## January 17, 2023

### The Tenfold Way (Part 8)

#### Posted by John Baez Last time I explained a wonderful one-to-one correspondence between the 10 Morita equivalence classes of Clifford algebras and Cartan’s 10 infinite families of compact symmetric spaces.

This time I’ll do something different. I’ll explain a wonderful one-to-one correspondence between the 10 Morita equivalence classes of Clifford algebras and Cartan’s 10 infinite families of compact symmetric spaces.

Yes, it’s different! Not only will the details of the construction look very different, it gives a different correspondence! And I hope you can help me figure out what’s going on.

I thank Claude Schochet for pointing out that these two constructions don’t match.

The construction of symmetric spaces from Clifford algebras I described last time is something I made up myself, though it’s so simple someone must have thought of it earlier. The one I’ll talk about now is nicely explained here:

Both Milnor and Dugger use this construction as part of a proof of Bott periodicity:

$\pi_{k+8}(\mathrm{O}(\infty)) \cong \pi_k(\mathrm{O}(\infty))$

where $\mathrm{O}(\infty)$ is the infinite-dimensional orthogonal group. The proof goes roughly as follows. First, you approximate $\mathrm{O}(\infty)$ by $\mathrm{O}(n)$, the group of orthogonal transformations of $\mathbb{R}^n$. Elements of $\pi_k(\mathrm{O}(n))$ are connected components of the $k$-fold loop space of $\mathrm{O}(n)$: that is, the space of loops in the space of loops in… $\mathrm{O}(n)$. Next comes a very interesting step: you can approximate this $k$-fold loop space by a much smaller space consisting of geodesics in a space of geodesics in… $\mathrm{O}(n)$. Milnor calls this space $\Omega_k(n)$. He describes $\Omega_k(n)$ algebraically using Clifford algebras! This lets him understand its set of connected components.

I’m going to sidestep most of this stuff, fascinating though it is, because I just want to get my hands on these spaces $\Omega_k(n)$ with a minimum of fuss. They are symmetric spaces! So they’ll give our second construction of symmetric spaces from Clifford algebras.

We’ll define these spaces $\Omega_k(n)$ recursively.

First, recall that a complex structure on $\mathbb{R}^n$ is a linear operator $J \colon \mathbb{R}^n \to \mathbb{R}^n$ such that $J^2 = - 1$. We say a complex structure is orthogonal if $J \in O(n)$.

Next, start picking orthogonal complex structures $J_1, J_2, \dots$ on $\mathbb{R}^n$, each of which anticommutes with all the previous ones:

$J_k J_\ell = - J_\ell J_k \quad \text{ if } \; \ell \lt k$

Let $\Omega_k(n)$ be the space of orthogonal complex structures $J_k$ that anticommute with all the previous ones.

Unfortunately this is a bit vague, because this space depends on all our previous choices $J_1, \dots, J_{k-1}$. Usually we’ll get isomorphic spaces $\Omega_k(n)$ no matter how we choose $J_{k-1}$. But at certain stages we’ll need to make a ‘good’ choice—which simply means any choice that makes $\Omega_k(n)$ have the largest possible dimension. This then determines $\Omega_k(n)$ uniquely up to isomorphism. I’ll say more about this later.

By definition we have

$\mathrm{O}(n) \supseteq \Omega_1(n) \supseteq \Omega_2(n) \supseteq \cdots$

and we might as well define $\Omega_0(n)$ to be $O(n)$ itself.

Milnor explicitly works out all these spaces $\Omega_k(n)$. He shows they’re all submanifolds of $\mathrm{O}(n)$. Since $\mathrm{O}(n)$ has a god-given Riemannian metric, the spaces $\Omega_k(n)$ become Riemannian manifolds. And he shows they’re all symmetric spaces!

Milnor works step by step, computing $\Omega_1(n), \Omega_2(n),$ and so on, and in a few pages he shows that

$\Omega_8(n) \cong \mathrm{O}(n/16)$

whenever $n$ is divisible by 16. This is a version of Bott periodicity!

What are these spaces $\Omega_k(n)$? The first couple are easy. $\Omega_1(n)$ is the space of orthogonal complex structures on $\mathbb{R}^n$ — or in other words, ways of making the real Hilbert space $\mathbb{R}^n$ into a complex Hilbert space. Notice this is empty unless $n$ is even.

Next, suppose $n$ is even and we’ve made $\mathbb{R}^n$ into a complex Hilbert space. Then $\Omega_2(n)$ is the space of ways of choosing a second orthogonal complex structure that anticommutes with the first. But this is the space of ways of ways of making our complex Hilbert space into a quaternionic Hilbert space! And this will be empty unless $n$ is divisible by 4.

It keeps on going like this, but it gets harder. On Mathstodon I talked my way through many of Milnor’s calculations, but here I’ll just state the results.

If $n$ isn’t a high enough power of $2$ then $\Omega_k(n)$ is just empty. The ones we care about, namely the ones up to $\Omega_8(n)$, are nonempty whenever $n$ is divisible by $16$. So I’ll just assume $n = 16r$ and state the results in that case. We get a very nice list of symmetric spaces:

• $\Omega_0(n) = \mathrm{O}(16r)$. This is the group of all orthogonal transformations of $\mathbb{R}^{16r}$.

• $\Omega_1(n) \cong \mathrm{O}(16r)/\mathrm{U}(8r)$. This is the space of orthogonal complex structures on $\mathbb{R}^{16r}$.

• $\Omega_2(n) \cong \mathrm{U}(8r)/\mathrm{Sp}(4r)$. This is the space of orthogonal quaternionic structures on $\mathbb{C}^{8r}$: that is, orthogonal complex structures on the underlying real Hilbert space of $\mathbb{C}^{8r}$ that anticommute with multiplication by $i$.

• $\Omega_3(n) \cong \bigsqcup_{0 \le d \le 4r} Sp(4r)/\mathrm{Sp}(d) \times \mathrm{Sp}(4r - d)$. This is the space of all quaternionic subspaces of $\mathbb{H}^{4r}$: a union of quaternionic Grassmannians.

• $\Omega_4(n) \cong Sp(2r)$. This is the group of all quaternionic unitary transformations of $\mathbb{H}^{2r}$, also known as the compact symplectic group.

• $\Omega_5(n) \cong Sp(2r)/U(2r)$. This is a complex Lagrangian Grassmannian: the space of all Lagrangian subspaces of a $4r$-dimensional complex symplectic vector space.

• $\Omega_6(n) \cong U(2r)/O(2r)$. This is a real Lagrangian Grassmannian: the space of all Lagrangian subspaces of a $4r$-dimensional real symplectic vector space.

• $\Omega_7(n) \cong \bigsqcup_{0 \le d \le 2r} O(2r)/\mathrm{O}(d) \times \mathrm{O}(2r - d)$. This is the space of all real subspaces of $\mathbb{R}^{2r}$: a union of real Grassmannians.

• $\Omega_8(n) \cong O(r)$.

What does this stuff have to do with Clifford algebras? Well, a bunch of anticommuting complex structures

$J_1, \dots , J_k : \mathbb{R}^n \to \mathbb{R}^n$

is exactly the same as a representation of the algebra $Cliff_k$ on $\mathbb{R}^n$, i.e. an algebra homomorphism

$\rho \colon Cliff_k \to M_n(\mathbb{R})$

But in the definition of $\Omega_k(n)$ we are also requiring that these complex structures be orthogonal. We can state this extra requirement using the $\ast$-algebra structure on $Cliff_k$ that I explained last time: it amounts to saying $\rho$ is a $\ast$-representation, meaning a representation with

$\rho(a^\ast) = \rho(a)^\ast$

(In other jargon, it’s a $\ast$-algebra homomorphism.)

So, we can describe the symmetric spaces $\Omega_k(n)$ recursively as follows:

$\Omega_k(n)$ is the space of ways of extending the already chosen $\ast$-representation of $Cliff_{k-1}$ on $\mathbb{R}^n$ to a $\ast$-representation of $Cliff_{k}$.

This description works for $k \ge 1$, and in some ways it’s very nice, but it involves a sequence of choices, so let me say a bit about that! When we get to

$\displaystyle{ \Omega_3(n) \cong \displaystyle{\bigsqcup_{0 \le d \le 4r} Sp(4r)/\mathrm{Sp}(d) \times \mathrm{Sp}(4r - d)} }$

this has many connected components, one for each dimension $d$, and we should pick a point in the component of highest dimension, namely $d = 2r$. Similarly, when we get to

$\displaystyle{ \Omega_7(n) \cong \bigsqcup_{0 \le d \le 2r} O(2r)/\mathrm{O}(d) \times \mathrm{O}(2r - d) }$

we should pick a point in the component of highest dimension, namely $d = r$. And if we continue on, we must do the same thing whenever $k = 3$ or $7$ modulo $8$.

But now for the main point!

We now have two different ways to build symmetric spaces: the way I just described and the way I described last time. We should compare them. To simplify notation, let’s work ‘stably’, taking the direct limit of the spaces $\Omega_k(n)$ as $n \to \infty$, and working with the infinite-dimensional Lie groups $\mathrm{O}, \mathrm{U}$ and $\mathrm{Sp}$ instead of their finite-dimensional incarnations as above.

So, today’s construction gives infinite-dimensional symmetric spaces

$\Omega_k = \lim_{n \to \infty} \Omega_k(n)$

depending only on $k$ mod 8. The notation $\Omega_k$ is nice because they are actually iterated loop spaces. And I’ll call the infinite-dimensional symmetric spaces we got last time $\Upsilon_k$ because this letter doesn’t get used enough. Let’s compare them:

$\begin{array}{ll} \Upsilon_0 \cong \mathrm{O}/\mathrm{O} \times \mathrm{O} \quad & \Omega_0 \cong \mathrm{O} \\ \Upsilon_1 \cong \mathrm{U}/\mathrm{O} & \Omega_1 \cong \mathrm{O}/\mathrm{U} \\ \Upsilon_2 \cong \mathrm{Sp}/\mathrm{U} & \Omega_2 \cong \mathrm{U}/\mathrm{Sp} \\ \Upsilon_3 \cong \mathrm{Sp} & \Omega_3 \cong \mathrm{Sp}/\mathrm{Sp} \times \mathrm{Sp} \\ \Upsilon_4 \cong \mathrm{Sp}/\mathrm{Sp} \times \mathrm{Sp} \quad & \Omega_4 \cong \mathrm{Sp} \\ \Upsilon_5 \cong \mathrm{U}/\mathrm{Sp} & \Omega_5 \cong \mathrm{Sp}/\mathrm{U} \\ \Upsilon_6 \cong \mathrm{O}/\mathrm{U} & \Omega_6 \cong \mathrm{U}/\mathrm{O} \\ \Upsilon_7 \cong \mathrm{O} & \Omega_7 \cong \mathrm{O}/\mathrm{O} \times \mathrm{O} \end{array}$

And look! The second list is just the first list turned upside down!

So, the question is why.

A similar thing happens for complex Clifford algebras, by the way. Last time we got two infinite-dimensional symmetric spaces from those, and Milnor also gets two. Using the obvious notation we have

$\begin{array}{ll} \Upsilon_0^{\mathbb{C}} \cong \mathrm{U}/\mathrm{U} \times \mathrm{U} \quad & \Omega_0^{\mathbb{C}} \cong \mathrm{U} \\ \Upsilon_1^{\mathbb{C}} \cong \mathrm{U} & \Omega_1^{\mathbb{C}} \cong \mathrm{U}/\mathrm{U} \times \mathrm{U} \end{array}$

One nice thing about explaining a problem in detail in a blog article is that it gives me time to think about it. So I now have some thoughts about what’s going on here. But I’d also like to hear yours!

By the way, the symmetric spaces that are actually groups stand out as odd in the charts above, but they’re not really so odd because in both constructions they naturally as quotients:

$\mathrm{O} \cong \mathrm{O} \times \mathrm{O}/\mathrm{O}, \qquad \mathrm{U} \cong \mathrm{U} \times \mathrm{U}/\mathrm{U}, \qquad \mathrm{Sp} \cong \mathrm{Sp} \times \mathrm{Sp}/\mathrm{Sp}$

So, while I’m drawing big charts, let me draw one using this notation:

$\begin{array}{ll} \Upsilon_0 \cong \mathrm{O} /\mathrm{O} \times \mathrm{O} \quad & \Omega_0 \cong \mathrm{O} \times \mathrm{O}/\mathrm{O} \\ \Upsilon_1 \cong \mathrm{U}/\mathrm{O} & \Omega_1 \cong \mathrm{O}/\mathrm{U} \\ \Upsilon_2 \cong \mathrm{Sp}/\mathrm{U} & \Omega_2 \cong \mathrm{U}/\mathrm{Sp} \\ \Upsilon_3 \cong \mathrm{Sp} \times \mathrm{Sp} / \mathrm{Sp} & \Omega_3 \cong \mathrm{Sp}/\mathrm{Sp} \times \mathrm{Sp} \\ \Upsilon_4 \cong \mathrm{Sp}/\mathrm{Sp} \times \mathrm{Sp} \quad & \Omega_4 \cong \mathrm{Sp} \times \mathrm{Sp} / \mathrm{Sp} \\ \Upsilon_5 \cong \mathrm{U}/\mathrm{Sp} & \Omega_5 \cong \mathrm{Sp}/\mathrm{U} \\ \Upsilon_6 \cong \mathrm{O}/\mathrm{U} & \Omega_6 \cong \mathrm{U}/\mathrm{O} \\ \Upsilon_7 \cong \mathrm{O} \times \mathrm{O}/\mathrm{O} & \Omega_7 \cong \mathrm{O}/\mathrm{O} \times \mathrm{O} \end{array}$

Now you see the second list is the first turned upside down in two completely different senses of ‘turned upside down’. You can flip the whole first list upside down, or you can take the reciprocal of each ‘fraction’ on the list.

It also works like this in the complex case:

$\begin{array}{ll} \Upsilon_0^{\mathbb{C}} \cong \mathrm{U}/\mathrm{U} \times \mathrm{U} \quad & \Omega_0^{\mathbb{C}} \cong \mathrm{U} \times \mathrm{U}/\mathrm{U} \\ \Upsilon_0^{\mathbb{C}} \cong \mathrm{U} \times \mathrm{U}/\mathrm{U} & \Omega_0^{\mathbb{C}} \cong \mathrm{U}/\mathrm{U} \times \mathrm{U} \end{array}$

Posted at January 17, 2023 11:05 PM UTC

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### Re: The Tenfold Way (Part 8)

These have been some of my favourite calculations, at times; …

One Thing that itches at the back of my brain in response to the last two notes is: neither of these really describe a correspondence between the 10 Morita Classes and the Symmetric Families. Each symmetric family seems to have TWO Morita classes involved. What they seem do both seem to describe is a correspondence between [a particular cycle of morphisms] and the Symmetric Spaces. I think one cycle is generated by ⊗Cliff₁, and the other by … ⊗Cliff₋₁, if you know what I mean.

Posted by: Jesse C. McKeown on January 20, 2023 7:18 PM | Permalink | Reply to this

### Re: The Tenfold Way (Part 8)

Yes, you’re right—your dissatisfaction echos Allen Knutson’s here.

The ‘principled’ thing to do is treat the Clifford algebras as superalgebras. Then, as Todd so nicely worked out in Part 4, they give all the Morita equivalence classes of superalgebras over $\mathbb{R}$ and $\mathbb{C}$ that are invertible modulo Morita equivalence. That is, real (resp. complex) superalgebras $A$ such that there exists a superalgebra $B$ such that $A \otimes B \simeq k$, where $\otimes$ is the tensor product of superalgebras, $\simeq$ is Morita equivalence, and $k$ is $\mathbb{R}$ (resp. $\mathbb{C}$). As I explained in Part 3, these are just the ‘central simple’ superalgebras, meaning those that have only trivial graded ideals and have trivial supercenter.

But I don’t know a deep reason from the viewpoint of symmetric spaces why $\mathrm{O}(16)/\mathrm{O}(8) \times \mathrm{O}(8)$ is better than $\mathrm{O}(16)/\mathrm{O}(6) \times \mathrm{O}(10)$ or $\mathrm{O}(20)/\mathrm{O}(10) \times \mathrm{O}(10)$. I just know the explanations I’m giving here.

Posted by: John Baez on January 20, 2023 8:20 PM | Permalink | Reply to this

### Dan

Posted by: Allen Knutson on January 20, 2023 9:03 PM | Permalink | Reply to this

Congratulations on being the first person here to deliberately change the subject header to someone’s name. And thanks! This was a weird slip on my part, since I know Dugger as one of the “Dans” — a homotopy group including Dan Christensen and others.

Posted by: John Baez on January 20, 2023 9:24 PM | Permalink | Reply to this

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