Clifford Algebras and Bott PeriodicityThe Atiyah–Jänich theorem

## 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
[Proof (cf. )] 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