amenability and property T
- amenable groups cannot have kazhdan property (T)!
- the point is that amenability is really about lack of expansion:
- for any finite set $K$, there is a set in your folner sequence which has small boundary w.r.t. $K$
- while kazhdan property (T) is explicitly about expansion
Proof (sketch)
- Let $Q$ any finite set in $T$ and $\eps > 0$ any choice
- The goal is to show that there's some representation $\pi$ and vector $v$ s.t. for every $g\in Q$ $|\pi(g)v-v|\leq \epsilon|v|$
- Well, by amenability, you can find a $F_n$ in your folner sequence which has an isoperimetry less than $\eps$. That is:
$$\forall g\in Q, |gF_n\Delta F_n| \leq \eps |F_n|$$ - Now, let your $\pi$ be the overall regular representation on the group (which satisfies having no nontrivial invariant vector). And $v$ be the indicator for $F_n$. Then, amenability tells you that none of your group elements will take $v$ far! This contradicts property (T)