# zeroes of forms on Riemann surfaces

Let $P$ be a point on a Riemann surface.

Does there exist a non-trivial differential form $\omega$ on $X$ such that $\omega$ vanishes at $P$?

Does there exist a non-constant rational function $f$ on $X$ such that the (rational) differential form $df$ vanishes at $P$?

What if we replace the set $\{P\}$ by a finite set of points?

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What about $\omega=0$ and $f=0$? – celtschk Jul 2 '12 at 21:15
I want to exclude this case. Thanks for pointing it out. – Harry Jul 2 '12 at 21:17
I don't understand the second question. – user18119 Jul 3 '12 at 9:19

Just to complete Georges's answer: let $S$ be an effective divisor on $X$ (i.e. a finite sum of closed points), the Riemann-Roch formula (with Serre duality) is $$h^0(Ω_X(−S))=h^0(O_X(S))+g-1-deg S\ge 1+g-1-\deg S=g-\deg S.$$ So the answer to your third question is positive when $\deg S\le g-1$ (this includes the case $S=P$ and $g=2$).

Things become more interesting when $\deg S=g$. The above arguments show that $h^0(\Omega_X(-S))> 0$ if and only if $h^0(O_X(S))\ge 2$. One can identifiy the set of effective divisors of degree $g$ on $X$ to the symetric product $X^{(g)}$ (which is $X^g$ quotient by the symetric group in $g$ elements acting on $X^g$ by permutation of the coordinates). Fix a point $x_0\in X$, then we have a morphism to the Jacobian of $X$ $$f: X^{(g)} \to J(X), \quad S \mapsto [S-gx_0].$$ It is well known that $f$ is birational. But what is the exceptional locus of $f$ ? It consists in those $S$ such that $\dim f^{-1}(f(S))>0$. As $f^{-1}(f(S))=|S|$ the linear system of effective divisors linearly equivalent to $S$ and has dimension (as projective variety) $h^0(O_X(S))-1$, we see that

when $\deg S=g$, $h^0(\Omega_X(-S))> 0$ if and only if $S$ belongs to the exceptional locus of $f$.

Edit Answer to Question 2. Let $t$ be a rational function which is an uniformizing element of $X$ at $P$, let $f=t^2$. Then $df=2tdt$ has a simple zero at $P$.

Remark on the divisor of a rational section (to respond to a question raised in the comments). Let $L$ be an invertible sheaf on $X$, let $s$ be a non-zero rational section of $L$ (i.e. $s\in L(U)$ for some non-empty open subset $U$ of $X$ and $s\ne 0$). Then the divisor $\mathrm{div}_L(s)$ is a Cartier divisor such that $O_X(\mathrm{div}_L(s))\simeq L$.

This is can be check by writing $L$ on an open covering $\{ X_i\}_i$ of $X$ as $L|_{X_i}=e_i O_{X_i}$. So $s=e_if_i$ for some $f_i\in K(X)$. The Cartier divisor $\mathrm{div}_L(s)$ is represented by $\{ (X_i, 1/f_i)\}_i$ and we have $sO_X(\mathrm{div}_L(s))=L$.

In the particular case when $L=\Omega_X$, for any rational differential form $s$, we have $\mathrm{div}_L(s)\simeq K_X$ for any canonical divisor $K_X$ on $X$.

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Wow, that was quick! Thanks for that great, general (with $S$ containing several points) answer, QiL. (And it solves the case $g=2$ and $S=$ a point in an embarrassingly easy way!) – Georges Elencwajg Jul 3 '12 at 9:37
This is indeed wonderful. The question about rational functions remains open, but I guess there's no simple answer. – Harry Jul 3 '12 at 16:18
@Harry: I still don't understand your question about rational functions. The sheaf of rational (meromorphic) differential forms is constant, so it doesn't make sense to talk about a section of the sheaf vanishing at some point. If you just want $df$ to belong to $\Omega_X(-P)_P$, it is easy. – user18119 Jul 3 '12 at 21:43
@QiL. I think I might confusing some things. I always thought that, if $f$ is a rational function on $X$, then the divisor of zeroes of $df$ is a canonical divisor, i.e., a rational section of $\Omega^1$. So assuming this is correct, (Is it?) I'm asking for a rational function such that $P$ is in the support of the divisor of zeroes of $df$. Does this make sense? If I understand your comment correctly, the answer is that this is indeed possible. – Harry Jul 5 '12 at 6:18
@Harry, OK now I understand your question. Thanks. The whole divisor of rational differential is a canonical divisor. I will put some details in the answer. – user18119 Jul 5 '12 at 22:24

For $g=\text {genus}(X)\geq 3$, yes there exist non-zero differential forms vanishing at an arbitrarily given point $P\in X$ : this results from Riemann-Roch applied to $\Omega_X(-P)$.
Indeed, writing $h^i=dim_\mathbb C H^i$, we have
$$h^0(\Omega_X(-P))=h^1(\Omega_X(-P))+1-g+\text {deg} \:\Omega_X(-P)=h^1(\Omega_X(-P))+1-g+2g-3 \\\geq 0+1-g+2g-3=g-2$$ so that $h ^0(X, \Omega_X(-P))\gt 0$ for $g\geq 3$, and thus there exists a nonzero differential form $\omega\in H^0(X, \Omega_X(-P))$ vanishing at $P$.

For $g=0$ no such form exists because only the zero form exists on $X$.
For $g=1$ a non zero differential form does not vanish, so your question has a negative answer .
For $g=2$, I don't know what happens .

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Dear Georges, $g=2$, the answer is also yes. – user18119 Jul 3 '12 at 9:04
Ah, great, thanks a lot @QiL! I suppose it is because a curve of genus $2$ is hyperelliptic and we know explicitly what differntial forms look like for hyperelliptic curves. However I would be very grateful to you if you sketched the calculation, as a a comment or as a new answer. – Georges Elencwajg Jul 3 '12 at 9:11
Georges, see my answer below. – user18119 Jul 3 '12 at 9:18
when $g=2$, if we want to use explicit computations, we can say that $\Omega_X \sim O_X(P+\bar{P})$ for any $P$ (where $\bar{P}$ is the hyperelliptic conjugate of $P$). So $\Omega_X(-P)\simeq O_X(\bar{P})$ and $h^0(\Omega_X(-P))=1$. – user18119 Jul 3 '12 at 9:39
Thanks for this other proof, @QiL. – Georges Elencwajg Jul 3 '12 at 10:00