Recent MathOverflow Questions

Enriched cartesian closed categories

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:56

Let $V$ be a complete and cocomplete cartesian closed category. Feel free to assume more about $V$ if necessary; in my application $V$ is simplicial sets, so it is a presheaf topos and hence has all sorts of nice properties you might want (except that its underlying-set functor $V(1,-) : V \to \rm Set$ is not faithful or conservative).

Let $C$ be a complete and cocomplete $V$-enriched category with powers and copowers (a.k.a. cotensors and tensors), hence all $V$-enriched weighted limits. And suppose that the underlying ordinary category $C_0$ is cartesian closed, i.e. we have natural isomorphisms of hom-sets

$$ C_0(X\times Y, Z) \cong C_0(X, Z^Y). $$

Is it necessarily the case that $C$ is also cartesian closed in the $V$-enriched sense, i.e. we have $V$-natural isomorphisms of $V$-valued hom-objects

$$ C(X\times Y, Z) \cong C(X, Z^Y)? $$

I can't decide whether I think this is likely to be true or not. I used to believe that it was true, and I believe I have implicitly used it (or more precisely its generalization to the locally cartesian closed case, which should follow from the simple one applied to enriched slice categories) in a published paper or two. However, right now I can't see how to prove it, and when stated clearly it sounds unlikely, since usually it is an extra condition on an adjunction between enriched categories to be an enriched adjunction (even when one of the functors is known to be an enriched functor, as is the case here for the cartesian product). On the other hand, it is true when $C=V$ itself: a cartesian closed category, defined by an isomorphism of hom-sets, automatically enhances to an isomorphism of hom-objects $Z^{X\times Y} \cong (Z^X)^Y$.

Does there exist a curve which avoids a given countable union of small subsets?

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:53

Let $X$ be a projective variety over $\mathbb{C}$. Let $X_1, X_2, \ldots$ be proper closed subsets. Then $\cup_i X_i \neq X(\mathbb{C})$. However, I am interested in a stronger statement.

Assume that, for all $i$, we have that $\mathrm{codim}(X_i)\geq 2$.

Then, does there exist a smooth projective curve $C\subset X$ such that $C$ and $\cup_i X_i$ are disjoint?

The condition on the codimension is clearly necessary. (Take $X_1$ to be an ample divisor to see this.)

Rewriting integral via higher derivatives

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:34

Let $f $ be an infinitely smooth function on $(0,\infty)$. Suppose further that $f $ and all of its derivatives are bounded. I have the following integral $$b\int_0^1f (a+bx)\; dx, $$ where $ a$ and $ b$ are real numbers. The number $b $ is of interest and I seek bounded, smooth function $g $ such that $$b\int_0^1f (a+bx)\; dx = b^2\int_0^1 f^{(\ell)}(a+bx)g (x) \; dx,$$ where $f^{(\ell)}$ is some derivative of $f$.

For example, I have used integration by parts and a linear $g (x)= x-1$ on the quantity $\int f(a+bx)g(x)$ to get $$b\int_0^1f (a+bx)\; dx =-bf(a)+ b^2\int_0^1 f'(a+bx)g (x) \; dx.$$

But I would like to have the intgral by itself. I haven't made an attempt for higher powers. I thought I'd ask here to see if anyone knows are results of this type, or can give me some pointers. Thanks

Objects with trivial automorphism group

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:31

Suppose I am working with a category of objects such that each object $X$ in the category has as a trivial automorphism group. An example of such an object would be genus zero curves with $n$-punctures, i.e., $(\mathbb{P}_k^1, (s_i)_{i=1}^n \in k )$.

Let $F(S)/{\sim}$ represent families of these objects over an affine scheme $S=\operatorname{Spec} A$ where $A$ is a $k$-algebra and $k$ an algebraically closed field. As an example, consider families of genus zero curves with $n$-punctures over $S$ and fibers $X= X_s \cong (\mathbb{P}_k^1, (s_i: k \to \mathbb{P}_k^1)_{i=1}^n) $. Then $H^1(X, TX)=0$ and $Aut(X)=0$.

Motivation for question: It seems that since both $H^1(X, TX)=0$ and $Aut(X)=0$ we can conclude that every family of genus zero curves with $n$-punctures is trivial, i.e any family over $S$ is isomorphic to the trivial family $(\mathbb{P}_A^1, (s_i: A \to \mathbb{P}_A^1)_{i=1}^n)$.

Question: Now going back to the beginning, suppose my family of objects over $S$ is such that all fibers are isomorphic to some objet $X$, $H^1(X, TX)=0$ and $Aut(X)=0$, does this imply that all families are trivial?

Or even more generally, suppose $H^1(X, TX)=n$ and $Aut(X)=0$, does this imply that there exists $n$ distinct families $F(S)/{\sim}$?

I know there are probably many conditions on $X$ I should include, however, just assume that my object $X$ has any property that makes it "nice" to work with.

When is it okay to intersect infinite families of proper classes?

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:14

For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.

Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1\implies ba=1$ for $a,b\in \mathbb{M}_n(R)$ (the $n\times n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $n\geq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.

So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?

Phrased a little differently my question is the following. Suppose we have an $\mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $\land_{n\in \mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?

[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]

Is this consequence of the invariant subspace problem known?

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:59

An interesting fact popped out of a paper I'm writing: if the invariant subspace problem for Hilbert space operators has a positive solution, then every $A \in B(\mathcal{H})$ can be made "upper triangular" in the sense that there is a maximal chain of closed subspaces of $\mathcal{H}$, each of which is invariant for $A$.

The proof is quite easy, almost trivial, yet I had never heard of it before. (It isn't mentioned in the answers to this question, for example.) It was a surprise to notice that the ISP has such a strong consequence for the structure of arbitrary operators because one thinks of is as merely the first, most basic question on that topic.

Surely this is known?

Marginal Distribution of Partition Matrix

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:56

Assume that $X\sim IW_{p}(n,I_p)$ has an inverse Wishart distribution, which probability density function is $$f(X\mid n)\propto |X|^{-\frac{n+p+1}{2}}exp\Big(-\frac{1}{2}tr( X^{-1})\Big),~~\qquad (1)$$ Partition the matrices $X$ with $$X=\pmatrix{X_{11}&X_{12}\\X_{21}&X_{22}},$$ where $X_{ij}$ is $p_i\times p_j$ matrices. From the definition of IW, we know $X_{11}\sim IW_{p_1}(n-p_2,I_{p_1}).$

Qustion: If random $p\times p$ positive matrice $Y$ satisfies $$f(Y\mid n)\propto |Y|^{-\frac{n+p+1}{2}}exp\Big(-\frac{1}{2}tr( Y^{-1})\Big)\Big[\prod_{i=1}^p(\lambda_i-\lambda_j)\Big]^{-1},~~\qquad (2)$$ where $\lambda_1>\lambda_2>\cdots>\lambda_p>0$ are the ordered eigenvalues.

Partition the matrices $Y$ with $$Y=\pmatrix{Y_{11}&Y_{12}\\Y_{21}&Y_{22}}.$$ What is the marginal distribution of $Y_{11}???$

Help with signal analysis question [on hold]

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:44

I am having trouble understanding the question and would like some help please,

Q) determine the number of terms required such that the ratio of the power of the non-zero nth harmonic to the power of the fundamental is < 1%.

Answer is 11

but how do you do it I have no idea.

see attached pics of the question also. square wave form

amplitude spectrum

"Almost planar graphs"

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:39

A graph $G$ is "Almost Planer" if it has a subgraph $H$ and an embedding in the plane such that $H$ is planar in this embedding and all the crossing edges are cords of faces of $H$ (all crossings and crossing edges are contained inside the faces of $H$.)
Is there any literatures on this class of graphs? Are they classified any other name? What kind of graphs are in this class? Are there graphs which are not Almost Planar?

Any information on this will be appreciated.

How many exceptional conductors are there?

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:00

We say that a conductor $q$ is exceptional if there is a primitive quadratic character $\chi$ modulo $q$ such that $L(s,\chi)$ has a real zero $\beta$ such that $\beta > 1-c/\log q$ (where $c$ is some tiny absolute constant).

We know from a result of Landau that if there is a sequence of such $q$, say $(q_i)$, then $q_{i+1}> q_i^C$ for some $C$ (another absolute constant).

It follows that the number of exceptional conductors $q$ in $[1,N]$ is $O(\log\log N)$.

Is any better (unconditional) upper bound than this known, on the number of exceptional conductors in the interval $[1,N]$?

It is, of course, conjectured that there are no exceptional conductors at all.

Pre-garbling does not improve capacity of a channel

Math Overflow Recent Questions - Wed, 01/30/2019 - 11:35

Suppose ABC=B for some column stochastic matrices A, B, and C. Can the following implication be made without further restrictions: There necessarily exists a column stochastic matrix D such that DB=BC?

I think this is implicated by lemma 1 of Rauh et al. - Coarse-graining and the Blackwell order and the theorem of Blackwell, Sherman and Stein. Unfortunately the paper contains just a very limited proof. It uses the statement that the capacity of a pre-garbling is bounded by the original experiment/channel and states this as a well-known fact.

My current efforts to prove this special step by myself failed. I would therefore appreciate any assistance.

Counter example to lifting contractibility of a topological space

Math Overflow Recent Questions - Wed, 01/30/2019 - 09:24

I'm looking for a simple example of an open proper continuous map between topological spaces $\varphi:X\to Y$ such that :

  • $Y$ is contractible and locally contractible ;
  • for any $y\in Y$, $\varphi^{-1}(\{y\})$ is contractible ;
  • $X$ is not contractible.

I have an example which is a little bit complicated but I wonder if there exists a simple one.

Is the set of points in the irreducible decompositions of this C$^{*}$ -algebra's representations closed?

Math Overflow Recent Questions - Wed, 01/30/2019 - 08:44

Suppose $X$ and $Y$ are compact Hausdorff spaces. Let $\varphi\colon C(X)\to M_{n}(C(Y))$ be any $*$-homomorphism. If $\pi$ is an irreducible representation of $M_{n}(C(Y))$, then $\pi$ is unitarily equivalent to a point evaluation $\textrm{ev}_{y}$. The $*$-homomorphism $\textrm{ev}_{y}\circ\varphi\colon C(X)\to M_{n}(\mathbb{C})$ is a representation of $C(X)$. Since it's a finite-dimensional representation, we can find a unitary $u_{y}\in M_{n}(\mathbb{C})$ and a set of points $X_{y}=\{x^{y}_{1},\ldots,x^{y}_{n}\}\subset X$ such that for all $f\in C(X)$, $$ (\varphi\circ f)(y)=(\textrm{ev}_{y}\circ\varphi)(f)=u_{y} \begin{pmatrix} f(x^{y}_{1}) & 0 & \cdots & 0\\ 0 & f(x^{y}_{2}) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & f(x^{y}_{n}) \end{pmatrix} u_{y}^{*}. $$

My question is:

Is the set $\widetilde{X}:=\bigcup_{y\in Y}X_{y}$ closed in $X$?

This question is in a similar vein to one of my earlier questions: Closeness of points in the irreducible decomposition of a C$^{*}$-algebra representation

tangent cone and local picture

Math Overflow Recent Questions - Wed, 01/30/2019 - 06:52

Let $X\subset \mathbb P^n_{\mathbb C}$ be a closed algebraic subset and $x\in X$. Suppose tha tangent cone of $X$ at $x$ is the union of (say) a plane and a line (meeting at the origin). Can we conclude that $X$ is around (in the euclidian/étale topology) $x$ a union of a smooth surface and a smooth curve meeting transversally?

Is a vertex- and edge-transitive polytope already a uniform polytope?

Math Overflow Recent Questions - Wed, 01/30/2019 - 06:47

I want to consider (convex) polytopes $P=\mathrm{conv}\{p_1,...,p_n\}\subset\Bbb R^d$ which are both, vertex- and edge-transitive (or maybe stronger: 1-flag-transitive).

Question: Is every such polytope already a uniform polytope?

I know only a few polytopes with such symmetries, all of them are uniform, probably also Wythoffian. Here are some:

The faces of uniform polytopes are uniform again. So far, all I can say about the faces of vertex- and edge-transitive polytopes is, that they are polytopes with all vertices on a sphere and all edges of the same length. While this means that all 2-faces are uniform, it does not immediately follow for the 3-faces (e.g. Pseudorhombicuboctahedron is not uniform but could be a face).

I know that after all, the Wythoffian uniform polytopes are the most well understood. Also, I do not know whethe there is any non-Wythoffian vertex- and edge-transitive polytope. So as a first step, I might ask:

Question: Is every Wythoffian polytopes with such symmetries already uniform?

Chromatic number of a connected Hausdorff space

Math Overflow Recent Questions - Wed, 01/30/2019 - 04:57

Let $(X,\tau)$ be a topological space such that $\tau$ contains no singleton. We say that a map $c:X\to \kappa$, where $\kappa$ is a cardinal, is a coloring for $(X,\tau)$, if for every $U\in \tau\setminus \{\emptyset\}$ the restriction $c|_U$ is non-constant. (Note that this coloring notion comes from hypergraph coloring.)

The chromatic number $\chi(X,\tau)$ of a space $(X,\tau)$ is the smallest cardinal $\kappa$ such that there is a coloring $c:X\to \kappa$.

We have $\chi(\mathbb{R})=2$ when $\mathbb{R}$ is endowed with the Euclidean topology: color $\mathbb{Q}$ with $0$ and $\mathbb{R}\setminus\mathbb{Q}$ with $1$. This works for all spaces having a dense set $D$ such that $X\setminus D$ is also dense. Note that not all connected $T_2$-spaces contain a dense subset with this property.

Given an integer $n>2$ is there a connected Hausdorff space $(X,\tau)$ such that $\chi(X,\tau) = n$?

Distribution of point knowing target in optimal matching

Math Overflow Recent Questions - Wed, 01/30/2019 - 03:50

I am a young PhD student in statistics. Recently, papers (Ambrosio, Stra and Trevisan; Talagrand; Ledoux to cite but a few) tackled the problem of finding the expected cost in an optimal matching, which is a first-order global understanding of what is going on.

I am wondering if anything is known (I found no paper about this) about the following problem.

Given an isotropic n by n grid and n² points drawn from a uniform over the square, what is the p.d.f. of a point knowing that it was mapped (through an optimal matching with the squared Euclidean distance as cost function) to a particular grid point ?

Simulations suggest that a normal r.v. should asymptotically appear for points in the middle; what happens at the boundary is unclear.

Some thoughts suggests that a combinatorial approach may yield results but it becomes very rapidly intractable. (I managed computing the exact distribution for small n) Trying to maximise entropy given the information we have seems too sloppy. Central limit theorem is obviously not applicable and Stein's method (à la Chatterjee, for instance) looks of little help.

Thank you in advance.

Have a great day,

Gilles

The Law of sine and the law of cosine were the only proof verification,but they're not [on hold]

Math Overflow Recent Questions - Wed, 01/30/2019 - 03:49

Simplyfied triangle with base1For a $\triangle ABC$ there exist three sides, $a, b, c$ and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa. At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.

The following equations represent a schematic of all triangles to obtain the result needed to find 3 sides of triangles.

$(c\times\sin B)^2+(b\times\cos C)^2=b^2$

$(c\times\sin A)^2+(a\times\cos C)^2=a^2$

$(c\times\sin B)^2+(c\times\cos B)^2=c^2$

$(c\times\sin A)^2+(c\times\cos A)^2=c^2$

$(b\times\cos C)+(c\times\cos B)=a$

$(a\times\cos C)+(c\times\cos A)=b$

$(a\times\cos B)+(b \times\cos A)=c$

$(c\times\cos B)+(b \times\cos C)=a$

$(c\times\cos A)+(a\times\cos C)=b$

and to find the altitudes for all triangles:

$h_a=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$

$h_b=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$

$h_c=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$

I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa without the Law of cosine even though I have three sides without three angles.

$(\sin A)^2+(\sin B)^2+(\sin C)^2+(\cos A)^2+(\cos B)^2+(\cos C)^2=3$ for all triangles.

And the law of sine states when one of the base of the triangle is one then:

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}=$ where the reciprocal of one of the diameter is $\angle C$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ where d is the diameter.

Let $\triangle ABC$ be sides $a$, $b$, and $c$. I are using that the length of the side opposite vertex $A$ is called $a$, $B$ for $b$ and $C$ for $c$.

If $\theta=\angle C$. Then the Cosine says that

$c^2=a^2+b^2-2ab\cos \theta.$ Since we know $a$, $b$, and $c$, we can use the the formula for all three altitudes or above formula to calculate $\cos\theta$.

We can use the Law of Cosine for all three to get the three angles.

A "nice" (but non-definite) quadratic programme

Math Overflow Recent Questions - Wed, 01/30/2019 - 03:42

For integers $n\geq k>0$, let $f$ be the following quadratic form: $$f(x_1,\ldots,x_n)=\sum_{i=1}^n\sum_{j=0}^{k-1}x_ix_{i+j\bmod n}.$$ Is it true that the minimum of $f$ over the unit simplex is attained at $(1/n,\ldots,1/n)$? Where the unit simplex is the set $\{x\in\mathbb R^n:x_i\geq 0\forall i,\ \sum x_i=1\}$.

A certain ratio condition for polynomials with real coefficients

Math Overflow Recent Questions - Tue, 01/29/2019 - 17:01

Let $p:\mathbb{C} \longrightarrow \mathbb{C}$ be a polynomial with real coefficients and suppose that $p$ satisfies \begin{equation} \frac{p(y)}{y} \le \frac{p(x)}{x} \tag{*} \label{ratcond} \end{equation} whenever $y < 0$ and $x + y \ge 0$.

Is there anything known about such polynomials? Is \eqref{ratcond} related to the derivative of $p$ in some way or equivalent to another well-known condition?

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