Homogeneous bundles and Mumford surjectivity

Work first over an algebraic closure. The Poincare family on $A\times B$, where $B={\mathrm{Pic}}^0=A^{\vee }$, gives through [F1] a morphism $A\times B\to B$; it is zero on $A\times \{0\}$ and on $\{0\}\times B$. The proper-factor rigidity lemma [F2] makes it zero everywhere, as an identity of morphisms. Thus every family classified by $B$ has zero Mumford map, even on nonreduced tests. The normalized square defines the map for any bundle as in [F1]. For a bundle over the ground field it is a homomorphism: the square identity establishes addition on geometric points, and the two resulting morphisms from the reduced $A\times A$ to separated $B$ therefore agree. For a test family, locally its classifying map lands in a component of the full Picard scheme. Each component is a translate of $B$, and its universal family is a fixed bundle tensored with the Poincare family. Tensor product adds normalized-square maps, so the preceding vanishing makes this family map the base change of the fixed bundle's homomorphism. Consequently the construction gives homomorphisms on all tests and commutes with base change.

Let $M$ be a ground-field bundle with ${\phi }_M=0$; by the normalized-square universal property its square family is trivial, giving $m^{\ast }M\cong p_1^{\ast }M\otimes p_2^{\ast }M$ after trivializing the constant identity fibre. Pulling back along $(\mathrm{id},-1)$ gives $[-1{]}^{\ast }M\cong M^{-1}$. If $M$ has a nonzero section, inversion gives a nonzero section of $M^{-1}$; their product is a nonzero scalar by integrality and [F3], so $M$ is trivial. A nontrivial $M$ therefore has $H^0(M)=0$. If $i>0$ is the least degree with $H^i(M)\mathrm{\ne }0$, multiplication pullback followed by restriction along $(\mathrm{id},0)$ is the identity on $H^i(M)$, but Kunneth identifies the intermediate group with ${⨁}_{a+b=i}{H}^a(M)\otimes H^b(M)=0$. This contradiction proves vanishing in every degree. Flat field base change gives the same vanishing over the original field.

Over an algebraic closure suppose $M$ has zero Mumford map but is not $t_x^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-1}$ for any $x$, and put $Q=\mathrm{\Lambda }(\mathcal{L})\otimes p_2^{\ast }{M}^{-1}$. On a $p_1$-fibre it is the nontrivial bundle $t_x^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-1}\otimes M^{-1}$, whose Mumford map is zero by step 1.1. Step 2.1 and the universal cohomology complex give $R{p}_{1,\ast }Q=0$, hence $H^{\ast }(Q)=0$. On a $p_2$-fibre its class is $t_y^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-1}$, since the other factors are constant lines; it has zero cohomology away from the finite kernel $K(\mathcal{L})$ supplied by [F3]. All $R^i{p}_{2,\ast }Q$ thus have finite support. They have no higher cohomology, so the second Leray sequence identifies their global sections with $H^i(Q)=0$ and makes every direct image zero. Derived base change then makes every fibre cohomology zero, contradicting the trivial bundle on the fibre at $y=0$. Therefore every such $M$ is a Mumford translate for the fixed ample $\mathcal{L}$.

A zero-Mumford test family has, by step 3.1 on each geometric fibre, all its fibre classes in the open identity component $B$ of the represented Picard scheme. Its classifying map therefore factors through $B$, including its nilpotent structure; no reduced-test argument is used. Conversely, step 1.1 makes every $B$-classified family have zero Mumford morphism. Thus the kernel sheaf is exactly ${\mathrm{Pic}}^0$ on all tests over an algebraic closure. Rigidified bundle descent and the field-compatible dual of [F1] descend this equality to $k$. This proves the asserted criterion, vanishing and geometric surjectivity.