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
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$\begingroup$ Are you allowing $\delta$ to depend on both $\epsilon$ and $x$? Or do you want $\delta$ to depend only on $\epsilon$ and $\Vert x \Vert$? $\endgroup$– Yemon ChoiJan 10, 2011 at 0:47
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2 Answers
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The present question seems very close to the following one, which has been well-studied in the literature:
Let $A$ be a C*-algebra. Does there exist a constant $K>0$ such that, for each $x\in A$, the distance from $x$ to $Z(A)$ is bounded above by $$ K \sup\{ \Vert xy-yx \Vert \;:\; y\in A, \Vert y\Vert \leq 1\} ?$$
A useful discussion of what is known can be found in Section 1 of
[MR2274022] R. J. Archbold, D. W. B. Somerset, Measuring noncommutativity in C*-algebras. JFA 242 (2007) no. 1, 247--271
According to this paper, the answer turns out to be "yes" if $A$ is a von Neumann algebra, or one of several natural classes of C*-algebra; but the answer is "no" in general, by an example in
[MR0215111] R. V. Kadison, E. C. Lance, J. R. Ringrose, Derivations and automorphisms of operator algebras. II. JFA 1 1967 204–221.
I haven't been able to read the Kadison-Lance-Ringrose paper, but a similar example seems to be discussed in Example 3.3 of
[MR0482236] R. J. Archbold, On the norm of an inner derivation of a C*-algebra. Math. Proc. Cambridge Philos. Soc. 84 (1978), no. 2, 273–291.
answered Jan 10, 2011 at 1:04
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$\begingroup$ @Yemon: The original question is a bit weaker than the boxed statement here, and is effectively asking whether, for each $\epsilon > 0$, there is a $K(\epsilon)$ such that the distance of any point from the center is bounded by $\epsilon\Vert x\Vert + K(\epsilon)\sup_{\Vert y\Vert\le 1}\Vert xy-yx\Vert$. Your boxed statement is asking whether $K(\epsilon)$ is bounded above as $\epsilon\to0$, which is rather stronger (but maybe you can reduce it to the same thing). $\endgroup$ Jan 10, 2011 at 2:32
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$\begingroup$ ...and replacing $\Vert x\Vert$ by $\inf\Vert x-z\Vert$ for $z$ in the center does reduce it to the same thing, for all $\epsilon < 1$. So, they are in fact the same statement. $\endgroup$ Jan 10, 2011 at 2:35
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$\begingroup$ This is exactly what I was looking for. Thank you guys! $\endgroup$– QingyunJan 10, 2011 at 17:29
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$\begingroup$ I'd encourage people to vote up George Lowther's answer, since unlike me he used thought rather than MathSciNet and actually gave an explicit, informative counter-example $\endgroup$ Jan 10, 2011 at 17:38
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1$\begingroup$ It seems that there is quite a bit of research about on this problem. E.g., if the algebra has trivial center then $K\in\{\frac12,1,\frac12+\frac{1}{\sqrt{3}}\}\cup[\frac32,\infty]$. And lots more surprising results besides that. Eg, plms.oxfordjournals.org/content/88/1/225.abstract $\endgroup$ Jan 10, 2011 at 23:05
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In general, the answer to this has to be "no". To construct a counterexample, let $S$ be a dense subset of the unit interval (0,1], and let $\mathcal{H}=\ell^2(S)$ be the Hilbert space with orthonormal basis $(e_s)_{s\in S}$. For each positive integer $n$, let $\mathcal{H}_n$ be the subspace of $\mathcal{H}$ generated by linear combinations of $e_s$ over $1/(n+1) < s\le 1/n$. We can define a $C^*$ sub-algebra $\mathcal{A}$ of the bounded linear operators $B(\mathcal{H})$ as follows: $A\in B(\mathcal{H})$ is in $\mathcal{A}$ if and only if the following are satisfied
- $A$ maps $\mathcal{H}_n$ into $\mathcal{H}_n$.
- There exists a sequence of complex numbers $a_n$ such that, for each fixed $n$, $$\Vert Ae_s-a_ne_s\Vert+\Vert A^*e_s-\bar a_ne_s\Vert\to0$$ as $s\to 1/n$.
It is easily seen that this is a sub-$C^*$-algebra. I'm thinking of $\mathcal{A}$ intuitively as being those elements of the direct product (sum?) $\prod_nB(\mathcal{H}_n)$ "joined continuously at the edges", and roughly as continuous functions $S\to\mathbb{C}$ but allowing some non-commutativity within the intervals $(1/(n+1),1/n]$.
If operator $A$ is in its center then it must restrict to the center of each $B(\mathcal{H}_n)$, so must be a constant on each of these. That is, $Ae_s=\lambda_se_s$ for some $\lambda_s\in\mathbb{C}$, which is independent of $s$ over $1/(n+1) < s\le 1/n$. Also, $s\mapsto\lambda_s$ must be continuous at $s=1/n$, from which we see that $s\mapsto\lambda_s$ is constant. So, $\mathcal{A}$ has trivial center.
Fixing a positive integer $m$, define an operator $A\in\mathcal{A}$ by $Ae_s=\min(ms,1)e_s$. Then, $\Vert A\Vert = 1$ and ${\rm dist}(A,{\rm Center}(\mathcal{A}))=1/2$. On the other hand, the restriction of $A$ to each $\mathcal{H}_n$ is within a distance $1/(2(m+1))$ from its center. This gives $\Vert AB - BA\Vert \le \Vert B\Vert / (m+1)$ for all $B\in\mathcal{A}$.
Choosing $m$ as large as we like, this contradicts your claim.
answered Jan 10, 2011 at 1:57