Infinite Dimensional Lie Algebras March 7, 2005

Gabriel C. Drummond-Cole

March 7, 2005

Sorry I’m late. Okay, so, anyone keeps track of lecture numbers? So, let me remind you what we are doing. We’re studying affine Lie algebra 𝔤̂ = 𝔤 ⊗ ℂ[t, t^{−1}] ⊕ ℂk ⊕ ℂd. We have [xt^{n}, yt^{m}] = [x, y]t^{m+n} +nδ_{n,−m}(x, y)k. d is the grading operator so it acts on t^{m} by m. From now on I will have 𝔤 a simple algebra, and (, ) is normalized by (α, α) = 2 for long roots.

The main result last time was 𝔤̂ = 𝔥̂ ⊕ (⊕_{α̂∈R̂}l̂agα̂) where R̂ = R̂_{re} ∪ R̂_{im} and R̂_{re} = {α + nδ, α ∈ R, n ∈ ℤ}.

I should have said 𝔥̂ = 𝔥 ⊕ ℂk ⊕ ℂd and 𝔥̂^{∗} = 𝔥^{∗} ⊕ ℂλ_{0} ⊕ ℂδ.

So we have 𝔤̂_{α+nδ} = 𝔤_{α}t^{n}, and then R̂_{im} = {nδ, n ≠ 0} so that l̂ag_{nδ} = 𝔥t^{n}.

The case we discussed was 𝔰̂𝔩(2) where you have

[figure: ⋮; ⋮; ⋮; f[1]; h[1]; e[1]; δ; f[1]; h[1]; e[1]; 𝔥^{∗}; f[1]; h[1]; e[1]; ⋮; ⋮; ⋮]

Note: R̂ generates a subspace of codimension one in 𝔥̂^{∗}, in that all have zero coefficient in front of λ_{0}. This is the adjoint representation so the central element acts by zero.

The second point is that (, )|_{𝔥^{∗}⊕ℂδ} is degenerate with kernel ℂδ.

It is positive semidefinite on 𝔥_{ℝ}^{∗} ⊕ ℝδ. It is positive definite on the left summand. This makes life a whole lot different from the finite dimensional case.

Now I’m going to try to develop the same standard theory. We studied the properties of the root system, including symmetries, generating sets, and then move to matrices or Dynkin diagrams.

The first thing would be to see whether the set of roots is invariant under some set of reflections. I’ll imitate what we did in the finite dimensional case. Whenever you have a root α, you can find an 𝔰𝔩(2) triple corresponding to it. Let’s see if we can do the same thing here.

Lemma 1 Let α̂ = α + nδ ∈ R̂_{re}. Then

1. (α̂, α̂) ≠ 0.

2. Choose eα̂ ∈ 𝔤α̂, fα̂ ∈ 𝔤_{−α̂} such that (eα̂, fα̂ = \frac{2}{(α,α)} and let hα̂ = [eâlpha, fα̂]. Then eα̂, fα̂, hα̂ satisfy the relations of 𝔰𝔩(2).

Proof.

1. (α + nδ, α + nδ) = (α, α) ≠ 0.

2. We can write eα̂ = e_{α}t^{n}, fα̂ = f_{α}t^{−n}. Then hα̂ = [eâlpha, fα̂] = [e_{α}, f_{α}] + n\frac{2}{(α,α)}k = h_{α} + n\frac{2}{(α,α)}k.

If I commute with eα̂ or fα̂ the second term gives zero. So we get [hα̂, eα̂] = [h_{α}, eα̂] = 2eα̂. Similarly check the relation for f and you get the same.

By the way, this fails if you take an imaginary root. It would not give you 𝔰𝔩(2), but instead the finite dimensional Heisenberg algebra. I’m not going to use that much. Let me think of real roots first.

Lemma 2 Define for α̂ ∈ R̂_{re}, sα̂ : 𝔥̂^{∗} → 𝔥̂^{∗} by sα̂(β̂) = β̂ − \frac{2(β̂,α̂)}{(α̂,α̂)} α̂.

Then sα̂(R̂) = R̂.

You should recall that (α̂, α̂) is never zero by the last lemma; for imaginary roots this just does not make sense.

I can do this by looking at the actions explicitly since I know what all of my roots are, but I think it is more instructive to modify the finite dimensional proof.

Proof. Consider 𝔤̂ as a module over 𝔰𝔩(2) generated by eα̂, fα̂, hα̂. If I have x ∈ 𝔤̂β̂, if I have a module over 𝔰𝔩(2) then the weights are symmetric. So x has weight equal to ⟨β̂, hα̂⟩, and it is an easy calculation, basically one we already did, that this is ⟨β, h_{α}⟩ = \frac{2(β,α)}{(α,α)}.

Okay. You remember how the proof goes for the normal finite dimensional case. If you have such a module and you start with weight λ, you can apply f and get weight −λ. This will be nonzero, so the set of weights is symmetric. Thus f n ; α̂x ≠ 0, ∈ 𝔤̂β̂−nα̂ = 𝔤̂_{sα̂(β̂)} for n = \frac{2(β̂,α̂)}{(α̂,α̂)}.

This is the same theory as in finite dimensional Lie algebras, but with a twist. I actually cheated a little bit. Can anyone see where I cheated? I said that in every 𝔰𝔩(2) module the roots are symmetric. That’s not true, it’s only true in finite dimensional modules. Unfortunately, 𝔤̂ by itself is not finite dimensional.

So now what I want to do is to split 𝔤̂ into the direct sum of finite dimensional 𝔰𝔩(2) modules. A simple argument is that the weights add. I slice my original set with slices parallel to whatever my original one is. As long as the slope is finite, each slice will be of finite size. So it is still the direct sum of finite dimensional modules.

The key argument is that this is what people call “locally finite dimensional,” so that any piece I take will be in a finite dimensional module, so I can ignore the rest and take what I want from the normal theory of finite dimensional modules over 𝔰𝔩(2). I should also have said that each of the fractions is an integer. The same thing as before but I replace “finite dimensional” with “locally finite dimensional.”

It is interesting to study what kind of group these reflections generate. They are reflections of a very funny nature.

1 Affine Weyl group

Let me start with the example of 𝔰̂𝔩(2). So 𝔥̂^{∗} = 𝔥^{∗} ⊕ ℂΛ_{0} ⊕ ℂδ, where 𝔥^{∗} = ℂα. Then 𝔥̂_{0} = 𝔥^{∗} ⊕ ℂδ.

If I take α̂ = α + nδ, then where does this map λ + aδ? We have

sα̂(λ + aδ = λ + aδ − \frac{2(λ + aδ, α + nδ)}{(α, α)}(α + nδ) =

(λ + nδ) − \frac{2(λ, α)}{(α, α)}(α + nδ) = s_{α}(λ) + (aδ − \frac{2(λ, δ)}{(α, α)}nδ),

and if I normalize so that (α, α) = 2 I get something even easier. Since the dimension is one the only reflection in 𝔥^{∗} sends λ to −λ. So this is −λ + (aδ − (λ, α)nδ).

For n = 0 we have sα̂ : λ + aδ ↦ −λ + aδ. For n = 1 it is sα̂ : λ + aδ ↦ −λ + aδ − (λ, α)δ.

Can you visualize the two dimensional linear operator at work here? It reflects in one of the directions and then shears.

If we identify 𝔥^{∗} with ℂ so that α ⟺ 2 then s_{α} : λ ↦ −λ and s_{α+δ} : λ+aδ ↦ −λ+aδ−λδ.

This is a very funny operation. So in the same example as 𝔰𝔩(2) let’s see what kind of group they generate. This is Ŵ = W_{aff}, the affine Weyl group.

Note τ = s_{α} ∘ s_{α+δ} : λ + aδ ↦ λ + aδ − λδ. Then τ^{2} : λ + aδ ↦ λ + aδ − 2λδ. From here you can see that τ is generating a group isomorphic to ℤ.

Theorem 1 For 𝔤̂ = 𝔰𝔩(2), Ŵ = ℤ_{2} ⋉ ℤ.

These come from s_{α} and τ respectively. Let’s check the commutation relations to be s_{α}τs_{α} = τ^{−1}.

Here we have a group which is infinite and generated by two reflections or a reflection and a shear, an element that acts freely. This is different than the usual Weyl groups because it includes shears.

Okay.

So more generally 𝔥̂^{∗} = ℂα ⊕ ℂΛ_{0} ⊕ ℂδ and

s_{α+nδ}(λ + kΛ_{0} + aδ) = (λ + kΛ_{0} + aδ) − (λ + kΛ_{0} + aδ, α + nδ)(α + nδ) =

λ + kΛ_{0} + aδ − ((λ, α) + kn)(α + nδ) = (λ − (λ, α)α − knα) + kΛ_{0} + (a − ((λ, α) + kn))δ.

Let me make some observations and then stop. We have not changed the Λ_{0} coefficient. Each of these affine subspaces is preserved by the action. Even in the pure finite dimensional part I don’t get the usual reflection. I get the usual reflection but also a knα term which is not linear in λ, so a translation. Also there is a rather complicated action in the third direction.

Actually this is not really so messy. Let me tell you one thing.

How Ŵ acts on 𝔥̂_{k}^{∗} = {λ+kΛ_{0}+aδ}. The statement is that each of these preserves it. How does s_{α} act? s_{α} : λ+kΛ_{0} +aδ ↦ −λ+kΛ_{0} +aδ. If n = 1 I get s_{α+δ} : λ+kΛ_{0} +aδ ↦ −λ+2k+· · · . So on the finite dimensional part, I get that s_{α+δ} is reflection around k instead of 0. So τ : λ + ⋯ ↦ λ + 2k + ⋯

So τ generates again a group isomorphic to ℤ. So either you say it’s generated by two reflections, or by a reflection and a translation.

Next time I will discuss this group in full generality. It will include the ordinary Weyl group but also something corresponding to the translations or shears.