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\begin{document}

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\begin{slide}

\color{Black}

\begin{center}
\Large A quick introduction to\\
       Lie and vertex algebras
    
William J. Cook \\
\color{YellowOrange}
Appalachian State University\\
\texttt{cookwj@appstate.edu}
       
\color{Black}
Friday, January 30, 2009
\end{center}
\end{slide}

\begin{slide}
{\large\bf\color{blue} Outline:}

\begin{description}
\item[{\color{blue}The Classical}] Associative and Lie Algebras
\item[{\color{blue}The Non-Classical}] Vertex Algebras
\item[{\color{blue}A Non-Trivial Example}] the $\hat{\mathfrak{sl}}_2$-module $L(\Lambda_0)$
\item[{\color{blue}Identities}] A character formula and combinatorial identities
\end{description}
\end{slide}

\begin{slide}
\begin{center}
{\large\bf\color{blue} Classical Algebras:\\
                       Associative Algebras\\
                       {\normalsize and}\\
                       Lie Algebras}
\end{center}
\end{slide}


\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{F}$):}

An {\color{blue}Algebra}, $\mathcal{A}$, is a vector space (over some field $\mathbb{F}$) 
equipped with a multiplication map: $m :\mathcal{A} \times \mathcal{A} \rightarrow \mathcal{A}$
such that $m$ is {\color{blue}bilinear}.  

Multiplication is {\it usually} denoted by juxtaposition:\\ $m(u,v)=uv$.

\vfill

{\large\bf\color{red}Examples:}

\small
\begin{description}
   \item[{\color{blue}Real Matrix Algebras}] Let $\mathcal{A} = \mathbb{R}^{n \times n}$ be all
        $n \times n$ matrices with real entries. 
   \item[{\color{blue}Polynomial Algebras}] Let $\mathcal{A} = \mathbb{R}[x,y]$ be all
        polynomials in two indeterminants ($x$ and $y$) with real coefficients.
   \item[{\color{blue}Cross Product Algebra}] Let $\mathcal{A} = \mathbb{R}^3$ here we multiply
        vectors by taking their cross product: $u \times v$.
\end{description} 

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Special Properties:}

Let $\mathcal{A}$ be an algebra (over some field $\mathbb{F}$).

\small
\begin{description}
\item[{\color{blue}Associative}] $\mathcal{A}$ is {\color{blue}Associative} if
     $u(vw)=(uv)w$ for all $u,v,w \in \mathcal{A}$.
\item[{\color{blue}Unital}] $\mathcal{A}$ is {\color{blue}Unital} or an
     algebra with {\color{blue}identity} if there exists some ${\bf 1} \in \mathcal{A}$
     such that ${\bf 1}v = v{\bf 1} = v$ for all $v \in \mathcal{A}$.
\item[{\color{blue}Commutative}] $\mathcal{A}$ is {\color{blue}Commutative} if
     $uv=vu$ for all $u,v \in \mathcal{A}$.
\end{description}

\vfill

{\large\bf\color{red}Examples:}

\begin{itemize}
\item $\mathcal{A}=\mathbb{R}[x,y]$ is a commutative associative unital algebra.
      It's identity is the polynomial $1$.
\item $\mathcal{A}=\mathbb{R}^{n \times n}$ is an associative unital algebra, but
      it is {\it not} commutative (unless $n=1$). It's identity is the {\it identity
      matrix} $I_n$.
\item $\mathcal{A}=\mathbb{R}^3$ equipped with the cross product is a non-associative
      non-commutative algebra and has no identity. So what kind of algebra is this?
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

Let $L$ be an algebra (over some field $\mathbb{F}$). Instead of using
juxtaposition, let's denote multiplication with a {\it bracket}: $m(u,v)=[u,v]$.
$L$ is called a {\color{blue}Lie Algebra} if the following axioms hold:

\small
\begin{description}
\item[{\color{blue}Skew-Commutative}] $[v,v]=0$ for all $v \in L$.
\item[{\color{blue}Jacobi Identity}] $[[u,v],w]+[[v,w],u]+[[w,u],v]=0$ for all $u,v,w \in L$.
\end{description}

\vfill

{\large\bf\color{red}Examples:}

\begin{itemize}
\item $\mathcal{A}=\mathbb{R}^3$ equipped with the cross product is a Lie algebra.
      Remember that $v \times v = 0$ (the cross product of parallel vectors is zero).
      A tedious calculation shows that the Jacobi identity holds as well.
\item If we give the matrix algebra $\mathbb{R}^{n \times n}$ a different multiplication,
      called the {\color{blue}commutator bracket}, defined by $[A,B]=AB-BA$, it becomes
      a Lie algebra. To remind ourselves that we are using a different ``multiplication''
      we call this algebra $\mathfrak{gl}(n,\mathbb{R})$.
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]\color{white}=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]          
      \color{white} Therefore, $[u,v] = -[v,u]$ (almost commutative!) 

\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ 0=[[u,v],w]+[[v,w],u]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]\color{white}=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]          
      \color{white} Therefore, $[u,v] = -[v,u]$ (almost commutative!) 

\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ 0=[[u,v],w]+[[v,w],u]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}
\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]\color{white}=[u,v]+[v,u] \]          
      \color{white} Therefore, $[u,v] = -[v,u]$ (almost commutative!) 

\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ 0=[[u,v],w]+[[v,w],u]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]          
      \color{white} Therefore, $[u,v] = -[v,u]$ (almost commutative!) 

\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ 0=[[u,v],w]+[[v,w],u]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, ${\color{blue}[u,v] = -[v,u]}$ (almost commutative!) \color{white}         

\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ 0=[[u,v],w]+[[v,w],u]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ 0=[[u,v],w]+{\color{blue}[[v,w],u]}+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ {\color{blue}-[[v,w],u]} = [[u,v],w]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ {\color{blue}[u,[v,w]]} = [[u,v],w]+[[w,u],v] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ [u,[v,w]] = [[u,v],w]+{\color{blue}[[w,u],v]} \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ [u,[v,w]] = [[u,v],w]{\color{blue}-[v,[w,u]]} \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ [u,[v,w]] = [[u,v],w]+[v,{\color{blue}-[w,u]}] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

      \[ [u,[v,w]] = [[u,v],w]+[v,{\color{blue}[u,w]}] \]
\end{itemize}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Lie Algebras:}

What do the axioms {\it really} say?

\small
\begin{itemize}
\item The first axiom, $[v,v]=0$, implies the following:\\
      \[ 0=[u+v,u+v]=[u,u+v]+[v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] \]
      Therefore, $[u,v] = -[v,u]$ (almost commutative!)
         
\item The Jacobi identity says much more. Using the above property we can re-write
      the Jocobi identity as follows:\\

\[ \begin{array}{rcccc}
   [u,[v,w]] & = & [[u,v],w] & + & [v,[u,w]] \vspace{0.2in} \\ \color{blue}
   \frac{d}{dt}\left[f(t)g(t)\right] & \color{blue} = & \color{blue} 
   \frac{d}{dt}\left[f(t)\right]g(t) 
   & \color{blue} + & \color{blue} f(t)\frac{d}{dt}\left[g(t)\right] 
\end{array} \]

Let $\mathcal{A}$ be an algebra and $\partial : \mathcal{A} \rightarrow \mathcal{A}$.
If $\partial(uv) = \partial(u)v+u\partial(v)$ for all $u,v \in \mathcal{A}$, then
we say that $\partial$ is a {\color{blue}derivation} of $\mathcal{A}$. 

The Jacobi identity simply says,\\
``The multiplication operators of $L$ are derivations.''
\end{itemize}

\vfill

\end{slide}

\begin{slide}
\begin{center}
{\large\bf\color{blue} Non-Classical Algebras:\\
                       Vertex (Operator) Algebras}
\end{center}
\end{slide}

\begin{slide}
{\large\bf\color{blue} Origins of Vertex Operator Algebras:}

\begin{description}
\item[\color{blue}1970s] Vertex operators appear in the study of {\color{blue} String Theory}.
\item[\color{blue}1980s] Mathematicians use vertex operators to study certain representations 
                         of {\color{blue} affine Lie algebras}.
\item[\color{blue}1984] I. Frenkel, J. Lepowsky, and A. Meurman construct $V^{\natural}$.
\item[\color{blue}1986] R. Borcherds introduces a set of axioms for a notion which he calls a 
                        ``{\color{blue}vertex algebra}''.
\item[\color{blue}1992] R. Borcherds proves the Conway-Norton conjectures.
\end{description}
\end{slide}

\begin{slide}
\begin{center}
{\large\bf\color{blue} The Definition of a Vertex Algebra}
\end{center}
\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

Let $V$ be a vector space over $\mathbb{C}$. 

\vfill

{\large\bf\color{blue}Vertex Algebras:}

Let $V$ be a vector space over $\mathbb{C}$.

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

Equip $V$ with a bilinear map $\cdot:V \times V \rightarrow V$ (a multiplication map).

$(u,v) \mapsto u \cdot v$


\vfill

{\large\bf\color{blue}Vertex Algebras:}

Equip $V$ with {\color{blue} infinitely many} bilinear maps $n:V \times V \rightarrow V$ (where $n \in \mathbb{Z}$).

$(u,v) \mapsto u_n v$

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

Equip $V$ with a bilinear map $\cdot:V \times V \rightarrow V$ (a multiplication map).

$(u,v) \mapsto u \cdot v$


\vfill

{\large\bf\color{blue}Vertex Algebras:}

Equip $V$ with a bilinear map $Y(\cdot,x):V \times V \rightarrow V[[x,x^{-1}]]$\\
($V[[x,x^{-1}]]$ are Laurent series with coefficients in $V$).

$\displaystyle{(u,v) \mapsto {\color{blue}Y(u,x)}v = \sum_{n \in \mathbb{Z}} {\color{blue}u_n} v \, x^{-n-1}}$

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Notation:}

\vspace{-0.6in}
\begin{description}
\item[{\color{blue}$V[x]$}] polynomials in $x$ with coefficients in $V$. \vspace{-0.6in}
\item[{\color{blue}$V[x,x^{-1}]$}] Laurent polynomials in $x$ with coefficients in $V$. \vspace{-0.6in}
\item[{\color{blue}$V[[x]]$}] power series in $x$ with coefficients in $V$. \vspace{-0.6in}
\item[{\color{blue}$V((x))$}] lower truncated Laurent series with coefficients in $V$. \vspace{-0.6in}
\item[{\color{blue}$V[[x,x^{-1}]]$}] Laurent series in $x$ with coefficients in $V$. 
\end{description}


{\bf\color{blue}WARNING: $\mathbb{C}[[x,x^{-1}]]$ is not an algebra!} Sometimes mutiplication isn't well defined.

{\bf Example:} $\delta(x) = \sum_{n \in \mathbb{Z}} x^n$ is the {\color{blue}formal delta function}.
Notice that $(\delta(x))^2$ is undefined.

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

Equip $V$ with a bilinear map $\cdot:V \times V \rightarrow V$ (a multiplication map).

$(u,v) \mapsto u \cdot v$


\vfill

{\large\bf\color{blue}Vertex Algebras:}

Equip $V$ with a bilinear map $Y(\cdot,x):V \times V \rightarrow {\color{blue}V((x))}$\\
($V((x))$ are lower truncated Laurent series with coefficients in $V$).

$\displaystyle{(u,v) \mapsto Y(u,x)v = \sum_{n \in \mathbb{Z}} u_n v \, x^{-n-1}}$

where $u_n v = 0$ for $n \gg 0$.

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

To be a {\color{blue}Lie algebra} $V$'s multiplication must be
skew-symmetric and satisfy the {\color{blue}Jacobi identity}.

$(uv)w+(vw)u+(wu)v=0$

\vfill

{\large\bf\color{blue}Vertex Algebras:}

$V$'s vertex operators must satisfy the {\color{blue}Jacobi identity}.

{\small \[ x_0^{-1} \delta \left( \frac{x_1 -x_2}{x_0} \right) Y(u,x_1)Y(v,x_2)w 
 - x_0^{-1} \delta \left( \frac{x_2 -x_1}{-x_0} \right) Y(v,x_2)Y(u,x_1)w \]
\[ = x_2^{-1} \delta \left( \frac{x_1 -x_0}{-x_2} \right) Y(Y(u,x_0)v,x_2)w \] }

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

To be a {\color{blue}Lie algebra} $V$'s multiplication must be
skew-symmetric and satisfy the {\color{blue}Jacobi identity}.

$(uv)w+(vw)u+(wu)v=0$ \hspace{-4in} {\color{BrickRed}$XXXXXXXXXXXXXXX$}\\
$u(vw) = (uv)w + v(uw)$ (the {\color{blue}product rule}) 

\vfill

{\large\bf\color{blue}Vertex Algebras:}

$V$'s vertex operators must satisfy the {\color{blue}Jacobi identity}.

{\small \[ x_0^{-1} \delta \left( \frac{x_1 -x_2}{x_0} \right) Y(u,x_1){\color{blue}\left(\color{black}Y(v,x_2)w\color{blue}\right)} = \]
\[ x_2^{-1} \delta \left( \frac{x_1 -x_0}{-x_2} \right) Y({\color{blue}Y(u,x_0)v},x_2)w 
 + x_0^{-1} \delta \left( \frac{x_2 -x_1}{-x_0} \right) Y(v,x_2){\color{blue}\left(\color{black}Y(u,x_1)w\color{blue}\right)} \] }

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

To be a {\color{blue}unital algebra} $V$ must have an {\color{blue}identity vector ${\bf 1}$}.

\[ {\bf 1}u=u{\bf 1}=u \]

\vfill

{\large\bf\color{blue}Vertex Algebras:}

A vertex algebra has a {\color{blue}vaccuum vector ${\bf 1}$}.

\[ Y({\bf 1},x)u = u \qquad \mathrm{and} \qquad Y(u,x){\bf 1} = e^{x\mathcal{D}}u \]

In particular, $Y(u,0){\bf 1}=u$. \hspace*{2in} {\small [Note: $\mathcal{D}(u) = u_{-2} {\bf 1}$]}

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

$V$ is a {\color{blue}commutative algebra} if...

\[ uv=vu \]

\vfill

{\large\bf\color{blue}Vertex Algebras:}

A vertex operators satisfy a property called {\color{blue}locality}.

\[ (x_1-x_2)^N Y(u,x_1)Y(v,x_2) = (x_1-x_2)^N Y(v,x_2)Y(u,x_1) \]

for some $N \gg 0$

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{blue}Algebras (over $\mathbb{C}$):}

$V$ is an {\color{blue}associative algebra} if...

\[ (uv)w = u(vw) \]

\vfill

{\large\bf\color{blue}Vertex Algebras:}

A vertex operators satisfy a property called {\color{blue}weak associativity}.

\[ (x_1-x_2)^N Y({\color{blue}(}Y(u,x_1)v{\color{blue})},x_2)w = (x_1+x_2)^N Y(u,x_1+x_2){\color{blue}\left(\color{black}Y(v,x_2)w\color{blue}\right)} \]

for some $N \gg 0$

\vfill

\end{slide}

\begin{slide}

{\color{blue}\bf Summary:} vertex algebras generalize both Lie algebras and commutative associative unital algebras.

\vspace{1in}

\color{white}
...which is interesting since $0=[1,1]=1$ so that... 
\[ u = [u,1] = [u,0] = 0 \] 
Classically speaking, these structures don't go together at all!

\end{slide}

\begin{slide}

{\color{blue}\bf Summary:} vertex algebras generalize both Lie algebras and commutative associative unital algebras.

\vspace{1in}

...which is interesting since $0=[1,1]=1$ so that... 
\[ u = [u,1] = [u,0] = 0 \] 
\color{white}
Classically speaking, these structures don't go together at all!

\end{slide}

\begin{slide}

{\color{blue}\bf Summary:} vertex algebras generalize both Lie algebras and commutative associative unital algebras.

\vspace{1in}

...which is interesting since $0=[1,1]=1$ so that... 
\[ u = [u,1] = [u,0] = 0 \] 
{\color{red}Classically speaking, these structures don't go together at all!}

\end{slide}

\begin{slide}
{\large\bf\color{red}A ``Trivial'' Example:}

{\color{blue}Holomorphic Vertex Algebras} 

\small
Let $\mathcal{A}=\mathbb{C}[t]$ be the commutative associative unital algebra of complex polynomials and let $D:\mathcal{A} \rightarrow \mathcal{A}$ be the derivation 
$\displaystyle{D = \frac{d}{dt}}$.

Define $Y(u,x) = e^{xD}u$. That is:
\[ Y(f(t),x) = e^{x\frac{d}{dt}}f(t) = f(t+x) \]
So...
\[ \color{blue} f(t)_{-n-1}g(t) = \mathrm{\;coefficient\;of\;}x^n\mathrm{\;in...} \]
\[ {\color{blue}Y(f(t),x)g(t) = f(t+x)g(t)}. \]

$\mathcal{A}=\mathbb{C}[t]$ equipped with this vertex operator map, $Y(\cdot,x)$,
is a vertex algebra with vacuum vector $1$.

\vfill

\end{slide}

\begin{slide}
{\large\bf\color{red}A ``Trivial'' Example:}

{\color{blue}Holomorphic Vertex Algebras} 

\small
Let $\mathcal{A}$ be a commutative associative unital algebra (whose identity is ${\bf 1}$)
and let $D:\mathcal{A} \rightarrow \mathcal{A}$ be a derivation of $\mathcal{A}$.

Define $Y(u,x) = e^{xD}u$. That is:
\[ Y(u,x)v = \sum_{n=0}^{\infty} u_{-n-1} v x^n \qquad \mathrm{where} \qquad
   u_{-n-1} v = \frac{1}{n!}D^n(u)v \]

Then, $\mathcal{A}$ equipped with this vertex operator map, $Y(\cdot,x)$,
becomes a vertex algebra with vacuum vector ${\bf 1}$.

{\it Note:} If $D=0$, we simply have $Y(u,x)v=uv$. That is:\\
{\color{blue}All commutative associative unital algebras are vertex algebras.}

\vfill

\end{slide}

\begin{slide}
\begin{center}
{\large\bf\color{blue} An Example: $\hat{\mathfrak{sl}}_2$ and the $\hat{\mathfrak{sl}}_2$-module $L(\Lambda_0)$}
\end{center}
\end{slide}

\begin{slide}
{\large\color{blue} The 3-dimensional simple Lie algebra $\mathfrak{sl}_2$:}

\[ \mathfrak{sl}_2(\mathbb{C}) = \left\{ A \in \mathbb{C}^{2 \times 2} 
   \,\large|\, \mathrm{tr}(A)=0 \right\} \]

The vector space $\mathfrak{sl}_2$ becomes a Lie algebra when given the commutator bracket
${\color{blue}[A,B] = AB-BA}$. 

Consider the following basis for $\mathfrak{sl}_2$:
\[ E = \begin{bmatrix} 0 & 1 \\ 0 &  0 \end{bmatrix}, \quad
   F = \begin{bmatrix} 0 & 0 \\ 1 &  0 \end{bmatrix}, \quad \mathrm{and} \quad
   H = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \]

The commutator brackets among the basis elements are:
\[ [E,F] = H \qquad [H,E]=2E \qquad [H,F]=-2F \]
(use skew-symmetry to find the rest).

\end{slide}

\begin{slide}
{\large\color{blue} Affine $\mathfrak{sl}_2$:}

The infinite dimensional Lie algebra $\hat{\mathfrak{sl}}_2$ is a central extension of the 
loop algebra over $\mathfrak{sl}_2$.

As a vector space, {\color{blue}$\hat{\mathfrak{sl}}_2 = \mathfrak{sl}_2 [t,t^{-1}] \oplus \mathbb{C}c$}.

The element $c$ is central: $[c,x]=0$ for all $x \in \hat{\mathfrak{sl}}_2$.

Let $a,b \in \mathfrak{sl}_2$ and $m,n \in \mathbb{Z}$ 
\[ [a\,t^m, b\,t^n ] = [a,b]\,t^{m+n} + m\,\mathrm{tr}(ab)\delta_{m+n,0}c \]

Let {\color{blue}$\mathfrak{b} = \mathfrak{sl}_2[t] \oplus \mathbb{C}c$}. Notice that $\mathfrak{b}$ is a subalgebra
of $\hat{\mathfrak{sl}}_2$.
\end{slide}

\begin{slide}

{\large\color{blue} The $\hat{\mathfrak{sl}}_2$-module $L(\Lambda_0)$.}

Let $\mathbb{C}_1 = \mathbb{C}{\bf 1}$ be a 1-dimensional $\mathfrak{b}$-module with module action 
defined as follows: \vspace{-0.3in}
\begin{itemize}
\item $c \cdot {\bf 1} = {\bf 1}$ \vspace{-0.3in}
\item $a\,t^n \cdot {\bf 1} = 0$ for all $a \in \mathfrak{sl}_2$ and $n \in \mathbb{Z}_{\geq 0}$
\end{itemize}

We define the $\hat{\mathfrak{sl}}_2$-module $V$ to be
\[ \color{blue} V = U(\hat{\mathfrak{sl}}_2) \otimes_{U(\mathfrak{b})} \mathbb{C}_1. \]
$V$ is an example of a {\color{blue}Verma module}.

Finally, let {\color{blue}$L(\Lambda_0)=V/J$} where $J$ is the maximal proper submodule of $V$
(this implies that $L(\Lambda_0)$ is an irreducible module).

Note: We will denote the action of $a \, t^n$ on $L(\Lambda_0)$ by {\color{blue}$a(n)$}.

\end{slide}

\begin{slide}

Collect all operators $a(n)$ (where $a \in \mathfrak{sl}_2$ and $n \in \mathbb{Z}$) together in a 
generating function: \vspace{-0.2in}
\[ a(x) = \sum\limits_{n \in \mathbb{Z}} a(n) x^{-n-1} \]

\vspace{-0.3in}
For each $n \in \mathbb{Z}$ define the following {\color{blue}$n^{th}$-product}:
\[ a(x)_n b(x) = \mathrm{Res}_{x_1} \left( (x_1-x)^n a(x_1)b(x) - (-x+x_1)^n b(x)a(x_1) \right) \]

Finally define a linear map $Y(\cdot,x)$ as follows:
{\small \[ Y(a^{(1)}(n_1)a^{(2)}(n_2)...a^{(r)}(n_r)\mathbf{1},x) 
                 = a^{(1)}(x)_{n_1}a^{(2)}(x)_{n_2}...a^{(r)}(x)_{n_r}\mathbf{1} \] }
for $r \geq 0$, $a^{(i)} \in \hat{\mathfrak{sl}}_2$ and $n_i \in \mathbb{Z}$.

Notice that for $a \in \mathfrak{sl}_2$ we have:
\[ \color{blue} Y(a,x) = a(x) = \sum_{n \in \mathbb{Z}} a(n)x^{-n-1} \]
\end{slide}

\begin{slide}

{\bf Thm:} $V$ is a vertex (operator) algebra,
           $J$ is an ideal of $V$, and ${\color{blue}L(\Lambda_0)} = V/J$ is a {\color{blue}simple vertex (operator) algebra}.

\end{slide}

\begin{slide}
\begin{center}
{\large\bf\color{blue} Combinatorial Identities}
\end{center}
\end{slide}

\begin{slide}

Define $\displaystyle{\Delta(H,x) = x^{H(0)} \mathrm{exp} \left( \sum\limits_{k=1}^{\infty}
                                  \frac{H(k)}{-k}(-x)^{-k} \right)}$.
				  
{\bf Thm [Li]:} $(L(\Lambda_0),Y(\cdot,x))$ is isomorphic to
           $(L(\Lambda_0),Y({\color{blue} \Delta(H,x)} \cdot,x))$ as an 
	   $L(\Lambda_0)$-module.

{\bf Notation:} Let $v \in L(\Lambda_0)$. 
                    \[ Y(\Delta(H,x)v,x) = \sum\limits_{m \in \mathbb{Z}} v_{(H)}(m)x^{-m-1} \]
		and
		    \[ Y(\Delta(H,x)\omega,x) = \sum\limits_{m \in \mathbb{Z}} L_{(H)}(m)x^{-m-2} \]    

\end{slide}

\begin{slide}

{\bf Def:} The {\color{blue} character} of $L(\Lambda_0)$ is given by:
\[ \chi(x;q) = \mathrm{tr}_{L(\Lambda_0)} x^{\frac{1}{2}H(0)} q^{L(0)} \]

{\bf Lemma:} \vspace{-0.2in}
\begin{eqnarray*}
\left(\frac{1}{2}H\right)_{(H)}(0) & = & \frac{1}{2}H(0) + 1 \\
& & \\
L_{(H)}(0) & = & L(0) + H(0) + 1 \\
           & = & L(0) + 2\left(\frac{1}{2}H(0)\right) + 1 
\end{eqnarray*}

{\bf Recurrence Thm:} 
\[ \chi(x;q) = (xq) \chi(xq^2;q) \]

\end{slide}

\begin{slide}

Solving the recurrence relations obtained from our Recurrence Theorem and 
plugging in an initial condition found by analyzing a ``lattice'' vertex
algebra construction of $L(\Lambda_0)$, we obtain the following formula for the
character of $L(\Lambda_0)$:

{\bf Cor:} 
\[ \chi(x;q) = \prod\limits_{j=1}^{\infty} (1-q^j)^{-1}\sum\limits_{n \in \mathbb{Z}} q^{n^2} x^n \]

\end{slide}

\begin{slide}

The {\color{blue}principal character} is easily obtained from the ``full'' character.\\
{\bf Prop:} $\chi^P(q) = \chi(q^{-1};q^2)$.

{\bf Cor:} We have the following {\color{blue}multi-sum} principal character formula... 
\[ \chi^P(q) = \prod_{j \geq 1} (1-q^{2j})^{-1} \sum_{n \in \mathbb{Z}} q^{2n^2 - n} \]

\end{slide}

\begin{slide}

Using the theory of affine Lie algebras Kac obtains the following
{\color{blue}product} principal character formula...
\[ \chi^P(q) = \prod_{j \geq 1} \frac{(1-q^{2j})}{(1-q^j)} \]

Combining this with our principal character formula, we obtain:
\[ \prod_{j \geq 1} \frac{(1-q^{2j})}{(1-q^j)} = 
   \prod_{j \geq 1} (1-q^{2j})^{-1} \sum_{n \in \mathbb{Z}} q^{2n^2 - n} \]

Rearranging terms a little bit gives us the following identity (credited to Gauss):
\[ \prod_{j \geq 1} \frac{(1-q^{2j})^2}{(1-q^j)} = \sum_{n \in \mathbb{Z}} q^{ 2n^2 - n} \]

\end{slide}

\end{document}

