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If $X$ is a compact Riemann surface, then any holomorphic differential on $X$ has $2g-2$ zeros.

I would like to know how to prove this. If possible, without some "heavy machinery" like divisors and the Riemann-Roch theorem, which I don't understand very well yet. I really think that we can use the Riemann-Hurwitz to do this, but I am not sure how.

Gabriel
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  • Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. https://math.stackexchange.com/questions/3176319/the-sum-of-orders-of-zeros-of-a-holomorphic-function-in-a-genus-g-riemann-surf – reuns Apr 08 '19 at 06:57
  • Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $\int_\gamma \frac{du}{u}=0$ for $\gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = \Bbb{CP^1}$ and $X=\Bbb{C/(Z+iZ)}$) – reuns Apr 08 '19 at 07:06
  • @reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that. – Gabriel Apr 08 '19 at 07:15
  • I'll do some thinking here and try to understand what you mean. Thanks a lot for your help – Gabriel Apr 08 '19 at 07:16
  • holomorphic functions are constant, an easy consequence of "compact" and that analytic functions don't have local maximum. the whole topic is complicated even for those having a good background in complex analysis, you need to accumulate a lot of texts to find what is happening – reuns Apr 08 '19 at 07:16
  • @reuns how can I choose a path enclosing the whole surface if it is compact? Would that be the path that we get by doing the product of all generators of the fundamental group? For example, for $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ that would be the image of the path that goes from $1$ to $0$ to $1$ by the quotient map? – Gabriel Apr 08 '19 at 07:32
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    For $X=\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $\overline{D}$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $\partial D$ then look at $\gamma = \partial D$ and show the edges of $\gamma$ cancel so $\int_\gamma \frac{du}{u}=0$. For more complicated $X$ you'll have $\gamma= \bigcup_{j=1}^J \Gamma_j\bigcup_{j=1}^J \Gamma_{\sigma(j)}^-$ for some curves $ \Gamma_j$ and some permutation $\sigma$ – reuns Apr 08 '19 at 07:48

1 Answers1

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If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form.

Thomas
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  • I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_\alpha :\mathrm{d}z_\alpha$ for each chart. – Gabriel Apr 08 '19 at 06:55
  • @Thomas See my answer there – reuns Apr 08 '19 at 06:58