# Degree of meromorphic function and meromorphic 1-form on compact Riemann surface

Let $$X$$ be compact Riemann surface, given a nonconstant meromorphic function $$f$$ on it, it's well known that the $$\deg \text{div}f = 0$$ (due to the fact that degree is constant on each fiber), where $$\text{div}$$ denotes taking divisor of $$f$$.

And we also have meromorphic 1-form $$\alpha = fdz$$, the result is that $$\deg \text{div} \alpha = 2g -2$$( $$g$$ is the genus of the surface). (for example in Donaldson's book, take a smooth 1-form such that multiplicity of zeros equals to the degree of $$\alpha$$ and can be proved multiplicity of zeros for smooth 1-form is topological invariant $$2g-2$$)

Why they are different? why degree of the divsor of the meromorphic function is zero while for meromorphic 1-form is 2g-2?Do I misunderstand somewhere?

• The meromorphic 1-form is $df$ (in local coordinate it is $f'(z)dz$) unclear what you mean with $f dz$. Oct 24, 2022 at 16:54
• Hi , @reuns . I found a post asked this question and they define the meromorphic 1 - form as I did : math.stackexchange.com/a/3817196/360262 . And this definition is much more natural. Oct 25, 2022 at 0:28

Well, a function is just a different thing from a 1-form. You may be used to working in $$\mathbb{C}$$ where a 1-form is essentially the same thing as a function, since any function $$f$$ gives a 1-form $$fdz$$ and conversely. On a Riemann surface, though, there is no "$$dz$$" which you can just multiply any function by to get a 1-form with the same zeroes and poles. Locally near any point you can define a 1-form "$$dz$$" using a holomorphic chart, but these will not necessarily glue together to give a well-defined 1-form on the entire surface. (If a Riemann surface $$X$$ does have a holomorphic nowhere vanishing 1-form $$\omega$$, then there is a bijection between meromorphic functions and meromorphic 1-forms given by $$f\mapsto f\omega$$ which preserves zeroes and poles. However, a compact Riemann surface turns out to have such an $$\omega$$ only if its genus is $$1$$.)
The example of $$X=\mathbb{C}\cup\{\infty\}$$ is perhaps instructive. On $$\mathbb{C}$$, there is the usual 1-form $$dz$$, which has no zeroes or poles. However, when you extend it to $$\infty$$, it actually has a double pole at $$\infty$$. Indeed, a local coordinate near $$\infty$$ is given by $$w=1/z$$, so $$dz$$ turns into $$d(1/w)=-dw/w^2$$, which has a double pole at $$w=0$$ (corresponding to $$z=\infty$$). So on $$\mathbb{C}\cup\{\infty\}$$ there is the constant holomorphic function $$1$$ which has degree $$0$$, but if you turn it into a 1-form $$1\cdot dz$$, this 1-form actually has degree $$-2$$ since it has a double pole at $$\infty$$.