Suppose $X$ and $Y$ are algebraic varieties over a field $k$, and $f:X \to Y$ is proper smooth. Then for a vector bundle $E$, and any $i\ge 0$, do we have $R^if_*(E)$ locally free? I know we need the dimension of $H^i(X_y,E_y)$ to be constant, but I don't know whether proper smooth ensures it.
4
-
2$\begingroup$ No, it does not. Let $Y = E$ be an elliptic curve and let $X = E\times E$ with $f\colon E\times E\to E$ the first projection. Consider the line bundle $L = \mathcal{O}_X(\Delta+(E\times 0))$ where $\Delta$ is the diagonal. Its restriction to the fiber above a point $P$ is the line bundle $\mathcal{O}_E(P - 0)$ of degree zero. It has a section if and only if it is trivial, i.e. for $P = 0$. So the dimension is not constant for $i=0$ or $i=1$. $\endgroup$– Piotr AchingerCommented Nov 11 at 7:11
-
2$\begingroup$ Correction: it should of course be $\Delta - (E\times 0)$. This divisor has self-intersection $-2$, and hence $\chi(L) = 1$ by Riemann-Roch. It follows that $R^i f_* (E)$ cannot be locally constant, because if so, they would all be zero, and then $\chi(L)=0$ by the Leray spectral sequence. $\endgroup$– Piotr AchingerCommented Nov 11 at 7:17
-
$\begingroup$ @PiotrAchinger could you post that as an answer? then it can be accepted! $\endgroup$– TimCommented Nov 11 at 9:39
-
1$\begingroup$ However, in the derived category of quasi-coherent sheaves of $Y$, the complex $Rf_*(E)$ will be perfect. Therefore, if $Y$ has an ample family of line bundles (e.g. if it is regular or affine), then $Rf_*(E)$ will be quasi-isomorphic to a bounded complex of vector bundles on $Y$. $\endgroup$– D.-C. CisinskiCommented Nov 11 at 12:05
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Posting comments as an answer.
The answer is "no". For a counterexample, let $Y=E$ be an elliptic curve and let $X=E\times E$ with $f\colon E\times E\to E$ the first projection. Consider the line bundle $L = \mathcal{O}_X(\Delta - (E\times 0))$ where $\Delta\subseteq E\times E = X$ is the diagonal. The restriction of $L$ to the fiber above a point $P\in E$ is the line bundle $\mathcal{O}_E(P-0)$ of degree zero. It has a section if and only if it is trivial, i.e. for $P=0$. So the dimension is not constant for $i=0,1$.
Note that the divisor $\Delta - (E\times 0)$ has self intersection $-2$ as $$ \Delta^2 = \chi_{\rm top}(X) = 0, \quad \Delta\cdot (E\times 0) = 1, \quad (E\times 0)^2 = 0. $$ Therefore, by Riemann-Roch for abelian surfaces $\chi(L) = -c_1(L)^2/2 = 1$.
Suppose the $R^if_* L$ were locally free. They both vanish over the open subset $E\setminus \{0\}$, and hence they are both zero. Thus, $\chi(L) = \chi(Rf_* L) = 0$, contradiction.
As D.-C. Cisinski points out in his comment, $Rf_* L$ is a perfect complex. This does not need that the map is smooth, only that the sheaf is flat over the base. The standard reference which (implicitly) shows this is Mumford's abelian varieties, the section about cohomology and base change.
answered Nov 11 at 12:43
-
1$\begingroup$ For the last paragraph: I think that you need that $f$ is of finite presentation. The flatness could be weakened to be of finite Tor-amplitude. This seems to be a consequence of stacks.math.columbia.edu/tag/0A1H $\endgroup$– Z. MCommented Nov 11 at 15:50
-
$\begingroup$ @Z.M $f$ is a proper morphism between varieties over a field, and hence is of finite presentation. $\endgroup$ Commented Nov 11 at 18:19