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Aug 27, 2013 at 19:25 answer added Ravi Vakil timeline score: 4
Jan 13, 2013 at 6:06 comment added anon @Piotr: What you suggest fails for $X = A \times A^t$, $Y = A^t$, $f$ the natural projection, and $F$ being the Poincare line bundle for an abelian variety $A$ of dimension $g$. Indeed, $R^i f_*(F) = 0$ for $i < g$, but the same is not true for the fibre over $0$.
Jan 9, 2013 at 21:35 answer added Angelo timeline score: 5
Dec 12, 2012 at 20:28 answer added jlk timeline score: 2
Dec 12, 2012 at 0:50 comment added Piotr Achinger You can still hope that the following holds: suppose $f:X\to Y$ and $\mathcal{F}$ are like in your first sentence. Then if $R^i f_*(\mathcal{F})$ is locally free on $Y$, then $R^i f_*(\mathcal{F})$ commutes with base change. This I think is true and follows from the discussion of cohomology and base change in Mumford's "Abelian varieties".
Dec 11, 2012 at 21:09 comment added Damian Rössler See mathoverflow.net/questions/114238/…
Dec 11, 2012 at 15:12 history asked HNuer CC BY-SA 3.0