User mahmood alaghmandan - MathOverflow

Norm functionals of $B(H)$ restricted to sub ven-Neumann algebras

2012-02-06T06:56:34Z

Let $H$ be a Hilbert space, we know that weak topology over $B(H)$, operator algebra of bounded linear operators from $H$ into $H$, is the topology generated by ${\langle \cdot \xi,\eta\rangle:\; \xi,\eta\in H}$.

So naturally, I think about the norm of $\langle \cdot \xi,\eta\rangle$ as a linear functional over $V$ a von Neumann subalgebra of $B(H)$. And I guess that $\|\langle\cdot \xi,\eta\rangle\|=\inf\{\|\xi'\|_H \|\eta'\|_H:\; s.t.\;\langle T \xi',\eta'\rangle = \langle T \xi,\eta\rangle\; \forall T\in V\}$. But I am not sure how can I show that. Indeed I am wondering whether this is correct or not even!

Are all of compact support functions of $A(G)$ in its abstract Segal algebras?

2011-08-17T09:48:13Z

Let $G$ be a locally compact group. We know that if $G$ is abelian and $\cal F$ implies the Fourier transform, for every Segal algebra of $G$ say $S^1(G)$, ${\cal F}S^1(G)$ is an abstract Segal algebra with respect to $A(\widehat{G})$ (since $S^1(G)$ is an abstract Segal algebra of $L^1(G)$). Reiter in [1, Proposition 6.2.5] has shown that $C_c(\widehat{G})\cap A(\widehat{G}) \subseteq {\cal F}S^1(G)$.

My Question is: Can we show that for an aribtarary locally compact group $G$ (not necessarily abelian) every abstract Segal algebra of $A(G)$ contains all functions in $A(G)$ that have compact support?

[1] H. Reiter, and J. D. Stegeman, Classical harmonic analysis and locally compact groups, 2nd edn, London Mathematical Society Monographs, New series 22, Oxford university press, New York, 2000.

Answer by Mahmood Alaghmandan for Solving the equation $xax=b$ in a C*-algebra.

2011-07-20T23:03:16Z

In the case of $\Bbb{M}_n(\Bbb{C})$, you should diagonalize $a$, say $d(\lambda_1,\cdots,\lambda_n)$. Then $a^{1/2}$ is $d(\sqrt{\lambda_1},\cdots,\sqrt{\lambda_n})$. Easily you can find $a^{-1/2}$ (since $a^{1/2}$ is of course invertible). The rest is just following the previous answer:

$x=a^{-1/2}(a^{1/2}ba^{1/2})^{1/2}a^{-1/2}.$

Then you can return the basis to the previous one (i.e. the basis before diagonalization of $a$).

Some special characters of finite groups

2011-07-20T05:08:39Z

Let $G$ be a finite group, for each irreducible character $\chi$, we define ${\bf Z}(\chi)$ to be the set of all $x\in G$ such that $|\chi(x)|=\chi(e)$ when $e$ is the identity of the group. For every irreducible charcter we know that

$\frac{|G|}{\chi(e)^2} \leq \frac{1}{\chi(e)} \sum_{x\in G} |\chi(x)| \leq \frac{|G|}{\chi(e)}$.

Now suppose that ${\bf Z}(\chi)$ is trivial i.e. it includes only $e$. Can we find a better bound for $\frac{1}{\chi(e)} \sum_{x\in G} |\chi(x)|$?

My firs clue about this question was in observance of affine $p$-groups: they satisfy this condition: (${\bf Z}(\chi)$ is trivial for only non-linear character $\chi_\pi$ of affine $p$-groups). On the other hand, for this non-linear character,

$\frac{1}{\chi_\pi(e)} \sum_{x\in G} |\chi_\pi(x)| =2$

for every $p$.

Subsequently, I tried some more character tables for finite groups that some of their characters, say $\chi$, satisify this condition and I observed that $\frac{1}{\chi(e)} \sum_{x\in G} |\chi(x)|$ is so close to $\frac{|G|}{\chi(e)^2}$ than $\frac{|G|}{\chi(e)}$.

Some infinite products related to prime numbers.

2011-07-17T08:40:15Z

Let $P$ be the set of all odd prime numbers. I am looking for all $s\in(1,\infty)$ for them

$ A=\prod_{p\in P} (1+\frac{1}{(p-1)^s})^{p-1} $

exists (i.e. is finite). I know that it should be somehow related to Riemann zeta function but I was not sure how can I pursue the calculations.

If I use natural logarithm I will get:

$ \ln(A)=\sum_{p\in P} (p-1) \ln(1+ \frac{1}{(p-1)^s})$

whcih I am not sure is useful of not!

Answer by Mahmood Alaghmandan for Suggestions for a good abstract algebra book

2011-05-21T08:06:10Z

I recently passed my qualifying exam and for that I studied "Abstract Algebra" by Dummit and Foote. It was an interesting book which explains everything from basic with several interesting examples. The only thing is that it is some kind of encyclopedia so you should not cover everything for your studying; I believe that you should get the book and talk with a professor about the parts that could be helpful for you.

Answer by Mahmood Alaghmandan for characters on a finite group with `extremal' behaviour

2011-05-18T08:02:52Z

@ Professor Landisch: I am wondering if you please give me the the name of Gallagher's paper which you have mentioned in the answer; moreover, what is the well-known terminology for $1$-minimal groups?

Comment by Mahmood Alaghmandan

2013-05-19T12:04:10Z

@Marc: This is true and you actually solved my question. Thank you.

Comment by Mahmood Alaghmandan

2013-05-18T21:32:24Z

Yes, exactly. Matrix coefficient functions corresponding to irreducible unitary matrices.

Comment by Mahmood Alaghmandan

2012-02-06T16:44:26Z

I had forgotten to mention a sub von Neumann algebra of $B(H)$.

Comment by Mahmood Alaghmandan

2012-02-06T16:41:21Z

Yes, you are right. But let me change my question somehow that do not have all elements in $B(H)$. So above I restrict myself to a von Neumann subalgebra of $B(H)$.

Comment by Mahmood Alaghmandan

2011-08-17T17:47:01Z

@ Yemon: I could not find any suggestion in Reiter's books for amenable groups. About your second comment: If I can show that a general version of "Wiener-Levy Theorem" is correct for $A(G)$ when $G$ is not abelian, then easily I can follow Reiter's proof. Therefore, I can rewrite the question as this: Is there any extension of "Wiener-Levy Theorem" for other locally compact groups (non-abelian ones)?

Comment by Mahmood Alaghmandan

2011-07-21T23:25:02Z

@ Martin: In the last part of the question Andre says:"Already in the case $A=M 2 (\Bbb{C})$ , I don't know how to solve this." This is the answer: " using unitary matrices switch the basis to one appropriate one and the follow the previous answer, then using the inverse of that unitary matrix return the base: The $x$." So I used the accepted answer in Matrix Analysis to find $x$! Nothing more!

Comment by Mahmood Alaghmandan

2011-07-20T08:56:02Z

Dear Geoff Robinson, First of all, the first inequality sounds a bit strange for me; but I get the last one which says $\|\chi\|_1\leq \sqrt{m(\chi) |G|}$. I have seen that theorem which says that a non-linear character will vanish at least on one conjugacy class. But have you seen ever a theorem that somehow gives an estimation about the numer of $m(\chi)$ or numer of conjugacy classes on them $\chi$ is zero.

Comment by Mahmood Alaghmandan

2011-07-17T17:59:12Z

Yes, I believe that this estimation is enough.

Comment by Mahmood Alaghmandan

2011-05-18T08:06:47Z

Sorry, since it was my first activity here, I did not know how I should write a comment; therefore, I wrote a comment in the answer. Even now I am not sure whether I have any chance to erase this or not! However, I would be greatful if you please answer to my question.