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\title{Infinite Dimensional Lie Algebras\\March 28, 2005}
\author{Gabriel C. Drummond-Cole}
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Okay, so, let me begin, I'm afraid I lost track of the lecture number, but I don't think it's that important.  So this is lecture $n+1.$

Last several weeks we've started with an affine Lie algebra $\hat{\la{g}}$ and obtained from that the affine root system $\hat{R}$ and then the affine Weyl group $\hat{W}$ and $\hat{\Pi}=\{\alpha_0,\ldots,\alpha_r\}.$  The theorem we proved last time is that if we introduce $e_i\in\hat{\la{g}_{\alpha_i}},f_i\in\hat{\la{g}_{-\alpha_i}},h_i\in\la{h},$ then \begin{thm}
\begin{enumerate}
\item they generate $\tilde{\la{g}}=\la{g}\otimes\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}k$ (Everything but $d$)
\item They satisfy the relations
\begin{enumerate}
\item $[h_i,h_j]=0.$
\\ $[h_i,e_j]=a_{ij}e_j,\qquad [h_i,f_j]=-a_{ij}f_j$
\item $[e_i,f_j]=\delta_{ij}h_i$
\item $(ad\ e_i)^{1-a_{ij}}e_j=0.$
\\ $(ad\ f_i)^{1-a_{ij}}f_j=0.$
\end{enumerate}
\end{enumerate}
\end{thm}
The $a_{ij}$ are nonpositive integers and this was not hard to prove.  A real root gave a triple isomorphic to $\la{sl}(2)$.  The last relations come from considering the whole Lie algebra as a module over $\la{sl}(2).$  We can easily find that applying $e_i$ enough times to the lowest weight vector gives the highest weight vector.  

So the only problem is that we're missing the grading element $d.$

Let $E$ be the span of the $\alpha_i$ and $\tilde{\la{h}}$ the span of $h_i.$  We have a pairing between $E$ and $\tilde{\la{h}}$ given by $\langle h_i,\alpha_j\rangle=a_{ij}.$  Since this is singular, having a one dimensional kernel, this won't be a perfect pairing, but we can fix it.  We want to extend these to get a perfect pairing.  This is purely linear algebra

Now, let us write $E=\ker\ A\oplus E'.$  I'll give an explicit example for an affine Lie algebra.  Let $\la{h}(A)=\tilde{\la{h}}\oplus (\ker\ A)^*.$  Then we have $\la{h}(A)\otimes E\to \mathbb{C}.$  These are now of a different size, but I can say now that every linear functional on $E$ can be implemented in $\la{h}(A).$  So we can modify the theorem to let $\la{h}(A)$ be as before.  Then
\begin{enumerate}
\item $\la{h}(A)=\hat{\la{h}}=\la{h}\oplus \mathbb{C}k\oplus\mathbb{C}d$
\item $\hat{\la{g}}$ is generated by $e_i,f_i,h\in \la{h}(A)$ with the relations
\\$[h,e_i]=\langle h,\alpha_i\rangle e_i$
\\$[h,f_i]=-\langle h,\alpha_i\rangle f_i$
\\and the rest unchanged.
\end{enumerate}
Let's talk about the proof.  Okay, so what will be the span of $\alpha_i?$  This is the proof of the first part; the second part will be more or less immediate.

The span is the span of the normal Cartan and then also $\mathbb{C}\delta.$  Now $\ker\ A=\mathbb{C}\delta.$  What about $\la{h}(A)?$  This will be the span of $h_0,\ldots,h_r,$ which is $\la{h}\oplus\mathbb{C}k.$  Then we add by hand $\mathbb{C}d,$ something exactly dual to $\delta,$ that is, $\langle d,\delta\rangle=1.$  You can check that nothing changes.  If I chose a different space complementary to the kernel of $A$ I would get an isomorphic picture.

The second one is immediate because we know the definition of a root space.

So it's not a big problem to get all of $\hat{\la{g}}.$  What is the major point?  We don't know that this is the full list of relations, we just know that these relations are satisfied.  The answer is that this is the full list of relations.  This is not so easy to prove, so let me just write the statement of the theorem.  First maybe I will introduce a slightly more general version of it.

Let $A$ be an integer matrix such that 
\begin{enumerate}
\item $a_{ii}=2,a_{ij}\in\mathbb{Z}_-$ for $i\ne j.$
\item $A$ is symmetrizable: there exist $d_i>0$ such that $a_{ij}d_i=a_{ji}d_j,$ that is, $(AD)^t=AD.$
\end{enumerate}
Call these generalized Cartan matrices.  Define $\la{g}(A)$ to be the Lie algebra generated by elements $e_i,f_i,h\in\la{h}(A)=span(h_i)\oplus(\ker\ A)^*$ and the relations
\\$[h,e_i]=\langle h,\alpha_i\rangle e_i$
\\$[h,f_i]=-\langle h,\alpha_i\rangle f_i$
\\$[e_i,f_j]=\delta_{ij}h_i$
\\and the Serre relations. 

\begin{thm}
$\la{g}(A)$ is almost simple, the easiest way to explain its simplicity is to say it has no ideals which intersect $\la{h}(A)$ trivially.  That is, if I only allow ideals with trivial intersection with $\la{h}(A)$ there are no such ideals.
\end{thm}
So this is ``almost'' simple.  Unfortunately, the proof I am unable to give; it takes some time and relies on representation theory and integrable representations.  We'll discuss it later.  Let me just say that the key step of the proof is that this algebra is itself considered as a module over $\la{sl}(2),$ is locally finite dimensional, that is, every vector lives in a finite dimensional module inside the algebra.  This is where you need the Serre relation, because without it, this fails.
\begin{cor}
\begin{enumerate}\item If $A$ is a Cartan matrix of a finite dimensional complex semisimple Lie algebra $\la{g},$ then $\la{g}=\la{g}(A).$ 
\item
If $A$ is a Cartan matrix of an affine Lie algebra $\hat{\la{g}}$ then $\hat{\la{g}}=\la{g}(A).$
\end{enumerate}\end{cor}
The beauty of this way of doing it is that the proofs are the same. 

We know that all of the relations of $\la{g}(A)$ are satisfied in $\la{g}$ which means that $\la{g}=\la{g}(A)/I$ where $I$ are some relations.  More formally you can come up with a morphism taking $e_i$ to $e_i$ and so on. 

Moreover, restricting to the Cartan subalgebra, we see that $\la{h}=\la{h}(A)$ so $I\cap \la{h}(A)=0.$

Therefore, from the theorem, $I=0$ and that's it.  The same proof goes for the second part, word for word.  The only thing you really need to check is that all of the relations are satisfied in an affine Lie algebra.  Then we have to check the Cartan, but that does it.  The key thing is the property of the theorem, which is not easy to prove.

So now we have no loop algebras, no polynomials with values in $\la{g};$ just take some generators and relations and see the Lie algebra they generate.

It would be interesting to look at what happens when the generating matrix is not a Carta matrix but something more general.

For example, consider $\left(\begin{array}{cc}2&-4\\-1&2\end{array}\right).$  There is a general bit of terminology; such Lie algebras are called Kac-Moody Lie algebras.  I'll have root decomposition and the Weyl group.  Unfortunately, we cannot really say what the root system will look like, whether it will be finite or infinite, or what the multiplicity of the roots will be.

So here we can check what root system we will get, and we will do it later, but it turns out there are not too many matrices of this sort and we can prove a classification theorem.

A matrix is decomposable, if by permuting indices one can make it (nontrivially) block diagonal.  Otherwise it is indecomposable.  This is just like connectedness of a Dynkin diagram in the classical case.
\begin{thm}
Let $A$ be an indecomposable generalized Cartan matrix. Then the statement is that there are exactly three possibilities:
\begin{enumerate}
\item $AD$ is positive definite, where $D$ is a symmetrizing diagonal matrix.  Then $A$ is a Cartan matrix of a finite dimensional simple Lie algebra.

That's a relatively easy part.  If you like, this is how you classify simple Lie algebras.
\item $AD$ is positive semidefinite with one dimensional kernel; then $A$ is a Cartan matrix of a (possibly twisted) affine Lie algebra.
\item $AD$ is indefinite.
\end{enumerate}
\end{thm}
There is a nontrivial statement; if it is positive semidefinite then its kernel is rank one.  This is not hard to prove; you don't need to know anything about Lie algebras.  The thing which is tricky is to say that in the second case $A$ comes from an affine Lie algeba.

Then there is the third case where the matrix is indefinite, has positive and negative eigenvalues.  I'm not really going to discuss the third case; you can still define things, but it will look very very different from what we have in the finite or affine case; the multiplicity of roots will be unbounded.

Let me just briefly say the reason.  In the first case we have positive length roots.  In the second case we had zero length roots which caused a little bit of trouble.  The positive length roots give something looking like $\la{sl}(2);$ the zero length roots look like Heisenberg algebras.

In the third case you get the same picture but also some negative length roots, and I can't even describe in the language we've been using what the thing generated by these is.  It will look very different and I am not going to study it.  This is sometimes referred to as hyperbolic.

In the remaining ten minutes let me say a little about the twisted case and the example.

The example will be of type two.  It will be twisted, and let me explain what is meant by twisted.  

Suppose you start with a finite dimensional semisimple Lie algebra $\la{g}$ with some automorphism $\sigma$ of finite order $N.$

Then we can consider $\hat{\la{g}}$ which will be $\la{g}\otimes\mathbb{C}[t,t^{-}\oplus\mathbb{C}k\oplus\mathbb{C}d.$  So now let's extend $\sigma$ to $\hat{g}.$  Not the obvious way, but as follows: $\sigma(xt^k)=\sigma(x)(e^{2\pi i/N} t)^k.$  I leave it to you to check that this is still an automorphism.  Of course, it's still of the same finite order, namely $n.$

Consider $(\hat{\la{g}})^\sigma,$ the subalgebra of invariants.  This can be defined in a slightly more natural way; if you think of $\hat{\la{g}}$ as polynomials on the circle with values in $\la{g},$ but a multivalued function on the circle where going around once changes the value by $\sigma.$  This explains the word twisted but I don't want to get too deeply into it.

Let me show you one example, but just to show an example, suppose that you have $\la{g}=\la{sl}(3)$ and $\sigma$ interchanging $e_1$ and $e_2,\ f_1$ with $f_2,$ and $h_1$ with $h_2.$  So what will be this algebra?  First of all, what elements appear at level zero?  $\hat{\la{g}}^\sigma[0]=\{xt^0|\sigma(xt^0)=xt^0\}.$  It doesn't matter what I multiply it with, so this is just $\{x\in \la{g}|\sigma(x)=x\}.$  What are the invariants?  The obvious ones are the sums $e_1+e_2,f_1+f_2,$ and $h_1+h_2.$  What about at level one?  Here I have $\{xt|\sigma(x)=-x\}=\langle e_1-e_2,f_1-f_2,h_1-h_2, [e_1,e_2],[f_1,f_2]\rangle.$  So we have a five dimensional space at this level.  The picture I would make would be like this:
$$
\xymatrix{
&&\vdots&\vdots&\vdots
\\3&\bullet &\bullet &\bullet &\bullet &\bullet 
\\2&&\bullet &\bullet &\bullet 
\\1&\bullet &\bullet &\bullet &\bullet &\bullet 
\\0&&\bullet &\bullet\ar[r]\ar[ull] &\bullet 
\\&&\vdots&\vdots&\vdots
}$$

Most of what I say will generalize obviously to these but I won't say much about these specifically.  Next time we'll talk about representations.
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