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Let $C$ be a Riemann surface of genus $g$, $Sym^{n}C$ be the nth symmetric power of $C$ with $n\geq 2g$, and $JC$ denote the Jacobian of $C$.

Question: Is it generally true that $Sym^{n}C\cong JC\times\mathbb{P}^{n-g}$, where $\mathbb{P}^{n-g}$ is a projective bundle?

Thanks a lot

flag	asked Apr 28 2011 at 15:02 YOURS 203●6

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Not quite. It won't be a product. What is true is that via the Abel-Jacobi map $Sym^n C\to JC$ is a $\mathbb{P}^{n-g}$ bundle for large $n$. – Donu Arapura Apr 28 2011 at 15:42

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Let $P(z)$ and $q(z)$ be the poincare polynomial of $Sym^{n}C$ and $\mathbb{P}^{n-g}$ respectively。 Is it true that the poincare polynomial of $JC$ is the product of $P(z)$ and $q(z)$？ Could you please give me some reference to how $JC$ and $Sym^{n}C$ are related? Thanks a lot – YOURS Apr 28 2011 at 16:06

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@TL: Yes, it is true as the fibration $Sym^nC\to JC$ has structure group $GL$ (i.e., is associated to a vector bundle). – Torsten Ekedahl Apr 28 2011 at 16:08

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The book of Arbarello-Cornalba-Griffiths-Harris on algebraic curves has a lot of detail on the geometry of symmetric products, such as the Chern classes of the vector bundle over $JC$ whose projectivization is the (large) $n$th symmetric product. – Tim Perutz Apr 28 2011 at 16:11

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See McDonald, Topology 1, 1962. He computes the cohomology ring of $Sym^n C$ (additively isomorphic to that of the product by the Leray-Hisch theorem) and his computations show that there is no retraction Sym^n C \to P^{n-g}$. In the stable case, $Sym^{\infty} C$ is a product by Dold-Thom. – Johannes Ebert Apr 28 2011 at 18:31

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nth symmetric power of a Riemann surface and its Jacobian

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