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Feb 5, 2013 at 9:48 vote accept Martin Brandenburg
Apr 30, 2010 at 8:40 comment added naf @Thanos D. Papaïoannou: The first example of such a variety given by Igusa is a bielliptic surface in characteristic 2: Let E be an ordinary elliptic curve and X the quotient of E x E by the involution given by $(x,y) \mapsto (-x, y+a)$, where a is a point of order 2.
Apr 30, 2010 at 3:24 comment added Angelo Metha and Srinivas wrote a paper on this topic, "Varieties in positive characteristic with trivial tangent bundle", <archive.numdam.org/ARCHIVE/CM/CM_1987__64_2/CM_1987__64_2_191_0/…>.
Apr 29, 2010 at 22:23 comment added Thanos D. Papaïoannou @unknown: Could you please give an example of failure in positive characteristic, i.e. an example of a connected, smooth, proper scheme $X/k$ with trivial tangent bundle and a $k$-point, where $k$ is a field of positive characteristic, such that $X/k$ doesn't have a group law?
Apr 29, 2010 at 11:09 comment added naf If the generic points of S have residue characteristic 0 then perhaps the condition on Aut_S X is unnecessary since any connected smooth proper variety over a field of characteristic 0 with trivial tangent bundle (and a rational point) is an abelian variety. This is not true in positive characteristics, but does suggest that some weaker condition might suffice.
Apr 28, 2010 at 21:05 comment added Angelo By a transitive action I mean that the morphism $\underline{\rm Aut}_S X \to X$ coming from the section is scheme-theoretically surjective. Since the fibers are connected, this implies that the connected component of the identity must dominate each fiber, and this is enough to conclude.
Apr 28, 2010 at 20:15 comment added BCnrd Nice! By Theorem 6.14 in GIT (for which the projective hypothesis relaxes to properness by using Artin's results on Hilbert and Hom functors via algebraic spaces instead of schemes), the abelian scheme structure exists (uniquely) if it does so on geometric fibers. Thus, the proposed criterion above reduces to case when $S$ is spectrum of alg. closed field, in which case it follows by the suggested argument (using rational curves and Chevalley's theorem) provided we modify the hypothesis on Aut's to involve the identity component of the Aut-scheme on fibers (still geometric criterion).
Apr 28, 2010 at 19:50 comment added Mariano Suárez-Álvarez Triviality of $\Omega_{X/S}$ disposes of $\mathbb P^1$, at least!
Apr 28, 2010 at 19:29 history answered Angelo CC BY-SA 2.5