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?

  • $\begingroup$ You assume that $F_n$ cover $\Gamma$, I presume? $\endgroup$
    – Algernon
    Sep 6, 2015 at 10:28
  • $\begingroup$ It is not necessary. $\endgroup$ 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
  • 1
    $\begingroup$ For discrete groups, if $H_{\Gamma'}$ is always equal to $H_\Gamma$ then $\Gamma' = \Gamma$. You can just consider the representation $\ell^2(\Gamma/\Gamma')$. $\endgroup$ Sep 6, 2015 at 22:06

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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