15
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ Can you (directly) show that such C doesn't exist for, say, free group? $\endgroup$ Jan 10, 2011 at 23:04
  • $\begingroup$ 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. $\endgroup$ Jan 11, 2011 at 6:41
  • $\begingroup$ 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? $\endgroup$ Jan 11, 2011 at 10:59
  • $\begingroup$ 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. $\endgroup$ Jan 11, 2011 at 14:02
  • 1
    $\begingroup$ @Yemon: Yes, exactly. $\endgroup$ Jan 12, 2011 at 22:09

1 Answer 1

3
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Jan 10, 2011 at 20:16
  • $\begingroup$ 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 $sup ||\lambda(g)T-T\lambda(g)||$. $\endgroup$ Jan 10, 2011 at 20:40
  • $\begingroup$ also arxiv.org/abs/0910.1368 and references there are related $\endgroup$ Jan 10, 2011 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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