4
$\begingroup$

So the question is as follows. Prove that there is no meromorphic function $f$ such that at every $z\in \mathbb{C}$ we have $f(z)=f(z+1)$ and $f(z)=f(z+i)$ with only simple poles at the points $m+ni, m,n \in \mathbb{Z}$.

So, this looks like a holomorphic map from the torus to the Riemann sphere. Since the map is continuous, the image of the torus is compact, hence closed. On the other hand, holomorphic maps are open, hence the image is also open, so the map is surjective.

Now, if I knew that $f'(z) \neq 0$, then I would have a covering, and there are no coverings from the torus to the sphere. However, I can't make that statement.

I'm wondering, is there a proof that uses only arguments from complex analysis?

Thanks for any help!

$\endgroup$
5
  • 1
    $\begingroup$ We presume that you mean that the only poles are at the integer lattice points. $\endgroup$ – Ted Shifrin Jan 16 '15 at 2:14
  • $\begingroup$ Consider the contour integral of $f'/(f-c)$ along the boundary of a fundamental region (translated unit square in this case). Can you find the value of this contour integral? Can you figure what this result means in terms of the number of solutions of $f(z) = c$ on that region? $\endgroup$ – Sangchul Lee Jan 16 '15 at 2:18
  • $\begingroup$ Constant functions are meromorphic, so perhaps you mean there are no non-constant meromorphic functions... $\endgroup$ – robjohn Jan 16 '15 at 2:33
  • $\begingroup$ @robjohn: A constant meromorphic function doesn't have simple poles at $m+ni$. $\endgroup$ – Michael Albanese Jan 16 '15 at 2:39
  • $\begingroup$ @MichaelAlbanese: when I see "with only simple poles" I read it as it can only have simple poles. I don't necessarily read it as it must have simple poles. Perhaps mine is not the common interpretation. $\endgroup$ – robjohn Jan 16 '15 at 2:46
7
$\begingroup$

Consider $R = \{z \in \mathbb{C} \mid \operatorname{Re}(z), \operatorname{Im}(z) \in \left[-\frac{1}{2}, \frac{1}{2}\right]\}$. By the periodicity of $f$,

$$\int_{\partial R}f(z) dz = 0.$$

However, by the residue theorem,

$$\int_{\partial R}f(z) dz = 2\pi i\operatorname{Res}(f, 0).$$

As $f$ has a simple pole at $0$, $\operatorname{Res}(f, 0) \neq 0$. Therefore, no such $f$ exists.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.