Hi,

I'm currently trying to understand the Atiyah-Singer index theorem and its proof as presented in the book "Spin Geometry" by Lawson and Michelsohn.

I do not understand why the analytic index map $\operatorname{ind}\colon K_{cpt}(T^\ast X) \to Z$, as defined in chapter III.$13 in equation (13.8), agrees with the Fredholm index of an elliptic pseudo-differential operator.

Recall how the analytic index map is constructed (this is chapter III.§13 in the book): Given an element $u \in K_{cpt}(T^\ast X) \cong K(DX, \partial DX)$ we can represent it by Lemma III.13.3 via a triple $(\pi^\ast E, \pi^\ast F; \sigma)$, where $E$ and $F$ are vector bundles over $X$, $\pi\colon T^\ast X \to X$ is the bundle projection and $\sigma\colon \pi^\ast E \to \pi^\ast F$ is homogeneous of degree 0 on the fibres of $T^\ast X$ (i.e. $\sigma$ is constant on the fibres). Then for any $m$ there exists an elliptic, classical pseudo-differential operator $P \in \Psi CO_m(E,F)$ whose asymptotic principal symbol is $\sigma$ (in particular, the symbol class $[\sigma(P)] \in K_{cpt}(T^\ast X)$ equals u). Then we set $\operatorname{ind}(u) := \operatorname{Fredholm-ind}(P)$. Then it is proven in the book, that this is well-defined, i.e. independent of all choices.

Now given an elliptic pseudo-differential operator $D \in \Psi DO_m(E,F)$, we can construct its symbol class $[\sigma(D)] \in K_{cpt}(T^\ast X)$ as in chapter III.§1 in equation (1.7). Now I expect that the Fredholm index of $D$ coincides with the analytic index of $[\sigma(D)]$, but I do not see that this is proven in the book. I also can't prove it on my own. Going through the construction above we get an elliptic, classical pseudo-differential operator $P \in \Psi CO_m(E,F)$ with $[\sigma(D)] = [\sigma(P)] \in K_{cpt}(T^\ast X)$. But why do $D$ and $P$ have the same Fredholm index?

Why is $\operatorname{Fredholm-ind}(D) = \operatorname{ind}([\sigma(D)])$ for an elliptic, pseudo-differential operator?