Welcome to MathOverflow

Question

2

1

Let $a,b\in A$ be self-adjoint elements in $C^*$-algebra $A$ with equal centralizers, $\{x\in A; [a,x]=0\}=\{x\in A; [b,x]=0\}$ . Can we say anything about the correspondence between $a$ and $b$?

For example, if $A=B(H)$ for a separable Hilbert space $H$, then according to the double commutant theorem $b=f(a)$ for some Borel function $f$ on the spectrum of $a$.

edited Jan 5 2012 at 18:29

Jesse Peterson
2,532●7●14

asked Jan 4 2012 at 6:20

spelas
149●5

Regarding your last comment: are you assuming $a$ is normal? (Otherwise take $a$ to be quasinilpotent.) – Yemon Choi Jan 4 2012 at 6:32

4

What type of general statement are you looking for? Have you considered the case when $A$ is commutative? – Jesse Peterson Jan 4 2012 at 16:31

1

@Spelas: Since you want to apply Borel functional calculus this is really a question about von Neumann algebras rather than $C^*$-algebras. It might be more insightful to edit the question to ask which von Neumann algebras $N$ have the property that for all commutative von Neumann subalgebras $A, B \subset N$ we have $A' \cap N = B' \cap N \implies A = B$. – Jesse Peterson Jan 5 2012 at 18:24

3

@Yemon: This doesn't work for all von Neumann algebras. For instance if $\mathbb F_2 = \langle a, b \rangle$ is the free group, $A = \{ a \}''$, $B = \{ a^2 \}''$. Then it is known that $A' \cap L\mathbb F_2 = B' \cap L\mathbb F_2$. In fact, my first guess would be that this property actually characterizes $B(H)$. – Jesse Peterson Jan 5 2012 at 18:27

2

mathoverflow.net/questions/85092/… – Jesse Peterson Jan 6 2012 at 22:14

show 7 more comments

Martin Argerami · Answer 1 · 2012-02-05 05:46:28Z UTC

The condition \begin{equation} \{x\in A: [a,x]=0\}=\{x\in A: [b,x]=0\} \end{equation} implies that $a$ and $b$ commute.

(Edit: after spelas noted that there was a relation in the example I wrote, now I'm rewriting the answer with an example where there is no relation)

This is of course no characterization, as Jesse's example above shows. But I don't think much can be said in this generality: consider $A=L^\infty[0,1]\oplus \ell^\infty(\mathbb{N})$, $a=f\oplus 0_m$, $b=0_n\oplus g$, where $f,g$ are respectively the functions $f(t)=t$, $g(n)=1/n$. Then $\{a\}'\cap A=A$ , $\{b\}'\cap A=A$ . So for this $a,b$ we have the equality among the centralizers, but no "correspondence" (in the language of the original question) between the two elements; indeed, $W^*(b)=\mathbb{C}\,1\oplus\ell^\infty(\mathbb{N})$ , $W^*(a)=L^\infty[0,1]\oplus\mathbb{C}\,1$ .

There's nothing very particular about this example, any two selfadjoint elements in the centre of a C $^*$ -algebra $A$ will do.

Centralizers in C*-algebra

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.