# Parametrized Atiyah-Singer index theorem

Let $M$ be any smooth manifold (could be unorientable - I think). Let $E,F \to M$ be two complex vector bundles. Let $S$ be any compact space, and let $D_s:E\to F,s\in S$ be a continuous family elliptical pseudo-differential operator. Since $D_s$ is a Fredholm operator (choosing the right Sobolev spaces to represent sections) for each $s$ we must be able to associated an index bundle over $S$.

The Atiyah-Singer index theorem basically states that the dimension of this bundle at a point $s\in S$ can be computed using the formula explained here.

Question: How do you describe and prove what the entire map $S\to \mathbb{Z}\times BU$ which classifies the index bundle is?

Notes:

1) I tried googling different things, but whenever I tried a link it was always something different than the answer to this question, but maybe I missed an obvious link. It seems like this should be explained somewhere.

2) In this question Paul Siegel gives a K-theoretic description of the index formula. This leads me to believe that the answer might be something a long the line of: The symbol of $D_s$ defines an element of $K((S\times M)^{TM})$ (Thom space of pull back bundle to the product). We can use the Thom isomorphism to identify this with a $K$ class of the spectrum $(S\times M)^{-TM}$ (we subtract the complex vector bundle $\mathbb{C}\otimes TM = TM\oplus TM$ from the Thom construction). We then use that $M^{-TM}$ has a unit map from the sphere spectrum $\mathbb{S} \to M^{-TM}$ to compose:

$$S_+\wedge \mathbb{S} \to S_+\wedge M^{-TM} \cong (S\times M)^{-TM}$$

and then pullback the class. If this is correct I would like to known, and also get a reference to where this is proved.

• How is this different from Atiyah-Singer IV: jstor.org/stable/1970756 – Mohammed Abouzaid Sep 1 '15 at 17:46
• Looking at it it seems it is not. Didn´t know that it was contained in these. Thank you for the reference. – Thomas Kragh Sep 1 '15 at 19:12