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\title{Infinite Dimensional Lie Algebras\\April 4, 2005}
\author{Gabriel C. Drummond-Cole}
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\begin{document}
\noindent
\maketitle  
Last time we discussed the notion of highest weight module, which is a representation $V$ generated by a single vector of some wheight $\Lambda$ such that $e_iv=0$ for all $i.$

We discussed that every such representation, assuming it has a weight decomposition, is the direct sum $\bigoplus_{\mu\in \Lambda-Q_+}V[\mu]$ where $Q_+=\{\sum n_i\alpha_i,n_i\ge 0\}.$  We discussed that among other examples we have the so-called Verma modules, freely generated by the highest weight.  This may be reducible so we define $L_\Lambda,$ the irreducible quotient.  This works the same way for Kac-Moody algebras, finite dimensional algebras, and affine Lie algebras.  Remember for $\la{sl}(2)$ the Verma module is an infinite chain and when the highest weight is an integer you can find a submodule and break off a finite dimensional irreducible quotient which is a highest weight module.

Just considering these is too restrictive; it's not closed under product.  So we consider the category $\mathscr{O}$ which is representations $V$ with weight decomposition $V=\bigoplus_{\Lambda\in P(V)}V[\Lambda],$ where $P(V)=\{\Lambda\in \hat{\la{h}}^*|V[\Lambda]\ne 0\}.$ subject to
\begin{enumerate}
\item $\dim\ V[\Lambda]<\infty$
\item $P(V)$ is a subset of a finite union of sets of the form $\Lambda_i-Q_+.$
\end{enumerate}
Sometimes this is called the category of highest weight modules; this is not literally true since this includes sums.

Let me state some properties.
\begin{enumerate}
\item Every highest weight module is in $\mathscr{O}.$
\item $\mathscr{O}$ is closed under $\oplus,$ submodules, and quotients.
\item Define $\mathscr{O}_k=\{V\in\mathscr{O}|K|_V=k\ Id\}$ for $k\in\mathbb{C}.$  If $Kv=kv$ you want to show that $xv$ satisfies the same condition.  So $Kxv=xKv=kxv$ for all $x\in\la{g}.$  So it makes sense to talk about representations of a fixed level; these are called representations of level $k.$

Then $\mathscr{O}=\bigoplus \mathscr{O}_k.$  This means that a representation of $\mathscr{O}$ can be written as a direct sum of representations from the different levels, and $Hom$ between different levels is trivial.

So to study $\mathscr{O}$ we need only study what happens at $\mathscr{O}_k$ for arbitrary $k.$
\item If $V\in\mathscr{O}$ then $P(V)$ is ``bounded from above.''  What does this mean?
\begin{enumerate}
\item $Re\ (\Lambda,\hat{\rho})$ is bounded from above.  Let me remind you that $\hat{\rho}$ is defined so that $\langle \hat{\rho},\alpha_i^\surd\rangle=1,\ \hat{\rho}=\rho+h^\surd K.$  This is one way to make this notion precise.
\item If you take some weight $\Lambda\in P(V),$ and then you take $(\Lambda+Q_+)\cap P(V),$ this is finite.
\end{enumerate}
\end{enumerate}
The proof is rather obvious.  The second part (of the last one) follows from the first.  To go up you are adding things in $Q_+;$ this makes the inner product with $\hat{\rho}$ go up, so if it's bounded from above, we know we can only do this finitely many times.

For the first part, the argument is essentially that $\hat{\rho}$ has to be one of the highest weights, and there are only finitely many.

\begin{lemma}
Irreducible representations in $\mathscr{O}$ are exactly $L_\Lambda,\Lambda\in \hat{\la{h}}^*.$
\end{lemma}
  
This is easy to believe; here is a formal proof.  Assume $V$ is irreducible.  Since the set of weights is bounded from above, there exists a weight $\Lambda\in P(V)$ such that $\Lambda+\alpha_i\notin P(V).$  Therefore, if you take $v\in V[\Lambda],$ this will be a highest weight vector in the sense that $e_iv=0.$

Consider the submodule generated by $v.$  If the representation is irreducible, the vector must generate all of $V.$  So $V$ is a highest weight representation.  This must be unique because these are nonisomorphic for different $\Lambda.$

The structure of this category is very complicated.  $\la{sl}(2)$ gives you a description of the category.  A lot is known about it, but I don't want to.  I'd have to go into perverse sheaves and things.

This does, though, have a fairly simple subcategory, of finite dimensional modules.  This category gives you a fairly nice piece.  So what about for affine Lie algebras.  You cannot expect finite dimensional modules.  We want something essentially as nice; there is an appropriate notion.
\begin{defi}
A representation $V\in\mathscr{O}$ is integrable if for any $v\in V$ the $\la{sl}(2)_i$-submodule generated by $v$ is finite dimensional.  Here $\la{sl}(2)_i=\langle e_i,f_i,h_i\rangle.$
\end{defi}
Later we'll see a more invariant definition.  Let me make a picture for $\widehat{\la{sl}(2)}.$

[...]

The notation is that this category is $\mathscr{O}^{int}$ and similarly I can define $\mathscr{O}_k^{int}.$  Integrable modules almost always means integrable and from the category $\mathscr{O}.$  I didn't even define the others.

Let me write as a lemma that \begin{lemma}
$\mathscr{O}^{int}$ is closed under $\oplus,$ submodules, and quotients
\end{lemma}

First, I would like to explain, I don't have a single example of such a module.  Verma modules are not integrable, since $f_1$ acts freely on $v$ and is infinite dimensional.

We do have examples.  For $\la{g}$ then every finite dimensional module is integrable.  As a matter of fact, if $\la{g}$ is finite dimensional, integrables are always finite dimensional.

But examples for affien Lie algebras will have to wait.  Let me start with some properties.  This is probably the most important result I'm going to prove today.
\begin{thm}
If $V\in\mathscr{O}^{int}$ then the set of weights $P(V)$ is $\hat{W}$-invariant.
\end{thm}
{\it Proof.}  It suffices to show that $P(V)$ is invariant under simple reflections $s_i,i=0,\ldots,r.$  Now, if $v\in V[\Lambda],$ let's say $n_i=\langle Lambda,h_i\rangle,$ the weight of $v$ considered as a module over $\la{sl}(2)_i.$  For one thing it should be integer, since it generates a finite dimensional module over $\la{sl}(2).$

If $n_i\ge 0$ then consider $$f_i^{n_i}v\in V[\Lambda-n_i\alpha_i]=V[s_i\Lambda].$$

By well-known results for $\la{sl}(2)$ this gives an isomorphism $V[\Lambda]\cong V[S_i\Lambda].$

Let me now write a simple corollary, a couple, actually.
\begin{cor}
\begin{enumerate}
\item If $V\in \mathscr{O}^{int},v\in V$ then $\la{sl}(2)_\alpha$-module generated by $v$ is finite dimensonal, $\la{sl}(2)_\alpha=\langle e_\alpha,f_\alpha,h_\alpha\rangle$ for $\alpha\in \hat{R}_{re}.$

This shows in the definition of integrability I could have stated in terms of all the roots, not just the simple ones.
\item
For $V\in \mathscr{O}^{int},$
$$P(V)\subset P = \{\Lambda\in\hat{\la{h}}^*|\langle \Lambda,h_i\rangle\in\mathbb{Z}\}
$$$$
=\{\lambda+k\Lambda_0+a\delta|\langle \lambda,h_i\rangle\in\mathbb{Z}, i=1,\ldots,r,\ \langle \lambda+k\Lambda_0,-\theta^\surd+K\rangle\in\mathbb{Z}=
$$$$
\{\lambda+k\Lambda_0+a\delta|\lambda\in P,k\in \mathbb{Z}\}
$$
This reminds us of the finite dimensional case.  
\item For $\la{g},\ \mathscr{O}^{int}$ are precisely the finite dimensional modules.  This is one reason this name is not used in finite dimensional cases.
\end{enumerate}
\end{cor}
{\it Proof.}
Let $v\in V[\Lambda].$  Then $\la{sl}(2)_\alpha v\in \bigoplus_{k\in\mathbb{Z}}V[\Lambda+k\alpha].$

Assuming that $\alpha$ is positive, sooner or later the right hand side is zero for some $k$ greater than zero by assumption.  The same is true for negative roots by invariance under $S_\alpha.$

For the third one, one direction I already proved.  Well, there is nothing to prove.  A finite dimensional module is integrable.  The difficult part is proving the opposite direction.  It suffices to show that $P(V)$ is finite.  Let me draw a picture.

[...]

Since the set of weights is bounded from above, $P(V)\cap C_+$ is finite, where $C_+$ is the positive Weyl chamber.  By Weyl group invariance, there are only finitely many weights in any Weyl chamber.  Since there are only finitely many, we're done.

This fails for $\hat{\la{g}}$ because there are infinitely many Weyl alcoves, or we can say, because $|\hat{W}|=\infty.$

Recall that $\hat{W}=W\ltimes Q^\surd,$ where this latter is the lattice generated by $\alpha_i^\surd$ acting on $W$ by $$\tau_{\alpha^\surd}:\Lambda=\kambda+k\Lambda_0+a\delta\mapsto (\lambda+k\alpha^\surd)+k\Lambda_0+(a-(\lambda,\alpha^\surd)-\frac{(\alpha^\surd,\alpha^\surd)}{2}k)\delta.$$
So how does this look for $\widehat{\la{sl}(2)}?$  Here
$$\tau_{\alpha^\surd}:\Lambda=\kambda+k\Lambda_0+a\delta\mapsto (\lambda+2nk)+k\Lambda_0+(a-n\lambda-n^2k)\delta,$$
identifying $\la{h}^*$ with $\mathbb{C}.$

So starting with zero, I get $\lambda$ goes to $\lambda+(a-n^2k)\delta.$  A typical orbit is a parabola, so in particular has infinitely many points.

Let's continue next time.  Even thought the objects are infinite dimensional, they will be nicely behaved.  Some people would say that these make life more complicated, others would say it makes life more interesting (or richer).
\end{document}
