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Let $G$ be a countable group and $\lambda \colon G \to U(\ell^2 G)$ its left-regular representation. Suppose that there exists a constant $C>0$ such that for all $T \in B(\ell^2 G)$

$$\inf \lbrace\|T-S\| \mid S \in \lambda(G)' \rbrace \leq C \cdot \sup\lbrace \|\lambda(g)T- T \lambda(g) \| \mid g \in G\rbrace.$$

(Here $\lambda(G)'$ denotes the commutant of $G$ in $B(\ell^2 G)$.)

Question: Is $G$ amenable?

It is fairly easy to see that amenability of $G$ implies the existence of such a constant. Indeed, one may take $S$ to be some fixed point for the conjugation action on $\overline{\rm conv}\lbrace \lambda(g)T\lambda(g)^* \mid g \in G\rbrace$. I am asking for the converse of this statement.

EDIT: Since the derivation problem came up in Kate's comment, I want to clarify to what version of it my question is related. The inequality above holds for some $C$ if and only if the first bounded cohomology of $G$ with coefficients in $B(\ell^2 G)$ (with the conjugation action induced by $\lambda$) is reduced. This is a straightforward application of the open mapping theorem. Now, two things are unclear:

Question: Can $H^1_b(G,B(\ell^2 G))$ be reduced without being zero?

and

Question: Can $H^1_b(G,B(\ell^2 G))$ be zero without $G$ being amenable?

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Can you (directly) show that such C doesn't exist for, say, free group? –  Łukasz Grabowski Jan 10 '11 at 23:04
    
Lukasz, this follows from the work of Pytlik-Szwarc. The idea is roughly as follows. If $G$ admits such a constant $C$ as above, then the constant for any subgroup is at most $C$. Now, for $\mathbb F_n \subset \mathbb F_2$ you consider the operator $T$ on $\ell^2 \mathbb F_n$ which moves a vertex towards the root in a standard Cayley graph of $\mathbb F_n$. One can show directly that the size of $C(\mathbb F_n)$ which is enforced by $T$ tends to infinity. The details need some clever work, but that (I think) is at least the idea. –  Andreas Thom Jan 11 '11 at 6:41
    
You must have thought about it already, but for the sake of completeness: Maybe using Whyte's metric solution to von Neumann problem and imitating the T you described could help? –  Łukasz Grabowski Jan 11 '11 at 10:59
    
The problem is that Kevin Whyte's construction is not equivariant at all. Random forrests have been used by Nicolas Monod and Inessa Epstein to extend the result to non-amenable groups without free subgroups. –  Andreas Thom Jan 11 '11 at 14:02
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@Yemon: Yes, exactly. –  Theo Buehler Jan 12 '11 at 22:09
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1 Answer 1

Let $M=C^*_\lambda(G)''$ be group von Neumann algebra of $G$. The the condition above implies:

$d(T, M')\leq C ||ad(T)|_{M}||$ for every $T\in B(l^2 G)$. The last inequality is equivalent to saying that every derivation of $M$ into $B(l^2 G)$ is inner.

Edit: the above inequality is satisfied automatically (was clarified to me by Stuart White).

It is known that if $M\subset B(H)$ has a cyclic vector, then every bounded derivation from $M$ into $B(H)$ is inner [E. Christensen, Extensions of derivations II, Math. Scand., 1982]. Thus [Christensen, op cit, Cor 5.4] we have

$d(T,M')\leq 3/2\|(\mathrm{ad}| T)_{M}\|$

for every $T\in B(l^2 G)$.

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I do not understand how you conclude the inequality in your answer. Isn't it is known that all derivations from $M$ into $B(H)$ are inner? What is your conclusion? –  Andreas Thom Jan 10 '11 at 20:16
    
I should have added this as a comment, not as an answer.. Your inequality is stronger than the inequality in the answer and might actually imply amenability. the derivation problem seems to be still unknown. also $||ad(T)|_M||$ is norm of operator $ad(T)$ restricted to $M$, which is not less than $sup ||\lambda(g)T-T\lambda(g)||$. –  Kate Juschenko Jan 10 '11 at 20:40
    
also arxiv.org/abs/0910.1368 and references there are related –  Kate Juschenko Jan 10 '11 at 21:21
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