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My question relates to the proof of the Atiyah-Singer Index Theorem for families of elliptic operators, as presented in "The Index of Elliptic Operators: IV", M. F. Atiyah and I. M. Singer.

Let $A$ be a compact Hausdorff space and $q:A\times \mathbb{C}^n\rightarrow A$ be the projection, then we obtain the induced Thom isomorphism $q_!:K_{\text{cpt}}(A\times \mathbb{C}^n)\rightarrow K(A)$. The map $q_!$ and the analytic index $ind:K_{\text{cpt}}(A\times \mathbb{C}^n)\rightarrow K(A)$ coincide. According to Atiyah this follows from the case $Y$ is a point, and the fact that the analytical index is a homomorphism of $K(Y)$-modules. My question is why are these two properties enough to show the two maps coincide?



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1 Answer 1

up vote 5 down vote accepted

The answer should probably go something like this:

Both $q_!$ and $ind$ are $K(A)$-module maps. Since $q_!$ is an isomorphism, to check that these are the same maps, it suffices to check they are the same on a generator; if $u_A\in K_{Cpt}(A\times C^n)$ is the unique element such that $q_!(u_A)=1$, then we need to show that $ind(u_A)=1$ as well.

The class $1\in K(A)$ is in the image of the tautological map $\mathbb{Z}=K(point)\to K(A)$, induced by $f:A\to point$. Both $q_!$ and $ind$ are natural with respect to $f$, so to prove that $ind(u_A)=1$, it is enough to prove that $ind(u_{point})=1$, since $ind(u_A)=ind(f^*(u_{point}))= f^*(ind(u_{point}))$.

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Thanks for your help! Tristan –  Tristan Jan 19 '11 at 19:02
Tristan, if you think this answers your question, you should accept Charles' answer. –  Adam Hughes Jan 19 '11 at 19:28

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