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\begin{document}
\section{Introduction}
In this unit on elliptic functions, we'll see how two very natural lines of questions interact. The first, as we have met several times in Berndt's book, involves elliptic integrals. In particular, we seek complex functions which are solutions to the indefinite integral
$$ \int \frac{dz}{\sqrt{Az^3+Bz+C}} $$
called an elliptic integral of the first kind. The second concerns periodic functions. For functions of a real variable, the trigonometric functions $\cos(2 \pi n x)$ and $\sin(2 \pi n x)$ are basic examples of periodic functions with period $1/n$, that is, functions for which
$$ f(x + 1/n) = f(x) \quad \forall x \in \mathbb{R}. $$
By Fourier theory, any suitably nice function of period $1/n$ (e.g. piece-wise differentiable functions with finitely many discontinuities over any period) can be expressed in terms of these elementary functions. For functions of a complex variable,
we can again ask about periodic functions with period $\omega \in \mathbb{C}$, where again we mean:
$$ f(z + \omega) = f(z) \quad \forall z \in \mathbb{C}. $$
Once again, there is a Fourier theory for complex functions, described in terms of the elementary periodic functions $e^{2 \pi i z / \omega}$ with period $\omega$.
Let's try to do this somewhat rigorously. Suppose we are given a (non-empty) connected, open domain $D$ with the property that $z \in D$ implies $z + \omega$ and $z - \omega$ are in $D$. Then let $D'$ be the image of $D$ under $z \mapsto \zeta = e^{2 \pi i z / \omega}$. (Convince yourself that $D'$ is still open and connected under this map.) For example, if $D$ is the complex plane (our most common example) then $D'$ is the punctured plane $\mathbb{C} - \{ 0 \}$. If $f(z)$ is a meromorphic function in $D$ with period $\omega$, then let $F$ be the unique function on $D'$ defined by
$$ F(\zeta) = F(e^{2 \pi i z / \omega}) = f(z). $$
Note $F$ is well-defined as $f$ has period $\omega$, and is meromorphic since $f$ was assumed meromorphic. Then if $D'$ contains an annulus $r_1 < |\zeta| < r_2$ for which $F$ has no poles, then in this region $F$ has a Laurent expansion
$$ F(\zeta) = \sum_{n = -\infty}^{\infty} c_n \zeta^n $$
or equivalently
$$ f(z) = \sum_{n = -\infty}^{\infty} c_n e^{2 \pi i n z / \omega}, $$
valid in the horizontal strip in $\mathbb{C}$ which is the preimage of our annulus under the exponential. The coefficients $c_n$ are given by
$$ c_n = \frac{1}{2 \pi i} \int_{|\zeta|=r} F(\zeta) \zeta^{-(n+1)} d\zeta $$
where the region of integration may be any circle with radius $r$ with $r_1 < r < r_2$. (If you've seen complex integration and Cauchy's theorem before, then you know that the integral is independent of our choice of $r$ owing to the assumption that there are no poles in the annulus.) Translating back via change of variables, we may write
$$ c_n = \frac{1}{\omega} \int_{a}^{a+\omega} f(z) e^{-2 \pi i n z/ \omega} dz $$
where we may take $a$ to be any point in the horizontal strip and the integration over any path from $a$ to $a+\omega$ within the strip. In particular, if the initial function $f(z)$ was analytic in the whole complex plane, then this Fourier series description is valid for all points $z \in \mathbb{C}$.
But why settle for the simple periodicity property? Let's demand that our functions be periodic with respect to an arbitrary set of periods $S$. This set has structure, since if $\omega_1, \omega_2 \in S$ then by definition of a period, $m \omega_1 + n \omega_2 \in S$ for any $m, n \in \mathbb{Z}$ (for algebra lovers, this means $S$ is a $\mathbb{Z}$-module).
\bigskip
{\bf Exercise 1:} Determine what happens when you allow arbitrary sets of periods for functions of a real variable. (Hint: You might want to consider cases according to the difference of any two periods in your set $S$, and you'll probably want to assume your functions are ``sufficiently nice'' as in the assumptions for real variable Fourier theory above.)
\bigskip
For functions of a complex variable, the extra dimension gives a new collection of periodic functions we haven't yet mentioned. Meromorphic functions of a complex variable are quite restricted, and in particular, any such function which is constant on a sequence of complex numbers converging to a limit point must be the constant function. Thus, the set of periods $S$ must not contain such an accumulation point, or else the only such periodic function will be constant. We call such a module without an accumulation point ``discrete.''
\begin{theorem} A discrete $\mathbb{Z}$-module $M$ consists of either 0 alone, integer multiples of a single non-zero complex number $\omega$, or all linear combinations $\{ m \omega_1 + n \omega_2 \}$ for a pair of complex numbers $\omega_1, \omega_2$ with $\omega_2 / \omega_1 \not\in \mathbb{R}$.
\end{theorem}
\begin{proof}
If $M \ne \{0 \}$, then we may choose an element $\omega_1 \in M$ with smallest positive absolute value (there may be several with the same norm, and in fact you can show the only possibilities are 2,4, or 6, but no matter). Then $\{ n \omega_1 \; | \; n \in \mathbb{Z} \}$ is contained in $M$.
Suppose there exists a complex number $\omega \in M$ which is not an integer multiple of $\omega_1$. Again, take $\omega_2$ to be the complex number of smallest absolute value with this property. If $\omega_2 / \omega_1$ real, then there exists an integer $n$ such that
$$ n < \omega_2 / \omega_1 < n+1 \; \Rightarrow \; 0 < |n \omega_1 - \omega_2 | < |\omega_1| $$
which contradicts the fact that $\omega_1$ was chosen to have smallest positive absolute value. So it just remains to show that, assuming such an $\omega_2$ exists, $M= \{ n \omega_1 + m \omega_2 \; | \; m,n \in \mathbb{Z} \}.$
First, we show any complex number $\omega$ can be written in the form $\omega = \lambda_1 \omega_1 + \lambda_2 \omega_2$ with $\lambda_1, \lambda_2$ real. Equivalently, we seek a solution $(\lambda_1, \lambda_2)$ to the system of equations
\begin{eqnarray*}
\omega = \lambda_1 \omega_1 + \lambda_2 \omega_2
\bar\omega = \lambda_1 \bar\omega_1 + \lambda_2 \bar\omega_2
\end{eqnarray*}
Because $\omega_2 / \omega_1$ is not real, this system is non-singular (as functions of the two complex variables $\lambda_1, \lambda_2$). That is, the determinant $\omega_1 \bar\omega_2 - \omega_2 \bar\omega_1 \ne 0$. So there is a unique solution in complex numbers $\lambda_1, \lambda_2$. On the other hand, both pairs $(\lambda_1, \lambda_2)$ and $(\bar\lambda_1, \bar\lambda_2)$ are visibly solutions, so $\lambda_1$ and $\lambda_2$ must be real.
Given any $\omega \in M$, we finally show that we can find $k,n \in \mathbb{Z}$ with $\omega - k \omega_1 - n \omega_2 = 0$. Given any $\omega = \lambda_1 \omega_1 + \lambda_2 \omega_2$, with $\lambda_i$ real, we can find $m,n$ with
$$ | \lambda_1 - m | \leq 1/2, \quad | \lambda_2 - n | \leq 1/2 $$
Consider
$$ \omega' = \omega - m \omega_1 - n \omega_2. $$
Then $| \omega' | < \frac{1}{2} | \omega_1 | + \frac{1}{2} | \omega_2 | \leq | \omega_2 |$, where importantly the first inequality is strict, since $\omega_2$ is not a real multiple of $\omega_1$ (giving a degenerate version of the triangle inequality, as the quotient being real means they lie on a line in the complex plane). This inequality implies $\omega'$ is an integral multiple of $\omega_1$ since we chose $\omega_2$ as above.
\end{proof}
We now begin our study of doubly periodic complex functions -- functions whose period module is a two-dimensional integer lattice.
\section{Relations among Period Lattices}
We're assuming that our period module is generated by a pair of complex numbers $\{ \omega_1, \omega_2 \}$, so that any period $\omega \in M$ can be uniquely written in the form $\omega = m \omega_1 + n \omega_2$ for some $m,n \in \mathbb{Z}$, i.e. $\omega_1, \omega_2$ form a basis for our $\mathbb{Z}$-module.
Given another basis $(\omega_1', \omega_2')$, then since $(\omega_1, \omega_2)$ is a basis, we may write:
\begin{eqnarray*}
\omega_2' & = & a\omega_2+b\omega_1 \\
\omega_1' & = & c\omega_2+d\omega_1
\end{eqnarray*}
for some integers $a,b,c,d$. Note the same relations apply upon taking complex conjugates of both sides, and this allows us to formulate an identity entirely in terms of $2 \times 2$ matrices:
$$ \begin{pmatrix} \omega_2' & \bar\omega_2' \\ \omega_1' & \bar\omega_1' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \omega_2 & \bar\omega_2 \\ \omega_1 & \bar\omega_1 \end{pmatrix} $$
We can similarly express $\omega_1$ and $\omega_2$ in terms of the new basis $(\omega_1', \omega_2')$ to obtain:
$$ \begin{pmatrix} \omega_2 & \bar\omega_2 \\ \omega_1 & \bar\omega_1 \end{pmatrix} = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix} \begin{pmatrix} \omega_2' & \bar\omega_2' \\ \omega_1' & \bar\omega_1' \end{pmatrix} $$
for some integers $a',b',c',d'$. Thus, substituting,
$$ \begin{pmatrix} \omega_2 & \bar\omega_2 \\ \omega_1 & \bar\omega_1 \end{pmatrix} = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \omega_2 & \bar\omega_2 \\ \omega_1 & \bar\omega_1 \end{pmatrix}. $$
Just as in the proof of the above theorem, knowing that the ratio $\omega_2 / \omega_1$ is not real implies that the determinant $\omega_2 \bar\omega_1 - \bar\omega_2 \omega_1$ is non-zero, and thus the matrix on the left-hand side above is invertible. Multiplying on the right by the inverse yields:
$$ \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
Because determinants are multiplicative, our matrices above must have determinant $\pm 1$. We call such matrices ``unimodular'' since, as you may recall from linear algebra, the determinant's absolute value measures the factor by which volume increases under the associated linear transformation. In terms of the period lattice, it means that the parallelogram formed by $\{0, \omega_1, \omega_2, \omega_1+\omega_2 \}$ has the same area for any basis $\omega_1, \omega_2$.
\section{Canonical Bases}
We can use the action of unimodular integer matrices to try to obtain canonical representatives for our period lattice. The next theorem says that we can almost do this uniquely.
\begin{theorem}
There exists a basis $(\omega_1,\omega_2)$ with ratio $\tau = \omega_2 / \omega_1$ satisfying the following conditions:
\begin{enumerate}
\item $\textrm{Im}(\tau) > 0$
\item $\frac{-1}{2} < \textrm{Re}(\tau) \leq \frac{1}{2}$
\item $ | \tau | \geq 1$
\item $ \textrm{Re}(\tau) \geq 0$ if $| \tau | =1$
\end{enumerate}
This determines $\tau$ uniquely, which in turn, determines either 2, 4, or 6 possibilities for $(\omega_1,\omega_2).$
\end{theorem}
Note this domain for $\tau$ is essentially the one we defined earlier as a fundamental domain for the action of $SL(2, \mathbb{Z})$ on the upper half plane, and we've made a choice in condition (4) according to equivalent points in this domain. However, strictly speaking, the group we're working with here is $GL(2,\mathbb{Z})$, remembering that the determinant of these integer matrices must be $\pm 1$ in order that the inverse matrix remains integral.
\bigskip
\begin{proof}
Choosing $\omega_1$ and $\omega_2$ as in the proof of Theorem 1, we have
$$ | \omega_1 | \leq | \omega_2 |, \quad | \omega_2 | \leq | \omega_1 + \omega_2 |, \quad | \omega_2 | \leq | \omega_1 - \omega_2 | $$
which in turn implies that $| \tau | \geq 1$ and $| \textrm{Re}(\tau) | \leq \frac{1}{2}$. By choice of $(-\omega_1, \omega_2)$ or $(\omega_1,\omega_2)$, we may guarantee that $\textrm{Im}(\tau) > 0$. Finally, if $\textrm{Re}(\tau) = -\frac{1}{2}$, replace $(\omega_1,\omega_2)$ by $(\omega_1, \omega_1 + \omega_2)$ and if $|\tau|=1$ with $\textrm{Re}(\tau)<0$ then replace $(\omega_1,\omega_2)$ by $(-\omega_2,\omega_1)$. Thus $\tau$ will have all required properties of the theorem. (Note, in particular, that none of these subsequent changes to the basis mess up earlier desired properties of $\tau$.)
It remains to show that these conditions uniquely determine $\tau$ (and hence finitely many choices of basis as indicated). Indeed, given any other basis $(\omega_1', \omega_2')$, then one readily checks $\tau' = \omega_2' / \omega_1'$ satisfies
$$ \tau' = \frac{a \tau + b}{c \tau + d}, \qquad ad-bc = \pm 1. $$
This gives
$$ \textrm{Im}(\tau') = \frac{\pm \textrm{Im}(\tau)}{|c \tau+d|^2} \quad \text{with sign in the numerator matching $ad-bc$.} $$
It suffices to show that if $\tau$ and $\tau'$ satisfy the four conditions, then they must be equal. Condition (1) implies that the sign in the numerator is positive, so $ad-bc=1$. Without loss of generality, we may assume $\textrm{Im}(\tau') \geq \textrm{Im}(\tau)$ since $\tau$ and $\tau'$ play symmetric roles in the above equation. This implies $|c \tau + d| \leq 1$ and this greatly reduces the possible integer $c$, $d$.
\bigskip
{\bf Exercise:} Work out the remaining possible choices of $c,d$ for which $\tau$ and $\tau'$ are in the fundamental domain and show that each implies $\tau = \tau'$. You should be able to reduce to the two cases $c=0$ and $|c|=1$ first, and then treat these cases separately.
\bigskip
To finish, note we can always replace $(\omega_1, \omega_2)$ by $(-\omega_1, -\omega_2)$ without changing $\tau$. This gives at least two choices of basis for any value of $\tau$.
\bigskip
{\bf Exercise:} Using your work in the previous exercise, determine the values of $\tau$ for which there are 2,4, or 6 possible choices of basis and explain your answer.
\bigskip
\end{proof}
\begin{thebibliography}{99}
\bibitem{ahlfors} Lars Ahlfors, \emph{Complex Analysis}, 3rd. ed., McGraw-Hill (1979).
\end{thebibliography}
\end{document}