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\title{Infinite Dimensional Lie Algebras\\April 6, 2005}
\author{Gabriel C. Drummond-Cole}
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So let me begin, let me start as usual by reminding what we've done so far.  We have this category $\mathscr{O}$ which is roughly the highest weight modules.  This contains the Verma modulees $M_\Lambda$ and their irreducible quotients $L_\Lambda.$  We also discussed integrable modules $\mathscr{O}^{int},$ where $dim(U\la{sl}(2)_i)v<\infty.$  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 $V\in\mathscr{O}^{int}$ then $P(V)\subset P=\{\Lambda\in\hat{\la{h}}^*|\langle \alpha^\surd,\lambda\rangle\in\mathbb{Z}\},$ i.e., the roots are integrable.

Also the set with multiplicity is invariant under the affine Weyl group, $\hat{W}(P(V))=P(V).$  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.

\begin{defi}If $V\in\mathscr{O}$ then $v\in V$ is smooth if for all $i,\ \dim(U\la{sl}(2)_i)v<\infty.$\end{defi}
\begin{lemma}
$v\in V$ is smooth implies $f_jv$ is smooth for $j=0,\ldots,r.$
\end{lemma}
This implies that the set of smooth vectors is actually a subrepresentation.
{\it Proof.} Recall $(ad\ f_i)^N f_j=0.$ for some appropriate $N.$  This can be rewritten in some nicer way, $f_i^Nf_j-\binom{N}{1}f_i^{N-1}f_jf_i+\cdots+f_jf_i^N=0.$

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 $f_i(f_if_j-f_jf_i)-(f_if_j-f_jf_i)f_i=0.$  So these will be sums with one $f_j$ and $N\ f_i,$ and the only problem is matching the coefficients.  Here this is $f_i^2f_j-2f_if_jf_i+f_jf_i^2=0.$  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 $U\la{sl}(2)_iv,$ which is finite dimensional, spanned by the basis $v_1,\ldots, v_\ell.$  Then $f_jv\in \langle f_jv_a,f_if_jv_a,f_i^2f_jv_a,\ldots\rangle.$  Where do we stop?  The last one is $\ldots, f_i^{N-1}f_jv_a\rangle_{a=1,\ldots,\ell}$  Denote this vector space by $W.$  Then I claim that this is invariant under $\la{sl}(2)_i.$
\begin{enumerate}
\item If I apply $e_i$ to $f_i^mf_jv_a,$ I get $\sum h\ldots f_i^{m-1}f_jv_a,$ where the $h\in\hat{\la{h}},$ and another term $f_i^mf_je_iv_a.$  Since we assumed that $v_a$ was a basis, we can write $e_iv_a$ as a linear combination of $v_b.$
\item What about $f?$  We have $f_if_i^mf_jv_a$ is either $f_i^{m+1}f_jv_a$ if $m<N-1,$ or $\sum(-1)^{k+1}\binom{N}{k} f_i^{N-k}f_jf_i^k.$

So the Serre relation allows you to stop at some point.
\end{enumerate}
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 $L_\Lambda,$ 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.
\begin{thm}
$L_\Lambda$ is integrable if and only if $\Lambda\in \hat{P}_+=\{\Lambda\in\hat{\la{h}}^*|\langle \Lambda,\alpha_i^\surd\rangle\in\mathbb{Z}_+\}.$
\end{thm}
\begin{cor}
Isomorphism classes of irreducible integrable modules are thus in one to one correspondence with $\hat{P}_+.$
\end{cor}

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 $L_\Lambda$ is integrable.  Among other things, the highest weight vector $v_\Lambda$ generates a finite dimensional $\la{sl}(2)_i$-module.  Since it's a highest weight vector, $e_iv_\Lambda=0.$  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 $\langle Lambda,h_i\rangle\in\mathbb{Z}_+.$  This is precisely the statement we wanted to show.

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

Assume that $\langle \Lambda,\alpha_i^\surd\rangle=N_i\in\mathbb{Z}_+.$  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 $\la{sl}(2),$ say I have a highest weight vector of weight $N.$  Then $M_\Lambda$ and $L_\Lambda$ look like
$$
\xymatrix{f^{N+1}v&f^Nv\ar@/^/[l]&\cdots&f^2v\ar@/^/[l]&fv&v\ar@/^/[l]
\\f^{N+1}v&f^Nv&\cdots&f^2v\ar@/^/[l]&fv&v\ar@/^/[l]}
$$
We get a similar picture for $\la{sl}(3):$
$$
\xymatrix{f_1^{N+1}v_\lambda\ar@{-}[r]&v_\lambda\ar@{-}[d]
\\\ar@{-}[u]&f_2^{N+1}v_\lambda\ar@{-}[l]}
$$
So $e_jf_i^{N_i+1}v_\Lambda)=0,$ including $j=i.$  Thus weights of the $U\hat{\la{g}}$-submodule generated by $f_i^{N_i+1}v_\Lambda$ will be of the form $\Lambda-(N_i+1)\alpha_i-Q_+$ so this submodule doesn't contain $v_\Lambda.$  This is already a remarkable thing, since a vector chosen at random generates the whole thing.

Now, let $V_\Lambda=M_\Lambda/(f_i^{N_i+1}v_\Lambda)_{i=0,\ldots,r}.$  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 $L_\Lambda,$ let me write, $L_\Lambda=M_\Lambda/$ maximal proper submodule, so $L_\Lambda=V_\Lambda/?,$ maybe some further modules.  Later if time permits I will show that this is an equality.  In $L_\Lambda,$ all we need to know for now is that all of the vectors $f_i^{N+1}v_\Lambda=0.$  Every vector living in a proper submodule of $M_\Lambda$ is zero in $L_\Lambda.$

Now I am done because I claim, it is easy to see that $v_\Lambda\in V_\Lambda$ 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 $v_\Lambda,$ 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 $L_\Lambda=V_\Lambda.$

\begin{thm}
If $\Lambda\in \hat{P}_+$ then $L_\Lambda=V_\Lambda=M_\Lambda/(f_i^{N_i+1}v_\Lambda).$
\end{thm}
{\it Proof.} We need to show that $V_\Lambda$ 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 $\mathscr{O}$ has at least one highest weight vector.  Call this $v_{\Lambda'}.$  The submodule is then integrable since its parent $V_\Lambda$ is integrable.  Then by the previous result, $\Lambda'\in\hat{P}_+.$  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 $\hat{C}$ which acts on a highest weight module by $(\Lambda,\Lambda+2\hat{\rho}).$  It acts by a constant everywhere.  This forces $(\Lambda,\Lambda+2\hat{\rho})=(\Lambda',\Lambda'+2\hat{\rho}).$  We also know that $\Lambda,\Lambda'$ are both dominant weights, and $\Lambda=\Lambda'+\sum n_i\alpha_i.$  I claim that these three conditions together lead to a contradiction.

The proof is actually quite easy because, let me write $\sum n_i\alpha_i$ as $\nu.$  Then $(\Lambda'+\nu,\Lambda'+\nu+2\hat{\rho})=(\Lambda',\Lambda'+2\hat{\rho}),$ and then just use bilinearity.  The terms not involving $\nu$ will be the same on both sides, while the terms involving $\nu$ will be $(\nu,\Lambda'+2\hat{\rho})+(\Lambda',\nu)+(\nu,\nu)=0.$  But all of these are nonnegative, since the first two are between dominant weights and positive roots and the last is between positive roots.

This is a complicated theorem, but we got a complete description of the integrable irreducible modules.  So we got a lot.  

Up to this point, I couldn't have cared less about whether the Lie algebra was finite dimensional or affine.  Now, if we consider affine Lie algebras, what does the set of dominant weights look like?  For $\la{g}$ the set of dominant weights are within a cone, namely integer points in $\bar{C}_+,$ the positive Weyl chamber.

What will it look like in the affine case?  In general, any weight can be writted $\Lambda=\lambda+k\Lambda_0+a\delta.$  The condition that this thing is a dominant weight $\Lambda\in \hat{P}_+,$ if I let $i$ range from $1$ to $r,$ I get that $\lambda\in P_+,$ that $\lambda$ is a dominant weight.  But then you also have the condition that $\langle\Lambda,\alpha_0^\surd\rangle\in\mathbb{Z}_+,$ which is $\langle \Lambda,-\theta^\surd+K\rangle=-\langle \lambda,\theta^\surd\rangle+k\in\mathbb{Z}_+.$

I already know that $\langle \lambda,\theta^\surd\rangle$ is a nonnegative integer, so I need that $k$ is a nonnegative integer and $\langle \lambda,\theta^\surd\rangle\le k.$

As for $a,$ it doesn't really enter here, since that just says what degree you assign to the highest weight vector (this then determines the others uniquely).  So for all practical purposes the $a$ part is not really needed.

So what is the end result?  First, if I consider $P_+^k=\{\Lambda\in\hat{P}_+,\langle\Lambda,K\rangle=k\}/\mathbb{C}\delta,$ then these are the highest possible dominant weights for fixed $k.$  These will be $0$ for $k$ not a nonegative integer.  The second is that if I fix a level, say, five, and I ignore the grading, then there are only finitely many of them.

If I fix the level I only have finitely many possibilities.  Again, that makes sense.  These are integral weights inside the appropriate Weyl chamber, but now we want to look inside Weyl alcoves.

Let's do the example for $\la{sl}(2),$ or rather $\widehat{\la{sl}}(2).$  In this case, there is exactly one root so $P_+^k=\{\lambda\in P_+|\langle\lambda,\alpha^\surd\rangle\le k\}.$  Let $\la{h}^*\cong \mathbb{C},$ with $\alpha\mapsto 2.$  Then $P_+\mapsto\mathbb{Z}_+.$  Then $P_+^k=\{\lambda\in\mathbb{Z}_+|\lambda\le k\}.$  Then these will be $L_{0,k},\ldots,L_{k,k}.$  So for level one there are exactly two modules.  For level zero there is exactly one module, and we can guess what it is.  $L_{0,0}=\mathbb{C}$ and all the others are infinite dimensional.

Sometimes I will write $L_\lambda=L_{\lambda,k},\Lambda=\lambda+k\Lambda_0+a\delta.$

It turns out you can cook up a tensor category of integrable modules.  At the moment I have not discussed a tensor product, but definitely as an Abelian category, you have one with only finitely many simple objects.  As I'll show next time, this category is semisimple.  This fails if I leave out the integrability condition.  Let's continue on Monday.  On Friday, Justin will be talking, and he will start with simple things, representations of $GL(n),$ highest weights and partitions and all that.


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