Recent MathOverflow Questions

How to compute the probability of a multivariate Gaussian when its covariance matrix is singular?

Math Overflow Recent Questions - Thu, 09/21/2017 - 20:45

The probability of a multivariate Gaussian is: $$\frac{1}{(2\pi)^{D/2}|\Sigma|^{1/2}}\exp \{-\frac{1}{2}x^T\Sigma^{-1}x\}$$ where $D$ is the dimention of $x$.But I don't know how to compute it when $\Sigma$ is a singular matrix.For example, $$ \Sigma= \left( \begin{matrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1&2 \end{matrix} \right) $$ I think this kind of covariance matrix exists in real problems. So how to compute the probability?

Exponential Growth [on hold]

Math Overflow Recent Questions - Thu, 09/21/2017 - 18:11

Assume the world population will continue to grow exponentially with a growth constant k=0.0132 (corresponding to a doubling time of about 52 years),

it takes 1/2 acre of land to supply food for one person, and

there are 13,500,000 square miles of arable land in the world.

How long will it be before the world reaches the maximum population? Note: There were 6.06 billion people in the year 2000 and 1 square mile is 640 acres.

The maximum population will be reached some time in the year:__________

Hint: Convert .5 acres of land per person (for food) to the number of square miles needed per person. Use this and the number of arable square miles to get the maximum number of people which could exist on Earth. Proceed as you have in previous problems involving exponential growth.

How to read regular expressions? [on hold]

Math Overflow Recent Questions - Thu, 09/21/2017 - 17:40

Consider the following regular expressions

  1. letter = a + b + c + d

  2. LETTER = A + B + C + D

  3. digit = 4+5+6+7+9
  4. R1 = (LETTER + digit + !) • (letter • digit)∗• (letter + digit)
  5. R2 = (LETTER + digit + ?) • (letter + digit)∗• (LETTER + digit)
  6. R3 = (digit + ? + !) • (letter∗• digit∗)∗• (LETTER + letter)
  7. R4 = (digit • letter + ? + !) • (letter∗• digit∗) • (LETTER)

I'm not sure what some of the characters mean ('+' , '•' , '*')?

Do I read regex backwards when check if a string is an element of L(R1)?


R1 = (LETTER + digit + !) • (letter • digit)∗• (letter + digit)

!5aA is an element of L(R1) I would think, but that's just a guess.

If someone can point me in the right direction to read a regex like this it would be greatly appreciated.

Notation convention of norm of a space

Math Overflow Recent Questions - Thu, 09/21/2017 - 17:24

I have a notation convention question. So we can consider a space-time problem, i.e. if we consider $f(t,x)$, we consider $t>0$ and $x \in \mathbb R^n$. For $\alpha \in \mathbb R$, we define $S_{\alpha} = \{f \in L_{loc}^{1}: (1+|x|)^{\alpha}f \in L^{\infty}\}$. Then is $\|f\|_{S_{\alpha}}=\|(1+|\cdot|)^{\alpha}f\|_{\infty}$?

Alternative description of $K$-theory of locally compact spaces using sequences of bundles

Math Overflow Recent Questions - Thu, 09/21/2017 - 16:43

In this paper Aityah, Bott and Shapiro give an alternative definition of (relative) $K$-theory groups $K(X,Y)$ using sequences of bundles (this group is denoted by $L_n(X,Y)$ where $n$ is the length of such sequence). This treatment deal with compact pairs, i.e. it gives an alternative description of $K(X,Y)$ where $X$ is compact and $Y \subset X$ is closed. In locally compact setting, when $Y=\emptyset$, the $K$-theory group $K(X)$ is defined as $\tilde{K}(X^+)$ (reduced K-theory of one point compactification of $X$). In the book "Spin Geometry" by Lawson and Michelsohn one can find the appropriate definition of $L_n(X)$ when $X$ is only locally compact compact: this is formed using sequence of bundles which are exact out of some compact set.

Is it true that if $X$ is only locally compact then $K(X)$ and $L_n(X)$ are isomorphic? If so, do the standard "difference element" construction (explained in the above paper for compact pairs) do the job?

Semigroups of nondecreasing functions

Math Overflow Recent Questions - Thu, 09/21/2017 - 15:55

Consider some partially ordered set $(E,\leq)$. Assume either that it is countable with the discrete topology, or that it has some topology compatible with the order, preferably one that makes it into a Polish space. Now consider the semigroup of (continuous) order-preserving functions from $E$ to $E$. What is known about the structure of this thing?

That is the most general set-up. The most specific examples I'm interested in are $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{N}^n$, and maybe $\mathbb{R}^n$. So some restrictions which could also be interesting if they have better results:

  1. Assume it is totally ordered, not just partially ordered
  2. Maybe even well-ordered
  3. It could have a minimal element, which we require to be mapped to itself
  4. Perhaps restrict to the functions which in some suitable sense go to infinity
  5. The case where we require them to be strictly increasing is less interesting, but if there are any results there I'm all ears

The reason I care about these semigroups is because I want to consider probability measures and random walks on them, which is why I had the restriction of it having a nice topology. This also means the ideal result would look something like "all subsemigroups are [explicit description], they're closed and completely simple if [explicit condition]" -- or any other classification of its subsemigroups of various types.

How to prove or disprove a type of states form an overcomplete basis in the Hilbert space?

Math Overflow Recent Questions - Thu, 09/21/2017 - 15:19

I am a PhD student in Physics. Let us consider a vector in an infinite dimensional Hilbert space as

\begin{equation} |f\rangle\equiv \begin{bmatrix} 1 \\ z \\ z^2 \\ \vdots \end{bmatrix}, \end{equation}

where $z$ is a complex number. The norm of the vector $|f\rangle$ is \begin{equation} \left\lVert\langle f|f\rangle\right\rVert=1+|z|^2+|z|^4+\cdots=\frac{1}{1-|z|^2}. \end{equation}

Hence, the vector $|f\rangle$ is normalizable only when $|z|<1$, and the normalized vector should be $|z\rangle=\sqrt{1-|z|^2}|f\rangle$. The overlap between any two distinct vectors is \begin{equation} \langle z_2|z_1\rangle = \frac{\sqrt{1-|z_2|^2}\sqrt{1-|z_1|^2}}{1-z^*_2z_1}, \end{equation} which yields \begin{equation} |\langle z_2|z_1\rangle|^2=\frac{(1-|z_2|^2)(1-|z_1|^2)}{|z_1-z_2|^2+(1-|z_2|^2)(1-|z_1|^2)}\leq 1, \end{equation} Hence, any two distinct vectors are non-orthogonal, and the overlap of the two vectors is always small than one.

Now, my question is: can these vectors form a complete/overcomplete basis in the infinite dimensional Hilbert space?

For any two arbitrary vectors $|\psi\rangle$ and $|\phi\rangle$, the above statement is true when the following relation is valid \begin{equation} \langle\psi|\phi\rangle=\frac{1}{\pi}\int \langle\psi|z\rangle\langle z|\phi\rangle g(z)d^2z, \end{equation} where $g(z)$ is an unknown integration kernal. If we denote $z=re^{i\phi}$, the integration becomes \begin{align} \langle\psi|\phi\rangle&=\frac{1}{2\pi}\sum_{n,m}\langle\psi|m\rangle\langle n|\phi\rangle\int_0^1 g(r)r^{n+m+1}dr\int_0^{2\pi}e^{i(n-m)}d\phi\\ &=\frac{1}{2\pi}\sum_{n,m}\langle\psi|m\rangle\langle n|\phi\rangle\int_0^1 g(r)r^{n+m+1}dr\int_0^{2\pi}e^{i(n-m)}d\phi\\ &=\sum_{n}\langle\psi|n\rangle\langle n|\phi\rangle\int_0^1 g(r)r^{2n+1}dr, \end{align} where we have selected a kernal function $g(r)$ that is independent of the polar angle $\phi$. Hence, if we can prove that for any positive integer $n$ \begin{equation} \int_0^1 g(r)r^{2n+1}dr=1, \end{equation} then $|z\rangle$ for $|z|<1$ will form a overcomplete basis in the infinite dimensional Hilbert space. However, I can't find such $g(r)$. Therefore, I doubt that $|z\rangle$ do not form a complete basis.

Bounding the distance between two matrix power sequences

Math Overflow Recent Questions - Thu, 09/21/2017 - 14:35

Let $A,B$ be Hermitian matrices so that $0 \le A,B < I$ and also $(1-\varepsilon)(I-B)\le I - A \le (1+\varepsilon)(I-B)$.

For every $t \in \mathbb{N}$, consider the matrix $A_{t} = \sum_{i=0}^{t}A^{i}$ and likewise $B_{t} = \sum_{i=0}^{t}B^{i}$. I am interested in how "close" $A_{t}$ and $B_{t}$ are, given that we know $A_{\infty}$ and $B_{\infty}$ are close.

Formally, define $\gamma_{i}$ as the minimal $\gamma \ge 1$ satisfying $A_{i} \le \gamma B_{i}$ (provided it exists). We know that $\gamma_{0} = 1$ and that $\gamma_{t}$ approaches $1/(1-\varepsilon)$.

What about intermediate values? How does the sequence $\gamma_{0},\gamma_{1},\ldots$ behave? For scalars, this sequence is monotone, but I did find (numerically) examples of $4 \times 4$ matrices for which this sequence was not monotone, however the deviation from $\gamma_{\infty}$ was small.

Can we, under some reasonable assumptions, bound $|\gamma_{i}-\gamma_{\infty}|$ (with something smaller than a multiple of $i$)? Or, is there a natural example of $A$ and $B$ that behave badly in that sense?

Eigenvalues of the Laplacian in $L^1$ space

Math Overflow Recent Questions - Thu, 09/21/2017 - 13:55

The first eigenvalue of $-\Delta \colon W_0^{2,2} \rightarrow L^2$ is strictly positive. This is well known and proved by using the energy method. But I learned that $-\Delta$ can be extended up to $L^1$. In this case, is the first eigenvalue still strictly positive?

Invertible function and convolution

Math Overflow Recent Questions - Thu, 09/21/2017 - 13:52

Let $(u_{1}, u_{2},...)$ be vectors with length $k$ and $G$ be a $k \times n$ matrix ($1<k<n$) then function $F(.)$ is equal to: \begin{equation} F(u)= (u_{1}.G)\circledast (u_{2}.G) \end{equation} where $\circledast$ is circular convolution.

Are there any matrix $G$ and vectors $u$ such that function $F$ be invertible?

Does this sequence have a limit?

Math Overflow Recent Questions - Thu, 09/21/2017 - 13:33

I asked this question a few hours ago on MathStackExchange and there it received some attention but we still do not have a proof so I decided to ask it here also, in an unchanged form, and here it is:

Digit sums of numbers $3^m$ in base $10$ for $m=1,2,...,50$ are:


Ratios $\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$ for $m=1,2,...,49$ to three decimal places are:


Does there exist limit of the sequence $a(m)=\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$?

I cannot resist to note some kind of chebyshevness of this question (if there is one) because we know that Chebyshev proved that if limit in the prime number theorem exists then it must be equal to $1$. It could be that this is also the case here.

I also welcome any computational effort and results obtained from such an experimental work if the proof is out of reach.

Group homology $\mathrm{SL}_2$ acting on $\mathrm{Sym}^g$

Math Overflow Recent Questions - Thu, 09/21/2017 - 13:11

Let $k$ be a field. We write $\mathrm{Sym}^g(k^2)$ for the $g$-th symmetric power of the (a?) standard representation of $\mathrm{GL}_2(k)$ ($g\geq 0$ an integer). Here I consider $\mathrm{Sym}^g(k^2)$ as polynomials $P(X,Y)$ that are homogenous of degree equal to $g$ in two variables $X$ and $Y$. Then I equip $\mathrm{Sym}^g(k^2)$ with the right $\mathrm{GL}_2(k)$-action $$ P|_{\gamma}(X,Y) = P(dX-cY,aY-bX),\;\;\;\;\;\;\; \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{GL_2}(k). $$

Now suppose that $p$ is a prime and $k = \mathbf{F}_p$ is a finite field of size $p$.

Question: Is there a description of $H_0(\mathrm{SL}_2(k), \mathrm{Sym}^g(k^2))$?

To be clear, $H_0$ is the 0-th group homology; it is the largest quotient on which $\mathrm{SL}_2(k)$ acts trivially.

I would be happy with knowing the behavior of the dimension of this space (if that is easier).

I note two similar questions I know the answer to.

If instead we ask for the invariants $H^0(\mathrm{SL}_2(k),\mathrm{Sym}^g(k^2))$ then the answer is given by a theorem of Dickson at the start of the 20th century: any $\mathrm{SL}_2(k)$-invariant is a polynomial in $\theta = X^pY - Y^pX$ and $\psi = (X^{p^2}Y - XY^{p^2})/\theta$. So, you can just count the number of such polynomials of degree $g$ for any given $g$.

I observe though that in characteristic $p$ the $\mathrm{Sym}^g$'s are not self-dual (unlike in characteristic zero, cf. Is $Sym^n (V^*) \cong Sym^n (V)^\ast$ naturally in positive characteristic?). Thus knowing the dimension of the $H^0$ does not give me the dimension of the $H_0$. (And they really are different, e.g. if $g = 6$ and $p=5$.)

The second related computation I know how to do is compute $H_0(H,\mathrm{Sym}^g(k^2))$ where $H \subset \mathrm{SL}_2(k)$ is the upper-triangular matrices. The answer here is that it is $(0)$ for $g \not\equiv 0 \bmod p-1$, and if $g \equiv 0 \bmod p-1$ then it has dimension $1 + \lfloor{g/p\rfloor}$. Explicitly, this space of coinvariants is generated as a vector space by the images of the vectors $(Y^p-X^{p-1}Y)^iY^{g-pi}$ for $i = 0,1,2,\dotsc,\lfloor{g/p\rfloor}$.

This already implies an answer to my question, sometimes: if $g \not\equiv 0 \bmod p-1$, then $H_0(\mathrm{SL}_2(k),\mathrm{Sym}^g(k^2)) = (0)$. Preliminary calculations on a computer suggest that, for $p$ an odd prime and $g \equiv 0 \bmod p-1$, dimension of the $H_0$ grows by $1$ after replacing $g$ by $g + p(p+1)(p-1)/2$. Perhaps someone even sees how to bootstrap the result from $H$ to the whole $\mathrm{SL}_2$, but I don't.

(I added the number theory tag is there are people who study modular symbols that may know the answer.)

Gap Between Abscissae of Conditional Convergence and Holomorphicity for Dirichlet Series

Math Overflow Recent Questions - Thu, 09/21/2017 - 12:47

For a Dirichlet series, $D = \sum_n a_n n^{-s}$ we may define the abscissae, in (non-strictly) increasing order

  • $\sigma_c(D) = \inf\{\sigma : D \text{ converges in } \mathrm{Re} s > \sigma \}$, the abscissa of conditional convergence,
  • $\sigma_b(D) = \inf\{\sigma : D \text{ defines a bounded, holomorphic (in $s$) function in } \mathrm{Re} s > \sigma \}$, the abscissa of boundedness,
  • $\sigma_u(D) = \inf\{\sigma : D \text{ converges uniformly (in $s$) in } \mathrm{Re} s > \sigma \}$, the abscissa of uniform convergence, and
  • $\sigma_a(D) = \inf\{\sigma : D \text{ converges absolutely in } \mathrm{Re} s > \sigma \}$, the abscissa of absolute convergence.

H. Bohr [B1913] studies these, finding $\sigma_b(D) = \sigma_u(D)$. He then turns to the gap $\sigma_a(D) - \sigma_u(D)$. D. Carando and P. Sevilla-Peris [CS2014] review Bohr's methods, the result of H.F. Bohnenblust and E. Hille [BH1931] that the supremum of this gap over all $D$ is $\frac{1}{2}$, and then generalize the results. (This gap has previously appeared on MathOverflow.)

Is anything known about the "other" gap, $\sigma_u(D) - \sigma_c(D)$, and its relation with Bohr's gap (beyond the elementary: the sum of the two gaps is at most $1$)?


[B1913] H. Bohr. Über die gleichmäßige Konvergenz Dirichletscher Reihen. J. Reine Angew. Math., 143:203–211, 1913.

[BH1931] H. F. Bohnenblust and E. Hille. On the absolute convergence of Dirichlet series. Ann. of Math. (2), 32(3):600–622, 1931. MR 1503020.

[CS2014] D. Carando and P. Sevilla-Peris. On the convergence of some classes of Dirichlet series. Actas del XII Congreso Dr. Antonio Monteiro (2013), :57-66, 2014. URL

Self-reciprocal polynomials over finite fields

Math Overflow Recent Questions - Thu, 09/21/2017 - 10:38

Let $SRMI_q(2n)$ denote the number of self-reciprocal irreducible monic polynomials of even degree $2n$ over the finite field $\mathbf{F}_q$ with $q$ elements. Recall that a polynomial $p(x) \in \mathbf{F}_q[x]$ of degree $n$ is self-reciprocal if $p(x)=x^np(x^{-1})$. According to Thm 3.1.20 of Handbook of Finite Fields by Mullen and Panario we have that \begin{equation*} SRMI_q(2n) = \frac{1}{2n}\sum_{\text{odd $d \mid n$}} \mu(d)(q^{n/d}-1), \qquad SRMI_q(2n) = \frac{1}{2n}\sum_{\text{odd $d \mid n$}} \mu(d)q^{n/d} \end{equation*} depending on whether $q$ is odd or even. The sums range over all odd divisors of $n$. Now comes my question. Is it possible to write the infinite product \begin{equation*} \prod_{n \geq 1} \frac{1}{(1-x^{2n})^{SRMI_q(2n)}} \end{equation*} as a rational function?

What I have in mind is an analogue of the well-known formula \begin{equation*} \prod_{n \geq 1} \frac{1}{(1-x^n)^{I_q(n)}} = \frac{1}{1-qx} \end{equation*} where $I_q(n)$ is the number of irreducible monic polynomials over $\mathbf{F}_q$.

Russian Equivalent of Big Rudin

Math Overflow Recent Questions - Thu, 09/21/2017 - 10:38

Is there any Russian-authored Textbook on Analysis equivalent to Big Rudin( Real and Complex Analysis)?

I like Russian math textbooks a lot. I am looking for Russian-textbooks (either in English or Russian, preferably English) in analysis covering same material as Principles of Mathematical Analysis, Real & Complex Analysis and Functional Analysis By Rudin.

Surgery and Curvature on Foliation

Math Overflow Recent Questions - Thu, 09/21/2017 - 04:14

Let $X$ be an oriented closed smooth $4$-manifold. Suppose that $TM$ admits a foliation $\mathcal F$ of dimension two, and admits a positvescalar curvature.

Q: If we do the surgery on $X$ to reduce the $b_1$, would the surgery affect the positivity of the scalar curvature on $\mathcal F$.

Residual properties of iterated semidirect products

Math Overflow Recent Questions - Thu, 09/21/2017 - 04:07

Consider $G^{\rtimes k} := (G\dots ) \rtimes G)\rtimes G$ with diagonal action by inner automorphisms, $G^{\ltimes 1} = G$. Let $\mathcal P$ be a collection of groups. Is it true that

  • $G$ residually $\mathcal P$ $\Rightarrow $ $G^{\rtimes \geq 2}$ residually $\mathcal P$? (very unlikely, but I can't find a counterexample)
  • $G^{\ltimes 2}$ residually $\mathcal P$ $\Rightarrow $ $G^{\rtimes \geq 2}$ residually $\mathcal P$?

I'm mostly interested in first question with $\mathcal P$ a quasivariety and $G$ finitely generated, but anything goes.

Cartesian liftings in double categories

Math Overflow Recent Questions - Thu, 09/21/2017 - 02:24

The question: I wonder whether the following definition, or something similar, has appeared somewhere (see below for motivations). Any reference or pointer is welcome!

(In what follows, I denote horizontal composition by $\cdot$ and vertical composition by $\ast$).

Let $\mathcal D$ be a (strict) double category, let $f:a\to b$ be a horizontal morphism and $p:b'\to b$ a vertical morphism. Let us define a cartesian lifting of $f$ with respect to $p$ to be a cell $\alpha$

such that, for every cell $\alpha'$ whose vertical target is $p$ and whose horizontal target is $f\cdot g$ for some $g$, there exist unique cells

such that $\alpha'=(\alpha\ast\delta)\cdot\gamma$. Pictorially:

We may call "fibrational" a double category in which cartesian liftings always exist. This seems to be unrelated to the notion of fibrant double category or framed bicategory.

The name "fibrational" is justified by the fact that a functor $F:\mathcal A\to\mathcal B$ induces a double category $\mathcal D(F)$ on the disjoint union of $\mathcal A$ and $\mathcal B$ (vertical morphisms are relations $F(a)=b$ and cells are relations $F(f)=g$; vertical composition is degenerate) such that $\mathcal D(F)$ is "fibrational" iff $F$ is a fibration (the cell $\delta$ is always the identity because of the degeneracy of vertical composition).

My motivations: I am studying term rewriting from a categorical perspective: terms are objects, rewriting paths are arrows. In particular, I am interested in formulating infinitary term calculi as ideal completions of finitary calculi. It is very natural to equip both terms and rewrites with a partial order relation yielding a posetal double category, i.e., a category internal to $\mathbf{Pos}$, the category of posets and monotonic functions. These may be seen as double categories in which vertical arrows are order relations between objects and cells are order relations between horizontal arrows (whose source and target are related).

Let $\mathbf{Dcpo}$ be the category of directed-complete partial orders and Scott-continuous functions. It is well known that the forgetful functor $\mathbf{Dcpo}\to\mathbf{Pos}$ has a left adjoint, called ideal completion. By contrast, it is not hard to see that the forgetful functor $\mathbf{Cat}(\mathbf{Dcpo})\to\mathbf{Cat}(\mathbf{Pos})$ does not have a left adjoint, so a posetal double category does not admit an "ideal completion" in general. However, one may show that if a posetal double category $\mathcal D$ is such that $\mathcal D^{\mathrm{tcoop}}$ is "fibrational", then its ideal completion may be defined ($\mathcal D^{\mathrm{tcoop}}$ is the double category obtained from $\mathcal D$ by swapping horizontal and vertical arrows and then reversing their direction). In other words, the forgetful functor has a left adjoint once we restrict to the subcategories of $\mathbf{Cat}(\mathbf{Dcpo})$ and $\mathbf{Cat}(\mathbf{Pos})$ in which objects are "tcoop-fibrational". Term calculi, with their order, turn out to be "tcoop-fibrational", so all the constructions I am interested in fit within this framework.

Since my motivations are quite peculiar, I am unsure about the general mathematical significance of the above definition, and I am not surprised I can't find it anywhere. Still, I thought it would be worth asking.

Refining topologies to make them Hausdorff

Math Overflow Recent Questions - Wed, 09/20/2017 - 23:48

Is there an infinite, connected space $(X,\tau)$ such that the only Hausdorff topology $\tau_2$ with the property that $\tau\subseteq \tau_2$ is the discrete topology?

Convex caps with prescribed projection of edges

Math Overflow Recent Questions - Wed, 09/20/2017 - 20:30

Let $P$ be a convex polygon in the plane $R^2=R^2\times \{0\}$, and $E$ be the edge graph of some subdivision of $P$ into convex polygons, which is $3$-connected. Does there exist a convex polyhedral cap $C\subset R^3$ such that the boundary of $C$ coincides with that of $P$, and the orthogonal projection of the edges of $C$ into $R^2$ coincide with $E$?

A convex polyhedral cap is a portion of the surface of a convex polyhedron cut off by a plane which contains an interior point of the polyhedron.


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