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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?

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MathSciNet references the Grigorchuk paper as follows: Grigorchuk, R. I, Symmetrical random walks on discrete groups, Multicomponent random systems, pp. 285--325, Adv. Probab. Related Topics, 6, Dekker, New York, 1980. – HJRW Sep 22 '10 at 21:07
Thanks for this, Henry. – Jon Bannon Sep 22 '10 at 21:34
For the benefit of those who don't know, an i.c.c. group is a group all of whose conjugacy classes are infinite (excluding $\{e\}$). Brought to you by the power of google, I'd never seen this term before. – Victor Protsak Sep 23 '10 at 6:21
Sorry Victor et. al.! – Jon Bannon Sep 23 '10 at 11:40
up vote 12 down vote accepted

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.

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Very, very nice! Thanks! – Jon Bannon Dec 4 '10 at 2:02
What is $(A_1)^{x_1}$? I have not seen this notation before. – Owen Sizemore Mar 7 '11 at 20:34
@Owen: Conjugation by $x_{1}$. – Jon Bannon Mar 8 '11 at 13:19

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.

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Thanks, Owen. Hopefully someone can provide us with examples of nonamenable groups with the above property. – Jon Bannon Jun 6 '10 at 16:41
Tiny notational quibble: I call a group `virtually P' if it has a finite-index subgroup with property P (and I think this is pretty standard). – HJRW Sep 22 '10 at 20:57

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.

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It's too bad that Property (T) isn't preserved under direct limits! If it were, then taking the union here for n \geq 3 may allow us to construct something that is inner amenable but has Property (T). That would give a much nicer example of a non-Gamma II_1 factor coming from an inner amenable group. – Jon Bannon May 11 '12 at 13:41
Jon, icc, (T) and inner amenability are not compatible. Indeed, if $\ell^2(G \setminus \lbrace e\rbrace)$ has almost invariant vectors for the conjugation action, then it has an invariant vector. This means that there is a finite (non-trivial) conjugation invariant set and in particular, it contains a non-trivial finite conjugacy class. – Andreas Thom May 11 '12 at 14:27
Right, Andreas. What I wrote was a knee-jerk reaction to seeing SL(n) appear in your example. Still, (even though I haven't thought about it much yet) since one can tweak your example to work for n \geq 3 one can see from your comment that property T can't possibly behave well with respect to limits. That's a nice perk to this example! – Jon Bannon May 11 '12 at 15:34

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

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Please do not leave questions as comments; and if you do ask questions, please try to phrase them in a clear and relatively self-contained way, preferably with some indication of your motivation or previous efforts http// – Yemon Choi May 8 '12 at 10:41

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