Infinite Dimensional Lie Algebras May 2, 2005

Gabriel C. Drummond-Cole

May 2, 2005

Today I will do somewhat simpler stuff than what we did last time. I’ll go back to the character formula we had for integrable modules. We’ll talk about the MacDonald identities and the η function. Let me go back to the character formula. Recall that for L_{Λ} a highest weight module, we have the following formula for the characters, ch L_{Λ} = \frac{∑Ŵ e^{w.Λ}(−1)^{ℓ(w)}}{∏_{α̂∈R̂_{+}} (1−e^{−α̂})} .

Here w.Λ = w(Λ + ρ̂) − ρ̂ is the twisted action, where ρ̂ = ρ + h^{√}Λ_{0}.

I want to go back to the formula and consider the special case Λ = 0. We know the integrable module with highest weight 0; this is L_{Λ} = ℂ. The left hand side is one, so you get 1 = \frac{∑ e^{w.0}(−1)^{ℓ(w)}}{∏(1−e^{−α̂})} . Let’s see whether we can get something interesting out of this. We can already get that a product equals a sum, which is nice. But let me be more careful. First of all let me deal with the numerator. ∑Ŵ e^{w.0}(−1)^{ℓ(w)}. But Ŵ splits as the semidirect product

Ŵ = W ⋉ Q^{√}, where t_{α^{√}}(λ + kΛ_{0}) = λ + kα^{√} + kΛ_{0} − (⟨λ, α^{√}⟩ + k^{\frac{(α^{√},α^{√})}{2}})δ.

Let’s wee if I can use this. Instead of having a sum over Ŵ , I can have a double sum over W and Q^{√}. I should also note that ℓ(w) is even for all lattice elements. So the double sum is

∑_{w∈W} ∑_{α^{√}∈Q^{√}} e^{w.(t_{α^{√}} .0)}(−1)^{ℓ(w)}

So what is t_{α^{√}}.0? That’s

t_{α^{√}}(ρ̂) − ρ̂ = (ρ + h^{√}α^{√} + h^{√}Λ_{0} − (⟨ρ, α^{√}⟩ + h^{√} \frac{(α^{√}, α^{√})}{2})δ) − ρ + h^{√}Λ_{0}

= h^{√}α^{√} − \frac{1}{2h^{√}} (|ρ + h^{√}α^{√}|^{2} − (ρ, ρ))δ = h^{√}α^{√} − (\frac{|ρ_{h}^{√}α^{√}|^{2}}{2h^{√}} − \frac{|ρ|^{2}}{2h^{√}})δ.

The δ part of this is invariant under the W action, so the double sum becomes

∑_{α^{√}∈Q^{√}} e^{−(^{\frac{|ρ{√ \atop h}α^{√}|^{2}}{2h^{√}}} − \frac{|ρ|^{2}}{2h^{√}})δ} ∑_{w∈W} e^{w.(h^{√}α^{√})}(−1)^{ℓ(w)}

= ∑_{Q^{√}} q^{^{\frac{|ρ{√ \atop h}α^{√}|^{2}}{2h^{√}}} − \frac{|ρ|^{2}}{2h^{√}}} ∑_{w∈W} e^{w.(h^{√}α^{√})}(−1)^{ℓ(w)}

Why is this any better? In the bottom we have the product over all positive affine roots, which comprise two kinds of roots, the usual positive roots in the finite dimensional part with no component in δ, and then the whole thing with positive δ component; R̂_{+} = R_{+} ∪ ⋃_{n≥1}{α + nδ}.

So I have, moving part of this to the left hand side but leaving the rest as a denominator,

∏_{R̂_{+}\R_{+}}(1 − e^{−α̂}) = \frac{∑_{Q^{√}} q^{^{\frac{|ρ{√ \atop h}α^{√}|^{2}}{2h^{√}}} − \frac{|ρ|^{2}}{2h^{√}}} ∑_{w∈W} e^{w.(h^{√}α^{√})}(−1)^{ℓ(w)}}{∏_{R_{+}}(1 − e^{−α})} .

The second “half” of the right hand side is the usual finite dimensional character formula, and so the right hand side can be rewritten as

∑_{Q^{√}} q^{^{\frac{|ρ{√ \atop h}α^{√}|^{2}}{2h^{√}}} − \frac{|ρ|^{2}}{2h^{√}}} ch L_{h^{√}α^{√}}

This L is a module over 𝔤, not 𝔤̂.

Let me now do the following. Both sides are formal combinations of exponentials like q^{−n} and e^{α}. Let me replace each e^{α} by 1. If you view the α as functions on a space, allow me to say this is an evaluation at h = 0, so that e^{α} = 1.

If I take the character, replacing every exponential by 1, I get ∑ e^{λ} dim V [λ] →^{e^{λ}=1} dim V.

The left hand side does what? At level one I get (1 − e^{α} − δ)(1 − e^{−δ})(1 − e^{−α−δ}). So I get for each level under evaluation (1 − e^{−nδ})^{3}, so in general I get, where r is the dimension of the Cartan, (1 − q^{n})^{|R|+r}, but this exponent is just dim 𝔤.

So this equation is now, after applying the strange formula \frac{|ρ|^{2}}{2h^{√}} = \frac{dim 𝔤}{24}, the Madonald η function identities:

(q^{\frac{1}{24}} ∏_{n≥1}(1 − q^{n}))^{dim 𝔤} = ∑_{Q^{√}} q^{\frac{|ρ{√ \atop h}α^{√}|^{2}}{2h^{√}}} dim L_{h^{√}α^{√}}.

Let’s do an example. For 𝔰𝔩(2), let’s check the strange identity. We have ρ = \frac{α}{2}, and then |ρ|^{2} = \frac{|α|^{2}}{4} = \frac{1}{2}. Now h^{√} = 2 so we get \frac{|ρ|^{2}}{2h^{√}} = \frac{1}{8} = \frac{dim 𝔰𝔩_{2}}{24} .

The base on the left side is called Dedekind’s η function.

So now we get L_{h^{√}+kα} = L_{2+kα} = V_{4k} so the dimension is 4k + 1.

Next, η^{3} = ∑_{k∈ℤ} q^{\frac{| \frac{α}{2} +2kα|^{2}}{4}} (4k + 1). This exponent is a quarter of |\frac{4k+1}{2}α|^{2} = \frac{(4k+1)^{2}}{4} |α|^{2} =

\frac{(4k+1)^{2}}{2} . So this right hand side is ∑_{ℤ} q^{(4k+1)^{2}}8(4k + 1).

Let me conclude with this. There was a famous paper by the physicist Dyson, entitled “Missed opportunities.” It is not a paper about math or physics, but more the philosophy of them. He was working in quantum field theory, so knew the representation theory of Lie algebras. He was also interested in modular forms and the like. At some moment, he tried to see if he could get nice formulas for the powers of η functions. It was known how to do it as a sum. You could do the cube purely combinatorially and get this equation too, due to Jacobi. He was curious about other powers of the η function. You can always do it; sometimes it will be short and evident, and other times it will be long and messy.

It is not a mathematical criterion, but he decided to look at which ones were simpler and which more complicated. Physicists are never afraid of explicit calculation. He got the answer that nice answers exist for n = 1, 3, 8, 10, 14, 15, and he was quite lost. He got at least up to twenty. He tried to see if there was a pattern; he was unable. Some time later, after MacDonald proved the η function identities, he realized these were the dimensions of the simple Lie algebras. These correspond to the Heisenberg algebra for the first case, then 𝔰𝔩_{2}, 𝔰𝔩_{3}, 𝔰𝔬_{5}, G_{2}, and 𝔰𝔩_{4}. He was embarrassed because he had never considered whether these combinatorial numbers might be the dimensions of simple Lie algebras.

I think I’ll stop here. Next time I’ll probably, we have one more week of class. I don’t know the best way to spend it. I’ll try to say a little bit about the fusion product. I wouldn’t be able to do it in any great detail. To give the definition, I can do it in two classes. That’s it for today.