January 16, 2023

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:

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!

Posted at January 16, 2023 12:15 AM UTC

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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)?

Posted by: Allen Knutson on January 17, 2023 4:38 AM | Permalink | Reply to this

Re: The Tenfold Way (Part 7)

Thanks. You’re right, and actually we only get symmetric spaces whose sizes involve powers of two. For example, the inclusion

$Cliff_{2n+1} \hookrightarrow Cliff_{2n}$

gives the diagonal inclusion

$M_{2^n}(\mathbb{C}) \oplus M_{2^n}(\mathbb{C}) \hookrightarrow M_{2^{n+1}}(\mathbb{C})$

and thence the inclusion

$U(2^n) \times U(2^n) \hookrightarrow U(2^{n+1})$

and the symmetric space

$U(2^{n+1})/U(2^n) \times U(2^n)$

So we even don’t get all the symmetric spaces $U(2n)/U(n) \times U(n)$, much less all those of the form $U(m+n)/U(m) \times U(n)$.

This ‘power of two’ issue shows up for all the symmetric spaces I listed. Maybe this is more of an excuse to focus on the $n \to \infty$ limit, where in some sense we do get all 10 possibilities from Clifford algebras.

Posted by: John Baez on January 17, 2023 9:06 PM | Permalink | Reply to this
Read the post The Tenfold Way (Part 8)
Weblog: The n-Category Café
Excerpt: A second wonderful one-to-one correspondence between the 10 Morita equivalence classes of Clifford algebras and Cartan's 10 infinite families of compact symmetric spaces.
Tracked: January 18, 2023 10:41 PM

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