## The Tenfold Way

#### November 22, 2020

There are 10 of each of these things:
• Ways that Hamiltonians can get along with time reversal ($$T$$) and charge conjugation ($$C$$) symmetry.
• Morita equivalence classes of real and complex Clifford algebras.
• Associative real super division algebras.
• Classical families of compact symmetric spaces.

I'll summarize how it works, and then you can read these for more:

This is a terse summary of the simplest mathematical ideas:
• John Baez, The tenfold way, Notices Amer. Math. Soc. 67 10 (2020), 1599–1601.

Here I'll start by explaining the origin of this subject in physics, and then move on to more mathematical aspects.

### Ten kinds of matter The idea of the ten-fold way goes back at least to 1996, when Altland and Zirnbauer discovered that substances can be divided into 10 kinds.

The basic idea is pretty simple. Some substances have time reversal symmetry: they would look the same, even on the atomic level, if you made a movie of them and ran it backwards. Some don't — these are more rare, like certain superconductors made of yttrium barium copper oxide! Time reversal symmetry is described by an antiunitary operator $$T$$ that squares to $$1$$ or to $$-1$$: please take my word for this, it's a quantum thing. So, we get 3 choices, which are listed in the chart under $$T$$ as $$1, -1$$, or $$0$$ (no time reversal symmetry).

Similarly, some substances have charge conjugation symmetry, meaning a symmetry where we switch particles and holes: places where a particle is missing. The 'particles' here can be rather abstract things, like 'phonons' — little vibrations of sound in a substance, which act like particles — or 'spinons' — little vibrations in the lined-up spins of electrons. Basically any way that something can wave can, thanks to quantum mechanics, act like a particle. And sometimes we can switch particles and holes, and a substance will act the same way!

Like time reversal symmetry, charge conjugation symmetry is described by an antiunitary operator $$C$$ that can square to $$\pm 1$$. So again we get 3 choices, listed in the chart under $$C$$ as $$1, -1$$, or $$0$$ (no charge conjugation symmetry).

So far we have $$3 \times 3 = 9$$ kinds of matter. What is the tenth kind?

Some kinds of matter have neither time reversal nor charge conjugation symmetry, but they're symmetrical under the combination of time reversal and charge conjugation! You switch particles and holes and run the movie backwards, and things look the same!

In the chart they write $$1$$ under the $$S$$ when your matter has this combined symmetry, and $$0$$ when it doesn't. So, "$$0 0 1$$" is the tenth kind of matter (the second row in the chart).

This is just the beginning of an amazing story. Since then people have found substances called topological insulators that act like insulators in their interior but conduct electricity on their surface. We can make 3-dimensional topological insulators, but also 2-dimensional ones (that is, thin films) and even 1-dimensional ones (wires). And we can theorize about higher-dimensional ones, though this is mainly a mathematical game.

So we can ask which of the 10 kinds of substance can arise as topological insulators in various dimensions. And the answer is: in any particular dimension, only 5 kinds can show up. But it's a different 5 in different dimensions! This chart shows how it works for dimensions 1 through 8. The kinds that can't show up are labelled 0.

If you look at the chart, you'll see it has some nice patterns. And it repeats after dimension 8! In other words, dimension 9 works just like dimension 1, and so on.

If you read some of the papers I listed, you'll see that the $$\mathbb{Z}$$'s and $$\mathbb{Z}_2$$'s in the chart are the homotopy groups of the ten classical series of compact symmetric spaces. The fact that dimension $$n+8$$ works like dimension $$n$$ is called Bott periodicity.

Furthermore, the stuff about operators $$T$$, $$C$$ and $$S$$ that square to 1, -1 or don't exist at all is closely connected to the classification of associative real super division algebras. So, it all fits together.

### Super division algebras

In 2005, Todd Trimble wrote a short paper on the super Brauer group and super division algebras.

In it, he gave a quick way to classify the associative real super division algebras: that is, finite-dimensional associative real $$\mathbb{Z}_2$$-graded algebras having the property that every nonzero homogeneous element is invertible. The result was known; some of the key ideas go back to here:

• C. T. C. Wall, Graded Brauer groups, Journal für die Reine und Angewandte Mathematik 213 (1964), 187–199.
But I really enjoyed Todd's effortless proof.

However, I didn't think much about how there are exactly 10 of these guys. Now this turns out to be a big deal.

3 of them are purely even, with no odd part: the usual division algebras $$\mathbb{R}, \mathbb{C}$$ and $$\mathbb{H}$$.

7 of them are not purely even. Of these, 6 are Morita equivalent to the real Clifford algebras $$\mathrm{Cl}_1, \mathrm{Cl}_2, \mathrm{Cl}_3, \mathrm{Cl}_5, \mathrm{Cl}_6$$ and $$\mathrm{Cl}_7$$. These are the super algebras generated by 1, 2, 3, 5, 6, or 7 odd square roots of -1.

Now you should have at least two questions:

• What's 'Morita equivalence'? — and even if you know, why should it matter here? Two algebras are Morita equivalent if they have equivalent categories of representations. The same definition works for super algebras, though now we look at their representations on super vector spaces ($$\mathbb{Z}_2$$-graded vector spaces). In physics, we're especially interested in representations of algebras: the algebras describe observables and symmetries, and the representations say how these act on a specific Hilbert space. So it sometimes makes sense to count two algebras as 'the same' if they're Morita equivalent. We use representations on super vector spaces to keep track of bosons versus fermions.

• 1, 2, 3, 5, 6, and 7? That's weird — why not 4? Well, $$\mathrm{Cl}_4$$ is Morita equivalent to the purely even super division algebra $$\mathbb{H}$$. So we already had that one on our list. Similarly, why not 0? Because $$\mathrm{Cl}_0$$ is just $$\mathbb{R}$$. So we had that one too. And why not 8 or more? It turns out that $$\mathrm{Cl}_{n+8}$$ is Morita equivalent to $$\mathrm{Cl}_n$$, so there no point in looking at $$n \ge 8$$.
Representations of Clifford algebras are used to describe spin-1/2 particles, so it's exciting that 8 of the 10 associative real super division algebras are Morita equivalent to real Clifford algebras.

But I've already mentioned one that's not: the complex numbers, $$\mathbb{C}$$, regarded as a purely even algebra. And there's one more! It's the complex Clifford algebra $$\mathbb{C}\mathrm{l}_1$$.

Let me explain. In general, the complex Clifford algebra $$\mathbb{C}\mathrm{l}_n$$ is defined to be the complex super algebra you get by taking the purely even algebra $$\mathbb{C}$$ and throwing in $$n$$ odd square roots of -1. So, $$\mathbb{C}\mathrm{l}_1$$ is $$\mathbb{C} \oplus \mathbb{C}e$$ where $$e$$ is an odd element with $$e^2 = -1$$.

And as soon as you think about this, you'll notice that the purely even algebra $$\mathbb{C}$$ is $$\mathbb{C}\mathrm{l}_0$$. In other words, it's the super algebra you get by taking the purely even algebra $$\mathbb{C}$$ and throwing in no odd square roots of -1.

So, in fact all 10 real super division algebras are Morita equivalent, as super algebras, to either real or complex Clifford algebras!

Even better, every real or complex Clifford algebra is Morita equivalent to a real super division algebra.

### More connections

At this point things start fitting together:

• You can multiply Morita equivalence classes of algebras using the tensor product of algebras: $$[A] \otimes [B] = [A \otimes B]$$. Some equivalence classes have multiplicative inverses, and these form the Brauer group. We can do the same thing for super algebras, and get the super Brauer group, traditionally known as the Brauer–Wall group. The super division algebras Morita equivalent to $$\mathrm{Cl}_0, \dots , \mathrm{Cl}_7$$ serve as representatives of the super Brauer group of the real numbers, which is $$\mathbb{Z}_8$$. I explained this in "week211" and further in "week212". It's a nice purely algebraic way to think about real Bott periodicity!

• As we've seen, the super division algebras Morita equivalent to $$\mathrm{Cl}_0$$ and $$\mathrm{Cl}_4$$ are a bit funny. They're purely even. So they serve as representatives of the plain old Brauer group of the real numbers, which is $$\mathbb{Z}_2$$.

• On the other hand, the complex Clifford algebras $$\mathbb{C}\mathrm{l}_0 = \mathbb{C}$$ and $$\mathbb{C}\mathrm{l}_1$$ serve as representatives of the super Brauer group of the complex numbers, which is also $$\mathbb{Z}_2$$. This is a purely algebraic way to think about complex Bott periodicity, which has period 2 instead of period 8.

Meanwhile, the purely even $$\mathbb{R}, \mathbb{C}$$ and $$\mathbb{H}$$ underlie Dyson's 'three-fold way', which I explained in detail here:

Briefly, if you have an irreducible unitary representation of a group on a complex Hilbert space $$H$$, there are three possibilities:

• The representation is isomorphic to its dual via an invariant symmetric bilinear pairing $$g : H \times H \to \mathbb{C}$$. In this case it has an invariant antiunitary operator $$J : H \to H$$ with $$J^2 = 1$$. This lets us write our representation as the complexification of a real one.

• The representation is isomorphic to its dual via an invariant antisymmetric bilinear pairing $$\omega : H \times H \to \mathbb{C}$$. In this case it has an invariant antiunitary operator $$J : H \to H$$ with $$J^2 = -1$$. This lets us promote our representation to a quaternionic one.

• The representation is not isomorphic to its dual. In this case we say it's truly complex.

In physics applications, we can take $$J$$ to be either time reversal symmetry, $$T$$, or charge conjugation symmetry, $$C$$. Studying either symmetry separately leads us to Dyson's three-fold way. Studying them both together leads to the ten-fold way!

So the ten-fold way seems to combine in one nice package:

• real Bott periodicity,

• complex Bott periodicity,

• the real Brauer group,

• the real super Brauer group,

• the complex super Brauer group, and

• the three-fold way.

I could throw 'the complex Brauer group' into this list, because that's lurking here too, but it's the trivial group, with $$\mathbb{C}$$ as its representative.

### The ten associative real super division algebras

Just for the record, here are all 10 associative real super division algebras. 8 are Morita equivalent to real Clifford algebras:

• $$\mathrm{Cl}_0$$ is $$\mathbb{R}$$. This is a purely even division algebra.

• $$\mathrm{Cl}_1$$ is $$\mathbb{C}$$ given a certain $$\mathbb{Z}_2$$-grading. It's isomorphic to the super division algebra $$\mathbb{R} \oplus \mathbb{R}e$$, where $$\mathbb{R}$$ is even and $$e$$ is an odd element with $$e^2 = -1$$.

• $$\mathrm{Cl}_2$$ is $$\mathbb{H}$$ given a certain $$\mathbb{Z}_2$$-grading. It's isomorphic to the super division algebra $$\mathbb{C} \oplus \mathbb{C}e$$, where $$\mathbb{C}$$ is even and $$e$$ is an odd element with $$e^2 = -1$$ and $$e i = -i e$$.

• $$\mathrm{Cl}_3$$ is $$\mathbb{H} \oplus \mathbb{H}$$ given a certain $$\mathbb{Z}_2$$-grading. It's isomorphic to the super division algebra $$\mathbb{H} \oplus \mathbb{H}e$$, where $$\mathbb{H}$$ is even and $$e$$ is an odd element with $$e^2 = 1$$ and $$e i = i e, e j = j e, e k = k e$$.

• $$\mathrm{Cl}_4$$ is $$\mathbb{H}$$, the algebra of 2 × 2 quaternionic matrices, given a certain $$\mathbb{Z}_2$$-grading. This is Morita equivalent to the purely even division algebra $$\mathbb{H}$$.

• $$\mathrm{Cl}_5$$ is $$\mathbb{C}$$, the algebra of 4 × 4 complex matrices, given a certain $$\mathbb{Z}_2$$-grading. This is Morita equivalent to the super division algebra $$\mathbb{H} \oplus \mathbb{H}e$$, where $$\mathbb{H}$$ is even and $$e$$ is an odd element with $$e^2 = -1$$ and $$e i = i e, e j = j e, e k = k e$$. As an algebra, this super division algebra is isomorphic to $$\mathbb{C}$$.

• $$\mathrm{Cl}_6$$ is $$\mathbb{R}$$, the algebra of 8 × 8 real matrices, given a certain $$\mathbb{Z}_2$$-grading. This is Morita equivalent to the super division algebra $$\mathbb{C} \oplus \mathbb{C}e$$, where $$\mathbb{C}$$ is even and $$e$$ is an odd element with $$e^2 = 1$$ and $$e i = -i e$$. As an algebra, this super division algebra is isomorphic to $$\mathbb{R}$$.

• $$\mathrm{Cl}_7$$ is $$\mathbb{R} \oplus \mathbb{R}$$ given a certain $$\mathbb{Z}_2$$-grading. This is Morita equivalent to the super division algebra $$\mathbb{R} \oplus \mathbb{R}e$$, where $$\mathbb{R}$$ is even and $$e$$ is an odd element with $$e^2 = 1$$. As an algebra, this super division algebra is isomorphic to $$\mathbb{R} \oplus \mathbb{R}$$.

$$\mathrm{Cl}_{n+8}$$ is Morita equivalent to $$\mathrm{Cl}_n$$, so we can stop here if we're just looking for Morita equivalence classes, and there also happen to be no more super division algebras down this road. It is nice to compare $$\mathrm{Cl}_n$$ and $$\mathrm{Cl}_{8-n}$$: there's a nice pattern here.

The remaining 2 real super division algebras are complex Clifford algebras:

• $$\mathbb{C}\mathrm{l}_0$$ is the purely even division algebra $$\mathbb{C}$$.

• $$\mathbb{C}\mathrm{l}_1$$ is the super division algebra $$\mathbb{C} \oplus \mathbb{C} e$$, where $$\mathbb{C}$$ is even and $$e$$ is an odd element with $$e^2 = -1$$ and $$e i = i e$$. As an algebra it's isomorphic to $$\mathbb{C} \oplus \mathbb{C}$$.

In the last one we could also say "with $$e^2 = 1$$": we'd get something isomorphic, not a new possibility.

### Ten kinds of matter, revisited

What does all this stuff about Clifford algebras and super division algebras have to do with Altland and Zirnbauer's classification of substances into 10 kinds? We are now in a position to see how this works!

We start with a complex Hilbert space that may (or may not) have some operators on it called $$T, C$$ and $$S$$. $$T$$ stands for time reversal, $$C$$ stands for charge conjugation, and $$S$$ stands for the combination of charge conjugation and time reversal, which may be a symmetry even when $$C$$ and $$T$$ are not both symmetries. Below we write $$= 0$$ to mean that one of these three possible symmetries is not actually a symmetry of the given kind of substance.

$$T$$ and $$C$$ are antiunitary, for physical reasons I will not explain here. When you apply them twice, they equal the identity up to a phase, since physically you get back where you started when you reverse the direction of time twice or switch particles and holes twice, but multiplying every state by the same phase has no physically detectable effect. Thus, we have $$T^2 = \alpha$$ and $$C^2 = \beta$$ for some phases (unit complex numbers) $$\alpha$$ and $$\beta$$. This implies $$\alpha T = T^2 T = T T^2 = T \alpha,$$ but because $$T$$ is antiunitary we also have $$\alpha T = T \overline{\alpha}.$$ Thus we have $$\alpha = \overline{\alpha}$$, which forces $$\alpha = \pm 1$$. In short we must have $$T^2 = \pm 1$$ whenever we our substance has time reversal symmetry. The same argument shows $$C^2 = \pm 1$$ whenever it has charge conjugation symmetry.

It follows that when we have both $$C$$ and $$T$$ as symmetries, $$S = CT$$ is a unitary operator with $$S^2 = \pm 1$$. Indeed we shall assume $$S$$ is unitary with $$S^2 = \alpha$$ for some phase $$\alpha$$ even when it's not really built as the product of symmetries $$C$$ and $$T$$. But note that since $$S$$ is unitary, if $$S^2 = \alpha$$ we can redefine $$S$$ by dividing it by $$\alpha^{1/2}$$ and get $$S^2 = 1$$. We will always do this. This does not work for $$C$$ and $$T$$ because they are conjugate-linear: for example if $$C^2 = \alpha$$ for some phase $$\alpha$$, then $$(C \alpha^{-1/2})^2$$ is still $$\alpha$$.

Next, given a unitary $$S$$ with $$S^2 = 1$$, we can always chop the complex Hilbert space it acts on as the direct sum of two subspaces: the part consisting of vectors $$v$$ with $$Sv = v$$, and the part consisting of vectors $$v$$ with $$Sv = -v$$. This is important, because it makes our complex Hilbert space $$\mathbb{Z}_2$$-graded! (When we have no operator $$S$$ we will think of our Hilbert space as purely even.)

What do the two subspaces mean? It would be nice to think of them as describing 'particles' and 'holes'.

For this interpretation to be reasonable, we want $$C$$ to map vectors with $$Sv = v$$ to vectors with $$Sv = -v$$, and vice versa. At first this seems incorrect, since $$C$$ and $$T$$ commute, for reasons I don't want to explain here. This implies that $$S = CT$$ commutes with $$C$$, so that $$C$$ maps vectors $$v$$ with $$Sv = v$$ to other vectors obeying this same equation.

But remember that when we had $$S = CT$$ with $$S^2 = -1$$ we redefined $$S$$, multiplying it by $$i$$ to achieve $$S^2 = 1$$. In this case, even though our original $$S = CT$$ commuted with $$C$$, our new $$S = iCT$$ will anticommute with it, since $$C$$ is antiunitary. So in this case $$C$$ will indeed map vectors with $$Sv = v$$ to vectors with $$Sv = -v$$, and vice versa, just as we want!

When does this happen? It happens when $$(CT)^2 = -1$$. In other words, it happens when just one of $$C$$ or $$T$$ squares to $$-1$$.

Next we'll see that substances with various kinds of symmetry are described by representations of various Clifford algebras — namely, those that are associative real super division algebras!

Let's start with the 2 super division algebras that are complex Clifford algebras:

• $$\mathbb{C}\mathrm{l}_0: \; T = 0, C = 0, S = 0$$. This is matter with no special symmetries. We describe quantum states by an ordinary complex Hilbert space, which is a representation of $$\mathbb{C}\mathrm{l}_0$$, the purely even division algebra $$\mathbb{C}$$.
• $$\mathbb{C}\mathrm{l}_1: \; T = 0, C = 0, S^2 = 1$$. This is matter with only $$S$$ symmetry: that is, a symmetry where you switch particle and holes and run time backwards. We describe quantum states of such matter with a complex Hilbert space equipped with a unitary operator $$S$$ obeying $$S^2 = 1$$. This is the same as a complex Hilbert space written as a direct sum of two other Hilbert spaces, namely the space of vectors $$v$$ with $$Sv = v$$ and the space of vectors with $$Sv = -v$$. This is the same as a representation of $$\mathbb{C}\oplus \mathbb{C}$$, which is $$\mathbb{C}\mathrm{l}_1$$.
Then let's do the 8 real ones. In these cases, unlike the last case, we can always get $$S$$ with $$S^2 = 1$$ iff we have $$C$$ and $$T$$ with $$C^2 = \pm 1, T^2 = \pm 1$$, simply by taking $$S = CT$$ if $$(CT)^2 = 1$$ and $$S = iCT$$ if $$(CT)^2 = -1$$.

• $$\mathrm{Cl}_0: \; T^2 = 1, C = 0, S = 0$$. This is matter with time reversal symmetry described by an antiunitary operator $$T$$ with $$T^2 = 1$$. A complex Hilbert space equipped with such an operator is the same as the complexification of a real Hilbert space, namely the space of vectors $$v$$ with $$Tv = v$$. A real Hilbert space is a representation of $$\mathbb{R}$$, the purely even division algebra $$\mathrm{Cl}_0$$.

• $$\mathrm{Cl}_1 : \; T^2 = 1, C^2 = -1, S^2 = 1$$. This is matter with charge conjugation symmetry and time reversal symmetry described by commuting antiunitary operators $$T,C$$ with $$T^2 = 1$$ and $$C^2 = -1$$. A complex Hilbert space equipped with an antiunitary with $$T^2 = 1$$ is the same as the complexification of a real vector space. Then, on this real vector space we have a $$\mathbb{Z}_2$$-grading, coming from $$S$$, and an odd operator $$C$$ with $$C^2 = -1$$. This is the same as a representation of the superalgebra $$\mathrm{Cl}_1$$, taking $$C$$ to be the odd square root of $$-1$$.

• $$\mathrm{Cl}_2 : \; T = 0, C^2 = -1, S = 0$$. This is matter with charge conjugation symmetry described by an antiunitary operator $$C$$ with $$C^2 = -1$$. A complex Hilbert space equipped with such an operator is the same as a complex representation of the superalgebra $$\mathrm{Cl}_2 \cong \mathbb{C} \oplus \mathbb{C}e$$, where $$\mathbb{C}$$ is even and $$e$$ is an odd element with $$e^2 = -1$$ and $$e i = -i e$$. To see this, we just set $$e = C$$ and use the usual multiplication by $$i$$ in our complex Hilbert space.

• $$\mathrm{Cl}_3 : \; T^2 = -1, C^2 = -1, S^2 = 1$$. This is matter with charge conjugation and time reversal symmetry described by commuting antiunitary operators $$T,C$$ with $$T^2 = C^2 = -1$$. A complex Hilbert space equipped with such operators is the same as representation of the superalgebra $$\mathrm{Cl}_3$$. To see this, note that $$C, iC$$ and $$iCT$$ anticommute and each square to $$-1$$.

• $$\mathrm{Cl}_4 : \; T^2 = -1, C = 0, S = 0$$. This is matter with time reversal symmetry described by an antiunitary operator $$T$$ with $$T^2 = -1$$. A complex Hilbert space equipped with such an operator is a representation of the purely even superalgebra $$\mathbb{H}$$, with $$T$$ providing the action of $$j \in \mathbb{H}$$, $$i$$ acting as usual, and $$k$$ acting as $$iT$$. In other words, it is a quaternionic Hilbert space. $$\mathrm{Cl}_4$$ is Morita equivalent to $$\mathbb{H}$$, so the category of quaternionic Hilbert spaces is equivalent to the category of representations of $$\mathrm{Cl}_4$$.

• $$\mathrm{Cl}_5: \; T^2 = -1, C^2 = 1, S^2 = 1$$. This is matter with time reversal and charge conjugation symmetry described by commuting antiunitary operators $$T,C$$ with $$T^2 = 1, C^2 = 1$$. A complex Hilbert space equipped with such operators is the same as a representation of the Clifford algebra generating by three commuting square roots of $$1$$, taking these square roots of $$1$$ to be $$C, iC$$ and $$iCT$$. But this Clifford algebra is Morita equivalent to $$\mathrm{Cl}_5$$.

• $$\mathrm{Cl}_6: \; T = 0, C^2 = 1, S = 0$$. This is matter with charge conjugation symmetry described by an antiunitary operator $$C$$ with $$C^2 = 1$$.

• $$\mathrm{Cl}_7: \; T^2 = 1, C^2 = 1, S^2 = 1$$. This is matter with time reversal and charge conjugation symmetry described by commuting antiunitary operators $$T, C$$ with $$T^2 = C^2 = 1$$. A complex Hilbert space equipped with an antiunitary with $$T^2 = 1$$ is the same as the complexification of a real vector space. Then, on this real vector space we have a $$\mathbb{Z}_2$$-grading, coming from $$S$$, and an odd operator $$C$$ with $$C^2 = 1$$. This is the same as a representation of the Clifford algebra generated by one odd element $$e$$ with $$e^2 = 1$$, namely $$\mathbb{R} \oplus \mathbb{R}e$$. But this superalgebra is Morita equivalent to $$\mathrm{Cl}_7 \cong \mathbb{R} \oplus \mathbb{R}e$$.
All this information is summarized in the following chart, where $$\simeq$$ means 'Morita equivalent to'. Click on the chart to download a PDF that explains things is a bit more detail:

### Symmetric spaces

What's the relation to the 10 classical infinite families of compact symmetric spaces? For the full answer to that, I suggest reading Gregory Moore's Quantum symmetries and compatible Hamiltonians. But if you look at this chart by Ryu et al, you'll see these families listed at right in the column "Hamiltonian":

Here's my story about how this stuff works. Each real Clifford algebra $$\mathrm{Cl}_n$$ is a '$$\ast$$-algebra'. In other words, for any $$a \in \mathrm{Cl}_n$$ there's an element $$a^\ast \in \mathrm{Cl}_n$$ such that $$(a + b)^\ast = a^\ast + b^\ast, \; (ab)^\ast = b^\ast a^\ast$$ and $$(\alpha a)^\ast = \alpha a^\ast$$ for all $$\alpha \in \mathbb{R}$$. The star operation is determined by this: each square root of -1 in $$\mathrm{Cl}_n$$, say $$e_i$$, has $$e_i^\ast = -e_i.$$ So, for example, the star operation in $$\mathrm{Cl}_1 = \mathbb{C}$$ is just the usual complex conjugation.

Any $$\ast$$-algebra gives a Lie group consisting of the "unitary" elements $$a \in A$$, meaning those with $$aa^\ast = a^\ast a = 1$$ Thus, Clifford algebras give Lie groups! By the way, these groups are not the Spin groups that Clifford algebras are famously used to construct: those are subgroups of the groups I'm talking about now.

The group of unitary elements in $$\mathrm{Cl}_n$$ depends heavily on $$n$$ mod 8, thanks to Bott periodicity. We can simplify the story in some ways by taking the limit as $$n \to \infty$$ while keeping $$n$$ the same mod 8. We then get certain infinite-dimensional Lie groups as follows:

• $$n = 0$$ 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 $$aa^\ast = a^\ast a = 1$$.

• $$n = 1$$ 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 $$aa^\ast = a^\ast a = 1$$.

• $$n = 2$$ mod 8 gives the group $$\mathrm{Sp}$$. This is the direct limit of the compact symplectic groups $$\mathrm{Sp}(n)$$, consisting of $$n \times n$$ quaternionic matrices $$a$$ with $$aa^\ast = a^\ast a = 1$$.

• $$n = 3$$ mod 8 gives the group $$\mathrm{Sp} \times \mathrm{Sp}$$.

• $$n = 4$$ mod 8 gives the group $$\mathrm{Sp}$$.

• $$n = 5$$ mod 8 gives the group $$\mathrm{U}$$.

• $$n = 6$$ mod 8 gives the group $$\mathrm{O}$$.

• $$n = 7$$ mod 8 gives the group $$\mathrm{O} \times \mathrm{O}$$.
Next, we use the fact that each Clifford algebra sits inside the next one: as an algebra, though not as a superalgebra, $$C_n$$ is just the even part of $$C_{n+1}$$. So, each of the above groups is 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!

For example, after the 7th group $$\mathrm{O} \times \mathrm{O}$$ comes the 0th group $$\mathrm{O}$$. $$\mathrm{O}(n) \times \mathrm{O}(n)$$ 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 Grassmanians $$\mathrm{O}(m+n)/(\mathrm{O}(m) \times \mathrm{O}(n))$$, 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 Lagrangian Grasmannians $$\mathrm{U}(n)/\mathrm{O}(n)$$, 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 Grasmannians $$\mathrm{Sp}(n)/\mathrm{O}(n)$$, 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 $$\mathrm{Sp}(n)$$, which consist of all $$n \times n$$ quaternionic matrices with $$aa^* = a^* a = 1$$.

• $$n = 4$$ mod 8 gives $$\mathrm{Sp}/(\mathrm{Sp} \times \mathrm{Sp})$$. This contains all the quaternionic Grassmanians $$\mathrm{Sp}(m+n)/(\mathrm{Sp}(m) \times \mathrm{Sp}(n))$$, 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 $$aa^\ast = a^\ast a = 1$$.
The other two infinite families of compact symmetric spaces come from complex Clifford algebras. Complex Clifford algebras are periodic mod 2, in the sense that $$\mathbb{C}\mathrm{l}_{n+2}$$ is isomorphic as a super algebra to $$2 \times 2$$ matrices with entries in $$\mathbb{C}\mathrm{l}_n$$. Thus, the group of unitary elements in $$\mathrm{Cl}_n$$ 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 Grassmanians $$\mathrm{U}(m+n)/(\mathrm{U}(m) \times \mathrm{U}(n))$$, 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 $$aa^\ast = a^\ast a = 1$$.

Cartan gave his 10 infinite families of compact symmetric spaces funny names: A, AI, AII, AIII, BDI, C, CI, CII, D, and DII. People like to use these even though they're a bit hard to remember. You can see how they're related to the groups $$\mathrm{O}, \mathrm{Sp}$$ and $$\mathrm{U}$$ by looking at Ryu's chart again:

### The ten associative real super division algebras, revisited

It's not hard to prove there are exactly 10 associative real super division algebras. You shouldn't be intimidated by this fact, so let's see why it's true! Suppose $$A$$ is a real super division algebra. Then its even part $$A_0$$ is closed under multiplication, and the inverse of an even element must be even, so $$A_0$$ a division algebra. One possibility is that $$A_1 = \{0\}$$, and this gives three options:
• The super division algebra with $$A_0 = \mathbb{R}$$, $$A_1 = \{0\}$$.
• The super division algebra with $$A_0 = \mathbb{C}$$, $$A_1 = \{0\}$$.
• The super division algebra with $$A_0 = \mathbb{H}$$, $$A_1 = \{0\}$$.
In short, the three real division algebras can be seen as "purely even" super division algebras.

What if $$A_1 \ne \{0\}$$? Then we can choose a nonzero element $$e \in A_1$$. Since it is invertible, multiplication by this element sets up an isomorphism of vector spaces $$A_0 \cong A_1$$. In the case $$A_0 = \mathbb{R}$$, we can rescale $$e$$ by a real number to obtain either $$e^2 = 1$$ or $$e^2 = -1$$. This gives two options:

• The super division algebra with $$A_0 = A_1 = \mathbb{R}$$ and an odd element $$e$$ with $$e^2 = 1$$.
• The super division algebra with $$A_0 = A_1 = \mathbb{R}$$ and an odd element $$e$$ with $$e^2 = -1$$.
In the case $$A_0 = \mathbb{C}$$, note that the map $$a \mapsto eae^{-1}$$ defines an automorphism of $$A_0$$, which must be either the identity or complex conjugation. If $$eae^{-1} = a$$ then $$e$$ commutes with all complex numbers, so we can rescale it by a complex number to make $$e^2$$ be any real number we want, say $$e^2 = 1$$. If $$eae^{-1} = \overline{a}$$ then $$ea =\overline{a}e$$, so $$e(e^2) = (e^2)e$$ implies that $$e^2$$ is real. Rescaling $$e$$ by a real number, we can obtain either $$e^2 = 1$$ or $$e^2 = -1$$. So, we have three options:
• The super division algebra with $$A_0 = A_1 = \mathbb{C}$$ and an odd element $$e$$ with $$ei = ie$$ and $$e^2 = 1$$.
• The super division algebra with $$A_0 = A_1 = \mathbb{C}$$ and an odd element $$e$$ with $$ei = -ie$$ and $$e^2 = 1$$.
• The super division algebra with $$A_0 = A_1 = \mathbb{C}$$ and an odd element $$e$$ with $$ei = -ie$$ and $$e^2 = -1$$.
Finally, in the case $$A_0 = \mathbb{H}$$, the map $$a \mapsto eae^{-1}$$ defines an automorphism of $$A_0$$ — which, it turns out, must be conjugation by some invertible quaternion $$q$$: $$eae^{-1} = qaq^{-1}.$$ But this means $$q^{-1}e$$ commutes with all of $$A_0$$, so replacing $$e$$ by $$qe^{-1}$$ we can assume our automorphism is the identity. In other words, now $$e$$ commutes with all of $$A_0$$, so $$e^2$$ does as well. Thus $$e^2 \in \mathbb{H}$$ is actually real. Rescaling $$e$$ by a real number we can obtain either $$e^2 = 1$$ or $$e^2 = -1$$. So, we have two options:
• The super division algebra with $$A_0 = A_1 = \mathbb{H}$$ and an odd element $$e$$ that commutes with everything in $$A_0$$ and has $$e^2 = 1$$.
• The super division algebra with $$A_0 = A_1 = \mathbb{H}$$ and an odd element $$e$$ that commutes with everything in $$A_0$$ and has $$e^2 = -1$$.
That's all!

### Ten dimensions of superstring theory

Oh yeah — what about the 10 dimensions in superstring theory? Are they really related to the ten-fold way?

It seems weird, but I think the answer is "yes, at least slightly".

Remember, 2 of the dimensions in 10d string theory are those of the string worldsheet, which is a complex manifold. The other 8 are connected to the octonions, which in turn are connected to the 8-fold periodicity of real Clifford algebra:

So the 8+2 split in string theory is at least slightly connected to the 8+2 split in the list of associative real super division algebras.

This may be more of a joke than a deep observation. After all, the 8 dimensions of the octonions are not individual things with distinct identities, as the 8 super division algebras coming from real Clifford algebras are. So there's no one-to-one correspondence going on here, just an equation between numbers.

Still, there are certain observations that it would be absurd to resist mentioning.