Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

2011-06-24T02:13:14Z

I am interested in this claim:

The $n$th symmetric power $C^{(n)}$ of a genus $g$ curve $C$ is isomorphic to the projectivization $\mathbb{P}(E_n)$ of the sheaf $E_n := \pi_\ast(P_n)$ over the Jacobian $J(C)$, where $P_n$ is a degree $n$ Poincare bundle over $C \times J(C)$ and $\pi$ is the projection $C \times J(C) \to J(C)$.

Moreover, under this isomorphism, the standard line bundle $\mathcal{O}(1)$ over $\mathbb{P}(E_n)$ corresponds to the line bundle $\mathcal{O}(D)$, where $D$ is the divisor corresponding to the image of the map $C^{(n-1)} \hookrightarrow C^{(n)}$ given by $p_1 + \cdots + p_{n-1} \mapsto p_1 + \cdots + p_{n-1} + p$, where $p$ is some fixed point.

(Also, the isomorphism $\phi : C^{(n)} \to \mathbb{P}(E_n)$ is compatible with the Abel-Jacobi map $u: C^{(n)} \to J(C)$, that is, $u = p \circ \phi$, where $p : \mathbb{P}(E_n) \to J(C)$.)

My questions:

This is claimed on page 309 of the book "Geometry of Algebraic Curves" by Arbarello-Cornalba-Grifiths-Harris, for $n \geq 2g-1$ (so that $E_n$ is a vector bundle, by Riemann-Roch; for smaller $n$ it isn't necessarily a vector bundle and they don't address this case). Their proof is pretty sketchy. It basically just says that, since the fibers of $\mathbb{P}(E_n) \to J(C)$ correspond to effective degree $n$ divisors, it follows that $C^{(n)} \cong \mathbb{P}(E_n)$. But this seems to me to only prove a set theoretic bijection between the two. So, how do I prove that I actually have an isomorphism of varieties? Or, is there a(nother) reference?
I believe the claim should still be true for $n < 2g-1$. Again, how do I prove this? Is there a reference? The sheaf $E_n$ will no longer be locally free, so $\mathbb{P}(E_n)$ will no longer be a bundle of projective spaces, but one should still be able to take the projectivization of a sheaf...
On page 7 of the paper http://arxiv.org/abs/0805.3621 by Moonen and Polishchuk, they talk about the families version of this statement. To be precise, they consider a family $\pi: C \to S$ of curves, and everything is done relative to the base $S$ (take relative symmetric product, take relative Jacobian, and so on). In this case, we must have a section $s: S \to C$ of $\pi$, corresponding to picking a point in each fiber, in order for the map $C^{(n-1)} \hookrightarrow C^{(n)}$ and the divisor $D$ to make sense. Anyway, Moonen and Polishchuk claim that in this families situation, $\mathbb{P}(E_n)$ is still isomorphic to the symmetric product $C^{(n)}$, and that under this isomorphism the line bundle $\mathcal{O}(1)$ corresponds to the line bundle $\mathcal{O}(D + n\psi)$, where $\psi$ is given by $\psi = \pi^\ast s^\ast K$, where $K$ is the relative canonical class of $\pi$. But how do I prove these statements?

Answer by Vivek Shende for Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

2011-06-24T02:29:18Z

This is worked out in excruciating detail in the article Jacobians and Symmetric products by Schwarzenberger. I think the arguments there are perfectly good in the families setting as well.

Answer by Rex for Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

2011-06-24T13:31:31Z

One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_*(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity and Riemann-Roch, for $n$ large this is a vector bundle $E$. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^*u_*L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$.

But it is not clear to me why $E=\text{some line bundle}\otimes P_n$.

Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof - MathOverflow

Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

Answer by Vivek Shende for Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

Answer by Rex for Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof