In the usual setting of the Atiyah-Singer index theorem the situation is as follows: we have a closed smooth manifold $M$ without boundary and $D$ is some elliptic differential operator acting on sections of some vector bundle over $M$. In this situation $D$ turns out to be Fredholm and one can compute its index which is an integer. However I read that AS index theorem has various generalisations: one of them is about a whole family $(D_x)_{x \in X}$ of elliptic operators parametrized by some topological space. I read that in such a situation one can define the index of this family as an element of $K^0(X)$.

Why the index of such a family is not just an integer valued function? Why it is defined as an element of $K$-theory?


3 Answers 3


You can actually see the family index as an element of $K^0(X)$. First, assume that $\ker(D_x)_{x\in X}$ has constant dimension. Then it defines a vector bundle over $X$. Because the Fredholm index of $D_x$ is constant, the cokernel also defines a vector bundle. And $\mathrm{ind}(D)=[\ker(D_x)]-[\mathrm{coker}(D_x)]$ is an honest element of $K^0(X)$.

If $\ker(D_x)$ does not have constant dimension, assume that $X$ is compact. Then the dimension of $\mathrm{coker}(D_x)$ is bounded. Moreover, for some large $N$, the span of $N$ sections of the range of $D_x$ suffice to exhaust $\mathrm{coker}(D_x)$. Add $\mathbb C^N$ to the domain of all $D_x$ and map it to the range in such a way that the cokernel is killed. Let $\tilde D_x$ be the new family and put $\mathrm{ind}(D)=[\ker(\tilde D_x)]-N\in K^0(X)$. If $X$ is noncompact, you need a sufficiently nice description of $K$-theory to extend this method.

One can check that this construction is independent of the modifications you choose. Usually, one assumes more generally that one has a proper submersion $E\to X$ with closed fibre $M$. Your situation is a special case. See Lawson-Michelsohn or Berline-Getzler-Vergne for more explanations and a proof.


The space of Fredholm operators classifies $K$-theory. Thus, any continuous family of elliptic operators $(D_x)_{x\in X}$ canonically defines an element in the Abelian group $K(X)$. When $X$ consists of a single point $x_0$, then one can canonically identify $K(\{x_0\})$ with the group of integers.

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    $\begingroup$ Thank you for your answer. Could you please give me some reference where actually I can find the proof of the fact that the space of Fredholms operators classifies $K$-theory? Also, it is obvious that there aren't any problems with the domain of operators (I mean by this the fact, that usually "space of Fredholm operators" is understood as a subset of the space of bounded_operators acting on in the _single Hilbert space) $\endgroup$
    – truebaran
    Jan 4, 2016 at 1:05
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    $\begingroup$ Check this paper www3.nd.edu/~lnicolae/bofu.pdf and the references therein. There I explain how to deal with unbounded operators. $\endgroup$ Jan 4, 2016 at 1:09
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    $\begingroup$ You can always assume the operators act on the same Hilbert space any Hilbert bundle over a compact CW complex is trivial. This follows from the remarkable fact that the group of invertible bounded linear operators on a Hilbert space is contractible. This is not true for other Banach spaces, $\endgroup$ Jan 4, 2016 at 1:16

This is just an addendum to Sebastian Goette's excellent answer:

In fact, you can retrieve the integer-valued function that you mention from the $K$-theory class: Let $\iota_x \colon \{pt\} \to X$ be the inclusion that sends $pt$ to $x \in X$. Let $ind(D) \in K^0(X)$ be the $K$-theory class representing the family index of $(D_x)_{x \in X}$. The function you mention will be $$ X \to K^0(pt) \cong \mathbb{Z} \quad ; \quad x \mapsto \iota_x^*(ind(D)) $$ But this loses a lot of information. In fact, it will just recover the virtual dimension of the fiber over $x$ of the vector bundle Sebastian Goette is talking about in his answer. The index class $ind(D) \in K^0(X)$ also contains global information about "how non-trivial" the vector bundle is. The $K$-theory class $ind(D) \in K^0(X)$ is not a bug, it is a feature!


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