Clifford Algebras and Bott PeriodicityThe first part of the proof

3.2 The first part of the proof

We can now begin to prove Theorem A. In the remainder of this section, we shall reduce the problem to something easier via a chain of rather straightforward homotopy equivalences, and then in the next section, we shall do some work to finish it off.

The representation theory of Clifford algebras was easier to study with graded modules, but analysis is easier on ungraded things. For k≥1k \geq 1, we let

F~k(H0)={B∈F(H0):B∗=−B,  BJi+JiB=0 for i=1,…,k−1}, \tilde{\mathcal{F}}^k(H^0) = \{ B \in \mathcal{F}(H^0): B^* = -B,\; B J_i + J_i B = 0\text{ for }i = 1, \ldots , k - 1\} ,

and whenever k≡−1(mod4)k \equiv -1\pmod4, we set

F~∗k(H0)={B∈F~∗k(H0):J1J2⋯JkB is not essentially definite}. \tilde{\mathcal{F}}^k_*(H^0) = \{ B \in \tilde{\mathcal{F}}^k_*(H^0): J_1 J_2 \cdots J_kB \text{ is not essentially definite}\} .

Otherwise, we set F~∗k(H0)=F~k(H0)\tilde{\mathcal{F}}^k_*(H^0) = \tilde{\mathcal{F}}^k(H^0). Then B↦BJkB \mapsto BJ_k gives a homeomorphism F~∗k(H0)≅F∗k(H)\tilde{\mathcal{F}}^k_*(H^0) \cong \mathcal{F}^k_*(H). Similarly, we set F~∗0(H0)=F(H0)\tilde{\mathcal{F}}^0_*(H^0) = \mathcal{F}(H^0). Thus, from now on, we will write F∗k\mathcal{F}^k_* to mean F~∗k(H0)\tilde{\mathcal{F}}^k_*(H^0).

In this notation, we want to prove

Theorem A'

For k≥1k \geq 1, there is a homotopy equivalence

F∗k→Ω(F∗k−1). \mathcal{F}^k_* \to \Omega (\mathcal{F}^{k - 1}_*).
Note that our loops start from Jk−1J_{k - 1} to −Jk−1-J_{k - 1}, but that doesn't matter.

To prove the theorem, it is convenient to consider the subspace of unitary elements in our operator algebras. We make the following huge list of definitions. Let A=End⁡(H)\mathcal{A} = \operatorname{End}(H), and UU denote the unitary elements in a Banach algebra. Then define

Ak={T∈A:T∗=−T,  TJi+JiT=0 for i=1,…,k−1}\mathcal{A}^k = \{ T \in \mathcal{A} : T^* = -T,\; T J_i + J_i T = 0\text{ for }i = 1, \ldots , k - 1\}

K=\mathcal{K} = compact operators in A\mathcal{A}

Kk=A∩Ak\mathcal{K}^k = \mathcal{A} \cap \mathcal{A}^k

B=A/K\mathcal{B} = \mathcal{A}/\mathcal{K}

Bk=Ak/Kk\mathcal{B}^k = \mathcal{A}^k/\mathcal{K}^k

G=B×\mathcal{G} = \mathcal{B}^\times

Gk=G∩Bk\mathcal{G}^k = \mathcal{G}\cap \mathcal{B}^k

G∗k=Gk∩p(F∗k)\mathcal{G}^k_* = \mathcal{G}^k \cap p(\mathcal{F}^k_*)

G∗k=G∗k∩UG^k_* = \mathcal{G}^k_* \cap U

F∗k=p−1G∗k∩UF^k_* = p^{-1} G^k_* \cap U

L=A×=GL(H)\mathcal{L} = \mathcal{A}^\times = \mathrm{GL}(H)

L=L∩UL = \mathcal{L} \cap U

Lk=L∩AkL^k = \mathcal{L} \cap \mathcal{A}^k

Ωk={T∈Ak:T∗T=1,  T≡Jk−1(modK)}\Omega _k = \{ T \in \mathcal{A}^k: T^*T = 1,\; T \equiv J_{k - 1} \pmod{\mathcal{K}}\}

where p:A→A/Kp: \mathcal{A} \to \mathcal{A}/\mathcal{K} is the natural projection (recall that F=p−1(G)\mathcal{F} = p^{-1}(\mathcal{G})).

We need a theorem of Bartle and Graves.

Theorem (Bartle–Graves, [artlegraves52)

Let U,BU, B be Banach spaces and π:U→B\pi : U \to B be a bounded linear surjection. Then π\pi has a (not necessarily linear) section.

Lemma

All the arrows marked as ∼\sim in the diagram below are homotopy equivalences:

\begin{useimager} 
    \[
      \begin{tikzcd}[row sep=small, column sep=tiny]
        F^k_* \ar[rrrr, bend left, red] \ar[rr, "\alpha"] \ar[dd, "\sim"] \ar[rd, "\sim"] && P(L^{k - 1}_*, \Omega_{k - 1}) \ar[rd, dash, dashed, gray] \ar[dd, green!50!black] \ar[rr, green!50!black] & & \Omega_{k - 1}\\
        & \mathcal{F}^k_* \ar[rr, crossing over, dash, dashed] && \Omega \mathcal{F}^{k - 1}_* \ar[dd, "\sim"]\\
        G^k_* \ar[rr, "\beta", pos=0.6] \ar[rd, "\sim"] & & \Omega G^{k - 1}_* \ar[rd, "\sim"]\\
        & \mathcal{G}^k_* \ar[from=uu, crossing over, "\sim"', pos=0.3] \ar[rr, dash, gray, dashed] & & \Omega \mathcal{G}^{k - 1}_*
      \end{tikzcd}
    \]
  \end{useimager}

where α(T)t=Jk−1exp⁡(πtTJk−1)\alpha (T)_t = J_{k - 1} \exp (\pi t TJ_{k - 1}), and β\beta is defined by the same formula. Note that by assumption, since elements T∈G∗kT \in G_*^k satisfy T2=−1T^2 = -1 and anti-commute with Jk−1J_{k - 1}, it follows that (TJk−1)2=−1(TJ_{k - 1})^2 = -1 as well. So we can equivalently write

β(T)t=Jk−1cos⁡πt+Asin⁡πt, \beta (T)_t = J_{k - 1}\cos \pi t + A \sin \pi t,

which makes it clearer that the maps indeed land in where we claim they do.

The dashed lines are there to make the diagram look like a cube, as opposed to indicating the existence of certain morphisms. The theorem would then follow once we prove that the red and green arrows are homotopy equivalences.
Proof
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Proof
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