Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
The Jacobian represents a functor $Pic^0$ defined on the category of schemes over the base of $C$; if you believe that then the result follows relatively easily. One proof of this is in Bosch-Guentzer-Remmert. –  user30035 Feb 11 '13 at 21:04
    
If $C$ were initially given over a subfield $k \subset \mathbf{C}$ do you also want to know that this analytically defined map arises from an algebraic map over $k$ (using the algebraic theory of the Jacobian)? Do you need to know something similar for "algebraic families" of such curves? The argument given by wccanard is the "best" one for such purposes (except that it masks the necessity to prove that the analytic map you have written down is actually computing the same thing as a map built using the functorial perspective in the algebraic theory, up to a sign). –  user28172 Feb 12 '13 at 0:54
1  
I recommend pages 91-93 of James Milne's beautiful notes on Abelian Varieties. jmilne.org/math/CourseNotes/AV.pdf Basically the connection between the analytic and algebraic points of view is Abel's theorem, which equates the Albanese variety and the abelian integral maps to it, with the Picard variety and the tautological maps to it. This theorem in the modern version is a corollary of Poincare duality in both its integral and complex forms, including relation to the Hodge decomposition. –  roy smith Feb 12 '13 at 15:49

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer

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) http://en.wikipedia.org/wiki/Theta_divisor

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.