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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 $a$.

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Regarding your last comment: are you assuming $a$ is normal? (Otherwise take $a$ to be quasinilpotent.) –  Yemon Choi Jan 4 '12 at 6:32
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What type of general statement are you looking for? Have you considered the case when $A$ is commutative? –  Jesse Peterson Jan 4 '12 at 16:31
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@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 '12 at 18:24
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@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 '12 at 18:27
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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.

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But it holds a+b=1, so b=f(a). –  spelas Jan 10 '12 at 14:33
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(sorry I saw your comment several weeks later) True, but as I said, "any two selfadjoint elements in the centre of a C$^*$-algebra will do". If you want, I'll edit the answer to show an example where no relation exists. –  Martin Argerami Feb 2 '12 at 0:55
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