Let $C$ be a smooth projective curve over an algebraically closed field $k$, of genus $g$. It is well known that, after fixing a point $p_0$, the map $C^{(n)}\to J$ sending $\{a_1,\dots,a_n\}$ to $[a_1+\dots+a_n-np_0]$, from the n-th symmetric product of the curve to its Jacobian, is an algebraic projective bundle (for $n>2g-2$).

Consider the map $C^{(n)}\times C^{(n)}\to J$, sending $(\{a_1,\dots,a_n\},\{b_1,\dots,b_n\})$ to $[a_1+\dots+a_n-b_1-\dots-b_n]$.

Is this map also a fiber bundle?

  • $\begingroup$ Certainly not. If it was, the induced map on the Albanese variety would be an isomorphism. But $\operatorname{Alb}(C^{(n)}\times C^{(n)})\cong J\times J $. $\endgroup$ – abx Feb 4 '18 at 13:56
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    $\begingroup$ I also agree that this is not a "fiber bundle" in the sense of a morphism that, locally on the target, is isomorphic to projection of a product scheme to a factor. However, I do not see how to deduce this from the Albanese variety (perhaps the fiber itself has nontrivial Albanese isomorphic to $J$). Rather, I believe that this follows from the fact that the projective bundle $C^{(n)}\to \text{Pic}^n_{C/k}$ is not equivariant for the natural action of $\text{Pic}^0_{C/k}$. Of course if $k$ equals $\mathbb{C}$, then Ehresmann applies to the underlying differentiable manifolds. $\endgroup$ – Jason Starr Feb 4 '18 at 16:25
  • $\begingroup$ Sorry, I assumed the OP meant projective fiber bundle. I am not sure what he means by "fiber bundle". $\endgroup$ – abx Feb 4 '18 at 17:01
  • $\begingroup$ @abx Sorry for the intrusion, but why is $\mathrm{Alb}(C^{(n)}\times C^{(n)})\cong J\times J$? Is in general, $\mathrm{Alb}(C^{(n)})\cong J$? Where does it come from? $\endgroup$ – Alessio Jun 10 at 9:49
  • $\begingroup$ @Alessio: Yes, $\operatorname{Alb}(C^{(n)})\cong J $. Use the fact that the natural map $C^{(n)}\rightarrow J$ induces an isomorphism on $H_1(-,\mathbb{Z})$. $\endgroup$ – abx Jun 10 at 16:44

Let $J_n = Pic(C)_n$ --- the moduli space of line bundles of degree $n$ on $C$. Then there is a map $$ C^{(n)} \to J^n,\qquad \{a_1,\dots,a_n\} \mapsto O(a_1+\dots+a_n). $$ This is a slightly more canonical version of the map you considered, in particular it is a projective bundle for $n > 2g - 2$.

The map you are interested in can be written as the composition $$ C^{(n)} \times C^{(n)} \to J_n \times J_n \cong J_n \times J_{-n} \to J_0. $$ Here the first map is the product of to projective bundles (its fiber is a product of two projective spaces), the second is an isomorphism (given by dualization of a line bundle in the second factor), and the third is a (trivial) abelian fibration.

So, altogether, the fibers of your map are fiber products of two projective bundles over an abelian variety.

  • $\begingroup$ I see why the fiber is the product of two projective spaces, but why do we have local triviality? $\endgroup$ – user4231 Feb 4 '18 at 17:37
  • $\begingroup$ @user4231: Local triviality of what? $\endgroup$ – Sasha Feb 4 '18 at 17:42
  • $\begingroup$ Maybe I should clarify my question. Arthur Mattuck, in his article “Picard bundle”, showed that $C^{(n)}\to J$ is a projective fiber bundle in the sense of en.m.wikipedia.org/wiki/Fiber_bundle. My question is, is the map that I gave a fiber bundle in that sense? $\endgroup$ – user4231 Feb 4 '18 at 17:53
  • $\begingroup$ @user4231: I am not sure it is locally trivial, but each component, $C^{(n)} \times C^{(n)} \to J_n \times J_n$ and $J_n \times J_n \to J_0$, is locally trivial. $\endgroup$ – Sasha Feb 4 '18 at 18:23
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    $\begingroup$ That is not locally trivial. That is what my comment was about. If you pullback the projective bundle by a general translation, it is a non-isomorphic projective bundle. $\endgroup$ – Jason Starr Feb 4 '18 at 18:24

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