# Degree of a canonical divisor on a compact Riemann surface

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"?

• 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. – user90189 Nov 22 '18 at 4:40