Let $\pi:\mathcal{X} \to B$ be a flat, projective surjective morphism over $\mathbb{C}$. Assume that $B$ is a smooth quasi-projective curve. Let $\mathcal{F}$ be a coherent sheaf on $\mathcal{X}$, flat over $B$. Assume further that $\pi_*\mathcal{F}$ is locally free $\mathcal{O}_B$-module. Let $b \in B$ be a closed point and $k(b)$ be its residue field. Denote by $f$ the closed immersion of the fiber $\pi^{-1}(b)$ into $\mathcal{X}$. Is it possible that $\pi_*\mathcal{F} \otimes k(b) \cong H^0(f^*\mathcal{F})$? If we replace $\mathcal{F}$ by $\mathcal{O}_{\mathcal{X}}$, can this hold true?


1 Answer 1


I don't think this is always true. Consider the following situation:

Let $B=\mathbb{A}^1$ with coordinate $t$ and let $\mathcal{X}$ be the blowup of $\mathbb{A}^2$ at $(0, 0)$, $\pi:\mathcal{X}\to\mathbb{A}^2\to B$. Note that $\pi$ is flat. If $\mathcal{F}$ is a coherent sheaf on $\mathcal{X}$ flat over $B$ (for example, locally free), we have a short exact sequence $$ 0\to \mathcal{F} \stackrel{t}{\to} \mathcal{F} \to \mathcal{F}|_{X_0} \to 0, $$ where $X_0 = \pi^{-1}(0)$, hence a long exact sequence $$ 0 \to H^0(\mathcal{X}, \mathcal{F}) \stackrel{t}\to H^0(\mathcal{X}, \mathcal{F}) \stackrel{r}\to H^0(X_0, \mathcal{F}|_{X_0}) \stackrel{\delta}\to H^1(\mathcal{X}, \mathcal{F}) \stackrel{t}\to H^1(\mathcal{X}, \mathcal{F}) \to H^1(X_0, \mathcal{F}|_{X_0}) \to 0. $$ Your question boils down to asking whether $r$ is surjective if $H^0(\mathcal{X}, \mathcal{F})$ is torsion free, and we see from the above that the obstruction lies in $H^1$.

Consider now $\mathcal{F} = \mathcal{O}_{\mathcal{X}}(2E)$, where $E$ is the exceptional divisor. Then $\mathcal{F}|_E = \mathcal{O}_E(-2)$ (as $E^2 = -1$), hence $H^1(E, \mathcal{F}|_E)$ is nonzero. Thereofore also $H^1(X_0, \mathcal{F}|_{X_0})$ is nonzero (because $X_0$ is a union of $E$ with an affine line). Hence by the long exact sequence above $H^1(\mathcal{X}, \mathcal{F})$ is nonzero and torsion (by flat base change to $\mathbb{A}^1 \setminus \{0\}$), hence $t:H^1\to H^1$ cannot be injective, so $\delta$ is nonzero, hence $r$ is not surjective.

On the other hand, what you want is true if you ask moreover that $R^i \pi_* \mathcal{F} = 0$ for $i>0$. Just pick a system of parameters $t_1, \ldots, t_n$ at $b$ and apply the above short exact sequence inductively.

In particular, this holds if $\mathcal{X}$ is smooth and $\pi$ is birational.

  • $\begingroup$ Why is $\pi$ flat? $\endgroup$
    – user45397
    May 3, 2014 at 20:09
  • $\begingroup$ Good question. That's because $\mathcal{X}$ is integral and the base $B$ is a smooth curve (so "flat" means "torsion free"). See e.g. III 9.7 in Hartshorne's "Algebraic Geometry". $\endgroup$ May 3, 2014 at 20:41
  • $\begingroup$ @Achinger: Thanks for the answer. Could you also tell me why $\pi$ is projective? I understand the morphism from $\mathcal{X}$ to $\mathbb{A}^2$ is projective. $\endgroup$
    – user45397
    May 4, 2014 at 13:46
  • $\begingroup$ That's right, $\pi$ is of course not projective, but we can consider the blow-up of $B\times \mathbb{P}^1$ instead. $\endgroup$ May 4, 2014 at 18:27

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