Infinite Dimensional Lie Algebras January 31, 2005

Gabriel C. Drummond-Cole

February 7, 2005

Let me remind you what we know so far. Let me start with a harmonic oscillator, which we discussed before. There we had, classically, (p, q) ∈ ℝ^{2}, two numbers, position and momentum, and the Hamiltonian H = 1/2(p^{2} + ω^{2}q^{2}) where ω is frequency. If you do this quantum mechanically you get [P, Q] = −i, and then a = \frac{1}{\sqrt{2}}(ωQ + iP), a^{∗} = \frac{1}{\sqrt{2}}(ωQ − iP). Now [a, a^{∗}] = ω. Since now we don’t have commutativity, we replace H with a^{∗}a + ω/2 = aa^{∗} − ω/2. Instead of a particle, now we have ϕ(z) = ∑ ϕ_{n}z^{−n}, π(z) = ∑ π_{n}z^{−n} with S^{1} = {|z| = 1} = ℝ/2πℤ. So instead of two coordinates now we have countably many coordinates, twice over.

The Hamiltonian we have chosen is H = \frac{1}{2} ∫ π^{2}(z) + |ϕ′(z)|^{2}, where ∫ is either \frac{1}{2πi} ∫ dz/z or \frac{1}{2π} ∫_{0}^{2π} dθ. If you want to write things in coordinates, this is not so bad, it is \frac{1}{2}(∑ π_{n}π_{−n} + n^{2}ϕ_{n}ϕ_{−n}). This is almost positive definite; ϕ_{0} doesn’t enter here so it’s only semidefinite. This might create problems, but we will deal with them later.

To complete the picture we need analogues of the commutation relations. What will they be here? Let me answer, let me give you the answer. Suppose I quantize everything, but first let me describe the Poisson bracket. We don’t need to know exactly what it is, here it is {π(z), ϕ(w)}. So these are measuring your field at point w or the momentum of your field at point z. These are unrelated if z ≠ w. So whatever this is, it involves a delta function δ(z/w) or δ(θ − θ′) where z = e^{iθ}, w = e^{iθ′}. So then we write [π(z), ϕ(w)] = −iδ(z/w). I still don’t exactly understand what this means.

Let’s do this in a straightforward and stupid way. This is [∑ π_{n}z^{−n}, ∑ ϕ_{n}w^{−n}]. Then can we write the delta function as a power series? It has integral one over the circle and is zero except at one point. The Fourier series of this is δ(z) = ∑ δ_{n}z^{−n} where δ_{n} = ∫ δ(z)z^{n} = ∫ δ(e^{iθ})e^{inθ}, which will give you just the value of e^{inθ} at one, which is 1. So δ(z) = ∑ z^{n}.

Pretending that I am a physicist and do not care about convergence, I get that this is −i ∑(z/w)^{k}. This is very nice since I can write these two series as equal. If the power of z and power of w are not negatives of each other, the terms on the left vanish. So that gives [π_{n}, ϕ_{m}] equal to 0 if n ≠ −m, and −i if n = −m. So what do we have? What we have is the following. Forget about fields, integrals, and all that. Let me write the summary. The summary is that whatever the description, you should have a Hilbert space with operators π_{n}, ϕ_{n}, with commutation relations [π_{n}, ϕ_{−n}] = −i, and everything else commutes. We also know that the Hamiltonian should be \frac{1}{2}(∑ π_{n}π_{−n} + n^{2}ϕ_{n}ϕ_{−n}). So we want a representation of this. So this algebra effectively splits into a direct sum.

So you can write this as, you have a subalgebra generated by π_{±1}, ϕ_{±1}, with H_{1} = π_{1}π_{−1} + 1ϕ_{1}ϕ_{−1}, and then one generated by π_{±2}, ϕ_{±2}, with H_{2} = π_{2}π_{−2} + 4ϕ_{2}ϕ_{−2}, and so on. The Hamiltonian is the sum of these Hamiltonians.

This looks similarly to the original Hamiltonian we were looking at for the harmonic oscillator. A physicist would tell you that a free bosonic field is represented as a series of harmonic oscillators with positive integer frequencies. This is not exactly true because there are four operators, not two. But we will clear this up later.

To study representations of the whole thing. If you have a direct sum of Lie algebras, under reasonable assumptions the only representations of 𝔤 = 𝔤_{1} ⊕𝔤_{2} are ⊕V′_{i} ⊗V′′_{i}, where these are representations of 𝔤_{1}, 𝔤_{2}, respectively. Semisimplicity of 𝔤 would be a sufficient reasonable assumption.

We still don’t have exactly the harmonic oscillators, because we have four instead of two. So now, for every n, introduce a_{n} = \frac{1}{\sqrt{2}}(nϕ_{n} + iπ_{n}) and a_{n}^{∗} = \frac{1}{\sqrt{2}}(nϕ_{n} − iϕ_{n}).

Exercise 1 If we write a(z) = ∑ a_{n}z^{−n−1} and a_{n}^{∗}(z) = ∑ a_{n}^{∗}z̄^{−n−1} then

ϕ(z) = \frac{1}{\sqrt{2}}(∫ a(z)dz + ∫ a^{∗}(z̄)dz̄).

This should be easy, since the Fourier series makes a like the derivative of ϕ.

[in response to a question]: We’re going to have an adjoint condition that ϕ_{n}^{†} = ϕ_{−n}, and likewise for π, so that a_{n}^{†} = a_{−n}^{∗}.

So now, what do I have with this change of variables, so what?

Theorem 1 So defined, a_{n}, a_{n}^{∗} satisfy:

• [a_{n}, a_{m}^{∗}] = 0

• [a_{n}, a_{m}] = nδ_{n−m} = [a_{n}^{∗}, a_{m}^{∗}].

Let’s check the first property. [a_{n}, a_{k}^{∗}] is clearly zero unless k = −n, because subscript n commutes with everything except subscript −n. So that’s the only case to check. Then

[a_{n}, a_{−n}^{∗}] = \frac{1}{2}([nϕ_{n}, −iπ_{−n}] + [iπ_{n}, −nϕ_{−n}])

= \frac{1}{2}([iπ_{−n}, nϕ_{n}] − [iϕ_{n}, nϕ_{−n}])

= \frac{in}{2}([π_{−n}, ϕ_{n}] − [ϕ_{n}, ϕ_{−n}]) = 0

So to do the ones for a, you do the same thing.

[a_{n}, a_{−n}] = \frac{1}{2}([nϕ_{n}, iπ_{−n}] + [iπ_{n}, −nϕ_{−n}])

= \frac{1}{2}(−[iπ_{−n}, nϕ_{n}] − [iϕ_{n}, nϕ_{−n}])

= \frac{in}{2}(−[π_{−n}, ϕ_{n}] − [ϕ_{n}, ϕ_{−n}]) = −2i\frac{in}{2} = n.

At this point we say goodbye to physics and go to mathematics.

Definition 1 The Heisenberg algebra ℋ is generated by a_{n}, n ∈ ℤ, subject to the relations [a_{n}, a_{m}] = nδ_{n−m}.

If we are talking about a free massless bosonic field, such a thing is described by a pair of \underline{commuting} Heisenberg algebras, one denoted by a_{n}, the other by a_{n}^{∗}. If I forget about adjointness, all you need to do to study the representations of such a pair is to study representations of one, and then look at tensor products. The Hilbert space of the theory will be ⊕V_{i} ⊗ V̄_{i}. I’m hiding a lot, but I’m not giving a course on quantum field theory.

So from now on, I’m interested in representations of one Heisenberg algebra. As before, this Heisenberg algebra has a single part a_{0}, and then pairs a_{1}, a_{−1} and a_{2}, a_{−2}, and so on. After suitable rescaling this is the direct sum of harmonic oscillators. The easiest and only way of constructing irreducible representations of ℋ is to tensor irreducible representations of each of these together. And subject to a positive energy condition there is a unique representation. So for a_{1}, a_{−1} you take v with a_{1}v = 0, so that the representation is ℂ[a_{−1}]. For the next it is ℂ[a_{−2}].

So for the whole thing ℱ , you generate it by v_{0}, called the vacuum vector, such that a_{i}v_{0} = 0 for i > 0. How do negative ones act? As a vector space, this is ℂ[a_{−1}, a_{−2}, ⋯]v_{0}, with all of these variables commuting.

Definition 2 ℱ is called the Fock module over the Heisenberg algebra.

It is described in a very simple way. Its vector space structure is clear, and we see how a_{i} with negative indices act on it, by multiplication. How about the action of a_{i} with positive indices?

Lemma 1 1. a_{n} = n\frac{δ}{δa_{−n}} for n > 0.

2. ℱ is irreducible.

3. ℱ is graded, that is, ℱ = ⊕_{n≤0}ℱ_{n}, where deg v_{0} = 0 and deg a_{i} = i. So a_{−2}^{3}a_{−4}v_{0} has degree 3(−2) + (−4) = −10.

I should check that this is consistent. This uniquely determines the degree of all vectors in the space; now the degree of the positive ones comes from seeing how they change the grading. If I define degree in this way, then the degree of a_{i} will be i for all i. I leave this as an exercise for you.

I was not entirely careful because of a_{0}. The only irreducible representation of a one-dimensional algebra like that of a_{0} is a constant representation. So this acts by a constant, but there is no restriction, the constant can be anything. So a_{0}v = λv for some fixed λ.

Theorem 2 Any graded irreducible ℋ -module with the following positivity condition, V_{n} = 0 for n >> 0, where V_{n} is the space with grading n, is isomorphic to one of the Fock modules ℱ_{λ}.

Let me make you a picture of this Fock module. On top we have v_{0}.

\begin{matrix}0 & v_{0} & 1 \\ -1 & a_{−1} & 1 \\ -2 & a_{−2}, a_{−1}^{2} & 2 \\ -3 & a_{−3}, a_{−2}a_{−1}, a_{−1}^{3} & 3 \end{matrix}

I should stop around here. I didn’t prove the theorem, but it is similar to the other one. Take the highest degree vector, then it must be killed by a_{n} for positive n, and so on. Okay, we still have minor problems, we don’t really know, we have the choice of eigenvalue for a_{0}, which is related to the fact that the others have pairs. Next time I’ll try to tell you how you can fix that. The operator a_{0} forgot about some of the structure. We had a_{n} = nϕ_{n} + iπ_{n}, which is bad for n = 0 because ϕ_{0} enters with coefficient 0.

We might try to fix this, which we will do next time.

[What happens if you drop the positivity condition?]

You can interchange them; you could have a vector killed by positive odd and negative even i. Next time I will define the energy, the Hamiltonian, and then instead we can say, positive energy.

I will see you on Wednesday.