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Let $C$ be a compact Riemann surfaces of genus $g$. Let $p$ be a point, $\Delta$ a disc around $p$, and $\Delta^*$ the disc minus $p$. Let $\omega$ be a holomorphic one form defined on $\Delta^*$.

Given a meromorphic function $f$ on $C$, regular outside $p$, we can consider the residue of $f\omega$ at $p$.

If $\omega$ is the restriction of a regular form on $C\setminus p$ to $\Delta^*$, then this residue is zero by the classical Residue Theorem.

I read that also the converse is true. So, if the residue of $f\omega$ at $p$ is zero for every meromorphic function $f$ on $C$, regular outside $p$, then $\omega$ is the restriction of a regular form on $C\setminus p$ to $\Delta^*$.

In the book "Vertex algebras and Algebraic Curves" this goes under the name of strong residue theorem (9.2.9). I struggle to find a proof in the literature, this could be something relatively elementary to be found in some book on Riemann surfaces. It should follow from a variant of Serre duality.

I would like to have a proof and/or a reference about this result. I am interested both in the case $C$ smooth and $C$ nodal. The latter case I think makes really sense just when $C$ is irreducible, but I am not sure.

Thanks

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  • $\begingroup$ Do you mean that $C$ is compact? $\endgroup$ Apr 24 '17 at 15:27
  • $\begingroup$ Yes, I'll edit the question. Thanks for pointing this out. I'll clarify also the singularities. $\endgroup$
    – Giulio
    Apr 24 '17 at 16:03
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    $\begingroup$ The idea is to make meromorphic functions with prescribed order pole at $p$, and all other poles outside $\Delta$. This is not too hard; you are looking at the fact that high enough degree divisors are very ample, but in the funny way of thinking of sections of line bundles associated to divisors as being meromorphic functions with at worst poles of a certain order along the divisor. See Griffiths's book on algebraic curves (from lectures in China) for that point of view. $\endgroup$
    – Ben McKay
    Apr 24 '17 at 19:47
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Indeed, this follows from Serre duality: Take your favorite local holomorphic coordinate $z$ centered at $p$, and use this as the transition function $g_{0,\infty}=z$ of the line bundle $L(p)$ with global holomorphic section $s_p$. Consider the section $$\omega\otimes s_{-np}$$ and take a $C^\infty$ cut-off away from $p$ as follows: Let $\varphi$ be a function with support in $\Delta$ which is constantly 1 near $p$. This gives a global section $\tilde\omega_n=\varphi \omega\otimes s_{n-p}\in \Gamma(C\setminus\{p\},K\otimes L(-np))$ which is meromorphic near $p$. Apply the $\bar\partial$ operator to obtain a smooth section $$\bar\partial\tilde\omega_n$$ of $\bar K KL(-np)$ with support in an annulus $A\subset\Delta,$ $p\notin A$. Consider a section $s\in H^0(C,L(np)).$ Then $$\int_C \bar\partial(\tilde\omega_n) s=\int_C(\bar\partial \varphi) \omega s_{-np} s=\int_{\partial A}\varphi \omega s_{-np} s=-\int_\gamma \omega s_{-np}s=-2\pi i Res_p(f\omega),$$ where $\gamma$ is a small curve around $p$ along which $\varphi$ is 1, and $f$ is the meromorphic function $f=s_{-np}s.$ Therefore, by your assumption, the pairing of $\bar\partial\tilde\omega_n$ with any holomorphic section in $L(np)$ vanishes, and Serre duality yields a smooth section $s$ of $KL(-np)$ such that $$\bar\partial\tilde \omega_n=\bar\partial s_n,$$ hence $$\tilde{\omega}_n-s_n$$ is a global meromorphic section of $KL(-np),$ or equivalently, a meromorphic 1-form $\omega_n$ with prescribed behavior up to order $n$ around $p.$ For $n$ big enough ($\geq 2g-2$), all $\omega_n$ are the same (e.g., by the easy part of the Serre duality) and hence coincide with $\omega $ on $\Delta\setminus\{p\}$.

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  • $\begingroup$ Thanks! Do you have a reference? Let me ask you some questions please, I am not sure I understand. Is $\tilde{\omega}_n$ holomorphic outside $p$? I do not think so, but $\tilde{\omega}_n-s_n$ is holomorphic, right? $\endgroup$
    – Giulio
    Apr 25 '17 at 16:23
  • $\begingroup$ On $C\setminus p$, the line bundle $L(-np)$ is trivial, so $K$ and $K\otimes L(-np)$ are equal. I am not understanding your proof, sorry. How you use Serre+Stokes? Thanks $\endgroup$
    – Giulio
    Apr 25 '17 at 16:47
  • $\begingroup$ Quite similar things are in Donaldson's book on Riemann surfaces. $\tilde\omega_n$ is only holomorphic in a punctured neighborhood, but the difference is everywhere except at $p$. $\endgroup$
    – Sebastian
    Apr 25 '17 at 18:27
  • $\begingroup$ I have edited my answer. $\endgroup$
    – Sebastian
    Apr 25 '17 at 19:51
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Let $C$ be the Riemann sphere, $p=0$. Then $$\omega(z)=\left(\sum_{-\infty}^\infty c_nz^n\right)dz.$$ Here the part with negative powers converges for $|z|>0$, while the part with positive powers converges in some disk $|z|<r$.

Your functions $f$ are meromorphic on $C$, regular except at $0$, so they are of the form $p(1/z)$ where $p$ is a polynomial. Consider for example $f_n=z^{-n}$, $n\geq 0$. Then the condition $\mathrm{res}_0{f_n\omega}=0$ implies $c_n=0$ for all $n\geq-1$. So the Laurent series of $\omega$ has only negative powers, and must converge in $C\backslash 0$. So we see that your $\omega$ is holomorphic on $C\backslash \{ p\}$. (As the highest degree in Laurent series is $-2$, $\omega$ is holomorphic at $\infty$).

This proves your statement for the case of the sphere.

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  • $\begingroup$ Thanks! This is very interesting, but I would like to see the higher genus case $\endgroup$
    – Giulio
    Apr 25 '17 at 16:21
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Let $z$ be a standard coordinate on $\Delta$. Under your meromorphic hypotheses, $\omega(z) = w(z) dz$, where $w(z) = \sum_{k=-N}^\infty w_k z^k$ has a convergent Laurent series expansion. The residue of $\omega(z)$ at the origin of $\Delta$ is then just $w_{-1}$.

If you choose the holomorphic functions $f_i(z) = z^i$, the residue of $f_i(z) \omega(z)$ will be $w_{-i-1}$. Thus, by checking that all of these residues vanish for $i\ge 0$, you get that $w_{k} = 0$ for all $k<0$, meaning that $w(z)$ is actually holomorphic at the origin of $\Delta$.

If you are allowed to test residues also with meromorphic $f(z)$ and you know that they also vanish, you get a much stronger conclusion. That is, you are allowed to use $f_i(z)$ with $i<0$, which will let you conclude that $w_k = 0$ even for $k\ge 0$. In other words, the only possible conclusion then is that $\omega(z) = 0$.

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  • $\begingroup$ The function I am allowing are the restriction of regular functions on $C\setminus p$, so these $f_i$ are not allowed, at least for $i>0$, or $i=-1$. (Say that the genus is big, or at least bigger than 0) $\endgroup$
    – Giulio
    Apr 24 '17 at 12:33
  • $\begingroup$ Ah, sorry! I mentally substituted $C$ for $\Delta$ while reading the question. Perhaps there is a way to rescue the argument, with more detailed information about what kind of regular or meromorphic functions are allowed on $C$, but I don't know enough about it. $\endgroup$ Apr 24 '17 at 22:00
  • $\begingroup$ What do you mean by a function $z^i$? We are on a compact Riemann surface $C$. $\endgroup$ Apr 24 '17 at 23:17

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