# Kodaira dimension of symmetric products of curves

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]

• 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
Commented Mar 8, 2013 at 7:58
• 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. Commented Jun 25, 2018 at 12:27

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.