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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.

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

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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.

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  • $\begingroup$ 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
    Commented 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
    Commented Sep 11, 2016 at 21:37

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