It is known that a flasque sheaf on a topological space has trivial cohomology. Suppose that we are in a relative situation of a smooth fibration $\pi: X \to S$ and $F$ is a sheaf on $X$. Is there are weaker condition on $F$ which implies that the higher direct images to $S$ are trivial (but not necessarily the cohomology groups of $F$ on $X$).
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asked Feb 20, 2014 at 22:01
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$\begingroup$ One possible answer is in the title of your question: since the stalks of $R^i\pi_*$ are $H^i(\text{fiber})$, it would suffice to require that the restrictions to fibers are flasque. $\endgroup$– Alex DegtyarevCommented Feb 20, 2014 at 22:30
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$\begingroup$ @AlexDegtyarev in algebraic geometry, the stalks of $R^i \pi_*$ are not always $H^i(fiber)$. For example, take $i=0$ and $X =$ blowup of a surface $S$ at a point and $F=\mathcal{O}(-1)$. Then at that point, the fiber is $\mathbb{P}^1$ and so $F|_{fiber}$ has no sections, but $\pi_* \mathcal{O}(-1)$ is the ideal sheaf of the point. $\endgroup$– Piotr AchingerCommented Feb 20, 2014 at 22:56
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$\begingroup$ @PiotrAchinger The restriction in your example has lots of sections; it's just not a coherent sheaf. By restriction I mean what you get before tensoring by $\mathcal{O}$. (Flasqueness isn't quite an algebraic-geometric notion.) $\endgroup$– Alex DegtyarevCommented Feb 20, 2014 at 23:08
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$\begingroup$ @AlexDegtyarev Of course! I confused the two $f^*$. $\endgroup$– Piotr AchingerCommented Feb 20, 2014 at 23:25
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$\begingroup$ Happens all the time :) BTW, your definition looks pretty much like flasqueness of the restriction (in the topological sense :) $\endgroup$– Alex DegtyarevCommented Feb 20, 2014 at 23:48
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1 Answer
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Modulo quasi-compactness, the following seems to work:
Definition. Let $f:X\to S$ be a map of topological spaces and let $F$ be a sheaf on $X$. We call $F$ flasque over $S$ if for any opens $U\subseteq V\subseteq X$, any $x\in U$, and any $\sigma\in\Gamma(U, F)$ there exists an open $W$ containing $f(x)$ such that $\sigma|_{U\cap f^{-1}(W)}$ lies in the image of $$ \Gamma(V\cap f^{-1}(W), F) \to \Gamma(U\cap f^{-1}(W), F). $$
Proposition. Suppose that $X$ is quasi-compact and that $F$ is flasque over $S$. Then $R^i f_* F = 0$ for $i>0$.
This will follow from the fact that (1) injective sheaves are flasque, hence flasque over $S$, (2) quotient of a flasque sheaf by a subsheaf flasque over $S$ is flasque over $S$, (3) the following lemma:
Lemma. With the hypotheses of the Proposition, for any short exact sequence $$ 0\to F\to F'\to F''\to 0$$ the map $f_* F'\to f_* F''$ is surjective.
Proof. Let $s\in S$ and let $\sigma$ be a section of $f_* F''$ on an open $W_0$ containing $s$. We need to find an open $W\subseteq W_0$ containing $s$ such that $\sigma|_W$ comes from a section of $f_* F'$ on $W$.
By definition, $\sigma\in \Gamma(f^{-1}(W_0), F'')$. We can cover $f^{-1}(W_0)$ by finitely many opens $U_i$ and find sections $\tilde\sigma_i$ of $F'$ over $U_i$ whose image in $F(U_i)$ is $\sigma|_{U_i}$. Let $\tau_{ij} = \tilde\sigma_i-\tilde\sigma_j \in \Gamma(U_i\cap U_j, F)$.
By flasqueness of $F$ over $S$, can find $s\in W_{ij}\subseteq W_0$ and $\tau'_{ij} \in \Gamma(f^{-1}(W_{ij}, F)$ extending $\tau_{ij}|_{U_{ij}\cap f^{-1}(W_{ij}, F)}$. Since there were finitely many $i$, $j$, we can intersect them, obtaining the desired $W$.
answered Feb 20, 2014 at 22:51
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$\begingroup$ Thank you, Piotr, this does look like the weakest condition one can put, and it does remind a lot what Alex suggested with the restriction (actually, the restriction he uses is often denoted by $f^{-1}$ and $f^*$ is reserved for the tensor product of functions on the closed set, when the sheaves are modules over functions) $\endgroup$ Commented Feb 21, 2014 at 8:17