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It is well-known that on any compact Riemann surface $X$ every non-constant meromorphic function $f:X\to\mathbb{CP}^1$ has as many zeros as poles, where each is counted according to multiplicities. Consider the holomorphic complex "charts" (a possible abuse of terminology) $\phi_{\alpha}:U_{\alpha}\to V_{\alpha}$ where $U_{\alpha}\subset X$ and $V_{\alpha}\subset\mathbb{C}$ are compact. Then by the argument principle, if $w:=\phi_{\alpha}^{-1}(z)$ for $z\in\mathbb{C},$ $$\frac{1}{2\pi i}\int_{\partial U_{\alpha}}\frac{f'(w)}{f(w)}dw=\frac{1}{2\pi i}\int_{\partial U_{\alpha}}\frac{f'(\phi_{\alpha}^{-1}(z))}{f(\phi_{\alpha}^{-1}(z))}d\phi_{\alpha}^{-1}=\frac{1}{2\pi i}\int_{\partial U_{\alpha}}\phi_{\alpha}^*\left(\frac{f'(z)}{f(z)}dz\right)=\sum_{k=1}^n\alpha_k-\sum_{k=1}^m\beta_k$$ vanishes, where $f(z)$ is analytic on $\overline{I(\partial U_{\alpha})}$ except for poles in $I(\partial U_{\alpha})$ at the points $b_1,..,b_n,$ $f(z)$ has finitely many zeros $a_1,...,a_m$ in $I(\partial U_{\alpha})$, but none on $\partial U_{\alpha}$ itself, and $\alpha_k$ is the order of $a_k$ and $\beta_k$ is the order of $b_k.$

However, if $f$ is entire (and thus meromorphic) then it has no poles and $$\frac{1}{2\pi i}\int_{\partial U_{\alpha}}\frac{f'(w)}{f(w)}dw=\sum_{k=1}^n\alpha_k-\sum_{k=1}^m\beta_k=\sum_{k=1}^n\alpha_k=0,$$ i.e. the sum of the orders of the zeros of an entire function $f$ in $U_{\alpha}$ is zero (note that $U_{\alpha}$ is a compact Riemann surface since $X$ is compact). Is this true?

Thanks in advance!

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  • $\begingroup$ A chart cannot be compact. $\endgroup$ – user99914 Oct 14 '17 at 4:06
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  • $\mathbb{CP}^1$ is a compact Riemann surface. A holomorphic function $f:X\to Y$ between two compact Riemann surfaces doesn't have any zeros/poles, but only branch points, ie. where locally in a chart $\phi,\psi$ around $a,f(a)$ : $\psi \circ f \circ \phi^{-1}(z)=C z^n+\mathcal{O}(z^{n+1}),n\ge 2$ so that $f$ is not locally biholomorphic.

  • So you meant $f$ is meromorphic $X\to \mathbb{C}$.

    Then the maximum modulus principle is enough to show if $f$ is holomorphic then it is constant (take $a\in X$ where $|f(a)|$ is maximum, it exists since $X$ is compact)

    Also you can define $f'$ only in some local chart, ie. $f'(\phi(z)) \phi'(z)$ (this is only a notation, $\phi'$ doesn't make sense) thus $f'(w)$ doesn't make sense. What makes sense are the meromorphic differentials $df$ and $\frac{df}{f}$ (in local chart it is $d f(\phi(z))$ and $\frac{df(\phi(z))}{f(\phi(z))}$) which by definition integrate (locally) to $f$ and $\log f$ in every chart.

    For a simple closed curve $\gamma$ in $X$ enclosing every point only once, you need to show $\gamma= \gamma_{1,+} \cup \ldots \cup \gamma_{n,+}\cup \gamma_{\sigma(1),-} \cup \ldots \gamma_{\sigma(n),-}$ where $\sigma$ is a permutation of $1 \ldots n$ and $\gamma_{i,+},\gamma_{i,-}$ are the same curve in $X$ traversed in opposite direction (this is so because to obtain a compact Riemann surface we identity the opposite edges of the boundary of a fundamental domain).

Therefore $$\int_\gamma \frac{df}{f} = \sum_{i=1}^n\int_{\gamma_{i,+}}-\sum_{i=1}^n\int_{\gamma_{\sigma(i),+}} \frac{df}{f} = 0.$$

And hence $f$ has the same number of poles and zeros.

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  • $\begingroup$ Thank you! One last thing: suppose $\phi:U\to\overline{I(\gamma)}$ for $U\subset\mathbb{C}$ compact, then would $\int_{\gamma}\frac{df}{f}=\int{\partial U}\phi^*\left(\frac{df}{d}\right)=0?$ Also, how would one show show $\gamma= \gamma_{1,+} \cup \ldots \cup \gamma_{n,+}\cup \gamma_{\sigma(1),-} \cup \ldots \gamma_{\sigma(n),-}$ where $\sigma$ is a permutation of $1, \ldots, n$ and $\gamma_{i,+},\gamma_{i,-}?$ @reuns $\endgroup$ – Sergio Charles Oct 15 '17 at 4:49

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