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I'm reading Jürgen Jost's "Compact Riemann Surfaces" Springer textbook 3rd ed (a very good read!).

Jost defines the divisor of a meromorphic differential $\eta$ on a compact Riemann surface by \begin{align} (\eta) := \sum (\text{ord}_{z_{\nu}}\eta) z_{\nu}, \quad \quad \quad \quad (5.4.3)\end{align} where, in local coordinates $\eta = f \ dz$, ord$_{z}\eta :=$ ord$_{z}f$ and the sum extends overall poles and zeroes of $f$. Define the degree of a divisor $D = \sum s_{\nu} z_{\nu}$ ($s_{\nu} \in \mathbb{Z}$) on a compact Riemann surface as \begin{align} \text{deg} D := \sum s_{\nu}. \quad \quad \quad \quad (5.4.9) \end{align} Definition 5.4.2: a canonical divisor is the divisor $(\eta)$ of a meromorphic 1-form $\eta$ on a compact Riemann surface. Jost later proves the following results:

Lemma 5.4.1: If $g$ is a meromorphic function on a compact Riemann surface then deg$(g) = 0$.

Lemma 5.4.4: If $K$ is a canonical divisor on a compact Riemann surface of genus $p$ then deg$(K) = 2p-2$.

My misunderstanding: doesn't Lemma 5.4.1 imply deg$(K)=0$, for a canonical divisor $K$? This contradicts Lemma 5.4.4 and I realise it must be nonsense, perhaps I've misunderstood a definition or something?

My reasoning: \begin{align} \text{deg}(K) &:= \text{deg}(\eta), \text{ with $\eta$ meromorphic,} &&\text{Definition 5.4.2,} \\ &:= \text{deg}\left(\sum (\text{ord}_{z_{\nu}} \eta) z_{\nu}\right), &&\text{Eqn (5.4.3),} \\ &:= \text{deg}\left(\sum (\text{ord}_{z_{\nu}} f) z_{\nu}\right), &&\text{in local coordinates (below Eqn (5.4.3),} \\ &=: \text{deg}(f)=0, &&\text{Lemma 5.4.1.} \end{align}

Is my mistake that ord$_{z}\eta =$ ord$_{z}f$ only holds in local coordinates $\eta = f \ dz$? If so then that would be quite interesting, any intuitive answer why deg$(K)=0$ "locally" but deg$(K)=2p-2$ "globally"?

Thank you for any comments or answers!

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  • $\begingroup$ Lemma 5.4.1 is about a meromorphic function, not a meromorphic form, they are not the same. If you have two meromorphic 1-forms $\omega$ and $\eta$, then locally you can write $\omega = f\,dz$ and $\eta=g\,dz$, and locally $\omega/\eta=f/g$ defines a meromorphic function, hence $\text{deg}((\omega)-(\eta))=\text{deg}(\omega/\eta)=0$, and then every two 1-forms have the same degree. Note that a form and a function don't have the same transformation after a change of variables. $\endgroup$ – user90189 Nov 22 '18 at 4:40
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    $\begingroup$ @user90189 thank you for that clarification! I'd be happy to formally accept that if you want to copy it as an answer $\endgroup$ – analytic Nov 22 '18 at 14:22

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