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\begin{document}

\title{The hypergeometric function
}
 \author{Speaker: Daniel Moseley
 \\ Typist:  Emily Peters
 }
 \date{\today}
 \thanks{Available online at \texttt{http://math.mit.edu/$\sim$eep/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!

Want to relate $F_\mu$ and $G_\mu$ after analytic continuation.  Writing $F_\mu$s in terms of $G_\nu$s -- coefficients are "transport coefficients."


\begin{enumerate}
\item Hypergeometric function/equation
\item Compute transport coefficients for the ``Basic ODE''
\end{enumerate}

\begin{defn}
{\em Gauss's hypergeometric equation:}  second order ODE with 3 regular singular points  $\{ 0, 1, \infty \}$:
$$ z(1-z)f''+[c-(1+a+b)z]f'-abf=0.
$$
What's cool about this are its solutions, built from
$${}_2F_1\left( a,b; c; z \right) = \Sigma_{n \geq 0} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}
$$
with $(a)_n:=a(a+1) \cdots (a+n-1)$.

Rewrite differential equation as 
$$F'(z)=(\frac{A}{z} + \frac{B}{1-z})F(z)$$
with
$ A =\left( \begin{array}{cc}
0 & 1 \\
0 & 1-c
\end{array} \right)
$
and $B=\left( \begin{array}{cc}
0 & 0 \\
-ab & 1+a+b-c
\end{array} \right)$
and $F=\left( \begin{array}{c}
f(z)\\
z f'(z)
\end{array} \right)$

Replacing $A,B$ by n-by-n matrices, this same equation is called an {\em abstract hypergeometric equation}.  Here $(a)_n = \frac{\Gamma(a+n)}{\Gamma(n)}$ with $\Gamma(x)=\int_0^\infty t^{x-1} e^{-t} dt$; for natural numbers $x$, $\Gamma(x)=x!$.
\end{defn}

Notation and assumptions:  $f'(z)=\frac{Pf}{z} + \frac{Qf}{1-z}$, $f:\C \rightarrow V=\C^n$, $P,Q \in \operatorname{End}(V)$.  $P$ has eigenvalues $\lambda_i$ with differ mod 1, and eigenvectors $\xi_i$.  $Q$ is a nonzero multiple of a rank-one idempotents, so $Q^2 = \delta Q$ for some nonzero $\delta$; so $tr(Q)=\delta$.  Rank one implies that there's some $\phi \in V^*$ and $v \in V$ so that $Q(x)=\phi(x)v$, $\phi(v)=\delta$.

$Q$ is in general position w.r.t.~$P$:  $v=\Sigma \delta_i \xi_i$, with $\delta_i \neq 0$.

Choose eigenvectos so that $\phi(\xi_i)=1$.  Let $R=Q-P$ and suppose $R$ satisfies the same conditions as $P$ with respect to $Q$.  Let $(\zeta_j, -\mu_j)$ be the normalized eigenvectors/values.

Look at power series for the solutions around zero, call these $f_i$:
$f_i(z) = \Sigma_n \xi_{i,n} z^{\lambda_i + n}$; $\xi_{i,0}=\xi_i$.  This converges in $\{z:  \vert z \vert < 1, z \not\in[0,1) \}$.

Similarly the solutions around infinity, $g_j$:
$g_j(z)= \Sigma_n \zeta_{j,n} z^{\mu_j-n}$; $\zeta_{j,0}=\zeta_j$.

Extend analytically and compare:
$$f_i(z)= \Sigma_{j} c_{ij} g_j(z)
$$

Goal:  compute $c_{ij}$.

We'll show:
$$c_{ij}=e^{i \pi (\lambda_i - \mu_i)} \frac{
\Pi_{k \neq i} \Gamma(\lambda_i-\lambda_k+1) \Pi_{\ell \neq j}\Gamma(\mu_j-\mu_\ell)}
{\Pi_{\ell \neq j} \Gamma(\lambda_i - \mu_\ell + 1) \Pi_{k \neq i} \Gamma(\mu_j - \lambda_k)}
$$

\begin{fact}
The transport matrix depends only on the eigenvalues of $P$ and $P-Q$.  This delendence is holomoprhic.
\end{fact}

\begin{fact}
$\sigma, \tau \in S_N$; $c_{ij}(\lambda_1, \ldots, \lambda_N, \mu_1, \ldots \mu_N)=c_{\sigma(i), \tau(j)}(\lambda_\sigma(1), \ldots, \lambda_\sigma(N), \mu_\tau(1), \ldots \mu_\tau(N))$.
\end{fact}

Look at $\phi(f_1(z))$ because $\phi(f_1(z))= \sum_j c_{ij} \phi(g_j(z))$.  Recall power series for $f_1(z)$.  

$\Sigma_{n \geq 0} (n+\lambda_1) \xi_{1,n}z^n = \Sigma_{n \geq 0} P \xi_{1,n} z^n+Q(1+z+z^2 + \ldots)\Sigma_{n \geq 0} \xi_{1,n} z^n$.

$(n+\lambda_1-P)\xi_{1,n} = Q(\xi_{1,0} + \cdots + \xi_{1,n-1})$

Let $\alpha_{1,n}=\phi(\xi_{1,0} + \cdots + \xi_{1,n})$.

By normalization $\alpha_{1,0}=1$.

Black box (see Wasserman): $\alpha_{1,n}=\Pi_{j=1}^m \Pi_{m=1}^n \frac{m+\lambda_i-\mu_j}{m+\lambda_i - \lambda_j}$

Now get $\frac{\phi(f_1(z))}{z^\lambda_1(1-z)} = \Sigma_{n \geq 0} \alpha_{1,n} z^n = \Sigma_{n \geq 0} z^n \Pi_{j=1}^m \Pi_{m=1}^n \frac{m+\lambda_i-\mu_j}{m+\lambda_i - \lambda_j}$.

Restrict the $\lambda_i$'s, $\mu_j$'s to real numbers; $\lambda_i$'s, $\mu_j$'s are written in order $\lambda_i > \lambda_{i+1}$, $\lambda_1+1>\mu_j > \lambda_j$ for all $j$.

Apply identity 
$$ \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} = \int_0^1 t^{a-1} (1-t)^{b-1} dt
$$
(true when $a,b > 0$).

So,
$$\phi(f_1(z))=(1-z)z^{\lambda_1} k \int \int \cdots \int (1-zt_2 \cdots t_N)^{\mu_1 - \lambda_1 - a} \cdot \Pi_{j\neq1}t_j^{\lambda_1 - \mu_j}(1-t_j)^{\mu_j-\lambda_j-1} dt_j
$$
with $k=\Pi_{j \neq 1} \frac{\Gamma(\lambda_1-\lambda_j+1)}{\Gamma(\lambda_1-\mu_j+1)\Gamma(\mu_j-\Lambda_j)}$

Black box:
$\phi(g_j(z)) \approx \vert z \vert^{\mu_j} e^{\pi i \mu_j}$ is $z \in \R$ is large negative.

So $\phi(g_j(z)) \approx c_{11} \vert z\vert ^ {\mu_1} e^{\pi i \mu_1}$;

Thus $\phi(f_1(z)) \approx K e^{i \pi \lambda_1} \vert z \vert^{\mu_1} \Pi_{j\neq 1} \in_0^1 t_j^{\mu_1-\mu_j+1}(1-t_j)^{\mu_j-\lambda_j-1}dt_j $

and undoing the $\beta$ identity, 

$$c_{11}=e^{i \pi (\lambda_1-\mu_1)} \Pi_{j \neq 1} \frac{\Gamma(\lambda_1-\Lambda_j+1) \Gamma(\mu_1 - \mu_j)}{\Gamma(\lambda_1-\mu_j+1) \Gamma(\mu_1-\lambda_j)}$$

\end{document}

































