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\title{Infinite Dimensional Lie Algebras\\March 13, 2005}
\author{Gabriel C. Drummond-Cole}
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\maketitle  

The twisted action $\omega.\Lambda=\omega(\Lambda+\hat{\rho})-\hat{\rho},$ where $\hat{\rho}=\rho+h^\surd K,$ and $\langle \hat{\rho},\alpha_i^\surd\rangle=1$ for $i=0,\ldots,r.$  

Now $h^\surd,$ the dual Coxeter number [?] is $\langle \rho,\theta^\surd\rangle+1.$

\begin{enumerate}
\item
If $\Lambda$ has level $k$ then $\Lambda+\hat{\rho}$ has level $k+h^\surd.$
\item
The usual action of $\hat{W}$ on $\hat{\la{h}}^*_k/\mathbb{C}\delta$ is given by reflection around the hyperplanes $\langle \lambda,\alpha_i^\surd\rangle=0$ for $i=1,\ldots,r$ and $\lambda,\theta^\surd\rangle=k.$  The shifted action $\omega.\Lambda$ is generated by reflections around $\langle \lambda+\rho,\alpha_i^\surd\rangle=0.$  This should be easy to believe, and is equivalent to say that $\langle \lambda,\alpha_i^\surd\rangle=-1.$  What about the affine wall?  $\langle\lambda+\rho,\theta^\surd\rangle=k+h^\surd.$  If you go back to the original definitions, what does that give you?  $\langle \lambda,\theta^\surd\rangle=k+1.$  So the picture is of a slightly enlarged, by $1$ on each side, positive Weyl alcove.  For $\widehat{la{sl}(2)}$ the usual reflections are through $0$ and $k.$  So the shifted action is generated by reflections through $-1$ and $k+1.$
\end{enumerate}
Okay, this was all preliminaries, let me start with some new theory.  My goal today will be to discuss characters of modules, and in particular of integrable modules.  Character is just a formal way to keep track of the dimension of weighted components.  So we define $ch(V)=\sum_{\Lambda\in P(V)}e^\Lambda \dim\ V[\Lambda].$  So what are these things $e^\Lambda.?$  These are formal generators subject to $e^{\Lambda_1+\Lambda_2}=e^{\Lambda_1}e^{\Lambda_2}.$  I can choose a basis in the weight lattice, and then the formal expressions of this sort can be identified with the Laurent polynomials in finitely many variables.  So maybe I should do it just for $\widehat{la{sl}(2)}.$

There we can introduce notation.  $\hat{h}^*$ is three dimensional.  It has the usual Cartan and then $\delta$ and $\Lambda_0.$  If $e^{-\delta}=q$ and $e^\alpha=z^2,$ squared because the generator of the weight lattice is $\alpha/2,$ then basically formal expressions of this form become $ch(V)=e^{k\Lambda_0}\sum_{n,\ell\in\mathbb{Z}a_{n,l}q^nz^\ell.$

For most of what we are talking about this is an infinite sum, this is not a polynomial but some kind of series.  $ch(V)\in \mathscr{S},$ the space of possibly infinite expressions of the form $\sum a_\Lambda e^\Lambda.$

I could be more precise and explain that I want them to be infinite in one direction only, like when I defined category $\mathscr{O}.$  I don't want to do that, I just want to say, this is a suitably completed category of polynomials.

The easiest character is that of a Verma module, which we can find very explicitly.  If you have some basis, remember that the Verma module $M_\Lambda$ can be written $U_{\la{n}_-}v_\Lambda.$  So $ch(M_\Lambda)=\sum e^{\Lambda+wt(b)}$ where $b$ is a basis for $U_{\la{n}_-}.$  The dependence on $\Lambda$ is trivial, so this is $e^\Lambda\cdot 1/\delta,$ where $1/\delta=\sum e^{wt(b)}.$  This is where the Poincar\'e Birkoff Witt theorem helps.  Recall that a basis in the Lie algebra itself, then a basis in the universal enveloping is given by monomials $x_1^{n_1}\ldots x_\ell^{n_\ell}.$  So we can easily get the formula, assuming a basis in $\hat{\la{n}}_-.$  It is $1/\delta=\sum_{n_i}e^{wt(x_1)n_1+wt(x_2)n_2+\ldots}=\prod_{i=1}^\ell \sum_{j=0}^\infty e^{jwt(n_i)}.$

So this is $1/\delta=\prod_i \frac{1}{1-e^{wt(x_i)}}.$  The only thing remaining is to choose our basis.  This will be $\prod_{\hat{\alpha}\in\hat{R}_-}\frac{1}{1-e^{\hat{\alpha}}}.$  This denominator should be to the power of the multiplicity of $\hat{\alpha}.$

We usually rewrite this in terms of positive roots $\prod_{\hat{\alpha}\in\hat{R}_+}\frac{1}{(1-e^{-\hat{\alpha}})^{mult(\hat{\alpha})}.$  This is $ch(M_\Lambda)=e^\Lambda/\delta.$

What is the end result?  It is:
\begin{thm}(Weyl-Kac)
\\
Let $\Lambda\in\hat{P}_+.$  Then $Ch(L_\Lambda)=\frac{\sum(-1)^{\ell(w)}e^{\omega\Lambda}}{\delta}.$

Sometimes this is written as $\frac{\sum (-1)^{\ell(\omega)}e^{\omega(\Lambda+\hat{\rho})}{e^{\hat{\rho}}\delta}.$  There are ways of doing this without the BGG resolution, but this is the best for intuitive understanding, and for telling you why this should be true.

Note $Ch(L_\Lambda)$ is $\hat{W}$-invariant, since the set of weights with multiplicities is invariant under the Weyl group.  However, the formula doesn't look as though it is Weyl group invariant.  Instead it is antinvariant.  The denominator is also skew invariant.

What other general properties do we have?  For now let me show you what this formula looks like in the simple example of $\widehat{\la{sl}(2)}.$  As before I'll write $e^-\delta=q$ and $e^{\alpha_1}=z^2.$  Then what is the denominator?  All multiplicities are one, and then this is $$\prod (1-e^{-\hat{\alpha}})=\prod_{n\ge 0} (1-e^{-(\alpha_1+n\delta)})\prod_{n>0} (1-e^{-n\delta})\prod_{n>0} (1-e^{-(-\alpha_1+n\delta)})
$$$$
=\prod_{n\ge 0} (1-z^{-2}q^n)(1-q^{n+1})(1-z^2q^{n+1}).$$

The numerator depends on what $\lambda$ we take.  The first observation is that $\hat{W}=\mathbb{Z}_2\ltimes \mathbb{Z}.$  These are the simple reflection $s_1$ and the translation $\tau=s_0s_1.$  So the numerator $\sum_{w\in\hat{W}}(-1)^{\ell(w)}e^{w.\Lambda}=\sum_{n\in\mathbb{Z}}e^{\tau^n.\Lambda-\sum_{n\in\mathbb{Z}}e^{\tau^ns_1.\Lambda}.$

I did some calculations at home and if $\Lambda=\lambda+k\Lambda_0,$ then the answer you get after a little bit of computing is $\tau^n.\Lambda=\lambda+2n(k+2)-((\lambda+1)n+n^2(k+2))\delta+k\Lambda_0.$  

Everything else you can do as well as I can [sigh].  Then $\sum_{n\in\mathbb{Z}}e^{\tau^n.\Lambda}=e^{k\Lambda_0}\sum_{n\in\mathbb{Z}}z^{\lambda+2n(k+2)}q^{(\lambda+1)n+n^2(k+2)}.$  This might remind you of theta functions.  I'll come back to this later, for now just mention that it is not very unusual.  Let me now put everything together.

For example, what would be the character of $L_{0,0},$ that is, $\Lambda=0.$  This is $1.$  On the other hand, if I use the formula, I get $$\frac{\sum (-1)^{\ell(w)}e^{w.0}}{\delta},$$ which gives $\delta=\sum (-1)^{\ell(w)}e^{w.0}.$  This works in general, and is called the denominator identity.  What is this for $\la{sl}(2)?$  $\sum (-1)^{\ell(w)}e^{w.0}=e^{k\Lambda_0}\sum z^{4n}q^{2n^2+n}.$  That's not the end of the story yet.  What happens with the other term?  The subtracted term has $s_1\Lambda,$ which is reflection about $-1,$ so makes $\lambda=-2.$  So it is $-e^{k\Lambda_0}\sum z^{-2+4n}q^{-n+2n^2}.$  If you put these together you should get one.  I might as well erase $-e^{k\Lambda_0}$ since $k=0.$  So you get the formula
$$
\prod_{n\ge 0} (1-z^{-2}q^n)(1-q^{n+1})(1-z^2 q^{n+1})=\sum_{n\in\mathbb{Z}} z^{4n}q^{n(2n+1)}-z^{-2+4n}q^{n(2n-1)}.$$
After changing indices and so on this is $\sum_\ell\in\mathbb{Z}(-1)^\ell z^{2\ell}q^{\frac{\ell(\ell+1)}{2}}.$  This was the Jacobi triple product identity from the Heisenberg algebra.  We got them by computing the dimension of a space in two ways, with partitions and a Fermionic representation.  This is only the character of a module which is trivial.

In the remaining six minutes, let me get as far as I can on the case $ch(L_{0,1}).$  This is $\Lambda=\Lambda_0.$  It's rather easy to see that since its weight with respect to the finite dimensional part is zero, it will be generated by $f_1v_\Lambda$ and $f_0^2v_\Lambda.$  It is very difficult to see what happens below.  Let me write the formula.  Having seen how I did the last part you can probably guess how I do this: $\frac{\sum z^{6n}q^{n+3n^2}-z^{6n-2}q^{3n^2-n}}{\delta}.$  This doesn't help much.  What if we take it modulo $q^2?$  Which terms survive on top?  $1-z^-2.$  In the denominator, we only keep the terms $(1-z^{-2})(1-qz^{-2}(1-q)(1-z^2q).$  

When we multiply out and reduce we get $\frac{1}{1-q(z^{-2}+1+z^2)}.$

What does this tell us about the character of the module?  With respect to the grading, we have only one thing at level zero, and three at the next level, at $2,0,$ and $-2.$  

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