Infinite Dimensional Lie Algebras April 6, 2005

Gabriel C. Drummond-Cole

April 6, 2005

So let me begin, let me start as usual by reminding what we’ve done so far. We have this category 𝒪 which is roughly the highest weight modules. This contains the Verma modulees M_{Λ} and their irreducible quotients L_{Λ}. We also discussed integrable modules 𝒪^{int}, where dim(U𝔰𝔩(2)_{i})v < ∞. So far we don’t really know many examples. For finite dimensional Lie algebras, these are finite dimensional highest weight modules.

We don’t know any infinite dimensional ones, but we have proved a couple of things. If V ∈ 𝒪^{int} then P (V) ⊂ P = {Λ ∈ 𝔥̂^{∗}|⟨α^{√}, λ⟩ ∈ ℤ}, i.e., the roots are integrable.

Also the set with multiplicity is invariant under the affine Weyl group, Ŵ (P (V)) = P (V). But this doesn’t answer the basic question, how do we construct such modules? Today I will classify all such, but first I want to give you a technical lemma.

Definition 1 If V ∈ 𝒪 then v ∈ V is smooth if for all i, dim(U𝔰𝔩(2)_{i})v < ∞.

Lemma 1 v ∈ V is smooth implies f_{j}v is smooth for j = 0, …, r.

This implies that the set of smooth vectors is actually a subrepresentation. Proof. Recall (ad f_{i})^{N}f_{j} = 0. for some appropriate N. This can be rewritten in some nicer way, f_{i}^{N}f_{j} − \binom{N}{1}f_{i}^{N−1}f_{j}f_{i} + ⋯ + f_{j}f_{i}^{N} = 0.

Let me go through this for N = 2. The statement (ad f_{i})^{2}f_{j} = 0 is [f_{i}, [f_{i}, f_{j}]] = 0. So this is f_{i}(f_{i}f_{j} − f_{j}f_{i}) − (f_{i}f_{j} − f_{j}f_{i})f_{i} = 0. So these will be sums with one f_{j} and N f_{i}, and the only problem is matching the coefficients. Here this is f_{i}^{2}f_{j} − 2f_{i}f_{j}f_{i} + f_{j}f_{i}^{2} = 0. I don’t really need to show you how to prove the general thing by induction.

So how does this help to prove the result? Consider U𝔰𝔩(2)_{i}v, which is finite dimensional, spanned by the basis v_{1}, …, v_{ℓ}. Then f_{j}v ∈ ⟨f_{j}v_{a}, f_{i}f_{j}v_{a}, f_{i}^{2}f_{j}v_{a}, …⟩. Where do we stop? The last one is …, f_{i}^{N−1}f_{j}v_{a}⟩_{a=1,...,ℓ} Denote this vector space by W. Then I claim that this is invariant under 𝔰𝔩(2)_{i}.

1. If I apply e_{i} to f_{i}^{m}f_{j}v_{a}, I get ∑ h … f_{i}^{m−1}f_{j}v_{a}, where the h ∈ 𝔥̂, and another term f_{i}^{m}f_{j}e_{i}v_{a}. Since we assumed that v_{a} was a basis, we can write e_{i}v_{a} as a linear combination of v_{b}.

2. What about f? We have f_{i}f_{i}^{m}f_{j}v_{a} is either f_{i}^{m+1}f_{j}v_{a} if m < N−1, or ∑(−1)^{k+1}\binom{N}{k}f_{i}^{N−k}f_{j}f_{i}^{k}.

So the Serre relation allows you to stop at some point.

It’s a little technical proof, but it is not too bad. We ddn’t use anything but the Serre relations. We could thus repeat the proof for any Kac-Moody algebra. Now that we have the lemma, things will be much easier. Now we can formulate the main theorem of today.

I’m going to classify irreducible integrable modules. We know that irreducibles are L_{Λ}, and the only problem is to find which ones are integrable. In the finite dimensional case, it was just when the highest weight was a dominant integral weight.

Theorem 1 L_{Λ} is integrable if and only if Λ ∈ P̂_{+} = {Λ ∈ 𝔥̂^{∗}|⟨Λ, α_{i}^{√}⟩ ∈ ℤ_{+}}.

Corollary 1 Isomorphism classes of irreducible integrable modules are thus in one to one correspondence with P̂_{+}.

Okay, this theorem will give us the complete classification of irreducible integrable modules just as we got the classification in the finite dimensional case for irreducible finite dimensional modules.

The proof in one direction is quite obvious. Assume L_{Λ} is integrable. Among other things, the highest weight vector v_{Λ} generates a finite dimensional 𝔰𝔩(2)_{i}-module. Since it’s a highest weight vector, e_{i}v_{Λ} = 0. You have a vector killed by e and require that it is finite dimensional, when is this possible? The well-known answer is that this is possible only if ⟨Lambda, h_{i}⟩ ∈ ℤ_{+}. This is precisely the statement we wanted to show.

The tricky part is proving in the opposite direction. First of all, even for 𝔰𝔩(2), you might have a highest weight vector of integral weight, and you will be infinite dimensional. In M_{Λ}, this is not integrable. So we need to use irreducibility.

Assume that ⟨Λ, α_{i}^{√}⟩ = N_{i} ∈ ℤ_{+}. Consider this vector, and apply to it f_{i} raised to the power N_{i} −1. You can probably see why I am using this. In 𝔰𝔩(2), say I have a highest weight vector of weight N. Then M_{Λ} and L_{Λ} look like

\xymatrix{f^{N+1}v & f^{N}v \ar[l] & ⋯ & f^{2}v \ar[l] & fv & v \ar[l] \\ f^{N+1}v & f^{N}v & ⋯ & f^{2}v \ar[l] & fv & v \ar[l]}

We get a similar picture for 𝔰𝔩(3) :

f_{1}^{N+1}v_{λ}  v_{λ} f_{2}^{N+1}v_{λ}

So e_{j}f_{i}^{N_{i}+1}v_{Λ}) = 0, including j = i. Thus weights of the U𝔤̂-submodule generated by f_{i}^{N_{i}+1}v_{Λ} will be of the form Λ−(N_{i} +1)α_{i} −Q_{+} so this submodule doesn’t contain v_{Λ}. This is already a remarkable thing, since a vector chosen at random generates the whole thing.

Now, let V_{Λ} = M_{Λ}/(f_{i}^{N_{i}+1}v_{Λ})_{i=0,...,r}. Okay, so consider this.

My first observation is that it is nontrivial, exactly because the highest weight vector won’t be in the submodule being quotiented out. My second argument is that since this is the quotient by the largest possible submodule, then in particular L_{Λ}, let me write, L_{Λ} = M_{Λ}/ maximal proper submodule, so L_{Λ} = V_{Λ}/?, maybe some further modules. Later if time permits I will show that this is an equality. In L_{Λ}, all we need to know for now is that all of the vectors f_{i}^{N+1}v_{Λ} = 0. Every vector living in a proper submodule of M_{Λ} is zero in L_{Λ}.

Now I am done because I claim, it is easy to see that v_{Λ} ∈ V_{Λ} is a smooth vector. If you want to apply to the highest weight vectors f_{i}, then you get only a finite dimensional space. By the lemma, which was actually the key step, all of what you get by applying the f_{i} is also smooth, so since the whole module is generated by applying f to v_{Λ}, we’re done. Every vector is smooth so the module is integrable. This gives the classification of the finite dimensional irreducible modules over finite dimensional semisimple Lie algebras, something which I stated but did not prove last semester.

Maybe I have five minutes, let me try to explain why you have an equality L_{Λ} = V_{Λ}.

Theorem 2 If Λ ∈ P̂_{+} then L_{Λ} = V_{Λ} = M_{Λ}/(f_{i}^{N_{i}+1}v_{Λ}).

Proof. We need to show that V_{Λ} is irreducible. Assume that it is not. Then it has a proper submodule which then has a highest weight vector. Any module in the category 𝒪 has at least one highest weight vector. Call this v_{Λ′}. The submodule is then integrable since its parent V_{Λ} is integrable. Then by the previous result, Λ′ ∈ P̂_{+}. So the picture we want to show impossible would have two highest weight vectors, both dominant. This is possible in general. There is something else that we need. There is a nontrivial argument we need to use.

Recall that we had a Casimir element Ĉ which acts on a highest weight module by (Λ, Λ+2ρ̂). It acts by a constant everywhere. This forces (Λ, Λ + 2ρ̂) = (Λ′, Λ′ + 2ρ̂). We also know that Λ, Λ′ are both dominant weights, and Λ = Λ′ + ∑ n_{i}α_{i}. I claim that these three conditions together lead to a contradiction.

The proof is actually quite easy because, let me write ∑ n_{i}α_{i} as ν. Then (Λ′+ν, Λ′+ν +2ρ̂) = (Λ′, Λ′ + 2ρ̂), and then just use bilinearity. The terms not involving ν will be the same on both sides, while the terms involving ν will be (ν, Λ′ + 2ρ̂) + (Λ′, ν) + (ν, ν) = 0. But all of these are nonnegative, since the first two are between dominant weights and positive roots and the last is between positive roots.