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\begin{document}

\title{Morning Discussion
}
 \author{Speaker:  Andr\'{e} Henriques
 \\ Typist:  Emily Peters
 }
 \date{\today}
 \thanks{Available online at \texttt{http://math.mit.edu/CFTworkshop}.  Please email \texttt{eep@math.mit.edu} with corrections and improvements!}

\maketitle

\begin{abstract}
Notes from the ``Conformal Field Theory and Operator Algebras workshop," August 2010, Oregon.
\end{abstract}

%%% Start typing here!
There were some questions about the physics motivation for what we're doing.

Minkowski space

``everything" -- $H_0$.

Quantum fields.  So, what is a quantum field?  It's an perator valued distribution -- a map from smooth functions on Minkowski space to $End(H_0)$.  If it was bounded maps it would be nice, but in typical examples its' not.  This should be linear, although that's difficult to define.  

So, a quantum field theory is a choice of $H_0$ and a bunch of fields.  These fields are subject to some relations w.r.t.~``composition'' -- ie a multiplication in $End(H_0)$, which ipso facto is not well-defined.  This multiplication might be "operator product expansion?"  (Audience member:  No, it's not.  This is the Liebniz setting -- OPE is unnecesary/innapropriate here.)

This formalism is called ``Wightman fields."

We would like to take $\R^4$ as our Minkowski space -- but this is hard to work with.  Good examples are not known.

So we try an easier case -- $\R^2$ is the smallest spacetime.  

Symmetries of Minkowski space:  Poincare group including translations, rotations.  We also need to Poincase group to act on $H_0$; so It acts on both sides of the operator valued distribution (smooth function on Minkowski space, and $End(H_0)$.  

For {\em conformal field theory}, we add more symmetries:  look at conformal diffeomorphisms of $\R^2$.  (conformal means the metric is sent to a function times the metric).

We have the nice formula $Conf(Mink) = Diff_+(\R)  \times Diff_+(\R)$.  Why?

we have distinguished light rays in $\R^2$; these give us foliations.

\includegraphics[scale=.5]{Tuesday9amPicture1.jpg}

By defn of conformal, this foliation is preseved by any conformal map.  So our only freedom is to reparametrize in the NE or the NW directions.

A {\em chiral} conformal field theory is a QFT on $\R$ with symmetry group $Diff_+(\R)$.

Last step:  replace $\R$ by $\R \cup \{ \infty \} = S^1$, for the sake of compactness.  

Caution:  anomolies in $Diff(S^1)$, ie it doesn't really act on $H_0$ but it acts projectively.

\begin{example} Let's see an example of a field.  

In a Dirac field, generated by two fields $\psi$ and $\psi^*$, we have $\psi(z)$ ($z \in S^1$ for now) subject to relations
\begin{align*}
[\psi(z),\psi(w)]_+&=0 \\
[\psi^*(z),\psi^*(w)]_+&=0 \\
[\psi(z),\psi^*(w)]_+&= \delta(z-w)\\
\end{align*}
with $[a,b]_+ = ab+ba$.

Now, let $H_0$ be Fock space $\mathcal{F}_P = \Lambda(PH) \tensor \Lambda(PH^\perp)^*$ with $H=L^2(S^1)$ and $PH$= Hardy space, namely the (closure of the) span of things of the form $z^n$ with $n\geq 0$.

Now, 
\begin{align*}
\psi: \{ L^2 \text{ functions on } S^1\} & \rightarrow B(H_0) \\
f & \mapsto \psi(f):=a(f) \\
\end{align*}
With $a(f)$ being yesterday's creation operator.  $a(f)$ satisfies $[a(f),a^*(g)]_+= \left< f,g \right>$.

Formally, $\psi(z)=\psi(\delta_z)$ with $a(f)=\int f(z) \psi(z)$.  This is a ``smeared field.''  We can check that this still satisfies the right commutation relations.

i.e., 
\begin{align*}
[a(f),a^*(g)]_+ & = [\int f(z) \psi(z), g(w)^* \psi^*(w)] \\
& = \int \int f(z) \overline{g(w)}[\psi(z), \psi^*(w)]_+=\int f(z) \overline{g(z)}
\end{align*}


\end{example}

Note:  Fields that go with $SU(N)$ are ``currents."  Good luck writing down explicit commutation relations for smeared currents.  

What do the computations look like here?  $\g$, $\{ X_\alpha\}$ a basis of $\g$, and $[X_\alpha, X_\beta]=c_{\alpha \beta}^{\gamma} X_\gamma$.  The nonabelian currents are $$[J_\alpha(z), J_\beta(w)]=
\Sigma_\gamma c_{\alpha \beta}^\gamma J_\gamma(w) \delta(z-w) + 
 \ell \left< X_\alpha, X_\beta \right> 
 \frac{\partial}{\partial w} \delta(z-w)
$$.

$J_\alpha(f) = \int f(z) J_\alpha(z)$.  Now $f(z) X_\alpha \in L \g = C^\infty(S^1,\g)$.


\end{document}

































