## April 13, 2011

### The Three-Fold Way (Part 6)

#### Posted by John Baez

I started this tale by pointing out two ‘problems’ with real and quaternionic quantum theory:

• Apparently you can’t get observables from symmetries. More precisely: 1-parameter unitary groups on real or quaternionic Hilbert spaces don’t have self-adjoint generators, as they do in complex quantum theory.
• Apparently you can’t tensor two quaternionic Hilbert spaces. More precisely, you can’t tensor two quaternionic Hilbert spaces (which are one-sided $\mathbb{H}$-modules) and get another one.
Then I spent some time talking about how real, complex and quaternionic quantum mechanics fit together in a unified structure: the three-fold way. Today, I’d like to conclude the tale by saying how the three-fold way solves — or dissolves — these problems.

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 $\alpha_* : Hilb_\mathbb{R} \to Hilb_\mathbb{C}$ and $\beta^* : Hilb_\mathbb{H} \to Hilb_\mathbb{C} .$ 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 $H \otimes {{}_\mathbb{R}\mathbb{C}_{\mathbb{C}}}$. By abuse of notation let us still call the resulting operator $S$. Now we can write $S = i A$. 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 $c \in \mathbb{R}$, there is corresponding eigenvector with eigenvalue $-c$. This is easy to see. Suppose that $A v = c v$. By a theorem from last time, the complexification $H \otimes {{}_\mathbb{R}\mathbb{C}_{\mathbb{C}}}$ comes equipped with an antiunitary operator $J$ with $J^2 = 1$, and we have $S J = J S$. It follows that $J v$ is an eigenvector of $A$ with eigenvalue $-c$: $A J v = - i S J v = - i J S v = J i S v = - J A v = - c J v .$ (In this calculation I’ve reverted to the standard practice of treating a complex Hilbert space as a left $\mathbb{C}$-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 $S = i A$. 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 $S J = J S$, 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{aligned} \otimes : Hilb_\mathbb{R} \times Hilb_\mathbb{R} & \to & Hilb_\mathbb{R} \\ \otimes : Hilb_\mathbb{R} \times Hilb_\mathbb{H} & \to & Hilb_\mathbb{H} \\ \otimes : Hilb_\mathbb{H} \times Hilb_\mathbb{R} & \to & Hilb_\mathbb{H} \\ \otimes : Hilb_\mathbb{H} \times Hilb_\mathbb{H} & \to & Hilb_\mathbb{R} . \end{aligned} We can also tensor a real or quaternionic Hilbert space with a complex one and get a complex one: \begin{aligned} \otimes : Hilb_\mathbb{C} \times Hilb_\mathbb{R} & \to & Hilb_\mathbb{C} \\ \otimes : Hilb_\mathbb{R} \times Hilb_\mathbb{C} & \to & Hilb_\mathbb{C} \\ \otimes : Hilb_\mathbb{C} \times Hilb_\mathbb{H} & \to & Hilb_\mathbb{C} \\ \otimes : Hilb_\mathbb{H} \times Hilb_\mathbb{C} & \to & Hilb_\mathbb{C} . \end{aligned} Finally, of course we have: \begin{aligned} \otimes : Hilb_\mathbb{C} \times Hilb_\mathbb{C} & \to & Hilb_\mathbb{C} . \end{aligned}

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 $\int_G \tr(\rho(g^2)) \, d g$ where $d g$ 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. $J^2 = -1$)?

Posted at April 13, 2011 7:33 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2384

### Re: The Three-Fold Way (Part 6)

Neat! It took me a little while to figure out what your puzzle is asking, though. Are you asking whether there is some property X of endomorphisms of complex Hilbert spaces, such that for any $\mathbb{K} \in \{ \mathbb{R}, \mathbb{C}, \mathbb{H}\}$, the category of $\mathbb{K}$-Hilbert spaces is equivalent to the category of complex Hilbert spaces equipped with an endomorphism $J$ satisfying property X and such that $J^2$ is equal to the Frobenius–Schur indicator of complex representations that come from $\mathbb{K}$-representations?

In any case, the burning question in my mind after reading this is why? Why do the same three numbers $+1$, $0$, and $-1$ serve as markers for the three types of Hilbert space in the multiplication table, in the Frobenius–Schur indicator, and in the squares of structure operators?

Posted by: Mike Shulman on April 13, 2011 7:58 PM | Permalink | PGP Sig | Reply to this

### Re: The Three-Fold Way (Part 6)

Mike wrote:

It took me a little while to figure out what your puzzle is asking, though.

The egregious typo making the first sentence of my puzzle nonsensical probably didn’t help! I’ll fix that now.

Are you asking whether there is some property…

Yes. Of course for each case separately we know there’s a property X that does the job. So the fun part is finding a single nice property that handles all three cases.

In any case, the burning question in my mind after reading this is why?

I don’t know! That’s another good puzzle, that must be related to mine.

Posted by: John Baez on April 14, 2011 4:03 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 6)

This may be naive, but regarding the first point (symmetries from observables), what if you simply redefine what an observable is? In traditional QM, an observable corresponds to self adjoint operator. But this is due to the $i$ appearing in the equations.

If you absorb the $i$ into the operator, then observables correspond to skew adjoint operators and the problem disappears immediately.

Posted by: Eric Forgy on April 14, 2011 12:47 AM | Permalink | Reply to this

### Re: The Three-Fold Way (Part 6)

In the case of complex quantum theory, skew-adjoint or self-adjoint makes little difference: multiplying by $i$ turns one into the other, it doesn’t change the eigenvectors, and it multiplies the eigenvalues by $i$.

So, say I have a skew-adjoint operator $S$ on a real Hilbert space, e.g. $\mathbb{R}^n$ with its usual dot product. If we ‘measure’ it, what values should we get out? Such an operator may have no eigenvectors:

S = \left(\begin{aligned} 0 & 1 \\ -1 & 0 \end{aligned}\right)

If you solve this puzzle, there’s a good chance your solution will be essentially the same as the one I presented here.

Posted by: John Baez on April 14, 2011 3:56 AM | Permalink | Reply to this

Post a New Comment