most recent 30 from http://mathoverflow.net 2013-06-18T03:40:43Z http://mathoverflow.net/feeds/question/121527 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121527/why-the-abel-jacoby-map-is-algebraic-morphism Klim Puhov 2013-02-11T20:55:21Z 2013-02-11T21:13:36Z

<p>The Abel-Jacobi map from the algebraic curve $C$ to its Jacobian $J(C)$ is given analitically by $$p\to \left( \ldots, \int^{p}_{p_0} \omega_i,\ldots\right),$$ where $p_0$ is some point on $C$ and $\omega_i$ form a basis of $H^0(C,K)$. Why it is an algebraic morphism?</p>

http://mathoverflow.net/questions/121527/why-the-abel-jacoby-map-is-algebraic-morphism/121529#121529 Abhinav Kumar 2013-02-11T21:05:12Z 2013-02-11T21:05:12Z

<p>Maybe there are easier ways to see it, but Chow's theorem/GAGA certainly gives you the result, since you have an analytic morphism of projective analytic varieties.</p>

http://mathoverflow.net/questions/121527/why-the-abel-jacoby-map-is-algebraic-morphism/121530#121530 solbap 2013-02-11T21:13:36Z 2013-02-11T21:13:36Z

<p>This map is written thinking of J(C) as C^g/L where g = genus(C) and L is the lattice of periods. C^g/L can be made algebraic by embedding into a projective space via theta functions. These are section of a line bundle associated to some multiple of a theta divisor $\Theta$ on J(C) <a href="http://en.wikipedia.org/wiki/Theta_divisor" rel="nofollow">http://en.wikipedia.org/wiki/Theta_divisor</a></p> <p>Then it suffices to show that the restriction to $\Theta$ to C is an ample divisor; but J(C) is a smooth manifold and dim $\Theta$ + dim C = dim J(C) and so $C \cap \Theta$ is a finite number of points, hence ample.</p>