Strong (Inverse of) Residue Theorem

Question

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

Answer 1

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 $\Delta\setminus\{p\}$.

Answer 2

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.

Answer 3

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 $\omega(z) = 0$.