existence of charaterization of amenable groups by complementation? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:08:48Z http://mathoverflow.net/feeds/question/40795 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40795/existence-of-charaterization-of-amenable-groups-by-complementation existence of charaterization of amenable groups by complementation? BigBill 2010-10-01T21:50:06Z 2010-10-02T08:12:45Z <p>Recall that we say that a closed space \$F\$ of a Banach space \$E\$ is complemented if there exists a contractive projection \$P\$ from \$E\$ onto \$F\$.</p> <blockquote> Do you know a charaterization of discrete amenable groups by the existence of a complementation of a closed space \$F\$ of a Banach space \$E\$? </blockquote> <p>More precisely, the required charaterization is<br> For all discrete group \$G\$, there exists a Banach space \$E_G\$ and a closed space \$F_G\$ of \$E_G\$ such that \$G\$ is amenable if and only if \$F_G\$ is complemented in \$E_G\$.</p> http://mathoverflow.net/questions/40795/existence-of-charaterization-of-amenable-groups-by-complementation/40820#40820 Answer by Matthew Daws for existence of charaterization of amenable groups by complementation? Matthew Daws 2010-10-02T08:12:45Z 2010-10-02T08:12:45Z <p>Yes: a <em>discrete</em> group \$G\$ is amenable if and only if the reduced group C*-algebra \$C^*_r(G)\$ is nuclear, see E.C. Lance, On nuclear \$C^{\ast} \$-algebras. J. Functional Analysis 12 (1973), 157--176. This is then equivalent to <code>\$W^*(G) = C^*_r(G)^{**}\$</code> being an <em>injective</em> von Neumann algebra: which by definition means that if \$W^*(G) \subseteq B(H)\$ then there is a <em>contractive</em> projection from \$W^*(G)\$ to \$B(H)\$.</p> <p>I'm pretty sure you could look at the group von Neumann algebra \$VN(G)\$ instead, but I cannot recall the correct reference (but it's all in Runde's book "Lectures on Amenability"). Note that all this only works because \$G\$ is discrete.</p> <p>Now, the problem is that you do need ``contractive'' projection here: it's still a conjecture if just having a bounded projection is enough.</p> <p>Also, I'm sure there are other answers (and perhaps some that are easier: even a streamlined approach to all this uses a lot of operator algebra theory)...</p>