The ThreeFold Way (Part 3)
Posted by John Baez
Last time we classified irreducible unitary group representations into three kinds: real, complex and quaternionic. But what does this mean for physics?
Well, since elementary particles are often described using representations like this, particles must come in three kinds: real, complex and quaternionic!
Of course the details depend not just on the particle itself, but on the group of symmetries we consider. But still, it sounds pretty farout. What sort of particle is quaternionic?
This time we’ll look at the simplest example: an electron, regarded as a representation of $SU(2)$. People usually describe its state with a pair of complex numbers. But in fact, it makes a lot of sense to use a single quaternion!
We’ll see why in a while. But first, in case you fell asleep last time, let me remind you what we proved — we need it now. There are three choices for an irreducible unitary representation of a group $G$ on a complex Hilbert space $H$:

Our representation may not be isomorphic to its dual, in which case we call it truly complex.

It may be isomorphic to its dual thanks to an invariant antiunitary operator $J: H \to H$ with $J^2 = 1.$ In this case we call it real, because it’s the complexification of a representation on a real Hilbert space. And in this case there’s an invariant nondegenerate bilinear form $g : H \times H \to \mathbb{C}$ with $g(v,w) = g(w,v) ,$ also known as an orthogonal structure on $H$.
 It may be isomorphic to its dual thanks to an invariant antiunitary operator $J: H \to H$ with $J^2 = 1 .$ In this case we call it quaternionic, because it comes from a representation on a quaternionic Hilbert space. In this case there’s an invariant nondegenerate bilinear form $g : H \times H \to \mathbb{C}$ with $g(v,w) = g(w,v) ,$ also known as a symplectic structure on $H$.
This is the threefold way.
Let’s consider an example: $G = SU(2)$. In physics this group is important because its representations describe the ways a particle can transform under rotations. There is one irreducible representation for each ‘spin’ $s = 0, \frac{1}{2}, 1, \dots$ When $s$ is an integer, this representation describes the angular momentum states of a boson; when $s$ is a halfinteger (meaning an integer plus $\frac{1}{2}$) it describes the angular momentum states of a fermion.
For example, take any spin$\frac{1}{2}$ particle, and consider only its rotational symmetries. Then we can describe this particle using the obvious unitary representation of $SU(2)$ on $\mathbb{C}^2$. This is called the spin$\frac{1}{2}$ representation, and it’s quaternionic.
Why? Because we can think of a pair of complex numbers as a single quaternion, and $SU(2)$ as the group of unit quaternions. The obvious representation of $\SU(2)$ on $\mathbb{C}^2$ then turns out to be the action of unit quaternions on $\mathbb{H}$ via left multiplication. And since this commutes with right multiplication by quaternions, we’ve got a quaternionic representation!
In slogan form: qubits are not just quantum—they are also quaternionic!
More generally, all particles of halfinteger spin are quaternionic, while particles of integer spin are real — as long as we consider them only as representations of $SU(2)$.
We shall see why this is true shortly, but I can’t resist making a little table that summarizes the pattern:
$\begin{aligned} integer \; spin & bosonic & \quad & real & J^2 = 1 \\ halfinteger \; spin & fermionic & & quaternionic & J^2 = 1 \end{aligned}$
Why does it work this way?
We’ve seen it’s true for the spin$\frac{1}{2}$ representation. But we can build all the irreducible unitary representations of $SU(2)$ as symmetrized tensor powers of the spin$\frac{1}{2}$ representation. The $n$th symmetrized tensor power, $S^n(\mathbb{C}^2)$, is the spin$s$ representation with $s = n/2$. At this point a wellknown general result will help:
Theorem: Suppose $G$ is a Lie group and $H, H' \in Rep(G)$. Then:
 $H$ and $H'$ are real $\implies$ $H \otimes H'$ is real.
 $H$ is real and $H'$ is quaternionic $\implies$ $H \otimes H'$ is quaternionic.
 $H'$ is quaternionic and $H'$ is real $\implies$ $H \otimes H'$ is quaternionic.
 $H$ and $H'$ are quaternionic $\implies$ $H \otimes H'$ is real.
To see this, just notice that $H$ is real (resp. quaternionic) if and only if it can be equipped with an invariant antiunitary operator $J: H \to H$ with $J^2 = 1$ (resp. $J^2 = 1$). So, pick such an antiunitary $J$ for $H$ and also one $J'$ for $H'$. Then $J \otimes J'$ is an invariant antiunitary operator on $H \otimes H'$, and $(J \otimes J')^2 = J^2 \otimes {J'}^2 .$ This makes the result obvious.
In short, the ‘multiplication table’ for tensoring real and quaternionic representations is just like the multiplication table for the numbers $1$ and $1$… and also, not coincidentally, just like the usual rule for combining bosons and fermions!
Next, note that any subrepresentation of a real representation is real, and any subrepresentation of a quaternionic representation is quaternionic. It follows that since the spin$\frac{1}{2}$ representation of $\SU(2)$ is quaternionic, its $n$th tensor power is real or quaternionic depending on whether $n$ is even or odd, and similarly for the subrepresentation $S^n(\mathbb{C}^2)$. This is the spin$s$ representation for $s = n/2$. So, the spin$s$ representation is real or quaternionic depending on whether $s$ is an integer or halfinteger.
But what is the physical meaning of the antiunitary operator $J$ on the spin$s$ representation? In fact, it describes time reversal.
To see this, let’s call our representation $\rho : \SU(2) \to \U(H)$, where $H$ is the Hilbert space $S^n(\mathbb{C}^2)$. Choose any element $X$ in the Lie algebra $su(2)$ and let $S = d\rho(X)$. Then $J \, \exp(t S) = \exp(t S) \, J$ for all $t \in \mathbb{R}$, so differentiating gives $J S = S J$
But the operator $S$ is skewadjoint. As discussed earlier, in quantum mechanics observables are self adjoint. To turn $S$ into an observable, say $A$, we should write $S = i A$. This observable $A$ tells us the particle’s angular momentum along some axis.
Since $J$ anticommutes with $i$, we have $J A = A J$ So, for every state of our spin$s$ particle, say $v \in H$, we have a new state $J v$ where the expected value of the observable $A$ is exactly opposite: $\langle J v, A J v \rangle =  \langle J V, J A v \rangle =  \langle A v, v \rangle =  \langle v, A v \rangle$ So, the antiunitary operator $J$ reverses angular momentum! Since the timereversed version of a spinning particle is a particle spinning the opposite way, physicists call $J$ time reversal.
It seems natural that time reversal should obey $J^2 = 1$, as it does in the bosonic case; it may seem strange to have $J^2 = 1$, as we do for fermions. Shouldn’t applying time reversal twice get us back where we started? It actually does, in a crucial sense: the expectation values of all observables are unchanged when we multiply a unit vector $v \in H$ by $1$. Still, this minus sign should remind you of the equally curious minus sign that a fermion picks up when you rotate it a full turn. Are these signs related? Is there a nice way to see the connection?
It seems so, and we’ll talk about that next time…
Re: The ThreeFold Way (Part 3)
Again a little advertising for the nLab: One of the most famous “theorems” in quantum field theory (QFT) says that nature is invariant under the simultaneous change of time (T), all charges (C), and parity (P), it is often called PCT or CPT theorem.
Since there are several different frameworks, with different axioms, for quantum field theories, it would be more accurate to talk about PCT theorems instead of the PCT theorem. And sometimes physicists use the word “theorem” for a statement that is not a theorem in the strict sense of mathematics, but:
There are fully rigorous approaches to QFT where the PCT theorem is actually a theorem with a proof, more details and references can be found on the nLab:
PCT theorem