Let $C$ be a hyperelliptic curve of genus $g$ and let $S = C^{(2)}$ denote the symmetric square of $C$. Let $\nabla$ be the divisor on $C^2$ defined by $\{(P, \overline{P}) \mid P \in C\}$ where $\overline{P}$ denotes the image of $P$ under the hyperelliptic involution. Finally let $\nabla_S$ be the push-forward of $\nabla$ under the quotient map $C^2 \to S$.
If $g = 2$, then $S$ is the blowup of the Jacobian $J_C$ of $C$ at the origin with exceptional divisor $\nabla_S$. So, by results from Hartshorne Section V.3 on the relationships between blowups and the intersection pairing, we immediately conclude that $\nabla_S$ is a canonical divisor of $S$ and it has self-intersection $-1$. I would like to know how to do these calculations without the crutch of using blowups, and hence, with any luck, obtain the analogous results for when $g>2$. My question is therefore
How do I calculate a canonical divisor, and its self-intersection number, of the symmetric square of a hyperelliptic curve of genus $g>2$?
Best Answer
Let $\pi: C^2\to S$ be the canonical quotient morphism. It is étale outside of the diagonal $\Delta$ of $C^2$. Let $K$ be the base field. The canonical map of differentials $$ \pi^*\Omega_{S/K}^1\to \Omega^1_{C^2/K}$$ is then an isomorphism outside of $\Delta$, and induces an injective homomorphism $\pi^*\omega_{S/K}\to \omega_{C^2/K}$ of canonical sheaves. So $\pi^*\omega_{S/K}=\omega_{C^2/K}(-D)$ for some effective divisor $D$ on $C^2$, with support in $\Delta$, hence $D=r\Delta$ for some integer $r\ge 0$. The multiplicity $r$ can be computed locally for the Zariski topology, and even for étale topology on $C$. So we can work with Spec ($K[x]$) and find $r=1$: $$ \pi^*\omega_{S/K}=\omega_{C^2/K}(-\Delta).$$ This should be enough to describe a canonical divisor on $S$ interms of the pushforward of a canonical divisor of $C^2$ and of $\pi(\Delta)$. The self-intersection should be easily computed with the projection formula. Otherwise ask for more details.
Edit Computation of the multiplicity $r$. Let $\xi$ be the generic point of $\Delta$ and $\eta=\pi(\xi)$. Then $$ \omega_{S/K, \xi}\otimes O_{C^2,\xi}=(\pi^*\omega_{S/K})_{\xi}=\omega_{C^2/K}(-r\Delta)_{\xi}=\omega_{C^2/K,\xi}(-r\Delta).$$ This explains why $r$ can be computed Zariski locally. Let $U$ be a dense open subset of $C$. Then one can compute $r$ on $U^2\to U^{(2)}$. If we can write $U$ as an étale cover $U\to V\subseteq \mathbb A^1_K$, then the map $$ \pi^*\Omega_{U^{(2)}/K}^1\to \Omega^1_{U^2/K}$$ is just the pull-back of the map $$ \pi^*\Omega_{V^{(2)}/K}^1\to \Omega^1_{V^2/K}.$$ If you take a local basis $dx, dy$ for $\Omega^1_{V^2/K}$, then $d(x+y), d(xy)$ is a local basis for $\Omega^1_{V^{(2)}/K}$, and their pull-backs to $U^2$ (resp. $U^{(2)}$) are local bases, and $r$ can be computed with these local bases. Now $\omega_{V^2/K}$ is generated by $dx\wedge dy$, and $\omega_{V^{(2)}/K}$ is generated by $d(x+y)\wedge d(xy)$ who image in $\omega_{V^2/K}$ is $(x-y)(dx\wedge dy)$. As $x-y$ generates locally the ideal of $\Delta$, we see that $r=1$.