Is there an i.c.c. nonamenable simple group that is inner amenable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:42:18Z http://mathoverflow.net/feeds/question/27233 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable Is there an i.c.c. nonamenable simple group that is inner amenable? Jon Bannon 2010-06-06T12:26:12Z 2012-12-26T19:52:57Z <p>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 $g \in G.$ </p> <p>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 $g \in G.$</p> <p>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 $e$. </p> <p>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).</p> <p>Can the growth of G detect inner amenability?</p> <p>I'd like to know if there is an i.c.c. discrete nonamenable simple group that is inner amenable? </p> <p>On a related note, what about an answer to Owen's question below?</p> http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/27242#27242 Answer by Owen Sizemore for Is there an i.c.c. nonamenable simple group that is inner amenable? Owen Sizemore 2010-06-06T14:37:13Z 2010-06-06T14:37:13Z <p>Hey Jon</p> <p>So my initial thought would be no. </p> <p>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.</p> <p>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. </p> <p>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.</p> <p>Now a final way to maybe make something out of this is to ask </p> <p>"If $G$ is inner amenable and along with all of its quotients, then is there a growth contsraint."</p> <p>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.</p> <p>AS for this class.... I have no idea. </p> http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/48106#48106 Answer by Denis Osin for Is there an i.c.c. nonamenable simple group that is inner amenable? Denis Osin 2010-12-02T22:17:13Z 2010-12-03T00:10:16Z <p>Here is a construction of a countable i.c.c. nonamenable simple group that is inner amenable. First consider the following condition:</p> <p>(*) For every finite subset $S\subseteq G$, there exists $g\in G\setminus { 1}$ such that $[g,s]=1$ for every $s\in S$.</p> <p>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 $g$.</p> <p>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 $F_2$. </p> <p>Modifying the above argument one can also construct an i.c.c. simple inner amenable non-amenable group without nontrivial free subgroups.</p> http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/96319#96319 Answer by bah for Is there an i.c.c. nonamenable simple group that is inner amenable? bah 2012-05-08T10:13:14Z 2012-05-08T10:13:14Z <p>Is there a non inner amenable locally compact group [map]group</p> http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/96657#96657 Answer by Andreas Thom for Is there an i.c.c. nonamenable simple group that is inner amenable? Andreas Thom 2012-05-11T10:54:38Z 2012-12-26T19:52:57Z <p>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 $$g_n:= 1_n \oplus \left(\begin{matrix} 0 &amp; 1 \newline -1 &amp; 0 \end{matrix}\right) \oplus 1_{\infty}.$$ and let $$m_{n}(A) := \begin{cases} 1 &amp; g_n \in A \newline 0 &amp; g_n \not \in A \end{cases}.$$ 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.</p>