December 7, 2010

The Three-Fold Way (Part 2)

Posted by John Baez Last time I described some problems with real and quaternionic quantum theory — or at least, ways in which they’re peculiar compared to good old complex version of this theory.

This time I’ll tell you about the three-fold way, and you’ll begin to see how real and quaternionic Hilbert spaces are lurking in complex quantum theory.

The name ‘three-fold way’ goes back to Dyson:

• Freeman Dyson, The threefold way: algebraic structure of symmetry groups and ensembles in quantum mechanics, Jour. Math. Phys. 3 (1962), 1199–1215.

But the idea goes back much further, to a paper by Frobenius and Schur:

• F. G. Frobenius and I. Schur, Über die reellen Darstellungen der endlichen Gruppen, Sitzungsber. Akad. Preuss. Wiss. (1906), 186–208.

I’ll admit I haven’t read this paper, so I’m not quite sure what they did, but everyone cites this and mentions the ‘Frobenius–Schur indicator’ when discussing the fact that irreducible group representations come in three kinds.

And that’s what I’ll explain now. As you’ll see, the trinity of ‘real’, ‘complex’ and ‘quaternionic’ goes hand-in-hand with another famous trinity: ‘orthogonal’, ‘unitary’ and ‘symplectic’!

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:

THE THREEFOLD WAY
\begin{aligned} complex & \quad & H \ncong H^* & \quad & unitary \\ & & \\ real & & H \cong H^* & &orthogonal \\ & &J^2 = 1 \\ \\ quaternionic & & H \cong H^* & & symplectic \\ & &J^2 = -1 \end{aligned}

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:

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.

Posted at December 7, 2010 1:00 AM UTC

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Re: The Three-Fold Way (Part 2)

There’s a problem with the ‘polymathematics’ link. The http://math.ucr.edu/home/baez/Polymath.pdf after the “(” works.

Posted by: Peter Morgan on December 7, 2010 1:25 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Fixed.

Posted by: David Corfield on December 7, 2010 2:01 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Posted by: John Baez on December 7, 2010 2:41 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

A little bit of advertisement for the nLab, somewhat related to the role of symmetry in quantum physics:

topological group explains what an unitary representation of a topological group on a Hilbert space is, and

gauge group explains the concepts of local and global internal symmetry groups.

Posted by: Tim van Beek on December 7, 2010 2:50 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

It’s now realized that Dyson’s 3-fold way is part of a 10-fold way (name due to Zirnbauer), with 10=8+2 (8 coming from real K-theory and 2 from complex K-theory). So…if you’ll allude to Dyson, perhaps one should say why things aren’t more general in your 3-fold way.

Posted by: matt on December 7, 2010 10:15 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

In previous posts I listed various theorems that pick out $\mathbb{R}, \mathbb{C}$, and $\mathbb{H}$ as special — go here and follow the links to read those posts.

How about explaining to me more precisely how the 3-fold way fits into some “10-fold way”?

Posted by: John Baez on December 8, 2010 5:21 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Zirnbauer’s paper is

His ten-fold way stems from Cartan’s classification of symmetric spaces.

Shinsei Ryu and colleagues have written several papers which I don’t pretend to understand, apparently connecting Zirnbauer’s classification to the condensed-matter physics of topological insulators and to $K$-theory. Here is one in an open-access journal:

Posted by: Blake Stacey on December 8, 2010 4:34 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

The refs given by Blake are good ones. See also recent papers by Kitaev. The 10-fold way is a classification of Hamiltonians with various anti-unitary symmetries. For example, time reversal is such a symmetry, and depending whether it squares to -1 or +1 you get either Dyson’s symplectic or orthogonal ensembles. However, there are other possible anti-unitary symmetries one can impose too. Another way to think of it is that each of the classes in the 10-fold way is a linear vector space of Hamiltonians, such that if A, B, and C are in that space, then so is the double commutator [A,[B,C]]. (Of course, one could have other spaces which are closed under this double commutator by taking direct sums, but other than direct sums, these spaces are, I think, all that there is).

Posted by: matt on December 9, 2010 4:18 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Thanks, Matt and Blake! Clearly I have some reading to do. Luckily I sort of understand Cartan’s classification of symmetric spaces and its relation to Lie triple systems, which sounds like what $[A,[B,C]]$ is all about.

In a nutshell: you get a symmetric space whenever you have a Lie algebra $L$ with an automomorphism that squares to 1. This lets you give $L$ a $\mathbb{Z}/2$-grading $L = L_0 \oplus L_1$, where $L_k$ is the space on which that automorphism acts as multiplication by $(-1)^k$. If $G$ is the group with Lie algebra $L$ and $H$ is the subgroup with Lie algebra $L_0$, then $G/H$ is your symmetric space.

But the bracket in $L$ also yields a trilinear operation

$[[-,-],-] : L_1 \times L_1 \times L_1 \to L_1$

and this makes $L_1$ into something called a Lie triple system. To build your symmetric space all you need is the Lie triple system. So, there’s an analogy:

$Lie \, algebra \, : \, Lie group \, :: \, Lie \, triple \, system \, :\, symmetric \, space$

But Zirnbauer seems to be using ‘symmetric superspaces’, or (bad pun) ‘supersymmetric spaces’. Everything should work in a similar way here, and I bet Victor Kac, after classifying simple Lie supergroups, went ahead and generalized Cartan’s classification of symmetric spaces!

Posted by: John Baez on December 9, 2010 4:40 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

In case any ultra-observant readers out there noticed the switch from Matt’s $[A,[B,C]]$ to my $[[-,-],-]$, I did that because the standard definition of Lie triple system uses the operation $[[-,-],-]$, which is antisymmetric in the first two arguments. It’s no big deal but it affects the definition.

Posted by: John Baez on December 9, 2010 6:55 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

where we count a discrete group as a 0-dimensional Lie group.

Huh. What does this add? Is it reasonable to consider the product of an n-dimensional Lie group and a discrete group to be another n-dimensional Lie group?

Posted by: Aaron Denney on December 9, 2010 8:01 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

where we count a discrete group as a 0-dimensional Lie group.

Huh. What does this add? Is it reasonable to consider the product of an n-dimensional Lie group and a discrete group to be another n-dimensional Lie group?

Yes. What else would you do?

Posted by: Eugene Lerman on December 9, 2010 8:38 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Isn’t that basically what a Lie group with more than one component is?

Posted by: Tim Silverman on December 9, 2010 9:49 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Isn’t that basically what a Lie group with more than one component is?

In case this is not meant as a rhetorical question: the answer is Yes.

Posted by: Urs Schreiber on December 9, 2010 11:58 PM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

John wrote:

… where we count a discrete group as a 0-dimensional Lie group.

Aaron wrote:

It’s just a clarification. A discrete group is a 0-dimensional Lie group, because a discrete topological space is a manifold. (See extra fine print below.) But people in Lie theory often avoid counting discrete groups as Lie groups. I just wanted to reassure the reader that I’m not avoiding those.

So, for example, when people speak of the classification of simple Lie groups, they don’t include the finite simple groups — much less any infinite discrete simple groups. Indeed, you’ll see that in the study of simple Lie groups, they’re defined in a rather convoluted way, not as Lie groups lacking nontrivial normal subgroups (the obvious thing you’d guess at first), but as ‘connected non-abelian Lie groups which do not have nontrivial connected normal subgroups.’ There’s a very good reason for this: it makes the classification manageable and nice. One thing it does is eliminate 0-dimensional Lie groups from consideration. But beware: different people use different definitions of simple Lie group, depending on what they want.

So: I just wanted to say relax: I’m not eliminating discrete groups from consideration. There are interesting discrete symmetries in quantum theory, notably for systems like crystals or highly symmetric molecules like buckyballs. The results discussed here apply to those symmetries. In fact, all the results I mentioned in this particular blog entry apply not just to Lie groups but to arbitrary topological groups! But I thought physicists would enjoy the phrase ‘Lie group’ more than ‘topological group’. I wasn’t trying for maximum generality, just something easy to read.

Extra fine print: some people do not consider uncountable discrete topological spaces to be 0-dimensional manifolds, because they want every manifold to be second-countable.

Is it reasonable to consider the product of an $n$-dimensional Lie group and a discrete group to be another $n$-dimensional Lie group?

Yes: as Tim pointed out, every Lie group with $N$ components is the product of a connected Lie group and an $N$-element discrete group.

Posted by: John Baez on December 10, 2010 1:08 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Hi John.

You defined irreducible as not having proper direct summands - but as you know that’s indecomposable. Irreducible means no proper subobjects, which is why we can apply Schur’s Lemma.

Posted by: andrew hubery on December 14, 2010 9:42 AM | Permalink | Reply to this

Re: The Three-Fold Way (Part 2)

Hi! I’m talking about finite-dimensional continuous unitary representations of Lie groups. These are indecomposable iff they’re irreducible, so my definition is equivalent to the usual one in this context.

You might consider it perverse to use a nonstandard definition even in a context where it’s equivalent to the usual one. However, I’d like physicists to understand and enjoy what I’m saying. Many physicists know and love “irreducible” representations but have never heard about “indecomposable” representations — in part because quantum physics focuses on representations of the sort I’m considering, where there’s no difference. To make such physicists feel at home, I wanted to use the word “irreducible”.