Constructing rational functions with ramification locus the divisor of some $n$-form - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:35:02Z http://mathoverflow.net/feeds/question/91643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91643/constructing-rational-functions-with-ramification-locus-the-divisor-of-some-n Constructing rational functions with ramification locus the divisor of some $n$-form Harry 2012-03-19T17:36:54Z 2012-03-19T22:40:03Z <p>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.</p> <p>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 $\omega_X^{\otimes n}$.</p> <p>Now, let $D$ be the divisor of $\omega$ on $X$.</p> <p>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? </p> <p>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 $D$?</p> <p>As Francesco points out, this is not possible if $g=2$ and $n=1$</p> <p>Probably, if $g$ is small compared to $n$, the answer will be negative. </p> http://mathoverflow.net/questions/91643/constructing-rational-functions-with-ramification-locus-the-divisor-of-some-n/91647#91647 Answer by Francesco Polizzi for Constructing rational functions with ramification locus the divisor of some $n$-form Francesco Polizzi 2012-03-19T18:14:19Z 2012-03-19T18:31:17Z <p>The answer is <strong>no</strong> as the following simple example shows.</p> <p>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.</p> <p>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. </p> <p><strong>EDIT.</strong> 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. </p>