Recall the following table of Clifford algebras: $$\begin{array}{ccc} n & Cl_n & M_n/i^{*}M_{n+1}\\ 1 & \mathbb{C} & \mathbb{Z}/2\mathbb{Z} \\ 2 & \mathbb{H} & \mathbb{Z}/2\mathbb{Z} \\ 3 & \mathbb{H}+\mathbb{H} & 0 \\ 4 & \mathbb{R}(4) & \mathbb{Z} \\ 5 & \mathbb{C}(4) & 0 \\ 6 & \mathbb{R}(8) & 0 \\ 7 & \mathbb{R}(8)+\mathbb{R}(8) & 0 \\ 8 & \mathbb{R}(16) & \mathbb{Z} \\ \end{array}$$

Here $Cl_n=T(\mathbb{R}^{n})/(v\otimes v-q(v))$ where $q$ is the quadratic function associated to the standard inner product on $\mathbb{R}^{n}$. $M_k$ denotes the free abelian group on irreducible $\mathbb{Z}/2\mathbb{Z}$ graded modules (these are easy to work out as this group is isomorphic to the free abelian group of ungraded irreducible modules on the algebra of one dimension lower and these are all matrix algebras). Finally $i:Cl_n\rightarrow Cl_{n+1}$ is the inclusion. For example (using the above isomorphism) the first two groups correspond to $\mathbb{C}$ splitting as two copies of $\mathbb{R}$ as a real representation and $\mathbb{H}$ splitting as two copies of $\mathbb{C}$ as a complex representation. From here the table is periodic mod 8 (up to graded Morita equivalence).

The last column certainly looks eerily familiar, it is the "Bott song" i.e. the coefficients of $\mathrm{KO}$. In their classic paper "Clifford Modules", Atiyah, Bott and Shaprio construct a class in $KO(B^n,S^{n-1})$ from a graded irreducible $Cl_n$ module using the difference construction where the trivialisation on the boundary is given by Clifford multiplication. They use this as well as knowledge of $KO_*$ to show that the aforementioned map gives a ring isomorphism:

$$ \bigoplus_{k\geq 0} M_k/M_{k+1}\rightarrow KO_{*} $$

Here the ring structure on the left is given by the super tensor product of modules. This is all rather spectacular. It seems reasonable from the perspective of the construction of the ABS map (i.e. by taking the difference bundle of a $\mathbb{Z}/2\mathbb{Z}$ graded vector bundle) that some aspect of the representation theory of superalgebras over $\mathbb{R}$ (in particular the Brauer--Wall group) should be closely conceptually related to real K-theory. Another basic observation is that, via the clutching construction, one identifies the coefficient ring of real K-theory with the (shifted) homotopy groups of the stable orthogonal group. The group $Pin$ may be constructed as the subgroup generated by the unit sphere in $V$ inside the group of units of $Cl_n(V)$ and this is the universal cover of the orthogonal group. It is conceivable that these groups and their representation theory are of more fundamental importance in understanding the rôle Clifford algebras play.

In Karoubi's book "K-theory: An Introduction" he provides an explicit construction of real and complex K-theory in terms of the K-theory of the Banach category of vector bundles with a $Cl_n$ action (this K-theory group is very similar in spirit to the above quotients $M_k/M_{k+1}$) and uses this to prove real Bott periodicity. Others have followed the same path and constructed K-theory classes in terms of something like bundles of $\mathbb{Z}/2\mathbb{Z}$ graded Hilbert spaces with a $Cl_{n}$ action and a self adjoint operator switching degrees. Unfortunately I find these accounts rather technical and don't really intuit why such constructions are reasonable to expect.

Question: Is there a conceptual explanation for the tight relationship (or some aspect thereof) between $KO$ and the representation theory of Clifford algebras?

Finally the work of Douglas and Henriques aims to replicate the above for $TMF$. Here they replace Clifford algebras with conformal nets, the spin groups with the string groups etc. If there is any general philosophical perspective on what drives such a generalisation and in particular on which aspects are important in the classical picture, I would also be extremely interested.

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    $\begingroup$ Hi Callan. I wrote something about this at a previous MO post: mathoverflow.net/questions/129199/…. Probably you're well beyond the level where this could help you, but just in case I thought I'd mention it $\endgroup$ Aug 7, 2013 at 1:32
  • $\begingroup$ A similar question: mathoverflow.net/questions/85516/… $\endgroup$ Aug 7, 2013 at 11:59
  • $\begingroup$ @David Thank you for the link. The "magic" you explain in your answer is some of the inspiration for the question. $\endgroup$ Aug 7, 2013 at 15:10
  • $\begingroup$ Yeah, we covered that more thoroughly in the course (as you can imagine, Hovey knows a lot about this), but I got sick of writing in that answer at some point. Anyway, if this thread lingers for awhile without an answer I can go dig out my notes from that course and try to write a bit more $\endgroup$ Aug 7, 2013 at 15:21
  • $\begingroup$ @Callan: there's a sign issue somewhere near the beginning. If you want the defining relation to be $v \otimes v = q(v)$ then $\text{Cl}_1$ is not $\mathbb{C}$ but $\mathbb{R} \times \mathbb{R}$. $\endgroup$ Oct 28, 2014 at 23:34

2 Answers 2


Here is one conceptual description of the relationship. (I wouldn't call it an explanation; I don't really know why it's true, except that it's because Bott periodicity is true.)

KO-theory is the first Weiss-derivative of the $K$-theory of Clifford algebras.

More precisely: given a real inner product space $V$, we get a category $M(V)=\mathrm{Mod}(Cl(V))$ of finite dimentional, $\mathbb{Z}/2$-graded modules over the real Clifford algebra $Cl(V)$. We can consider the $K$-theory space $K(M(V))$ of the topological category $M(V)$; take this to mean something like group completion with respect to direct sum.

Thus, $$ M_n = \pi_0 K(M(\mathbb{R}^n)),$$ using the notation in your question.

We thus have a functor $$ V\mapsto K(M(V))\colon \mathcal{L} \to \mathrm{Top}_* $$ from the category $\mathcal{L}$ of finite dimentional inner product spaces and isometries, to pointed spaces.

Michael Weiss came up with an "orthogonal calculus", which produces a tower of functors approximating a functor $F\colon \mathcal{L}\to \mathrm{Top}_*$, whose $n$th layer takes the form $$V\mapsto \Omega^\infty( ((D_nF) \wedge S^{V\otimes \mathbb{R}^n})_{hO(n)}),$$ where $D_nF$ is some spectrum equipped with an action by the orthogonal group $O(n)$.

If we take $F(V)= K(M(V))$, then it turns out that

  • $D_nF\approx *$ for $n>1$, and
  • $D_1F\approx KO$.

So $KO$ is the first Weiss derivative of $F$.

Why is this true? Weiss gives an easy formula for $D_1F$. Define $$F^1(V) = \mathrm{ho.fib.}\bigl( F(V) \to F(V\oplus \mathbb{R})).$$ This turns out to come with a tautological map $$a_V\colon F^1(V) \to \Omega (F^1(V\oplus \mathbb{R})).$$ It turns out that $D_1F$ is exactly the spectrum associated to the prespectrum $\{F^1(V), a_V\}$.

In our case, the spaces $F^1(V)$ are the spaces of the $KO$-spectrum (as in Karoubi, I guess), and the maps $a_V$ turn out to be the Bott maps. Which are equivalences by Bott's theorem.

The higher derivatives $D_nF$ are computed using the fibers of $a_V$, but as these are already contractible, $D_2(V),D_3(V),\dots$ must vanish.

Replace $Cl(V)$ with $Cl(V)\otimes \mathbb{C}$ to get $KU$.

This seems like a neat fact, though I've never been able to figure out what it's good for.

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    $\begingroup$ Beautiful answer $\endgroup$ Oct 11, 2013 at 8:07

It may be worthwhile to vary the question ever so slightly to read

Question: Is there a conceptual explanation for the tight relationship (or some aspect thereof) between K-theory and superalgebra?

This seems to have a deep answer as follows:

  1. The $\mathbb{Z}_2$-grading in superalgebra/supergeometry is usefully identified as the low degree pieces of $\mathbb{S}$-grading for $\mathbb{S}$ the sphere spectrum;

  2. every $E_\infty$ ring spectrum such as $KO$ and $KU$ is canonically $\mathbb{S}$-graded, or at least its group of units is.

This is discussed in a bit more detail on the nLab at superalgebra - Abstract idea. It provides, I think, a conceptual explanation of what might seem an unreasonable effectiveness of superalgebra in generalized cohomology theory.

A few weeks back I had written the following advertizement of this fact, which maybe I can reproduce here:

As we see amplified these days, homotopy theory, instead of being the complicated edifice as which it was historically obtained, is actually simple in that it is foundational. That's the point of the new "univalent foundations":

homotopy theory follows from formal logic the moment you stop insisting that you can decide if two things are actually equal and admit that you can only provide an explicit equivalence that exhibits the equal-ness. (What physicists call a gauge equivalence.. )

Algebra in homotopy theory is homotopical algebra, and in a way the most fundamental object here is the sphere spectrum, the free commutative infinity-ring on a single element. The sphere spectrum is in homotopy foundations what the integers is in traditional foundations.

Recently Kapranov (based on some pre-history of thoughts that are hard for me to track down precisely) amplifies a simple but striking observation:

grading over the sphere spectrum is supersymmetry.

More in detail: a commutative oo-ring whose oo-group of units is graded over the sphere spectrum is a homotoy-analog/refinement of a superalgebra (see at superalgebra -- Abstract idea).

Hm, you think, so which commutative $\infty$-rings are graded in this way over the sphere spectrum?

Just about a year ago Sagave gives a striking reply to this: every single one is, and canonically so (see here).

In summary, there is a remarkable piece of magic that is happening here, magic in the sense that it opens our eyes to a truth that has been there all along without us noticing it: supersymmetry is right there in the new foundations. It's inevitable.

Hints of this have been seen all along of course. It is standard among mathematicians to praise the role that the idea of supersymmetry has played in pure mathematics, quite independently of its role in physics. Notably index theorems tend to have their most natural formulaton in superalgebra.

Hm, index theorems? What is an index theorem? An index theorem is the characterization of a push-forward in a cohomology theory. A cohomology theory, in turn, is just the theory of maps into a commutative infinity-ring. And there the circle closes.

For instance start with the abelian 2-group $B U(1)$ of ordinary line bundles. Its group ring (meaning: infinity-group infinity ring over the sphere spectrum) contains an element called the Bott element, quotienting that out yields the commutative infinity-ring known as KU, the one that gives the cohomology theory known as ordinary complex K-theory .

$KU = \mathbb{S}[BU(1)][Bott^-1]$

(This is Snaith's theorem ).

Now what is the $\infty$-group of units of KU? That's the 2-group of super line 2-bundles (on the 3-truncation).

In some disguise (see the references at the above link) this has been known for ages. This is why supergeometry is such a powerful tool in K-theory and index theory. But here we see that this nice "technical trick" as it may seem is but a shadow of something very deep, with "very deep" in the technical sense: just a handful of lines of code above the very univalent foundations of mathematics.

Kapranov combined with Sagave shows us that commutative algebra in homotopy theory is automatically and necessarily, in a "god given" way: superalgebra.

  • $\begingroup$ Dear Urs, this seems extremely insightful, I will have to look more thoroughly through your links and the attached references. Am I to understand Kapranov as suggesting that the free Picard n-category (whatever this is) should be in some sense (presumably via an appropriate nerve construction) given by the truncation of the sphere spectrum? Many Thanks, $\endgroup$ Aug 7, 2013 at 15:18
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    $\begingroup$ Yes, that much is clear: the sphere spectrum, as an $E_\infty$-ring, is free on a single element, which means that it is the "free abelian $\infty$-group" on a single element, which means that its n-truncation is the "free Picard n-category" on a single generator. $\endgroup$ Aug 7, 2013 at 15:22

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