Let $X$ be a compact Riemann surface, $\omega$ a meromorphic differential on $X$ and $f$ a meromorphic function on $X$ with poles only over the points $P_1,\dots,P_d$. The product $\;f\cdot\omega\;$ is still a meromorphic differential, so by the residue theorem we know that $$ \sum_{P\in X} res_P(\omega) = \sum_{P\in X} res_P(\;f\cdot\omega) = 0. $$

My question is:

what can we say about $ \sum_{i=1}^{d} res_{P_i}(\;f\cdot\omega)$ ?

Is this sum also zero, or is it there any condition on $\omega$ that forces it to be zero?


$$ \heartsuit \quad \heartsuit \quad \heartsuit $$


Remark (to make it more interesting)

If we think of $X$ as an algebraic curve over $\mathbb{C}$ and we set $D=P_1+\dots+ P_d$, we can consider the exact sequence $$ H^0(\mathcal{O}_X(D)) \to H^0(\mathcal{O}_X(D)\otimes \mathcal{O}_D)\to H^1(\mathcal{O}_X) $$ where the first map is the restriction map $\alpha$ and the second one is the coboundary $\delta$ coming from the Mittag-Leffler sequence of invertible sheaves $$ \mathcal{O}_X \to \mathcal{O}_X(D)\to \mathcal{O}_X(D)\otimes \mathcal{O}_D\;. $$

In this language the above meromorphic $f$ is an element of $\mathcal{O}_X(D)$ and, if $v\in H^0(\mathcal{O}_X(D)\otimes \mathcal{O}_D)$ is such that $\alpha(f)=v$, then we have $\delta v = 0$.

One can define a perfect pairing $$ \langle \bullet, \bullet \rangle : H^1(\mathcal{O}_X) \otimes H^0(K) \to \mathbb{C} $$ by setting $$ \langle \delta v, \omega \rangle := \sum_{P\in D} res_P (\omega\cdot v_P). $$

My question arises from the fact that I would expect $\delta v = 0$ to imply that $\langle \delta v, \omega \rangle = 0$ for every differential $\omega$.


The answer to your first question is, "no." Let $X$ be the Riemann sphere $\mathbb{C}P^1$ with the meromorphic differential $\omega = dz/z = -dw/w$, where $w=1/z$. Let $f$ be the meromorphic function $f(z) = z/(z-z_0)$ for some $z_0 \neq 0$. The only pole of $f$ is at $z_0$, and the residue of $f\omega$ at that pole, i.e., the residue of $dz/(z-z_0)$ at $z_0$, is nonzero.

  • $\begingroup$ So what about the Remark? Is that definition of the pairing just a nonsense then? $\endgroup$ – Abramo Nov 8 '13 at 20:20

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