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What can one say about the Chern class $c_1(\mathcal{L})$ of a line bundle $\mathcal{L}$ on an Abelian variety $A$ inducing a (principal) polarisation $A \to A^\vee$?

Why am I asking this? My situation is as follows: Let $A/k$ be an Abelian variety with $k$-endomorphism $f: A \to A$, $\mathcal{L} \in \mathrm{Pic}(A)$ a line bundle inducing a (principal) polarisation $A \to A^\vee$. Now for simplicity let us assume $A$ to be a surface. I want the expression $\mathrm{deg}(((f,c_\mathcal{L})^*\mathcal{P}_A \cup c_1(\mathcal{L})) \in \mathbf{Z}$, where $\mathcal{P}_A$ denotes the Poincaré bundle on $A \times_k A^\vee$ and $c_\mathcal{L}$ the polarisation induced by $\mathcal{L}$, to be independent of the choice of the ample line bundle $\mathcal{L}$.

Edit: In the form stated above, it is wrong: $c_{\mathcal{L} \otimes \mathcal{M}} = c_{\mathcal{L}} + c_{\mathcal{M}}$ and $c_1({\mathcal{L} \otimes \mathcal{M}}) = c_1({\mathcal{L}}) + c_1({\mathcal{M}})$, so if you replace $\mathcal{L}$ by $\mathcal{L}^{\otimes n}$, the term is multiplied by $n^2$.

Perhaps one has to replace $\mathrm{deg}(((f,c_\mathcal{L})^*\mathcal{P}_A \cup c_1(\mathcal{L}))$ by $\mathrm{deg}(((f,c_\mathcal{L})^*\mathcal{P}_A \cup c_1(\mathcal{L})^{-1})$, but this still does not work with all tensor products $\mathcal{L} \otimes \mathcal{M}$.

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    $\begingroup$ Dear Timo -- Could you say more precisely what you are looking for? $\endgroup$
    – Jason Starr
    Commented Jun 26, 2012 at 12:47
  • $\begingroup$ I will extend my question in a minute. $\endgroup$
    – user19475
    Commented Jun 26, 2012 at 13:07
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    $\begingroup$ @Francesco: What if there is more than one principal polarization? For instance, a product of two isomorphic elliptic curves will always have many principal polarizations. $\endgroup$
    – Will Sawin
    Commented Jun 26, 2012 at 17:48
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    $\begingroup$ If $L$ is symmetric and ample then $(f,c_{\cal L})^*{\cal P}_A\simeq L^{\otimes 2}$ so ${\rm deg}((f,c_{\cal L})^*{\cal P}_A\otimes L^{\otimes -2})$ is independent of the choice of $\cal L$, provided $\cal L$ is symmetric. $\endgroup$
    – Damian Rössler
    Commented Jun 28, 2012 at 6:11
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    $\begingroup$ Well, up to a translation (which does not change the numerical class) one can assume that $\mathcal{L}$ is symmetric $\endgroup$
    – Francesco Polizzi
    Commented Jun 28, 2012 at 10:06

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