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Mar 14, 2016 at 15:42 vote accept Lisa S.
Mar 14, 2016 at 5:22 comment added nfdc23 The motivation for Mumford's approach is a property that doesn't refer to divisors (and so better-suited to the relative situation): inspired by the dictionary between quadratic forms and symmetric bilinear forms, Mumford's notion of polarization $f:A \rightarrow A^{\vee}$ is a map symmetric with respect to double duality such that the pullback of the Poincare bundle $P_A$ along $(1,f):A \rightarrow A \times A^{\vee}$ is ample (akin to positive-definte quadratic form). For a divisor $D$, $t_a^{\ast}(O(D))=O(t_{-a}(D))$. So the sign is caused by the language of divisors.
Mar 14, 2016 at 3:33 history answered abx CC BY-SA 3.0