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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

$$ M_k \to \widetilde{KO}(S^k)$$ 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 $KO(B_k, S^{k-1} ) = \widetilde{KO}(S^k)$.

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?

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Nice question. Can I object mildly to a point of terminology? It seems that the modules in question are not "graded", which I think usually conotes something like "$\mathbb Z$-graded", but more like "super" or "$\mathbb Z/2$-graded", since you just want two of them. More broadly, you probably know that Clifford algebras are most naturally part of superalgebra. A Clifford algebra has a $\mathbb Z/2$-grading, which usually does not lift to a $\mathbb Z$-grading, given by putting the generators in odd degree. (From this point of view, a Clifford algebra is exactly the Weyl algebra of a ... –  Theo Johnson-Freyd Jan 12 '12 at 20:05
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... purely odd presymplectic vector space, and this "explains" why the classical limit = associated graded is the symmetric algebra of said purely odd vector space.) Anyway, then the representations of such an algebra ought to be super, and I think this is what you really mean. That said, I'm now confused by why you don't also use the Clifford multiplication $N_1 \to N_0$; it seems that the natural module structure has both, with composition = $\pm r^2$ times identity at radius $r$. –  Theo Johnson-Freyd Jan 12 '12 at 20:14
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If you'll let me, I'm now going to ramble a little on something tangentially related, more to the title than the question. There's something I wish I understood further, which is the following. Given any reasonable symmetric monoidal category $C$, like the category of supervector spaces, you can form a ring whose elements are associative algebras in $C$ modulo Morita equivalence, $\times = \otimes$, and $+ = \oplus$. (Really I mean elements are reasonable $C$-module categories, with $\otimes$ and $\oplus$ of categories.) Then you can take the group of units in this ring. This is the ... –  Theo Johnson-Freyd Jan 12 '12 at 20:17
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... Brauer group of $C$. (One categorical dimension down, the same construction builds the Picard group of a commutative ring $C$.) Anyway, the statement is that for $C = $supervect, the representatives of Brauer are certain Clifford algebras, and over $\mathbb R$ this "super" Brauer group is a $\mathbb Z/8$ and is the same $\mathbb Z/8$ that shows up in Bott periodicity and KO-theory. But I've never really understood (in a deep way) why all of these are the same. Anyway, you probably know all this, and I sadly have nothing useful to say to your actual question. –  Theo Johnson-Freyd Jan 12 '12 at 20:20
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@Theo; I don't know how much it helps, but there are Baez's copy of Trimble's notes, math.ucr.edu/home/baez/trimble/superdivision.html which goes a little further, that the super-Brauer classes are all represented by superdivision algebras. Hmmm... since I've said "goes further", I probably ought to check that superdivision algebras are also clifford algebras. Does it seem plausible, at least? –  some guy on the street Jan 22 '12 at 3:14

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