# Any nonzero meromorphic $1$-form on a compact Riemann surface has degree $2g-2$

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.

• $f$ 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 $Div(h dg) = Div(h) + Div(dg) , \deg(Div(h)) = 0, \deg(Div(h dg) ) =2g-2$ – reuns Jun 12 at 17:41
• @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? – Larsson Jun 12 at 17:50
• What do you get when applying Riemann-Roch to $D = K=Div(dg)$ ? – reuns Jun 12 at 17:58
• @reuns it's not exactly clear to me...can you pls write a detailed answer maybe? – 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 $$K_X=2g-2$$.