TITLE: Numerical invariants of symmetric products of curves
QUESTION [5 upvotes]: Let $n\ge 2$ be an integer and $C$ be a smooth projective curve of genus $g>n$. The $n$-the symmetric product $C(n)$ is a smooth variety of general type.
If $n=2$ then $C(2)$ is a minimal surface ad it is not hard to compute its numerical invariants. I would like to know whether $C(n)$ is minimal also for $n>2$ and what is the volume of the canonical bundle. 
There is famous article by MacDonald
http://www.sciencedirect.com/science/article/pii/0040938362900198
about symmetric products. In it I found the formula for $\chi(K_{C(n)})$, but I was not able to find the answer to my question above, although it may be  there.

REPLY [4 votes]: Claim:  If $C$ is a curve of genus $g \geq 2,$ the canonical class of $C^{(d)}$ is nef and big if and only if $1 \leq d \leq g-1.$
Proof:  Let $\theta \in {\rm NS}(C^{(d)})$ be the class of the pullback of the theta-divisor of ${\rm Jac}(C)$ via the Abel map, and let $x \in {\rm NS}(C^{(d)})$ be the class of the divisor $\{D + p_{0} : D \in C^{(d-1)}\},$ where $p_0 \in C$ is a given point.  (Moving $p_{0}$ keeps us in the same algebraic equivalence class, so $x$ is independent of $p_{0}.$)  These classes are linearly independent, by the formulas for $\theta^{j} \cdot x^{d-j}$ given in ACGH.  Since the theta-divisor on ${\rm Jac}(C)$ is ample, $\theta$ is nef.    
As abx points out, $C^{(d)}$ is a projective bundle over ${\rm Jac}(C)$ for $d \geq 2g-1;$ it is the subspace projectivization of a Picard bundle.  In this case $x$ is the class of the relative $\mathcal{O}(1).$  The Chern class formulas for the Picard bundle given in ACGH, together with repeated application of the adjunction formula to the embedding $C^{(d)} \hookrightarrow C^{(d+1)}$ induced by adding a point, imply that 
$$K_{C^{(d)}} = \theta + (g-d-1)x$$
A similar argument using the ampleness of the dual of the Picard bundle implies that the divisor class $x$ is ample.
The nefness of $\theta$ and the ampleness of $x$ imply that $d \leq g-1$ is sufficient for $K_{C^{(d)}}$ to be nef and big.  For necessity, note that when $d \geq g,$ the class $\theta$ spans a boundary of the nef cone since the Abel map contracts a positive-dimensional locus; if $K_{C^{(d)}}$ is nef and big, then $\theta = K_{C^{(d)}}-(g-d-1)x$ is ample, which is absurd.  This concludes the proof.
NOTE:  The previous argument shows that $K_{C^{(d)}}$ is ample if and only if $1 \leq d \leq g-2.$