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For group homomorphisms $f,g:A\to B$, this Picard group class is symmetric and additive in both arguments. The Theorem of the Cube says precisely that the third additive difference of $h\mapsto [h^{\ast }L]$ vanishes, proving additivity of the cross-effect. The Poincaré line bundle expresses it as $(f,{\varphi }_{L}g{)}^{\ast }{P}_B$, so it depends only on the Néron-Severi group class of $L$. Its associated homomorphism is $\hat{f}{\varphi }_{L}g+\hat{g}{\varphi }_{L}f$.

For $M\in {\mathrm{Pic}}^0(B)$, translations preserve its line bundle class, so ${\varphi }_M=0$. Equivalently, the kernel of $L\mapsto {\varphi }_L$ is the Identity component of the Picard group. Applying the preceding Poincaré line bundle formula to $M$ yieldssince the normalized Poincaré line bundle is trivial on $B\times \{0\}$. Pullback and tensor products of sheaves also give $D_{L\otimes M}=D_L\otimes D_M$. Thus $D_L$ depends only on $[L]\in \boxed{\mathrm{NS}(B)}$, the Néron-Severi group.

For a line bundle $L$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square givesso ${\varphi }_L$ is a homomorphism. Pullback distributes over the tensor product of sheaves, and thereforeIterating the homomorphism law in $x$ gives

Suppose $L^{\otimes n}\in {\mathrm{Pic}}^0(X)$. Then ${\varphi }_L(nx)$ is trivial for every $x$. The multiplication-by-n morphism on an abelian variety is surjective, so ${\varphi }_L$ is trivial and $L\in {\mathrm{Pic}}^0(X)$. Thus the Néron-Severi groupis torsion-free.

Finally, put $M={\varphi }_L(x)$. For every $y$,which is trivial by the Theorem of the square. Hence