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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:

  1. 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?

  2. 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...

  3. 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?

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This could be complete bs, but maybe you should view their opaque sentence as telling you that the two spaces represent the same functors... both of these spaces have pretty clear functorial interpretations. The symmetric product of curves represents families of effective divisors of degree d. The projective bundle P(E) represents a functor by definition, which is equivalent to the data of a line bundle of degree d on the curve with a section(presumably this is what they explain), which gives us an effective divisor. This might make 3 more obvious as well... just a thought –  Daniel Pomerleano Jun 24 '11 at 10:56
About 2: you need to assume $n>g$. –  rita Jun 25 '11 at 15:45
I recommend: Abelian Integrals, by George Kempf, published by the Univercidad Nacional Autonoma de Mexico, 1983. –  roy smith Jun 26 '11 at 15:44

2 Answers 2

up vote 6 down vote accepted

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.

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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 $E=\text{some line bundle}\otimes P_n$.

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