The n-Category Café

Some aspects of quantum theory become more visible when we introduce symmetry. This is especially true when it comes to the relation between real, complex and quaternionic quantum theory. So, instead of bare Hilbert spaces, let’s consider Hilbert spaces equipped with a representation of a group. For simplicity suppose that $G$ is a Lie group, where we count a discrete group as a 0-dimensional Lie group. Let $Rep(G)$ be the category where:

• An object is a finite-dimensional complex Hilbert space $H$ equipped with a continuous unitary representation of $G$, say $\rho : G \to \U(H)$.
• A morphism is an operator $T : H \to H'$ such that $T \rho(g) = \rho'(g) T$ for all $g \in G$.

To keep the notation simple, I’ll often call a representation $\rho : G \to \U(H)$ simply by the name of its underlying Hilbert space, $H$.

Many of the operations that work for finite-dimensional Hilbert spaces also work for $Rep(G)$, with the same formal properties: for example, direct sums, tensor products and duals. This lets us formulate the three-fold way as follows.

If an object $H \in Rep(G)$ is irreducible — not a direct sum of other representations in a nontrivial way — then there are three mutually exclusive choices:

• The representation $H$ is not isomorphic to its dual. In this case we call it complex.
• The representation $H$ is isomorphic to its dual and it is real, meaning that we can get it from a representation of $G$ on a real Hilbert space $H_\mathbb{R}$: $$H = H_\mathbb{R} \otimes \mathbb{C}$$
• The representation $H$ is isomorphic to its dual and it is quaternionic, meaning that we can get it from a representation of $G$ on a quaternionic Hilbert space $H_\mathbb{H}$:
  $H =$ the underlying complex representation of $H_\mathbb{H}$

This is the three-fold way. But where do these three choices come from?

Well, suppose that $H \in Rep(G)$ is irreducible. Then there is a 1-dimensional space of morphisms $f : H \to H$, by Schur’s Lemma. Since $H \cong H^*$, there is also a 1d space of morphisms $T : H \to H^*$, and thus a 1d space of morphisms $$g : H \otimes H \to \mathbb{C} .$$ We can also think of these as bilinear maps $$g : H \times H \to \mathbb{C}$$ that are invariant under the action of $G$ on $H$.

But the representation $H \otimes H$ is the direct sum of two others: the space $S^2 H$ of symmetric tensors, and the space $\Lambda^2 H$ of antisymmetric tensors: $$H \otimes H \cong S^2 H \oplus \Lambda^2 H$$ So, either there exists a nonzero $g$ that is symmetric: $$g(v,w) = g(w,v)$$ or a nonzero $g$ that is antisymmetric: $$g(v,w) = -g(w,v)$$ One or the other, not both!—for if we had both, the space of morphisms $g : H \otimes H \to \mathbb{C}$ would be at least two-dimensional.

Either way, we can write $$g(v,w) = \langle J v, w \rangle$$ for some function $J: H \to H$. Since $g$ and the inner product are both invariant under the action of $G$, this function $J$ must commute with the action of $G$. But note that since $g$ is linear in the first slot, while the inner product is not, $J$ must be antilinear, meaning $$J(v x + w y) = J(v)x^* + J(w)y^*$$ for all $v,w \in V$ and $x,y \in \mathbb{K}$.

The square of an antilinear operator is linear. Thus, $J^2$ is linear and it commutes with the action of $G$. By Schur’s Lemma, it must be a scalar multiple of the identity: $$J^2 = c$$ for some $c \in \mathbb{C}$. We wish to show that depending on whether $g$ is symmetric or antisymmetric, we can rescale $J$ to achieve either $J^2 = 1$ or $J^2 = -1$. To see this, first note note that depending on whether $g$ is symmetric or antisymmetric, we have $$\pm g(v,w) = g(w,v)$$ and thus $$\pm \langle J v, w\rangle = \langle J w, v \rangle .$$ Now choose $v = J w$. This gives $$\pm \langle J^2 w,w\rangle = \langle J w , J w \rangle \ge 0.$$ It follows that $\pm J^2 = c$ is positive. So, if we divide $J$ by the positive square root of $c$, we get a new antilinear operator — let’s again call it $J$ — with $J^2 = \pm 1$.

Rescaled this way, $J$ is antiunitary: it is an invertible antilinear operator with $$\langle J v, J w\rangle = \langle w, v \rangle .$$ Now consider the two cases:

• If $g$ is symmetric, $H$ is equipped with a real structure: an antiunitary operator $J$ with $$J^2 = 1 .$$ It follows that $$H_{\mathbb{R}} = \{ v \in H \colon \; J v = v \}$$ is a real Hilbert space whose complexification is $H$. And since $J$ commutes with the action of $G$, there is a unitary representation of $G$ on $H_{\mathbb{R}}$ whose complexification gets us back the representation we started with! $H$.
  
  
• If $g$ is antisymmetric, $H$ is equipped with a quaternionic structure: an antiunitary operator $J$ with $$J^2 = -1 .$$ Whenever a complex Hilbert space has such a structure on it, we can make it into a quaternionic Hilbert space in exactly one way such that multiplication by $i$ is what it was before and multiplication by $j$ is the operator $J$. Let’s call this quaternionic Hilbert space $H_\mathbb{H}$. Since $J$ commutes with the action of $G$, there is a unitary representation of $G$ on $H_\mathbb{H}$ whose underlying complex representation is the one we started with!

In both these cases, $g$ is nondegenerate, meaning $$\forall v \in V \;\; g(v,w) = 0 \quad \implies \quad w = 0 .$$ The reason is that $g(v,w) = \langle J v, w \rangle$, and the inner product is nondegenerate, while $J$ is one-to-one.

We can compare real versus quaternionic representations using either $g$ or $J$. In the following statements, we don’t need the representation to be irreducible:

• Given a complex Hilbert space $H$, a nondegenerate symmetric bilinear map $g : H \times H \to \mathbb{C}$ is called an orthogonal structure on $H$. So, a representation $\rho : G \to \U(H)$ is real iff it preserves some orthogonal structure $g$ on $H$. It’s also easy to check that $\rho$ is real iff there is a real structure $J : H \to H$ that commutes with the action of $G$.
  
  
• Similarly, given a complex Hilbert space $H$, a nondegenerate skew-symmetric bilinear map $g : H \times H \to \mathbb{C}$ is called a symplectic structure on $H$. So, a representation $\rho : G \to \U(H)$ is quaternionic iff it preserves some symplectic structure $g$ on $H$. It’s also easy to check that $\rho$ is quaternionic iff there is a quaternionic structure $J : H \to H$ that commutes with the action of $G$.

This pattern is so cool that I can’t resist summarizing it in a cute little chart, suitable for printing out and putting in your wallet:

For more on how this pattern pervades mathematics, see Dyson’s paper on the three-fold way, and also Arnold’s paper on mathematical ‘trinities’, which you can easily get online:

• Vladimir I. Arnold, Symplectization, complexification and mathematical trinities, in The Arnoldfest: Proceedings of a Conference in Honour of V.I. Arnold for His Sixtieth Birthday, edited by E. Bierstone, B. Khesin, A. Khovanskii and J. E. Marsden, AMS, Providence, Rhode Island, 1999.

Also see Arnold’s paper on ‘polymathematics’ and Lieven le Bruyn’s discussion of this paper. Arnold does a pretty good job of grabbing the reader’s attention at the start here:

All mathematics is divided into three parts: cryptography (paid for by CIA, KGB and the like), hydrodynamics (supported by manufacturers of atomic submarines) and celestial mechanics (financed by military and by other institutions dealing with missiles, such as NASA).

But he gets into some very deep waters!

Next time, I’ll say more about what the three-fold way means for quantum physics.