Doubly periodic function
A function $f:\mathbb C\to\mathbb C$ is said to be periodic with period $\omega$ if $f(z)=f(z+\omega)$ for all $z\in\mathbb C$. Intuitively, the values of $f$ "repeat" in intervals of $\omega$. If $f$ has period $\omega$, we can divide the complex plane into countably infinitely many "strips" so that the values taken by $f$ are the same in each strip:
If $\omega$ is a period, then so are $2\omega,3\omega,$ and more generally $n\omega$ for all $n\in\mathbb N$. A fundamental period of $f$ is a period which is not an integer multiple of another period with smaller magnitude. A doubly periodic function is a function which is periodic with two distinct fundamental periods that are not opposites (for if $\omega$ is a fundamental period, then so is $-\omega$). These periods are often denoted $\omega_1,\omega_2$. If these two periods are not real multiples of each other, then the complex plane can be divided into parallelograms, called fundamental parallelograms, in which the values of $f$ repeat:
In the image above, the function $f$ takes the same value at all of the green points. The purple and blue vectors represent the fundamental periods $\omega_1$ and $\omega_2$.
As it happens, the holomorphic doubly periodic functions $f$ have some very nice properties that make them easy to classify. For instance:
- If $f,g$ are doubly periodic with periods $\omega_1,\omega_2$, then the functions $f+g$, $fg$, and $f/g$ are also doubly periodic. The derivative $f'$ is also doubly periodic.
- The ratio $\omega_1/\omega_2$ cannot be real. This is because their ratio must be irrational if both are to be fundamental periods, but this means that the integer combinations $m\omega_1+n\omega_2$ are dense on the line ${r\omega_1:r\in\mathbb R}$, and therefore $f(r\omega_1)=f(0)$ for all $r\in\mathbb R$, making $f$ a constant function. But constant functions do not have fundamental periods.
- The function $f$ must have poles in the finite portion of $\mathbb C$. For if $f$ had no poles, then it would be bounded on the fundamental parallelogram (which is compact) meaning that it would be bounded in all of $\mathbb C$. But by Liouville's Theorem , this would make $f$ constant, which is impossible. Hence, we cannot find such $f$ that is holomorphic everywhere, and we must weaken the requirement to meromorphic functions.
- The function $f$ must have at least two poles (counting multiplicity) in a fundamental parallelogram. For if we choose a fundamental parallelogram with no poles on its boundary and integrate $f$ around it counterclockwise, the value of this integral equals $2\pi i$ times the sum of the residues of $f$ at these poles. If $f$ has only one pole, then this integral must be nonzero. But the portions of the integral along opposite sides of the fundamental parallelogram must cancel each other, so the integral must vanish, meaning that $f$ cannot have exactly one pole.
- The number of poles and zeros of $f$ in a fundamental parallelogram must be equal (counting multiplicity). If we consider the logarithmic derivative $f'/f$ and integrate this function around the boundary of a fundamental parallelogram which does not contain any zeros or poles, each zero will contribute $+2\pi i$ to the value of this integral, and each pole will contribute $-2\pi i$. However, we have seen that this integral must equal $0$.
- The function $f$ must take on all values $a\in\mathbb C$ an equal number of times (counting multiplicity) in a fundamental parallelogram. To see why this is true, consider the function $f-a$, which has the same poles as $f$, and with the same orders. The zeros of this function, however, are the solutions to $f(z)=a$.
- A nonconstant doubly periodic function $f$ can be conceived of as a conformal mapping from a torus to the Riemann sphere .