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.

  • $\begingroup$ Why doesn't the connection with Clifford algebras count as an "a priori" reason? $\endgroup$ – Paul Siegel Feb 2 '12 at 17:55
  • 3
    $\begingroup$ @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? $\endgroup$ – Kevin Walker Feb 2 '12 at 18:17
  • 2
    $\begingroup$ 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. $\endgroup$ – André Henriques Feb 2 '12 at 21:23
  • 3
    $\begingroup$ @André: That sounds interesting. Do you know where I can read more about the Brauer group grading? $\endgroup$ – Kevin Walker Feb 2 '12 at 21:31
  • $\begingroup$ The fact that the super Braur group is given by classes of Clifford Algebras is specific to $\mathbb{R}$ and $\mathbb{C}$. Ultimately it comes from the (ungraded) Brauer groups of $\mathbb{R}$ and $\mathbb{C}$ that are given by (ungraded) Clifford algebra : the non trivial one being the quaternion algebra $\mathbb{H}$. You Can find details at page 115 in the IAS, Quantum fields ans strings, a course for mathematicians. $\endgroup$ – InfiniteLooper Sep 12 '17 at 13:56

I'm a bit late to the party, but here's what I suspect the answer should look like. Forgive me for working somewhat to very heuristically throughout.

To star things off, here's a silly question: why is cohomology $\mathbb{Z}$-graded? Starting from a spectrum $E$ and a space $X$ we can consider the set $[X, E]$ of homotopy classes of maps from the suspension spectrum of $X$ to $E$. This only gives the zeroth part $E^0(X)$ of the $E$-cohomology of $X$. To get more interesting information one thing we can do is smash $E$ with invertible spectra. These are precisely the shifts $S^n, n \in \mathbb{Z}$ of the sphere spectrum, and

$$X \mapsto [X, S^n \wedge E]$$

recovers $E$-cohomology as ordinarily understood. So we should think of $\mathbb{Z}$ here as being the Picard group $\text{Pic}(\text{Sp})$ of the symmetric monoidal $\infty$-category of spectra. That is:

Cohomology in general is graded by the Picard group of invertible spectra.

In more complicated situations we get more interesting Picard groups that we can grade by. For example, $\infty$-local systems of spectra over a space $X$ describe cohomology theories for spaces over $X$, and these cohomology theories are therefore graded by the Picard group $\text{Pic}(\text{Loc}_{\text{Sp}}(X))$. The fiberwise suspension spectrum of a spherical fibration over $X$ is in particular an example of an element of this Picard group. This should be relevant to André Henriques' observation in the comments.

One of the more complicated situations we could be in is the following. Let $R$ be an $E_{\infty}$ ring spectrum and let $E$ be an $R$-module spectrum. Rather than settling for smashing $E$ with invertible spectra, we can now smash $E$ with invertible $R$-module spectra over $R$ to get more interesting information of the form

$$X \mapsto [X, A \wedge_R E]$$

where $A$ is invertible with respect to $\wedge_R$; that is, $E$-cohomology acquires a grading by $\text{Pic}(R\text{-ModSp})$, which I will abbreviate to $\text{Pic}(R)$. In particular, letting $R = E = KO$:

Real K-theory is graded by the Picard group of invertible $KO$-module spectra.

The punchline should now be that there is a natural group homomorphism from the super Brauer group of $\mathbb{R}$ to the Picard group of $KO$; possibly this is even an isomorphism. The construction should go something like this:

  • Start with a finite-dimensional superalgebra $A$ over $\mathbb{R}$.
  • Construct the category of finite-dimensional (projective?) right $A$-supermodules, which is symmetric monoidal with respect to direct sum and enriched over supervector spaces. Topologize its homs as finite-dimensional real vector spaces.
  • Take the core of the above, which is a symmetric monoidal topological groupoid, and hence which presents a symmetric monoidal $\infty$-groupoid.
  • Take the group completion of the above, which is a grouplike $E_{\infty}$ $\infty$-groupoid, or equivalently an infinite loop space.

But this isn't quite right, because we only get a connective spectrum this way. I don't know how to fix this. Maybe we should take chain complexes or something.

Anyway, let's pretend I did the right thing, and let's call the resulting spectrum $K(A)$. Let's also pretend that $K(\mathbb{R})$ ended up being $KO$. The categories I constructed above are all tensored over supervector spaces, or equivalently are all module categories over supervector spaces, so the resulting spectra should all be $KO$-module spectra. Moreover, taking tensor products should give natural equivalences

$$K(A) \wedge_{K(\mathbb{R})} K(B) \cong K(A \otimes_{\mathbb{R}} B)$$

and since $K(A)$ only depends on the category of $A$-modules, any Morita-invertible $A$ should give an invertible $K(\mathbb{R})$-module spectrum $K(A)$ in a way that respects products.

So real K-theory has two gradings, one by $\text{Pic}(S)$ ($S$ the sphere spectrum) and one by $\text{Pic}(KO)$. The content of the Clifford-algebraic proofs of Bott periodicity should be that these gradings collapse into a single grading, and in particular that all of the $K(A)$ end up being shifts of $K(\mathbb{R})$ as $K(\mathbb{R})$-module spectra.

Of course we can replace $\mathbb{R}$ by $\mathbb{C}$ throughout. Actually it seems like we really ought to replace $\mathbb{R}$ by an arbitrary ring spectrum $R$ throughout, and try to relate the Brauer group of $R$ and the Picard group of $K(R)$; this might make it easier to figure out the right statements.

  • $\begingroup$ Thanks Qiaochu, that's a nice story. Despite the gaps, I'll give it the green check mark (unless some expert later points out that there are serious problems with your outline). $\endgroup$ – Kevin Walker Jul 29 '14 at 16:23
  • $\begingroup$ The part I'm most worried about is that I didn't do anything analogous to the killing of the trivial objects in Rezk's answer. Maybe incorporating that suitably fixes everything... $\endgroup$ – Qiaochu Yuan Jul 29 '14 at 17:39

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.

  • 2
    $\begingroup$ "... 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]. $\endgroup$ – André Henriques Feb 3 '12 at 10:11
  • $\begingroup$ @ 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.) $\endgroup$ – Charles Rezk Feb 3 '12 at 16:00

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.