Representing KO-theory using Clifford algebras I'm trying to understand a statement Segal makes in this book:
Let $C_q$ be the real Clifford algebra associated to the standard negative definite form on $\mathbb{R^q}$ and let $\Phi_q(n)$ be the space of symmetric unitary (with respect to the automorphism of $C_q$ induced by $e_i \mapsto -e_i$) $n\times n$-matrices over $C_q$.
Segal then claims that $\Phi_q=\cup_n \Phi_q(n)$ represents $KO^q$ for $q\geq 1$ and that this is a reformulation of Bott periodicity.
Can someone indicate how this works or is there any good reference for this ?
 A: I learned about this in a course taught by Mark Hovey, but we didn't use a book and his lecture notes are not available online. As I recall, you first understand how the signature of a quadratic form classifies quadratic forms of given degree. Then you use this to classify Clifford algebras and notice the algebraic periodicity which will later be revealed to be Bott periodicity. If you want to be more concrete, let $q_{r,s}(x_1,\dots,x_n) = x_1^2 + \dots + x_r^2 - x_{r+1}^2 - \dots - x_{r+s}^2$ and then $C_{r,s}$ is the corresponding Clifford Algebra. You can reduce the classification to classifying $C_q = C_{q,0}$ by noting how $C_{r,s}$ splits up via tensor products. You seem to already understand this part of the process, so I won't say any more about it. The table classifying $C_q$ can be found here and it's 8 periodic.
Now we look at representations, and that's where your $\Phi_q$ comes in. Let $M_n$ be the Grothendieck group of representations of $C_n$ and note that $M_n$ is $\mathbb{Z}$ generated by some $\mathbb{R}^m$ if $n$ is not $3$ or $7$ mod $8$, and for all other $n$ we have $M_n\cong \mathbb{Z}\oplus \mathbb{Z}$ generated by $\mathbb{R}^m$ and $\mathbb{R}^m$. This is just observing which of the Clifford algebras $C_n$ in the table above split and $m$ is just the dimension of the corresponding vector space in the table considered as a vector space over $\mathbb{R}$. Now consider the standard inclusion $i:C_n\to C_{n+1}$ and note that you get $i^*:C_{n+1}-mod \to C_n-mod$ which induces $i^*:M_{n+1}\to M_n$.
The Atiyah-Bott-Shapiro Theorem says that there's a natural map $M_{n-1}/i^*M_n \to KO^{-n}(\ast) \cong \tilde{KO}(S^n)$
You can then compute all the quotients very simply, using the fact that you know the generators of $M_n$. For example, $M_0/i^*M_1$ is $\mathbb{Z}/2$ because $M_0$ is $\mathbb{Z}$ generated by $C_0 = \mathbb{R}$ and $M_1$ is $\mathbb{Z}$ generated by $C_1=\mathbb{C}$, and the map $i^*$ is just multiplication by 2 because it takes $\mathbb{C}\mapsto \mathbb{R}^2$. So the quotient is $\mathbb{Z} / 2\mathbb{Z}$. I recommend working out the rest of the $M_j/i^*M_{j+1}$ on your own. It's a good way to check your understanding of Clifford algebras and you'll see the Bott groups appear as if by magic.
