8
$\begingroup$

I repost here from stackexchange, as I was not given an answer there. (https://math.stackexchange.com/questions/2956530/distance-between-subalgebras-and-commutants-in-matrix-algebras)

This is a question on how to relate two different distances in the matrix setting. Everywhere below, $M_n$ denotes the square matrices $n\times n$ whose entries are in $\mathbb C$. We consider the operator norm $\|\cdot\|$ on $M_n$.

By a subalgebra we mean a $C^*$-subalgebra of $M_n$. If $A\subseteq M_n$ is a subalgebra and $x\in M_n$, define the distance from $x$ to $A$ as the real number $d(x,A)=\inf_{a \in A}\|x-a\|$.

The question is the following: suppose that $A$ is a unital $C^*$-subalgebra of $M_n$ and that $x\in M_n$ is a positive contraction with $d(x,A)\geq\frac{1}{2}$. Does there exist $u\in A'$ (the commutant of $A$) such that $\|[x,uxu^*]\|\geq\frac{1}{16}$ (or, for what matter, $\frac{1}{64}$, the only important thing is that this number doesn't depend on the choice of $n$, $A$, or $x$)?

Beware, I am not asking whether I can find a $u$ with $\|[u,x]\|\geq\frac{1}{16}$. This is clearly possible since $y=\int_{\mathcal U(A')}uxu^*d\mu(u)$, when I am integrating over the Haar measure on the unitary group of $A'$, is the conditional expectation onto $A''=A$. Since $y\in A$, we have $\|x-y\|\geq\frac{1}{4}$, and thefore there must be a unitary $u\in \mathcal U(A')$ such that $\|x-uxu^*\|\geq\frac{1}{16}$, or in other words, such that $\|[x,u]\|\geq\frac{1}{16}$.

Thanks and best,

$\endgroup$
  • $\begingroup$ Trivial typo: you presumably want $\Vert u\Vert =1$. Indeed, do you mean that $u$ should be a unitary element of $A'$? $\endgroup$ – Yemon Choi Feb 4 at 21:30
  • $\begingroup$ $u$ is a unitary (almost) by definition, of course! But any contraction would do, as any element is a combination of four or them. $\endgroup$ – Alessandro Vignati Feb 4 at 21:46
4
$\begingroup$

YES. Let $x=x^*$ and put $C:=d(x,A)$. As you observed, there is a unitary element $v\in A'$ such that $\| [x,v] \|\geq C$. By decomposing $v$ into the real and imaginary part, one finds a self-adjoint contraction $h\in A'$ such that $\|[x,h]\|\geq C/2$. Since $h$ is a convex combination of $p - p^\perp$, $p$ spectral projections of $h$, there is a projection $p\in A'$ such that $\|[x,p]\|\geq C/4$. (This means that the hyperreflexive constant of a finite-dimensional C*-algebra is at most $4$---I think the optimal constant is $2$, but didn't find a reference). Note that $\|[x,p]\|=\|p^\perp x p\|$. Put $u := p+\sqrt{-1}p^\perp \in U(A')$. Then, $$p (x uxu^* - uxu^* x) p = pxuxp - pxu^*xp = 2\sqrt{-1} pxp^\perp xp$$ and so $\| [x,uxu^*]\| \geq 2\|p^\perp x p\|^2\geq C^2/8.$ (This proof is probably not optimal, but I like working on operator matrices. The displayed formula has a better looking in the operator matrix form.)

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.