Recent MathOverflow Questions

Spectrum of self-adjoint operator

Math Overflow Recent Questions - Sun, 06/04/2017 - 16:11

As a non functional analyst, I stumbled over the following question:

Given a self-adjoint Operator $T:D(T) \subset H \rightarrow H.$ Assume we know that $T$ has some eigenvalue $\lambda$ which is isolated in the spectrum of $T$. The eigenvalue may be non-degenerate.

We do have an approximating sequence of eigenfunctions for $(T-\lambda),$ i.e. $(T-\lambda)x_n \rightarrow 0$ for $x_n \in D(T)$ normalized.

Now consider a closed restriction $S \subset T.$ Assuming $x_n \in D(S)$ and $(S-\lambda)x_n \rightarrow 0$ as well.

Does this imply that there is an eigenfunction to $T$ with eigenvalue $\lambda$ that is also in $D(S)$?

Grothendieck toposes and logic

Math Overflow Recent Questions - Sun, 06/04/2017 - 14:27

I am searching results in which one can extract logic information from a topological (Grothendieck topos) perspective (such as Gödel's Completeness Theorem and Deligne's Theorem ("theorem by P. Deligne about coherent topoi (coherent topoi have enough points) and how this theorem is equivalent to Gödel's completeness theorem for first order logic").

The ideal answer would be a result already contained in SGA IV (as Deligne theorem), a real scenario would be a Grothendieck topos theorem (conjecture) that extract an important aspect of logic.

Kodaira embedding theorem for rigid analytic varieties

Math Overflow Recent Questions - Sun, 06/04/2017 - 09:48

Kodaira embedding theorem can be regarded as a vast generalizaton of the projectivity criterion for complex tori: indeed, the Riemann conditions essentially say that the line bundle defined by the polarisation is positive.

If $k$ is a complete field endowed with non-Archimedean absolute value, then Riemann conditions for analytic tori $\mathbb{G}_m^n(k)/\mathbb{Z}^n$ can also be given, polarisation becomes a map $\varphi: \mathbb{Z}^n \to \mathrm{Hom}(\mathbb{G}_m^n, \mathbb{G}_m)$ such that $\varphi(\lambda)(\lambda')=\varphi(\lambda')(\lambda)$ for any $\lambda, \lambda' \in \mathbb{Z}^n$, and the positivity condition now reads as that the form $\sigma(\lambda, \lambda')=-\log |\varphi(\lambda)(\lambda')|$ is positive definite (all this is nicely exposed in Fresnel and van der Put's book "Rigid analytic geometry and its applications", Section 6.5).

It is then natural to ask: does there exist a criterion for a line bundle over a rigid analytic space to be ample?

While there is a notion of a metrized line bundle over a rigid analytic space (a metric on a line bunde $L$ is just a map of sheaves $h: L \to C(X^{an},\mathbb{R})$, where $C(X^{an},\mathbb{R})$ is the sheaf of continuous $\mathbb{R}$-valued functions on $X^{an}$, such that for local sections $s$ and local functions $f$, $h(f \cdot s) = |f|\cdot h(s)$), is there a notion of positive curvature for such a metric?

Homotopy equivalence vs gauge equivalence

Math Overflow Recent Questions - Sun, 06/04/2017 - 09:18

Let $(\mathfrak{g},[-,-])$ be a pronilpotent Lie algebra (considered of degree zero). We can consider $(\mathfrak{g},[-,-])$ as a differential graded Lie algebra endowed with the $0$-differential. Let $A$ be a commutative non-negatively graded differential algebra. Then $A\widehat{\otimes }\mathfrak{g}$ is a differential graded Lie algebra whit bracket

$[a\otimes V,b\otimes W]:=ab\otimes[V,W] $

and where the differential is the obvious differential in the tensor product. We denote such a differential with $d$. Let $MC(A\widehat{\otimes}\mathfrak{g})$ be the set of maurer-Cartan elements, i.e solutions of the equations

$d\alpha +\frac{1}{2}[\alpha,\alpha]=0$.

There are two type of equivalences on this differential graded algebra.

1) gauge equivalences

The set $(A\widehat{\otimes }\mathfrak{g})^{0}$ acts on the set of maurer-Cartan elements via the action

$d+\alpha\mapsto e^{u}(d+\alpha)e^{-u}=d+e^{ad_{u}}\alpha+\frac{1-e^{ad_{u}}}{ad_{u}}du$.

Two maurer-Cartan elements $\alpha_{0},\alpha_{0}$ are gauge equivalent if for there exists an $u$ as above connecting them.

2)homotopy equivalence Let $\Omega(1)$ be the differential graded algebra of polynomial forms on $[0,1]$. Then $(\Omega(1)\otimes A)\widehat{\otimes }\mathfrak{g}$ is again a differential graded algebra. Two maurer-Cartan elements $\alpha_{0},\alpha_{0}$ are homotopy equivalent if there exists a Maurer-Cartan element $\alpha$ in $(\Omega(1)\otimes A)\widehat{\otimes }\mathfrak{g}$ such that $\alpha(0)=\alpha_{0}$ and $\alpha(1)=\alpha_{1}$.

Here my question: Under which condition of $A$ is 1 equivalent to 2? Do I need some special condition on $A$?

Edit: here some observations: Notice that if $A$ is finite dimensional then $A\widehat{\otimes }\mathfrak{g}$ is again pronilpotent and the two definitions are equivalent. On the other hand if $A$ is not finite dimensional, then $A\widehat{\otimes }\mathfrak{g}$ can be written as the projective limit of non necessarlily finite dimensional Lie algebras. So my question can be writtens as: what happen if $A$ is not finite-dimensional?

How to prove $\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}$?

Math Overflow Recent Questions - Sat, 06/03/2017 - 22:35

Numerical calculation suggests that for prime $p\ge 5$, \begin{align*} \sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}. \end{align*}

How can we arrive at this congruence?

How can I show the principal symbol is not elliptic?

Math Overflow Recent Questions - Sat, 06/03/2017 - 01:44

Let us assume the principal symbol of a nonlinear differential operator $E$ at a point $p$ is $$\sigma‎_{‎E‎}(p):‎\Gamma (‎T^*M‎)\times \Gamma (T^*‎M‎)\to \Gamma (T^*‎M‎)‎$$‎ which acts as follows: ‎‎‎$$ ‎\sigma_{E}(‎p‎) ‎(a,\eta)‎ ‎=‎‎‎‎ \sum_{k,l=1}^n ‎‎‎‎\Big(‎ a_k a_l‎\alpha_i‎\eta_j ‎+ a_k a_l ‎\eta‎_i\alpha‎_j -a_i a_k\alpha_l‎\eta_j- a_i a_k\eta‎_l\alpha‎_j$$ ‎$$‎-a_ja_k‎‎\alpha_l \eta_i-a_j a_k‎\eta‎_l\alpha_i ‎+‎‎a_i a_j\alpha_k ‎\eta_l ‎+a_i a_j\eta‎_k\alpha‎_l \Big)‎ \xi^j $$‎ $$‎-‎‎ 2\rho \Big(‎a^t‎ a_t ‎\alpha_k‎ ‎\eta_l‎ +‎a^t ‎a_t ‎‎\eta_k ‎\alpha_l‎ -‎a^ta^s ‎\alpha_t \eta_s -‎a‎^ta^s ‎\alpha_s‎ \eta_t‎‎ ‎\Big)\alpha_i‎ $$ In above the Einstein summation convention is used and $\rho$ is a real scalar.

The Problem is: I want to show this equation is not strictly parabolic.

I put $a=(1,0,\cdots ,0)$ and $\eta=(1,0,\cdots ,0)$. With this assumptions I only can show the first part is equal to zero.

Any suggestion is highly appreciated.

Counting block-equivalent permutations

Math Overflow Recent Questions - Fri, 06/02/2017 - 10:57

Consider the group $\mathfrak{S}_n$ of permutations on the letters $\{1,2,\dots,n\}$.

We say two permutations are b-equivalent, $\pi_1\,\pmb{\sim^b}\,\pi_2$, if one can be determined from the other by reversing a block of $b$ consecutive integers. For example, $617\pmb{5432}\,\pmb{\sim^4}\,617\pmb{2345}$.

Question. Is this true? The number $h_b(n)$ of $(b+1)$-equivalent classes is given by $$h_b(n)=\sum_{j=0}^{\lfloor\frac{n}{b+1}\rfloor}(-1)^j(n-bj)!\binom{n-bj}j.$$

The special case $b=1$ recovers Theorem 2.2 of this paper.

The first eigenfunction of Dirac operator for surface

Math Overflow Recent Questions - Fri, 06/02/2017 - 08:20

Let $M$ be a spherical oriented surface with Riemannian metric and with trivial spin structure. We know that the equation $$D \phi = \rho \phi$$, where $\rho: M\rightarrow \mathbb{R}$ is a real scalar function, corresponds to a conformal immersion $f:M\rightarrow \mathbb{R}^3$ (See Friedrich 03). It follows that the eigenvalue problem $D \phi =\lambda_i \phi$, where $\lambda_i\in \mathbb{R}$, gives some special immersion of this surface $f_i:M\rightarrow \mathbb{R}^3$. My questions is, if $f_1$, the immersion corresponding to the first eigenvalue, has some special meaning? Actually I'm highly interested in the Willmore energy of this immersion. I guess the this immersion would have a small Willmore energy from the inequality in (Bär 98): $$\lambda_1^2\leq \frac{\int_M H^2}{\mathrm{area}(M)}$$ However it might be too optimistic to expect that this immersion is a round sphere. Anyone has some insights for this immersion? Thank you very much.

Distribution of $0-1$ matrices

Math Overflow Recent Questions - Fri, 06/02/2017 - 07:39

Consider $n\times n$ matrices with entries in $\{0,1\}$. The determinants of these ranges from $0$ to the Hadamard bound $\frac{(n+1)^{\frac{n+1}2}}{2^n}$. Assume $n$ is large enough.

What does the distribution of the determinants look like? Is it normal or skewed?

What proportion of such $n\times n$ determinants is singular?

Topology of the Hamel basis in a TVS

Math Overflow Recent Questions - Fri, 06/02/2017 - 07:22

Let $V$ be a complex topological vector space, and let $I$ be a Hamel basis of it. Then as a subset $I\subset V$ acquires an induced topology, becoming a topological space. For a topological space $X$ consider the space $C_\#(X,\mathbb{C})$ of finitely supported complex functions on $X$. Then $C_\#(I,\mathbb{C})$ is algebraically isomorphic to $V$ by definition of a Hamel basis, $$ \forall v\in V,\quad v=\sum_{e\in I}\lambda_e^v e,\quad I\ni e\mapsto\lambda_e^v\in\mathbb{C}. $$

Question: Does there exist a topology on $C_\#(X,\mathbb{C})$, such that $C_\#(I,\mathbb{C})$ is isomorphic to $V$ as topological vector spaces?

In finite dimensions $I$ is finite and hence discrete, so that all reasonable topologies on $C_\#(I,\mathbb{C})=\mathbb{C}^I$ coincide and give a positive answer. The question therefore pertains to infinite dimensions.

I am mostly interested in separable Hausdorff $V$. Thank you.

Is there a concrete way to show the existence of canonical model for non-modular Shimura curves?

Math Overflow Recent Questions - Fri, 06/02/2017 - 07:11

I am trying to read Carayol's article on the construction of Galois representations associated to Hilbert modular forms ( The main geometric ingredient is a study of the reduction of (non-modular) Shimura curves associated to a quaternion algebra over a totally real number field $F$ over $Q$, where $F \neq Q$.

In that article the existence of a canonical model (in the sense of Deligne) is deduced from Deligne's general results about general Shimura varieties ( If I remember correctly Deligne's proof is very general and works by showing the result for entire classes of groups $G$ at once.

I was wondering if there was an easier way to show existence in this particular case ?


Legendre expansion of $r(x) = f(x)/g(x)$ using a finite number of samples from $f(x)$ and $g(x)$

Math Overflow Recent Questions - Fri, 06/02/2017 - 07:04

I have two finite sets of events $\{x_1, ..., x_N\}$ and $\{y_1, ..., y_N\}$ that are sampled from the PDFs $f(x)$ and $g(x)$, respectively. I want to estimate the Legendre expansion of $r(x) = f(x)/g(x)$ using these events. What is the best method?

Deciding equality in free models of a (generalized) Lawvere theory

Math Overflow Recent Questions - Fri, 06/02/2017 - 06:10

Let $F : \mathcal{C} \rightarrow \mathcal{D}$ be functor of Lawvere theories $\mathcal{C}, \mathcal{D}$ (i.e. cartesian categories where every object is isomorphic to some power of a chosen object) that preserves finite limits and the chosen object. Precomposition $- \circ F$ turns every $\mathcal{D}$-model $A : \mathcal{D} \rightarrow \underline{\text{Set}}$ into a $\mathcal{D}$-model $A \circ F : \mathcal{C} \rightarrow \underline{\text{Set}}$ in a functorial way. This functor preserves limits and directed colimits (because they are computed componentwise---use the fact that directed colimits commute with finite limits in $\underline{\text{Set}}$ for the latter). From Adamek and Rosicky's book on locally presentable categories, we know that

  • Every Lawvere theory's category of set-theoretic models is locally presentable. (Thm. 1.46)
  • A functor between locally presentable categories is right adjoint iff it preserves $\lambda$-directed colimits for some regular cardinal $\lambda$. (Thm. 1.66)

Therefore, the above-mentioned functor $- \circ F$ admits a left adjoint $G$. Now suppose that we have a model $A : \mathcal{C} \rightarrow \mathcal{D}$ and a way to decide whether two elements in its carrier are equal, and we can compute all functions induced by the structure as a model of $\mathcal{C}$. When can we construct $G$ in such a way that we can also decide equality in $G(A)$?

For example, this seems to be the case where $\mathcal{C}$ is the trivial theory whose models are sets, and $\mathcal{D}$ the theory of monoids, because we can easily decide whether two lists (elements of the free monoid) are equal if we can compare its elements.

The situation is less clear to me when $\mathcal{C}$ is the theory of monoids and $\mathcal{D}$ the theory of groups. This seems related to the world problem, but easier.

I'm also very much interested in the generalization to multisorted Lawvere/algebraic theories (i.e. with $\mathcal{C}, \mathcal{D}$ arbitrary cartesian categories), and to essentially algebraic theories (where $\mathcal{C}$ and $\mathcal{D}$ are finitely complete categories and $F$ preserves finite limits). For example, the free category over a graph has decidable equality, while I'm again unsure about the free groupoid over a category.

Answers to special cases are also welcome.

Showing that the Frechet space is not globally convex [on hold]

Math Overflow Recent Questions - Fri, 06/02/2017 - 05:26

For any open set $D \subset \mathbb{C}^n$, the ring $\mathscr{C}_D$ of continuous, complex valued functions in $D$ has a natural topology, the topology of uniform convergence on compact subsets. That is, for any compact set $K \subset D$ and any $\epsilon > 0$, let $$U(K,\epsilon): = \{ f \in \mathscr{C}_D \ \vert \ \left| f(z) \right| < \epsilon \ \ \ \forall z \in K \}$$ form a basis of the topology.

Note that a Hausdorff topological vector space over $\mathbb{R}$ or $\mathbb{C}$ in which every neighbourhood of the zero element contains a convex neighbourhood of the zero element is said to be locally convex. Therefore, to show that the Frechet space $\mathscr{C}_D$ is locally convex, we need only show that the set $$U(K,\epsilon) : = \{ f \in \mathscr{C}_D \ \vert \ \left| f(z) \right| < \epsilon \ \ \ \forall z \in K \}$$ is convex. To this end, let $f,g \in U(K,\epsilon)$ and consider the chord formed between them. That is, $tf + (1-t)g$ for $0 \leq t \leq 1$. To see that the entire line segment is contained in $U(K,\epsilon)$, we observe that \begin{eqnarray*} \left| tf(z) + (1-t)g(z) \right| & \leq & t \left| f(z) \right| + (1-t) \left| g(z) \right| \\ & < & \epsilon t + (1-t) \epsilon = \epsilon. \end{eqnarray*} This shows that $\mathscr{C}_D$ is locally convex.

Can someone justify why $\mathscr{C}_D$ is not globally convex?

Bounding a sum of products of binomials

Math Overflow Recent Questions - Fri, 06/02/2017 - 05:15

A problem led me to the following arithmetic counting problem: Let $n$ be a fixed number, $k\in\{1,\ldots,n\}$ and $Q_{k}$ the set of integer tuples $(t,u,v,w)$ of "compositions" of $k$, meaning $k=t+u+v+w$, but where contrary to the standard definition,

  • we allow $t,u,v,w\geqslant0$ such that, e.g., $k=(k-1)+1+0+0$ and $k=(k-1)+0+0+1$ count as two different "compositions" and not as one, $k=(k-1)+1$
  • we assume the constraints $n-k\geqslant u$ and $u\geqslant v$}.

(So the first bullet increases the number of possibilities compared to compositions of $k$, while the second one decreases again the number of possibilities.)

Question 1: What techniques would help me to obtain a good upper bound on the size of $Q_{k}$?

Question 2: Let \begin{alignat*}{1} f(k):=\sum_{(t,u,v,w)\in Q_{k}}\binom{k}{t}\cdot\binom{n-k}{u}u!\cdot\binom{u}{v}v!\,\cdot\,!w\cdot\left(\frac{1}{2}\right)^{t}\left(\frac{1}{2n-2}\right)^{k-t} & , \end{alignat*}

where, as usual, we denote with $!w$ the number of derangements of $w$-many objects. With what (general) techniques can I achieve some upper bounds, in closed-form, on $f$?

I am not well versed in combinatorics so the easier the potential technique, the better. I'd also be very grateful of the technique were citeable somewhere.

(Subquestion 2.5: There is a general technique to evaluate similar sums of products involving binomials, by Egorychev, in his book "Integral Representation and the Computation of Combinatorial Sum", translated from Russian. It would probably take me weeks to really understand his techniques (my complex analysis is rather rusty and he uses quite a lot of it), so can you tell me, if his technique is likely to help me establish an upper bound for $f$?)

Linear cost traveling salesman heuristic

Math Overflow Recent Questions - Fri, 06/02/2017 - 03:44

I have a sequence of points $y_1,\ldots,y_n\in \mathbb{R}^3$ and want to approximately minimise $$ \sum_{i=1}^{n-1}|y_i-y_{i+1}|. $$ I have a budget of $\mathcal{O}(n\log(n)^p)$ for some $p\in\mathbb{N}$. What would be the best heuristic for this endeavour?

Maximality with respect to having no marriage

Math Overflow Recent Questions - Fri, 06/02/2017 - 03:10

Let $A,B\neq \emptyset$ be disjoint and suppose $G = (A\cup B, E)$ is bipartite where for all $e\in E$ we have $e\cap A \neq \emptyset\neq e\cap B$. For $a\in A$ we set $N_G(a) = \{b\in B: (\exists e\in E)\{a,b\}\in e\}$, and for $S\subseteq A$ let $N_G(S) = \bigcup\{N_G(a):a\in S\}$.

A matching is a set $M$ of pairwise disjoint edges, and a marriage of $A$ is a matching such that $A\subseteq \bigcup M$. For finite bipartite graphs, Hall's marriage theorem says that there is a marriage if

(Hall's condition:) for all $S\subseteq A$ we have $|S| \leq |N_G(S)|$.

Let $A,B$ be infinite and disjoint and set $[A,B]_2 = \big\{\{a,b\}: a\in A, b\in B\big\}$. Consider the collection ${\cal E}$ of sets $E\subseteq [A,B]_2$ such that Hall's condition holds for $(A\cup B,E)$ but $(A\cup B,E)$ has no marriage.

Question. Does every element of the poset $({\cal E},\subseteq)$ lie below a maximal element?

Constructing ramified covers with prescribed multiplicities at ramification points

Math Overflow Recent Questions - Fri, 06/02/2017 - 03:01

Let $Y$ be a smooth projective curve defined over number field $K$. Let $P_1,\dots ,P_m$ be some $K$-points to which we will associate "multiplicities" $m_i\in\{ 2,3,\dots \}$ for $i=1,\dots ,m$. The other $K$ points (and even those not defined over $K$ ) are considered with "multiplicities" $1$. We want to find a smooth projective curve $X$ defined over some number field $M$, and a morphism $\phi :X\to Y$ such that $e_{T}=m_{\phi (T)}$ for all closed points $T$ of $X$. Here $e_T$ is the ramification at the point $T$. We assume the "characteristics" of $Y$ is negative, that is $2-2g-\sum _P(1-\frac{1}{m_P})<0$. Here $g$ is the genus.

In Serre's Topics in Galois theory, he solves the problem for the Riemann surfaces. He doesn't require that the first curve is projective. The universal cover is $\mathbb{H}$, the Poincaré half-plane, if the "characteristics" is negative.

The cover here is infinite. Of course, I need a finite. Therefore somehow I should quotient it by some group and get a smooth projective curve. Then I know how to use GAGA to finish the problem.

Any ideas how to make the missing step?

Contracted product of torsors

Math Overflow Recent Questions - Fri, 06/02/2017 - 02:57

Given a group $G$ and a left $G$-set $X$, then we can make $X$ a right $G$-set defining the action as $xg:=g^{-1}x$ , or if you prefer we are considering the opposite group $G^{op}$ to make the left action a right action.

Now, talking about $G$-torsors, there is the notion of contracted product, i.e. a product of torsors. In the literature the definition of the product is:

Given X a $(G,H)$-bitorsor and $Y$ a $(H,G)$-bitorsor then $X \times^H Y=\frac{X \times Y}{\Delta(H)}$ is a $G$-bitorsor

Here $\Delta(H)$ is the diagonal action, or equivalently the equivalence relation is defined by $(xh,y) \sim (x,hy)$. One can show that this product is associative.

Now, my question is, here there is a definition based on bitorsors, but what if, given $X,Y$ left $G$-torsor, i take the quotient of the fibre product of $X,Y$ by the equivalence relation $(gx,y) \sim (x,gy)$? Why this is not a left $G$-torsor?

In the case where $G$ is abelian, we can note that every $G$-torsor is a $G$-bitorsor and the product will be a $G$-bitorsor as well. But in the non abelian case? I suppose that there could be a problem of associativity with this product, but I don't know if it is the case and why.

Open/closed immersion and quotient stacks

Math Overflow Recent Questions - Fri, 06/02/2017 - 02:22

I'm quite new to stacks, so this might be very easy. In particular, if there is a canonical reference I can consult for these things, please feel free to point it out.

Let $f:X\to Y$ be a $G$-equivariant morphism of schemes, for $G$ an affine group scheme (please feel free to relax the assumptions if it is natural, but this is the case I care about). Under what conditions on $f$ is the map between quotient stacks $[X/G]\to [Y/G]$ an open/closed immersion? Does it suffice for $f$ to be open/closed? In case not, does that become true after adding some reasonable assumptions on $G$?


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