Let $\phi:X \to Y$ be a projective/proper, birational morphism between complex algebraic varieties, with connected fibers and $\phi_*\mathcal{O}_X \cong \mathcal{O}_Y$. Suppose further that $X$ is a non-singular. Let $F$ be a subsheaf of a free $\mathcal{O}_X$-module (not just locally-free) such that the quotient is torsion-free. Is the natural morphism $$\phi^*\phi_*F \to F$$ injective or surjective?

  • $\begingroup$ It is certainly not surjective: take for $\phi$ the blowing up of $Y$ at a smooth point, let $E\subset X$ be the exceptional divisor, and let $F=\mathcal{O}_X(E)$. It is easy to see that $\phi_*F\cong \mathcal{O}_Y$, so the homomorphism $\phi^*\phi_*F\rightarrow F$ is not surjective. $\endgroup$
    – abx
    Oct 31 '19 at 5:20
  • 1
    $\begingroup$ @abx But $\mathcal{O}_X(E)$ is not a subsheaf of a direct sum of $\mathcal{O}_X$. $\endgroup$
    – Ron
    Oct 31 '19 at 7:17
  • $\begingroup$ Oops, sorry, I overlooked that hypothesis. $\endgroup$
    – abx
    Oct 31 '19 at 7:26

The answer to both questions (injectivity ans surjectivity) is no without further hypothesis.

Surjectivity : Let $Y$ be a smooth projective variety and let $\phi : X \longrightarrow Y$ be the blow-up of $Y$ along a smooth subvariety. Denote by $L = \mathcal{O}_{X}(-E) \otimes \phi^*\mathcal{O}_{Y}(1)$, where $E$ is the exceptional divisor and $\mathcal{O}_{Y}(1)$ is an ample line bundle on $Y$. The line bundle $L$ is ample so a sufficiently high power of $L$ is globally generated. That is, we have an exact sequence: $$0 \longrightarrow F \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow L^{\otimes m} \longrightarrow 0,$$ for some $m,r \in \mathbb{N}$, when $m$ is big enough. Dualizing this sequence, we get:

$$0 \longrightarrow L^{\otimes -m} \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow F^* \longrightarrow 0,$$

with $F^*$ being torsion-free as the dual of a coherent sheaf. We have: $$\phi_* L^{-m} = \mathcal{O}_{Y}(m) \otimes \phi_*(\mathcal{O}(mE)) = \mathcal{O}_{Y}(m),$$ by the projection formula and the fact that $\phi_* \mathcal{O}_{X}(mE) = \mathcal{O}_{Y}$ for $m \geq 0$. As a consequence, the map $\phi^* \phi_* L^{-m} \longrightarrow L^{-m}$ has cokernel equal to $\phi^*\mathcal{O}_{Y}(-m) \otimes \mathcal{O}_{mE}(mE)$, it is not surjective.

injectivity : Let $Y$ be a smooth projective variety and $Z$ be a smooth codimension $3$ subvariety of $Y$. Let $J_{Z}$ be the ideal sheaf of $Z$ and let $\mathcal{O}_{Y}(1)$ be an ample line bundle on $Y$. For $m$ big enough, the sheaf $\mathcal{J}_{Z}(m)$ is globally generated, so that we have an exact sequence:

$$0 \longrightarrow F \longrightarrow \mathcal{O}_{Y}^{\oplus r} \longrightarrow \mathcal{J}_{Z}(m) \longrightarrow 0,$$ for some $m,r \in \mathbb{N}$, when $m$ is big enough. Let $\phi : X \longrightarrow Y$ be the blow-up along $Z$. Pulling back the above exact sequence along $\phi$, we get a long exact sequence:

$$ 0 \longrightarrow Tor^1(\mathcal{J}_Z(m),\mathcal{O}_{X}) \longrightarrow \phi^*F \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow \phi^*\mathcal{J}_Z(m) \longrightarrow 0.$$ There are no higher $Tor$ because $Z$ has codimension $3$ in $Y$. Denote by $G$ the cokernel of: $$0 \longrightarrow Tor^1(\mathcal{J}_Z(m),\mathcal{O}_{X}) \longrightarrow \phi^*F.$$ We then have an exact sequence:

$$0 \longrightarrow G \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow \phi^* \mathcal{J}_{Z}(m) \longrightarrow 0.$$

Let's push forward this exact sequence by $\phi_*$, we have a long exact sequence:

$$0 \longrightarrow \phi_* G \longrightarrow \mathcal{O}_{Y}^{\oplus r} \longrightarrow \phi_* \phi^* \mathcal{J}_{Z}(m) \longrightarrow R^1 \phi_* G \longrightarrow 0.$$

Let's show that $\phi_* \phi^* \mathcal{J}_{Z}(m) = \mathcal{J}_{Z}(m)$. We have an exact sequence:

$$ 0 \longrightarrow Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) \longrightarrow \phi^* \mathcal{J}_Z \longrightarrow \mathcal{O}_{X}(-E) \longrightarrow 0.$$ Since $Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) = \Omega_{E/Z}(-E)$, (where $\Omega_{E/Z}$ is the bundle of relative differentials for the map $E \longrightarrow Z$), The relative version of Bott's vanishing Theorem implies that: $$\phi_* Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) = R^1 \phi_* Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) = 0.$$ We deduce that $\phi_* \phi^* \mathcal{J}_Z(m) = \mathcal{J}_Z(m)$.

But the map $\mathcal{O}_{Y}^{\oplus r} \longrightarrow \mathcal{J}_{Z}(m)$ is the one we started with and it is surjective. We thus get $R^1 \phi_* G = 0$ and $\phi_* G = F$. Finally, we get that the map: $$ \phi^* \phi_* G \longrightarrow G$$ is not injective as its kernal is: $$Tor^1(J_Z(m),O_{X}) = Tor^{2}(\mathcal{O}_{Z}(m),\mathcal{O}_{X}) = \bigwedge^2\Omega_{E/Z}(-2E) \otimes \phi^* \mathcal{O}_{Y}(m).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.