20
$\begingroup$

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 $\kappa(C^{(d)})=-\infty$ in this case. (Said in another way, $C^{(d)}$ is uniruled.)

Is something known for $2\leq d\leq g-1$? (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).

[Edited to correct an incorrect assertion, as pointed out by @pbelmans]

$\endgroup$
  • 12
    $\begingroup$ 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. $\endgroup$ – M P Mar 8 '13 at 7:58
  • 3
    $\begingroup$ It's probably worth noting that $\kappa(C^{(d)})=-\infty$ for $d>g$, not 0: these symmetric powers are uniruled. One has $\kappa(C^{(g)})=0$ though, being birational to an abelian variety. $\endgroup$ – pbelmans Jun 25 '18 at 12:27
23
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I am currently working on a related topic and I need this exact result, so is there a paper, book or publication having this proof that I can reference? $\endgroup$ – pjox Jul 25 '18 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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