5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.