Is $SU(\infty)$ amenable? 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 $Sp(\infty)$.
 A: 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 $SU(\infty)$.
