Recent MathOverflow Questions

Correspondence between notions of Gerbe on a Manifold and Gerbe on a Stack

Math Overflow Recent Questions - Thu, 08/02/2018 - 05:28

There are two notions of gerbes. Gerbe on a topological space and gerbe on a stack.

A gerbe on a topological space $X$ is a stack $\mathbb{G}$ of groupoids on $X$ that is

  1. Locally non empty i.e., i.e., given $x\in X$ there is an object/open set $U$ containing $x$ such that $\mathbb{G}(U)$ is non empty.

  2. Locally connected i.e., given $a,b\in\mathbb{G}(U)$ and $x\in U$ there exists an open subset $V$ of $U$ containing $x$ such that $a|_V$ is isomorphic to $b|_V$.

Above definition is from

A stack $\mathfrak{R}$ endowed with a morphism $\mathfrak{R}\rightarrow \mathfrak{X}$ is called a gerbe over $\mathfrak{X}$ if both $\mathfrak{R}\rightarrow \mathfrak{X}$ and $\mathfrak{R}\rightarrow \mathfrak{R}\times_{\mathfrak{X}}\mathfrak{R}$ are epimorphisms.

The above definition is from

Given a smooth manifold $M$, there is a stack associated to it, denoted by $\underline{M}$. This can be found in Orbifolds as Stacks by Eugene Lerman.

So, I am expecting that

  • given a Gerbe $\mathbb{G}$ on a Manifold $M$, there is a Gerbe over stack $\underline{M}$ and
  • given a Gerbe over stack $\underline{M}$, there is a Gerbe over manifold $M$ and

that these two ways are something like inverse to each other.

Has any one looked at this before?

My previous question says that for a gerbe over a stack $\mathfrak{R}\rightarrow \mathfrak{X}$, it means that the functor is locally surjective on objects and morphisms.

Given a stack $\mathfrak{R}\rightarrow \mathfrak{X}=\underline{M}$ I am trying to construct gerbe over $M$. I define $\mathbb{G}(U)=\mathfrak{R}(U)$.

I have to check that it is locally non empty and locally connected.

Given $a\in M$, there is an open set $U$ containing $a$. I claim that $\mathbb{G}(U)$ is non empty. First condition in my answer here is that given an object $x\in \mathfrak{X}(U)$ there is an object $y\in \mathfrak{R}(U)$ and an isomorphism $F(y)\rightarrow x$. As $\mathfrak{X}=\underline{M}$, $\mathfrak{X}(U)$ is just the collection of smooth maps $U\rightarrow M$. There is an obvious map we can go for, i.e., the inclusion map $x=i:U\rightarrow M$. For this, there is an object $y\in \mathfrak{R}(U)$ and an isomorphism $F(y)\rightarrow x$. This in particular says that $\mathfrak{R}(U)\neq \emptyset$.

Let $y,y'\in \mathbb{G}(U)=\mathfrak{R}(U)$. Then, we have $F(y),F(y')\in \mathfrak{X}(U)=\underline{M}(U)$. Assuming there is an arrow (I could not figure out why there has to be any yet) $F(y)\rightarrow F(y')$ in $\mathfrak{X}(U)$, second condition in my answer says that there is an arrow $y\rightarrow y'$ (locally atleast) in $\mathfrak{R}(U)=\mathbb{G}(U)$ that maps to the arrow $F(y)\rightarrow F(y')$. This says $\mathbb{G}(U)$ is locally connected.

Thus, we have a Gerbe on manifold $M$ given a gerbe over stack $\underline{M}$.

Given a Gerbe $\mathbb{G}$ on a manifold $M$, I am trying to come up with a gerbe on stack $\underline{M}$. At the least, I have to associate $\mathfrak{R}(U)$ for each manifold $U$. In case when $U$ is an open subset of $M$, I would declare $\mathfrak{R}(U)$ to be just $\mathbb{G}(U)$. This does not say what should be the functor $\mathfrak{R}(U)\rightarrow \underline{M}(U)$ has to be..

Any comments are welcome.

Has any one done something similar to this before.

Is there a slight chance of having such correspondence between notions of Gerbe over a Manifold $M$ and Gerbe over stacks of the form $\mathfrak{X}=\underline{M}$ or Is it just some coincidence of names? I strongly feel there is a correspondence.

Dominating powers of a random matrix

Math Overflow Recent Questions - Thu, 08/02/2018 - 04:57

Let $A_n$ be a (sequence of) random matrix such that $ A_n = (a_{ij})_{1 \leq i,j \leq n}$ and the $a_{ij}$ are iid, $\mathbb E\left[ a_{ij}\right] = 0 \quad E\left[ \vert a_{ij}\vert^2\right] = 1$.

For a problem of mine I need to control this term :

$$R_n =\sum_{k=2}^\infty \left(\frac{1}{\sqrt{2\log n}}\right)^{k-1}\left(\frac{A_n}{\sqrt n}\right)^k \mathbb 1_n $$

Where $\mathbb 1_n$ is the vector of $\mathbb{R}^n$ whose components are all 1's.

$R_n$ is a vector of $\mathbb{R}^n$ and it seems to me we should have :

$$\forall i \in \mathbb N, \quad R_n(i) = [R_n]_i \overset{\mathcal P}{\underset{n\rightarrow \infty}{\longrightarrow}} 0$$

Indeed it is known that for instance, if ones add a 4th moment hypothesis : $E\left[ \vert a_{ij}\vert^4\right] < \infty$ then the matrix $\left(I_n - \frac{A_n}{\sqrt 2n} \right)$ is almost surely invertible when n goes to infinity, and hence the above series exists. Here the additional $\left(\frac{1}{\sqrt {\log n}}\right)^{k-1}$ should ensure the result I need right ?

One can add this 4th moment hypothesis, it is no problem for my purpose.

reuleaux triagnle based problem [on hold]

Math Overflow Recent Questions - Thu, 08/02/2018 - 04:42

I need a formula that returns a degree of the arch QP (from 0 to 120) with the following parameters: the radius r of the circle k with center O and the length form the circle k's center O to the point B, or a single parameter giving their relation. The triangles ABC and ACD are equilateral. The arch l1 (BCD) is 120 degrees and it is centered around A and the arch l2 (BAD) is 120 degrees and is centered around C.

Here's a diagram:

Thank you in advance and excuse my bad English.

The Riemann hypothesis as a problem in analysis

Math Overflow Recent Questions - Thu, 08/02/2018 - 02:50

The recent post("Long-standing conjectures in analysis ... often turn out to be false") prompted me to think about a question which I have not given much though before: to what extent the Riemann hypothesis (RH) may be regarded as a problem in analysis. It may actually be not as silly as it sounds.

The particular side of it I am curious about is the following. The

Theorem : $\zeta(s)\neq 0$ for $\Re s>1$.

may be considered a very weak consequence of RH. However, this statement is only trivial in the context of number theory. One may ask, is it possible to give it a "purely analytic" proof, without using Euler product and other stuff related to the primes? Apparently, it is not possible to formulate this as a precise mathematical question because all the theories used to formalize analysis contain a good deal of arithmetic, but it should be precise enough for practical purposes. (You know number theory when you see it.)

Most likely such a proof does not exist yet, but some readers may know more about the relevant things then I do. I am honestly curious because while it definitely looks tough, it may be not entirely implausible. If such a proof is found, it might give us a fresh look on the old problem.

How to classify a plane complex curve?

Math Overflow Recent Questions - Thu, 08/02/2018 - 01:49

Let $p_1, p_2, t_1, t_2, a \in \mathbb{C}$ be constants. Consider the following plane complex curve in $\mathbb{C}^2$ ($c_1, c_2$ are indetermniates) \begin{align} & {p_1}^2 {p_2}^2 c_1 {t_1}^2 t_2 + {p_1}^2 p_2 {c_1}^2 c_2 {t_1}^2 t_2 + {p_1}^2 p_2 c_2 {t_1}^4 + {p_1}^2 c_1 {c_2}^2 {t_2}^3 \\ & + 2 p_1 p_2 {c_1}^3 c_2 {t_2}^3 + p_1 p_2 {c_1}^2 {c_2}^2 {t_1}^2 t_2 + p_2 {c_1}^3 {c_2}^2 {t_2}^3 = a {p_1}^2 p_2 c_1 c_2 {t_1}^2 t_2. \end{align} According to the article, the genus of the curve is the number of integer lattice points in the interior of the Newton polygon of the curve.

The Newton polygon of the curve is the convex hull of the points $(1,0)$, $(2,1)$, $(0,1)$, $(1,2)$, $(3,1)$, $(2,2)$, $(3,2)$, $(1,1)$. There are two integer lattice points in the Newton polygon: $(1,1)$, $(2,1)$. Therefore the genus of this curve is $2$. Is this correct? How to check that whether or not this curve is smooth? Is this curve an elliptic curve? Thank you very much.

Prove that if a group $G$ is generated by all cyclic subnormal subgroups, then every cyclic subgroup is subnormal

Math Overflow Recent Questions - Thu, 08/02/2018 - 01:28

I have already found two definitions for a Baer group.

  1. $G$ is a Baer group if it is generated by all cyclic subnormal subgroups.

  2. $G$ is a Baer group if every cyclic subgroup is subnormal.

I want to prove the equivalence of the two definitions. Obviously, (2) implies (1). Please help me with the converse.

Derive unique denominators from number [on hold]

Math Overflow Recent Questions - Wed, 08/01/2018 - 22:01

I'm wondering if it is possible to run a set of numbers ('target numbers') through a function and get out a number that when queried against in some way (with a 'target' number as all/part of the query) would return a true / false as to whether or not that number exists in the output of the function. Tough for me to verbalize so I'll show an example

say my list of numbers is ListOfNumbers = [1,5,100,46736,3]

Can I run it through some function f(ListOfNumbers) = outputNumber

So that I can then run another function f(queryNumber,outputNumber) = true/false

also, the outputNumber (or one of the functions) has to be such that if I did something like this..

f(46736,outputNumber) = true but (23368,outputNumber) = false <-- it's a factor of 46736 but should not return true, only the numbers in the list return true.

Also, if it's not possible with one function, I wonder if it is possible with a couple different functions and if you get two or 'x' trues and no falses it is in the list.

Reference request for weak solutions of an Elliptic PDE

Math Overflow Recent Questions - Wed, 08/01/2018 - 20:19

Edit : I just learned that all weak solutions are $C^\infty$, so this question, by Willie, seems more appropriate than the current one.

I want to find weak, non trivial, continuous, solutions of $$\Delta u - \lambda u = 0$$ for a square domain in $\mathbb{R}^N$, $N \ge 2$, under periodic boundary conditions, and under an added constraint that, the weak solutions $u$ should take given values, at a given finite set of points in the interior of the domain. $u(x_i) = d_i$, $x_i$ lie in the interior of the domain, and $d_i$ are reals.

Reference request, if someone already solved it, or partially solved it or any relevant work. I am trying to solve it and I want to know if it makes sense, and I am not re-inventing, or barking up the wrong tree.

PS : Solving, I mean, having a numerical solution that converges pointwise, to the actual solution.

Finding a characteristic for which the zero-locus of an ideal is not empty

Math Overflow Recent Questions - Wed, 08/01/2018 - 15:08

I have a set of polynomials $f_1, \dots, f_m \in \mathbb{Z}[x_1, \dots, x_n]$ and I am interested in finding if these polynomials have a common root inside either $\mathbb{C}[x_1, \dots, x_n]$ or $\overline{\mathbb{F}_p}[x_1, \dots, x_n]$ for some prime $p$. One way to do this is to calculate the Gröbner basis of the ideal $$I = (f_1, \dots, f_m) \subseteq \mathbb{Z}[x_1, \dots, x_n].$$ If this basis contains some integer $k > 1$, then we can deduce that the ideal is not trivial inside $\overline{\mathbb{F}_p}[x_1, \dots, x_n]$ for any prime $p$ dividing $k$. If this basis contains a $1$, we find that $I$ is trivial in any characteristic, while if the basis does not contain an integer, we find that $I$ is non-trivial characteristic $0$. Since the computation of a Gröbner basis can be very difficult, I was wondering if there is another more direct way of computing this integer $k$ (whether or not it is $0$,$1$ or $>1$). Maybe there is another way altogether to determine whether or not a set of equations has a root in some characteristic.

I am also very interested in software that is capable of doing such a computation. So far I've used Mathematica and Sage for Gröbner basis computations, but I am not sure if these packages are the most well suited for the job.

Visual proof of convergence for Steiner's symmetrization

Math Overflow Recent Questions - Wed, 08/01/2018 - 08:48

I want to find a visual proof of the following fact:

For any convex figure in the plane there is a sequence of Steiner's symmetrizations that makes it arbitrary close to a circular disc.

All proofs I know require some integral estimates. I would prefer a more visual proof (even if it is more involved).

A kind of exponential concavity for polynomials?

Math Overflow Recent Questions - Wed, 08/01/2018 - 08:17

Is there $C > 0$ such that the inequality $$ \prod_{n\in\mathbb{N}} p(n)^{a_n} \leq p\left(C\prod_{n\in\mathbb{N}} n^{a_n}\right) $$ holds for all finitely supported sequences $(a_n)$ with $a_n\geq 0$ and $\sum_n a_n = 1$ and polynomials with nonnegative coefficients $p\in \mathbb{R}_+[X]$?

Is it even possible to take $C=1$?

It's clear that the inequality trivially holds with $C = 1$ whenever $p$ is a monomial.

Induction principle on proving an inequality

Math Overflow Recent Questions - Tue, 07/31/2018 - 18:52

If $P(z)=\sum_{k=0}^na_kz^k, (a_n=1)$ having no zeros in $|z|<1,$ I am trying to prove $$\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}\leq \frac{n-1}{2}+\frac{1}{1+|a_0|}.$$ The result is true for $n=1.$ Can I apply induction principle to prove this?

Attempt at a proof: let us try to show that the inequality holds by induction on the degree $n$ of the polynomial $P(z)$.

If $n=1$ then $P(z)=z-w$ with $|w|\geq 1$, and we have
$$\displaystyle\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}=\frac{1}{1+|w|}\leq \frac{1}{1+1},$$ which is nothing but the given inequality when $n=1$.

Let $Q(z)=(z-w)P(z)$ with $|w|\geq 1$, where $P(z)$ is a polynomial of degree $n$ having all its zeros in $|z|\geq 1$. $$\displaystyle\frac{\max_{|z|=1}|Q'(z)|}{\max_{|z|=1}|Q(z)|}=\frac{\max_{|z|=1}|(z-w)P'(z)+P(z)|}{\max_{|z|=1}|(z-w)P(z)|}$$ \begin{equation}\label{p1}\displaystyle\leq\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}+\frac{1}{\max_{|z|=1}|z-w|} (?).\end{equation}

By induction hypothesis we will have then $$\displaystyle\frac{\max_{|z|=1}|Q'(z)|}{\max_{|z|=1}|Q(z)|}\leq \frac{n-1}{2}+\frac{1}{1+|a_0|}+\frac{1}{1+|w|}\leq \frac{n}{2}+\frac{1}{1+|a_0||w|},$$ which is true follows from the fact that $$-\frac{1}{2}+\frac{1}{1+|a_0|}+\frac{1}{1+|w|}-\frac{1}{1+|a_0||w|}=\frac{(1-|a_0|)(1-|w|)(1-|a_0w|)}{2(1+|a_0|)(1+|w|)(1+|a_0w|)}\leq 0.$$

Thus the proof by induction will be complete once we establish (?) part of the above.

Expressive power of FO with $\mu$

Math Overflow Recent Questions - Tue, 07/31/2018 - 15:58

Let us consider the first-order logic extended with the least fixed point operator (FO+LFP). That is, together with the usual first-order formulas, we also have formulas of the form:

$$\mu X[\overline{y}] . \phi(X, \overline{y})$$

where $X$ (must occur positively in $\phi$) is a "predicate" variable of arity equal to the length of sequence of "parameters" $\overline{y}$. The semantic of this formula (in a given algebraic structure) is the least set $X^*$ such that:

$$X^*(\overline{y}) \Leftrightarrow \phi(X^*, \overline{y})$$

For example, if $R$ is a binary relation symbol, then:

$$\mu X[y_1, y_2] . R(y_1, y_2) \vee (\exists_z R(y_1, z) \wedge X(z, y_2))$$

defines the transitive closure of $R$.

If $A$ is an algebraic structure, let us write $\mathit{Th_{lfp}}(A)$ for the first-order theory of $A$ extended with the least fixed point operator (i.e. the set of all FO+LFP sentences that are true in $A$).

Does there exist an algebraic structure $A$ such that both of the following hold:

  • $\mathit{Th_{lfp}}(A)$ is decidable
  • FO+LFP is strictly more expressive than FO over $A$ (i.e. there is a FO+LFP formula that is not equivalent to FO formula over $A$)?

An example of a structure that satisfies the first property (but does not satisfy the second) is the structure of rational numbers with the natural ordering $\langle\mathcal{Q}, \leq\rangle$.

An example of a structure that satisfies the second property (but does not satisfy the first) is the structure of natural numbers with the natural ordering $\langle\mathcal{N}, \leq\rangle$.

One intuition is that there should be no such structure $A$ --- if $A$ defines arbitrary long well-founded orders, then the theory of $A$ should be undecidable; and if it does not define, then LFP seems to be pretty useless.

Another intuition is that there might be such a structure, because the above intuition is difficult to formalize, thus may contain essential holes.

Do choice principles in all generic extensions imply AC in $V$?

Math Overflow Recent Questions - Tue, 07/31/2018 - 11:58

It's well-known that not all choice principles are preserved under forcing, e.g. in this answer Asaf shows the ordering principle can hold in $V$ and fail in a generic extension. Indeed, the standard proof for preservation of AC is based on the fact that well-orderability is preserved under surjection, a fact that doesn't seem to have any nice generalization for weaker choice principles at all. So I wonder if we can get any results in the opposite direction.

Are there any known results of the form "If all generic extensions satisfy [some weak choice principle] then [some stronger choice principle] holds in $V$"?

I take choice principles to include e.g. AC, DC, AC$_{\omega}$, the selection principle, "all infinite sets are Dedekind-infinite," and "(strongly) amorphous sets don't exist." Two conjectures I want to focus on are:

Plausible conjecture: AC$_{\omega}$ in all generic extensions implies AC in $V$ (the idea here is that if there's a set in $V$ without a choice function, maybe there's a way to collapse its cardinality to $\omega$ without adding a choice function),


Ridiculous conjecture: If every generic extension has no strongly amorphous sets, then AC holds in $V$ (I can't believe this is true, but I also have no idea what property $V$ can have to prevent forcing amorphous sets).

If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?

Math Overflow Recent Questions - Mon, 07/30/2018 - 22:49

Let $E\neq \{0\}$ be a Banach space. For each $p\in[1,\infty), $ we define $$E\oplus_p E = \{(x,y): x\in E, y\in E, \|(x,y)\| = \sqrt[p]{\|x\|^p + \|y\|^p}\}.$$ Let $F$ be another Banach space. By $E\cong F,$ I mean that $E$ and $F$ are isometrically isomorphic.

Question: Suppose that $p,q\in [1,\infty).$ If $$E\oplus_p E \cong E\oplus_q E\,$$ then is it true that $p=q$?

If $E$ is of finite-dimensional, then the question is affirmative. However, I do not know what will happen if $E$ is of infinite-dimensional. I would be glad to see a proof if it is true or a counterexample if it is false.

We say that a norm $\phi:\mathbb{R}^2\to\mathbb{R}$ is normalized if $$\phi(0,1) = \phi(1,0) = 1.$$

Also, $\phi$ is monotone if for $|a_1|\leq |b_1|$ and $|a_2|\leq |b_2|,$ then $$\phi(a_1,a_2) \leq \phi(b_1,b_2).$$

We define $$E\oplus_\phi E = \{(x,y): x\in E, y\in E, \|(x,y)\| = \phi(\|x\|, \|y\|) \}.$$

A more general question:

Suppose that $\phi,\psi:\mathbb{R}^2\to \mathbb{R}$ are norms that satisfy normalization and monotonicity. Assume that $\phi$ and $\psi$ are not $\ell^\infty$ norm. If $$E\oplus_\phi E \cong E\oplus_\psi E,$$ then is it true that $\phi = \psi?$?

compare N(f,a,r) with T(f,r)

Math Overflow Recent Questions - Mon, 07/30/2018 - 09:06

I'm reading William Cherry and Zhuan Ye's book 'Nevanlinna's theory of value distribution, the second main theorem and its error terms'. In Section 1.12, they explains why $N$ and $T$ is used in Nevanlinna theory instead of $n$ and $A$, where $A(f,r)=\int_{D(t)}f^\ast\omega$. Then they gave some results on the comparison of $n(f,a,r)$ and $A(f,r)$. For example,

Gol'dberg in 1978 constructed an entire function $f$ such that for every $a\in \mathbb{C}$ such that $$\limsup_{r\to \infty}\frac{n(f,a,r)}{A(f,r)}=\infty.$$

In another direction, Hayman and Stewart in 1954 formulate a theorem

Let $f$ be a non-constant meromorphic function on $\mathbb{C}$ and set $n(f,r)=\sup_{a\in \mathbb{P}^1}n(f,a,r)$. Then $$1\le\liminf_{r\to \infty}\frac{n(f,r)}{A(f,r)}\le e.$$

My question is: what can we say about $\frac{N(f,a,r)}{T(f,r)}$?

Firstly, by FMT we know $\frac{N(f,a,r)}{T(f,a,r)}\le 1$.

I also checked some elementaty functions. For examples, exponential function $f(z)=e^z$. By a simple calculation, $N(f,\infty,r)=0,N(f,a,r)=\frac{r}{\pi}+O(\log r),T(f,r)=\frac{r}{\pi}$. So $$\frac{N(f,\infty,r)}{T(f,r)}=0,\frac{N(f,0,r)}{T(f,r)}=1.$$ When does this ratio nonzero for a general meromorphic function?

Any reply or reference is appreciated.

reference request about COVERING SETS OF THE INTEGERS

Math Overflow Recent Questions - Mon, 07/30/2018 - 08:43

I am looking for the following reference:

Krukenberg, Claire Emil, COVERING SETS OF THE INTEGERS. Thesis (Ph.D.)–University of Illinois at Urbana-Champaign. 1971. 104 pp.

I searched the internet but no success. I would appreciate the link for downloading this reference.


References for the study of parameter dependent symbols $s(t,x,\xi)$ having low regularity in parameter($t$)

Math Overflow Recent Questions - Mon, 07/30/2018 - 07:24

I am currently studying parameter dependent symbols, $s(t,x,\xi)$, where $t\in [0,1],x\in \Omega, \xi \in \mathbb{R^n}$. I wanted to know how the low regularity (for example, $s$ is just continuous w.r.t. $t$) of symbol w.r.t. parameter affects further study of symbols and the corresponding operators.

To be specific, I need this to study hyperbolic operators which are low regular in time. All the books which I have referred till now assume smoothness w.r.t. parameter.

Thanking you in advance.

Differentiability of a stochastic process depending on a spatial parameter

Math Overflow Recent Questions - Mon, 07/30/2018 - 07:19


  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $T>0$
  • $I:=(0,T]$
  • $d\in\mathbb N$
  • $M:\Omega\times\overline I\times\mathbb R^d\to\mathbb R$ such that $M(\;\cdot\;,\;\cdot\;,x)$ is $\mathcal A\otimes\mathcal B(\overline I)$-measurable for all $x\in\mathbb R^d$

Now, let $i\in\left\{1,\ldots,d\right\}$ and $$N(\omega,t,x,\theta):=\frac{M(\omega,t,x+\theta e_i)-M(\omega,t,x)}\theta$$ for $(\omega,t,x,\theta)\in\Omega\times\overline I\times\mathbb R^d\times\left(\mathbb R\setminus\left\{0\right\}\right)$.

Let $\delta\in(0,1]$. Assuming that for all $p\ge2$, there is a $C>0$ such that $$\int\sup_{t\in\overline I}\left|N(\omega,t,x,\theta)-N(\omega,t,y,\vartheta)\right|^p\operatorname P\left[{\rm d}\omega\right]\le C\left(|x-y|^{\delta p}+|\theta-\vartheta|^{\delta p}\right)\tag1$$ for all $(x,\theta),(y,\vartheta)\in\mathbb R^d\times\left(\mathbb R\setminus\left\{0\right\}\right)$, how can we conclude that $M(\omega,t,\;\cdot\;)$ is partially differentiable with respect to the $i$th variable for $\operatorname P$-almost all $\omega\in\Omega$ for all $t\in\overline I$?

This should be an application of a Kolmogorov-type theorem, but which version of that theorem do we need and how do we need to apply it exactly?

Covering system of congruences with specific properties?

Math Overflow Recent Questions - Mon, 07/30/2018 - 07:00

A family of residue classes $a_i (\mod n_i)$ with $2\leq n_1\leq\cdots\leq n_r$ is called a covering system of congruences if every integer belongs to at least one of the residue classes, that is, every integer satisfies at least one of the congruences $a_i (\mod n_i)$. The known examples are:

$0 (\mod 2),\ 0 (\mod 3),\ 1 (\mod 4),\ 5 (\mod 6),\ 7 (\mod 12)$

$0 (\mod 2),\ 0 (\mod 3),\ 1 (\mod 4),\ 3 (\mod 8),\ 7 (\mod 12),\ 23 (\mod 24)$

My question is that is it possible to construct a covering system of congruences for non perfect square odd integers $>1$ such that

$$ \left(\frac{a_i}{n_i}\right)=-1,\qquad \text{for}\ \ 1\leq i\leq r $$

where the parenthesis is Legendre (or Jacobi ) symbol.


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