### The Tenfold Way (Part 7)

#### Posted by John Baez

Last time I reviewed a bit of Bott periodicity. Now I want to start leading up to a question about it. It will take a while.

So, this time, I will 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.

Then, next time, I will 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.

Unfortunately these correspondences are *not the same!* And my question will be: *why?* Or, perhaps better: what’s the relationship between them?

The **Clifford algebra** $Cliff_n$ is the free real algebra on $n$ anticommuting square roots of $-1$, say $e_1, \dots, e_n$. We can make this algebra into a **$\ast$-algebra**, meaning that it has a ‘star’ operation such that

$(a + b)^\ast = a^\ast + b^\ast, \; (a b)^\ast = b^\ast a^\ast$

and also

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

for all $\alpha \in \mathbb{R}$. The star operation on $Cliff_n$ is determined by decreeing that each $e_i$ has

$e_i^\ast = -e_i$

So, for example, the star operation in $\mathrm{Cliff}_1 \cong \mathbb{C}$ is just the usual complex conjugation, and the star operation in $\mathrm{Cliff}_2 \cong \mathbb{H}$ is ‘quaternionic conjugation’.

Any $\ast$-algebra $A$ gives a group consisting of the **unitary** elements $a \in A$, meaning those with

$a a^\ast = a^\ast a = 1$

If $A$ is finite-dimensional this is a Lie group.

Thus, Clifford algebras give Lie groups! By the way, these are not the spin groups that Clifford algebras are famously used to construct: the spin groups are subgroups of the groups I’m talking about now.

The group of unitary elements in $\mathrm{Cliff}_n$ depends heavily on
$n$ mod 8, thanks to Bott periodicity. We can simplify the story in
some ways by taking the direct limit of these groups as $n \to \infty$ while keeping $n$
the same mod 8. We then get 8 different *infinite-dimensional* Lie groups,
which I’ll list. But if you don’t like infinite-dimensional Lie groups, you can
keep $n$ finite and get Lie groups that depend on $n$.

When I list these infinite-dimensional Lie groups, I’ll also remind you from last time what the Clifford algebras $\mathrm{Cliff}_n$ are for $n = 0$ to $n = 7$. This together with Bott periodicity should make the list very believable.

For example, when $n = 1$ we have $\mathrm{Cliff}_1 = \mathbb{C}$, so Bott periodicity says $\mathrm{Cliff}_{8n+1}$ is an algebra of square matrices with entries in $\mathbb{C}$. Those matrices with $a a^\ast = a^\ast a = 1$ turn out to be just the unitary matrices, as you might expect, so the Lie group we get is $U(k)$ for some $k$ that depends on $n$. In fact $k = 16^n$, but the exact formula doesn’t matter here: the main thing is that as $n$ increases, so does $k$, so taking the $n \to \infty$ limit we get the infinite-dimensional Lie group $O(\infty)$.

Okay, here goes:

$Cliff_0 \cong \mathbb{R}$, so $n = 0$ mod 8 gives the group $\mathrm{O}$. This is the direct limit of the orthogonal groups $\mathrm{O}(n)$, consisting of $n \times n$ real matrices $a$ with $a a^\ast = a^\ast a = 1$.

$Cliff_1 \cong \mathbb{C}$, so $n = 1$ mod 8 gives the group $\mathrm{U}$. This is the direct limit of the unitary groups $\mathrm{U}(n)$, consisting of $n \times n$ complex matrices $a$ with $a a^\ast = a^\ast a = 1$.

$Cliff_2 \cong \mathbb{H}$, so $n = 2$ mod 8 gives the group $\mathrm{Sp}$. This is the direct limit of the compact symplectic groups $\mathrm{Sp}(n)$, also known as quaternionic unitary groups, consisting of $n \times n$ quaternionic matrices $a$ with $a a^\ast = a^\ast a = 1$.

$Cliff_3 \cong \mathbb{H} \oplus \mathbb{H}$, so $n = 3$ mod 8 gives the group $\mathrm{Sp} \times \mathrm{Sp}$.

$Cliff_4 \cong M_2(\mathbb{H})$, so $n = 4$ mod 8 again gives the group $\mathrm{Sp}$.

$\Cliff_5 \cong M_4(\mathbb{C})$, so $n = 5$ mod 8 again gives the group $\mathrm{U}$.

$\Cliff_6 \cong M_8(\mathbb{R})$, so $n = 6$ mod 8 again gives the group $\mathrm{O}$.

$Cliff_7 \cong M_8(\mathbb{R}) \oplus M_8(\mathbb{R})$, so $n = 7$ mod 8 gives the group $\mathrm{O} \times \mathrm{O}$.

We’ll get symmetric spaces by taking a quotient of each group by the previous one. To do this,
we need a way to stick each Clifford algebra inside the next one. Each Clifford algebra is a $\mathbb{Z}$/2-graded algebra, or **superalgebra**, in such a way that the elements $e_1, \dots , e_n \in Cliff_n$ are odd. Then there’s a nice fact: $Cliff_n$ is isomorphic to the even part of $Cliff_{n+1}$.

Using this each of the above groups becomes a subgroup of the next one, in a way that cycles around mod 8. We can thus take the quotient of each one by the previous one! These are our symmetric spaces.

For example, after the 7th group $\mathrm{O} \times \mathrm{O}$ comes the 0th group, which conveniently happens to be called $\mathrm{O}$. $\mathrm{O}(n) \times \mathrm{O}(n)$ sits inside $\mathrm{O}(n)$ as block diagonal matrices so $\mathrm{O} \times \mathrm{O}$ is a subgroup of $\mathrm{O}$. The quotient $\mathrm{O}/\mathrm{O} \times \mathrm{O}$ is an interesting space: you can think of it as the space of all infinite-dimensional real subspaces of an infinite-dimensional real vector space that’s ‘twice as big’.

The ‘real Grassmannian’ $\mathrm{O}(m+n)/\mathrm{O}(m) \times \mathrm{O}(n)$ is the set of all $m$-dimensional subspaces of $\mathbb{R}^{m+n}$. It’s a Riemannian manifold that’s so symmetrical that for every point there’s a symmetry called inversion about that point, which fixes that point and sends each tangent vector $v$ to that point to $-v$. Such a Riemannian manifold is called a symmetric space.

Cartan discovered that there are 10 infinite families of compact
symmetric spaces and also 17 exceptions. Here I’m including compact
Lie groups, since these really *are* compact symmetric spaces, even
though most people don’t include them. For the precise rules behind this
classification, go here:

- Classification of Riemannian symmetric spaces, Wikipedia.

The 10 infinite families are closely connected to the tenfold way! For example, one of these families consists of the real Grassmannians $\mathrm{O}(m+n)/\mathrm{O}(m) \times \mathrm{O}(n)$, all of which sit in $\mathrm{O}/\mathrm{O} \times \mathrm{O}$ in a nice way.

Let’s see all 10 infinite families of compact symmetric spaces. We get 8 from the real Clifford algebras, in the way I’ve just described:

$n = 0$ mod 8 gives $\mathrm{O}/(\mathrm{O} \times \mathrm{O})$. This contains all the real Grassmannians $\mathrm{O}(m+n)/(\mathrm{O}(m) \times \mathrm{O}(n))$, which consist of all $m$-dimensional subspaces of $\mathbb{R}^{m+n}$.

$n = 1$ mod 8 gives $\mathrm{U}/\mathrm{O}$. This contains all the real Lagrangian Grassmannians $\mathrm{U}(n)/\mathrm{O}(n)$, which consist of all Lagrangian subspaces of a $2n$-dimensional real symplectic vector space.

$n = 2$ mod 8 gives $\mathrm{Sp}/\mathrm{U}$. This contains all the complex Lagrangian Grassmannians $\mathrm{Sp}(n)/\mathrm{O}(n)$, which consist of all Lagrangian subspaces of a $2n$-dimensional complex symplectic vector space.

$n = 3$ mod 8 gives $(\mathrm{Sp} \times \mathrm{Sp})/\mathrm{Sp} \cong \mathrm{Sp}$. This contains all the compact symplectic groups $\mathrm{Sp}(n)$, also known as quaternionic unitary groups, which consist of all $n \times n$ quaternionic matrices with $a a^\ast = a^\ast a = 1$.

$n = 4$ mod 8 gives $\mathrm{Sp}/(\mathrm{Sp} \times \mathrm{Sp})$. This contains all the quaternionic Grassmannians $\mathrm{Sp}(m+n)/(\mathrm{Sp}(m) \times \mathrm{Sp}(n))$, which consist of all $m$-dimensional subspaces of $\mathbb{H}^{m+n}$.

$n = 5$ mod 8 gives $\mathrm{U}/\mathrm{Sp}$. This contains all the spaces $\mathrm{U}(2n)/\mathrm{Sp}(n)$, which consist of all quaternionic structures on $\mathbb{C}^{2n}$ compatible with its usual complex Hilbert space structure.

$n = 6$ mod 8 gives $\mathrm{O}/\mathrm{U}$. This contains all the spaces $\mathrm{O}(2n)/\mathrm{U}(n)$, which consist of all complex structures on $\mathbb{R}^{2n}$ compatible with its usual real Hilbert space structure.

$n = 7$ mod 8 gives the group $(\mathrm{O} \times \mathrm{O})/\mathrm{O} \cong \mathrm{O}$. This contains all the orthogonal groups $\mathrm{O}(n)$, which consist of all $n \times n$ real matrices with $a a^\ast = a^\ast a = 1$.

The other two infinite families of compact symmetric spaces come from complex Clifford algebras. We can copy the whole theory above in the complex case, using complex $\ast$-algebras. Complex Clifford algebras are periodic mod 2, in the sense that $\mathbb{C}\mathrm{liff}_{n+2}$ is isomorphic as a complex $\ast$-algebra to $2 \times 2$ matrices with entries in $\mathbb{C}\mathrm{liff}_n$. Thus, the group of unitary elements in $\mathrm{Cliff}_n$ depends heavily on $n$ mod 2. We can take the limit of these groups by letting $n \to \infty$ while keeping $n$ the same mod 2. We then get two infinite-dimensional Lie groups:

$n = 0$ mod 2 gives the group $\mathrm{U}$.

$n = 1$ mod 2 gives the group $\mathrm{U} \times \mathrm{U}$.

$\mathrm{U}$ sits inside $\mathrm{U} \times \mathrm{U}$ as the elements of the form $(g,g)$. $\mathrm{U} \times \mathrm{U}$ sits inside $\mathrm{U}$ as the block diagonal matrices. We thus get the remaining two infinite families of compact symmetric spaces:

$n = 0$ mod 2 gives $\mathrm{U}/(\mathrm{U} \times \mathrm{U})$. This contains all the complex Grassmannians $\mathrm{U}(m+n)/(\mathrm{U}(m) \times \mathrm{U}(n))$, which consist of all $m$-dimensional subspaces of $\mathbb{C}^{m+n}$.

$n = 1$ mod 2 gives $(\mathrm{U} \times \mathrm{U})/\mathrm{U} \cong \mathrm{U}$. This contains all the unitary groups $\mathrm{U}(n)$, which consist of all $n \times n$ complex matrices with $a a^\ast = a^\ast a = 1$.

You might wonder how we know all these spaces are symmetric spaces. There’s a beautiful answer, at least for those that are actually of the form

$\text{unitary elements of } \; Cliff_{n+1} / \text{unitary elements of } \; Cliff_{n}$

or

$\text{unitary elements of } \; \mathbb{C}\!\; liff_{n+1} / \text{unitary elements of } \; \mathbb{C}\!\; liff_{n}$

The reason is that a symmetric space is the same as quotient of Lie groups $G/H$ where the Lie algebra $\mathfrak{g}$ admits a $\mathbb{Z}/2$-grading for which $\mathfrak{h} \subset \mathfrak{g}$ is the even part! And since we’re embedding $Cliff_n$ as the even part of the superalgebra $Cliff_{n+1}$, this is exactly what we get!

## Re: The Tenfold Way (Part 7)

That’s a great observation at the end. But it suggests that the bigraded family of spaces U(m+n)/U(m)xU(n) doesn’t naturally appear in this story — only some singly-graded subfamily. I assume it’s U(2n)/U(n)xU(n)?