5 added 52 characters in body

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...)

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: 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...)

For a general curve $C$, 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)$H^2(J;\mathbb{Z})$ 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 degree $g-1+1=g$. Hence there is 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: 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 $C$...