26
$\begingroup$

The Chern character is a ring homomorphism from complex K-theory to the usual cohomology.

1) I wonder if there are "Chern character"-like ring homomorphisms from other generalized cohomology theories to the usual cohomology. Are they related with Atiyah-Hirzebruch?

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?

When I think about it, I cannot even recall seeing anything like this in real K-theory, but that is probably because I don't really know real K-theory at all.

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ I think, your question is related to the orientability of cohomology theories. There are many multiplicative transformations between bordism theories and other (co)homology theories such as oriented bordism to singular homology, spin bordism to K-theory, string bordism to TMF. As you see, the natural home for such a morphism is not always singular cohomology, but eg in the case of spin bordism you can, of course, compose the morphism to K-theory with the chern character. $\endgroup$
    – Lennart Meier
    Commented Jun 23, 2010 at 8:39

2 Answers 2

Reset to default
31
$\begingroup$

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.

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.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Would it even make sense if one had to choose that isomorphism? How can one get the Chern character or the Pontryagin character if there is an ambiguity in a choice of isomorphism? $\endgroup$
    – user137162
    Commented Jun 20, 2019 at 9:36
  • $\begingroup$ Yet I do not see why the splitting of rational spectra into EM spectra is canonical. $\endgroup$
    – user137162
    Commented Jun 20, 2019 at 10:19
  • 2
    $\begingroup$ If $E$ and $F$ are spectra whose homotopy groups are rational, then for any map of graded groups $\pi_\ast E\to \pi_\ast F$ there is a unique map of spectra $E\to F$ inducing it. The essential point is that rational homology of EM spectra has nothing subtle about it. $\endgroup$
    – Tom Goodwillie
    Commented Jun 20, 2019 at 12:15
31
$\begingroup$

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 $E_\mathbb{Q}^*(X) \simeq H^*(X;\pi_{-*}(E)\otimes \mathbb{Q})$, and the rationalisation gives a map $$ch_E : E^*(X) \longrightarrow H^*(X;\pi_{-*}(E)\otimes \mathbb{Q}).$$

For complex K-theory this gives the Chern character, and for real K-theory it gives the Pontrjagin character.

Of course, if $E$ is a ring spectrum so is $E_\mathbb{Q}$, and one must identify the induced ring structure on $H^*(X;\pi_{-*}(E)\otimes \mathbb{Q})$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .