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

Let $C$ be a smooth projective curve over some field (algebraically closed if necessary), ${\rm Pic}^{d}(C)$ the Picard variety of degree $d$ divisor classes on $C$, $C^{(d)}$ the $d$th symmetric power of $C$. We want to define an Abel map $\alpha_{d}:C^{(d)} \rightarrow {\rm Pic}^{d}(C)$, which intuitively sends a point on $C^{(d)}$ thought of as an unordered $d$-tuple $x_1,...,x_d$ of points in $C$ to the class of the divisor $x_1+...+x_d$.

Questions:

  1. Is $C^{(d)}$ a fine moduli space of effective degree $d$ divisors and ${\rm Pic}^{d}(C)$ a fine moduli space of degree $d$ divisor classes (or degree $d$ line bundles), so that there are appropriate universal families over them? Does the field of definition matter here?

  2. I've heard that for $d \leq g$, the Abel map $\alpha_{d}$ is generically injective. Does this mean that it is injective on a Zariski open subset of $C$?

  3. Consider the case $d=1$, so that the Abel map is $\alpha_1: C \rightarrow {\rm Pic}^{d}(C)$. Can one describe the image and fibres explicitly? What if $C$ is hyperelliptic? Then can we give such a description in terms of the map $C \rightarrow \mathbb{P}^{1}$?

share|improve this question

1 Answer 1

  1. I don't know offhand, would have to look up. The standard reference is Arbarello, Cornalba, Griffiths, Harris, Geometry of Algebraic Curves, Springer. Great book.

  2. Yes.

  3. For $d=1$ (and $g>0$), the map is an isomorphism to its image. In general, a fiber of $\alpha_d$ is a projective space consisting of effective divisors in the same linear equivalence class.

Hyperellipticity plays no role when $d=1$. For $d=2$, if the curve is not hyperelliptic, the map is an isomorphism (again to its image if $g > 1$). If the curve is hyperelliptic, the map is injective outside the set of divisors of the form $(x,y)+(x,-y)$ and crushes this set to one point.

Edit: Clarified point raised in comments. Isomorphism to its image.

share|improve this answer
    
"For $d = 1$ (and $g > 0$), the map is an isomorphism." Que?? Unless my head is screwed on especially loosely this morning, this is not true unless $g = 1$. –  Pete L. Clark Oct 25 '10 at 14:22
    
I think it is meant that the map is an isomorphism onto its image. –  A. Pascal Oct 25 '10 at 14:29
    
If $\alpha_{1}$ is indeed injective for $g \geq 1$, this means that two distinct points $x$ and $y$ on $C$ are never linearly equivalent as divisors, or equivalently, that the line bundle $\mathcal{O}(x-y)$ is always non-trivial. Is this easy to see? –  A. Pascal Oct 25 '10 at 14:38
1  
Oh, okay. I didn't realize that kind of terminology was still in play, but sure, it's an embedding: if for distinct points $P$ and $Q$, $[P] \sim [Q]$, then $C$ admits a degree $1$ map to $\mathbb{P}^1$. –  Pete L. Clark Oct 25 '10 at 14:39
1  
@Pete L. Clark. Here is my attempt to provide more details: if $[P] \equiv [Q]$, then $\mathcal{O}(P−Q)$ is trivial and has one non-zero section up to scalar. Then the short exact sequence $0 \rightarrow \mathcal{O}(P-Q)\rightarrow \mathcal{O}(P) \rightarrow \mathcal{O}_{Q}\rightarrow 0$ shows that $\mathcal{O}(P)$ has a two-dimensional space of global sections. Also, I suppose you can show it is globally generated. Then the complete linear system $|P|$ gives the desired map to P1. Sounds good? –  A. Pascal Oct 25 '10 at 15:03

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.