Integration in equivariant K-theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:48:25Z http://mathoverflow.net/feeds/question/6412 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6412/integration-in-equivariant-k-theory Integration in equivariant K-theory Michael Ortiz 2009-11-21T22:11:27Z 2010-06-21T02:55:30Z <p>Let F be a smooth classifying space for K-theory (ordinary or equivariant). If X is a smooth compact manifold and W is a real vector space of dimension n, there is an integration map from the compactly supported K-theory of X x W to that of X. It can be represented on the level of classifying maps by regarding a map f from X x W into F as a map from X into the nth loop space of F.</p> <p>Does a similar picture exist in equivariant K-theory, where X is a G-space and W is a group representation? Of course it does if W is simply a trivial representation. But what if W is some other representation? Is there an integration map, and can one represent it on the level of classifying maps? Is there a good reference for this?</p> http://mathoverflow.net/questions/6412/integration-in-equivariant-k-theory/6453#6453 Answer by yeshengkui for Integration in equivariant K-theory yeshengkui 2009-11-22T10:08:44Z 2009-11-22T10:08:44Z <p>What do you mean by "smooth classifying space for K-theory"? The "integration" for equivarient case may be not not so easy if X is just a G-space.</p> http://mathoverflow.net/questions/6412/integration-in-equivariant-k-theory/9327#9327 Answer by Michael for Integration in equivariant K-theory Michael 2009-12-18T22:51:41Z 2009-12-18T22:51:41Z <p>I don't know about the homotopy-theoretic picture (and would like to learn more about it), but maybe the following also helps: If I understand you correctly, the integration maps of K-theory are what is often called the Gysin/shriek maps (the case you mention would be the Gysin map induced by the zero section of a trivial G-vector bundle). I think they only exist if the map is K-oriented; in that case they can be implemented by right multiplication with functorial KK-elements. The canonical reference for the non-equivariant case would be Connes-Skandalis and for the equivariant case Kasparov-Skandalis.</p> http://mathoverflow.net/questions/6412/integration-in-equivariant-k-theory/10064#10064 Answer by Chris Schommer-Pries for Integration in equivariant K-theory Chris Schommer-Pries 2009-12-29T20:55:13Z 2009-12-29T20:55:13Z <p>It sounds like what you are after is a theory of genuine equivariant K-theory (Which exists!)</p> <p>If you think of a cohomology theory as a sequence of functors to abelian groups together with some properties and suspension isomorphism, then a <em>genuine</em> G-equivariant cohomology theory can be similarly thought of as a sequence of functors (indexed on the representation ring of G) together with "suspension isomorphisms" where you suspend by any G-representation, i.e. take the G-sphere with is the one point compactification of W (with its G-action) and smash with it. This is only the rough picture. The real theory is somewhat technical and develops the theory of genuine G-spectra. There were some comments about this <a href="http://mathoverflow.net/questions/3154/1-categorical-description-of-equivariant-homotopy-theory" rel="nofollow">here</a> with some references. The punchline is that yes K-theory is an example of a genuine G-spectrum.</p> <p>The next step is that you need to identify your K-theory with compact supports as the reduced K-theory of the suspension. I don't think this is too hard. Then your integration map reduces to the suspension isomorphism:</p> <p>$\tilde{K}^*<em>G( \Sigma^W X) \cong \tilde{K}^{</em> -W}_G(X)$</p> <p>As Michael and Kevin mention, under a suitable equivariant K-theory orientation hypothesis there will be similar integration maps for more general vector bundles, but that's another story.</p> http://mathoverflow.net/questions/6412/integration-in-equivariant-k-theory/28905#28905 Answer by Ezra Getzler for Integration in equivariant K-theory Ezra Getzler 2010-06-21T02:55:30Z 2010-06-21T02:55:30Z <p>Atiyah gave a very precise answer to this question: an equivariant vector bundle V/M is K-orientable (satisfies the Thom isomorphism for K-theory, i.e. K(B(V),S(V)) is a rank-one free module over K(M)) if and only if V has an equivariant Spin_c-structure. Atiyah's proof definitely involves analysis (the family index). The proof by Kasparov is a little more technical, since the definition of the product in Kasparov's KK theory is rather technical. (It is a generalization of the family index theorem of Atiyah and Singer.) But all of these proofs come down to the same thing: the study of families of equivariant Fredholm operators.</p>