We can write the finitary special unitary group $SU(\infty)$ as the direct limit $\varinjlim SU(n)$ of ordinary special unitary groups. These groups $SU(n)$ are compact, thus amenable. In other words each of them has an invariant mean (in this case, from Haar measure). Is $SU(\infty)$ amenable? Of course one then asks the same question for the full unitary group $U(\infty)$, the special orthogonal group $SO(\infty)$ and the unitary symplectic (quaternion unitary) group $Sp(\infty)$.
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$\begingroup$ I think that it contains free groups, isn't it? $\endgroup$– Valerio CapraroCommented Oct 16, 2011 at 19:44
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$\begingroup$ @Valerio Capraro Certainly it does, but that only makes it non-amenable as a discrete group. $\endgroup$– John Wiltshire-GordonCommented Oct 16, 2011 at 21:12
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7$\begingroup$ Sorry if this is silly but what exactly do you mean by amenability for non locally compact groups such as $SU(\infty)$? I'm not at all an expert on this but one definition I guess could be that if $G$ acts on a compact space then there is an invariant probability measure. With that definition I think any direct limit of amenable (in that sense) groups is amenable by weak compactness of probability measures on a compact space. $\endgroup$– Vitali KapovitchCommented Oct 16, 2011 at 21:53
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1 Answer
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The answer is that $G=SU(\infty)$ (with the direct limit topology of the usual Hilbert-Schmidt topologies) is extremely amenable. This means (by definition) that every continuous action of $G$ on a compact set has a fixed point.
This was proved as an application of the isoperimetric inequality by Gromov and Milman
M. Gromov and V.D. Milman, A topological application of the isoperimetric inequality, Amer. J. Math. 105 (1983), 843–854.
Since $SU(\infty)$ is not locally compact, various characterizations of amenability have to be adapted. One way to define amenability for such groups is to say that there exists a $G$-invariant mean on the algebra of bounded uniformly continuous real-valued functions on $G$. Extreme amenability then gives even the existence of a $G$-invariant character on this algebra. This is somewhat unintuitive, since obviously no compact group can admit such a character on the algebra of continuous functions on it.
Extreme amenability is a concept which is related to phenomena of measure concentration. Gromov and Milman proved (as an application of lower bounds on the Ricci curvature) that for every sequence of measurable subsets $A_n \subset SU(n)$ with $\liminf_{n \to \infty} \mu_n(A_n) \neq 0$, one has $$\lim_{n \to \infty} \mu_n(A_{n,\epsilon}) = 1.$$ Here, $\mu_n$ denotes the normalized Haar measure and $A_{n,\varepsilon}$ the $\varepsilon$-neighborhood of $A_n$ in the unnormalized Hilbert-Schmidt metric. This concentration phenomenon can be used to prove extreme amenability of $SU(\infty)$.
answered Oct 17, 2011 at 17:23
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1$\begingroup$ I found another affirmative answer in Example 3.4 of a paper "Functional Analytic Background for a Theory of Infinite-Dimensional Reductive Lie Groups" by Daniel Beltita. He shows that if $G$ is a topological group and there is a directed system $\{G_\alpha\}_{\alpha \in A}$ of amenable topological subgroups whose union is dense in $G$ then $G$ is amenable. Here "amenable" means that there is a left-invariant mean. $\endgroup$ Commented Oct 25, 2011 at 18:20