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Recall that the Hilbert space version of von Neumann's mean ergodic theorem says the following.

Let $\{F_n\}_{n=1}^\infty$ be a right Følner sequence of a countable discrete amenable group $\Gamma$ and $\pi:\Gamma\to B(H)$ be a unitary representation of $\Gamma$ on a Hilbert space $H$. Then $$\lim_{n\to\infty}\frac{1}{|F_n|}\sum_{s\in F_n} \pi(s)y=Py$$ for every $y\in H$, where $P$ is the orthogonal projection from $H$ onto $H_\Gamma=\{x\in H\,|\,\pi(s)x=x \, {\rm for\, all}\, s\in\Gamma\}.$

Question: Is the converse true?

More precisely, suppose a countable discrete group $\Gamma$ has a sequence of finite subsets $\{F_n\}_{n=1}^\infty$ such that for every unitary representation $\pi:\Gamma\to B(H)$, one have $$\lim_{n\to\infty}\frac{1}{|F_n|}\sum_{s\in F_n} \pi(s)y=Py$$ for every $y\in H$. Here $P$ is the orthogonal projection from $H$ onto $H_\Gamma$.

Is $\Gamma$ amenable?

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  • $\begingroup$ You assume that $F_n$ cover $\Gamma$, I presume? $\endgroup$
    – Algernon
    Sep 6, 2015 at 10:28
  • $\begingroup$ It is not necessary. $\endgroup$
    – Huichi Huang
    Sep 6, 2015 at 19:49
  • $\begingroup$ I might have misunderstood something, but if $F_n$ form a Følner sequence of an amenable subgroup of $\Gamma$, then the condition is satisfied, even if the group $\Gamma$ itself is not amenable. $\endgroup$
    – Algernon
    Sep 6, 2015 at 20:19
  • $\begingroup$ The corresponding claim for continuous unitary actions of Lie groups is false due to the Moore ergodicity theorem (or Mautner phenomenon). For instance, $SL_2(R)$ is not amenable, but averaging along a Folner sequence along the unipotent upper triangular subgroup $U^+(R)$, which is amenable, will converge to $P_{U^+(R)} = P_{SL_2(R)}$. Similarly, as per Algernon's comment, if $\Gamma$ is nonamenable but contains an amenable subgroup $\Gamma'$ for which $H_{\Gamma'}$ is always equal to $H_\Gamma$, this would be a discrete counterexample. $\endgroup$
    – Terry Tao
    Sep 6, 2015 at 20:32
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    $\begingroup$ For discrete groups, if $H_{\Gamma'}$ is always equal to $H_\Gamma$ then $\Gamma' = \Gamma$. You can just consider the representation $\ell^2(\Gamma/\Gamma')$. $\endgroup$
    – Jesse Peterson
    Sep 6, 2015 at 22:06

1 Answer 1

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If $\Gamma$ is a property (T) group with infinite conjugacy classes, e.g., $\Gamma = PSL_3(\mathbb Z)$, then $\Gamma$ will have such a sequence. The ``spectral gap'' criterion for property (T) shows that there exists $c > 0$ and $F \subset \Gamma$ finite such that for any unitary representation $\pi$ we have $\left\| \frac{1}{|F|} \sum_{\gamma \in F} \pi(\gamma) - P \right\| < 1 - c$. Since $\Gamma$ has infinite conjugacy classes a short argument shows that there exists a sequence $\{ \gamma_n \}_{n = 1}^\infty \subset \Gamma$ so that $F_n = ( \gamma_n F \gamma_n^{-1}) ( \gamma_{n-1} F \gamma_{n-1}^{-1} ) \cdots (\gamma_1 F \gamma_1^{-1} )$ satisfies $| F_n | = | F |^n$ we then have $$ \left\| \frac{1}{|F_n|} \sum_{\gamma \in F_n} \pi(\gamma) - P \right\| = \left\| \left(\frac{1}{|F|} \sum_{\gamma \in F} \pi(\gamma_n\gamma\gamma_n^{-1}) - P\right )\cdots \left( \frac{1}{|F|} \sum_{\gamma \in F} \pi(\gamma_1\gamma\gamma_1^{-1}) - P \right)\right\| $$ $$ \leq (1 - c)^n \to 0. $$

Hence, this sequence of operators also converges in the strong operator topology.

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