Recent MathOverflow Questions

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.

Dixmier traces, Wodzicki residue and residues of zeta functions

Math Overflow Recent Questions - Sat, 11/17/2018 - 20:08

Let $M$ be an $n$ dimensional closed manifold and consider an elliptic, pseudodifferential operator $P$ of order $-n$. Here are some facts which I had learned so far:
1. There exists a density defined in terms of $-n$-th homogenous part of the symbol of $P$ such that integrating this density one obtains a trace on the algebra of all (classical) pseudodifferential operators. This is the so called Wodzicki residue, to be denoted by $Wres$. 2. If $n>1$ this is unique such trace.
3. If $P$ is as above that $P$ belongs to the Dixmier trace, is measurable (i.e. the Dixmier trace does not depend from the choice of a state) and thus Dixmier trace $Tr^+P$ is defined.
4. We have and equality $Tr^+P=Res_{s=1}(Trace(P^s))$ and thus $P \mapsto Res_{s=1} (Trace(P^s))$ is also a trace.
5. Also the function $P \mapsto Res_{s=0}(Trace(P\Delta^{-\frac{s}{2}}))$ is a trace.
Finally the last fact: 6. There is a universal constant $c>0$ (depending only on the dimension of $M$) such that $Res_{s=0}(Trace(P\Delta^{-\frac{s}{2}}))=c Wres(P)$ (note that this is slightly more which can be deduced from the uniqueness of the trace since this constant does not depend on the choice of $M$ provided that the dimension remains the same).

To have the full picture of these issues I would like to know

  1. How to determine (in the easiest way, provided we know all the above facts) this universal constant $c$ (it should be $\frac{1}{n(2\pi)^n}$).
  2. Is it clear that $Res_{s=1}(Trace(P^s))$ coincides with $Res_{s=0}(Trace(P\Delta^{-\frac{s}{2}}))$? Is it clear that the former is a trace?

I would be happy in any solution to these two problems (i.e. answering 2. once we know the answer for 1. etc).

Limits, colimits and universes

Math Overflow Recent Questions - Sat, 11/17/2018 - 02:02

For many purposes in category theory, we consider limit and colimits of diagrams $F\colon\mathsf{J\to C}$ where $\mathsf{J}$ is small category, that is, a category where the classes of objects and morphisms are actually sets.

However, what if instead of sets and classes we want to adopt the foundational system with Grothendieck universes? Do we even need the smallness condition anymore?

For instance, in the framework of classes (without universes) we say that a category $\mathsf{C}$ is complete (resp., cocomplete) if limit (resp., colimits) of all diagrams indexed by small categories exist. What are the appropriate versions of these notions in the framework of universes?

If Goldbach's Conjecture is eventually true, is it necessarily true?

Math Overflow Recent Questions - Fri, 11/16/2018 - 23:30

We have all heard that if Goldbach's conjecture is independent, then it is true. This is because if GC is false then there is an even number which is not the sum of two primes, and hence a finite proof. What if this number is very large? Maybe beyond our reach?

If one proves that it is independent that eventually every even number is the sum of two primes, can we conclude that eventually every even number is the sum of two primes? Even if we may never know the $e$ from which it is Goldbach's conjecture, all the way, into infinity?

If we know that there is an $e$ from which every even number after and including $e$ is the sum of two primes, is independent, can we conclude that $e$ and every even number beyond, is the sum of two primes?

Even if we may never know what $e$ is?

Let $e$ be the least even integer such that $e$ is the sum of two primes and for every $k=2n$ where $n \in \mathbb{N}$ where $k$ is greater than $e$, then $k$ is also the sum of two primes. Suppose it is independent of ZFC that $e$ exists. Does it follow that $e$ exists?

Literature request: Function that depends on a linear optimization problem

Math Overflow Recent Questions - Fri, 11/16/2018 - 07:41

my question is about functions of the following form:

$$ f(t) = \max_{\mathbf{x}}~ \mathbf{c^T x} ~ {\rm s.t. \mathbf{Ax} +t \cdot \mathbf{a} \leq \mathbf{b}}, $$ where $\mathbf{x},\mathbf{b}, $ and $\mathbf{a} $ are vectors and $\mathbf{A} $ is matrix.

Here, the evaluation of $f(t)$ requires to compute the solution of linear optimization problem. I wonder if there is literature available on such functions. In particular I would like to have information about

  • the form and properties
  • extrema

of this function. Moreover, I would like to know if there is an simple way to obtain function evaluations that do not require to solve an optimization problem.

Embedding of $CP^2/CP^1$ into euclidean space

Math Overflow Recent Questions - Thu, 11/15/2018 - 21:06

It is a standard exercise in embedding theory to show that $S^3 \to \mathbb{R}^4$ given by $(x,y,z) \mapsto (x^2-y^2,xy,xz,yz)$ induces an embedding $\mathbb{R}P^2 \to \mathbb{R}^4$. Since $\mathbb{R}P^2/\,\mathbb{R}P^1 \cong \mathbb{R}P^2$, the previous map gives an embedding of $\mathbb{R}P^2/\,\mathbb{R}P^1$ into $\mathbb{R}^4$.

Is there a nice embedding of $\mathbb{C}P^2/\,\mathbb{C}P^1$ into $\mathbb{R}^8$?

Graph with at most 2 degrees of separation between every node, but minimal average degree

Math Overflow Recent Questions - Thu, 11/15/2018 - 20:44

Is there a simple way to construct such a graph? For example a fully connected graph obviously has degree of separation between every vertex of 1 but has maximal total degree. If we only wanted to minimise the total degree then I think the answer would be a star graph. But I want the average degree to be smallest rather than just relying on a single high degree vertex to be the common neighbour for all vertices. I can sort of see an algorithm starting with a cycle5 graph and adding nodes until the degree of separation between each pair of nodes is <= 2, but not sure if this would be optimal.


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