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?
Best Answer
The Fredholm index of an elliptic operator only depends on the symbol class. Here is the proof (which I memorize from Lawson-Michelsohn and Atiyah-Singer).
If $D: \Gamma(E_0) \to \Gamma(E_1)$ has order $k \neq 0$, pick a connection $\nabla$ on $E_0$. Then $A=(1+\nabla^{\ast}\nabla)$ is a self-adjoint invertible operator of order $2$ and $D \circ A^{-k/2}$ has the same Fredholm index as $D$ and the same symbol class; but it has order $0$. So for any operator, there is an order $0$ operator with the same symbol class and the same index; and this reduces the problem to the order $0$ case.
It has been mentioned that the index of an operator only depends on the homotopy class of its symbol (by the way, the proof of this in Lawson-Michelssohn is incomplete. In the proof of Theorem 7.10, they say ''to construct a family of operators $P_t$ with $\sigma(P_t)=\sigma_t$, [...] is evidently possible locally (in coordinates)''. One needs to know that in $R^n$, the operator norm of a pseudo-DO can be estimated by the symbol. You can see this by going through the proof of Prop. III.3.2 of L.-M.).
Recall the (modified) definition of $K^0 (TX,TX-0)$: it is the group of all equivalence classes of $(E_0,E_1,f)$; $E_i \to X$ vector bundles and $f: \pi^{\ast} E_0 \to \pi^{\ast} E_1$ a bundle map that is an isomorphism away from the zero section and homogeneous of order $0$ outside the zero section. The equivalence relation is generated by (1) homotopy, (2) isomorphism and (3) addition of things of the form $(E,E,id)$. (1) and (2) preserve the index. (3) also preserves the index; since a pseudo-DO with symbol $(E,E,id)$ is e.g. the identity operator, which has index $0$.