The setting is quite simple: Let $X$ be a compact Riemann surface and $\omega$ is a holomorphic 1-form i.e. $\omega \in \Omega_X^1$. Then we can write $\omega = df$ for a holomorphic function $f:X \to \mathbb{C}$.

  1. Question 1: since $\omega \in \Omega_X^1$ is it ok to say that $[\omega]\in H^1(X,\mathbb{C})$? That is it is an exact holomorphic form.
  2. I read that there is an injective map $H^0(X, \Omega_X^1) \to H^1(X,\mathbb{C})$. The classes of the latter cohomology give equivalence classes of holomorphic 1-forms. What are the classes of the former cohomology $H^0(X, \Omega_X^1)$?

I see that the coefficients must be holomorphic one forms but I cannot understand what a representative of this class looks like (locally or globally). Usually the zero cohomology gives us information about the global sections so what can we understand from the information above?

  • $\begingroup$ There seem to be several points of confusion in your question, but perhaps the first is that in general we can't write $\omega$ in the form $df$ (contrary to your assertion). $\endgroup$ – tracing Dec 29 '16 at 0:16
  • $\begingroup$ @tracing I assume $\omega$ is exact. Sorry for not mentioning. $\endgroup$ – Marion Dec 29 '16 at 1:44
  • $\begingroup$ If $X$ is a connected compact Riemann surface then there are no non-constant holomorphic functions, so in your set-up, $f$ is constant and so $df = 0$. In fact it is a general result that a non-zero holomorphic one-form on a connected compact Riemann surface is never exact, which is why the map $H^0(X,\Omega^1_X) \to H^0(X,\mathbb C)$ is injective. On the other hand, this map is not surjective (unless the genus is $0$, in which case both sides are $0$) and so it is not true that the $H^1(X,\mathbb C)$ gives equivalence class of holomorphic $1$-forms. $\endgroup$ – tracing Dec 29 '16 at 4:25
  • $\begingroup$ @tracing thanks. So what you are saying is that in the setting above the only globally defined holomorphic functions are the constant functions and as a result $df=0$. Then in the globally defined holomorphic 1-forms $H^0(X,\Omega_X^1)$ map to globally defined holomorphic functions $H^0(X,\mathbb{C})$. Is this a correct interpretation? (I am a physics major so I am sloppy with the coefficients, actually this is part of the question, what are elements of each cohomology group). By the way I got this question from arxiv.org/pdf/0812.1803v3.pdf page 70. $\endgroup$ – Marion Dec 29 '16 at 17:22
  • $\begingroup$ In your last comment you wrote $H^0(X,\mathbb C)$, which means the locallly constant complex-valued functions on $X$. When $X$ is connected (as it is in your set-up) this is just one-dimensional. What you wrote in your question, and what I wrote in my comment, is $H^1(X,\mathbb C)$ --- this is cohom. in degree $1$ with complex coefficients, which can be computed in various ways, e.g. via a triangulation of $X$, or (more immediately relevant here) via de Rham theory: it is closed smooth $1$-forms (allowing complex numbers as scalars) modulo exact smooth $1$-forms. $\endgroup$ – tracing Dec 29 '16 at 22:13

Your Answer

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

Browse other questions tagged or ask your own question.