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Is there some a priori reason why we should expect the Brauer group of real [complex] super vector spaces to be closely related to periodicity in real [complex] K-theory? By "a priori" I mean a proof that does not involve computing that both are Z/8 [Z/2] and does not involve noticing that both are related to Clifford algebras.

Theo Johnson-Freyd posed a similar question in the comments of this MO question.

"Brauer group" above means the group of Morita equivalence classes of Morita-invertible super algebras over $\mathbb R$ [$\mathbb C$].

I suspect the answer is some form of "No", since several bouts of googling have failed to turn anything up. But I would be happy to be proved wrong.

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Why doesn't the connection with Clifford algebras count as an "a priori" reason? –  Paul Siegel Feb 2 '12 at 17:55
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@Paul: I guess it's because the relation between Brauer groups and Clifford algebras seems contingent to me. Does the relation hold for any field, or is it a special fact about $\mathbb R$ and $\mathbb C$? Is it possible to show that all elements of the Brauer group are represented by Clifford algebras without calculating what the Brauer group is? –  Kevin Walker Feb 2 '12 at 18:17
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K-theory is Brauer-group-graded, and also Z-graded. In other words, it's (Brauer-group x Z)-graded. The surprising thing is that this grading collapses to a cyclic-group grading. If you look at K-theory as a cohomology theory on the category of spaces over X (where X is some arbitrary space), then K-theory will be Brauer(X)-graded, which is no longer a cyclic group. –  André Henriques Feb 2 '12 at 21:23
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@André: That sounds interesting. Do you know where I can read more about the Brauer group grading? –  Kevin Walker Feb 2 '12 at 21:31
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2 Answers

To see why $K$-theory should be Brauer graded, it may help to see how, for a superalgebra $A$, the associated $K$-group only depends on the category $\mathcal{M}_A$ of finite dimensional $Z/2$-graded modules, which is endowed with a superstructure: it is enriched over the category of $Z/2$-graded vector spaces.

Given an object $X$ in such a $Z/2$-graded category $\mathcal{M}$, we can consider the set of odd degree endomorphisms $F: X^\pm\to X^\mp$. Say that $X$ is trivial if there exists such an $F$ for which $F^2=\mu\;\mathrm{Id}$, where $\mu$ is real and positive; I'll call $F$ a trivialization of $X$. Now define $$K(\mathcal{M}) := (\text{Grothendieck group of $\mathcal{M}$} / \text{trivial objects}).$$ For modules over $\mathrm{Cliff}(n)$, this gives the expected group $K^n(\text{point})$ or $KO^n(\text{point})$, depending on your base field. Note that $K(\mathcal{M})$ is clearly a Morita-equivalence invariant.

There's a pairing $$K(\mathcal{M}_A) \otimes K(\mathcal{M}_B) \to K(\mathcal{M}_{A\otimes B}),$$ given by tensor product of modules. To see that this is well-defined, note that given odd endomorphisms $F_X\colon X^\pm\to X^\mp$ and $F_Y\colon Y^\pm\to Y^\mp$, you obtain an odd endomorphism $$F=F_X\otimes \mathrm{Id}+\mathrm{Id}\otimes F_Y$$ of $X\otimes Y$, and that $$F^2 = (F_X)^2\otimes \mathrm{Id} + \mathrm{Id} \otimes (F_Y)^2,$$ by the magic of sign conventions. Thus, if both $F_X$ and $F_Y$ are "trivializations" of $X$ and $Y$, or if just one of the two is a trivialization and the other is set to $0$, then $F$ is also a trivialization.

This suggests how $K$-theory should be Brauer-group graded, without explicitly invoking Clifford algebras.

This sort of thing seems "well-known", but I don't really know a reference. I learned this from Michael Joachim's paper "A Symmetric Ring Spectrum representing K-theory" (Topology, 2001), though he uses spaces of operators rather than finite dimensional modules.

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"... though he uses spaces of operators rather than finite dimensional modules": if one tries to implement word-for-word what you wrote with finite dimensional modules, one ends up with a version of K-theory that fails the excision axiom. Going over to an infinite dimensional settings in one of the possible fixes. The other possible fix (I think!) is to sheaffify the presheaf [opens in spec(Z(A))] |--> [K-theory spectrum of the localized algebra]. –  André Henriques Feb 3 '12 at 10:11
    
Thanks Charles, that's helpful. –  Kevin Walker Feb 3 '12 at 14:27
    
@ Andre: I'm sure you're right. I'm only claiming to produce the K-theory of a point here. It would be nice to see this version worked up carefully, without going to infinite dimensions. (I simplified things a bit, because I didn't want to introduce a metric. If I'd done so, I would have, following Joachim, made $F$ a self-adjoint operator, rather than using a naive positivity condition. This makes it easier to glue things, because the space of self-adjoint operators is convex, though I don't know if that fully repairs the problem with excision.) –  Charles Rezk Feb 3 '12 at 16:00
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I don't know if the following is exactly what you're looking for. There is a theorem of Madsen-Snaith-Tornehave from 1977 that says that $gl_1 KO$ is equivalent to $K(\mathbb{Z}/2,0)\times K(\mathbb{Z}/2,1)\times bso_{\otimes}$, where $gl_1 KO$ is the infinite loopspace of automorphisms of the spectrum $KO$, $K(\pi,n)$ denotes an Eilenberg-MacLane space, and $bso_\otimes$ denotes some $3$-connected infinite loopspace that classifies virtual real vector bundles of rank $1$ with the tensor product group structure. The classifying space $Bgl_1 KO$ then has a factor $K(\mathbb{Z}/2,1)\times K(\mathbb{Z}/2,2)$. So, over a space $X$, each class $\alpha$ of the cohomology group $H^2(X,\mathbb{Z}/2)$ gives rise to a twist $KO(X)_{\alpha}$ of real $K$-theory. On the other hand, this product of cohomology groups is also exactly the real super Brauer group of $X$. There is an similar story for complex $K$-theory for algebraic $K$-theory as well.

Glancing at the paper I mentioned above, I don't see any sign of them using Clifford algebras, so the link I describe here is just through the fact that the same cohomology group arises in both places.

There is a more algebraic reason this comes up as well. Given a Brauer class $\alpha$, there is a category of $\alpha$-twisted real vector bundles. Over a finite CW-complex $X$, the $K$-theory of this category is naturally a module over $K(X)$. This immediately gives the Brauer graded structure, and it does so without reference to Clifford algebras. Again, there are complex and algebraic $K$-theory versions of this.

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That's interesting -- thanks! –  Kevin Walker Jun 28 '12 at 18:23
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