I'd like to have an expression for the (or some) line bundle on the Jacobian $J$ of a smooth complex projective curve $C$ with genus $g >1$ which pulls back to a chosen spin bundle (theta characteristic) $\kappa$ on $C$ via the Abel-Jacobi map $\alpha\_c$ based at $c \in C$.


I know (from looking at Birkenhake and Lange 11.3) that by Riemann's Theorem and the Seesaw principle 


$\alpha\_c^\*\mathcal{O}\_J(\Theta \_{\kappa}) = \kappa \otimes \mathcal{O}\_C(c)$, 


where $\Theta \_{\kappa}$ is the (symmetric) theta divisor that $\kappa$ determines by $\alpha \_{\kappa}^\* \Theta \_{\kappa} = W \_{g-1} \subset Pic^{g-1}$. So what I guess I'm looking for is a way to move the $c$ dependence over to the left side of this equation. For what it's worth, my hope is to be able to lift the vector bundle $\kappa \oplus \kappa^{-1}$ to $J$, which I'm assuming is just a matter of lifting $\kappa$.


This doesn't seem difficult but I don't have a lot of experience with Abelian varieties and I'm not sure how to get started.



Pulling back a line bundle on the Jacobian to a spin bundle on the curve
For a general curve $C$ of genus $g$, it is a fact that the Neron-Severi group of the Jacobian $J$ of $C$ is generated by the class $\theta$ corresponding to the divisor $\Theta$. (I am not very strong in algebraic geometry, so I guess that I would rather prefer to work with the probably equivalent statement: The subgroup of $H^2(J;\mathbb{Z})$ generated by first Chern classes of algebraic line bundles on $J$ is generated by $\theta$.) I don't know the proof of this, but the reference seems to be Arbarello-Cornalba-Griffiths-Harris, volume II...


So, by the formula you cite, it follows that for any algebraic line bundle $L$ on $J$, the degree of $\alpha\_c^\ast L$ must be an integer multiple of $g-1+1=g$. Hence there can be no $L$ such that $\alpha\_c^\ast L = \kappa$, since $\kappa$ is of degree $g-1$. 


Err, hmm, well, actually, if $g=1$ then it's possible, since then $\kappa$ is of degree zero: for example, put $\kappa = \mathcal{O}\_C$ and $L = \mathcal{O}\_J$, and then we do have $\alpha\_c^\ast L = \kappa$. But that's kind of trivial, anyways.


At least this all seems to work for a general curve $C$... I have no idea about a curve for which the above statement about the Neron-Severi group doesn't hold.


(As for $\kappa \oplus \kappa^{-1}$, this argument doesn't rule out the possibility of an $E$ such that $\alpha\_c^\ast E = \kappa \oplus \kappa^{-1}$. But $\kappa \oplus \kappa^{-1}$ is degree zero, so such an $E$ would have to be degree zero...)


