A takehome exam problem for my Riemann Surfaces class, which used Griffith's Introduction to Algebraic Curves, was the following:

Show that $S=\{(x,y)\in \mathbb C^2|y^2=\sin x\}$ is not interior of a compact Riemann Surface with boundary.

(the exam is over now, so don't worry about plagiarism!) I tried to prove this but to no avail. My strategy was that:

  1. The solution set $S$ has "infinitely many holes", i.e. it has infinite dimensional first homology group $H_1(S,\mathbb Z)$. I am guessing that this may be shown by supplying a holomorphic form on $S$ with nonvanishing loop integrals over countably many loops on $S$, with each loop integral yielding a different value.
  2. A compact set can't have infinite dimensional homology group.
  3. If $S$ is interior of a Riemann Surface with boundary $T$, then $S$ having infinitely many non-equivalent 1-cycles (in homology group $H_1$) also supplies $T$ with infinitely many non-equivalent 1-cycles in $H_1$, and now this should give a direct contradiction with #2.

I am not sure about whether each step works... any ideas? This problem seems really interesting at any case.

  • $\begingroup$ This is also a problem in Donaldson's Riemann Surfaces. $\endgroup$ – Michael Albanese Jan 5 '15 at 18:19
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    $\begingroup$ $S$ is the Riemann surface of $\sqrt{\sin(z)}$. The latter is obtained by making infinitely many slits in the complex plane (connecting zeroes of $\sin(z)$ in pairs) and gluing two copies of the resulting surface. From this description, you can see that each slit gives you a 1-cycle in $S$ and these cycles are linearly independent (use Mayer-Vietoris to check this claim). $\endgroup$ – Moishe Kohan Jan 28 '15 at 17:10

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