# Is the genus of biholomorphic Riemann surfaces the same?

Is the statement above true for $$X \cong_{bihol} Y$$? I would say yes, since I can transform any holomorphic function on a open set in $$X$$ to one in $$Y$$ and vice versa.

• Your second sentence suggests you meant the geometric genus defined as the dimension of the vector space of holomorphic 1-forms, so you need to inject those 1-forms from X to Y – reuns Feb 23 at 20:51
• Yes, I meant $g:= dim H^1(X, \mathcal{O})$. Why 1-forms? – User1 Feb 23 at 21:37
• I'm sorry, I don't quite get it yet, $\mathcal{O}$ is the sheaf of holomorphic functions, why do I need to look at 1-forms at all? – User1 Feb 24 at 8:36
• Fix some $x_0\in X$, let $R_X$ be the space of functions analytic around $x_0$ having an analytic continuation over any curve $\in X$ (it is the same as the analytic functions on the universal cover). Let $O_X$ be the globally analytic functions on $X$, and $S_X$ those $\in R_X$ whose continuation over any closed loop differs by some element of $O_X$, and $\Omega_X$ the analytic 1-forms. If $X$ is a compact Riemann surface then $O_X$ is just the constant functions and $f\mapsto df$ is an isomorphism $H^1(X,O)=S_X/O_X\to\Omega_X$. If $X,Y$ are biholomorphic those things stay the same on $Y$ – reuns Feb 25 at 3:58
• Now I see that in your first comment you wrote "vector space of holomorphic one forms", I kind of overread that but that's not my definition of genus, its the dimension of the first cohomology group for the golomorphic functions (not 1- forms) – User1 Feb 25 at 9:34

Biholomorphic $$\implies$$ homeomorphic...