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Hi, my question concerns the Picard bundles in the special case of an elliptic curve $E$, say over the complex numbers. Let $p$ and $q$ be the projections of $E \times E$.

One defines the $n$-th Picard bundle ($n$ an integer) as follows: take the line bundle associated to the divisor $n$ times O (the point zero weighted with $n$) on $E$, pull it back via $p$, tensor it on $E \times E$ with the Poincaré bundle, and then take the direct image of this under $q$.

Does anybody know whether one can say explicitly what the result of this construction is in the special case of an elliptic curve? (one usually defines the Picard sheaves for general curves and uses the jacobian, see the book of Birkenhake Lange, Complex abelian varieties)

Thanks a lot!

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Let $J^d$ be the degree $d$ Jacobian variety, parameterizing degree $d$ line bundles. Let $P^0$ be a degree $0$ Poincare bundle on $E \times J^0$.

Let $x$ be a point of $E$. The bundle $p^\ast (\mathcal{O}_E(nx)) \otimes P^0$ on $E \times J^0$ is the pullback of a degree $n$ Poincare bundle $P^n$ on $E \times J^n$ along the map $\operatorname{id}_E \times (- \otimes \mathcal{O}_E(nx)) : E \times J^0 \to E \times J^n$. This map is clearly an isomorphism.

By Proposition 2.1 on page 309 of Arbarello-Cornalba-Griffiths-Harris [for a more comprehensive treatment of this result, see Schwarzenberger's "Jacobians and Symmetric Products"], for $n \geq 1$, (the total space of) the projectivization of the sheaf $q_\ast(P^n)$ is isomorphic to the $n$th symmetric power of $E$, and under this isomorphism, the map from this projective bundle to $J^0 \cong E$ is the Abel-Jacobi map. Hence the same is true for the projectivization of $q_\ast(p^\ast (\mathcal{O}_E(nx)) \otimes P^0)$.

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Thanks a lot, Kevin. What I am aiming at is an explicit description of the last bundle. For example, can it be that it is isomorphic to the vector bundle associated to the constant system $H^1_{DR}(E)$? –  Descartes Aug 1 '11 at 19:54
    
I'm not sure what you mean by "the constant system $H^1_{dR}(E)$"... In any case, it's a rank $n$ bundle... whereas $H^1_{dR}(E)$ is rank $2$... –  Kevin H. Lin Aug 1 '11 at 19:59
    
in the case n=2 I mean –  Descartes Aug 1 '11 at 20:32
    
So if our isomorphism $E \cong J^0$ is given by $y \mapsto \mathcal{O}(y-x)$, then I think our bundle over $E$ is such that its fiber over $y \in E$ is given by $H^0(E; \mathcal{O}(nx + y - x)) = H^0(E; \mathcal{O}(y + (n-1)x))$. Not sure how we might proceed from this. If the elliptic curve is given in some standard form like $Y^2 = X^3 + aX + b$, then maybe there is some standard way to write down these sections, and then we can see how these vary with $y$... –  Kevin H. Lin Aug 1 '11 at 20:39
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Let's just do it with divisor classes. The Poincare divisor (class) on $E\times E$ is $((O)\times E)+(E\times(O))$, or equivalently, $p^*((O))+q^*((O))$. You want to add this divisor class to $np^*((O))$ and then push down via $q_*$. So you're computing $$q_*((n+1)p^*((O))+q^*((O)),$$ which seems to me should just equal $(O)$, since doesn't $q_*p^*((O))=0$? (I'm also going to add an "elliptic-curve" tag to your question.)

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Shouldn't the Poincare class also include the diagonal: p^*((o))+q^*((o))-Diagonal ? –  t3suji Aug 1 '11 at 19:03
    
I don't think so; one knows that these sheaves are zero for n<O but locally free of rank n if n>o. The divisor class of the Poincaré bundle is: antidiagonal minus EtimesO minus zerotimesE. –  Descartes Aug 1 '11 at 19:10
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It seems to me that in this case the Picard sheaf coincides with the Fourier-Mukai transform of $\mathcal{O}_E(no)$. Since $h^i(\mathcal{O}_E(no))=0$ for $i >0$, it must be a semistable vector bundle of rank $h^0(\mathcal{O}_E(no))=n$. See [Birkenhake-Lange, complex abelian varieties, Corollary 14.3.10 p. 453] –  Francesco Polizzi Aug 1 '11 at 19:17
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@Joe Silverman: direct image of a line bundle is not the same as image of some divisor representing it. –  t3suji Aug 1 '11 at 19:33
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@t3suji: Oops, knew I was missing something here. I should probably delete my answer, but maybe I'll leave it up for a bit so that others will realize there's no shame in making a mistake!! –  Joe Silverman Aug 1 '11 at 20:25
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