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Jun 10, 2017 at 19:55 comment added user21574 See 2.3. Proposition, and 3.9. Corollary math1.unice.fr/~hoering/articles/a8-pos-geom.pdf
Jun 10, 2017 at 19:32 comment added user21574 In general Fujita-Kawamata semi-positivity say that Let $π : X → Y$ be an algebraic fiber space which satisfies Unipotent Reduction Condition. Then $π_∗ω_{X/Y}$ is a locally free sheaf and semi-positive. For example , If $π : X → Y$ is semi-stable family,then the Unipotent Reduction Condition holds automatically
Nov 6, 2015 at 18:06 comment added Tong @nfdc23 Thank you for your answer. In my case, there could be some multiple fibers.
Nov 6, 2015 at 18:04 comment added Tong @KarlSchwede Thank you very much for your reference. I will read that.
Nov 6, 2015 at 15:09 comment added nfdc23 Correction: for my comment above I should have also assumed the geometric fibers to be reduced (or more generally that $O_{X_y}(X_y) = k(y)$ for each $y \in Y$); that was implicit in knowing the fibral trace is an isomorphism.
Nov 6, 2015 at 15:00 comment added Karl Schwede Some statements along these lines are in section 7 of Kollár-Kovács (click here) but they are probably not quite what you want.
Nov 6, 2015 at 14:55 comment added nfdc23 For $n=1$ (so ${\rm{R}}^1f_{\ast}$ is right-exact on coherent sheaves), it suffices that $f$ has geometrically connected fibers. Indeed, fibral trace ${\rm{H}}^1(X_y \omega_{X_y/y})\rightarrow k(y)$ is then an isomorphism, so ${\rm{R}}^1f_{\ast}(\omega_{X/Y})$ is locally monogenic by Nakayama. Thus, the base-change compatible trace to $\mathscr{O}_Y$ is a fiberwise isomorphism, so an isomorphism. The formation of ${\rm{R}}^1f_{\ast}(\omega_{X/Y})$ thus commutes with base change to fibers, so $f_{\ast}\omega_{X/Y}$ is a vector bundle compatible with any base change.
Nov 6, 2015 at 14:45 history edited Tong CC BY-SA 3.0
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Nov 6, 2015 at 14:39 history edited Tong CC BY-SA 3.0
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Nov 6, 2015 at 14:32 history asked Tong CC BY-SA 3.0