I have a question about the Proof of Bott Periodicity in twisted K-theory by Atiyah and Segal in their paper Twisted K-theory.

Following their notation, to prove Bott periodicity in this context it is enough to provide a $U(H)$-equivariant homotopy equivalence $$ Fred^{(0)}(H)\to \Omega^{2}Fred^{0}(H). $$ One may assume that all the spaces in sight have the norm topology for simplicity. This is done in two steps.

Step 1. Take $S_{n}$ an irreducible graded module for the complexified Clifford algebra $C_{n}$. Then for $n$ even, tensoring with $S_{n}$ gives an isomorphism $$ Fred^{0}(H)\to Fred^{n}(S_{n}\otimes H). $$ This map is clearly $U(H)$-equivariant.

Step 2.There is a map $$ Fred^{n}(S_{n}\otimes H)\to \Omega^{n}Fred^{0}(S_{n}\otimes H). $$ which was constructed explicitly by Atiyah and Singer and it is easy to see that it is a $U(H)$-equivariant homotopy equivalence.

However, one would like to get back to $\Omega^{n}Fred^{0}(H)$. The spaces
$$ \Omega^{n}Fred^{0}(S_{n}\otimes H) \text{ and } \Omega^{n}Fred^{0}(H) $$ are homotopy equivalent but all the maps I seem to be able to construct don't preserve $U(H)$-equivariance and this is taken as granted in the proof by Atiyah and Segal.

Can anyone tell me what I am missing?


2 Answers 2


Twisted K-theory is just a particular case of K-theory of Banach algebras. Therefore, Bott periodicity is a consequence of general results. See for instance Max Karoubi. Twisted K-theory old and new.


I attempted the following baby version. Namely, I asked myself whether there exists a unitary map $S:H\oplus H\rightarrow H$ such that the induced map $Fred(H\oplus H)\rightarrow Fred(H)$ is $U(H)$-equivariant, where $U(H)$ acts in the usual way on $H$ and diagonally on $H\oplus H$. Let $S$ be represented by two operators (not necessarily isomorphisms) $S_i:H\rightarrow H$, so that we can write $$S=\begin{pmatrix} S_1 & S_2 \end{pmatrix}. $$ Then, we are asking whether or not we can find $S$ such that $S\circ(T\oplus T)=(T\oplus T)\circ S$ for all unitary opators $T$ on $U$. This forces $T\circ S_1=S_1\circ T$ and $T\circ S_2=S_2\circ T$ for all $T$. This in fact forces $S_i$ to be a unitary isomorphism. Indeed, suppose that $v$ is in the kernel of $S_1$, then $e_1$ is in the kernel of $T\circ S_1$. Let $w$ be an element orthogonal to $v$ and not in the kernel of $S_1$, and let $T$ be the unitary operator that switches $w$ and $v$. Then, $v$ is not in the kernel of $S_1\circ T$. Therefore, $S_i$ has no kernel. The same argument using the adjoints $S_i^*$ show that $S_i$ has no cokernel. Therefore, each $S_i$ is a unitary isomorphism. Then, the commutativity restraint implies that each $S_i$ is diagonal $S_i=\lambda_i Id_{H}$. But, then, $S$ clearly has a non-trivial kernel, which is a contradiction.

Thus, I think that if there is a $PU(H)$-equivariant equivalence as asked in the question, it seems that it cannot come from $PU(H)$ acting diagonally on $S_n\otimes H$ and on $H$.


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