3.1 The last part of the proof

We can now return to Clifford algebras and Bott periodicity. The idea of Atiyah and Singer [3] was to extend the Atiyah–Jänich theorem to provide explicit descriptions of the loop spaces $\Omega ^k \mathcal{F}(H)$. It turns out we can interpret this space (up to homotopy equivalence, of course) as Fredholm homomorphisms of Clifford algebra representations.

Note that $C_k$ naturally comes with an involution defined by $e_i^* = - e_i$, and we shall require our representations to respect this. Let $H = H^0 \oplus H^1$ be a $\mathbb {Z}_2$-graded Hilbert space with a simultaneous graded action of all $C_k$, i.e. there are bounded linear operators $J_1, J_2, \ldots$ of degree $1$ such that

$$J_i J_j + J_j J_i = 0,\quad J_i^2 = -1,\quad J_i^* = -J_i,$$

for all $i \not= j$. We will focus on skew-Hermitian operators. Since we have a graded Hilbert space, we require our skew-Hermitian operators to have degree $1$. So that $B$ takes the form

$$\begin{pmatrix} 0 & -T^* \\ T & 0 \end{pmatrix}.$$

After picking an arbitrary isomorphism $H^0 \cong H^1$ of vector spaces, sending $B \in \hat{\mathcal{F}}(H) = \mathcal{F}^0(H)$ to $T \in \mathcal{F}(H^0)$ as above gives a homeomorphism $\mathcal{F}(H^0) \to \mathcal{F}^0(H)$. Hence $\mathcal{F}^0$ represents $KO = KO^0$.

It turns out these $\mathcal{F}^k(H)$ are not what we want. To understand this, suppose $B \in \mathcal{F}^k(H)$ is unitary (which we can achieve by restricting to $\ker B^\perp$ and then rescaling), so that $B^2 = -B^* B = -1$. Then $B$ “acts as” $J_{k + 1}$ on $H$, and this gives $H$ a new $C_{k + 1}$ action. If $k \not\equiv -1 \pmod4$, then there is a unique irreducible $C_{k + 1}$ module, hence the structure of $H$ as a $C_{k + 1}$-module is completely determined. However, if $k \equiv -1 \pmod4$, then we want to ensure $H$ has infinitely many copies of each irreducible module, for it to be well-behaved.

Recall that to count how many copies of each each irrep we've got, we are supposed to count the eigenvalues of $w = e_1 e_2 \cdots e_k$ after turning $H^0$ to a $C_k$-module. The process of turning $H^0$ into a $C_k$-module involves the inclusion $C_k \hookrightarrow C_{k + 1}$, which sends $e_i$ to $e_i e_{k + 1}$. In our case, $e_{k + 1} = B$, and $e_i = J_i$ for $i = 1, \ldots , k$. Thus, we are counting the eigenvalues of

$$w = (J_1 B) (J_2 B)(J_3 B) \cdots (J_k B).$$

Since $k \equiv -1 \pmod4$ and $B^2 = -1$, this is equal to

$$w = J_1 J_2 \cdots J_k B.$$

We then want to require that the eigenvalues $\pm 1$ to both have infinite multiplicity. In general, for an arbitrary $B$, it is not unitary, or even injective. The desired “niceness” property is now that when we restrict $B$ to the $\ker B^\perp$, and then rescale $B$ so that it is unitary, the multiplicity of $\pm 1$ in $w$ are both infinite. Equivalently, without doing the restriction and rescaling business, we want $J_1 J_2 \cdots J_k B$ to have infinitely many eigenvalues of each sign, counted with multiplicity.

Observe that these form a component of $\mathcal{F}^k(H)$.

The main theorem of Atiyah and Singer is the following:

We first see how this proves that $KO^{-k}(*) = A_k$, and in particular implies Bott periodicity.

Let $M$ be an irreducible $C_8$-module. Then there is an isomorphism

$$\begin{aligned} \mathcal{F}_*^k(H) & \cong \mathcal{F}_*^{k + 8}(H \mathbin {\hat\otimes }M)\\ A & \mapsto A \otimes I, \end{aligned}$$

and there is a contractible space of $C_{k + 8}$-module isomorphisms $H \cong H\mathbin {\hat\otimes }M$.

This in fact gives us an isomorphism of rings.

For convenience, we drop the $(H)$ in $F^k_*(H)$ when it is clear.

We leave the $k = 0$ case for the reader. Note that $C_0 = \mathbb {R}$ is concentrated in degree $0$, and it has two irreducible graded representations given by $0 \oplus \mathbb {R}$ and $\mathbb {R}\oplus 0$.

For $k > 0$, we first show that the kernel map is continuous, so that it factors through $\pi _0(\mathcal{F}_*^k)$. Let $B \in \mathcal{F}^k_*$. Note that $B^2$ is self-adjoint and negative, since

$$\langle B^2 x, x\rangle = -\langle Bx, Bx\rangle \leq 0.$$

Hence we know that $\sigma (B^2) \subseteq (-\infty , 0]$. In fact, since $B$ is Fredholm, we know $0$ is an isolated point in the spectrum, and by scaling $B$, we may assume

$$\sigma (B^2) \subseteq (-\infty , -1] \cup \{ 0\} .$$

Since the spectrum depends continuously on $B^2$, we can pick a small neighbourhood of $B$ in $\hat{\mathcal{F}}^k_*$ such that whenever $C$ is in the neighbourhood,

$$\sigma (C^2) \subseteq (-\infty , -1 + \varepsilon ] \cup [-\varepsilon , 0],$$

and further that $\| B^2 - C^2\| < \varepsilon < \frac{1}{2}$.

Let $E$ be the spectral projection to $[-\varepsilon , 0]$. We claim that $E$ is isomorphic to $\ker B$. Indeed, consider the orthogonal projection $P$ from $E$ to $\ker B$.

• $P$ is injective. If not, suppose $x \in E \cap (\ker B)^\perp$, and wlog assume $\| x\| = 1$. Since $\sigma (B^2|_{\ker B^\perp }) \subseteq (-\infty , -1]$, we have

  $$|\langle (B^2 - C^2)x, x\rangle | = |\langle B^2 x, x\rangle - \langle C^2 x, x\rangle | \geq 1 - \varepsilon ,$$

  a contradiction.

• $P$ is surjective. If not, suppose $x \in \ker B \cap E^\perp$, and again assume $\| x\| = 1$. Then

  $$\langle (B^2 - C^2)x, x\rangle = -\langle C^2 x, x\rangle \geq 1 - \varepsilon ,$$

  a contradiction.

Now $B^2$ and $C^2$ commute with $C_k$, so the orthogonal projection is in fact a $C_k$-module isomorphism. We write $E = \ker C \oplus \ker C^\perp$. Then

$$\operatorname{idx}B - \operatorname{idx}C = [\ker C^\perp ].$$

But the restriction of $C$ to $\ker C^\perp$ is non-singular and skew-Hermitian, so we can use the action of $C$ to turn $\ker C^\perp$ into a $C_{k + 1}$ module. Morally, we should be able to just “scale” $C$, and the correct thing to write down is

$$J_{k + 1} = C (-C^2)^{-1/2}.$$

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Surjectivity is trivial (for $[M] \in A_k$, pick an isomorphism $H \cong H \oplus M$ and use $J_{k + 1} \oplus 0 \in \mathcal{F}^k_* (H \oplus M)$).

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Injectivity follows from the following sequence of observations:

By restricting to the complement of $\ker B$, we may assume $\ker B$ is trivial. Note that the operator $B(-B^2)^{-1/2}$ is unitary. So we can use the path $B(t - (1 - t)B^2)^{-1/2}$.

We may assume $B$ is unitary on the complement of $\ker B$. Since $B \in \mathcal{F}_*^k$, we know each $C_{k + 1}$ module appears as a direct summand of $\ker B^\perp$ when $B$ acts as $J_{k + 1}$. So we can decompose

$$H = \ker B \oplus V \oplus \text{remaining},$$

and then linearly scale $B$ to vanish on $V$.

Thus, if $B, C \in \mathcal{F}^k_*$ are such that $[\ker B] = [\ker C]$, then we can deform $B$ and $C$ so that their kernels are in fact isomorphic.

Again wlog assume $B$ and $C$ are unitary on the complement, so that they act as $J_{k + 1}$. Then the complements are isomorphic as $C_{k + 1}$ modules, since both have infinitely many copies of each irrep. Hence there must be a $C_k$-module isomorphism that sends $B$ to $C$.

We can then conclude the theorem if we can find a path from $T$ to the identity.

By Schur's lemma, this group is isomorphic to $\mathrm{GL}(H)$, hence is in fact contractible.