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?

  • 1
    $\begingroup$ Oh ... the tags "pseudo-differential" and "operators" should be one tag. Sorry. $\endgroup$
    – AlexE
    Aug 24, 2011 at 15:42

2 Answers 2


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 $0$.

  • $\begingroup$ You have shown that the analytic index map $K^0(TX, TX-0) \to Z$, defined by mapping the class $u = (E_0, E_1; f)$ to the Fredholm index of F (a classical $\Psi DO$ with asymptotic symbol f), is well-defined. But I don't see how it follows that given an arbitrary elliptic operator D of order 0, the operator F we get from $[\sigma(D)] = [(E_0, E_1; f)]$ has the same Fredholm index as D. For this, we would need to show, e.g., that the symbol $\sigma(D)$ is regularly homotopic to one which is homogeneous of order 0 outside the zero section - and I don't see this yet. $\endgroup$
    – AlexE
    Sep 5, 2011 at 20:38
  • $\begingroup$ But this is shown in Lawson-Michelssohn, Lemma 13.3, page 245. $\endgroup$ Sep 6, 2011 at 8:44
  • $\begingroup$ Ok, now I see it, you are right. First I was confused, because you put the "homogeneous of order 0" directly into the definition of the K-groups. But Lemma 13.3 tells us that we get the same groups if we put this "homogeneous-thing" into the definition or not. Thanks. $\endgroup$
    – AlexE
    Sep 8, 2011 at 11:30

I can't give you an answer that is well-adapted to Lawson and Michelsohn's formulation of the pseudodifferential calculus. But here's how this sort of argument is supposed to go: two elliptic pseudodifferential operators with the same principal symbol differ by a smoothing operator, and smoothing operators are compact (one often proves this by showing they are Hilbert-Schmidt). By Atkinson's theorem, two Fredholm operators which differ by a compact operator have the same Fredholm index.

So the real point is to understand why the symbol of an operator characterizes it up to smoothing operators. The problem is that you can have a lot of flexibility in exactly what class of operators you decide to allow to be called "pseudodifferential", and so your proof of this fact has to be well-adapted to your specific construction. But I don't think that there is much controversy over the statement that whatever algebra of operators you consider to be the algebra of pseudodifferential operators there should be a symbol map whose kernel consists only of smoothing operators (or, at worst, compact operators). I unfortunately can't help you verify that Lawson and Michelsohn's choice meets this criterion.

EDIT: I see that I did not quite interpret your question correctly. However, showing that the index is determined by the symbol is most of the work toward showing that the index is determined by the symbol class. What I wrote above tells you how to show that there is a well defined index map from the set of symbols to the integers, so as Johannes Ebert points out you only need to check that this map is compatible with the relations which define $K(T^\ast M)$. The relations at the symbol level (homotopy, direct sum) pass to corresponding relations at the level of the operators which do not change the index.

  • $\begingroup$ That the Fredholm-index of a pseudo differential operator depends only on the regular homotopy class of its principal symbol is shown in the book. But the analytic index map first constructs an operator with the same principal symbol class and then takes its Fredholm-index, i.e. I need the (stronger?) statement that the Fredholm-index of an operator depends only on its symbol class. Is this true? $\endgroup$
    – AlexE
    Aug 31, 2011 at 16:49

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