most recent 30 from http://mathoverflow.net 2013-05-25T13:10:50Z http://mathoverflow.net/feeds/question/85516 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85516/why-do-clifford-algebras-determine-ko-and-k-theory Akhil Mathew 2012-01-12T19:17:40Z 2012-01-12T23:22:15Z

<p>In the paper "Clifford modules" by Atiyah-Bott-Shapiro, they construct a family of Clifford algebras $C_k$ over the real numbers, so that $C_k$ is the algebra associated to a negative definite form on $\mathbb{R}^k$. Let $M_k$ be the Grothendieck group of $\mathbb{Z}/2$-graded $C_k$-modules. The authors define a map</p> <p>$$ M_k \to \widetilde{KO}(S^k)$$ and similarly in complex $K$-theory. The map is as follows: given a $\mathbb{Z}/2$-graded Clifford module $(N_0, N_1)$, one considers the constant vector bundles $\pi^*N_0, \pi^*N_1$ on the unit ball $B_k$ and the map $\pi^* N_1 \to \pi^*N_0$ which is given by Clifford multiplication by the vector in $B_k$. This is an isomorphism on $S^{k-1}$, so the "difference bundle" construction yields an element of $KO(B_k, S^{k-1} ) = \widetilde{KO}(S^k)$.</p> <p>This map factors through the image of $M_{k+1}$ in $M_k$, and one thus gets a map of graded rings $$\bigoplus M_k/M_{k+1} \to \bigoplus \widetilde{KO}^{-k}(\ast),$$ which they prove is an isomorphism, by using Bott periodicity and a bit of computation: one has to work out what the (purely algebraic) Grothendieck groups are, and then figure out which elements in $KO$-theory they correspond to. They remark that it would be nice to have a direct proof that the last map is an isomorphism; in particular, then one would be able to work out the $KO^\bullet(\ast)$ ring purely algebraically. Can this be done? </p>