The n-Category Café

We’ve seen how on a real, complex or quaternionic Hilbert space, any continuous one-parameter unitary group has a skew-adjoint generator $S$. In the complex case we can write $S$ as $i$ times a self-adjoint operator $A$, which in physics describes a real-valued observable. Alas, in the real or quaternionic cases we cannot do this.

However, as we saw last time, real and quaternionic Hilbert spaces can be seen as a complex Hilbert space with extra structure. This solves this problem. Indeed, we’ve seen faithful embeddings $${\alpha }_{*}:{\mathrm{Hilb}}_{ℝ}\to {\mathrm{Hilb}}_{ℂ}$$ and $${\beta }^{*}:{\mathrm{Hilb}}_{ℍ}\to {\mathrm{Hilb}}_{ℂ}.$$ And these are ‘dagger-functors’, so they send skew-adjoint operators to skew-adjoint operators.

So, given a skew-adjoint operator $S$ on a real Hilbert space $H$, we can apply the functor ${\alpha }_{*}$ to reinterpret it as a skew-adjoint operator on the complexification of $H$, namely $H\otimes {}_{ℝ}{ℂ}_{ℂ}$. By abuse of notation let us still call the resulting operator $S$. Now we can write $S=iA$. But resulting self-adjoint operator $A$ has an interesting feature: its spectrum is symmetric about $0$!

In the finite-dimensional case, all this means is that for any eigenvector of $A$ with eigenvalue $c\in ℝ$, there is corresponding eigenvector with eigenvalue $-c$. This is easy to see. Suppose that $Av=cv$. By a theorem from last time, the complexification $H\otimes {}_{ℝ}{ℂ}_{ℂ}$ comes equipped with an antiunitary operator $J$ with $J^2=1$, and we have $SJ=JS$. It follows that $Jv$ is an eigenvector of $A$ with eigenvalue $-c$: $$AJv=-iSJv=-iJSv=JiSv=-JAv=-cJv.$$ (In this calculation I’ve reverted to the standard practice of treating a complex Hilbert space as a left $ℂ$-module.) In the infinite-dimensional case, we can make an analogous but more subtle statement about the continuous spectrum.

Similarly, given a skew-adjoint operator $S$ on a quaternionic Hilbert space $H$, we can apply the functor ${\beta }^{*}$ to reinterpret it as a skew-adjoint operator on the underlying complex Hilbert space. Let us again call the resulting operator $S$. We again can write $S=iA$. And again, the spectrum of $A$ is symmetric about $0$. The proof is the same as in the real case: now, by another theorem from last time, the underlying complex Hilbert space $H$ is equipped with an antiunitary operator $J$ with $J^2=-1$, but we again have $SJ=JS$, so the same calculation goes through.

So, that takes care of one problem . The second ‘problem’ is that we cannot take the tensor product of two quaternionic Hilbert spaces and get another quaternionic Hilbert spaces.

But we’ve seen that there’s a strong analogy between bosons and real Hilbert spaces, and between fermions and quaternionic Hilbert spaces. This makes it seem rather odd to want to tensor two quaternionic Hilbert space and get another one. Just as two fermions make a boson, the tensor product of two quaternionic Hilbert spaces is naturally a real Hilbert space.

After all, we saw last time that a quaternionic Hilbert space can be identified with a complex Hilbert space with an antiunitary $J$ such that $J^2=-1$. If we tensor two such spaces, we get a complex Hilbert space equipped with an antiunitary $J$ such that that $J^2=1$. And we’ve seen that this can be identified with a real Hilbert space.

Further arguments of this sort give four tensor product functors: $$\begin{array}{rlr}\otimes :{\mathrm{Hilb}}_{ℝ}\times {\mathrm{Hilb}}_{ℝ}& \to & {\mathrm{Hilb}}_{ℝ}\\ \otimes :{\mathrm{Hilb}}_{ℝ}\times {\mathrm{Hilb}}_{ℍ}& \to & {\mathrm{Hilb}}_{ℍ}\\ \otimes :{\mathrm{Hilb}}_{ℍ}\times {\mathrm{Hilb}}_{ℝ}& \to & {\mathrm{Hilb}}_{ℍ}\\ \otimes :{\mathrm{Hilb}}_{ℍ}\times {\mathrm{Hilb}}_{ℍ}& \to & {\mathrm{Hilb}}_{ℝ}.\end{array}$$ We can also tensor a real or quaternionic Hilbert space with a complex one and get a complex one: $$\begin{array}{rlr}\otimes :{\mathrm{Hilb}}_{ℂ}\times {\mathrm{Hilb}}_{ℝ}& \to & {\mathrm{Hilb}}_{ℂ}\\ \otimes :{\mathrm{Hilb}}_{ℝ}\times {\mathrm{Hilb}}_{ℂ}& \to & {\mathrm{Hilb}}_{ℂ}\\ \otimes :{\mathrm{Hilb}}_{ℂ}\times {\mathrm{Hilb}}_{ℍ}& \to & {\mathrm{Hilb}}_{ℂ}\\ \otimes :{\mathrm{Hilb}}_{ℍ}\times {\mathrm{Hilb}}_{ℂ}& \to & {\mathrm{Hilb}}_{ℂ}.\end{array}$$ Finally, of course we have: $$\begin{array}{rlr}\otimes :{\mathrm{Hilb}}_{ℂ}\times {\mathrm{Hilb}}_{ℂ}& \to & {\mathrm{Hilb}}_{ℂ}.\end{array}$$

In short, the ‘multiplication’ table of real, complex and quaternionic Hilbert spaces matches the usual multiplication table for the numbers $+1,0,-1$!

This is in fact nothing new: it goes back to a very old idea in group theory: the Frobenius–Schur indicator, which was developed in a joint paper by these two authors in 1906.

(If you were at Oxford when I gave my lecture on this stuff, you’ll probably remember that after my talk, or maybe during it, someone we all know and love raised his hand and said “Isn’t this just the Frobenius-Schur indicator?” And I said something like “yes”.)

What’s the Frobenius–Schur indicator? It’s a way of computing a number from an irreducible unitary representation of a compact group $G$ on a complex Hilbert space $H$, say $$\rho :G\to U(H)$$ where $U(H)$ is the group of unitary operators on $H$. Any such representation is finite-dimensional, so we can take the trace of the operator $\rho (g^2)$ for any group element $g\in G$, and then perform the integral $${\int }_{G}tr(\rho (g^2))dg$$ where $dg$ is the normalized Haar measure on $G$. This integral is the Frobenius–Schur indicator. It always equals $1$, $0$ or $-1$, and these three cases correspond to whether the representation $\rho$ is ‘real’, ‘complex’ or ‘quaternionic’ in the sense we’ve explained before.

The moral, then, is not to fight the patterns of mathematics, but to notice them and follow them.

Puzzle. - Is the category of real (resp. complex, resp. quaternionic) Hilbert spaces equivalent to the category of complex Hilbert spaces $H$ equipped with a operator $J:H\to H$ having some nice property and obeying the equation $J^2=+1$ (resp. $J^2=0$, resp. $J^2=-1$)?