When is non amenablity witnessed by a single non measurable set? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:18:23Z http://mathoverflow.net/feeds/question/60247 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60247/when-is-non-amenablity-witnessed-by-a-single-non-measurable-set When is non amenablity witnessed by a single non measurable set? Justin Moore 2011-04-01T01:16:53Z 2011-05-25T12:11:35Z <p>Suppose $G$ is a finitely generated discrete group and that there is a subset $E$ of $G$ such that if $\mu$ is a finitely additive probability measure on $G$, then there is a $g$ in $G$ such that $\mu(E \cdot g) \ne \mu(E)$. Certainly $G$ is non amenable. Can more be said about $G$? Must $G$ contain $\mathbb{F}_2$?</p> <p>It should be noted that the above situation can happen: let $E$ be all elements of $\mathbb{F}_2$ which begin'' with $a$ or $a^{-1}$. Then both $E$ and its complement have infinitely many disjoint translates (by powers of $b$ and $a$, respectively).</p> http://mathoverflow.net/questions/60247/when-is-non-amenablity-witnessed-by-a-single-non-measurable-set/60319#60319 Answer by Justin Moore for When is non amenablity witnessed by a single non measurable set? Justin Moore 2011-04-01T20:41:29Z 2011-05-25T12:11:35Z <p>The answer is that the above is equivalent to non amenability. Fix a group $(G,*)$. Since $(G,*)$ is non amenable if and only if every finitely generated subgroup is non amenable, we may assume that $G$ is finitely generated.</p> <p>If $\mu$ and $\nu$ are finitely supported probability measures on $G$, define $$\mu * \nu (Z) = \sum_{x * y \in Z} \mu (\{x\}) \nu (\{y\})$$ Observe that $g * \nu (E) = \nu ( g^{-1} * E)$. If $S$ is a subset of $S$, let $P(S)$ denote all probability measures on $S$ (which are identified with probability measures on $G$ which are supported on $S$). I will identify $G$ with the point masses in $P(G)$.</p> <p>If $A$ and $B$ are subsets of $G$ and $A$ is finite, we say that $B$ is $\epsilon$-Ramsey with respect to $A$ if for every $E \subseteq B$, then there is a $\nu$ in $P(B)$ such that $P(A) * \nu \subseteq P(B)$ and $$|\mu * \nu (E) - \nu (E)| &lt; \epsilon$$ for all $\mu$ in $P(A)$. Notice that in some sense $E$ is defining a partition of $P(B)$ and we are postulating the existence of a copy of $P(A)$ in $P(B)$ which is homogeneous for $E$ up to an error of $\epsilon$.</p> <p>It can be shown with an argument similar to the one below that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ there is a $\nu$ in $P(B)$ such that $$|f(\mu * \nu) - f(\nu)| &lt; \epsilon$$ where $f$ has been extended linearly to $P(B)$.</p> <p>We say that $(G,*)$ is Ramsey if for every finite subset $A \subseteq G$ and every $\epsilon > 0$, there is a finite subset $B$ of $G$ with is $\epsilon$-Ramsey with respect to $A$. Notice that if $B$ satisfies that for every $E \subseteq B$ there is a $\nu$ in $P(B)$ such that $$|g * \nu (E) - \nu (E)| &lt; \epsilon$$ for all $g$ in $A$, then $B$ is contained in a finite set which is $\epsilon$-Ramsey (we need only to replace $B$ by $A * B \cup B$).</p> <p>To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite $A \subseteq G$ and $\epsilon > 0$. I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. Let $B_n$ $(n &lt; \infty)$ be an increasing sequence of finite sets covering $G$. Let $T_n$ be the set of all subsets $E$ of $B_n$ which witness that $B_n$ is not $\epsilon$-Ramsey with respect to $A$. Observe that if $E$ is in $T_{n+1}$, then $E \cap B_n$ is in $T_n$. Otherwise there would be a $\nu$ in $P(B_n)$ such that $g * \nu$ is in $P(B_n)$ for each $g$ in $A$ and $$|g * \nu (E \cap B_n) - \nu (E \cap B_n)| &lt; \epsilon$$ Such a $\nu$ would also witness that $E$ is not in $T_{n+1}$. Define $T = \bigcup_n T_n$ and order $E \leq_T E'$ if $E = E' \cap B_m$ where $E$ is in $T_n$. This order makes $T$ into an infinite finitely branching tree. By K&ouml;nig's lemma, $T$ has an infinite path whose union is some $E \subseteq G$. If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates in $A$. But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ and $\nu$ would witness that $E \cap B_n$ was not in $T$.</p> <p>Now the claim is that the Ramsey property of a discrete group is equivalent to its amenability. That amenability implies the Ramsey property follows from F&oslash;lner's characterization of amenability. Also observe that $G$ is amenable provided that for every $\epsilon > 0$, every finite list $E_i$ $(i &lt; n)$ of subsets of $G$, and $g_i$ $(i &lt; n)$ in $G$, there is a finitely supported $\mu$ such that $$|\mu (g_i * E_i) - \mu (E_i) | &lt; \epsilon.$$ Set $B_{-1} = \{1_G\} \cup \{g^{-1}_i :i &lt; n\}$ and construct a sequence $B_i$ $(i &lt; n)$ such that $B_{i+1}$ is $\epsilon/2$-Ramsey with respect to $B_i$.</p> <p>Now inductively construct $\nu_i$ $(i &lt; n)$ by downward recursion on $i$. If $\nu_j$ $(i &lt; j)$ has been constructed, let $\nu_i \in P(B_i)$ be such that $$|\mu * \nu_{i} * \ldots * \nu_{n-1} (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| &lt; \epsilon/2$$ for all $\mu$ in $P(B_{i-1})$. Set $\mu = \nu_0 * \ldots * \nu_{n-1}$. If $i &lt; n$, then since $\nu_0 * \ldots * \nu_{i-1}$ and $g_i^{-1} * \nu_0 * \ldots * \nu_{i-1}$ are in $P(B_{i-1})$, $$|g_i^{-1} * \mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| &lt; \epsilon/2$$ $$|\mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| &lt; \epsilon/2$$ and therefore $|\mu (g_i * E_i) - \mu (E_i)| &lt; \epsilon$.</p>