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 $x_1+...+x_d$.


  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 $C \rightarrow \mathbb{P}^{1}$?


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.

  • $\begingroup$ "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 $g = 1$. $\endgroup$ Commented Oct 25, 2010 at 14:22
  • $\begingroup$ I think it is meant that the map is an isomorphism onto its image. $\endgroup$
    – A. Pascal
    Commented Oct 25, 2010 at 14:29
  • $\begingroup$ 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? $\endgroup$
    – A. Pascal
    Commented Oct 25, 2010 at 14:38
  • 1
    $\begingroup$ 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 $\mathbb{P}^1$. $\endgroup$ Commented Oct 25, 2010 at 14:39
  • 1
    $\begingroup$ @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? $\endgroup$
    – A. Pascal
    Commented Oct 25, 2010 at 15:03

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