I'm still busy learning the theory of linear systems for compact Riemann surfaces. If the answer to the following question is negative, then there might not be any point in continuing.

Let $X$ be a compact connected Riemann surface and let $\omega$ be an $n$-form on $X$. That is, $\omega$ is a global section of the canonical sheaf $\omega_X^{\otimes n}$.

Now, let $D$ be the divisor of $\omega$ on $X$.

Can we construct a morphism $X\to \mathbf{P}^1$ such that the support of the ramification locus equals the support of $D$ for some choice of $n$? If yes, the degree of such a morphism equals the degree of $\omega_X^{\otimes n}$, right?

Slightly weaker: can we construct a morphism $X\to \mathbf{P}^1$ such that the support of the ramification locus is contained in the support of $D$?

As Francesco points out, this is not possible if $g=2$ and $n=1$

Probably, if $g$ is small compared to $n$, the answer will be negative.

  • $\begingroup$ Based on your comment on Francesco Polizzi's answer, you have stated this wrong. You should say something like "Can we construct a morphism $X \to \mathbb P^1$ such that the ramification locus is the divisor of some $n−form$ on $X$? Do you want to count ramification with multiplicity or without? $\endgroup$ – Will Sawin Mar 19 '12 at 19:35
  • $\begingroup$ You're right! Without multiplicity at first. I'll take care of multiplicities later. $\endgroup$ – Harry Mar 19 '12 at 20:58
  • $\begingroup$ Why did you delete this question? $\endgroup$ – S. Carnahan Mar 20 '12 at 10:35
  • $\begingroup$ It's better if I figure it out by myself maybe. I did not mean to offend and am grateful to Francesco for his helpful answer. $\endgroup$ – Harry Mar 20 '12 at 10:45

The answer is no as the following simple example shows.

Assume $g(X)=2$ and take $n=1$, i.e. $D$ is the divisor of a holomorphic $1$-form. Then $\deg D=2$, so if your morphism $f \colon X \to \mathbf{P}^1$ exists, it is ramified at two points.

Consequently, $f$ is branched at at most two points, so at exactly two points since $\mathbf{P}^1$ minus a point is simply connected. But any cover of $\mathbf{P}^1$ branched at two points is still $\mathbf{P}^1$, a contradiction.

EDIT. For completeness, let me show my assertion that if the cover $f \colon X \to \mathbf{P}^1$ is branched at two points, then $ X \cong \mathbf{P}^1$. In general, if $f$ has degree $d$, the branch points are $b_1, \ldots ,b_n$ and the permutation $\sigma_i$ giving the local monodromy at $b_i$ is the product of $k_i$ disjoint cycles, then $$g(X)=1 + \frac{(n-2)d-\sum_{i=1}^nk_i}{2},$$ see for instance [Miranda, Algebraic curves and Riemann surfaces, page 93]. In particular if $n=2$ the only possibility is $k_1=k_2=1$ and $g(X)=0$, so $X$ is isomorphic to the projective line.

  • 2
    $\begingroup$ Yes, but that's why I allow one to take powers of $\omega$. So what if we take for example a section of $\omega^{1/2 g(g+1)}$? Then the degree of the divisor is $g^3-g$. So according to your answer, it should ramify at $g^3-g$ points (counted with multiplicity?). So the number of branch points is bounded by the genus again, but now the bound is big. By the way, what is the relation between the ramification divisor (so ramification locus with proper multiplicities) and the divisor of our section in this example? $\endgroup$ – Harry Mar 19 '12 at 18:18

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