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May 9, 2023

Symmetric Spaces and the Tenfold Way

Posted by John Baez

I’ve finally figured out the really nice connection between Clifford algebra and symmetric spaces! I gave a talk about it, and you can watch a video.

I gave my talk in Nicohl Furey’s series Algebra, Particles and Quantum Theory on Monday May 15, 2023. This talk is a followup to an earlier talk, also about the tenfold way.

You can see a video of my new talk here and see my slides here. The slides have material on category theory that I didn’t get around to talking about.

Abstract: The tenfold way has many manifestations. It began as a tenfold classification of states of matter based on their behavior under time reversal and charge conjugation. Mathematically, it relies on the fact that there are ten super division algebras and ten kinds of Clifford algebras, where two Clifford algebras are of the same kind if they have equivalent super-categories of super-representations. But Cartan also showed that there are ten infinite families of compact symmetric spaces! After explaining symmetric spaces, we show how they arise naturally from forgetful functors between categories of representations of Clifford algebras.

The final upshot is this:

Let the Clifford algebra Cliff n\mathrm{Cliff}_{n} be the free real or complex algebra on nn anticommuting square roots of 1-1. Let

F:Rep(Cliff n)Rep(Cliff n1) F \colon \mathsf{Rep}(\mathrm{Cliff}_{n}) \to \mathsf{Rep}(\mathrm{Cliff}_{n-1})

be the forgetful functor between representation categories. Then the essential fibers of this functor are disjoint unions of compact symmetric spaces. Moreover we get all the compact symmetric spaces in the ten infinite families this way!

(There are also finitely many exceptional ones, which all arise from the octonions.)

Posted at May 9, 2023 8:55 PM UTC

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U(g)

This sounds great! I had some questions straightaway, but I am beginning to find answers to them in the slides.

What are the ways to make a universal enveloping algebra U(g) a *-algebra?

Posted by: Allen Knutson on May 10, 2023 12:24 AM | Permalink | Reply to this

Re: U(g)

OK I have a question I didn’t see in the notes. Where does the loop space story (that’s supposed to rotate these clocks) come in, in this picture?

Posted by: Allen Knutson on May 10, 2023 12:27 AM | Permalink | Reply to this

Re: U(g)

Do you really need the *-rep stuff? If you skip it, allowing non-*-reps, I’m sure your fibers will be different – but will they be different homotopically?

Posted by: Allen Knutson on May 10, 2023 12:53 AM | Permalink | Reply to this

Re: U(g)

Allen wrote:

What are the ways to make a universal enveloping algebra U(𝔤)U(\mathfrak{g}) a *\ast-algebra?

The most reasonable way to make the universal enveloping algebra U(𝔤)U(\mathfrak{g}) of the Lie algebra 𝔤\mathfrak{g} of a compact Lie group GG into a *\ast-algebra is to make all the elements of 𝔤\mathfrak{g} skew-adjoint. Then a finite-dimensional unitary representation of GG gives a finite-dimensional complex *\ast-representation of U(𝔤)U(\mathfrak{g}) and vice versa.

Where does the loop space story (that’s supposed to rotate these clocks) come in, in this picture?

Good question — it’s been cleverly absorbed into the algebra! If you read Milnor’s book Morse Theory or my distillation here you’ll see he approximates the kk-fold loop space of O(n)\mathrm{O}(n) by a space Ω k(n)\Omega_k(n), which is a certain space geodesics in a certain space of geodesics in a certain space of geodesics… in O(n)\mathrm{O}(n). And he cleverly sets up these spaces of kk-fold iterated geodesics in such a way that you can describe them using *\ast-representations of Cliff k\mathrm{Cliff}_k on the real Hilbert space n\mathbb{R}^n.

How does this work? Milnor explains it with his usual clarity, but the core fact is that a unit-speed geodesic loop of length 2π2\pi in O(n)\mathrm{O}(n) exactly corresponds to what Milnor and I call a complex structure on n\mathbb{R}^n, namely a linear operator J: n nJ : \mathbb{R}^n \to \mathbb{R}^n with

J 2=1,J =J 1 J^2 = -1, \quad J^\dagger = J^{-1}

The first equation is what most people mean when they say ‘complex structure’, but Milnor and I want to restrict attention to orthogonal operators JJ, which the second equation accomplishes. These two equations are necessary and sufficient for

exp(tJ) \exp(t J)

to be a 1-parameter subgroup of O(n)\mathrm{O}(n) that first gets back to the identity when t=2πt = 2\pi.

So, all the hard topology in Bott periodicity comes from using Morse theory to show that the tractable space Ω k(n)\Omega_k(n) has the same fundamental group as the kk-fold loop space of O(n)\mathrm{O}(n) when kk is small enough compared to nn.

I want to avoid all that stuff, since to me the spaces Ω k(n)\Omega_k(n) are more interesting! The way I do things, which is slightly different than how Milnor does it, these spaces become essential fibers of forgetful functors — and it’s easy to show they are disjoint unions of symmetric spaces, and it turns out we get all the compact symmetric spaces in Cartan’s ten infinite families this way. (You may recall that when I followed Milnor, I only got some.)

Do you really need the *\ast-rep stuff?

I need it to get compact symmetric spaces!

If you skip it, allowing non-*\ast-reps, I’m sure your fibers will be different – but will they be different homotopically?

If I skipped the *\ast-stuff I’d get real affine varieties, not compact Riemannian symmetric spaces. I haven’t carefully checked, but I believe they would be homotopy equivalent. It’s clear in a bunch of cases. In some vague sense the *\ast-stuff lets me pick out the ‘maximal compacts’ in what I’d get if I skipped the *\ast-stuff. But I’m not sure what that means. For example, in one case I get

O(m+n)/O(m)×O(n) \mathrm{O}(m+n)/\mathrm{O}(m) \times \mathrm{O}(n)

if I use the *\ast-stuff, but

GL(m+n,)/GL(m,)×GL(n,) GL(m+n,\mathbb{R})/GL(m,\mathbb{R}) \times GL(n,\mathbb{R})

if I don’t. These are isomorphic as real affine varieties, but I think only the former comes with a god-given Riemannian metric!

Posted by: John Baez on May 10, 2023 1:31 AM | Permalink | Reply to this

Re: U(g)

They’re not isomorphic as real affine varieties; the real Grassmannian is the quotient of GL(m+n) by block upper triangular matrices, not by block diagonal. But in particular the spaces are indeed homotopy equivalent (one being an affine bundle over the other).

I’m used to the interest in these spaces being their homotopy. (And when I move from homotopy to geometry, I’d personally rather acquire algebraic geometry than metrics, but I understand not everyone agrees.) Do you have plans to do anything with the metrics?

Posted by: Allen Knutson on May 10, 2023 1:08 PM | Permalink | Reply to this

Re: U(g)

Allen wrote:

They’re not isomorphic as real affine varieties; the real Grassmannian is the quotient of GL(m+n) by block upper triangular matrices, not by block diagonal.

Oh, whoops! That’s actually good: I was wanting to say that requiring compatibility with the *\ast-structure picks out a ‘compact real form’ of a noncompact variety, so I was puzzled by how in this case I seemed to be getting the same thing either way.

Do you have plans to do anything with the metrics?

So far I was just trying to understand Cartan’s classification of compact symmetric spaces using Clifford algebras, and the metrics are part of this story. People in condensed matter physics do a lot with these compact symmetric spaces: they show up as ‘spaces of Hamiltonians for free fermion systems’ but also in other ways, in a complicated story that associates several different compact symmetric spaces to each of the 10 kinds of Clifford algebra. But I don’t understand any of this as well as I’d like, which is why I’d been seeking a simpler story. Maybe now I can go back and figure out more, like what the metric is good for.

Posted by: John Baez on May 10, 2023 6:38 PM | Permalink | Reply to this

Re: U(g)

In general I expect the *-structure to pick out the real points under some antiholomorphic involution, and I’m not sure that it will always be a deformation retract. But maybe these cases are specific enough to make that happen.

I’m used to symmetric spaces (you say compact, I say complex, let’s call the whole thing homotopic) being of the form G/K where K is the fixed point set of an involution. If you have an A-module M and a homomorphism A->B of algebras, what is the essential fiber over M of Rep(B)->Rep(A)? When should one expect it to be a homogeneous space? And when symmetric?

Posted by: Allen Knutson on May 10, 2023 10:48 PM | Permalink | Reply to this

Re: U(g)

Allen wrote something homotopic to:

If you have an AA-module MM and a homomorphism f:ABf: A \to B of algebras, what is the essential fiber over MM of the restriction functor f *:Rep(B)Rep(A)f^\ast : Rep(B) \to Rep(A)? When should one expect it to be a homogeneous space?

I have theorems about this near the end of my talk slides.

In general the essential fiber over MM just is what is: the category of BB-modules XX equipped with isomorphism α:f *(X)M\alpha: f^\ast (X) \to M.

But this essential fiber is a mere poset when f *f^\ast is faithful: that is, when it’s injective on each hom-set. This happens, for example, when ff is an inclusion of algebras.

And the essential fiber is a mere groupoid when f *f^\ast is conservative. This means that when f *f^\ast applied to some morphism is an iso, that morphism had to be an iso to begin with. This also happens whenever ff is an inclusion of algebras.

When both these conditions hold, the essential fiber is both a poset and a groupoid, so it’s essentially a set!

Let’s assume that happens. Now, when is this set a homogeneous space?

This set always acted on by the group Aut(A)Aut(A), so it’s always a disjoint union of orbits, which are homogeneous spaces for Aut(A)Aut(A). In my examples, there is often just one orbit. Then the essential orbit is itself a homogeneous space. But sometimes there are many orbits.

I state a condition under which the essential fiber consists of just one orbit: f *f^\ast is injective on isomorphism classes. In other words, if two BB-modules restrct to isomorphic AA-modules, they had to be isomorphic to begin with.

This happens with for inclusion \mathbb{R} \hookrightarrow \mathbb{C}, but not for the diagonal inclusion \mathbb{R} \hookrightarrow \mathbb{R} \oplus \mathbb{R}.

Finally: when is the essential fiber a disjoint union of abstract symmetric spaces — that is, sets Aut(A)/KAut(A)/K where the subgroup KK consists of all the fixed points of some involution on Aut(A)Aut(A)? This always happens in my examples, and the reason is that all my homomorphisms

f:Cliff nCliff n+1 f: Cliff_n \to Cliff_{n+1}

are algebra inclusions where you get the bigger algebra by throwing in an extra square root of 1-1. The involution business arises simply from (1) 2=1(-1)^2 = 1. I explain how in my slides. It would also work if we threw in a square root of +1+1.

Posted by: John Baez on May 11, 2023 1:07 AM | Permalink | Reply to this

Re: U(g)

One of the many ways to get a hold of the ring of symmetric functions (or more precisely, a biRees algebra thereof) is to apply H_* to the disjoint union of all Grassmannians Gr(a,a+b). This space Gr(,) has a map Gr(,)^2 -> Gr(,) called “direct sum”.

Do each of your ten spaces come with such a thing? If so, do you know what the analogue is of the (biRees algebra of the) ring of symmetric functions?

Posted by: Allen Knutson on May 14, 2023 11:09 PM | Permalink | Reply to this

Re: U(g)

I don’t know about *-algebra structures on U(g) in general, but if instead we ask about *-Hopf algebra structures then they’re the same thing as real forms of g.

Posted by: Alexander Shamov on May 10, 2023 6:28 AM | Permalink | Reply to this

Re: U(g)

That’s a nice result, Alexander. And I guess it makes sense. Over any field there’s an equivalence between affine algebraic group schemes and cocommutative Hopf algebras. So, we can take the concept of ‘real form’ for a complex affine algebraic group scheme GG and transport it to get some concept of ‘real form’ for a complex cocommutative Hopf algebra. And it’s not too surprising that this concept would amount to a *\ast-structure on that Hopf algebra. But I hadn’t thought about how the *\ast-structure needs to be compatible with all the operations and co-operations in the Hopf algebra, not just the algebra operations!

Posted by: John Baez on May 10, 2023 6:30 PM | Permalink | Reply to this

Re: U(g)

Even in the non-cocommutative case there is a nice notion of Hopf *-algebras. The involution reverses the order in the product but NOT in the coproduct - instead, the coproduct is a morphism of *-algebras. Despite this asymmetry, the theory is self-dual: the involution on the dual of, say, a finite-dimensional Hopf *-algebra is twisted by the antipode: x *,y:=x,S(y) *¯\langle x^\ast, y \rangle := \overline{ \langle x, S(y)^\ast \rangle }. In the setting of C*-algebras this leads to a very satisfying (but much more sophisticated) theory of “locally compact quantum groups” enjoying perfect Pontryagin duality.
Posted by: Alexander Shamov on May 10, 2023 11:58 PM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

I think there is a typo in the list of Clifford algebras (slide 12-13). You skipped Cliff4. The description as matrix ring over R,C,H seems to not skip the Cliff4, but the Rep do, so there is a mismatch.

Posted by: Malo Tarpin on May 10, 2023 8:36 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

Thanks very much for catching that numbering mistake! That would have been very annoying to correct while giving my talk. I also found another and fixed that too.

Posted by: John Baez on May 10, 2023 7:20 PM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

I am used to Clifford algebra being defined with arbitrary signature (p square root of 1 and q square root of -1). There exist formula relating Clifford algebra at constant/varying signature which can be seen to imply Bott periodicity. Have you tried playing the same game but adding or removing square root of 1 instead to see which kind of symmetric space appears?

Posted by: Malo Tarpin on May 10, 2023 11:36 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

“constant/varying signature”

that was not clear at all.

I mean, fixing p+qp+q and varying pqp-q, or fixing pqp-q and varying p+qp+q.

For example, I have in mind

Cliff p+1,q+1=M 2(Cliff p,q) \text{Cliff}_{p+1,q+1} = M_2\left( \text{Cliff}_{p,q}\right)

for any pqp-q except 1 mod 41\text{ mod }4 (the split ones).

Posted by: Malo Tarpin on May 10, 2023 1:57 PM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

Malo wrote:

I am used to Clifford algebras being defined with arbitrary signature (pp square roots of 1 and qq square roots of -1).

Right, me too! But if you look at my slides, you’ll see I’m getting the compact symmetric spaces not directly from the Clifford algebras but from their categories of representations, and the forgetful functors between these. Thanks to the fact you mentioned:

Cliff p+1,q+1M 2Cliff p,q) \mathrm{Cliff}_{p+1,q+1} \cong \mathrm{M}_2\mathrm{Cliff}_{p,q})

we have an equivalence of categories

Rep(Cliff p+1,q+1)Rep(Cliff p,q) \mathsf{Rep}(\mathrm{Cliff}_{p+1,q+1}) \simeq \mathsf{Rep}(\mathrm{Cliff}_{p,q})

This means we can simplify the situation to the case where either p=0p = 0 or q=0q = 0. Then we have

Rep(Cliff p,0)Rep(Cliff 0,8p) \mathsf{Rep}(\mathrm{Cliff}_{p,0}) \simeq \mathsf{Rep}(\mathrm{Cliff}_{0,8-p})

so we might as well just consider Clifford algebras with only square roots of 1-1.

I think square roots of 1-1 are more important geometrically than +1+1, but to some extent that’s a matter of taste. I could have used square roots of +1+1. Then we would go around this ‘clock’ in the opposite direction, counterclockwise:

Posted by: John Baez on May 10, 2023 6:48 PM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

Ah! So actually the identities I was thinking about are enough to reduce the classification of categories of representations to only those 8, I did not realise it.

Posted by: Malo Tarpin on May 11, 2023 8:52 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

Are Clifford algebras the only way to get a universal enveloping algebra?

Posted by: a on May 10, 2023 11:59 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

I’m not sure what you mean by that, since I don’t know how to use Clifford algebras to get universal enveloping algebras.

Posted by: John Baez on May 11, 2023 12:19 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

While writing an earlier comment I thought of a question. Suppose you have a ring homomorphism f:ABf: A \to B. Restriction along ff gives a functor

f *:Mod(B)Mod(A) f^\ast: Mod(B) \to Mod(A)

and this has a left adjoint, often called extension along ff:

f !:Mod(A)Mod(B) f_!: Mod(A) \to Mod(B)

Concretely we have

f !(M)=B AM f_!(M) = B \otimes_A M

My question: when is the functor

f *:Mod(B)Mod(A) f^\ast: Mod(B) \to Mod(A)

monadic?

The reason I care is that this implies f *f^\ast is faithful and conservative: two conditions I cared about in my previous comment.

I feel I should learn the answer, because I’m getting interested in monadic descent. I should probably figure it out, but I’m hoping someone here knows!

Posted by: John Baez on May 11, 2023 2:07 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

In fact f *f^\ast is always monadic. On the Category Theory Community Server we saw how this can be shown using Beck’s monadicity theorem, and also that it’s a special case of a much more general theorem for arbitrary single-sorted Lawvere theories… but Mike Shulman wrote:

I often find it easier to show a functor is monadic by directly comparing its domain to the category of algebras for the induced monad, rather than by mucking around with split or reflexive coequalizers. In this case an f *f !f^*f_!-algebra is an AA-module with a map B AMMB\otimes _A M \to M satisfying some conditions, or equivalently an AA-bilinear map (B,M)M(B,M) \to M satisfying some conditions. If you forget about AA-bilinearity and just regard this map as \mathbb{Z}-bilinear, the conditions are precisely those that make it define a BB-module, and then the extra AA-bilinearity is equivalent to saying that the given AA-module structure is the one induced from this BB-module structure via ff.

Posted by: John Baez on May 14, 2023 1:42 AM | Permalink | Reply to this

Re: Symmetric Spaces and the Tenfold Way

David Roberts made a nice comment elsewhere:

It seems to me that it’s perfectly reasonable to have a disconnected compact symmetric space, since the definition doesn’t rely on connectedness, though it is included. The important thing is that for each point there is an involution fixing that point and acting as -1 on the tangent space there.

Clearly for the purposes of classification, getting the connected ones is enough. And if I’m not mistaken, each connected component of one of these “disconnected compact symmetric spaces” should be a symmetric space, too.

In the slides, the example where the essential fibre F 1( n)F^{-1}(\mathbb{H}^n) is a disjoint union is where one has a two-parameter family, with the second parameter can be taken to have a finite range. I presume this is true for the other cases where one has these two parameter families (namely the real and complex Grassmannians)? One might make an argument that among the ten infinite families, they really should only have one parameter, just that three of those families consist of disconnected symmetric spaces (obviously with finitely many components).

I replied:

It is indeed perfectly possible to have a disconnected symmetric space, but the usual definition requires connectedness so I didn’t want to mess with that in this talk. A really nice abstract definition of symmetric space would be ‘involutory quandle object in the category of smooth manifolds’, but then to recover Cartan’s classification we need to impose conditions like compactness, preservation of a Riemannian metric by the quandle operation xx \triangleright -, a condition saying that each point xx is an isolated fixed point of xx \triangleright -, and even more (some ‘irreducibility’ and simply-connectedness properties I didn’t want to talk about).

Anyway, all your guesses are right. There are three two-parameter families of compact symmetric spaces, the Grassmannians, ultimately because three Morita equivalence classes of Clifford algebras are not Morita equivalent to division algebras (whose representations are classified by a single natural number). Let me talk about the real case for simplicity. If you take the manifold of all subspaces of n\mathbb{R}^n, it is a disjoint union of Grassmannians, but the whole thing is a quandle object, because it makes sense to reflect any subspace WW of n\mathbb{R}^n across any subspace VV, by applying the transformation that preserves VV and multiplies vectors orthogonal to VV by 1-1. So, it’s very nice — it just happens to be disconnected, as you noted.

Posted by: John Baez on May 12, 2023 9:13 PM | Permalink | Reply to this

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