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Everything below is defined over $\mathbb C$.

Let $T$ be a smooth affine variety, $\pi : \mathscr X \to T$ be a smooth family of smooth projective varieties, and $\mathcal F$ be a locally free coherent sheaf on $\mathscr X$. As usual write $\mathscr X_t$ for the fiber of $\pi$ over $t$ and $\mathcal F_t$ for the restriction of $\mathcal F$ at it.

Suppose $h^0(\mathscr X_t, \mathcal F_t) >0 $ for every $t \in T$ and fix $t_0 \in T$. Can I conclude the existence of a non-zero section of $\mathcal F_{t_0}$ which extends to a section of $\mathcal F$ on a neighborhood of $\mathscr X_{t_0}$?

Clarification: By a smooth family of smooth projective varieties, I mean that $\mathscr X$ is smooth, $\pi$ is a submersion, and the fibers are smooth and connected. In particular, the family is topologically trivial. Also, I do not need a section at an affine neighborhood of $\mathscr X_{t_0}$: an analytic neighborhood would suffice.

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up vote 7 down vote accepted

No, this is not true. Here is the simplest example. Take $T = \mathbb A^2$, $X = \mathbb A^2 \times \mathbb P^1$. The morphims $\pi \colon X \to T$ is the projection. We will denote by $\cal O(d)$ the pullback of $\cal O_{\mathbb P^1}(d)$ to $X$. Let $x$ and $y$ be the coordinates on $\mathbb A^2$, and $u$ and $v$ the homogeneous coordinates on $\mathbb P^1$. Set $R = \mathbb C[x,y]$.

Let $F$ be the generic extension of ${\cal O}(1)$ by ${\cal O}(-2)$. We have an isomorphism $Ext^1({\cal O}(1), {\cal O}(-2)) \simeq R^2$; if $a$, $b$ denotes a basis of $Ext^1({\cal O}(1), {\cal O}(-2))$ as an $R$-module, then $F$ corresponds to the element $xa+yb$. By restricting to the fiber over $t$, we get the split exension for $t = (0,0)$, and a non-split extension for $t \neq (0,0)$. It is easy to see that $h^0(F_t)$ is 2 for $t = (0,0)$ and 1 otherwise. We get an exact sequence $$ 0 \longrightarrow H^0(F) \longrightarrow H^0({\cal O}(1)) \longrightarrow H^1({\cal O}(-2)) $$ that is $$ 0 \longrightarrow H^0(F) \longrightarrow R^2 \longrightarrow R; $$ the image of the boundary map $R^2 \longrightarrow R$ is the maximal ideal $(x,y)$. Hence, the composite $H^0(F) \to H^0({\cal O}(1)) \to H^0({\cal O}(1)_{t_0})$ is 0. Since $H^0(F_{t_0})$ maps isomorphically onto $H^0({\cal O}(1)_{t_0})$, we conclude that the restriction map $H^0(F) \to H^0(F_{t_0})$ is 0. This gives the counterexample.

This can be usefully viewed using the result of Grothendieck saying that, if $Y = \mathop{\rm Spec}R$, there is a finite $R$-module $Q$ such that for every finite $R$-module $M$ there is a canonical isomorphism of $R$-modules $H^0(X, F\otimes_R M) \simeq Hom_R(Q,M)$, which is functorial in $M$. In the counterexample above, $Q$ is the maximal ideal $(x,y)$. When $Y$ is 1-dimensional, we can use the structure theorem for modules over Dedekind rings, and conclude that $Q$ is the direct sum of a projective module and a torsion module. If $h^0(F_t) \neq 0$ for all $t$, then the projective summand must be non-zero, so the statement does hold.

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Thanks for the answer. –  jvp Oct 7 '10 at 12:25
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