# Strong ergodicity and spectral gap

Suppose that $\Gamma \curvearrowright (X,\mu)$ is a measure-preserving action of a discrete, finitely generated group $\Gamma = \left\langle S \right\rangle$ on a probability space $(X,\mu)$. Let $\pi: \Gamma \to L^{2}(X,\mu)$ be the associated Koopman representation $(\pi(\gamma)f)(x) := f(\gamma^{-1}x)$. I've come across two definitions of a spectral gap for such an action and it seems to me that they are not equivalent and this is what I would like to ask about.

1. There exists a constant $\kappa>0$ such that for every measurable set $A \subset X$ we have $\max_{s \in S} \mu(A \Delta sA) \geqslant \kappa \mu(A)(1-\mu(A))$.

2. There exists a constant $\kappa'>0$ such that for every $v\in L^{2}_{0}(X,\mu)$, the subspace of mean-zero functions, we have $\max_{s\in S}\|v - \pi(s)v\|_2 \geqslant \kappa' \|v\|_2$.

The first definition implies that the action is strongly ergodic, i.e. if $(A_n)_{n \in \mathbb{N}}$ is a sequence of measurable sets such that $\lim_{n\to \infty} \max_{s\in S}\mu(A_n \Delta sA_n)=0$ then $\lim_{n\to \infty} \mu(A_n)(1 - \mu(A_n))=0$. Here comes the first question:

• Is the first definition equivalent to strong ergodicity?

I know that there do exist strongly ergodic actions that do not admit spectral gap, so an affirmative answer would show that the definitions are not equivalent. If the answer is negative, one has to ask the following question.

• Is the first definition equivalent to the second one?

These things appear naturally in Popa's deformation/rigidity theory, hence the choice of tags.

EDIT: I realised that in the context of expanders the way to obtain a spectral gap from a property of indicator functions only is to use Cheeger's inequality. Maybe the right question to ask is: is there a Cheeger-type inequality in this case that shows that the two definitions introduced above are equivalent?

Yes, conditions 1 and 2 are equivalent (as you already noticed).

Let me start by replacing both definitions with their negations, and replace definition 2 with its $L^1$-version. A and B below are the negations of 1 and 2 correspondingly, slightly rewritten:

A. There exists a sequence of measurable subsets $\emptyset,X\neq A_n\subset X$ such that for every $s\in S$, $$(*)\quad \frac{\|(\chi_{A_n}-\mu(A_n)\cdot 1)-s(\chi_{A_n}-\mu(A_n)\cdot 1)\|_1}{\|(\chi_{A_n}-\mu(A_n)\cdot 1)\|_1} = \frac{\mu(A_n \Delta sA_n)}{\mu(A_n)(1-\mu(A_n))}\to 0.$$

B. There exists a sequence of vectors $0\neq v_n$ in $L^1_0(X)$ such that for every $s\in S$, $$(**) \quad \frac{\|v_n-sv_n\|_1}{\|v_n\|_1} \to 0.$$

To see that A is the negation of 1 is straightforward. To see that B is the negation of 2, recall that the map $L^2\ni u\mapsto \text{sgn}(u)\cdot |u|^2\in L^1$ is uniformly continuous equivariant map on the unit sphere. Use it and a projection mod constants. I am not elaborating, as this is well known. From now on we work in $L^1$ exclusively, thus $\|\cdot\|=\|\cdot\|_1$.

Clearly A $\Rightarrow$ B. The rest of this post is devoted to proving the application B $\Rightarrow$ A. Thus, we assume that $v_n$ is a sequence of unit vectors satisfying $(**)$ and argue to provide a sequence $A_n$ satisfying $(*)$.

Let me denote by $B$ the unit ball in $L^1_0(X)$. The idea, as usual, is to map $B$ continuously into a compact space and take a limit point of the image of the sequence $v_n$. A first candidate for such a map that comes to mind is the embedding into the unit ball of $L^1(X)^{**}\simeq L^\infty(X)^*$. This idea will lead us eventually to prove the existence of an invariant mean, as in the answer by Mateusz Wasilewski, but justifying everything down this road seems longer than the path we take below in which we use another map.

Consider the space of probability measures on $\mathbb{R}$, $P$. Endow $P$ with the weakest topology for which integration against every function in $C_c(\mathbb{R})$ is continuous. Observe that the map $B\to P$, $f\mapsto f_*\mu$ is continuous and that the image is precompact (use Prokhorov's theorem). Let me denote $\nu_n=(v_n)_*\mu\in P$ and assume as we may that $\nu_n\to \nu$. We consider two different cases: either $\nu$ is a point mass or not.

In case $\nu$ is NOT a point mass we can find and fix $t\in \mathbb{R}$ such that $\nu(-\infty,t),\nu(t,\infty)>0$ and then, defining $A_n=v_n^{-1}(t,\infty)$, it easy to check that $(*)$ is satisfied (use the continuity of $B\to P$).

From now on we assume $\nu$ is a point mass, that is $\nu=\delta_t$ for some $t\in \mathbb{R}$. By the fact that $\int v_n=0$ and $\int|v_n|=1$ it is easy to conclude that $|t|\leq 1/2$ and it follows that $1/2\leq \|v_n-t\|<3/2$. It is also easy to see that the sequence $|v_n-t|$ is almost invariant in $L^1(X)$ and that $|v_n-t|_*\mu \to \delta_0$. We normalize this sequence, setting $u_n=|v_n-t|/\|v_n-t\|$.

To summarize: we found a new almost invariant sequence of positive unit vectors $u_n\in L^1(X)$ such that $(u_n)_*\mu\to \delta_0$. We assume as we may that $\sum_{s\in S}\|su_n-u_n\|< \|u_n\|/n$. The construction of the sequence $A_n$ will follow by the "layer cake decomposition" method: for every positive function $f\in L^1(X)$, $\int_X f=\int_0^\infty f_*\mu(t,\infty)dt$. Applying this decomposition to the functions $|su_n-u_n|$ and $u_n$ we obtain $$\sum_{s\in S}\int_0^\infty |su_n-u_n|_*\mu(t,\infty)dt=\sum_{s\in S}\int_X |su_n-u_n|=\sum_{s\in S}\|su_n-u_n\|$$ $$< \|u_n\|/n = 1/n \int_X u_n =1/n \int_0^\infty (u_n)_*\mu(t,\infty)dt$$ and deduce the existence of $t>0$ for which $$\sum_{s\in S} |su_n-u_n|_*\mu(t,\infty) < 1/n (u_n)_*\mu(t,\infty).$$ We set $A_n=u_n^{-1}(t,\infty)$ and conclude $\sum_{s\in S} \mu(A_n \Delta sA_n) < 1/n \mu(A_n)$. By the strict inequality we get $A_n\neq \emptyset$. Since $(u_n)_*\mu\to \delta_0$ we have $1-\mu(A_n)\to 1$, thus we may assume $A_n\neq X$, and indeed we get $$\frac{\mu(A_n \Delta sA_n)}{\mu(A_n)(1-\mu(A_n))} \to 0.$$

I think I can answer my own question, although in a rather convoluted way. The two definitions are equivalent. Since the first definition is just the second one restricted to indicator functions, we just need to show that 1. implies 2. Assume that 2. does not hold. It follows that integration against $\mu$ is not the only $\Gamma$-invariant mean on $L^{\infty}(X,\mu)$ (this is discussed in "Amenability, Kazhdan's property T, strong ergodicity and invariant means for ergodic group-actions" by K. Schmidt). Schmidt points out to a result due to Rosenblatt (Theorem 1.3 in "Uniqueness of Invariant Means for Measure-Preserving Transformations"), which says that if the invariant mean is not unique then there exists a sequence of measurable sets $(A_n)_{n \in \mathbb{N}}$ such that $\mu(A_n)>0$, $\lim_{n\to \infty} \mu(A_n) = 0$ and $\lim_{n\to \infty} \frac{\mu(A_n \Delta \gamma A_n)}{\mu(A_n)}=0$ for any $\gamma \in \Gamma$; this violates 1.

Unfortunately, this is a very indirect argument and I don't feel satisfied with it.

• If you read carefully Schmidt's argument you see that it does give a quite direct answer to your question. Given an a.i sequence of unit vectors $v_n$ in $L^2_0$ you get an a.i sequence $u_n$ in $L^1_0$ (projecting $\text{sgn}(v_n)\cdot |v_n|^2$). Consider the measures $\sigma_n=(u_n)_*\mu$ on $\mathbb{R}$ - this a precompact family that you can assume to converge to $\sigma$. Pick $t$ such that $\sigma(t,\infty)\in(0,1)$ and define $A_n=u_n^{-1}(t,\infty)$. This process fails if $\sigma$ is a point mass, but you can overcome it by deforming first your sequence $u_n$. The details are in p. 227 Nov 22 '16 at 9:23
• @UriBader, thanks a lot! Would it be possible for you to extend this comment to an answer? Nov 22 '16 at 19:24
• Done. I extracted the proof from Schmidt, slightly simplifying. I hope it helps. Nov 23 '16 at 16:17