Doubly-periodic function

In mathematics, a doubly periodic function is a function "f" defined at all points "z" in a plane and having two "periods", which are linearly independent vectors "u" and "v" such that

:$f(z)\; =\; f(z\; +\; u)\; =\; f(z\; +\; v).,$

The doubly periodic function is thus a two-dimensional extension of the simpler singly periodic function, which repeats itself in a single dimension. Familiar examples of functions with a single period on the real number line include the trigonometric functions like cos"x" and sin"x". In the complex plane the exponential function "e"^"z" is a singly periodic function, with period 2"πi".

We can construct a doubly periodic step function on the complex plane with relatively little effort. For example, assume that the periods are 1 and "i", so that the repeating lattice is the set of unit squares with vertices at the Gaussian integers 0, ±"i", ±2"i", ..., ±1, ±1±"i", ±1±2"i", ..., etc. Clearly we can divide the prototype square (open on two adjoining sides) into any finite number of pieces and define a two-dimensional step function on that square by assigning a single value to each piece. A doubly periodic step function results when that prototype square is replicated in every square on the lattice.

The step function can be replaced with a piecewise continuous function on the prototype square to obtain a slightly more general doubly periodic function. If the vectors 1 and "i" in this example are replaced by linearly independent vectors "u" and "v" the prototype square becomes a prototype parallelogram, which still tiles the plane. And the "origin" of the lattice of parallelograms does not have to be the point 0; the lattice can start from any point. In other words, we can think of the plane and its associated functional values as remaining fixed, and mentally translate the lattice to gain insight into the function's characteristics.

What happens if we try to construct a "smooth" (or analytic) doubly periodic function "f" as a function of a complex variable? Quite a bit of information about such a function "f" can be obtained by applying some basic theorems from complex analysis.

* To be of much interest, the "smooth" doubly periodic function "f" cannot be bounded on the prototype parallelogram. For if it were it would be bounded everywhere, and therefore constant by Liouville's theorem.

* If the function "f" has no essential singularities ("ie", if it is meromorphic), then its poles are isolated and a translated lattice that does not pass through any pole can be constructed. But then the contour integral around any parallelogram in the lattice must vanish, because the values assumed by the doubly periodic function "f" along the two pairs of parallel sides are identical, and the two pairs of sides are traversed in opposite directions as we move around the contour. By the Residue theorem the function "f" cannot have a single simple pole inside each parallelogram – it must have at least two simple poles within each parallelogram (Jacobian case), or it must have one or more poles of order greater than one (Weierstrassian case).

* A similar argument can be applied to the function "g" = 1/"f". Clearly the zeroes of "f" become the poles of "g" under this inversion. So the meromorphic doubly periodic function "f" cannot have one simple zero lying within each parallelogram on the lattice – it must have at least two simple zeroes, or it must have at least one zero of multiplicity greater than one.

See elliptic function for a more complete account of doubly periodic functions that are meromorphic on the complex plane, and fundamental pair of periods for an account of the lattices involved. Also see Jacobi's elliptic functions and Weierstrass's elliptic functions.