In this post, I’ll use the Riemann-Roch Theorem to prove that the topological genus and the geometric genus agree. I will work with compact Riemann surfaces, but since these are the same as smooth algebraic curves over , these results are valid in the algebraic setting as well.
Recall the statement of Riemann-Roch: If is a compact Riemann surface with genus , and is a divisor on , then
Define the to be the quantity appearing above, which is intuitively the number of holes that has (by the classification of compact surfaces and the fact that the complex structure induces orientability, it follows that the underlying real -manifold of is the connected sum of tori). Define the to be the dimension of the space of holomorphic -forms .
The geometric genus and topological genus agree.
If is the empty divisor, then is just the space of all holomorphic functions. Now, the space of holomorphic functions on any compact Riemann surface is by the maximum principle, i.e. one-dimensional. Putting into the Riemann-Roch formula, alongside the fact that , we have that . Now, suppose that for some meromorphic -form . For any , the -form will not have any poles: if it did, would not be everywhere nonnegative. That is, is a holomorphic -form. Conversely, suppose is a holomorphic -form; we can write , for some meromorphic function . Since , it follows that . This gives a linear isomorphism between and , and since is -dimensional, so too is .
There is a proof using Hodge theory as well: we can decompose into holomorphic and antiholomorphic forms respectively. Since is -dimensional, the space of holomorphic 1-forms is -dimensional.