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Let $G$ be a non-amenable countable discrete group. How can I show that the group von Neumann algebra $L(G)$ has no injective direct summand?

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I must be missing something. Why can't you take an abelian vN subalg of L(G) and then project onto it? –  Yemon Choi Jul 4 '12 at 17:08
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There is no reason that this subalgebra will be a direct summand –  Owen Sizemore Jul 4 '12 at 17:36
    
@Owen: right, for some reason I had "direct summand as a Banach space" in my mind –  Yemon Choi Jul 4 '12 at 23:52
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Why not accept Jesse Peterson's answer? –  Yemon Choi Jul 9 '12 at 23:32
    
For that matter, why haven't you accepted any answers? –  S. Carnahan Jul 11 '12 at 4:49

1 Answer 1

up vote 8 down vote accepted

Here is an adaptation of the standard proof that $G$ is amenable if $LG$ is injective. (I believe for instance that it is contained in the book of Brown and Ozawa).

Suppose $p \in LG$ is a non-zero central projection such that $p LG$ is injective. Thus, there exists a conditional expectation $E: \mathcal B(p \ell^2 G) \to p LG$. If we view $\ell^\infty G \subset \mathcal B(\ell^2 G)$ as diagonal multiplication operators (for $f \in \ell^\infty G$ and $\xi \in \ell^2 G$ we set $(M_f \xi)(\gamma) = f(\gamma) \xi(\gamma)$), and if we denote by $\tau$ a tracial state on $pLG$ then we can construct a state $\varphi$ on $\ell^\infty G$ by the formula $\varphi(f) = \tau \circ E(p M_f p)$. If $\gamma \in G$ then we have $$ \varphi( f \circ \gamma) = \tau \circ E(p M_{f \circ \gamma} p) $$ $$ = \tau \circ E(p \lambda_{\gamma^{-1}} M_f \lambda_{\gamma} p) = \tau( (p\lambda_{\gamma^{-1}}p) E(p M_f p) (p \lambda_{\gamma}p) ) = \varphi(f). $$ Thus $\varphi$ is an invariant mean for $G$ and so $G$ is amenable.

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thanks for the nice answer. –  m07kl Jul 5 '12 at 16:48

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