are archived at http://eulerarchive.maa.org/hedi/.
Again, back to the source--the mathemage, Euler--in "Euler and his work on infinite series" by Varadarajan (https://www.ams.org/journals/bull/2007-44-04/S0273-0979-07-01175-5/S0273-0979-07-01175-5.pdf). See also https://mathoverflow.net/questions/19201/summation-methods-for-divergent-series and https://mathoverflow.net/questions/45811/use-of-everywhere-divergent-generating-functions/45868#45868.
And "Euler's 1760 paper on divergent series" by Barbeau and Leah https://www.sciencedirect.com/science/article/pii/0315086076900306
"Convergence from Divergence" by Costin and Dunne https://arxiv.org/abs/1705.09687
Also a translation by Aycock of Euler's "On divergent series" https://arxiv.org/abs/1808.02841
"Transseries for beginners" by G. A. Edgar https://arxiv.org/abs/0801.4877
On the iconic divergent power series with $a_n = n!$, see refs in https://oeis.org/A003319 and the blog posts https://qchu.wordpress.com/2015/11/03/the-answer-to-the-puzzle/, https://qchu.wordpress.com/2015/11/04/the-categorical-exponential-formula/, and https://tcjpn.wordpress.com/2014/12/14/the-hirzebruch-criterion-fo-the-todd-class/.
Related: https://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1 and discussion on formal inversion (multiplicative and compositional) of and composition with divergent series.
See also the book Continued Fractions by Jones and Thron.
"On Numbers, Germs, and Transseries" by Aschenbrenner, van den Dries, van der Hoeven https://arxiv.org/abs/1711.06936 and "Asymptotics and Borel Summability" by Costin
"On the Specialness of Special Functions (The Nonrandom Effusions of the Divine Mathematician)" by Batterman and the refs therein.
"Differential Equations: A Dynamical Systems Approach" by Hubbard and West. See the Appendix: Asymptotic Development.
Another book by Balser not mentioned in Wikipedia: "Formal Power Series and Linear Systems of Meromorphic Differential Equations." Wikipedia and MathWorld links are https://en.wikipedia.org/wiki/Asymptotic_analysis, https://en.wikipedia.org/wiki/Divergent_series, https://en.wikipedia.org/wiki/Category:Summability_methods, https://en.wikipedia.org/wiki/Category:Asymptotic_analysis, https://mathworld.wolfram.com/AsymptoticSeries.html, https://mathworld.wolfram.com/DivergentSeries.html
Fairly comprehensive intro: Quantum Field Theory II: Quantum Electrodynamics by Zeidler
"Polynomial expansions of Analytic Functions" by Boas and Buck
"From Resurgence to BPS States" by Marino https://member.ipmu.jp/yuji.tachikawa/stringsmirrors/2019/2_M_Marino.pdf
Despite the exhortations against divergent series / formal power series by Abel, since at least Laplace, such series have found important applications in physics. See, e.g., "Divergent series: past, present, future . . . " by Christiane Rousseau (https://arxiv.org/pdf/1312.5712.pdf).
See also applications in Operational Calculus (2nd ed.) by Bremmer and van der Pol.
Informal intro with some history: the 2023 video by Sir Michael Berry - Divergent Series: From Thomas Bayes's Bewilderment to Today's Resurgence (https://youtu.be/73CLvdrdKuU?si=FVt1Lq1-dCl1Fo9E)
'Summability, Tauberian theorems, and Fourier series' a blog post at https://www.victorchen.org/2019/10/21/summability-tauberian-fourier-series/
Recent article using Borel summation: "Semiclassical expansion for exactly solvable differential operators" by Borrego-Morell and Shapiro https://arxiv.org/pdf/2402.19087
"The Development of the Theory of Summable Divergent Series from 1880 to 1925" by
John Tucciarone
Some more history in "The First Modern Definition of the Sum of a Divergent Series: An Aspect of the Rise of 20th Century Mathematics" by Giovanni Ferraro
As far as on-line-available things go, I've attempted to modernize some arguments and give examples of asymptotics of integrals (both Watson's Lemma and easy Laplace/saddle-point examples), as well as asymptotics for ordinary differential equations, both regular and certain irregular singular points. On-line, as well as a chapter in my Cambridge Univ Press book of 2018 (http://www.math.umn.edu/~garrett/m/v/current_version.pdf). For earlier, separate treatments, see http://www.math.umn.edu/~garrett/m/mfms/notes_2019-20/05e_asymptotics_of_integrals.pdf, http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/11b_reg_sing_pt.pdf, and http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/11c_irreg_sing_pt.pdf. Those notes (and the book, on-line or not) have substantial bibliographic/historical references.
Nice lists of references. I'm wondering if you could contribute to the somewhat related MO-Q https://mathoverflow.net/questions/397238/renormalization-in-physics-vs-dynamical-systems.
@TomCopeland, I'll take a look... :)
@TomCopeland, although rigorous notions of "asymptotic expansion" might put that other question and its answers into a broader context, it doesn't seem that that's the main issue in that question.
|
2025-03-21T14:48:30.930848
| 2020-05-10T04:33:36 |
359885
|
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|
Stack Exchange
|
Splitting formulas for spectral flows
I'm asking if there are splitting formulas for equivariant spectral flow and higher spectral flow (of Dai-Zhang) for paths of Dirac operators, concerning gluing together two smooth compact Spinc manifolds along their common boundary.
Thank you
|
2025-03-21T14:48:30.930915
| 2020-05-10T04:36:39 |
359886
|
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|
Stack Exchange
|
For a given value of $n$ and $m$, find $\text{fib}(n)$ $\text{mod } m$ where $n$ is very huge. (Pisano Period)
Input
Integers $'n'$ (up to $10^{14}$) and $'m'$(up to $10^3$)
Output
$\text{Fib}(n)$ $\text{modulo}$ $m$
My questions
For example : Why $\text{fib}(n=2015)$ $\text{mod}$ $3$ is equivalent to $\text{fib}(7)$ $\text{mod } 3$? (for $ = 3$ the period is $01120221$ and has length $8$ and $2015=251*8 + 7$)
In general, after getting the remainder sequence, how (mathematical proof) it is used for computing $\text{Fib}(n)$ $\text{mod } m$?
The wikipedia article is quite enlightening on the Pisano period, but from the algorithmic viewpoint, it shows that you only need to compute the period for $n$ a prime power $p^k,$, and in that case it divides either $p^{k-1}(p-1)$ or $p^{k-1}2(p+1).$ For your range of $n$ brute force will tell you the period quickly, by computing the multiplicative order of $\begin{pmatrix}0&1\\1&1\end{pmatrix}$ modulo $n.$
Thanks! @Igor, I had figured how to compute "period" but I have difficulties on how to use period length to get fib(n) mod m. Also, I am not able to understand the relation between fib(n) and n (as mentioned in the question I don't understand why fib(2015) mod 3 = fib(7) mod 3 ). More code-oriented question is on StackOverflow(https://stackoverflow.com/questions/61706899/for-a-given-value-of-n-and-m-find-fibn-mod-m-where-n-is-very-huge-pisano-pe)
Wikiwand looks attractive but is it secure? When I attempted to add wikiwand to Chrome then I got a confusing message about wikiwand changing my data, etc. Does it do so under my control or without it?
Don’t use wikiwand. Not only is it unethical to drive traffic to a website that (albeit legally) just republishes content created by others with a dubious facelift instead of to the originating website, but more importantly, it defeats one of the major purposes of Wikipedia, “the free encyclopedia that anyone can edit”: no one cannot edit the copies at wikiwand.
@EmilJeřábek You are wrong. If you hover on "Wikipedia Tools" in the top right, one of these tools is "Edit". Clicking on it sends you to the standard Wikipedia editor. As for "unethical", I would like you to apologize - Wikiwand clearly makes no money, and neither do I.
All right, so they do provide links to edit pages on Wikipedia. However, my first point stands. Wikiwand is a private for-profit company that (apart from angel investors) makes money from advertising on content they copy wholesale from Wikipedia (with which they are unaffiliated). I have nothing to appologize for.
Note that I didn’t call you unethical. I wrote that driving traffic to Wikiwand instead of Wikipedia is unethical. In fact, I assumed you linked to it by mistake. But if you’d intentionally deceive unsuspecting readers by posting links to Wikiwand labelled as “Wikipedia”, then yes, I’d call that unethical. Money has nothing to do with this.
From an algorithmic viewpoint, you can compute $F_n\bmod m$ efficiently in time $\tilde O\bigl((\log n)(\log m))$ [or $O\bigl((\log n)(\log m)^2)$ when employing a naive schoolbook multiplication algorithm] by computing
$$\begin{pmatrix}1&1\\1&0\end{pmatrix}^n\begin{pmatrix}1\\0\end{pmatrix}\bmod m$$
where the matrix power is evaluated by repeated squaring modulo $m$. Stated in a different way, this amounts to using the recurrences
$$\begin{align*}
F_{2n-1}&=F_n^2 + F_{n-1}^2,\\
F_{2n}&=(2F_{n-1}+F_n)F_n
\end{align*}$$
modulo $m$.
In contrast, I don’t think there is any known method to compute the Pisano period faster than factorizing $m$ (which takes exponential time $O\bigl(2^{(\log m)^\alpha}\bigr)$ for some $\alpha>0$).
Thanks, @Emil, I have computed period and length of period, as question says, how can I find fib(n) mod m using that period? and why fib(2015) mod 3 = fib(7) mod 3(for m=3 the period is 01120221 and has length 8 and 2015=251∗8+7)?
I can read myself, thank you. I am telling you it is more efficient to not compute the period.
your answer doesn't explain why fib(2015) mod 3 = fib(7) mod 3(for m=3 the period is 01120221 and has length 8 and 2015=251∗8+7)
|
2025-03-21T14:48:30.931197
| 2020-05-10T05:05:28 |
359887
|
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|
Stack Exchange
|
A question in elementary differential geometry
Let $M$ be a finite dimensional manifold of constant curvature $\kappa$. Consider a solution of the Hamilton--Jacobi equation
$$ \partial_t u + |\nabla u|^2 = 0. $$
Can we give a precise estimate of a modulus of semi-concavity, that is an estimate of the form
$$ D^2 u_t \leq \lambda_t \mbox{id} ? $$
In case $\kappa=0$, we have
$$ \lambda_t \leq \frac{C}{t}, $$
an estimate which does not depend on the initial function $u_0$.
We will be however interested in the case where $\kappa<0$.
Will the following formula be correct? It involves a sectional curvature. $$ D^2 (\frac{1}{2}|\nabla u|^2) [\xi,\xi] = (\nabla u \cdot \nabla D^2u)[\xi,\xi] + |D^2u \xi|^2 + R(\nabla u,\xi, \nabla u, \xi). $$
|
2025-03-21T14:48:30.931270
| 2020-05-10T05:16:44 |
359888
|
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|
Stack Exchange
|
Motive to motive via Langlands
I have been tying myself in knots trying to straighten out this speculative circle of connections, no doubt because I am just a novice in these matters, so I thought I'd pop the question here and learn if it might not be arrant nonsense:
Take $E$, a pure motive over $\mathbb{Q}$. Let $G_M(E)$ be its motivic Galois group.
$G_M(E)$ is a reductive group. Assume it is connected.
Conjecturally to each $L$-algebraic automorphic representation of $G_M(E)$ is attached a $p$-adic Galois representation, $\varphi$, into the $L$-group of $G_M(E)$.
Suppose $\varphi$ is geometric. Then, $\varphi$ is a realization of a motive $E_\varphi$.
How are the motives $E$ and $E_\varphi$ related - directly, geometrically? Or indeed, how is the entire collection of $E_\varphi$'s, one for each geometric Galois representation attached to an algebraic automorphic representation of the motivic Galois group of $E$, related to $E$?
Are you sure this is right? I think the correct data for $\phi$ is a representation of $G_M(E)$ in the sense of algebraic groups, not an automorphic rep -- so e.g. $G_M(E)$ could be something like $GL_2$ and $\phi$ could be something like $Sym^2$.
|
2025-03-21T14:48:30.931383
| 2020-05-10T06:43:47 |
359892
|
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|
Stack Exchange
|
Is the square root of a monotonic function whose all derivatives vanish smooth?
Let $g:[0,\infty] \to [0,\infty]$ be a smooth strictly increasing function satisfying $g(0)=0$ and $g^{(k)}(0)=0$ for every natural $k$.
Is $\sqrt g$ is infinitely (right) differentiable at $x=0$?
I know that $\sqrt g \in C^1$ at zero*, and that in complete generality, one cannot expect for $\sqrt g$ to be even $C^2$. However, in the counter-example given in the linked question, $g$ was not monotonic.
Does this additional assumption of (strict) monotonicity save us? I tried to look at the literature, but did not find a treatment of this particular case.
*The proof that $\sqrt g \in C^1$ goes by rewriting $g(x)=x^2h(x)$ where $h \ge 0$ is smooth (this is possible since $g(0)=g'(0)=0$).
Edit:
As pointed out by Igor Rivin, it seems that theorem 2.2 (on page 639) here (pdf) does the job. It states that
any square root of $f$ "precised up to order $m$" is of class $C^m$. (The definition of a "square root precised up to order $m$" is Definition 1.1 on page 636).
This certainly settles the issue.
However, I think it would still be nice to find a simpler approach, since here we assume much more-the strict monotonicity is a much stronger assumption than those assumed in the paper.
Comment:
If we assume that $g''>0$ in a neighbourhood of zero (which implies that $g'>0$), then $\sqrt g \in C^2$. (details below).
I think that there is a chance for smoothness under the additional assumption that $g^{(k)}>0$ in a neighbourhood of zero for every $k$, but I am not sure. The calculations become quite messy even when trying to establish $\sqrt g \in C^3$.
A proof $\sqrt g \in C^2$ when $g',g''>0$ near zero: (We use these assumptions when applying L'Hôpital's rule).
$$\sqrt{g}'' = \frac{g''}{2\sqrt{g}} - \frac{(g')^2}{4g^{3/2}}.$$
Thus it is enough to prove that $(g'')^2/g\to 0$ and $(g')^4/g^3\to 0$.
$$
\lim_{x\to 0^+} \frac{(g'')^2}{g} = \lim_{x\to 0^+} 2\frac{g''g^{(3)}}{g'} = \lim_{x\to 0^+} 2\frac{g''g^{(4)}+(g^{(3)})^2}{g''} = 0,
$$
where in the last equality we applied $\frac{(h')^2}{h}\to 0$ above for $h=g''$.
$$
\lim_{x\to 0^+} \frac{(g')^4}{g^3} = \lim_{x\to 0^+} \frac{4(g')^2g''}{3g^2} = \lim_{x\to 0^+} \frac{8(g'')^2 + 4g' g^{(3)}}{6g} = \lim_{x\to 0^+} \frac{2g' g^{(3)}}{3g} = \lim_{x\to 0^+} \left(\frac{2g^{(4)}}{3} + \frac{2g''g^{(3)}}{3g'}\right)=\lim_{x\to 0^+} \frac{2g''g^{(3)}}{3g'} = \lim_{x\to 0^+} \frac{2g^{(4)}}{3}+\frac{2(g^{(3)})^2}{3g''} = 0,$$
where in the first row we used the first calculation, and in the second we again applied $\frac{(h')^2}{h}\to 0$ to $h=g''$.
Just a comment: if all derivatives of $g$ are nonegative in a common neighborhood of zero, then $g$ can be expanded in a power series, by Bernstein theorem, and would be equal to 0. However, the question changes if the neighborhhod is allowed to depend on the derivative.
The answer is yes, by the results of
Bony, Jean-Michel; Colombini, Ferruccio; Pernazza, Ludovico, On square roots of class (C^m) of nonnegative functions of one variable, Ann. Sc. Norm. Super. Pisa, Cl. Sci. (5) 9, No. 3, 635-644 (2010). ZBL1207.26004.
Here is the math review:
Clearly the condition is fulfilled in the OP (for any $m$).
The DOI link doesn't work for me; the journal however released the back issues on numdam for free, so here's a direct link http://www.numdam.org/article/ASNSP_2010_5_9_3_635_0.pdf
@WillieWong Thanks!
Thank you. I see that theorem 2.2 there does the job. It says "that
any square root of $f$ precised up to order $m$ is of class $C^m$". (The relevant definition of a "square root precised up to order $m$" is Definition 1.1 on page 636).
Link to Numdam abstract, rather than straight to PDF: Bony, Colombini, and Pernazza - On square roots of class $C^m$ of nonnegative functions of one variable. (The DOI link works for me; just to have it here.)
Extend the domain of the function $g$ to $\mathbb R$ by letting $g(x):=0$ for real $x<0$. The resulting function, which we shall still denote by $g$, is $C^\infty$ on $\mathbb R$.
Theorem 3.5 on page 144 implies that a nonnegative $C^4$ function $f$ on $\mathbb R$ has a $C^2$ square root if for any minimum $x_0$ of $f$ we have $f(x_0)=0$.
This latter condition obviously holds for our function $g$ in place of $f$ -- because $g$ is strictly increasing on $[0,\infty)$ and thus has no minima in $(0,\infty)$, and $g=0$ on $(-\infty,0]$.
Therefore, we can conclude that $\sqrt g$ is $C^2$ on $\mathbb R$, even without assuming that $g''>0$ in a neighborhood of zero.
Yet, this conclusion falls short of your main goal, to show that $\sqrt g$ is $C^\infty$. Looking at the proof of the mentioned Theorem 3.5, this task may be too big for a usual MO answer and may require a full-blown paper.
"... may require a full-blown paper." Evidently Bony and collaborators thought the same :-)
Thank you. Actually I see now that theorem 3.1 on page 142 states exactly your corollary.
|
2025-03-21T14:48:30.931716
| 2020-05-10T06:56:18 |
359894
|
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|
Stack Exchange
|
A recommendation for a book on perverse sheaves
I would like to learn about perverse sheaves.
I will be grateful if someone could recommend me a book with the following structure.
Introduction to basic homotopy theory (derived category and t-structure)
Introduction to sheaves
Introduction to perverse sheaves
The book by Dimca, Sheaves in Topology (Universitext, Springer), does that — and many other things.
Although not a book, Mark Goresky has some notes on his webpage which does all of 1,2 and 3 in the question
https://www.math.ias.edu/~goresky/math2710/index.html.
Pramod Achar is working on a book on perverse sheaves and applications in representation theory. It's a great book!
EDIT (2021): The book has now been published by the AMS:
Perverse Sheaves and Applications to Representation Theory.
Is it available anywhere?
It seems that one can obtain a copy of this book by emailing the organizers of this seminar: http://www.math.toronto.edu/jkamnitz/seminar/perverse/perverse.html
That’s wonderful news!!
I suggest instead emailing Pramod to ask him for copy. When I wrote "Please contact the organizers if you want the link to the book", it was intended just for the participants in the seminar.
With due apologies to everyone else, Pramod is the one person I would trust to write a comprehensible-to-me book on this subject.
It's a great book! One thing that you should know is that the book is written with coefficients in arbitrary fields, which makes things a bit more complicated, especially if you are learning the subject for the first time.
I'm currently taking a course on perverse sheaves and we are using Kashiwara & Schapira's Sheaves on Manifolds (published by Springer). It has all the things you mention and I've found it very readable!
Here are two possibilities:
Topological Invariants of Stratified Spaces by Markus Banagl
Intersection Homology & Perverse Sheaves: with Applications to Singularities by Laurenţiu G. Maxim
|
2025-03-21T14:48:30.931897
| 2020-05-10T07:03:07 |
359895
|
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Stack Exchange
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High sum of fractional parts
Let $n\geq 2$ and $x_1,\ldots,x_n > 0$ be such that $x_1+\cdots+x_n =1$. Is it true that there must exist a positive integer $k$ such that $$\{x_1k\}+\cdots+\{x_nk\} = n-1?$$
This looks closely related to the density of the fractional part.
Note that the quantity $\{x_1k\}+\cdots+\{x_nk\}$ is always an integer, since it equals $k-\lfloor x_1k\rfloor - \dots - \lfloor x_nk\rfloor$. Also, as each term is strictly less than one, $n-1$ is the highest value the sum can take.
If the $x_i$ are rational, take the lcm of the denominators and decrease it by $1$.
If they are not necessarily rational, act similarly: using, e.g., Kronecker's theorem, take a $k$ such that all the $kx_i$ are sufficiently close to integers, and decrease that $k$ by $1$.
I've put a reference, although it is an overkill: one may merely use the pigeonhole principle.
... another way of saying it is that the curve $t\mapsto (tx_1,\ldots ,tx_n)$ is dense in the torus $\mathbb{R}^n/\mathbb{Z}^n$, thus becomes as close as you want to $0$ for some $t>0$.
@abx: this is incorrect as soon as $1,x_1,x_2,\dots,x_n$ are linearly dependent over $\mathbb Q$. That's why Kronecker's theorem at my reference looks a bit... involved. In our case, they are dependent, as the $x_i$ sum up to $1$!
Oh, right of course. Still the curve is dense in some nontrivial subtorus (if at least one of the $x_i$ is irrational). But that makes the proof a bit complicated.
@IlyaBogdanov what does "as soon as" mean?
You may replace it with "whenever". But, in fact, in that concrete case it is "iff".
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2025-03-21T14:48:30.932040
| 2020-05-10T07:18:52 |
359898
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|
Stack Exchange
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$\frac{\partial}{\partial x}\int_{\mathbb{R}}\frac{1}{\sqrt{2 \pi \varepsilon}}e^{-\frac{(x-y)^2}{2\varepsilon}}l(y)dy\leq C\frac{1}{x}$
Let $l$ be a continuous bounded function ($l$ is not differentiable). I want to prove for $x$ large enough that
$$\frac{\partial}{\partial x}\int_{\mathbb{R}}\frac{1}{\sqrt{2 \pi \varepsilon}}e^{-\frac{(x-y)^2}{2\varepsilon}}l(y)dy\leq C\frac{1}{x}. $$
Where $C$ is a positive constant, and uniformly in $\varepsilon$.
Doesn't hold. For $l(x)=\cos x$ the integral is equal to $e^{-\varepsilon/2} \cos (x)$.
Maybe $C/\sqrt{\epsilon}$.
No I don't want $C/\sqrt{\epsilon}$.
This cannot hold true, at least not without further assumptions on $l$ (more precisely, on the decay of $l'(x)$ at infinity).
Indeed, let $l_\epsilon:=\Gamma_\epsilon \ast l$ be the $\epsilon$-mollification as in your statement (here $\Gamma_\epsilon$ is the heat kernel at time $\epsilon$). You are asking whether you can contol $|l_\epsilon'(x)|\leq \frac{C}{x}$ for large $x$, uniformly in $\epsilon$.
But it is well-known that, if for example $l$ is smooth enough (say $C^1$) then $l_{\epsilon}'=\Gamma_\epsilon\ast (l')$, which converges at least pointwise to $l'$ as $\epsilon\to 0$.
So, roughly speaking, if $l'$ does not decay at infinity at least as $\frac{1}{|x|}$ you cannot expect this estimate to be true.
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