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\title{Infinite Dimensional Lie Algebras\\February 14, 2005}
\author{Gabriel C. Drummond-Cole}
\newtheorem{lemma}{Lemma}
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\begin{document}
\noindent
\maketitle  
Okay, let's see what's going on.  We have $\mathscr{H}=\langle a_n,n\in\mathbb{Z}\rangle,$ subject to the commutation relations $[a_n,a_m]=n\delta_{n,-m}$ and then $a(z)=\sum a_nz^{-n-1}$ with relations $[a(z),a(w)]=z^{-1}\delta_w\delta(w/z).$

[something about vector fields on $S^1$]

Then there are $L_n$ with $[L_n,a_k]=-ka_{n+k}$ and $L(z)=\sum L_n z^{-n-2}=\frac{1}{2}:a^2(z):$ and then we have $[L(z),a(w)=z^{-1}a(w)\delta_w\delta(w/z).$  

We want to be able to think of $a(z),L(z)$ as operator valued functions, i.e., $a(z)$ as a function on $\mathbb{C}^\times$ with values in $End(\mathscr{F}).$

So $a(z)\ket{0}=\sum a_n\ket{0}z^{-n-1}=\sum_{n<0}a_n\ket{0}z^{-n-1}.$  This is bad because it's not in $\mathscr{F}.$  We have $\mathscr{F}=\mathbb{C}[a_{-1},a_{-2},\cdots ]\ket{0}.$  We want to work with $\bar{\mathscr{F}}=\mathbb{C}[[a_{-1},a_{-2},\cdots]]\ket{0},$ that is, infinite linear combinations.  So this is $\sum v_k$ where the degree of $v_k$ goes to negative infinity.  So $a(z)$ is a function on $\mathbb{C}^\times$ with values in $Hom(\mathscr{F},\bar{\mathscr{F}}).$

There is a problem, as usual.  Such operators cannot be multiplied.  E.g., can $a(z)a(w)\ket{0}\in\bar{\mathscr{F}}$ be defined?
So $a(z)a(w)\ket{0}=a(z)\sum a_{-n}w^{n-1}=\sum_{n>0}n z^{-n-1}w^{n-1}\ket{0}$ plus other terms.  This last part can be written $z^{-1}w^{-1}\sum n(\frac{w}{z})^n.$  Recall that $a_na_{-n}\ket{0}=n\ket{0}$ for $n>0.$

\begin{thm}
\begin{enumerate}\item
For $|w|<|z|\  a(z)a(w)=:a(z)a(w):+\frac{1}{(z-w)^2}$ where the fraction is expanded in a power series in $w/z.$
\item $:a(z)a(w):$ is a well-defined operator $\mathscr{F}\to\bar{mathscr{F}}$ for any $z,w\in\mathbb{C}^*.$
\end{enumerate}\end{thm}
\begin{cor}
$a(z)a(w)$ can be defined as a function of $z,w\in \mathbb{C}^*$ for $|w|<|z|$ with values in $Hom(\mathscr{F},\bar{\mathscr{F}}).$
\end{cor}
So for the proof of the theorem, for the first part we have
$$
a(z)a(w)=\sum a_na_mz^{-n-1}w^{-m-1};$$$$
:a(z)a(w):=\sum :a_na_m:z^{-n-1}w^{-m-1}.$$
The only places these differ are when the first has $a_na_{-n}z^{-n-1}w^{n-1}$ and the second $a_{-n}a_nz^{-n-1}w^{n-1}.$
So $$a(z)a(w)-:a(z)a(w):=\sum_{n>0} nz^{-n-1}w^{n-1}
$$$$
=
z^{-1}\sum_{n\ge 0}\frac{nw^{n-1}}{z^n}=z^{-1}\delta_w(\sum_{n\ge 0}(\frac{w}{z})^n)
$$$$=z^{-1}\delta_w(\frac{1}{1-\frac{w}{z}})=\delta_w\frac{1}{z-w}=\frac{1}{(z-w)^2}.$$

\begin{cor}
Matrix coefficients of $a(z)a(w)$ can be analytically continued to $(\mathbb{C}^*)^2\backslash diagonal$ and they will have a second order pole at $z=w.$  Physicists write this as $a(z)a(w)\sim \frac{1}{(z-w)^2}.$
\end{cor}
So this is $a(w)a(z)=:a(w)a(z):+\frac{1}{(z-w)^2}$ for $|z|<|w|$ and $a(z)a(w)=:a(z)a(w):+\frac{1}{(z-w)^2}$ for $|z|>|w|.$  Now, this means that $a(z)a(w)-a(w)a(z)$ is nowhere defined.  As formal power series this would be the expansion of $\frac{1}{(z-w)^2}$ for $|z|<|w|$ less the expansion of the same function for $|w|>|z|.$

So without analytic continuation $[a(z),a(w)]$ is nowhere convergent; with analytic continuation $[a(z),a(w)]=0.$  The moral is that all information about $[a_n,a_k]$ can be obtained from the polar part of $a(z)a(w)\sim\frac{1}{(z-w)^2}.$

\begin{example}
$a(z)L(w)=:a(z)L(w):+\frac{a(w)}{(z-w)^2}$ for $|w|<|z|.$
\end{example}
\begin{defi}
$:a_nL_k:=a_nL_k$ for $n<0,\ L_ka_n$ for $n\ge 0.$
\end{defi}
So 
$$a(z)L(w)-:a(z)L(w):=\sum_{n\ge 0,k\in\mathbb{Z}}[a_n,L_k]z^{-n-1}w^{-k-2}
$$$$=\sum na_{n+k}z^{-n-1}w^{-k-2}
=\sum nz^{-n-1}w^{n-1}a_{n+k}w^{-n-k-1}
$$$$
=(z^{-1}\sum_{n\ge 0}\frac{nw^{n-1}}{z^n})(\sum a_lw^{-l-1})=\frac{1}{(z-w)^2}a(w).
$$
So $a(z)L(w)=:a(z)L(w):+\frac{a(w)}{(z-w)^2}$ for $|w|<|z|$ and $L(w)a(z)=:a(z)L(w):\frac{a(w)}{(z-w)^2}$ for $|w|>|z|.$  Then $a(z)L(w)\sim \frac{a(w)}{(z-w)^2}.$
\end{document}
  