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\title{Infinite Dimensional Lie Algebras\\February 9, 2005}
\author{Gabriel C. Drummond-Cole}
\newtheorem{lemma}{Lemma}
\newtheorem{exercise}{Exercise}
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\newtheorem{example}{Example}
\newtheorem{defi}{Definition}
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\begin{document}
\noindent
\maketitle  
Last time: we say that a lattice is integral if $(L,L)\in\mathbb{Z}.$ What is the order of the set $\{\lambda_0+L,(\lambda_0,L)\in\mathbb{Z}\}?$  That was the question.  So $L=\oplus\mathbb{Z}e_i,\ A=(e_i,e_j),$ and $|L^*/L|=\det A.$


For example, let $L=\langle e_1,e_2\rangle$ be the root lattice for $A_2.$  Then $L^*$ is the weight lattice, $A=\left(\begin{array}{cc}2&1\\1&2\end{array}\right)$ and $|L^*/L|=3.$

Okay, so $\hat{\la{h}}=\la{h}\oplus\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}$ subject to the relations $[xt^n,yt^m]=n(x,y)\delta_{n,-m}.$  Here $xt^n=x[n].$  So when $\la{h}=\mathbb{C}$ then $a[n]=a_n.$

Now $W=Vect(S^1)\otimes \mathbb{C}=\langle d_i\rangle.$  Here $d_n=-t^{n+1}\frac{d}{dt}.$  The relations are $[d_n,d_m]=(n-m)d_{n+m}.$

So $W$ acts on $\hat{\la{h}}$ by $d_n x[k]=-kx[k+n].$ 

A question: can we define the action of $W$ in representations of $\hat{\la{h}}?$  So $d_n\to L_n\in End(\mathscr{F}_\lambda).$  Here $[L_n,x[k]]=d_nx[k]=-kx[k+n].$  For $h=\mathbb{C}$ we have $L_n=\frac{1}{2}\sum_{k+l=n}:a_ka_l:.$  Oh, $L_0=H.$

So why this formula?  Well, acting by $a$ decreases the number of $a$ by one so it should be quadratic.  By degree considerations, $L_n$ must be of degree $n.$  So $k+l=n.$  So it must be $\sum c_{kl}:a_ka_l:$ and then solving for constants gives this formula.

Now 
$$[L_n,L_m]=\frac{1}{2}[L_n,\sum :a_ka_l:]=\frac{1}{2}\sum([L_n,a_k]a_l+a_k[L_n,a_l])
$$$$=\frac{1}{2}\sum (-ka_{n+k}a_l-la_ka_{l+n})
.$$
Call this *.  Now, if $n-m\ne 0$ this becomes
$$\frac{1}{2}\sum -ka_{n+k}a_l+\frac{1}{2}\sum -la_ka_{l+n}$$$$
=\frac{1}{2}\sum_{i+j=m+n}(n-i)a_ia_j+\frac{1}{2}\sum (i-m)a_ia_k$$$$
=\frac{1}{2}\sum_{i+j=m+n}(n-m)a_ia_j=(n-m)L_{m+n}.$$

When $m+n=0$ * makes sense since for all but finitely many terms we get cancellation.  There we have 
$$\frac{1}{2}\sum ((n-i)a_ia_{-i}+ia_{i-n}a_{n-i})
$$ which we turn into $$
=\frac{1}{2}\sum :((n-i)a_ia_{-i}+ia_{i-n}a_{n-i}):+
\frac{1}{2}\sum (n-i)ix(i)+i(i-n)x(i-n)$$$$
=2nL_0+\frac{1}{2}\sum_{i=0}^n (n-i)i=2nL_0+\frac{1}{12}(n^3-n).$$
Here $x(i)$ is zero for negative $i$ and $1$ otherwise so that $:a_ia_{-i}:+ix(i)=a_{i}a_{-i}.$
So $[L_n,L_m]=(n-m)L_{n+m}+\delta_{n,-m}\frac{n^3-n}{12}.$
\begin{defi}$Vir=W\oplus \mathbb{C}c=\langle L_n,c\rangle$ with commutation relations $[L_n,L_m]=(n0m)L_{n+m}+\delta_{n,-m}\frac{n^3-n}{12}c$ and $[c,L_n]=0.$
\end{defi}
\begin{thm}
The action of $W$ on $\hat{\la{h}}$ extends to an action of $Vir$ on $\mathscr{F}_\lambda$ such that $c$ acts by $\dim\ \la{h}.$ [i.e., a representation with (Virasoro) central charge $\dim\ \la{h}$]
\end{thm}
Instead of writing commutation relations between $a_n,L_n,$ consider $a(z)=\sum_{n\in \mathbb{Z}}a_nz^{-n-1},$ that is, what physicists call functions in $z$ with operator values.
Then $$[a(z),a(w)]=\sum [a_n,a_m]z^{-n-1}w^{-m-1}=\sum_{n\in\mathbb{Z}}nz^{-n-1}w^{n-1}=z^{-1}\delta_w(\sum_{n\in\mathbb{Z}}(w/z)^n)$$
where this last sum is the Fourier series for the $\delta$ function at $w/z.$  Then this is $z^{-1}\delta_w\delta(w/z),$ where $\delta(z)=\sum z^n.$
\begin{exercise}
For $P$ a Laurent polynomial in $z$ and $w,\ P(z,w)\delta(z/w)=P(z,z)\delta(z/w)=P(w,w)\delta(z/w).$  This is similar to $f(x)\delta(y)=f(y)\delta(x).$
\end{exercise}
Now $L(z)=\sum L_n z^{-n-2}=\frac{1}{2}\sum_{k,l}:a_ka_l:z^{-(k+l)-2}=\frac{1}{2}:a^2(z):.$
The commutation relations are $$[L(z),a(w)]=\sum_{n,k}[L_n,a_k]z^{-n-2}w^{-k-1}
$$$$
=\sum -ka_{n+k}z^{-n-2}w^{-k-1}=\sum a_i(\sum_{n+k=i}-kz^{-n-2}w^{-k-1})$$
Try to figure out where to go from here.

Okay, now
$$
\frac{1}{2}[:a^2(z):,a(w)]=\frac{1}{2}[a^2(z),a_w)]=\frac{1}{2}([a(z),a(w)]a(z)+a(z)[a(z),a(w)])
$$$$
=\frac{1}{2} (z^{-1}\delta_w\delta(w/z)a(z)+a(z)z^{-1}\delta_w\delta(w/z))
$$$$
=z^{-1}a(z)\delta_w\delta(w/z).$$
As homework, try to write $[L(z),L(w)].$  Here is a reference: E. Frenkel and B. Zvi, ``Vertex Algebras and Algebraic Curves,'' and V. Kac, ``Vertex Algebras for Beginners.''

The operators ``almost'' commute except when $w=z.$


\end{document}
  