Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 18 / 3

The bundle is on $A$, so the printed, undefined target $\mathrm{Pic}(X)$ must be read as the Picard group $\mathrm{Pic}(A)$. Write this Picard group additively and set $q(h)=[h^{\ast }L]$. The zero morphism pulls back $L$ to a constant one-dimensional vector space tensored with ${\mathcal{O}}_A$, so $q(0)=0$. The Theorem of the Cube, pulled back along $(f,g,h):A\to B^3$, givesThis is the bilinear cross-effect of a line bundle on an abelian variety: set $d(f,g)=q(f+g)-q(f)-q(g)$. Substitution into this identity givesThe definition gives $d(f,g)=d(g,f)$, so additivity holds in both arguments. Additivity includes negative multiples and the zero morphism. Hence $D_L$ is a symmetric bilinear map with values in $\mathrm{Pic}(A)$. The equalities refer to isomorphism classes of line bundles; without chosen rigidifications they are not asserted to be specified canonical isomorphisms of bundles.

Let $m:B\times B\to B$ be addition, and let $p_1,p_2$ be the projections. Consider the line bundleIts restriction to $B\times \{y\}$ has Picard group class $[T_y^{\ast }L\otimes L^{-1}]={\varphi }_L(y)$; the factor $L_y^{-1}$ is a constant line and has trivial class. Its restriction to $\{0\}\times B$ is also trivial. Choose a trivialization of the fiber $L_0$ and use it to normalize along the zero sections. The defining universal property of the normalized Poincaré line bundle then identifiesMore explicitly, both sides have the same restrictions to all fibers of the second projection, and both are normalized on $\{0\}\times B$; the Seesaw theorem makes their quotient trivial. Here ${\varphi }_L:B\to \hat{B}$ is the homomorphism associated to a line bundle on an abelian variety, and $\hat{B}$ is the dual abelian variety. Pull back by $(f,g)$ to obtain the required identityThe choice of fiber trivialization is immaterial to this identity in the Picard group.

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 group homomorphism $h:A\to B$, the dual morphism $\hat{h}:\hat{B}\to \hat{A}$ is pullback of algebraically trivial line bundles. Translations satisfy $h\circ T_x=T_{h(x)}\circ h$, and thereforeAlso ${\varphi }_{L\otimes M}={\varphi }_L+{\varphi }_M$ and ${\varphi }_{L^{-1}}=-{\varphi }_L$. Dualizing is additive: $\widehat{f+g}=\hat{f}+\hat{g}$, since algebraically trivial line bundles obey the Theorem of the square. Apply these identities to the three factors defining $D_L$. The two diagonal terms cancel in the expansionHence $\boxed{{\varphi }_{D_L(f,g)}=\hat{f}\circ {\varphi }_L\circ g+\hat{g}\circ {\varphi }_L\circ f}$.