# Can a doubly periodic function be locally univalent?

I am looking for a meromorphic doubly periodic function such that the function is locally univalent.

A standard meromorphic doubly periodic funtion is the Weirestrass $$\wp$$ function, defined as $$\wp(z):=\frac{1}{z^2}+\sum_{\lambda \in \Lambda \setminus\{0\}}\left(\frac{1}{(z-\lambda)^2}-\frac{1}{\lambda^2}\right),$$where $$\Lambda$$ is the lattice generated by two linearly independent complex numbers over $$\mathbb{R}$$. For such function, $$\wp'=0$$ on lattice points and hence $$\wp$$ is not locally univalent.

I have a hard time in trying to find out other concrete examples of meromorphic doubly periodic functions. I know that a meromorphic doubly periodic function cannot be analytic eveywhere, due to Liouville's Theorem. Also, in each prototype parallelogram, such function can either have at least two simple poles, or at least one pole with order greater than $$1$$. These are what I know so far, but based on these I cannot construct the desired example.

Any comments and ideas are fully appreciated!

• Think about holomorphic maps from smooth elliptic curves to $P^1$. May 24 at 3:05

If $$f$$ is a doubly-periodic meromorphic function on $$\Bbb C$$ then $$f'$$ necessarily has zeros – otherwise $$1/f'$$ would be an entire doubly-periodic function and therefore constant. (More precisely, the number of zeros equals the number of poles in the fundamental parallelogram, if counted with multiplicity. For a derivative, that number is at least three.)
So $$f$$ can not be locally univalent everywhere.