MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the Kodaira dimension of symmetric products of curves? That is, given a projective smooth, connected complex curve $C$, what is the Kodaira dimension of $C^{(d)}=C^d/\mathfrak S_d$?

When $d> g$, the genus of $C$, then $C^{(d)}$ is a bundle in projective spaces over the Jacobian of $C$, hence all pluriforms on $C^{(d)}$ vanish on the fibers of this fibration and $\kappa(C^{(d)})=0$ in this case.

Is something known for $2\leq g\leq g-1$? (This question is prompted by this other post.) I suspect that the answer will strongly depend on fine properties of the curve $X$ (gonality, Brill-Noether properties) and there might not be a general and neat answer.

Perhaps surprisingly, the analogous question in higher dimensions is quite different since if $X$ is a projective smooth connected variety of dimension $>1$, the Kodaira dimension of (any desingularization) of $X^{(d)}$ is equal to $d \kappa(X)$, where $\kappa(X)$ is the Kodaira dimension of $X$ (D. Arapura, S. Archava, Kodaira dimension of symmetric powers, Proc. AMS, 2003).

share|cite|improve this question
If the Jacobian of the curve is simple, then all its proper subvarieties are of general type; in particular the symmetric product of the curve is of general type, until the Abel-Jacobi map is surjective. By deformation, I would guess that the same is true for all curves, not just the ones that have simple Jacobian. – M P Mar 8 '13 at 7:58
up vote 16 down vote accepted

Let $C$ be a smooth projective connected complex curve of genus $\geq 2$. Let me show that $C^{(d)}$ is of general type if $1\leq d\leq g-1$.

Equivalently, one needs to show that the image $W_d$ of $C^{(d)}$ in the jacobian $J(C)$ is of general type, because $C^{(d)}\to W_d$ is birational. If $W_d$ were not of general type, then, by [Ueno, Classification of algebraic varieties I, Theorem 3.10], there would be a non-trivial abelian variety $A\subset J(C)$ such that $W_d$ is stable by translation by $A$ (this is the argument in MP's comment above). But then, $W_{g-1}$ would also be stable by translation by $A$. Now choose a point $x$ outside of $W_{g-1}$ and consider the orbit $A.x$ : it is a positive-dimensional variety avoiding $W_{g-1}$. This is a contradiction because $W_{g-1}$ is an ample divisor: the theta divisor.

share|cite|improve this answer
Thanks a lot, Olivier! (I should have known that...) – ACL Mar 8 '13 at 13:59

Your Answer


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.