# Pulling back a line bundle on the Jacobian to a spin bundle on the curve

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.

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What exactly do you mean "lift $\kappa \oplus \kappa^{-1}$ to $J$"? – Kevin H. Lin Jul 8 '11 at 8:26
Sorry, I just meant I'd like to find a bundle $V$ with $\alpha_c^*(V) \cong \kappa \oplus \kappa^{-1}$... – aaron gerding Jul 8 '11 at 22:35
Ok. Typically I think that when one says "lift $E$ to $X$", one means that there is a map $f: X \to Y$ with $E$ living on $Y$, and one wishes to find some $F$ living on $X$ which pushes down via $f$ to $E$, or such that $E$ pulls back to $F$ via $f$. – Kevin H. Lin Jul 8 '11 at 23:03
btw, I asked a somewhat related question on MO recently... mathoverflow.net/questions/69146/… – Kevin H. Lin Jul 8 '11 at 23:43
Yes I had actually found your question and was trying to relate it to mine by pulling back again via $x \mapsto (x,x)$ or maybe $x \mapsto (x,c)$ but couldn't work it out. Saying that $V$ would "lift" $\kappa \oplus \kappa^{-1}$ was motivated I guess by the thought that a diagram chase might get something like what I'm looking for from the $g-1$ Poincare bundle, which would live on $C \times J^{g-1})$ (maybe) and restrict to some version of $\kappa$ on the Abel image in the fibers over $C$. – aaron gerding Jul 9 '11 at 1:09

## 1 Answer

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...)

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I think this is exactly the point I was missing. Thanks a lot! I guess seeing $\kappa \oplus \kappa^{-1}$ as a pull-back is more subtle, if even possible at all... – aaron gerding Jul 8 '11 at 22:41