13
$\begingroup$

There is a general (abstract) index theorem in noncommutative geometry: you take a K-theory class and K-homology class (which is represented by a triple $(A,H,F)$) and you pair them together. This pairing is computed as an index of certain operator. There is a notion of (noncommutaive analog of) Chern character in this context which takes values in cyclic cohomology and cyclic homology. Therefore you can apply this Chern character to both: K-theory class and K-homology class obtaining two classes, in cyclic cohomology and homology. In this context you have a natural pairing between cohomology and homology. The remarkable result is that this equal to the previous pairing between K-theory and K-homology (I would like to omit all technical details involving precise definitions: detailed discussion can be found in the book "Basic Noncommutative Geometry" by Masoud Khalkhali). My question is:

Can one deduce the 'usual' Atiyah Singer theorem from this abstract index theorem?

If so, how to proceed? For example, one problem is that in the data defining K-homology cycle the operator $F$ is bounded which is not the case for order $>0$ differential operators.

$\endgroup$
1
  • 4
    $\begingroup$ You can replace an unbounded Fredholm operator $D$ by a bounded one of the same index $D\circ(1+D^*D)^{-1/2}$. $\endgroup$ Mar 8, 2016 at 17:42

1 Answer 1

3
$\begingroup$

No.

What you describe is purely analytic (the definitions of the groups, of the pairings, and of the Chern-Connes characters), but the Atiyah-Singer index theorem has also a topological part.

By the way, I wouldn't call what you describe an "index theorem" (because, as I said, you are missing completely the topological part). What you have is just the compatibility of the Chern-Connes characters with the pairings.

$\endgroup$
2
  • $\begingroup$ But isn't it so, that the fact that analitical index is well defined is some sort of topological information? I mean that it factors to K-theory and K-homology (i.e. it depends only of K-classes). $\endgroup$
    – truebaran
    Mar 9, 2016 at 16:35
  • $\begingroup$ Cyclic homology of smooth functions on a space gives de Rham cohomology and the pairing with cyclic cohomology should be given by a current, i.e., by an integral. The commutativity gives the equality and the cyclic part of the diagram is the topological data. $\endgroup$
    – vap
    Sep 11, 2016 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.