Recent MathOverflow Questions

Measurability of the product on particular topological vector spaces

Math Overflow Recent Questions - Sat, 04/21/2018 - 07:22

Let $X$ be a topological vector space. Let us say that $X$ has property P if there exists a sequence of closed subsets $\{X_n\}$ such that

1- $X=\bigcup X_n$

2- The relative topology is both metrizable and second countable on $X_n$'s.

Q. Assume $X$ satisfying P property. Let $m: X\times X\to X$ be an associative multiplication. Is $m$ measurable?

Open Gromov-Witten invariants via lozalization with $\mathbb{C}^{*}$ (not $S^1$) action

Math Overflow Recent Questions - Sat, 04/21/2018 - 07:16

Amplitudes of open A-model on a Calabi-Yau 3-fold $X$ with branes are given by the open Gromov-Witten invariants of $X$. It is known how to compute them if there is a toric action on a manifold, which preserves all the branes and the image of the curve (see Katz, Liu, Enumerative Geometry of Stable Maps with Lagrangian Boundary Conditions and Multiple Covers of the Disc and Graber, Zaslow, Open-String Gromov-Witten Invariants: Calculations and a Mirror "Theorem" for examples). It is done via equivariant localization on the fixed points of the toric action.

All the references I read, consider the toric action with integer weights (thus $S^1$), while the brane configuration I have is preserved only by certain $\mathbb{C}^{*}$ action, which rotates some of the coordinates by $\mathrm{e}^{\mathrm{i} b}$, while the others by $\mathrm{e}^{\mathrm{i/b}}$, for irrational $b$. If I try to follow the usual procedure used for the cases of $S^1$ action, I get the result which contradicts my physical expectations. I suspect that the reason is that this procedure must be modified appropriately in order to deal with irrational weights.

Could anybody recommend any reference in which such a problem is discussed?

Polynomials dense with primes

Math Overflow Recent Questions - Sat, 04/21/2018 - 06:32

Let $p(n)$ be a polynomial with integer coefficients. Define $\Delta( p(n) )$, the prime density of $p(n)$, to be the limit of the ratio with respect to $n$ of the number of primes $p(k)$ generated when the polynomial is evaluated at the natural numbers $k=1,2,\ldots,n$: $$ \Delta( p(n) ) \;=\; \lim_{n \to \infty} \frac{ \textrm{number of } p(k), k \le n, \textrm{that are prime}} {n} $$ For example, Euler's polynomial $p(n)=n^2+n+41$ starts out with ratio $1$, but then diminishes beyond $n=39$:           And it continues to diminish ...           ... and by $n=10^7$ has reached $\Delta=0.22$.

Q. What is the largest known $\Delta( p(n) )$ over all polynomials $p(n)$?

In particular, are there any polynomials known to have $\Delta > 0$?

Maybe these questions can be answered assuming one or more conjectures?

Notation for an upper bound [on hold]

Math Overflow Recent Questions - Sat, 04/21/2018 - 06:20

What is the meaning of this notation? In particular the top-cut square brackets. To put it in some context: the upper-bound of $\alpha$ is $\left \lfloor \sqrt[4]{\beta_{0}/4} \right \rfloor$

Growth order of numbers whose prime factors are all congruent to +1 or -1 modulo 8

Math Overflow Recent Questions - Sat, 04/21/2018 - 05:27

Here are the numbers whose prime factors all congruent to $\pm 1\pmod 8$:

My questions are:

(1) What is the order of growth of these numbers? That is, what is the order of magnitude of the $n$th smallest among them?

(2) For the numbers whose prime factors are all congruent to $\pm 3\pmod 8$, is the order of growth the same as in (1)?

Surjective group homomorphism from $\text{Sym}(X)$ onto $\mathbb{Z}$ [on hold]

Math Overflow Recent Questions - Sat, 04/21/2018 - 02:08

For any non-empty set $X$ let $\text{Sym}(X)$ denote the group of bijections $f:X\to X$ with composition.

Is there an infinite set $X$ and a surjective group homomorphism $\pi: \text{Sym}(X)\to \mathbb{Z}$?

Is there a Feynman-Kac formula for vector-valued Schrödinger operators?

Math Overflow Recent Questions - Fri, 04/20/2018 - 19:59

Given a vector function $$f=(f_1,\ldots,f_n)\in L^2(\mathbb R,\mathbb R^n)$$ (for some $n\in\mathbb N$), let us define $$\Delta f:=(\Delta f_1,\ldots,\Delta f_n),$$ where $\Delta$ is the Laplacian operator, and let $Q:\mathbb R\to\mathbb R^{n\times n}$ be a potential taking values in symmetric $n\times n$ matrices. I'm interested in vector Schrödinger operators of the form $$Hf=-\Delta f+Qf,\qquad f\in L^2(\mathbb R,\mathbb R^n).$$

Question. Is there a Feynman-Kac type formula known for $H$'s semigroup in the case where $Q(x)$ is not necessarily diagonal?

(Note: I specify abote that I'm interested in the case where $Q$ is not diagonal; if $Q=\mathrm{diag}(Q_1,\ldots,Q_n)$, then $$Hf=(-\Delta f_1+Q_1 f_1,\ldots,-\Delta f_1+Q_1 f_1),$$ in which case we can simply apply the one-dimensional Feynman-Kac formula to each component.)

Proving tail bound of a random projection

Math Overflow Recent Questions - Fri, 04/20/2018 - 19:27

Let $R \in \mathbb{R}^{n,d}$ be a random Gaussian matrix comprised of independent $\operatorname{N}(0,\frac{1}{n})$ entries, let $\textbf{w}$ and $\textbf{x}$ be vectors in $\mathbb{R}^d$, and let $\epsilon \in (0,1)$. In Shi et al. 2012, it is claimed in Lemma 10 that $$\operatorname{Pr}(1-\epsilon \leq \frac{\|R\textbf{x}\|^2}{\|\textbf{x}\|^2} \leq 1 + \epsilon) \geq 1 - 2\exp(-\frac{n}{2}(\frac{\epsilon^2}{2}-\frac{\epsilon^3}{3}))$$

I don't understand how this lower bound was obtained. It seems to me that when you want to find a lower bound on a probability in a situation like this, you should try to minimize $$\operatorname{Pr}\bigg(\frac{\|R\textbf{x}\|^2}{\|\textbf{x}\|^2} \leq 1 + \epsilon \bigg)$$ and maximize $$\operatorname{Pr}\bigg(\frac{\|R\textbf{x}\|^2}{\|\textbf{x}\|^2} \leq 1 - \epsilon \bigg)$$

If that's what Shi et al. did in this situation, I don't understand how they did it.

Numerical range of a pair of operators

Math Overflow Recent Questions - Fri, 04/20/2018 - 13:26

Let $M\in \mathcal{B}(F)^+$ and $S_1,S_2\in \mathcal{B}(F)$ where $F$ is an infinite-dimensional complex Hilbert space. We consider $$W(S_1,S_2)=\{(\langle MS_1x, x\rangle,\langle MS_2x, x\rangle);\;\;x\in F,\;\langle Mx,x \rangle=1\}.$$

Assume that $S_1$ and $S_2$ are $M$-self adjoint (i.e. $MS_1=S_1^*M$ and $MS_2=S_2^*M$) such that $S_1S_2=S_2S_1$ and $MS_k=S_kM,\,k=1,2.$ It is possible to show that $W(S_1,S_2)$ is convex?

Since $S_1$ and $S_2$ are commuting and $M$-self adjoint such that $MS_k=S_kM,\,k=1,2$, then $\exists\,(X,\mu)$, $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that $$U(MS_k)U^*h=\varphi_kh,\;\forall h\in L^2(\mu),\,k=1,2.$$

Let $\lambda=(\lambda_1,\lambda_2)$, $\eta=(\eta_1,\eta_2)$ be any pair of point in $W(S_1,S_2)$, then there exist $f,g \in L^2(X, \mu)$ such that for all $k= 1,2$ we have $$\lambda_k=\displaystyle\int_X \varphi_k|f|^2d\mu\;\;\text{and}\;\;\eta_k=\displaystyle\int_X \varphi_k|g|^2d\mu.$$ Let $\xi=t\lambda+(1-t)\eta,\;t\in[0,1]$. So $$\xi_k=t\displaystyle\int_X \varphi_k|f|^2d\mu+(1-t)\displaystyle\int_X \varphi_k|g|^2d\mu.$$

It is possible to show that $\xi\in W(S_1,S_2)$?

A cohomology associated to a Riemannian manifold

Math Overflow Recent Questions - Fri, 04/20/2018 - 12:59

Let $N$ be a compact Riemanian manifold and $G$ be its isometry group. Let $M=\chi^{\infty}(N)$ be the space of smooth vector fields on $N$. There is a natural right action of $G$ on $M$ with $X.g=g^*(X),\; g\in G, X\in M$, the push forward of $X$ under the isometry $g$. So $M$ is a $G$ module.

How can one express the group cohomologies $H^n(G,M)$ explicitely?Is there a reference which contain such computations? What can be said about a riemannian manifold whose all cohomology groups $H^n(G,M)$ vanish?

Edit: according to the comment of Neal I understand the following part of the previous version is a trivial question:

Does this sequence of cohomologies determine the geometry of $N$? Namely is it true to say that two nonisometric metrics on $N$ give two different cohomology sequence?

Counting primitive lattice points

Math Overflow Recent Questions - Fri, 04/20/2018 - 07:49

In Lemma 2 of [1], Heath-Brown proves the following (I state a simplified version of a more general result):

Let $\Lambda \subset \mathbb{Z}^2$ be a lattice of determinant $d(\Lambda)$. Then $$\# \{ (x_1,x_2) \in \Lambda: \max_i |x_i| \leq B, \gcd(x_1,x_2) = 1\} \leq 16\left (\frac{B^2}{d(\Lambda)} + 1\right).$$

My question is whether this generalises to arbitrary dimensions.

Does an analogous result hold for lattices in $\Lambda \subset \mathbb{Z}^n$? Namely, is $$\# \{ (x_1,\dots, x_n) \in \Lambda: \max_i |x_i| \leq B, \gcd(x_1,\dots,x_n) = 1\} \leq C_n\left (\frac{B^n}{d(\Lambda)} + 1 \right)$$ for some constant $C_n$?

If it helps, I'm primarily interested in the case $n=3$.

Obviously I'm aware of standard lattice point counting techniques, but these usually give an error term of the shape $O(\text{boundary of the region/first successive minima})$, and I don't know how to control this in my case. So I'm just looking for uniform upper bounds where this term doesn't appear.

[1] Heath-Brown - Diophantine approximation with Square-free numbers

Homology of $\mathrm{PGL}_2(F)$

Math Overflow Recent Questions - Fri, 04/20/2018 - 03:58

Here is a hopelessly naive question. Please point me to the relevant literature!

Let $F$ be any field. The Cartan subgroup of $\mathrm{PGL}_2(F)$ is $F^\times\rtimes \Sigma_2$. Let $X_2(F)$ be the homotopy cofiber of $B(F^\times\rtimes\Sigma_2)\to B\mathrm{PGL}_2(F)$, so there is a long exact sequence $$ \ldots\to H_i(F^\times\rtimes\Sigma_2)\to H_i(\mathrm{PGL}_2(F))\to H_i(X_2(F))\to \ldots $$ (I'm not sure if I should write $H_i(G)$ or $H_i(BG)$...). If I am not making a stupid mistake, then the known computations of the homology of $\mathrm{GL}_2(F)$ going up to the first unstable group $H_3(\mathrm{GL}_2(F))$ can be summarized by saying that $H_i(X_2(F))=0$ for $i=1,2$, and $H_3(X_2(F))=\mathfrak p(F)$ is the pre-Bloch group of $F$. (The pre-Bloch group is the quotient of $\mathbb Z[F^\times]$ by the ``$5$-term relations''.) (I might be off by some $2$-torsion.) By Hurewicz (and the check that $\pi_1 X_2(F)=0$), the same holds true for the homotopy groups of $X_2(F)$.

As far as I could find, little is known about the homology of $\mathrm{GL}_2(F)$ or $\mathrm{PGL}_2(F)$ in degrees $>3$. The rational structure could be determined by the following very naive question.

Question. Are the homotopy groups $\pi_i X_2(F)$ bounded torsion for $i\neq 3$?

I might even expect the implicit bound to be independent of $F$ (but of course depend on $i$; the order of magnitude should be $i!$).

The only evidence I have is that a back-of-the-envelope calculation seems to suggest that this holds true for finite fields. In that case the order of $\mathrm{PGL}_2(\mathbb F_q)$ is the product of $q-1$, $q$ and $q+1$. Localized at $q-1$, the homology agrees with the homology of the Cartan; localized at $q$, it is only in degrees larger than $q$ (which is OK as the bound on torsion may depend on $i$), and localized at $q+1$, the homology seems to agree with the homology of a $K(\mathbb Z/(q+1),3)$ (up to bounded torsion in each degree), which is roughly $\mathbb Z/(q+1)$ in all degrees $\equiv 3\mod 4$ (and is bounded torsion in other degrees). The pre-Bloch group $\mathfrak p(\mathbb F_q)$ is also $\mathbb Z/(q+1)$ up to $2$-torsion, so things add up.

The case of $F=\mathbb Q$ might be amenable to computations, but I was unable to do those. Are there any relevant results about $H_4(\mathrm{GL}_2(F))$ that might shed light on $\pi_4 X_2(F)$?

A final remark: I expect that it is critical that $F$ is a field. The similar statement should be false already for discrete valuation rings, or $\mathbb Z[\tfrac 1n]$ (which is why the case of $F=\mathbb Q$ is not so easy to compute; I presume one would try to first compute for all $\mathbb Z[\tfrac 1n]$ and then pass to a colimit, and the desired structure should only appear in the colimit).

Infinite-time Turing machines and the formal Church's thesis

Math Overflow Recent Questions - Thu, 04/19/2018 - 12:16

Infinite-time Turing machines are known to either halt or loop in countable time.

In the spirit of double-negation translation, is there a statement which is: classically equivalent to this; consistent with the formal Church's thesis, that all total functions on the integers are recursive; and (this part is of course subjective) which actually tells us something "meaningful or informative" beyond the vacuous sense in which any such translation does?

Stabilizer of two short exact sequences at the same time

Math Overflow Recent Questions - Thu, 04/19/2018 - 07:51

For two short exact sequences of say, finitely generated modules of some ring, $0\rightarrow N\xrightarrow{a} R\xrightarrow{b} M\rightarrow0, 0\rightarrow K\xrightarrow{a'}R\xrightarrow{b'}L\rightarrow0.$ Let $\phi \in Aut(R)$ and $\phi$ acting on these two short exact sequences by $\phi.a=\phi\circ a, \phi.b=b\circ\phi^{-1},\phi.a'=\phi\circ a', \phi.b'=b'\circ\phi^{-1}.$ My question is that the book I am reading says that by some diagram chasing, the cardinality of stabilizer of this action of $Aut(R)$ is the same as the cardinality of $Hom(Coker\space b'a, Ker\space b'a).$ I have no idea how to do explain it, can somebody help me?

What, mathematically speaking, does it mean to say that the continuation monad can simulate all monads?

Math Overflow Recent Questions - Thu, 04/19/2018 - 04:47

In various places it is stated that the continuation monad can simulate all monads in some sense (see for example In particular, in it is claimed that

any monad whose unit and extension operations are expressible as purely functional terms can be embedded in a call-by-value language with “composable continuations”.

I was wondering what (Category-theoretic) mathematical content these claims of simulation have, and what precisely they show us about monads and categories in mathematical terms. In what ways is the continuation monad special, mathematically, compared to other monads, if at all? (I seem to remember some connection between the Yoneda embedding and continuations which might be relevant, although I don't know)

One other relevant fact might be that the continuation monad is the monad which takes individuals $\alpha$ to the principal ultrafilters containing them (that is it provides the map $\alpha \mapsto (\alpha \rightarrow \omega) \rightarrow \omega$) )

Edit: I have been asked to explain what I mean by the continuation monad. Suppose we have a monad mapping types of the simply typed lambda calculus to types of the simply typed lambda calculus (the relevant types of this calculus are of two kinds: (1) the basic types, (I.e, the type $e$ of individuals type belonging to domain $D_e$ and the type of truth values belonging to domain $D_t$) and, (2) for all basic types $\alpha, \beta$, the type of functions between objects of type $\alpha$ and objects of type $\beta$ belonging to domain $D_{\beta}^{D_{\alpha}}$). Let $\alpha, \beta$ denote types and $\rightarrow$ a mapping between types. Let $a : \alpha \hspace{0.2cm}$ (or $b: \beta)$ indicate that $a\hspace{0.2cm}$ (or $b$) is an expression of type $\alpha \hspace{0.2cm}$ (or $\beta)$. Let $\lambda x. t$ denote a function from objects of the type of the variable $x$, to objects of the type of $t$, as in the simply typed lambda calculus. Then a continuation monad is a structure $\thinspace(\mathbb{M}, \eta, ⋆)\thinspace$, with $\mathbb{M}$ an endofunctor on the category of types of the simply typed lambda calculus, $\eta$ the unit (a natural transformation) and ⋆ the binary operation of the monoid) such that:

$$\mathbb{M} \thinspace α = (α → ω) → ω, \hspace{1cm} ∀α$$ $$η(a) = λc. c(a) : \mathbb{M} \thinspace α \hspace{1cm} ∀a : α $$ $$m ⋆ k = λc. m (λa. k(a)(c)): \mathbb{M}\thinspace β \hspace{1cm} ∀m : \mathbb{M}\thinspace α, k : α → \mathbb{M}\thinspace β . $$

The continuation of an expression $a$ is $\eta(a) = (a \rightarrow \omega) \rightarrow \omega$.

Edit 2: let $ \omega$ denote some fixed type, such as the type of truth values (i.e, $\{ \top, \bot\}$)

Bredon cohomology of $\mathbb{S}^\sigma$

Math Overflow Recent Questions - Wed, 04/18/2018 - 12:17

I tried to compute Bredon cohomology of $\mathbb{S}^\sigma$, where $\sigma$ is a sign representation of $\mathbb{Z}/2$, following first chapter and first construction of cohomology from Bredon's "Equivariant cohomology theories". Could somebody please verify it, at least the result?

Throughout $G$ means $\mathbb{Z}/2$.

So I assume the following: $G$-CW structure is obvious, given by two points with trivial action as 0-cells and two arcs with swapping action as 1-cells. I am using simple coefficient system $\mathcal{L}$ on it, that is my functor from "cellular category" to abelian groups factors through some coefficient system $M$. $M$ consists of two groups $M(*)$ - trivial $G$-module and $M(G)$ - $G$-module, and an equivariant map $\epsilon:M(*)\rightarrow M(G)$.

$C^0(\mathbb{S}^\sigma;\mathcal{L})$ consists of the functions $f:\{e^0_1,e^0_2\}\rightarrow M(*)$. Since action of G is trivial on 0-cells, induced action on 0-chains is also trivial, therefore $C^0(\mathbb{S}^\sigma;\mathcal{L})=C^0_G(\mathbb{S}^\sigma;\mathcal{L})=M(*)^2$. $C^1(\mathbb{S}^\sigma;\mathcal{L})$ consists of the functions $f:\{e^1_1,e^1_2\}\rightarrow M(G)$. Action on 1-cells is non-trivial (even free), so $C^1_G(\mathbb{S}^\sigma;\mathcal{L})$ consists of equivariant $f$'s. Thus $C^1_G(\mathbb{S}^\sigma;\mathcal{L})=M(1)$.

The only non-trivial differential is $\delta :C^0\rightarrow C^1$ and is given by $(\delta f)(\tau)=\pm\epsilon(f(e^0_1))\mp\epsilon(f(e^0_2))$. Here $\tau$ of course means any of two 1-dimensional cells.

So $H^0_G(\mathbb{S}^\sigma;\mathcal{L})=M(*)$ and $H^1_G(\mathbb{S}^\sigma;\mathcal{L})=M(G)/M(G)^G$ - but for this I have to assume that $\epsilon$ is an iso on $M(G)^G$.

If this is not "mathoverflow" question, I can ask it also on MathStack.

Can we deduce the sign of this integral which includes cosine transform?

Math Overflow Recent Questions - Wed, 04/18/2018 - 11:50

Let's $v(t)$ be a negative real function and consider the following integral with $f(t)$ complex (rapidly decreasing at infinity):

$$A= \int\limits_{0}^\infty \frac{1}{x} \int\limits_{x}^\infty \Big(v(t)f'(t) + \frac{1}{2} v'(t)f(t) \Big) \overline{f(t)} + \overline{\Big(v(t)f'(t) + \frac{1}{2} v'(t)f(t) \Big) }f(t) dt dx$$

Providing the integral is well defined, it is not hard to see that $A$ is always positive ($v(t)$ is negative by hypothesis) by an integration by parts we directly have:

$$A= \int\limits_{0}^\infty \frac{1}{x} [v(t) f(t) \overline{f(t)} ]_{x}^\infty dx = - \int\limits_{0}^\infty \frac{1}{x} v(x) f(x) \overline{f(t)} dx$$

Now, can we say something about the sign of the same expression where we consider the cosine transform of precedent functions ? (we note $\mathcal{F_c}$ the cosine transform)

$$B=\int\limits_{0}^\infty \frac{1}{x} \int\limits_{x}^\infty \mathcal{F_c}\Big(v(t)f'(t) + \frac{1}{2} v'(t)f(t) \Big) \overline{\mathcal{F_c}(f(t))} + \overline{\mathcal{F_c}\Big(v(t)f'(t) + \frac{1}{2} v'(t)f(t) \Big) } \mathcal{F_c}(f(t)) dt dx$$

I do not manage to use Parseval's equality. May be we cannot say anything about the sign of $B$ ? Any idea on how to investigate this sign ? Or conditions on $v$ and $f$ to have it also positive ?

(Note that to have integral $A$ converging we need : $$\int\limits_{0}^\infty \Big(v(t)f'(t) + \frac{1}{2} v'(t)f(t) \Big) \overline{f(t)} + \overline{\Big(v(t)f'(t) + \frac{1}{2} v'(t)f(t) \Big) }f(t) dt =0$$)

How does raising the zeta function to 1/lnx transform the properties of the zeta function? [on hold]

Math Overflow Recent Questions - Wed, 04/18/2018 - 11:43

How does pre-exponentiating the zeta function by 1/lnx changes the function in terms of its properties. How are the zeros changed, if at all?

The stalk of the sheaf of sheaves

Math Overflow Recent Questions - Wed, 04/18/2018 - 11:40

For $X$ a reasonable topological space denote by $Sh(X)$ the derived (stable $\infty$-) category of sheaves of vector spaces (over a fixed field $k$) on $X$.

Consider the filtered diagram whose whose objects are $Sh(U)$ for all $0 \in U \subset \mathbb{R}^n$ open subsets. And morphisms are restriction functors $j^*:Sh(V) \to Sh(U)$ for every inclusion $j : U \to V$.

Question: What is the ($\infty$-)colimit of this diagram?

In slightly less precise terms i'm looking for the stalk at $0$ of the sheaf of categories on $\mathbb{R}^n$ which assigns to any open set the category of sheaves on that open.

I have a hunch that this colimit is equivalent to the category of $\mathbb{R}_{>0}$-equivariant sheaves on $\mathbb{R}^n$ but I have no idea how to prove this...

Optimization with bounds on the control and its derivative

Math Overflow Recent Questions - Wed, 04/18/2018 - 11:26

I would like to understand the following optimization problem. Let $F(t,x)$ be a continuous function defined on $[0,1]\times [0,1]$, which is increasing in $t$ and convex in $x$ (I have in mind $F(t,x)=x(t-\frac{x}{2})$), the goal is to solve: \begin{align*} \min_{\alpha \in \mathcal{C}_A} \int_0^1F(t,x_t)dt \end{align*} subject to : \begin{align*} x_t \in \arg \max_{x \in [0,1]} F(t,x)-\alpha(x)1\!\!1_{x\neq 0} \end{align*} and $\mathcal{C}_A$ is the class of differentiable functions such that: \begin{align*} \int_0^1\alpha(x_t)dt &\leq A,\\ \forall x\in [0,1], 0 \leq \alpha(x) & \leq K, \text{and }\vert \alpha'(x)\vert \leq K, \end{align*} where $K> 0$ is a fixed constant.

What annoys me the most is the boundness conditions on the function $\alpha$ and its derivative. Would you have any hint or reference which would help me to get started ?


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