Recent MathOverflow Questions

a family of varieties with nonempty and smooth intersections

Math Overflow Recent Questions - Mon, 11/19/2018 - 22:14

I would like to know if there are interesting examples of affine varieties $S_1, \dots, S_m\subseteq \mathbb{A}^n$ with the following property: for any $J\subseteq [m]$, $\cap_{i\in J}S_i$ is non-empty and smooth.

For example, if $S_i$'s are linear spaces, then any intersection is also a linear space, and non-empty because of the origin point.

I understand that when $m$ is a constant, it should be easy to construct many families. I would be more interested in the case when $m$ is large, say polynomial in $n$.

For what sets does the Lebesgue Diffferentiation Theorem hold in one dimension?

Math Overflow Recent Questions - Mon, 11/19/2018 - 00:39

Lebesgue's differentiation theorem states that if $x$ is a point in $\mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a Lebesgue integrable function, then the limit of $\frac{\int_B f d\lambda}{\lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.

But my question is, what is known about what collections of sets the Lebesgue Differentiation Theorem holds for in one dimension? We know it holds for bounded intervals. Does it hold for finite unions of bounded open intervals? What about bounded Borel sets? What about bounded Lebesgue measurable sets in general?

Or are all these open problems?

Does the Lebesgue Differentiation Theorem hold for regular polytopes?

Math Overflow Recent Questions - Mon, 11/19/2018 - 00:09

Lebesgue's differentiation theorem states that if $x$ is a point in $\mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a Lebesgue integrable function, then the limit of $\frac{\int_B f d\lambda}{\lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.

It does hold for squares, though, and more generally for $n$-dimensional cubes. So my question is, is it known if it holds for all regular polygons? And more generally, is it known if it hold for all regular polyhedra and regular polytopes?

I ask "is it known" because a lot of problems in this topic tend to be open.

m-ary tree question!

Math Overflow Recent Questions - Mon, 11/19/2018 - 00:07

A chain letter starts when someone sends a letter to five other people. Everyone who receives the letter will send it back to five other people who have never received it or did not send the letter at all. For example, it was known that there were 10,000 people who had forwarded the chain letters before finally stopping and no more people received the letter. How many people receive the letter and how many do not send it out?

My attempt :

I used n = mi + 1 where i is number of internal vertices, m is the arry ( 5-arry in this case) and n is number of vertices.

Now the number of people who received the letter, n = mi+1 n = (5)(10000) + 1 = 50001

Now as far as the number of internal vertices is concerned, i have counted the root as a internal vertex.Thats how the book defined the root.But the solution posted on this website excludes the root and i dont understand why.Plus if you look at the example given on this section (section 10.3),the author has done a similar problem in which he counted the root as an internal vertex.May i know why the root should not be counted ,if thats the correct way ?

For the second part, i used l = n-i,l represents leaves.(number of people who did not send out letters).

l = 50001 - 10000 = 40001. So this answer also turns out to be wrong since i used the vertex i got in part a.

I would appreciate your feedback. I would really like an expert to answer this question. If you are not sure, please don't attempt it.

If $\text{dim}(X \times X) = 2\text{dim}(X)$, does $\text{dim}(X^n) = n\text{dim}(X)$?

Math Overflow Recent Questions - Sun, 11/18/2018 - 21:36

I have been learning some (topological) dimension theory and have gotten through most of the basic material, at this point, and am about to start looking at papers. In particular, I want to get familiar with the standard counterexamples regarding dimension of products, but I haven't noticed the question in the title addressed.

So my question would be what conditions on $X$ (e.g. separable metric, compact metric, continuum, homology manifold) are sufficient to ensure that if $\text{dim}(X \times X) = 2\text{dim}(X)$, then $\text{dim}(X^n) = n\text{dim}(X)$? The initial assumption could also be weakened from $\text{dim}(X \times X) = 2\text{dim}(X)$ to $\text{dim}(X^m) = m\text{dim}(X)$ for some $1 < m \leq n$, or for various subsets of $\lbrace 2, 3, \dots, n \rbrace$.

In particular, have the dimensions of all the powers of the Pontryagin Surface been computed?

Algebraic proof for closed form for nested summation of powers of first n natural numbers

Math Overflow Recent Questions - Sun, 11/18/2018 - 21:22

Consider the following two definitions and the formula:

Definition 1:

$$\sum^{(m)} n^k = \begin{cases} n^k, &\quad\text{if m=0}\\ \sum\limits^{(m-1)} 1^k +\sum\limits^{(m-1)} 2^k + \cdots + \sum\limits^{(m-1)} n^k &\quad\text{if $m \ge 1$}\\ \end{cases} $$

Recurrence Definition 2 :

$$\mu[k,r] = \begin{cases} \text{1,} &\quad\text{if r=0}\\ (r)* \mu[k-1,r-1]+(r+1)*\mu[k-1,r] &\quad\text{if $1 \le r \le k$}\\ \text{0,} &\quad\text{r > k}\\ \end{cases} $$

Formula :

$$\sum^{(m)} n^k = \sum_{i=0}^k \binom{n+m-1}{m+i} \mu[k,i]$$

How to prove the formula algebraically for all values of $m$?

Related : Pre-print claiming the formula, Math SE question

modules whose every submodule is a homomorphic image

Math Overflow Recent Questions - Sun, 11/18/2018 - 21:10

Let $R$ be a commutative ring with unity. Let us say that an $R$-module $M$ satisfies property $\mathcal P$ if every submodule of $M$ is a homomorphic image of $M$.

Can we characterize all Noetherian rings $R$ such that for every $R$-module $M$, the $R$-module $\prod_{\mathfrak p \in Spec (R)} M_{\mathfrak p}$ satisfies property $\mathcal P$ ?

commutative ring satisfying descending chain condition on radical ideals

Math Overflow Recent Questions - Sun, 11/18/2018 - 21:04

Let $R$ be a commutative ring with unity which satisfies d.c.c. on radical ideals. then does $R$ satisfy a.c.c. on radical ideals ? If this is not true in general, then what happens if we also assume $R$ is a local domain and/or that $R$ has finite Krull dimension ?

Roots of unity, vanishing sums and derivatives

Math Overflow Recent Questions - Sun, 11/18/2018 - 20:48

Fix integers $n\geqslant1$ and $k\geqslant 0$. For an integer $i$, the $k$-fold derivative of $x^i$ can be denoted by $i^{\underline{k}}x^{i-k}$ where $i^{\underline{k}}$ means $i(i-1)\cdots(i-k+1)$ if $k>0$, and 1 if $k=0$. Let $\alpha$ be an integer satisfying $\alpha^n\equiv1\pmod n$ and $f(x)=\sum_{i=0}^{n-1}x^i$. We proved that the following rather weak conditions on $(k,n,\alpha)$ are equivalent to the $k$-fold derivative $f^{(k)}(\alpha)\equiv0\pmod n$:
(a) $k+1\not\in\{4,p\}$ where $p$ is prime, or
(b) $k+1=4$ and $4\nmid n$, or
(c) $k+1$ is a prime, call it $p$, and either $p\nmid n$ or $\alpha\not\equiv1\pmod p$.

Assume $k\leqslant n$, otherwise $f^{(k)}(x)$ is the zero polynomial, and the conclusion is vacuously true.

Questions: Is this easily deduced from known results? (Our proof is 3 pages long!) Have you seen this before? Is it of interest? (We needed the case $k=1$ in a paper, that's how we came up with it.)

Is the ratio of $\pi$/$e$ transcendental?

Math Overflow Recent Questions - Sun, 11/18/2018 - 20:34

Is the fraction $\frac{\pi}{e}$ a transcendental number? Is its reciprocal transcendental?

Monoidal equivalence of categories of modules in different models of higher algebra

Math Overflow Recent Questions - Sun, 11/18/2018 - 20:10

A result of Shipley states the following (in the language of model categories):

  • For a differential graded algebra $A$, the $\infty$-category of (dg)-modules over $A$ agrees with the $\infty$-category of module spectra over the corresponding $E_1$ ring spectrum $HA$.
  • For a $H\mathbb Z$-algebra spectrum $B$, the $\infty$-categories of module spectra over $B$ and dg-modules over the corresponding dg-algebra $\Theta B$ are equivalent.

Similar equivalence seems to exist between simplicial modules over a simplicial ring and connective versions of the above rings and modules.

My question is, do these equivalences respect the tensor product of modules over higher rings (here I restrict my attention to ''commutative'' rings over $\mathbb Q$)? I think they should, but I cannot find a reference for this. My reasoning is that the relative tensor product over $A$ is a coequalizer of a diagram $A \otimes M \otimes N \stackrel{\to}{\to} M \otimes N$, and these identifications do respect the absolute tensor product. Is this at least morally correct?

Permutations $\pi\in S_n$ with $\sum_{k=1}^n\frac1{k+\pi(k)}=1$

Math Overflow Recent Questions - Sun, 11/18/2018 - 20:08

Let $S_n$ be the symmetric group of all the permutations of $\{1,\ldots,n\}$. Motivated by Question 315568 (, here I pose the following question.

QUESTION: Is it true that for each integer $n>5$ we have $$\sum_{k=1}^n\frac1{k+\pi(k)}=1$$ for some odd (or even) permutation $\pi\in S_n$?

Let $a_n$ be the number of all permutations $\pi\in S_n$ with $\sum_{k=1}^n(k+\pi(k))^{-1}=1$. Via Mathematica, I find that \begin{gather}a_1=a_2=a_3=a_5=0,\ a_4=1,\ a_6=7, \\ a_7=6,\ a_8=30,\ a_9=110, \ a_{10}=278,\ a_{11}=1332.\end{gather} For example, $(1,4,3,2)$ is the unique (odd) permutation in $S_4$ meeting our requirement for $n=4$; in fact, $$\frac1{1+1}+\frac1{2+4}+\frac1{3+3}+\frac1{4+2}=1.$$ For $n=11$, we may take the odd permutation $(4,8,9,11,10,6,5,7,3,2,1)$ since \begin{align}&\frac1{1+4}+\frac1{2+8}+\frac1{3+9}+\frac1{4+11}+\frac1{5+10}\\&+\frac1{6+6}+\frac1{7+5}+\frac1{8+7}+\frac1{9+3}+\frac1{10+2}+\frac1{11+1}\end{align} has the value $1$, we may also take the even permutation $(5, 6, 7, 11, 10, 4, 9, 8, 3, 2, 1)$ to meet the reuirement.

I conjecture that the question has a positive answer. Your comments are welcome!

PS: After my initial posting of this question, Brian Hopkins pointed out that A073112($n$) on OEIS gives the number of permutations $p\in S_n$ with $\sum_{k=1}^n\frac1{k+p(k)}\in\mathbb Z$, but A073112 contains no comment or conjecture.

Criterion for acyclicity of flag complexes

Math Overflow Recent Questions - Sun, 11/18/2018 - 18:53

Let $\Delta$ be a flag complex on $n$ vertices. Let $r$ be the smallest size of the facets of $\Delta$. Suppose that $2r>n$. Must $\Delta$ be acyclic?

If you can approximate vectors well, can you do so parametrically?

Math Overflow Recent Questions - Sun, 11/18/2018 - 18:47

Let $B$ be a Banach space, and $V$ a topological vector space with topology induced by some translation-invariant metric $d$, equipped with a continuous linear map $p: B \to V$ with dense image. Interpreting this as saying that we may approximate elements of $V$ arbitrarily well by elements of $B$, I would like to know if the same is true parametrically, for compact Hausdorff parameter space.

Precisely, given a compact Hausdorff space $X$, a continuous map $g: X \to V$ and $\epsilon > 0$, can we find a continuous map $\tilde g: X \to B$ so that $$d\left(g(x), p\tilde g(x)\right) < \epsilon$$ for all $x \in X$?

If $V$ is a normed space, this is true: pick a countable dense set $x_i \in X$, and choose elements $b_i$ so that $\|g(x_i) - p(b_i)\| < \epsilon$; then there is a neighborhood $U_i$ of $x_i$ on which this remains true. Choosing a partition of unity $\rho_i$ subordinate to this cover, set $\tilde g(x) = \sum \rho_i(x) b_i$. Then one sees $$\|\tilde g(x) - p(g(x))\| = \|\sum_i \rho_i(x) p\big(b_i - g(x)\big)\| = \sum_i \rho_i(x) \|p(b_i - g(x))\| < \epsilon.$$

This falls apart when working with a translation-invariant metric when one tries to pull the $\rho_i(x)$ out of the distance, and it was unclear to me how to resolve this.

If this is true as stated, I would also be interested in knowing how little we need to assume on $V$ to make it true. (Is it true for locally convex topological vector spaces, in the sense that the image of $\text{Map}(X, B) \to \text{Map}(X, V)$ is dense in the compact-open topology?)

Most assumptions on $X$ and $B$ are also likely superfluous: the only properties of compact Hausdorff used here were second-countability and paracompactness, and the only assumption we used on $B$ was that it was a topological vector space (partitions of unity are locally finite, so no infinite sums are necessary). I state these assumptions above because this is true in the case I care about, and maybe they're useful in establishing this for more general $V$.

Exchangeable Bernoulli random variables with bounded summation implies negative correlation?

Math Overflow Recent Questions - Sun, 11/18/2018 - 13:38

Let $\big\{X_1, X_2, ..., X_n \big\}$ be $n$ jointly exchangeable Bernoulli random variables, i.e., exchanging the order of these random variables does not change the joint distribution. If we know that $$\sum_{i=1}^{n} X_i \leq m < n$$ holds for sure, does this imply that any $X_i$ and $X_j$ are negatively correlated? This is quite intuitive to me because these random variables are symmetric and their sum is bounded from above...

upper bound on power of neyman-pearson hypothesis test

Math Overflow Recent Questions - Sun, 11/18/2018 - 13:27

Let $H_0$ and $H_1$ be two distributions. The Neyman-Pearson lemma says that of all rejection regions $R$ with fixed probability $\alpha$ under $H_0$, the one with maximal probability under $H_1$ is the set of the form $R = \{x: \frac{p_1(x)}{p_0(x)} \ge c\}$ with $c$ chosen such that $\mathbb{P}_{x \sim H_0}(x \in R) = \alpha$. The power of the test is then $\mathbb{P}_{x \sim H_1}(x \in R)$.

In my case, $H_0 = \mathcal{N}(0, \tau^2 I_d)$ and $H_1 = \mathcal{N}(\mu, \sigma^2 I_d)$ with $\tau > \sigma$. The dimension is high (hundreds of thousands).

For this choice of $H_0$ and $H_1$, computing the rejection region in closed form does not appear to be possible, so I'd like to compute an upper bound on the power of the NP test at level $\alpha$ without actually computing the rejection region.

Are there any generic methods to compute an upper bound on the power of a Neyman-Pearson test at level $\alpha$? I'm looking for an exact bound, not an approximate bound based on e.g. the CLT.

In textbooks and papers, I've seen many ways to upper-bound the power as a function of $c$ (the likelihood ratio threshold), but none to upper-bound the power as a function of $\alpha$. That said, if I had a right tail bound on $\frac{p_1(x)}{p_0(x)}$ under $H_1$ and a left tail bound on $\frac{p_1(x)}{p_0(x)}$ under $H_0$, I could combine those to get an upper bound on the power as a function of $\alpha$.

A "boundary map" for the algebraic equivalence relation of cycles

Math Overflow Recent Questions - Sun, 11/18/2018 - 09:55

In what follows by projective variety I will mean the zero locus of homegeneous polynomials in some projective space. Anyway, feel free to deal with a sufficiently good scheme if you want.

Let $X$ be a projective variety. Denote by $\mathcal{Z}_{p}( X)$ the group of the $p$-algebraic cycles of $X$. We say that a $p$-cycle $\gamma\in\mathcal{Z}_p(X) $ is algebraically equivalent to zero if there exist a non-singular irreducible projective curve $C$, two points $s,t\in C$ and a finite number of $(p+1)$-prime cycles (irriducible subavrieties of $C\times X$) $V_i\in \mathcal{Z}_{p+1}(C\times X)$ such that $$\gamma=\sum_i[V_i(s)]-[V_i(t)],$$ where by $[V_i(c)]$ we mean the cycle associated to the fiber of the restriction to $V_i$ of the projection $C\times X\longrightarrow C$. I am trying to understand if this definition can be given by mean of a "boundary map". Just to clarify what I mean let me consider the case $C=\mathbb P^1$ (rational equivalence). In this case we can introduce the boundary map $$\partial\colon \mathcal{Z}_{p+1}(\mathbb P^1\times X)\longrightarrow \mathcal{Z}_p(X)$$ defined as follows. If $W\subseteq \mathbb P^1\times X$ is a irreducible projective variety whose image via the projection $\mathbb P^1\times X\longrightarrow \mathbb P^1$ is the whole $\mathbb P^1$ then $$\partial W=[W(s)]-[W(t)],$$ otherwise $\partial W=0$. Now the map $\partial$ extends by linearity on all of $\mathcal{Z}_{p+1}(\mathbb P^1\times X)$. A cycle $\gamma \in\mathcal{Z}_p(X)$ is said to be rationally equivalent to the zero if it lies in the image of $\partial$.

In this case the image of $\partial$ does not depend on the choice of the points $s,t\in\mathbb P^1$. Indeed, for any pair of points $s',t'\in\mathbb P^1$ we can find an automorphism $\phi$ of $\mathbb P^1$ taking $s'$ to $s$ and $t'$ to $t$ in such a way that $[W(s')]-[W(t')]=[f(W(s))]-[f(W(t))]$, where $f= \phi\times id\colon \mathbb P^1\times X\longrightarrow \mathbb P^1\times X$.

Question: Is it possible to introcuce a boundary map $\partial_C$ for any non-singular irreducible projective curve $C$ in such a way that the image of $\partial_C $ does not depend on the choice $s,t\in C$?

Do solenoids embed into Möbius strips?

Math Overflow Recent Questions - Sun, 11/18/2018 - 08:06

I found a strange attractor which looks a lot like a solenoid. The attractor continuum is the closure of a continuous line which limits onto itself, and it is locally homeomorphic to Cantor set times Reals. It sits in the Möbius strip.

Does the Dyadic solenoid embed into the Möbius strip? What about solenoids in general? Could the strange attractor above actually be a solenoid?

Supremum of an almost surely continuous random process

Math Overflow Recent Questions - Sun, 11/18/2018 - 07:12

I was learning this proposition

and now I have a question, how to prove it for an almost surely continuous process? I would be very grateful for any tips.

Irreducible surface singularity that is not a local set-theoretical complete intersection

Math Overflow Recent Questions - Sat, 11/17/2018 - 23:49

I have been looking for a criterion for the germ of an irreducible complex surface singularity $(X,x)$ to be a set-theoretical complete intersection.

A germ $(X,x)$ of an isolated complex singularity of dimension $k$ is said to be an isolated complete intersection singularity if for some embedding $(X,x) \hookrightarrow \mathbb{C}^{k+s}$, the ideal of functions that vanish on $(X,x)$ can be generated by $f_1, \ldots, f_s$ holomorphic functions of in $\mathbb{C}^{k+s}$.

A germ $(X,x)$ of an isolated complex singularity of dimension $k$ is said to be a set theoretical isolated complete intersection if the analytic set defined by the embedding of $X$ in some euclidean complex space can be written as the zeros of as many holomorphic functions as the codimension of the space in that euclidean space.

I have noticed that often the difference between "complete intersection" and "set-theoretical complete intersection" is blurred in the literature. And a lot of papers on the topic, do not define the object (as it is assumed to be widely known in the area) or they define it casually.

Sometimes it is said that a a singularity $(X,x)$ of pure dimension $r$ is a complete intersection if it can be expressed as the zeroes of $n-r$ polynomials in $\mathbb{C}^n$. But this is the weaker notion of set-theoretical complete intersection. For example, in, C.T.C. Wall says that an ICIS is a surface defined by $n$ equations in $\mathbb{C}^{n+2}$ with an isolated singularity. Next he shows an example of a surface singularity which is not a complete intersection (end of page 2 and page 3). The singularity is the one generated by the polynomials $$f_1 = xz-y^2, f_2 = xw-yz, f_3 = yw-z^2$$ in $\mathbb{C}^4$. But one can very easily see that $V(f_1) \cap V(f_2) \subset V(f_3)$. Hence, the singularity $X$ is a set theoretical complete intersection defined by the $2$ first equations in $\mathbb{C}^4$!

I know the typical example of two planes intersecting at a point as a surface singularity which is not a set-theoretical complete intersection (and also not irreducible).

By irreducible I mean, that the germ is an irreducible complex analytic germ.

My question is, what are examples (if any) of irreducible complex surface singularities that are not set-theoretical complete intersections? And if it is studied somewhere the criterion for (the germ of) an irreducible analytic space to be a (local) set-theoretical complete intersection?

I have found tons of examples of varieties that are local complete intersection but not global complete intersections. And of set-theoretical (local and global) complete intersections that are not complete intersections in the ideal sense of the number of generators of their defining ideals. But not much in the necessary conditions for an algebraic set to be a set-theoretical intersection near a given point.


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