Recent MathOverflow Questions

Classification of a system of two second order PDEs with two dependent and two independent variables

Math Overflow Recent Questions - Sat, 07/07/2018 - 02:11

If we have a second order quasilinear PDE of the form

$A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+ lower\,\, order \,\, terms=0$

where $A,B,C$ are functions of $x,y,u$,

then the equation is called elliptic if $det=\begin{vmatrix}A &C \\C & B\end{vmatrix}>0$, parabolic if $det=0$ and hyperbolic if $det<0$.

Now what happens if we have a system of two coupled PDEs of the form

$A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+D\frac{\partial^2 v}{\partial x^2}+E\frac{\partial^2 v}{\partial y^2}+2F\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0 \\ G\frac{\partial^2 u}{\partial x^2}+H\frac{\partial^2 u}{\partial y^2}+2K\frac{\partial^2 u}{\partial x\partial y}+L\frac{\partial^2 v}{\partial x^2}+M\frac{\partial^2 v}{\partial y^2}+2N\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0$

with $A,B,C,...$ being functions of $x,y,u,v$.

Does it make sense to construct the determinant

$det=\begin{vmatrix}A &C & D&F\\C & B&F&E\\G &K & L&N\\K &H & N&M\end{vmatrix}$

and investigate its sign, or something like that?

--------- Update ----------

In the chapter 10 of the book [Seiler, Werner M., Involution. The formal theory of differential equations and its applications in computer algebra] the author states that since ellipticity is a property defined at points and depends only on the principal symbol it then suffices to study it for linear equations. I understand that the same procedure holds for quasilinear ones, the only difference is that the coefficients are allowed to be functions of the dependent variables. Then regarding the specific example above if one implement the theory for linear equations the principal symbol should be

$\tau[\xi,\eta]=\begin{bmatrix}B \eta^2 + 2 C \eta\xi + A \xi^2& E\eta^2 + 2 F \eta\xi + D \xi^2\\ H\eta^2 + 2 K \eta\xi + G \xi^2 & M\eta^2 + 2 N \eta\xi + L \xi^2\end{bmatrix}$

Ellipticity requires $det(\tau)\neq 0$ that is

$(B M-E H) \eta^4 + 2 (C M + B N-F H - E K ) \eta^3 \xi + (B L + A M + 4 C N-E G - D H - 4 F K ) \eta^2 \xi^2 + 2 ( C L + A N-F G - D K ) \eta \xi^3 + (A L-D G ) \xi^4\neq 0$

well, when this condition is to be satisfied? Is it true that it is hyperbolic in direction $x$ or $y$ direction, if

$(B M-E H) \eta^4 + 2 (C M + B N-F H - E K ) \eta^3 \xi + (B L + A M + 4 C N-E G - D H - 4 F K ) \eta^2 \xi^2 + 2 ( C L + A N-F G - D K ) \eta \xi^3 + (A L-D G ) \xi^4 = 0$

possesses only simple real roots?

Which primes $\mathfrak{p} \in \mathbb{Z}[i]$ can be represented as $\mathfrak{p} = x^2 + 2y^2$?

Math Overflow Recent Questions - Fri, 07/06/2018 - 09:22

Consider the quadratic form $x^2 + 2y^2$ over the ring $\mathbb{Z}[i]$. It is irreducible, so I wanted to know which primes could be represented by this quadratic form.

$$ \mathfrak{p} = x^2 + 2y^2 $$

There's two slightly different questions here. One about the integer $\mathfrak{p} \in \mathbb{Z}[i]$ and one about the ideals $(\mathfrak{p}) \subseteq \mathbb{Z}[i]$. This is still a PID so I can write just one element.

This hopefully in analogy to Fermat's theorem on primes as the sum of two squares, where it is solved by a congruence condition. The analogous question for $x^2 + y^2 = (x+iy)(x-iy)$ is degenerate.

A conjecture on simplex

Math Overflow Recent Questions - Fri, 07/06/2018 - 01:45

Let $A_0A_1...A_n$ be a simplex in $\Bbb E^n.$ Let $B_{ij}$ be the points on the edges $A_iA_j,\ 0\le i\not=j\le n.$

Denote by $(\beta_i)$ the hyperplane passing through points $B_{i0},$ $B_{i1},$ $B_{ii-1},$ $B_{ii+1},$ $...,$ $B_{in}.$

Assume that the reflections of $A_i$ in hyperplane $(\beta_i)$ lies on the hyperplane $(A_0A_1...A_{i-i}A_{i+1}...A_n).$

My question. Can we show that $B_{ij}$ must be midpoints of the edges $A_iA_j?$

About the product of finite subsets of a torsion free group

Math Overflow Recent Questions - Fri, 07/06/2018 - 01:41

Let $G$ be a torsion free group with identity $e$. For a subset $X$ of $G$, denote by $X^\#$ the set $X\setminus\{e\}$. Let $A$ be a finite subset of $G$ containing $e$. Is there a finite subset $B$ containing $e$ such that $$A\subset B^\#A\quad\text{and}\quad B\subset BA^\#$$ It is obvious there is no such $B$ when $A=\{e,x\}$; this follows from the right inclusion and that $G$ is torsion free.

distribution of supremum of quadratic form

Math Overflow Recent Questions - Fri, 07/06/2018 - 00:39

For any given vector $w$, we have score test statistic $\frac{w^{T}Aw}{\sigma^{2}}$, A is some symmetric matrix, from asymptotic theory, we know this test statistic asymptotically follows $\chi^{2}$ distribution with some degree of freedom,e.g. d, my question is: what is the distribution of $sup_{w}\frac{w^{T}Aw}{\sigma^{2}}$?

Cyclic quadrilateral in metric space

Math Overflow Recent Questions - Fri, 07/06/2018 - 00:29

Consider a metric space $(\Bbb M,d).$

If $X,Y,Z\in \Bbb M.$ We define cosin of angle by

$$\cos(\angle YXZ)=\frac{d(X,Y)^2+d(X,Z)^2-d(Y,Z)^2}{2d(X,Y)\cdot d(X,Z)}.$$

If we have four points $A,$ $B,$ $C$ and $D$ in $\Bbb M$ satify the indentity

$$d(A,B)\cdot d(C,D)+d(A,D)\cdot d(B,C)=d(A,C)\cdot d(B,D).$$

Then we say that $A,$ $B,$ $C,$ $D$ are concyclic.

My Question 1. If $A,$ $B,$ $C$ and $D$ are concyclic then can we show that $$\cos(\angle BAC)=\cos(\angle BDC).$$

My Question 2. If $A,$ $B,$ $C$ and $D$ are concyclic then can we show there is a point $O$ such that $$d(O,A)=d(O,B)=d(O,C)=d(O,D).$$

My Question 3. If two questions 1 and 2 are not true then what is condition of $\Bbb M$ such that these are true?

Higher degree of Hilbert's irreducibility theorem

Math Overflow Recent Questions - Fri, 07/06/2018 - 00:23

A basic form of Hilbert's irreducibility theorem can be formulated as follows:

Let $f(t,x)\in\mathbb{Q}[t,x]\setminus\mathbb{Q}[t]$ be an irreducible polynomial. There exist infinitely many linear polynomials $p(t)\in\mathbb{Q}[t]$, $p(t)=t-t_0$ such that $(p(t),f(t,x))$ is a maximal ideal of $\mathbb{Q}[t,x]$.

My question is: can we drop the "linear" assumption above and have polynomials $p(t)$ of arbitrarily high degree? More precisely, is the following true:

Let $f(t,x)\in\mathbb{Q}[t,x]\setminus\mathbb{Q}[t]$ be an irreducible polynomial. Let $n$ be a positive integer. There exists a polynomial $p(t)\in\mathbb{Q}[t]$ such that $\deg p(t)>n$ and $(p(t),f(t,x))$ is a maximal ideal of $\mathbb{Q}[t,x]$.

Of course $p(t)$ must be irreducible itself, but that's far from enough.

Decomposition of $\widehat{k^{\times}}$ occuring in local class field theory

Math Overflow Recent Questions - Thu, 07/05/2018 - 23:37

Let $k$ be a finite extension of $\mathbb{Q}_p$ very often we use the isomorphism that $Gal(\overline{k}/k)^{ab} \simeq \hat{(k^{\times})}$ given by local class field theory. My question would be do we have (and if yes how can I prove it) $ \hat{O_k^{\times}} \times \hat{\mathbb{Z}} \simeq \hat{(k^{\times})}$ ? and is the image of the inertia subgroup of $Gal(\overline{k}/k)$ in $Gal(\overline{k}/k)^{ab}$ isomorphic to $ \hat{O_k^{\times}}$ ? I imagine that the proof of the first point can come from the fact that $\hat{k^{\times}} \simeq \widehat{O_k^{\times} \times \mathbb{Z}}$ using the decomposition with an uniformizer. But then can I split the product ? Do we have $\widehat{O_k^{\times}} = O^{\times}_k$ ?

representing an uncountable free group as a union of an increasing sequence of countable subgroups

Math Overflow Recent Questions - Thu, 07/05/2018 - 23:18

Let $(G_\alpha)$ and $(K_\alpha)$ $(\alpha<\aleph_1)$ be strictly increasing chains of countable sets such that if $\alpha$ is a limit, then $G_\alpha=\bigcup_{\beta<\alpha}G_\beta$ and $K_\alpha=\bigcup_{\beta<\alpha}K_\beta$. Assume $\bigcup_{\alpha<\aleph_1}G_\alpha=\bigcup_{\alpha<\aleph_1}K_\alpha$. Does there exist a club $C$ such that $K_\gamma=G_\gamma$ for all $\gamma\in C$?

In the paper ``The Abelianization of Almost Free Groups" (the end of the proof of Lemma 2.5 on page 1801), this assertion is made where the $G_\alpha$ and $K_\alpha$ are subgroups of a free group with some additional properties, but the author makes the assertion I want without any explanation, so I assume the reason must be simple and may not depend on the group theory.

Infinite connected graphs isomorphic to their line graph

Math Overflow Recent Questions - Thu, 07/05/2018 - 23:14

For any simple, undirected graph $G$, let $L(G)$ denote its line graph.

$G=(\mathbb{Z}, E)$ with $E = \{\{k, k+1\}:k\in \mathbb{Z}\}$ has the property that $G\cong L(G)$.

Is there a connected infinite graph $G$ such that every vertex has more than $2$ neighbors, and $G\cong L(G)$?

Can $S_n$ be partitioned into subsets containing an involution and satisfying $∀σ≠τ, ∃j$ s.t. $σ(j)≠τ(j),σ^{−1}(j)=τ^{−1}(j)$?

Math Overflow Recent Questions - Thu, 07/05/2018 - 23:09

Let $\sigma, \tau \in S_n$. We will say that $\sigma$ and $\tau$ are locally orthogonal and write $\sigma \perp \tau$ if there exists $j \in \{1, 2, \ldots, n\}$ such that $\sigma(j) \neq \tau(j)$ but $\sigma^{-1}(j) = \tau^{-1}(j)$. (I will explain this terminology in the Motivation section below.) We will call a subset $S \subset S_n$ exclusive if $\sigma \perp \tau$ for any $\sigma, \tau \in S$, $\sigma \neq \tau$. Observe that any exclusive subset of $S_n$ can contain at most one involution.

We say $\sigma, \tau \in S_n$ are Knuth equivalent if they have the same insertion tableau under the Robinson-Schensted correspondence. This is the same as $\sigma^{-1}$ and $\tau^{-1}$ being dual Knuth equivalent, i.e., having the same recording tableau. Involutions are precisely those permutations that have the same recording and insertion tableaux, so each Knuth equivalence class contains exactly one involution, as does each dual Knuth equivalence class.

  1. Does there exist a partition of $S_n$ into exclusive subsets such that each subset contains an involution?

  2. Does either Knuth equivalence or dual Knuth equivalence imply local orthogonality? Either would imply a positive answer to Q1.

  3. Has this notion of local orthogonality for elements of $S_n$ been defined anywhere before (presumably under some other name)?

I have verified Q1 by brute force for $n \leq 11$.


The notion of local orthogonality appears in a more general context as part of a necessary condition for multipartite quantum correlations. A positive answer to Q1 would imply that in the variant of the 100 prisoners problem (sometimes called the locker puzzle) in which $n$ prisoners may only open 2 drawers, the classical best solution cannot be improved upon using shared quantum entanglement between the prisoners. (Note that this is different from the quantum version of the game considered by Avis and Broadbent in which the prisoners are allowed to open a superposition of drawers.)

Cancellation in this exponential sum?

Math Overflow Recent Questions - Thu, 07/05/2018 - 22:53

I would like to know whether it is possible to obtain cancellation in the sum

$$\sum_{p \leq X} e^{{2\pi iX}/{p}}$$

where $X$ is a real number that goes to $\infty$, and $p$ denotes a prime number.

Pascal's theorem for spherical hexagon

Math Overflow Recent Questions - Thu, 07/05/2018 - 22:01

I draw a cyclic spherical hexagon and I check by geogebra that Pascal's theorem is true in this case.

My question 1. Is there simple proof for this?

My question 2. Can we change the circle on sphere by the curve like "conic" in plane?

My question 3. If we use orthogonal projection to project this configuration onto a plane. Which do the plane problem we obtain?

Is there any charactrization for lifting clopen subsets

Math Overflow Recent Questions - Thu, 07/05/2018 - 21:01

Let $Y$ be a subset of a topological space $X$, we say that clopen subset of $Y$ lift to $X$ whenever $L$ is a clopen subset of $Y$ then there exists a clopen subset $H$ of $X$ such that $H\cap Y=L$.

Let $X$ be a compact and $T_0$-space and $Y$ be a closed of $X $, I am looking for equivalent conditions under which the clopen subsets of $Y $ lift to $X $.

Von Neumann's theorem on realizing automorphisms of the measure algebra

Math Overflow Recent Questions - Thu, 07/05/2018 - 20:17

I'm looking for a proof, in English, of the following theorem due to von Neumann (which apparently originates in the paper Einige Sätze über Messbare Abbildungen, Ann. of Math, 1932):

Every automorphism of the Boolean algebra of (Lebesgue) measurable subsets of $[0,1]$ modulo null sets is realized by a Borel measurable bijection on $[0,1]$.

Obviously, $[0,1]$ can be substituted with any standard finite measure space.

I know that this is generalized by results of Mackey, Maharam, and others, but I'm really just interested in the original result and its proof. Alas, I cannot read German.

Is $G$ non-solvable?

Math Overflow Recent Questions - Thu, 07/05/2018 - 19:33

Let $G$ be a finite group of order $2^7\cdot3^3\cdot5^2\cdot7$. Let $\mathrm{Irr}(G)$ be the set of all the irreducible $\mathbb{C}$-characters. Suppose that

(1) there is a character $\chi\in\mathrm{Irr}(G)$ such that $2^5\cdot7|\chi(1)$;

(2) there is a character $\theta\in\mathrm{Irr}(G)$ such that $5^2\cdot7|\theta(1)$;

(3) there is a character $\xi\in\mathrm{Irr}(G)$ such that $3^3\cdot7|\xi(1)$;

Question: Is $G$ non-solvable?

Coloring circles in plane

Math Overflow Recent Questions - Thu, 07/05/2018 - 16:47

We assume that all the circles in the plane are each colored with one of two colors: red or blue.

My question 1. Does there always exist an equilateral triangle such that its circumcircle and its incircle have the same color?

My question 2. Does there always exist an equilateral triangle such that its incircle and its three excircles have the same color?

My question 3. Does there always exist a regular polygon such that its circumcircle and its incircle have the same color?

fractional Brownian Motion driven stochastic integrals

Math Overflow Recent Questions - Thu, 07/05/2018 - 16:39

We consider a stochastic process $\left(X_{t}\right)_{t\geq 0}$, defined as an integral process, s.t. $$X_{t}=\int_{0}^{t}u_{s}\,dB_{s}^{H}.$$ With a fractional Brownian motion $B^H_{t}$. If $H\neq\frac{1}{2}$, the stochastic integral can not be defined in the classical Itô sense, due to Bichteler-Dellacherie theorem.

Using the classical Young theory, $X_{t}$ is well defined, if the trajectories of $u_{t}$ has finite $q$ variation, if $q<\frac{1}{1-H}$.

Question 1: Is it possible to define $X_{t}$ in such a way, that the trajectories of $u_{t}$ don't have to be restricted w.r.t. there regularity? As far as I know, it is possible to use Rough Path Theory, to extend Young's classical result. Up to which extent, is it possible to extend Young's theory by rough path theory?

Question 2: Which classical stochastic analysis tools are available using the rough path approach? More precisely are there substitutes of the following classical tools?

  • Itô formula
  • Burkholder inequality (Upper bounds for moments of $X^{*}_{t}=\underset{s\leq t}{\text{sup}}\,X_{s}$ )

Question 3: Is it possible to extend Young's approach using other tools?

  • Regularity structures
  • Malliavin Calculus (Skorohod integral)
  • White Noise Analysis
  • ...

Example of tensor category with non-simple unit $J\to \mathbb{1} \to Q$ and suitably extension $Q\to M\to J$

Math Overflow Recent Questions - Thu, 07/05/2018 - 15:24

Edit: Thanx very much to Neil Strickland for quickly explaining to us that the following cannot be realized over finite commutative $\mathbb{C}$-algebras, as I had originally asked.

I know that there is a finite tensor category (from minimal models) with the following relations that seem rather strange to me. In particular it seems now one cannot realize it as modules over a ring. Does anyone know an algebraic situation, where something similar occurs? (maybe in a derived category?).

A non-simple unit object $\mathbb{1}$

$$0\to J\to \mathbb{1} \to Q \to 0$$ $$J\otimes Q=\{0\}$$ $$Q\otimes Q=Q$$ (for modules over a ring $R,\otimes_R$, this means $J$ is an ideal with $J^2=J$, thus $R$ often splits, see below).

such that $J\otimes J$ is an extension the-other-way-around $$M:=J\otimes J$$ $$0\to Q\to M \to J \to 0$$ (for modules over a ring $R,\otimes_R$ the product $J\otimes J$ cannot be larger then $J$)

and which acts somewhat like a second identity $$M\otimes M=M $$ $$M\otimes Q=\{0\}$$ $$M\otimes J= M$$

Any hints what some of these situations are called in literature are also very welcome.

Thanx very much for your help in advance!

Example of a ring with non-finitely generated unit group?

Math Overflow Recent Questions - Thu, 07/05/2018 - 14:08

The well known Dirichlet's unit theorem states that the unit group of a maximal order in a quadratic number field is finitely generated of rank blah blah blah. I think it's pretty naive to expect a most radical generalization to hold, namely:

Generalized Dirichlet unit theorem. The unit group of an associative unital ring with finitely generated additive group is finitely generated.

Is there any concrete counterexample?


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