User qingyun - MathOverflow

system of homogeneous matrix equations

2013-02-01T07:42:04Z

Edit: Maybe I should make it clear that by a solution I mean a pair of matrices $(A,B)$ such that the identity below holds for any complex numbers $x,y$.

One of my friend asked me the following question (with some variation): Let $A$,$B$ be two $n\times n$ complex matrices, let $x$ and $y$ be two complex variables. Suppose that $$ (xA+yB)^n=(x^n+y^n)I_n $$ For all $x$ and $y$, Where $xA$ is the scalar multiplication of $A$ by $x$. $I_n$ is the identity matrix.

Question: Can we get explicit solutions when $n$ is small? Can we at least say that the solutions forms a manifold of certain dimension for arbitrary $n$?

You can expand both sides, and move the right side to the left, getting a polynomial whose coefficients are matrices which is identical to $0$. This means the coefficients are all $0$, thus we get a system of homogeneous equations on $A$ and $B$. We can observe that the solution set is stable under taking conjugation by the same invertible matrix. Since $A^n=I_n$, we know that it's diagonalizable, whose eigenvalues are $n$-th roots of unity. Hence we can assume that $A$ is a diagonal matrix. By this simplification, using elementary methods, I was able to find the explicit solution for n=2 and 3. But as $n$ getting large, even for $n=5$, the redundant method requires one to solve a system of polynomial equations of 25 variables. I actually tried it using mathematica, but my poor computer became unresponsive very soon. Any suggestion is appreciated.

Interesting examples of minimal action on torus

2013-01-29T17:49:58Z

Edit 1:This is a cross post on MSE. See math.stackexchange.com/q/289595/12952

Edit 2:I originally asked for finite group actions as I thought that will be easier. But as pointed out by Victor minimal action does not exist for finite groups. So I am asking for general discrete group actions. What I really want to know is some interesting examples of minimal actions (not just a single homeomorphism) on suitable nice topological space. I just read an article on Furstenburg transformation am I was guessing the construction could be generalized to give minimal actions.

For a n-torus $\mathbb{T}^n$, A Furstenburg transformation $\phi$ is defined by: $$ \phi(\xi_1,\xi_2,\dots,\xi_n)=(e^{2\pi i\theta}\xi_1, f_1(\xi_1)\xi_2,\dots,f_{n-1}(\xi_1,\dots,\xi_{n-1})\xi_n) $$ Where $\theta\in \mathbb{R}$ and for each $i$, $f_i$ is a real continuous function on $\mathbb{T}^i$.

It is known that when $\theta$ is irrational, (Edit3: and all functions $f_i$ are in suitable homotopy classes not containing the constant functions) Furstenburg transformation defines a minimal dynamic system.

My question is, are there any interesting examples of minimal actions of other discrete groups on n-torus? I am thinking something similar to Furstenburg transformation, but any other examples are welcome too.

Finite projection in Von Neumann algebra

2012-09-28T16:26:39Z

I had the following question when I am learning von Neumann algebras: Let p be a finite projection in a finite von Neumann algebra $M$, let $p>p_1>p_2>\cdots$ be a decreasing sequence of projections such that $p_i\neq p_{i+1}$. Is such a sequence necessarily of finite length? Any help or recommendations on books/papers are greatly appreciated.

Group C* algebras of finite groups

2012-05-08T20:58:09Z

I am learning group C* algebras for my graduate research. I've known that C*(G)=C(\hat{G}), if G is abelian. What can we say if the group is not abelian? Do we have explicit description of C*(G) if G is finite non-abelian? Any suggestions or recommendations on books/papers will be greatly appreciated.

positive elements in tensor products

2010-10-22T07:25:24Z

Let $A \otimes B$ be the algebraic tensor of two $C^{\ast}$ -algebras, and an element x in $A\otimes B$ is positive if $x=yy^{\ast}$. Then is it always possible to write x in the form $x=\sum a_i\otimes b_i$, where $a_i$ and $b_i$ are positive elements?

Integral interpolation by polynomial

2011-04-28T21:37:41Z

This question arises from a discussion with my friends on a commonly encountered IQ test questions: "What's the next number in this series 2,6,12,20,...". Here a "number" usually means an integer. I was wondering whether there is a systematical way to solve such problems.Let us call a point on a plane integer point if all the components of it are integers. I want to know the following:

Give a finite set of integer points, Can we always find a corresponding polynomial that passes all these points and maps integers to integers?

A Perturbation problem for U(n)

2011-03-26T08:54:59Z

Let G be a finite subgroup of U(n), the unitary group acts on $\mathbb{C}^n$. If there is a unit vector $x$ in $\mathbb{C}^n$ such that g(x) is almost orthogonal to x, for all $g\in G$ except the identity, can we perturb x so that g(x) is exactly orthogonal to x, for all $g\in G$ except the identity? More precisely, can we find a very small number $\epsilon>0$, so that if there exist a unit vector $x$ and the inner product $|(g(x),x))|<\epsilon$ for all $g\in G$ \ {1}, then we can find another unit vector $y$, such that $(g(y),y)=0$ for all $g\in G$ \ {1}? Is it possible to further require that $||x-y||$ be small too?

positive element in C* tensor product

2011-03-11T18:54:23Z

Let A, B be two C*-Algebras and $A\otimes B$ denote their minimal tensor product(I don't know whether C*-norm matters or not, but for simplicity we can assume that one of them is nuclear so all C*-norm coincide). Let x be a non-zero positive element in $A\otimes B$, can we always find a single tensor $0\neq x_1\otimes x_2$, where $x_1$ and $x_2$ are positive elements in A and B respectively, such that $x_1\otimes x_2\leq x$?

It's fairly easy to see that if both C*-algebras are communicative or one of them is a finite dimensional C*-Algebra(Sorry this is false), then the above assertion is true. So it's tempting to think that more general case should hold.

I asked a similar question before, where the stronger assertion that any positive element in tensor algebra is a sum of tensors of positive elements, is false. See the following link:

link text

Perturbation in C*-Algebra

2011-01-09T21:35:55Z

Let x be an element in a C*-algebra A, is it true that if x approximately commute with every element in A, then x is near the centre of A? More precisely, I want to know whether the following is true： Let x be an element in a C*-algebra A with norm 1. Then for any $\epsilon>0$, there exist a $\delta>0$ such that the following holds: $\forall y\in A,||y||=1,||xy-yx||<\delta\Rightarrow dist（x, Centre(A)）<\epsilon$. This is true if A is the matrix algebra, but I was wondering whether it can be generalized to any C*-algebra

Projection in Hereditary C* subalgebra

2010-10-29T07:48:06Z

This is actually something in a paper but the author claimed it without proof. Let x be a positive elment of norm 1 in a $C^*-$algebra A, and Her(x) is the hereditary subalegbra generated by x. Given $\epsilon>0$,let $f_\epsilon$ be thecontinuous function on R defined as follow:

$f_\epsilon \equiv 0 \quad on \quad [-\infty,\epsilon/2]$

$f_\epsilon \quad is \quad linear \quad on\quad [\epsilon/2,\epsilon]$

$f_\epsilon \equiv 1 \quad on\quad [\epsilon, +\infty]$

So $f_\epsilon$ increase to the identiy function on [0,1] when $\epsilon$ decrease to 0, and $Her(x)=\overline{\cup_{\epsilon>0} f_\epsilon(x)Af_\epsilon(x)}$. Let p be a projection in Her(x). then how do we know that there must exist a $\epsilon$ such that $p\in \overline{f_\epsilon(x)Af_\epsilon(x)}$? Or more generally, Let A be the inductive limit of {$A_n$}, and p is a projection in A,does it follow that p is actually in some $A_n$?

Comment by Qingyun

2013-02-02T23:25:49Z

I do not know the background of this equation, the person who asked me this problem is working in algebraic geometry which I know nothing about. You answer is very helpful, are these the only solutions?

Comment by Qingyun

2013-02-02T23:21:28Z

I see, as a corollary of your conclusion, there is no solution if $n$ is not dividing the size of matrices, and $xA+yB$ will have distinct eigenvalues if $n$ equal to the size of matrices, are I right?

Comment by Qingyun

2013-02-01T19:29:04Z

Sorry for not making the question clear, I am looking for matrices $A,B$ such that the identity holds for all $x,y$. I guess my terminology is incorrect.

Comment by Qingyun

2013-01-31T22:04:26Z

@ Alain Valette @ Michele Triestino Thanks!

Comment by Qingyun

2013-01-31T21:50:31Z

@Lee Mosher Yes you are right, thanks for pointing this out. The correct statement should be that the functions $f_i$ are in suitable homotopy classes other than the one containing constant functions. The details are in Theorem 2.1 of Furstenberg's paper STRICT ERGODICTICY AND TRANSFORMATION OF THE TORUS and the remark after it.

Comment by Qingyun

2011-04-01T18:14:54Z

Since F(n) is homogeneous degree 0, F(n) always has a minimum point. If we take the gradient and set it to 0, we get a system of homogeneous polynomial equations(k equations and k variables), it seems that we should be able to solve it and thereafter find the minimum of F(n), but I've no idea of how to deal with such a system.

Comment by Qingyun

2011-03-27T19:43:06Z

I was trying to restate the question in the following way: For any finite subgroup G of U(n), define $\lambda_G=$$inf_{x\in\mathbb{C}^n}$$\sum_{g\neq1}$|(gx,x)|.So you argument shows that $inf$ {$\lambda_G\neq 0$} is non-zero, but this lower bound depends on n. Am I understanding correctly?

Comment by Qingyun

2011-03-26T19:04:41Z

Things may be more complicated than you thought. If the dimension is not 3, then a rotation may not has an axis (or more than one axis?). For example, consider $\mathbb{R}^6=R^2\oplus R^2\oplus R^2$, let $g\in G$ be a rotation of the form $g_1\oplus g_2\ oplus g_3$, where $g_i$ is a rotation of $R^2$ by angle $\theta_i$, the i-th root of unity. If $x=(x_1,\dots,x_6)$, then (gx,x) is a convex combination of $cos(\theta_i)$, which can be arbitrary small without being 0.

Comment by Qingyun

2011-03-12T07:45:58Z

Got it, thanks!

Comment by Qingyun

2011-03-12T06:55:19Z

Right, I made a mistake

Comment by Qingyun

2011-01-10T17:29:29Z

This is exactly what I was looking for. Thank you guys!

Comment by Qingyun

2010-11-01T21:12:20Z

To Andreas Thom: Corrected. Thanks