Is there an i.c.c. nonamenable simple group that is inner amenable?

Question

A finitely presented, countable discrete group $G$ is amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gX)=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$

A countable discrete group $G$ is inner amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gXg^{-1})=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$

The growth $b:\mathbb{N} \rightarrow \mathbb{N}$ of $G$ (with respect to a given word length metric on $G$) is defined as the number of elements $b(n)$ in $G$ lying inside the ball of radius $n$ around $e$.

It is possible to detect the amenability of $G$ in terms of the growth of G (c.f. R. I. Grigorchuk, “Symmetric random walks on discrete groups”, UMN, 32:6(198) (1977), 217–218).

Can the growth of G detect inner amenability?

I'd like to know if there is an i.c.c. discrete nonamenable simple group that is inner amenable?

On a related note, what about an answer to Owen's question below?

Answer 1

Here is a construction of a countable i.c.c. nonamenable simple group that is inner amenable. First consider the following condition:

(*) For every finite subset $S\subseteq G$, there exists $g\in G\setminus \{ 1\} $ such that $[g,s]=1$ for every $s\in S$.

Using paradoxical decomposition for non-inner amenable groups, it is not hard to show that () implies inner amenability. Indeed let $$ G\setminus \{ 1\} =A_1\sqcup \ldots \sqcup A_k\sqcup B_1\sqcup \ldots \sqcup B_m $$ and let $x_1, \ldots x_k, y_1, \ldots y_m$ be elements of $G$ such that $$ G\setminus \{ 1\} =(A_1)^{x_1}\sqcup \ldots \sqcup (A_k)^{x_k}=(B_1)^{y_1}\sqcup \ldots \sqcup (B_m)^{y_m} $$ By () there exists $g\ne 1$ that commutes with all $x_i$ and $y_j$. Let $A_i^\prime =A_i\cap \langle g\rangle$, $B_i^\prime =B_i\cap \langle g\rangle$. Intersecting $\langle g\rangle$ with the above decompositions of $G$ and noting that $(A_i)^{x_i}\cap \langle g\rangle=A_i^\prime$ and similarly for $B_i$'s, we obtain $$ \langle g\rangle\setminus \{ 1\} = A_1^\prime \sqcup \ldots \sqcup A_k^\prime \sqcup B_1^\prime \sqcup \ldots \sqcup B_m^\prime = A_1^\prime \sqcup \ldots \sqcup A_k^\prime = B_1^\prime \sqcup \ldots \sqcup B_m^\prime. $$ This is impossible for nontrivial $g$.

Now let us construct a group $G$ by induction. Let $G_0=F_2$, the free group of rank $2$. For $n> 0$ let $G_n$ be a countable simple group that contains $G_{n-1}\times \mathbb Z$ (every countable group embeds in a countable simple group). Let $G$ be the union of the chain $G_0\subset G_1\subset \ldots $. Clearly $G$ is simple being a union of simple groups and satisfies (*) by construction. Hence $G$ is inner amenable. As $G$ is simple and infinite, it is i.c.c. Finally $G$ is non-amenable as it contains $F_2$.

Modifying the above argument one can also construct an i.c.c. simple inner amenable non-amenable group without nontrivial free subgroups.

Answer 2

Hey Jon

So my initial thought would be no.

First, in full generality every group is virtually inner-amenable. Meaning that for any group $G$, the group $G \times \mathbb{Z}/2\mathbb{Z}$ is inner amenable. In fact, any non-icc group is inner amenable just by taking the mean to be the counting measure on a finite conjugacy class, and 0 elsewhere.

Even if we restrict to icc groups then, for any icc group $G$, $G\times S_\infty$ (or just choose the second group to be anything inner amenable) is still inner amenable.

And because the group is formed as a direct product there is not any way for the generators of $S_\infty$ to sort of "slow down" the growth in the $G$ factor.

Now a final way to maybe make something out of this is to ask

"If $G$ is inner amenable and along with all of its quotients, then is there a growth contsraint."

This will get rid of the examples above. Amenable groups fall into this class, and I would be willing to bet that there are others as well (if anyone knows examples that would be nice) but I can't think of any on the spot.

AS for this class.... I have no idea.

Answer 3

The group $G:=SL_{\infty}(\mathbb Q) = \cup_n SL_n(\mathbb Q)$ is a concrete example. It is obviously simple and non-amenable. Let $g_n \in SL_{\infty}(\mathbb Q)$ be the matrix which is $$g_n:= 1_n \oplus \left(\begin{matrix} 0 & 1 \newline -1 & 0 \end{matrix}\right) \oplus 1_{\infty}.$$ and let $$m_{n}(A) := \begin{cases} 1 & g_n \in A \newline 0 & g_n \not \in A \end{cases}.$$ be the finitely additive probability measure associated with $g_n$. Now, for any non-principal ultrafilter $\omega \in \beta \mathbb N \setminus \mathbb N$, $$m(A) := \lim_{n \to \omega} m_n(A) \in [0,1]$$ is a conjugation invariant finitely additive probability measure on $G \setminus \{e\}$. Conjugation invariance follows since the each element in $G$ commutes with $g_n$ for $n$ large enough.

Answer 4

Is there a non inner amenable locally compact group [map]group