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:
- 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.
- A compact set can't have infinite dimensional homology group.
- 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.