The Tenfold Way
John Baez
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:
 Todd Trimble,
The
super Brauer group and super division algebras, April 27, 2005.

Alexei Kitaev,
Periodic table for topological insulators and superconductors,
AIP Conference Proceedings 1134 (2009), 22.

Shinsei Ryu, Andreas P. Schnyde, Akira Furusaki and
Andreas Ludwig, Topological insulators and
superconductors: tenfold way and dimensional hierarchy,
New J. Phys. 12 (2010), 065010.

Daniel S. Freed and Gregory W. Moore
Twisted equivariant
matter, Ann. Henri Poincaré 14 (2013),
1927–2023.

Gregory W. Moore,
Quantum symmetries and compatible Hamiltonians, December 15, 2013.
I've written a terse summary of the simplest mathematical ideas:

John Baez, The tenfold way,
Notices Amer. Math. Soc. 67 10 (2020), 1599–1601.
I've also given a talk on this stuff, and you
can see my slides and watch a video. But that talk doesn't cover everything
in 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 tenfold 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 linedup 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 3dimensional topological
insulators, but also 2dimensional ones (that is, thin films) and even
1dimensional ones (wires). And we can theorize about
higherdimensional 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 1996, as part of an Institute for Advanced Studies program on
quantum field theory, Pierre Deligne wrote some notes
on spinors which included a classification of the associative
real super division algebras: that is, finitedimensional
associative real \(\mathbb{Z}_2\)graded algebras having the property
that every nonzero homogeneous element is invertible. Later
this classification was very tersely handled in Todd Trimble's paper
on the super
Brauer group and super division algebras. Some of the key ideas
are much older:

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.
At the time 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 spin1/2
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 'threefold 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 threefold way.
Studying them both together leads to the tenfold way!
So the tenfold 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 threefold 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. In
fact, all 8 of these are real Clifford algebras if we expand
the definition a bit. I've talked about \(\mathrm{Cl}_n\), which is
the free superalgebra on \(n\) anticommuting square roots of \(1\).
But let's define \(\mathrm{Cl}_{n}\) to be the free superalgebra on
\(n\) anticommuting square roots of \(+1\). This is nice because with
this definition it's still true that adding 8 to the index of our
Clifford algebra gives a Morita equivalent superalgebra. We write
this as
$$ \mathrm{Cl}_{n+8} \simeq \mathrm{Cl}_n. $$
And we have these 8 super division 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 \(e i = i e, e j = j e, e k = k e\).

Both \(\mathrm{Cl}_4\) and \(\mathrm{Cl}_{4}\) are isomorphic
to \(\mathrm{M}_2(\mathbb{H}\), the algebra of 2 × 2
quaternionic matrices, given a certain \(\mathbb{Z}_2\)grading. This
is Morita equivalent to the purely even division algebra \(\mathbb{H}\).
 \(\mathrm{Cl}_{3}\) is \(\mathrm{M}_2(\mathbb{C})\), the algebra
of 2 × 2 complex matrices, given a certain
\(\mathbb{Z}_2\)grading. This is 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 \(e i = i e, e
j = j e, e k = k e\).

\(\mathrm{Cl}_{2}\) is \(\mathrm{M}_2(\mathbb{R})\), the algebra of 4 ×
4 real matrices, given a certain \(\mathbb{Z}_2\)grading. This is
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 \(e i = i e\).

\(\mathrm{Cl}_{1}\) is \(\mathbb{R} \oplus \mathbb{R}\) given a certain
\(\mathbb{Z}_2\)grading. This is isomorphic to the super
division algebra \(\mathbb{R} \oplus \mathbb{R}e\), where \(\mathbb{R}\) is
even and \(e\) is an odd element with \(e^2 = 1\).
Since \(\mathrm{Cl}_{n+8}\) is Morita equivalent to \(\mathrm{Cl}_n\)
both for positive and negative \(n\), 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.
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 conjugatelinear:
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 \(\mathrm{Cl}_{3}\) generating
by three anticommuting 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 \(C^2 = 1\). A superHilbert space with an odd antiunitary \(C\)
with \(C^2 = 1\) is the same as a representation of the Clifford algebra
\(\mathrm{Cl}_{2}\) generated by two anticommuting square roots of
\(1\), taking these to be \(C\) and \(iC\). But this Clifford algebra
is Morita equivalent to \(\mathrm{Cl}_6\).
 \(\mathrm{Cl}_7: \; T^2 = 1, C^2 = 1, S^2 = 1\). 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 \(\mathrm{Cl}_{1}\)
generated by one odd square root of 1. But this Clifford algebra is
Morita equivalent to \(\mathrm{Cl}_7\).
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 infinitedimensional 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 infinitedimensional real subspaces of an infinitedimensional
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 infinitedimensional 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 tenfold 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 8fold
periodicity of real Clifford algebra:

John Baez and John Huerta,
Division algebras and
supersymmetry I, in Superstrings, Geometry, Topology, and
C*Algebras, eds. Robert Doran, Greg Friedman and Jonathan
Rosenberg, Proc. Symp. Pure Math.
81, AMS, Providence, 2010, pp. 65–80.

John Baez and John Huerta,
Division algebras and
supersymmetry II, Adv. Math. Theor. Phys. 15 (2011),
1373–1410.
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 onetoone correspondence going on here,
just an equation between numbers.
Still, there are certain observations that it would be absurd to resist
mentioning.
© 2020 John Baez
baez@math.removethis.ucr.andthis.edu