I am reading "Compact Riemann Surfaces" by Raghavan Narashimhan. Say X be a compact Riemann surface; after proving that the degree of the canonical bundle $K_X$ is $2g-2$ (using Riemann-Roch), where $g$ is the genus, he just says that Equivalently if $w\neq 0$ is any meromorphic $1$-form, the degree of the divisor of $w$ is $2g-2$. I can't see it. How does it follow from the previous line? I might be missing something very obvious. Still an explanation would be very helpful.

  • $\begingroup$ $f$ a meromorphic function, for almost every $a$, the $d$ poles of $g=1/(f-a)$ are simples, $g$ is an holomorphic map $X \to P^1$ and Riemann-Hurwitz says it has $2d+2g-2$ branch points (counted with multiplicity) so $Div(dg) = B-2P$ where $B \ge 0, P \ge 0, \deg(B) = 2d+2g-2, \deg(P) = d$ and any other meromorphic one-form will be $h dg$ with $h$ meromorphic function and $Div(h dg) = Div(h) + Div(dg) , \deg(Div(h)) = 0, \deg(Div(h dg) ) =2g-2$ $\endgroup$ – reuns Jun 12 at 17:41
  • $\begingroup$ @reuns thanks for the comment. The book actually does Riemann-Hurwitz in the next chapter, is it possible to see it only from Riemann-Roch? $\endgroup$ – Larsson Jun 12 at 17:50
  • $\begingroup$ What do you get when applying Riemann-Roch to $D = K=Div(dg)$ ? $\endgroup$ – reuns Jun 12 at 17:58
  • $\begingroup$ @reuns it's not exactly clear to me...can you pls write a detailed answer maybe? $\endgroup$ – Larsson Jun 13 at 6:31

This is actually easy once one understands the definition of the canonical line bundle properly $: K_X$ is a line bundle such that for any open set $U\subset X, H^0(U,K_X)=\Omega_X(U)=$ space of holomorphic $1$-forms on $U$. Here meromorphic sections of $K_X$ correspond to meromorphic $1$-forms.

Now, the degree of $K_X$ is actually degree of any section of $K_X$ (holomorphic or meromorphic). Let $s$ be the meromorphic section of $K_X$ corresponding to $w$. So, degree of $w$ = degree of $s$ = degree of $K_X=2g-2$.


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