Does there exist a non effective divisor with positive degree? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:54:57Z http://mathoverflow.net/feeds/question/109577 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109577/does-there-exist-a-non-effective-divisor-with-positive-degree Does there exist a non effective divisor with positive degree? MZWang 2012-10-14T02:37:33Z 2012-10-25T05:32:19Z <p>Suppose $X$ be a smooth projective curve over $\mathbb{C}$. Let $D$ be a divisor with $\deg D>0$ on $X$. Is it possible that $l(D)=0$, i.e. D is not linearly equivalent to an effective divisor? </p> http://mathoverflow.net/questions/109577/does-there-exist-a-non-effective-divisor-with-positive-degree/109581#109581 Answer by Faisal for Does there exist a non effective divisor with positive degree? Faisal 2012-10-14T04:54:18Z 2012-10-14T15:04:02Z <p>Here's one way of seeing why Piotr's claim is true. Write <code>$g$</code> for the genus of <code>$X$</code> and begin with the following observation. If <code>$p_1, \ldots, p_g$</code> are points on <code>$X$</code>, then the divisor <code>$E=p_1+\cdots+p_g$</code> has <code>$l(E) = g + 1 - \rho$</code>, where <code>$\rho$</code> is the rank of the associated Brill–Noether matrix. This matrix is <code>$g\times g$</code> so for points <code>$p_1, \ldots, p_g$</code> in general poisiton, we'll have <code>$\rho=g$</code> and consequently <code>$l(E)=1$</code>. (For a more precise statement, see &sect;7c in Gunning, <em>Lectures on Riemann Surfaces</em>, PUP 1966.) Now take one such <code>$E$</code> and consider the divisor <code>$D=E-q$</code>. We have <code>$\deg D = g-1 &gt; 0$</code> and, for generic <code>$q$</code>, <code>$l(D)=l(E)-1=0$</code>.</p> <p>If <code>$g=0$</code> or <code>$1$</code> then it follows easily from Riemann–Roch that <code>$\deg D &gt; 0 \implies l(D)&gt;0$</code>.</p> http://mathoverflow.net/questions/109577/does-there-exist-a-non-effective-divisor-with-positive-degree/109584#109584 Answer by Sándor Kovács for Does there exist a non effective divisor with positive degree? Sándor Kovács 2012-10-14T06:13:06Z 2012-10-25T05:32:19Z <p>If your curve is not rational or elliptic, i.e., $g\geq 2$ as Piotr commented, then there is a simple way to give an example:</p> <p>Let $D=P_1+P_2-Q_1$ for some $P_1,P_2, Q_1\in X$ pairwise different points. Clearly $\deg D=1$, so if it were linearly equivalent to an effective divisor, which would have to be a point, say $Q_2\in X$. It follows that then there exists a rational function on $X$ with zeroes at $P_1$ and $P_2$ and poles at $Q_1$ and $Q_2$. In other words, there exists a degree $2$ morphism $X\to \mathbb P^1$ with $P_1$ and $P_2$ mapping to the same point. Hence this works already if $X$ is not hyperelliptic. </p> <p>If $X$ is hyperelliptic, then this morphism is given by the basepoint-free linear system $|K_X|$. In that case choose a new $P_2$ (say $P_2'$) that's different from the original $P_2$. If $P_1+P_2'\in |K_X|$, then $P_2\sim P_2'$, which would imply (similarly as above) that $X\simeq \mathbb P^1$ and if $P_1+P_2'\not\in |K_X|$, then the above construction gives an example as desired.</p> <p><strong>Remark</strong> actually it is not important that the basepoint-free linear system in the second paragraph is $|K_X|$, just that it is some linear system.</p> http://mathoverflow.net/questions/109577/does-there-exist-a-non-effective-divisor-with-positive-degree/109600#109600 Answer by Henri for Does there exist a non effective divisor with positive degree? Henri 2012-10-14T11:06:04Z 2012-10-14T11:06:04Z <p>Another possible approach is the following one, using Abel-Jacobi map : if $g(X)\geq 2$, then Abel-Jacobi map gives an isomorphism $\varphi:Pic^0(X) \simeq \mathbb C^g/\Lambda$ for some lattice $\Lambda$. Now you can translate the map by some point $p\in X$; more precisely, define $\psi(D):=\varphi(D-(p))$ for every divisor $D$ of degree $1$ on $X$.</p> <p>As $g\geq 2$, the image by $\psi$ of all points (seen as degree $1$ divisors) will be one-dimensional inside the $g$-dimensional variety $\mathbb C^g/\Lambda$. Therefore there exist on $X$ (a lot of) divisors of degree $0$ which cannot be written as $(q)-(p)$ for any point $q$. </p> <p>Now consider such a divisor $D$, and define $D'=D+(p)$. It has degree 1, but if it were effective, it would be equivalent to $(q)$ for some point $q\in X$. Therefore one would have $D \sim (q)-(p)$ which is impossible. </p>