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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.

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  • $\begingroup$ When you write "minimal", does that mean that the variety is a "minimal model", i.e., the canonical bundle is nef? $\endgroup$
    – Jason Starr
    Commented Dec 26, 2015 at 12:22
  • $\begingroup$ yes, I mean nef&big canonical bundle $\endgroup$
    – rita
    Commented Dec 26, 2015 at 12:28
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    $\begingroup$ Most of what I know about this topic, I learned from Yusuf Mustopa. If $n$ is at least as large as the gonality of $C$ (e.g., certainly $n\geq (g+2)/2$ suffices), and if $C(n)$ has Picard rank $2$ (as it does if $C$ is sufficiently general in moduli), then we can easily compute the nef cone: the pullback of the theta divisor by the Abel map gives one ray, and the "pairwise difference" morphism to the product of Kummers of the Jacobian gives a second contraction (contracting the small diagonal). So we should be able to determine where the canonical divisor class is relative to the two rays. $\endgroup$
    – Jason Starr
    Commented Dec 26, 2015 at 12:52
  • $\begingroup$ @JasonStarr: thank you. In fact $C$ general is what I am interested in, and $g$ large with respect to $n$ is also fine. $\endgroup$
    – rita
    Commented Dec 26, 2015 at 14:35
  • $\begingroup$ For $n=g$ the exceptional divisor of the birational contraction $C(g)\rightarrow JC$ is a canonical divisor, it is certainly not nef. $\endgroup$
    – abx
    Commented Dec 26, 2015 at 15:58

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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 $1 \leq d \leq g-2.$

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