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\title[Lecture 35]{Complex analysis: Lecture 35}
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\begin{document}
\maketitle

\PP{35.0}{Review} Let $\tau\in\C$ be such that $\Im(\tau)>0$. 
We defined Jacobi's theta function, denoted by $\theta(z)$ (or, by 
$\theta(z;\tau)$ if dependence on $\tau$
needs to be highlighted) in Lecture 34, \S 34.1, as follows. \\

$\theta(z)$ is the unique holomorphic function, defined on the entire
complex plane $\theta:\C\to\C$, which satisfies the following
three properties:\\

\begin{equation}
\tag{Periodicity}
\theta(z+1) = -\theta(z)\qquad\text{and}\qquad
\theta(z+\tau) = -e^{-\pi\iota\tau}e^{-2\pi\iota z}
\theta(z)
\end{equation}

\begin{equation}
\tag{Zeroes}
\theta(z)=0 \text{ if, and only if }
z\in\Lambda_\tau
\end{equation}
Moreover, the order of vanishing of $\theta(z)$
at the points $m+n\tau$ ($m,n\in\Z$) is $1$.

\begin{equation}
\tag{Normalization} \theta'(0)=1
\end{equation}

\begin{enumerate}
\item An infinite product expression for $\theta(z)$ was given in
Lecture 34, \S 34.4. From this we concluded that $\theta(-z)=-\theta(z)$.\\

\item We proved in Lecture 34, \S 34.2 that the three properties
listed above uniquely determine $\theta(z)$.\\

\item Finally, we showed that any doubly--periodic function can be
written in terms of theta function. More precisely, if $f(z)$ is
a doubly--periodic function, then we can find
$a_1,a_2,\ldots,a_N \in \C$ and $b_1,b_2,\ldots,b_N\in\C$
two collections of complex numbers satisfying
\[
a_1+a_2+\cdots+a_N = b_1+b_2+\cdots+b_N\ ,
\]
such that $\ds f(z) = C \prod_{k=1}^N \frac{\theta(z-a_k)}{\theta(z-b_k)}$,
where $C\in\C$ is a constant.\\
\end{enumerate}

\noindent {\bf A word on the method of proofs.}
In Lecture 34, Proposition 34.2 and Theorem 34.5 are proved using a 
very simple and elegant argument, which I would like to emphasize here. 
Assume that we want to prove
an identity of the form $f_1(z)=f_2(z)$. We can do this in the following three steps:
\begin{itemize}
\item Prove that $\ds F(z) = \frac{f_1(z)}{f_2(z)}$ is doubly--periodic.
\item Prove that $F(z)$ is holomorphic, by showing that zeroes and poles
of $f_1$ and $f_2$ cancel each other. At this point we invoke Theorem
33.2 to conclude that $F(z)=C$ is a constant.
\item Finally determine $C$ by evaluating $F(z)$ at a convenient point
$z=z_0$.\\
\end{itemize}

\PP{35.1}{What is and isn't in these notes} 
We will illustrate the method sketched
above, by proving a non--trivial identity (often called {\em Fay's trisecant
identity}
\footnote{J. Fay: {\em Theta functions on Riemann surfaces (1973)}.}
) involving theta function
in \S 35.2 below. In general, it is the method that is important, and
any such identity can be easily proved using it. But I will still write
a nice way to memorize Fay's identity, in \S 35.3. 
It has been conjectured that any relation
involving theta function can be deduced from Fay's trisecant identity
\footnote{D. Mumford: {\em Tata lectures on theta II (1984)}.}
but as far as I know, there is no proof of this.\\

In \S 35.4, I will illustrate a systemetic way of obtaining
a formula of an infinite sum which satisfies the exact
same periodicity property as $\theta(z)$. The resulting
infinite sum is shown to be uniformly convergent in
\S 35.5. Therefore, it defines an entire holomorphic function,
denoted by $\theta_1(z)$. In Theorem 35.5, we will show that
$\theta_1(z)$ has zeroes of order $1$ at $z=m+n\tau$
where $m,n\in\Z$. Hence, $\theta_1(z)$ and $\theta(z)$ must
be related by a constant\footnote{Historically, it is $\theta_1(z)$
that appeared first in the works of Jacobi.}.\\

The computation of this constant is {\bf not} in these notes,
though its value is given in \S 35.6.
It relies on a rather non--trivial identity
called {\em Jacobi's triple product identity}
(because it features three infinite products):\\

\begin{empheq}[box=\fbox]{gather*}
\lp\prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}\rp \rp \cdot
\lp\prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}e^{2\pi\iota z}\rp \rp \cdot
\lp\prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}e^{-2\pi\iota z}\rp \rp
\\=\sum_{k=0}^{\infty} (-1)^k e^{\pi\iota\tau k(k+1)} 
\frac{\sin((2k+1)\pi z)}{\sin(\pi z)}
\end{empheq}

This identity appeared in Jacobi's {\em Fundamenta Nova
Theoriae Functionum Ellipticarum (1829)}, and is one of the most
beautiful formulae in the theory of doubly--periodic functions. 
It gives rise to several interesting
combinatorial results. For instance, by letting $z\to 0$, and writing
$q = e^{\pi\iota\tau}$, we get:

\[
\sum_{k=0}^{\infty} (-1)^k (2k+1) q^{k(k+1)} = 
\prod_{n=1}^{\infty} (1-q^{2n})^3
\]

In turn, such combinatorial formulae go back to Euler (1775). 
I encourage you to read more about these,
for instance search for {\em Euler's pentagonal number theorem.}\\

\noindent {\bf On the proofs of Jacobi's triple product identity.}
I know two ways to prove this, none are elementary. 
One of these proofs will be written up as Optional Reading B,
and is the same as Jacobi's original proof. It involves
rather lengthy, but still elementary, 
computations with the heat equation (a differential equation satisfied
by theta function).\\

I.G. Macdonald obtained a generalization of this identity,
now known as Macdonald identities, in {\em Affine root systems
and Dedekind's $\eta$ function (Invent. Math. 1972)}. It was later
realized by V.G. Kac (1974) and R.V. Moody (1975) that all of these identities
can be obtained from the {\em Weyl--Kac character formula}. This is
the second proof I know, but I will leave it for you to learn it
in a course on representation theory.\\

\PP{35.2}{Fay's trisecant identity} This identity is a relation between
values of theta function at sums and differences of $4$ complex numbers.
It only looks scary, but is very easy to remember (see \S 35.3 below)
and prove.\\

\begin{thm}
Let $\alpha,\beta,\gamma,\delta$ be four complex numbers. Then:
\begin{align*}
\theta(\alpha-\gamma)\theta(\alpha+\gamma)\theta(\beta-\delta)\theta(\beta+\delta)
&=
\theta(\alpha-\beta)\theta(\alpha+\beta)\theta(\gamma-\delta)\theta(\gamma+\delta)\\
&\qquad +\theta(\alpha-\delta)\theta(\alpha+\delta)\theta(\beta-\gamma)\theta(\beta+\gamma)
\end{align*}
\end{thm}

\begin{proof}
Let us keep $\beta,\gamma,\delta\in\C$ fixed and view both sides
as functions of $z=\alpha$.
Let us consider the following function:
\[
F(z) = \frac{\theta(z-\beta)\theta(z+\beta)\theta(\gamma-\delta)\theta(\gamma+\delta)
+\theta(z-\delta)\theta(z+\delta)\theta(\beta-\gamma)\theta(\beta+\gamma)}
{\theta(z-\gamma)\theta(z+\gamma)}
\]
We want to show that $F(z)$ is a constant, given by
$\theta(\beta-\delta)\theta(\beta+\delta)$. To do this, we proceed with the method
outlined in \S 35.0 above.\\

\noindent {\em Step 1.} Prove that $F(z)$ is doubly--periodic.\\

It is easy to check that $F(z+1)=F(z)$. Now, using 
$\theta(x+\tau) = -e^{-\pi\iota\tau}e^{-2\pi\iota x}\theta(x)$, we get:
\[
\theta(z+\tau-\gamma)\theta(z+\tau+\gamma) = 
e^{-2\pi\iota\tau}e^{-4\pi\iota z}\theta(z-\gamma)\theta(z+\gamma).
\]
(same with $\gamma$ changed to $\beta$ or $\delta$).\\

Thus, upon replacing $z$ by $z+\tau$, both the numerator and the denominator
of $F(z)$ get rescaled by $e^{-2\pi\iota\tau}e^{-4\pi\iota z}$. Hence
we get $F(z+\tau)=F(z)$.\\

\noindent {\em Step 2.} Prove that $F(z)$ is holomorphic.\\

The apparent poles of $F(z)$ are at the zeroes of $\theta(z-\gamma)\theta(z+\gamma)$.
That is, at $z=\pm\gamma + m+n\tau$, where $m,n\in\Z$. By periodicity of $F(z)$ already
established, it is enough to show that $z=\pm\gamma$ are removable
poles of $F(z)$. Since the order of vanishing of $\theta(z-\gamma)$
at $z=\gamma$ is $1$, we only have to check that the numerator
of $F(z)$ also vanishes at $z=\gamma$ (same argument applies to
$z=-\gamma$ as well). Let us set $z=\gamma$ in the numerator
of $F(z)$, and use the fact that $\theta(-x) = -\theta(x)$:
\[
\theta(\gamma-\beta)\theta(\gamma+\beta)\theta(\gamma-\delta)\theta(\gamma+\delta)
+\theta(\gamma-\delta)\theta(\gamma+\delta)\theta(\beta-\gamma)\theta(\beta+\gamma)
\]
\[
= \theta(\gamma-\delta)\theta(\gamma+\delta)\theta(\beta+\gamma)
\lp \theta(-(\beta-\gamma)) + \theta(\beta-\gamma)\rp
\]
\[
=0.
\]

So, $F(z)$ is holomorphic and doubly--periodic, hence it must be a constant
$F(z)=C$. The last step is to figure out the constant, which we can by
specializing $z$ to a convenient value. For instance, let us set $z=\beta$:
\[
C=F(\beta) = \theta(\beta-\delta)\theta(\beta+\delta).
\]
The theorem is proved.
\end{proof}
\ \\

\PP{35.3}{How to memorize Fay's identity}
In year 130, Greek mathematician Ptolemy proved the following result
(called Ptolemy relation). Let $ABCD$ be a quadrilateral inscribed in
a circle (see Figure \ref{fig:ptolemy} below). Denoting
by $|PQ|$ the length of the line segment joining two points
$P$ and $Q$, the following relation must hold.\\

\[
\tag{Ptolemy relation}
\boxed{|AC|\cdot |BD| = |AB|\cdot |CD| + |AD|\cdot |BC|}
\]

\begin{figure}[h]
\includegraphics[width=2in]{ptolemy.pdf}
\caption{A quadrilateral $ABCD$ inscribed in a circle.}
\label{fig:ptolemy}
\end{figure}

There are several ways to prove this relation (using similar triangles,
trigonometric identities etc). I will omit a proof of this, since
this section is more of a memorization trick.\\

To get Fay's trisecant identity, simply replace
\[
|AC| \mapsto \theta(\alpha-\gamma)\theta(\alpha+\gamma),\qquad 
|BD| \mapsto \theta(\beta-\delta)\theta(\beta+\delta)
\]
and similarly for other sides. Upon this replacement, Ptolemy relation
becomes Fay's trisecant identity.\\

\PP{35.4}{Theta function as a Fourier series}
Let us consider the problem of finding a function, say $\theta_1(z)$,
which has the same periodicity property as $\theta(z)$:
\[
\theta_1(z+1)=-\theta_1(z) \qquad\text{and}\qquad
\theta_1(z+\tau) = -e^{-\pi\iota\tau}e^{-2\pi\iota z}\theta_1(z).
\]
From the first equation, we can guess that the function we are looking
for can be written as an infinite series in $e^{\pi\iota (2\ell+1)z}$, where
$\ell\in\Z$. This is because $e^{\pi\iota (2\ell+1)}=-1$ for every $\ell\in\Z$.
So we postulate:
\[
\theta_1(z) = \sum_{\ell\in\Z} c_{\ell} e^{\pi\iota(2\ell+1)z}\ .
\]
In order to figure out the coefficients ($c_{\ell}$, $\ell\in\Z$), 
we use the second periodicity
requirement.
\[
\theta_1(z+\tau) = \sum_{\ell\in\Z} c_{\ell} e^{\pi\iota(2\ell+1)\tau}
e^{\pi\iota(2\ell+1)z}
\]
\[
-e^{-\pi\iota\tau}e^{-2\pi\iota z}\theta_1(z)
= \sum_{\ell\in\Z} (-c_{\ell}e^{-\pi\iota\tau}) e^{\pi\iota(2\ell-1)z}
= \sum_{n\in\Z} (-c_{n+1}e^{-\pi\iota\tau}) e^{\pi\iota(2n+1)z}
\]
where, in the last step, I substituted $\ell=n+1$. Now comparing the coefficients
of $e^{2\pi\iota(2\ell+1)z}$, we obtain:
\[
-c_{\ell+1}e^{-\pi\iota\tau} = c_{\ell}e^{\pi\iota(2\ell+1)\tau}
\Rightarrow \boxed{c_{\ell+1} = -c_{\ell}e^{\pi\iota(2\ell+2)\tau}}
\]

Let us agree to normalize $c_0=1$, so that the above relation can be solved as:
\[
c_{\ell} = (-1)^{\ell} e^{\pi\iota\tau \ell(\ell+1)}.
\]

This computation motivates us to define a function:
\[
\boxed{
\theta_1(z) = \sum_{\ell\in\Z} (-1)^{\ell} e^{\pi\iota\tau \ell(\ell+1)}
e^{\pi\iota (2\ell+1)z}
}
\]
\ \\

\PP{35.5}{Convergence and zeroes of $\theta_1(z)$}
As always, our first task is going to be to prove that the infinite sum
written above converges uniformly on compact subsets of $\C$, hence
defines $\theta_1(z)$ as a holomorphic function $\theta_1:\C\to\C$.
We will also have to know the zeroes of $\theta_1(z)$.\\

In order to do this, it is convenient to group certain terms of
$\theta_1(z)$ together. Note that the coefficient $\ds e^{\pi\iota\tau\ell(\ell+1)}$
gives the same value for $\ell=k$ and $\ell=-k-1$, where $k\in\Z_{\geq 0}$.
This allows us to write $\theta_1(z)$ as an infinite sum
over $k\in\Z_{\geq 0}$, combining terms $(0,-1)$, $(1,-2)$, and so on.\\

\[
\boxed{
\theta_1(z) = \sum_{k=0}^{\infty} (-1)^k e^{\pi\iota\tau k(k+1)}
\lp e^{\pi\iota (2k+1)z} - e^{-\pi\iota (2k+1)z}\rp
}
\]

\begin{thm}
The infinite sum given above converges uniformly on compact subsets of $\C$.
Thus, $\theta_1(z)$ is an entire holomorphic function with
the following periodicity properties:
\[
\theta_1(z+1)=-\theta_1(z) \qquad\text{and}\qquad
\theta_1(z+\tau) = -e^{-\pi\iota\tau}e^{-2\pi\iota z}\theta_1(z).
\]
Moreover, $\theta_1(z)=0$ if, and only if $z=m+n\tau$, where $m,n\in\Z$.
Each of these zeroes is of order $1$.
\end{thm}

\begin{proof}
We begin by proving the uniform convergence
\footnote{This part of the proof is optional.}.
Let $D\subset\C$ be a compact subset. Choose $A\in\R_{>0}$ so that
$\ds -A < \Im(z) < A$ for every $z\in D$. Let us write
$\ds q=e^{\pi\iota\tau}$. Note that $|q| = e^{-\pi\Im(\tau)}<1$
since $\Im(\tau)>0$.\\

So, for every $z\in D$, and $k\in\Z_{\geq 0}$, we have:
\[
\left| e^{\pi\iota (2k+1)z}\right| = e^{-\pi (2k+1)\Im(z)} < \lp e^{\pi A}\rp^{2k+1}
\]
and similarly $\ds\left| e^{-\pi\iota (2k+1)z}\right| < \lp e^{\pi A}\rp^{2k+1}$.
Therefore we obtain:
\[
\left| e^{\pi\iota (2k+1)z} - e^{-\pi\iota (2k+1)z}\right| 
< 2\lp e^{\pi A}\rp^{2k+1}.
\]

Now, the ratio test implies that the following series, which dominates the series
defining $\theta_1(z)$, converges:
\[
2 \sum_{k=0}^{\infty} |q|^{k(k+1)} \lp e^{\pi A}\rp^{2k+1}
\]
Ratio of successive terms is $\ds |q|^{2(k+1)}e^{2\pi A} \to 0$ as $k\to\infty$,
since $|q|<1$.\\

Hence, the infinite series defining $\theta_1(z)$ converges uniformly in $z\in D$,
as we wanted.\\

The periodicity properties of $\theta_1(z)$ have already been verified
in \S 35.4 above. It is also easy to see that $\theta_1(0)=0$ from the
formula written above. By periodicity, we conclude that
$\theta_1(m+n\tau)=0$ for every $m,n\in\Z$. Thus, it remains to show
that these are the only zeroes of $\theta_1$,
and the order of vanishing of $\theta_1(z)$ at $z=0$ is $1$.\\

Let us compute the number of zeroes of $\theta_1(z)$ within a fundamental
parallelogram $R_t$ with vertices $t,t+1,t+\tau,t+1+\tau$
(see Figure \ref{fig:shifted} below).\\

\begin{figure}[h]
\includegraphics[width=3in]{shifted.pdf}
\caption{Contour $C$ is the counterclockwise boundary of
the parallelogram $R_t$.}
\label{fig:shifted}
\end{figure}

As we know by now (Problem 5 of Homework 7), if $N$ is the number of
zeroes of $\theta_1(z)$ within $C$, then: $\ds N = \frac{1}{2\pi\iota}
\int_C \frac{\theta_1'(z)}{\theta_1(z)}\, dz$. This integral
computes number of zeroes - number of poles, but we already know
$\theta_1(z)$ is holomorphic on $\C$, so it does not have any poles.

\[
N = \frac{1}{2\pi\iota} \lp\int_{L_1} \frac{\theta_1'(z)}{\theta_1(z)}\, dz +
\int_{L_2} \frac{\theta_1'(z)}{\theta_1(z)}\, dz +
\int_{L_3} \frac{\theta_1'(z)}{\theta_1(z)}\, dz + 
\int_{L_4} \frac{\theta_1'(z)}{\theta_1(z)}\, dz
\rp
\]

Using the periodicity properties of $\theta_1(z)$, we can compute:
\[
\frac{\theta_1'(z+1)}{\theta_1(z+1)} = \frac{\theta_1'(z)}{\theta_1(z)}
\qquad\text{and}\qquad
\frac{\theta_1'(z+\tau)}{\theta_1(z+\tau)} = -2\pi\iota + 
\frac{\theta_1'(z)}{\theta_1(z)}
\]

Hence the integral over $L_2$ cancels with the integral over $L_4$. Moreover,
\[
\int_{L_3} \frac{\theta_1'(z)}{\theta_1(z)}\, dz 
= -\int_{t+\tau}^{t+1+\tau}  \frac{\theta_1'(z)}{\theta_1(z)}\, dz
= -\int_t^{t+1} \frac{\theta_1'(z+\tau)}{\theta_1(z+\tau)}\, dz
\]
\[
= -\int_t^{t+1} \lp -2\pi\iota + \frac{\theta_1'(z)}{\theta_1(z)}\rp\, dz
= -\int_{L_1} \frac{\theta_1'(z)}{\theta_1(z)}\, dz + 2\pi\iota.
\]
This implies that
\[
\frac{1}{2\pi\iota}\int_C \frac{\theta_1'(z)}{\theta_1(z)}\, dz = 1.
\]
Since we already know $\theta_1(0)=0$, and there is only one zero
of $\theta_1(z)$ within $C$, it must be at $z=0$, with order of vanishing $1$,
and nowhere else. The theorem is proved.
\end{proof}
\ \\

\PP{35.6}{$\theta_1(z)$ vs $\theta(z)$}
In the previous two paragraphs, we proved that 
\[
\theta_1(z)=\sum_{k=0}^{\infty} (-1)^k e^{\pi\iota\tau k(k+1)}
\lp e^{\pi\iota(2k+1)z}-e^{-\pi\iota(2k+1)z}\rp
\]
satisfies two of the
three defining properties of 
\[
\theta(z) = \frac{e^{\pi\iota z}-e^{-\pi\iota z}}{2\pi\iota}
\cdot \prod_{n=1}^{\infty} \frac{\lp 1-e^{2\pi\iota n\tau}e^{2\pi\iota z}\rp
\lp 1-e^{2\pi\iota n\tau}e^{-2\pi\iota z}\rp}
{\lp 1-e^{2\pi\iota n\tau}\rp^2}.
\]
Hence, on general grounds (see Proposition 34.2 of Lecture 34)
they must be related by a constant, say $C$ (which depends on $\tau$,
but is independent of $z$): $\theta_1(z) = C \theta(z)$. Jacobi's
triple product identity
\footnote{a proof of this identity is given in Optional Reading B.}
computes this constant as:
\[
C = 2\pi\iota \prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}\rp^3
\]
Combining all of this, and writing $\ds
\frac{e^{\pi\iota (2k+1)z}-e^{-\pi\iota (2k+1)z}}{e^{\pi\iota z}-e^{-\pi\iota z}}
$ as $\ds \frac{\sin((2k+1)\pi z)}{\sin(\pi z)}$ we get
the following equation, which was written in \S 35.1 above.

\begin{empheq}[box=\fbox]{gather*}
\lp\prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}\rp \rp \cdot
\lp\prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}e^{2\pi\iota z}\rp \rp \cdot
\lp\prod_{n=1}^{\infty} \lp 1-e^{2\pi\iota n\tau}e^{-2\pi\iota z}\rp \rp
\\=\sum_{k=0}^{\infty} (-1)^k e^{\pi\iota\tau k(k+1)} 
\frac{\sin((2k+1)\pi z)}{\sin(\pi z)}
\end{empheq}


\end{document}
