Amenability and ultrafilters - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:46:19Z http://mathoverflow.net/feeds/question/94547 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94547/amenability-and-ultrafilters Amenability and ultrafilters Misha 2012-04-19T16:59:34Z 2012-04-19T17:15:13Z <p>Among hundreds of equivalent definitions of amenability (for discrete, countable, groups), I would like to discuss two which are most common:</p> <p>A1. A group \$G\$ is amenable if it admits a Folner sequence. </p> <p>A2. A group \$G\$ is amenable if it admits an invariant mean. </p> <p>(See e.g. <a href="http://en.wikipedia.org/wiki/Amenable_group" rel="nofollow">http://en.wikipedia.org/wiki/Amenable_group</a> or <a href="http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/" rel="nofollow">http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/</a>) </p> <p>However, proofs of equivalence that I know (even for \$G={\mathbb Z}\$) require either axiom of choice or, at least, existence of a nonprincipal ultrafilter on \${\mathbb N}\$. </p> <p>Question: Is there a proof that A1 \$\iff\$ A2 which uses only ZF axioms? Or, maybe \$A1\iff A2\$ implies existence of nonprincipal ultrafilters, maybe in a weakened form? </p> <p>This question was discussed a bit in <a href="http://mathoverflow.net/questions/12169/why-are-abelian-groups-amenable" rel="nofollow">http://mathoverflow.net/questions/12169/why-are-abelian-groups-amenable</a> and <a href="http://mathoverflow.net/questions/93897/why-groups-that-admit-folner-sequences-are-amenable" rel="nofollow">http://mathoverflow.net/questions/93897/why-groups-that-admit-folner-sequences-are-amenable</a> but not in the above form. </p> <p>Note: I am not a logician, but a geometric group-theorist and I frequently use ultrafilters. As the result I am often asked if the results could be proven without ultrafilters. For most proofs my answer usually is: "Yes, if you work much harder and write ugly and long proofs." However, I do not know the answer in the context of amenable groups. </p> http://mathoverflow.net/questions/94547/amenability-and-ultrafilters/94548#94548 Answer by Simon Thomas for Amenability and ultrafilters Simon Thomas 2012-04-19T17:09:56Z 2012-04-19T17:15:13Z <p>Of course, \$ZF\$ is enough to prove that \$\mathbb{Z}\$ has a Folner sequence. But, as you point out, \$ZF\$ is not enough to prove that \$\mathbb{Z}\$ has an invariant mean. Thus \$ZF\$ does not prove the equivalence of A1 and A2.</p> <p>On the other hand, the Hahn-Banach Theorem is enough to prove the equivalence of A1 and A2 for countable discrete groups and Pincus-Solovay have constructed a model of \$ZF\$ in which the Hahn-Banach Theorem is true but there are no nonprincipal ultrafilters on \$\mathbb{N}\$. Hence the equivalence of A1 and A2 for countable groups does not imply the existence of nonprincipal ultrafilters on \$\mathbb{N}\$.</p>