Does a "Chern character" exist for any generalized cohomology theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:27:23Z http://mathoverflow.net/feeds/question/29202 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29202/does-a-chern-character-exist-for-any-generalized-cohomology-theory Does a "Chern character" exist for any generalized cohomology theory? Bo Peng 2010-06-23T07:55:03Z 2010-06-24T10:25:01Z <p>The Chern character is a ring homomorphism from the complex K-theory to the usual cohomology.</p> <p>1) I wonder if there are "Chern character"-like ring homomorphisms from other generalized cohomology theory to the usual cohomlogy. Are they related with Atiyah-Hirzebruch?</p> <p>2) And if there are such nice homomorphisms, what is the "Todd genus" in these cases, making the generalization of that famous diagram in Grothendieck–Hirzebruch–Riemann–Roch commute?</p> <p>When I think about it, I cannot even recall seeing anything like this in real K-theory, but that is probably because I dont really know real K-theory at all.</p> <p>Thank you very much.</p> http://mathoverflow.net/questions/29202/does-a-chern-character-exist-for-any-generalized-cohomology-theory/29206#29206 Answer by Oscar Randal-Williams for Does a "Chern character" exist for any generalized cohomology theory? Oscar Randal-Williams 2010-06-23T08:53:57Z 2010-06-23T11:30:19Z <p>For any (connective) spectrum $E$ one may rationalise it to get a rational spectrum $E_\mathbb{Q}$, and a map $E \to E_\mathbb{Q}$. Now rational spectra split as wedges of Eilenberg-Mac Lane spectra, so one may choose an isomorphism <code>$E_\mathbb{Q}^*(X) \simeq H^*(X;\pi_{-*}(E)\otimes \mathbb{Q})$</code>, and the rationalisation gives a map <code>$$ch_E : E^*(X) \longrightarrow H^*(X;\pi_{-*}(E)\otimes \mathbb{Q}).$$</code></p> <p>For complex K-theory this gives the Chern character, and for real K-theory it gives the Pontrjagin character.</p> <p>Of course, if $E$ is a ring spectrum so is $E_\mathbb{Q}$, and one must identify the induced ring structure on <code>$H^*(X;\pi_{-*}(E)\otimes \mathbb{Q})$</code>.</p> http://mathoverflow.net/questions/29202/does-a-chern-character-exist-for-any-generalized-cohomology-theory/29223#29223 Answer by Tom Goodwillie for Does a "Chern character" exist for any generalized cohomology theory? Tom Goodwillie 2010-06-23T11:21:33Z 2010-06-23T11:21:33Z <p>In Oscar Randall-Williams' answer "connective" is unnecessary. Also, there is no need to choose that isomorphism (from a rational theory to the ordinary theory with the same coefficient groups); it is canonical.</p> <p>The generalization of the Todd genus or Todd class arises when the multiplicative theory $E$ has a "complex orientation": a multiplicatively well-behaved way of producing Thom isomorphisms in $E^*$-theory for all complex vector bundles. </p>