Clifford Algebras and Bott Periodicity: The Atiyah

2.3 The Atiyah–Jänich theorem

Equipped with Kuiper's theorem, we can finally prove what we wanted to.

Theorem (Atiyah, Jänich)

Let $H$ be an infinite-dimensional real Hilbert space, and $\mathcal{F}$ the space of all Fredholm operators on $H$ . Then for every compact space $X$ , there is a natural isomorphism

\operatorname{idx}: [X, \mathcal{F}] \to KO(X).

The same holds in the complex case with $KO$ replaced by $KU$ .

For $f: X \to \mathcal{F}$ , we would like to define $\operatorname{idx}(f)$ by setting the fiber at each $x \in X$ to be the formal difference between $\ker f(x)$ and $\operatorname{coker}f(x)$ . In general, this does not give a vector bundle, but it is a general fact (which we shall not prove) that we can homotope $f$ so that it does, and the difference $[\ker f] - [\operatorname{coker}f]$ is independent of the choice of homotopy. In particular, it depends only on the homotopy class of $f$ .

Proof

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[Proof (cf. [1])] First observe the following trivial lemma:

Lemma

Let $f: M \to G$ be a surjective homomorphism from a monoid to an abelian group with trivial kernel, i.e. $f^{-1}(\{ 0\} ) = \{ 0\}$ . Then $f$ is in fact injective, hence an isomorphism.

So it suffices to show our map is surjective and has trivial kernel.

To show the kernel is trivial, by Kuiper's theorem, it suffices to show that every map $T: X \to \mathcal{F}$ with index $0$ is homotopic to a map with image in $\mathrm{GL}(H)$ .

If $\operatorname{idx}T = 0$ , we can homotope $T$ so that $[\ker T] = [\operatorname{coker}T]$ . Thus, by definition of the $K$ -groups, we can find some large $m$ such that
$\ker T \oplus (X \times \mathbb {R}^m) \cong \operatorname{coker}T \oplus (X \times \mathbb {R}^m).$
Thus, by homotoping $T$ to vanish further on a trivial subbundle isomorphic to $X \times \mathbb {R}^m$ , we may assume that there is an isomorphism $\phi : \ker T \to \operatorname{coker}T \cong (\operatorname{im}T)^\perp$ . Then we have a homotopy
$(t\phi , T|_{(\ker P_n T)^\perp }) : \ker P_n T \oplus (\ker P_n T)^\perp \to H$
going from $T$ to an isomorphism. Then we are done.
To show surjectivity, since every element in $K(X)$ is of the form $[E] - [X \times \mathbb {R}^k]$ , it suffices to show that we can produce maps with index $[E]$ and $-[X \times \mathbb {R}^k]$ .

The case of $-[X \times \mathbb {R}^k]$ is easy — take the map $T$ to constantly be $\mathrm{shift}^{-k}$ .

To construct the one with $[E]$ , first pick some bundle $F$ such that $E \oplus F \cong X \times \mathbb {R}^N$ for some $N$ .

Now consider the bundle $(E \oplus F) \oplus (E \oplus F) \oplus (E \oplus F) \oplus \cdots$ , which is isomorphic to $X \times H$ . Then take the Fredholm operator to be
$(e_1, f_1, e_2, f_2, e_3, f_3, \dots ) \mapsto (e_2, f_1, e_3, f_2, e_4, f_3, \dots ).$

Proof

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