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most recent 30 from 2017-11-21T15:06:57Z

Convergence of an iterated sequence

Thu, 11/02/2017 - 15:32

Let $K=[0,1]^2$ be a square and $p\in (0,1)$ be a fixed number. We define a map $F: K^2\to K^2$ as follows.

For $(x_1,y_1), (x_2,y_2)\in K$, it follows by a straightforward computation that there exists a (unique) $\omega\in\mathbb R$ s.t. $\mathcal Les\big(\{(x,y)\in K:~ L_w(x,y)~\le~ 0\}\big)=p$, where $\mathcal Les$ denotes the Lebesgue measure and $L_w$ is defined by


Let $(x_1',y_1')$ and $(x_2',y_2')$ be respectively the centroids (barycentres) of $\{(x,y)\in K:~ L_w(x,y)~\le~ 0\}$ and $\{(x,y)\in K:~ L_w(x,y)~\ge~ 0\}$. Then define the map by


Then of course $F$ has a fixed point. My question is the following: Take an arbitrary $\big((x_1^0,y_1^0),(x_2^0,y_2^0)\big)\in K^2$, and construct the sequence by iteration $\big((x_1^{n+1},y_1^{n+1}),(x_2^{n+1},y_2^{n+1})\big)=F\big((x_1^n,y_1^n),(x_2^n,y_2^n)\big)$. Is this sequence convergent?

Any reply, remark and comment are highly appreciated! Thanks a lot!

Decomposing the automorphism group of relation preserving transformations

Thu, 11/02/2017 - 15:21

Working with functions and relations in a set theoretic manner (i.e. as just sets of ordered pairs) one finds for any function $f$ and arbitrary relations $R$ and $L$ that conjugacy with respect to relation composition can be used to concisely state: $$\forall x,y\in \text{dom}(f)\left(xRy\implies f(x)Lf(y)\right)\iff f^{-1}\circ L\circ f\supseteq R\cap \text{dom}(f)^2$$ $$\forall x,y\in \text{dom}(f)\left(f(x)Lf(y)\implies xRy\right)\iff f^{-1}\circ L\circ f\subseteq R\cap \text{dom}(f)^2$$ $$\forall x,y\in \text{dom}(f)\left(xRy\iff f(x)Lf(y)\right)\iff f^{-1}\circ L\circ f=R\cap \text{dom}(f)^2$$

Moreover writing the image of any set $S$ under a relation $R$ as $R[S]=\{y:\left(\exists x\in S:xRy\right)\}$ with the transitive closure of $R$ denoted by $R^{*}$, one can identify $R$ geometrically with a unique labeled digraph on the vertex set $\text{fld}(R)=\text{dom}(R)\cup \text{rng}(R)$ whose arcs are the elements of $R$ thus translating digraph properties into equivalent relational properties one has:

$$H\text{ is connected}\iff (H\cup H^{-1})^{*}=\text{fld}(H)^2$$ $$H\text{ is strongly connected}\iff H^{*}=\text{fld}(H)^2$$

While the connected and strongly connected components of $R$ are given respectively by:

$$\mathcal{C}(R)=\left\{R\cap \left((R\cup R^{-1})^{*}[\{v\}]\right)^2:v\in \text{fld}((R\cup R^{-1})^{*})\right\}$$

$$\mathcal{S}(R)=\left\{R\cap \left((R^{*}\cap (R^{*})^{-1})[\{v\}]\right)^2:v\in \text{fld}(R^{*}\cap (R^{*})^{-1})\right\}$$

Which gives the equivalences:

$$R\text{ is acyclic}\iff \mathcal{S}(R)=\emptyset\iff R^{-1}\cap R^{*}=\emptyset\iff R\text{ contains no cycles}\iff R^{*}\text{ is irreflexive}\iff R^{*}\text{ is a strict order}\iff R\text{ is a subset of some strict-order}$$

Thus if for any relations $R$ and $L$ we define a family of functional relations $\text{Iso}(R,L)$ as follows:

$$\left\{f\subseteq\text{fld}(R)\times \text{fld}(L):[f\circ f^{-1}=\text{Id}_{\text{fld}(L)}]\land [f^{-1}\circ f=\text{Id}_{\text{fld}(R)}]\land [f\circ R\circ f^{-1}=L]\right\}$$

One can see each $\phi\in \text{Iso}(R,L)$ corresponds with a relation perserving isomorphism from $R$ to $L$ so if we let $\text{Aut}(R)=\text{Iso}(R,R)$ denote the relation preserving automorphisms of $R$. While concurrently defining a binary operation $\sqcup$ between families of relations say $\mathcal{F}$ and $\mathcal{G}$ as follows:

$$\mathcal{F}\sqcup \mathcal{G}=\{R\cup L:(R,L)\in \mathcal{F}\times \mathcal{G}\}$$

We can then from here partition the connected components of any relation $R$ into their respective isomorphism classes to form a unique quotient set over $\mathcal{C}(R)$ induced by $\cong$ and written $\mathcal{C}(R)/\cong$ moreover if for every $E\in \mathcal{C}(R)/\cong$ we let $\Gamma_{E}$ be an arbitrary class representative of $E$ so that we have $[\Gamma_{E}]=E$ and then for each $L\in \mathcal{C}(R)$ we let $\psi_{L}$ be any element of $\text{Iso}(\Gamma_{[L]},L)$ we can then explicitly write out $\text{Aut}(R)$ in terms of a family of unions and "direct unions" of double sided cosets of automorphism groups or instead just up to isomorphism as I did at the end in terms of wreath products of permutation groups being naturally acted upon by the symmetric groups of each aforementioned equivalence class:

$$\text{Aut}(R)=\bigsqcup_{E\in \mathcal{C}(R)/\cong}\left(\bigcup_{\sigma\in \text{Sym}(E)}\left(\bigsqcup_{L\in E}\text{Iso}(L,\sigma(L))\right)\right)\\=\bigsqcup_{E\in \mathcal{C}(R)/\cong}\left(\bigcup_{\sigma\in \text{Sym}(E)}\left(\bigsqcup_{L\in E}\psi_{L}\circ \text{Aut}(\Gamma_{E})\circ \psi_{\sigma(L)}^{-1}\right)\right)\cong \prod_{E\in \mathcal{C}(R)/\cong}\text{Aut}(\Gamma_{E})\wr_{E}\text{Sym}(E)$$

Now letting $R^{\dagger}=R\cap [\text{fld}(R)\setminus \text{fld}(R^{-1}\cap R^{*})]^2$ so $R^{\dagger}$ is the unique maximal acyclic subset of $R$ sharing no vertices with any cycle. Then since $\text{Aut}(R)=\text{Aut}(R^{\dagger})\sqcup\text{Aut}(R\cap (R^{*})^{-1})$ we see partitioning out the strong components of $R$ into their respective isomorphism classes to form a unique quotient set over $\mathcal{S}(R)$ induced by $\cong$ and denoting it similarly as before by $\mathcal{S}(R)/\cong$ as well as defining for every $E\in \mathcal{S}(R)/\cong$ a relation $\xi_{E}$ as an arbitrary class representative of $E$ so that $[\xi_{E}]=E$ that if we again in the same manner denote for each $L\in \mathcal{S}(R)$ a relation $\phi_{L}$ to be an arbitrary element of $\text{Iso}(\xi_{[L]},L)$ then this proves $\text{Aut}(R)$ is a subgroup of a somewhat similar looking expression:

$$\text{Aut}(R)\subseteq \text{Aut}(R^{\dagger})\sqcup\bigsqcup_{E\in \mathcal{S}(R)/\cong}\left(\bigcup_{\sigma\in \text{Sym}(E)}\left(\bigsqcup_{L\in E}\text{Iso}(L,\sigma(L))\right)\right)\\=\text{Aut}(R^{\dagger})\sqcup\bigsqcup_{E\in \mathcal{S}(R)/\cong}\left(\bigcup_{\sigma\in \text{Sym}(E)}\left(\bigsqcup_{L\in E}\phi_{L}\circ \text{Aut}(\xi_{E})\circ \phi_{\sigma(L)}^{-1}\right)\right)\\\cong \left(\prod_{E\in \mathcal{C}(R^{\dagger})/\cong}\text{Aut}(\Gamma_{E})\wr_{E}\text{Sym}(E)\right)\times \left(\prod_{E\in \mathcal{S}(R)/\cong}\text{Aut}(\xi_{E})\wr_{E}\text{Sym}(E)\right)$$

Therefore at this point it appears the studying of the automorphism group of relation perserving functions for any binary relation reduces to the studying of the automorphism groups of connected acyclic relations and strongly connected relations. So now looking through books/papers online I found a special case of that expression I found above for symmetric relations (well to be more precise it was for the automorphisms of undirected graphs, but as each can be uniquely identified with a symmetric relation they end up being basically the same). In particular their formula was:

So basically I can identify their graph $X$ with a finite symmetric relation $R$ which then just corresponds to the first formula I wrote for decomposing $\text{Aut}(R)$ based upon its connected components. Though this was really all I could dig up on the subject, and I wasn't particularly interested in just finite symmetric relations or equivalently finite undirected graph automorphisms. I was looking for more general relations or equivalently digraphs with far fewer restrictions on them e.g. not having to be finite (in particular that was why I used that wacky "direct union" $\bigsqcup$ notation, so if dealing with say a relation possessing an uncountable number of pairwise non-isomorphic connected components one wouldn't have to haggle with some weird direct/wreath product generalization to account for the fact one can no longer express each of the automorphisms as tuples because the required number of coordinate projections needed for this would no longer be countable, regardless though I'm fine with weakening the problem to make it some what more workable e.g. assuming the relation is both locally-out/in finite so we are working with less crazy permutations). In any case my idea going forward from here was basically as I mentioned earlier, just now focus on categorizing the automorphism groups of connected acyclic relations and strong relations. In particular for any acyclic relation call it $R$ noting $\text{Aut}(R)$ is a subgroup of $\text{Aut}(R^{*})$ where $R^{*}$ is a strict order - that since for any relation $L$ the identity:

$$L\text{ is a strict-order}\iff \forall a,b\in \text{fld}(L)\left(aLb\implies L[\{b\}]\subset L[\{a\}]\right)\\\iff \forall a,b\in \text{fld}(L)\left(aLb\implies L^{-1}[\{a\}]\subset L^{-1}[\{b\}]\right)$$

Forces us to have $(v,\sigma(v))\in R^{*}\cup (R^{*})^{-1}\implies |R^{*}[\{v\}]|\neq |R^{*}[\{\sigma(v)\}]|$ for every single $\sigma\in \text{Aut}(R)\subseteq \text{Aut}(R^{*})$ and yet because two vertices in a common orbit of the corresponding labeled digraph of $R$ must always have the same out-degree cardinality this means every such permutation of vertices occurs between independent vertex sets, or in other words adjoining loops to the strict order $R^{*}$ so as to form a partial order on $\text{fld}(R)$, we get that the orbits in every permutation $\sigma\in \text{Aut}(R)$ will always be anti-chains of this newly formed poset. Yet not sure how much this simplifies things though... In any case looking back now at the automorphism groups for strong relations, I'm confident it shouldn't be hard to separate out the symmetric cycles (cycles with less then three vertices) in fact I kind of already had an idea on how to do this. Then from here one would just be working with decomposing the strong orientations of undirected graphs, perhaps this could be easier to deal with? But honestly I'm not sure how much more time I should devote to this, its interesting though I get a very strong sense its probably already either been explored with more advanced tools (perhaps explaining why I was unable to find documentation on the subject for maybe I was using the wrong keywords?) or maybe it has been reduced to some known hard problem in which case any further pursuit would surely be a waste of my time. Also I know I've kinda been vague with the word "decompose" here, but to put it in a some what less informal way - I just wanted to see how far I could keep partitioning out the automorphism groups into smaller automorphism groups each corresponding to subsets of the parent relation. Though seeing the second decomposition into strong components was able to be formed without using all the edges of $R$ this has me kinda scared this might be related to something like graph reconstruction, which if that is the case then I imagine I should probably abandon this right? So any advice here would be appreciated, as I'm not sure how much deeper I should try and go with this. Can I keep separating out those automorphism groups into smaller ones formed from subrelations of their parent relations? Are there any papers/prints on this sort of problem one can direct me to? I've searched online with keywords ranging over anything I thought could be involved from the obvious "relation perserving functions" to "digraph automorphisms" to "imprimitive permutations" etc. my background in group theory is not the strongest though I've really tried and I can't find what I want so to re-iterate one last time, any help would greatly be appreciated.

Update: From reading more about permutation groups, it appears I'm trying to express/decompose each automorphism group using mostly what are called "block systems" with the action of group conjugation on subsets of my parent relation. For example given any permutation $\sigma\in \text{Aut}(R)$ writing the action $\phi_{\sigma}(H)=\sigma\circ H\circ \sigma^{-1}$ one sees for each connected component $L\in \mathcal{C}(R)$ that this partition forms a block system for $R$ as we have:

$$R=\bigcup_{L\in \mathcal{C}(R)}L \text{ and either }\phi_{\sigma}(L)\cap L=\emptyset \text{ or }\phi_{\sigma}(L)\cap L=L$$

Anyway this seems to go pretty deep (a lot of it beyond my grasp right now, though I know not for most of the very talented users on this website) and spending time working on this problem is interfering with my required school work. So for now I'm going to leave this question open for maybe another week in case anyone has comments on the matter then I'll probably just close it.

Homotopy groups of smooth part of moduli space

Thu, 11/02/2017 - 10:41

Let $M_g$ be the moduli space of Riemann surfaces, as described for example in the book of Harris and Morrison - Moduli of curves. As a topological space, or better as orbifold, it has smooth points and singular points. Let $S\subseteq M_g$ be the subset of smooth points. In the article by Cornalba ("On the locus of curves with automorphisms") this subset is identified, in the case $g>3$, with the set of those curves which have just the trivial automorphism.

Is there any idea (or any reference) of how to compute the homotopy groups $$\pi_2(M_g,S) \qquad \mbox{and} \qquad \pi_1(S)?$$

Edit: I should have included a base point above, but I intentionally didn't because I wouldn't know which one to choose.

Submanifolds of $\mathbb{R}^N$ whose local charts have uniformly bounded derivatives

Thu, 11/02/2017 - 08:29

Working on a problem in Differential Geometry, which is quite far away from my area of expertise, I was recently led to consider the class of those smooth, $n$-dimensional embedded submanifolds $M \subset \mathbb{R}^N$ such that the following condition is satisfied:

$M$ possesses an oriented atlas $\{(U_{\alpha}, \, \phi_{\alpha})\}$, where $\phi_{\alpha} \colon U_{\alpha} \stackrel{\cong}{\longrightarrow} V_{\alpha} \subset \mathbb{R}^{n}$, such that, calling $i_{\alpha} \colon U_{\alpha} \hookrightarrow \mathbb{R}^N$ the inclusion map, the composition $$i_{\alpha} \circ \phi_{\alpha}^{-1} \colon \, V_{\alpha} \longrightarrow \mathbb{R}^N$$
has bounded partial derivatives up to order $k$, uniformly on $\alpha$.

My question is now the following:

Does the condition above have a name? In this case, is there any characterization of the manifolds $M$ satisfying it in terms of the usual Riemannian invariants (for instance, curvature)?

Any answer or reference to the relevant literature will be greatly appreciated.

Long time behaviour of the exterior Neumann problem for the heat equation

Thu, 11/02/2017 - 07:42

Let $\Omega \subseteq \mathbb{R}^n$ be open, bounded and connected such that $\Omega^- \equiv \mathbb{R}^n \setminus\mathrm{cl}\Omega$ is connecetd. Let $\Omega \in C^{1+\alpha}$, $g \in C_b^{\frac{\alpha}{2};\alpha}((0,+\infty) \times \partial \Omega),\, g(0,\cdot) = 0$. Let \begin{equation} \begin{cases} \partial_t u - \Delta u = 0 \qquad &\mbox{ in } \;(0,+\infty)\times \Omega^-,\\ \partial_\nu u = g & \mbox{ on }\;(0,+\infty) \times \partial \Omega,\\ u(0,\cdot) = 0 & \mbox{ in } \; \mathrm{cl} \Omega^-. \end{cases} \end{equation}

Is there some known results on the asymptotic behaviour of $u$ at $+\infty$ in time under some assumpiton on $g$? That is, under which assumptions on $g$ the solution $u$ is bounded or has a limit at $+\infty$?

How expressive can $\mathcal{L}_{\kappa,\kappa}$ be?

Wed, 11/01/2017 - 20:27

(Note that the logic systems described in this question only refer to logic systems restricted to the language $\in$)

1. Can $\mathcal{L}_{\kappa,\kappa}$ express $n$-th order finitary logic? It is clear that it can express first-order logic ($\Pi_{<\omega}^0$), but it seems unlikely that $\mathcal{L}_{\kappa,\kappa}$ can express even $\Pi_{<\omega}^1$.

By the answer to this question: Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem), it is known that $L^{\Pi_{<\omega}^1}=HOD$. Assuming $\mathcal{L}_{\kappa,\kappa}$ expresses $\Pi_{<\omega}^1$, it is then not difficult to show that $HOD\subseteq L^{\mathcal{L}_{\kappa,\kappa}}\subseteq L(V_\kappa)$. Thus, unless $L(V_\kappa)=V$ (which is true iff $L=V$), $V\neq HOD$.

Therefore, the existence of such a $\kappa$ is inconsistent with $V\neq L\land V=HOD$. This is very doubtful.

2. Can $\mathcal{L}_{\kappa,\kappa}$ express $\Pi_n^1$ for $n>0$? For $n=0$, it is once again easy to see that, yes, this is true ($\Pi_0^1=\Pi_0^0$) for every $\kappa$. Other than this, I have no information for this question.

What rectangles can a set of rectangles tile?

Wed, 11/01/2017 - 19:42

(I asked this question first on math.stackexchange, but did not get any responses so I thought I would try here.)

If we have a set of $p_i \times q_i$ rectangles ($p_i, q_i \in \mathbf{N}$), which $m \times n$ rectangles can be tiled with copies from the set? (No rotation allowed.)

I am particularly interested in the algorithm that realizes Theorem 4 below.

What I know so far:

Theorem 0

  • We need $mn = \sum p_iq_ic_i$ for some $c_i \geq 0$.
  • Considering how rectangles form the border, we need at least $m = \sum a_ip_i$ and $n = \sum b_iq_i$ for some $a_i \geq 0$ and $b_i \geq 0$.

Theorem 1 For two rectangles with $\gcd(p_1, p_2) = \gcd(q_1, q_2) = 1$, a tiling exists if and only if one of the following holds [2]:

  1. $p_1 \mid m$ and $q_1 \mid n$
  2. $p_2 \mid m$ and $q_2 \mid n$
  3. $p_1q_1 \mid m$ and $ap_2 + bq_2 = n$ for some integers $a, b$
  4. $p_2q_2 \mid n$ and $ap_1 + bq_1 = n$ for some integers $a, b$.

Theorem 2 For any number of rectangles, if any side of all rectangles share a common factor, then they can only tile a larger rectangle if one side has the same common factor [3].

(Between these first theorems dealing with sets of two rectangles is easy.)

Theorem 3 A set of three or more rectangles satisfying $\gcd(p_i, p_j) = \gcd(q_i, q_j) = 1$ , for $i \neq j$ there exist some $C$ such that the set will tile any rectangle with $m, n > C$ [4, 5].

How to find such a $C$ is given in [3]. Unfortunately, this $C$ can be quite large, and is not generally tight (there is a smaller $C$ that also works). So there is a whole bunch of rectangles for which it says nothing.

In addition, it says nothing about rectangles that do not satisfy the conditions. For example, it is hard to know much about which rectangles can be tiled by a set with a $6\times 6, 10\times 10$ and $15 \times 15$ rectangle. In this example, pairs of squares share a common factor, but we cannot reduce the tiling problem because there is not a common factor among all tiles.

Theorem 4 For every finite set of rectangular tiles, the tileability problem of an $m\times n$ rectangle can be decided in $O(\log mn)$ time.

This result is mentioned in [4] (and some others), but unfortunately it references a mysterious unpublished manuscript, and there is no details available; no proof, and no hint at what the algorithm might be.

(The unpublished manuscript is Tiling rectangles with rectangles by Lam, Miller and Pak. I also saw a reference to "Packing boxes with boxes", also unpublished, by the same authors, which I suspect is the same. I could find neither one :-/)

I also wrote a computer program to investigate some examples. My own optimized-but-still-exponential-time algorithm starts becoming unpractical around for $m, n >80$ with a set of only three tiles, so I have not been able to get much insight from it.

[2] Bower, Richard J.; Michael, T.S., When can you tile a box with translates of two given rectangular bricks?, Electron. J. Comb. 11, No. 1, Research paper N7, 9 p. (2004). ZBL1053.05027.

[3] de Bruijn, N.G., Filling boxes with bricks, Am. Math. Mon. 76, 37-40 (1969). ZBL0174.25501.

[4] Labrousse, D.; Ramírez Alfonsín, J.L., A tiling problem and the Frobenius number, Chudnovsky, David (ed.) et al., Additive number theory. Festschrift in honor of the sixtieth birthday of Melvyn B. Nathanson. New York, NY: Springer (ISBN 978-0-387-37029-3/hbk; 978-0-387-68361-4/ebook). 203-220 (2010). ZBL1248.11022.

[5] Pak, Igor; Yang, Jed, Tiling simply connected regions with rectangles, J. Comb. Theory, Ser. A 120, No. 7, 1804-1816 (2013). ZBL1314.05034.

Undecidable easy arithmetical statement

Tue, 10/31/2017 - 15:18

Is there a basic arithmetic statement which is known to be undecidable ?

By basic arithmetic statement I do mean an easy statement in the spirit of the Collatz conjecture . By the way is there some reasons to believe that the Collatz conjecture is undecidable ?

A questions about a function related to prime numbers [on hold]

Tue, 10/31/2017 - 13:58

Theorem $1$: Let $\mathbb{P}$ be the set prime numbers and $S$ is a set that has been made as below: put a point on the beginning of each member of $\Bbb{P}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $(0.1,1)$ of real numbers.

Theorem $2$: For each subinterval of $[0.1,1)$ like $(a,b)$ then $\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ such that $t\times 10^k\in \Bbb P$. $$\\$$ Suppose $r:\Bbb N \to (0,1)$ is a mapping given by $r(n)$ is obtained as put a point on the beginning of $n$ like $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,$ $r_k: \Bbb N \to (0,1)$ by $r_k(n)=10^{-k}\times r(n)$.

Based on theorem $2$ we can define $f:\{(c,d)\,|\, (c,d)\subseteq [0.1,1)\}\to\Bbb N$ is a function that $f((c,d))$ is the least $n\in\Bbb N$ that $\exists t\in (c,d)\,\exists k\in\Bbb N$ that $p_n=t\cdot 10^k$ that $p_n$ is $n$_th prime and $\forall m\ge f((c,d))$ $\exists u\in\color{red}{(c,d)}$ that $u\cdot 10^m\in\Bbb P$

and $g:(0,0.9)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.9)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\,|\, d-c=\epsilon,$ $(c,d)\subseteq [0.1,1)\})$. $$\\$$

Question: Isn't $g$ an injective function.

$\color{red}{Edit}$:Sorry, by mistake I'd written $[0.1,1)$ instead $(c,d)$! $$\\$$ I thank you in advance.

Are unirational K3 surfaces defined over finite fields?

Tue, 10/31/2017 - 05:52

Is every supersingular (thus unirational for ${\rm char }\ k = p\geq 5$, from Liedtke) $K3$ surface defined over a finite field? I guess this is true for Kummer surfaces, for example, since supersingular Kummer surfaces can be seen as the resolution of singularities of $(E\times E)/<-1>$, $E$ is a supersingular curve, so it is defined over a finite field but I don't know if there are examples of a supersingular surfaces defined over a function field over a finite field, say.

About the degree of character of $PSL(n,q)$

Tue, 10/31/2017 - 05:37

It is well known that for $n\geq2$ the group $PSL(n,q)$ is simple except for $PSL(2,2)=S_3$ and $PSL(2,3)=A_4$.

Let $G$ be one of the simple groups $PSL(n,q)$. From the ATLAS of finite group, we guess that for every prime $p\in\pi(G)$ there exists some $\mathbb{C}$-representation $\mathfrak{X}$ such that its degree $\chi(1)$ satisfies $\chi(1)_p=|G|_p$.

Now I want know some information about this conjecture.


Random projection increases the distance?

Tue, 10/31/2017 - 04:25

Consider two absolutely continuous random variables $X: \Omega \mapsto \mathbb{R}^d$ and $Y: \Omega \mapsto \mathbb{R}^d$ for probability spaces $(\Omega, \mathcal{F},p_X)$ and $(\Omega, \mathcal{F},p_Y)$. Define distance measure $d: \mathbb{R}^d\times \mathbb{R}^d \mapsto \mathbb{R} $ as $d(x,y) = (x-y)^T(x-y)$, where $T$ is the transpose. Consider a linear projection $Q: \mathbb{R}^d \mapsto \mathbb{R}^k$ for $k \gg d$.

Under which condition on $p_X$, $p_Y$, and $Q$, we can find a lower bound on the probability of the following event: $$\dfrac{d(X,E[X])}{d(Y,E[X])} \leq \dfrac{d(QX,QE[X])}{d(QY,QE[X])}$$ where $E[X]$ is the expected value of $X$.

A question on connected sum of compact manifolds

Tue, 10/31/2017 - 03:57

Let $M$ be a compact orientable manifold which is homeomorphic to its connected sum with itself $M\# M$. Must $M$ be homeomorphic to a sphere?

I will explain why I am interested (at the risk of being outright foolish or overambitious). The 'equation' $M = M\#M$ is the simplest conceivable equation in the category of compact topological manifolds, just as $x+x=x$ characterises $0$ in an abelian group. Although I am suspicious if $M$ could be a homology sphere, I hope to get a characterisation of spheres.

Thanks for any comments or suggested references.

Bounded holomorphic functions on hypersurfaces of $\Bbb C^n$

Tue, 10/31/2017 - 03:45

Is it true that every bounded holomorphic functions on a smooth analytic hypersurface $X$ of $\Bbb C^n$ is constant?

  • Remark that if $X$ is algebraic, the answer is yes.

Otherwise can you provide counterexamples?

Pre image of singletons under the rank function of finitely generated locally free modules

Tue, 10/31/2017 - 03:11

Let $R$ be a commutative ring with unity and $M$ be a finitely generated , locally free $R$-module (i.e. $M_p$ is free over $R_p$ for every prime ideal $p$ of $R$ ) . Consider the rank function $rk : Spec (R) \to \mathbb N \cup \{0\} $ as $rk (p)=dim_{R_p} M_p $. My question is : When can we say that $rk^{-1} \{n\} $ is Zariski-closed in $Spec R$ , for every integer $n \ge 0$ ?

Eigenspaces of equivariant matrices and irreps of finite groups

Tue, 10/31/2017 - 03:06

Let $T_1,T_2\in R^{d\times k}$ matrices and $G$, $\tilde{G}$ two finite unitary group of cardinality $N$. Indicating (a matrix representation of the) elements of $G$ (respectively $\tilde{G}$) with $g$ (respectively $\tilde{g}$), equivariant matrices can be written as: $$ C_{1}=\sum_{i=1}^{N} g_{i}T_1T_1^Tg_{i}^T $$

$$ C_{2}=\sum_{i=1}^{N} \tilde{g}_{i}T_2T_2^T\tilde{g}_{i}^T $$ In fact the matrix has the property that $$ [C_{1},g_{i}]=0\;\;\forall\;i,\;\;[C_{2},\tilde{g}_{i}]=0\;\;\forall\;i $$ which defines equivariance, and suppose $[C_1,C_2]=0$. I can prove that $G$ (respectively $\tilde{G}$) acts irreducibly on any eigenspace of $C_1$, (respectively $C_2$) and from this I can deduce that the representations of $G,\tilde{G}$ are conjugate, being irreducible on the same spaces, i.e. $g_=U^T\tilde{g}_iU,\forall\;i$, exists $U$.

I would like to be able to prove more i.e. $U=I$ the identity. It seems to me that I cannot from the hypotheses above. If this is the case? Any idea on which further conditions I need to impose to have $G=\tilde{G}$?

Thank you


Automorphisms of rationally connected varieties

Tue, 10/31/2017 - 03:01

Let $X$ be a smooth, rationally connected variety over an algebraically closed field of characteristic zero. Denote by $\mathrm{Aut}(X)$ the space of automorphisms of $X$ and for a given $\phi \in \mathrm{Aut}(X)$, denote by $\mathrm{Fix}(\phi)$ the fixed points of the automorphism $\phi$. I wanted to ask, if there is any known condition under which the intersection $F$ of all $\mathrm{Fix}(\phi)$ as $\phi$ ranges over all elements of $\mathrm{Aut}(X)$ is the emptyset?

Does $f \equiv 0 \ (mod \ p)$? [migrated]

Tue, 10/31/2017 - 02:53

Let $p$ is a prime number and $f(x) \in \mathbb{Z}_p[x]$, $deg f(x)=n$. If $f(x)$ has $n+1$ roots in $\mathbb{Z}_p$. Does $f \equiv 0 \ (mod \ p)$?

Why is Oka's coherence theorem a deep result?

Tue, 10/31/2017 - 01:00

This is a very naive question.

Let $X$ be a complex manifold. Let $\mathcal{O}_X$ be the structure sheaf of $X$, a sheaf of rings whose sections over opens $U\subset X$ are just the holomorphic functions $U\rightarrow\mathbb{C}$.

A sheaf of $\mathcal{O}_X$-modules $F$ is coherent if:

  1. It is locally finitely generated: Ie, there is an open cover $\{U_i\}$ of $X$ such that $F|_{U_i}$ admits surjections $\mathcal{O}_{U_i}\rightarrow F|_{U_i}$ for each $i$.
  2. For any open $V\subset X$, and any morphism $f : \mathcal{O}_V^s\rightarrow F|_V$, $\text{Ker}(f)$ is a locally finitely generated sheaf on $V$ (ie, satisfies condition 1).

Then, Oka's coherence theorem states that the structure sheaf $\mathcal{O}_X$ is coherent.

When $X$ is a scheme, this statement is almost tautological. What is it about the setting of complex manifolds that makes this theorem deep?

Interchanging sums and integrals in a specific instance

Tue, 10/31/2017 - 00:22

Suppose $f_n$ is a sequence of holomorphic functions taking $\mathbb{D} \to \mathbb{C}$ where $\mathbb{D}$ is the unit disk. Further, $f_n$ has a continuous extension to $\overline{\mathbb{D}}$. We can assume $\sum_{n=0}^\infty f_n$ converges normally on compact subsets of $\mathbb{D}$ to a holomorphic function $f$.

Additionally, for each $f_n$ we know $\int_C |f_n| < \infty$ for any contour $C \subset \overline{\mathbb{D}}$. Less strictly, so we don't have enough to use the monotone convergence theorem, we know that $\sum_{n=0}^\infty |\int_C f_n| < \infty$. But additionally, for any closed contour $C^*$ in $\overline{\mathbb{D}}$ we know $\int_{C^*}f_n = 0$.

Must it follow

$$\int_C f = \sum_{n=0}^\infty \int_C f_n$$

I'm asking because all the naive instances where the monotone convergence theorem fail are exempt from these criterion. I think there's something more subtle in this instance.

I've been able to strengthen the condition to a proof that

$$\int_C f = \sum_{n=0}^\infty \int_C f_n \,\,\Leftrightarrow\,\, \sum_{n=0}^\infty \sup_{C\subset \overline{\mathbb{D}}}|\int_Cf_n| < \infty$$

Or rephrase it to

$$\sum_{n=0}^\infty |\int_C f_n| < \infty\,\, \Leftrightarrow \,\,\,f \, \text{can be continuously extended to}\,\overline{\mathbb{D}}$$

Those are the two avenues I've taken. None have really given me an answer.

Any suggestions, comments, questions are greatly appreciated.