3
$\begingroup$

There is a homological Chern character $ch_\ast \colon K_\ast(X) \to H_\ast(X)$ for $X$ a smooth, compact manifold.

I found only one definition of it (in the paper "K-Homology and Index Theory" by Baum, Douglas), where it is defined (i) by the usual cohomological Chern character map and Poincare duality between $H^\ast(X)$ and $H_\ast(X)$, and (ii) by using the geometric definition of the K-homology groups $K_\ast(X)$ of $X$.

(The geometric definition of K-homology uses cycles $(M, E, \phi)$, where $M$ is a closed Spin$^c$ manifold, $E$ a complex vector bundle on it and $\phi \colon M \to X$ a continuous map. Then $ch_\ast (M, E, \phi) := \phi_\ast(ch^\ast(E) \cup Td(TM) \cap [M])$, where $ch^\ast$ is the cohomological Chern character.)

Especially the use of Poincare duality to reduce the definition to the usual cohomological Chern character bothers me.

Is there some direct definition of the homological Chern character?

With "direct" I mean not using Poincare duality and then the cohomological Chern character. It would also be nice if this direct definition would not use the geometric description of K-homology but either the one by Fredholm modules or through the dual algebra $\mathfrak{D}(C_0(X))$.

Maybe one has to go through cyclic (co-)homology or Hochschild homology, e.g., first define $K_\ast(X) \to HC$ and then $HC \to H_\ast(X)$. That would be ok for me.

Improve this question
$\endgroup$
3
  • $\begingroup$ Even when defining the usual Chern character doesn't one use (in one form or another) a wrong-way map? Sometimes this is done via the Thom isomorphism instead, but in cases of interest one can usually rephrase via Poincaré duality. I imagine in this case you can also replace any instance of a wrong-way map in homology with the Thom isomorphism. So, for example, the 'homological Euler class' could be defined as pushing forward the fundamental class along the zero section and the pulling back (wrong way!). $\endgroup$
    – Dylan Wilson
    Commented May 24, 2013 at 12:04
  • $\begingroup$ The usual Chern character $ch^\ast \colon K^\ast(X) \to H^\ast(X)$ is defined by taking appropriate Chern classes of the vector bundle. I don't see any wrong-way map or Thom isomorphism in this definition (maybe in the definition of the Chern classes itself somewhere). $\endgroup$
    – AlexE
    Commented May 24, 2013 at 12:52
  • $\begingroup$ @AlexE: Yes, the definition of Chern classes themselves usually has this in one form or another... I mean once you have the Euler class you're good via the splitting principle, but defining Euler classes without some version of a wrong-way map is... well I don't know what that would look like. Having a theory of Euler classes is roughly equivalent to having wrong-way maps which is roughly equivalent to having a Thom isomorphism. $\endgroup$
    – Dylan Wilson
    Commented May 24, 2013 at 19:02

1 Answer 1

Reset to default
2
$\begingroup$

I like this question! I think this problem (the Chern character of K-homology) has been studied and solved by Alain Connes in his paper "Noncommutative differential geometry" in 1985. He indeed used cyclic cohomology etc. Unfortunately he uses analytic description of K-homology instead of dual algebras but they are essentially the same.

You can also read the book "Elements of noncommutative geometry" by Gracia-Bondia, Varilly and Figueroa. In particular Chapter 10 of that book gives a detailed description of Connes' construction.

Improve this answer
$\endgroup$

You must log in to answer this question.

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