Symplectic, Quaternionic, Fermionic

John Baez

July 20, 2014

It used to confuse the bejeezus out of me that "symplectic group" was used to mean two completely unrelated things: the group of real matrices that preserve a symplectic structure, and the group of unitary quaternionic matrices.

Eventually I realized that these were not unrelated at all! They are both real forms of the same complex simple Lie group... and there really is a profound conceptual connection between symplectic structures and quaternions that's responsible for this "coincidence".

Even better, once we understand this connection, we'll see that the two basic types of particles in nature, fermions and bosons, correspond very natrually to the two kinds of Hilbert spaces other than the usual complex Hilbert spaces physicists love: namely, quaternionic and real Hilbert spaces!

To understand this stuff, we need some precise terminology. First define the groups O(n), U(n) and Sp(n) as follows:

Then define the groups Sp(2n,R), Sp(2n,C) and Sp(2n,H) as follows: Here the "Sp" also stands for "symplectic", but the meaning of "symplectic" here is — at least superficially — completely different from that in our earlier definition of the group Sp(n)!

What's the connection?

For starters, we have this fact: the intersection of U(2n) and Sp(2n,C) is isomorphic to Sp(n).

Why is this true?

Well, I'll say some stuff that implies it's true, but I won't give the proof that it's true — I'll let you work that out.

Here's how I think about this stuff. Suppose H is an irreducible unitary representation of some group. And supppose H is is isomorphic to its dual. Then we get following amazing fact:

There is a conjugate-linear isomorphism j: H → H, and either:

A)         j2 = 1


B)         j2 = -1

In case A), j is called a "real structure", since it makes H into the complexification of the real Hilbert space

{x ∈ H: jψ = ψ}

which is also a representation of our group. The point is that j has all the usual properties of complex conjugation.

In case B), j is a "quaternionic structure", since it makes H into a quaternionic Hilbert space, which is also a representation of our group. Why? Well, acting on H we have the usual complex number i, and also this operator j, satisfying i2 = j2 = -1, ij = -ji. This means we can define k = ij and get an action of the quaternions on our space H.

Case A) happens precisely when we have a nondegenerate symmetric bilinear pairing

f: H × H → C

which gives an intertwining operator from H tensor H to C. In other words, when H is equipped with an invariant "orthogonal structure".

Case B) happens precisely when we have a nondegenerate antisymmetric bilinear pairing

f: H × H → C

which gives an intertwining operator from H tensor H to C. In other words, when H is equipped with an invariant "symplectic structure".

All these facts are easy to prove — for a sketch of the proofs, try this.

This stuff is incredibly important! We've got a marvelous analogy:

unitary : orthogonal : symplectic :: complex : real : quaternionic

with (amusingly) the case of representations on complex Hilbert spaces being the basic one, and the real and quaternionic cases showing up as the two ways a representation can be isomorphic to its dual!

Now, after I wrote the above, Toby Bartels complained a little, noting that is was all very well to rhapsodize over how profound it is that

U(2n) ∩ Sp(2n,C) = Sp(n)

but then why isn't this fact equally profound:

O(2n) ∩ Sp(2n,R) = U(n) ?

In fact, this result seems quite obvious compared to the other. But if the first equation says there's a profound analogy

unitary : symplectic :: complex : quaternionic in a complex setting

then the second one should give a profound analogy

orthogonal : symplectic :: real : complex in a real setting

giving a profound relationship between symplectic structures and the complex numbers!

So, is there such a relationship?

And I replied:

There is.

Here's the proof that O(2n) ∩ Sp(2n,R) = U(n):

Take an n-dimensional complex vector space with an inner product. The real part of this inner product is an inner product on the underlying 2n-dimensional real vector space, so the group of real-linear transformations that preserve that is O(2n). The imaginary part of this inner product is a symplectic structure on the underlying 2n-dimensional real vector space, so the group of real-linear transformations that preserve that is Sp(2n,R). The transformations that preserve both lie in the intersection of these two groups, so U(n) is the intersection of O(2n) and Sp(2n,R).

I deliberately skipped one crucial little step of the proof here; I leave it as a puzzle to spot this hole and fill it in.

But the moral is clear: the cool thing about a complex inner product is that it's built from a real inner product and a symplectic structure.

It might be nicer to say the same thing with different terminology: a unitary structure is built from an orthogonal structure plus a symplectic structure.

(Here a "unitary structure" is my name for a complex inner product, and an "orthogonal structure" is my name for a real inner product.)

In previous discussions, I've rhapsodized endlessly over the profound implications this has for physics! A symplectic structure is just what the phase space of a boson has, and just what need to write down the canonical commutation relations. An orthogonal structure is just what the phase space of a fermion has, and just what we need to write down the canonical anticommutation relations. A complex Hilbert space has both! To get from one to the other, we need a compatible complex structure. And so on, and so on... I wrote a 291-page book about this, so I won't go on.

(By the way, I just gave away the answer to the puzzle I posed earlier in this post. At least if you read carefully....)

(And if you dig a wee bit deeper, you'll see what the connection between symplectic structures and complex numbers is — the one that Toby wanted.)

Then, following up on all these supposedly profound relationships, Toby asked:
"So, the reals have to do with bosons and the quaternions with fermions?"
And I replied: "I guess that's what the math gods are trying to tell us!"

But then I thought harder and discovered a nice way to make this idea precise. Fermions are indeed related to quaternions — and not just because they rhyme.

Consider the Hilbert space of a nonrelativistic spin-1/2 particle. This is a 2-dimensional complex vector space H equipped with angular momentum operators satisfying the usual commutation relations:

[J1,J2] = J3 and cyclic permutations thereof

In math lingo, what we've got here is the spin-1/2 representation of SU(2).

Now suppose we want an operator T: H → H to describe the effect of time reversal on our spin-1/2 particle. Time reversal should switch the sign of the angular momentum, so we want:

<Tψ, JiTψ> = -<ψ, Jiψ>

for every state ψ in H.

Does such an operator T exist?

It turns out no unitary operator with these properties exists. But an antiunitary operator with these properties does exist, and is unique up to phase. It satisfies T2 = -1.

Now what have we got? We've got a Hilbert space H. Acting on this we have operators we can call

i           (multiplication by i)

j = T           (time reversal)


k = ij

and it follows from what I've said that

i2 = j2 = -1


ij = -ji

so H becomes a 1-dimensional quaternionic Hilbert space!

So: spin-1/2 particles in nonrelativistic quantum mechanics are naturally quaternionic if we take time reversal into account!

The 2-dimensional complex Hilbert space of spinors, H, is secretly a 1-dimensional quaternionic Hilbert space! And the group SU(2) is secretly just Sp(1): the group of unit quaternions.

If you examine the above argument carefully, you'll see it's just a special case of the general argument I gave earlier about the relation between the symplectic group and quaternions. What's secretly going on is that the spin-1/2 representation of SU(2) is isomorphic to its dual. The general argument shows that when this happens, we get an antiunitary intertwining operator T from the representation to itself. We must have T2 = 1 or T2 = -1, depending on whether the representation is real or quaternionic. In this case T2 = -1, so the representation is quaternionic.

Now you should guess what happens with particle with other spins!

Don't calculate — guess!

Now, let me continue the tale by quoting some discussion between Toby and myself on the newsgroup sci.physics.research. Remember how I wrote

<Tψ, JiTψ> = -<ψ, Jiψ> ?

Well, figured out a quicker way to say this, namely:

T* Ji T = -Ji.

And I had to admit he was right. Why didn't I just come out and say this in the first place? Well, since T turns out to be conjugate-linear, I didn't want to freak people out by talking about its adjoint T*. Many people get a bit queasy when they start taking adjoints of conjugate-linear operators — which is a healthy reaction until you go through and check that the concept really does make sense.

By the way, here's something that's mildly amusing if you didn't notice it already: the map

T: H → H

is a conjugate-linear intertwining operator so it satisfies

T g = g T

for any element g of SU(2). It commutes with rotations! How then does it manage to reverse the sign of the angular momentum? Simple: consider rotations around the x axis, for example. These are described by the operators

g = exp(iθJ1)

so we have

T exp(iθJ1) = exp(iθJ1) T

Differentiating, we get

iJ1 T = T iJ1

So the skew-adjoint operator iJ1 commutes with T. But T is conjugate-linear, so this implies

J1 T = -T J1

or in other words

T* J1 T = -J1

The minus sign comes in from the fact that T anticommutes with i!

And of course this is completely general: if we have two unitary group representations, any conjugate-linear intertwining operator from one to the other will commute with the group action and thus commute with the skew-adjoint generators, but it will anticommute with the self-adjoint generators that physicists prefer to work with.

Now remember where I wrote: Now you should guess what happens with particles with other spins! Don't calculate — guess! Well, Toby guessed:

Since I'm not allowed to calculate, I can rattle this off quickly: If the spin is a half integer, it's just like above; if the spin is an integer, T2 = 1 is a real structure, making the Hilbert space the complexification of a real Hilbert space. Is this correct?
Right! All the representations of SU(2) are self-dual. The integer spin ones are real representations, and the half-integer ones are quaternionic.
Apparently, I'm not allowed to check.
No, I'm just testing your instincts. You've got the right instincts: bosons are real, fermions are quaternionic.

It's pretty easy to check, too:

Let's say a unitary rep H of a group G is "real" if it has a conjugate-linear intertwiner j: H → H with j2 = 1, and let's say it's "quaternionic" if it has one with j2 = -1.

By this definition, it's clear that if we tensor two quaternionic representations of a group we get a real one. Tensoring two real reps also gives a real rep. On the other hand, tensoring a real rep and a quaternionic rep gives a quaternionic rep.

Every integer-spin rep of SU(2) sits inside an even tensor power of the spin-1/2 rep, while every half-integer rep sits inside an odd tensor power of the spin-1/2 rep.

Presto! Fermions are quaternionic, bosons real.

Even better, it turns out that the same stuff applies to representations of the Poincare group: the reps corresponding to fermions are quaternionic, while the reps corresponding to bosons are real - and the operator j turns out to be nothing other than the CPT operator! For more details, check out the end of week156 of This Week's Finds. There are also fascinating connections to Toby's theory that "time is imaginary", and our discussions of quaternionic quantum mechanics, which will eventually be written up as a paper. For more on these, try Toby's papers. To dig deeper, try these:

The threefold way turns out to be part of something deeper:

Probably all these connections would have been clarified long ago, if quantum physicists had not been hampered by a prejudice in favor of complex and against real numbers. - Freeman Dyson

© 2010 John Baez