# Locally free sheaves on affine bundles

Let $$f:Y \to X$$ be an affine bundle (for example, the total space of a vector bundle) and $$s: X \to Y$$ be a global section. Given any locally free sheaf $$E$$ on $$Y$$, is $$E$$ isomorphic to $$f^*s^*E$$? Roughly, I would think this is true as there should be a morphism from $$E$$ to $$f^*s^*E$$ (locally given by the canonical map from $$M$$ to $$M \otimes_{A[X_1,....,X_n]} A \otimes_A A[X_1,...,X_n]$$ given by $$m \mapsto m \otimes 1$$, where the first morphism $$A[X_1,...,X_n] \to A$$ sends $$X_i$$ to 0 and the second morphism $$A \to A[X_1,...,X_n]$$ is the canonical inclusion). Furthermore, applying $$- \otimes k(y)$$ to this morphism will keep it injective, which will imply that the cokernel of $$E \to f^*s^*E$$ is flat. By rank argument this would imply the two locally free sheaves are isomorphic. Is this correct? Probably, there could be some gluing issue.

NB. If necessary, assume that $$X$$ is non-singular and the underlying field is $$\mathbb{C}$$.

In general this is not true. Consider $$X=\mathbb{P}^1$$ and $$Y=X\times \mathbb{A}^1$$, the trivial bundle. Since $$\operatorname{Ext}^1(O_X(1),O_X(-1))=k$$ (say an algebraically closed field), we can identify elements of $$k$$ with closed points of the affine line. This gives a vector bundle $$E$$ on $$Y$$ which fits into an exact sequence $$0\to O_X(-1) \to E\to O_X(1)\to 0$$. Restricted to $$X\times \{0\}$$, $$E=O_X(-1)\oplus O_X(1)$$, but for $$t\neq 0$$, restricted to $$X\times \{t\}$$, it is just $$O_X\oplus O_X$$. Thus, it cannot be of the form you suggest.

• Do you mean that the first and the last terms of the short exact sequence is the pull-back of $\mathcal{O}_X(-1)$ and $\mathcal{O}_X(1)$ to $Y$ (the middle term of the short exact sequence is on $Y$ but the 2 other terms are on $X$)? Jul 3, 2021 at 16:34
• Thanks for the nice example. Can we at least say that there is a non-trivial morphism from $E$ to $f^*s^*E$? Jul 3, 2021 at 16:43

Question: "Furthermore, applying $$−⊗k(y)$$ to this morphism will keep it injective, which will imply that the cokernel of $$E→f^∗s^∗E$$ is flat. By rank argument this would imply the two locally free sheaves are isomorphic. Is this correct?"

Answer: Assume $$\pi: Y:=Spec(B) \rightarrow X:=Spec(A)$$ has a section $$s:X \rightarrow Y$$ with $$\pi \circ s =Id_X$$. It follows we get maps of rings

$$A \rightarrow^f B \rightarrow^g A$$

with $$g \circ f =Id_A$$ hence the map $$g:B \rightarrow A$$ is surjective with $$I:=ker(g)$$ and $$A \cong B/I$$. The map $$\phi:= f \circ g$$ satisfies

$$\phi^2 =\phi.$$

Hence you get an idempotent endomorphism of rings $$\phi: B \rightarrow B$$. The pull back $$\phi^*E$$ satisfies

$$\phi^*(E):=B\otimes_{B/I} E/IE$$

and since $$I \subseteq ann(\phi^*(E))$$ it follows $$B/I\otimes_B \phi^*(E)=\phi^*(E)$$. We get

$$\phi^*(E) \cong B/I\otimes_B \phi^*(E) \cong B/I\otimes_B B\otimes_{B/I} E/IE \cong E/IE$$

hence if $$(0)\neq I$$ and $$I \subsetneq ann(E)$$ it follows $$\phi^*(E) \neq E$$.

Question: "Thanks for the nice example. Can we at least say that there is a non-trivial morphism from $$E$$ to $$f^∗s^∗E$$?"

There is a canonical map

$$q:E \rightarrow \phi^*(E)\cong E/IE$$

and $$q$$ is an isomorphism iff $$I \subseteq ann(E)=(0)$$ - If $$E$$ is projective $$ann(E)=(0)$$.

• Thanks for the detailed answer. Jul 3, 2021 at 17:24
• @user43198 - Note: If $\phi: A \rightarrow B$ is any map of commutative rings and if $E$ is a projective $A$-module, it follows $B\otimes_A E$ is projective. Hence there appears to be something "paradoxical" in the above argument but I do not immediatley see what it is - The module $\phi^*(E)$ should be projective. Jul 4, 2021 at 8:27
• I think "I $\subset Ann(\phi^*(E))$" is incorrect. This is because, the natural morphism from $B \otimes_B \phi^*(E)$ to $\phi^*(E)$ is injective. The subtle point (as you write) is $\phi^*(E) \cong B \otimes_{B/I} E/IE$ and the natural map from $B \otimes_B (B \otimes_{B/I} E/IE)$ to $(B \otimes_{B/I} E/IE)$ is given by $b \otimes (b' \otimes m) \mapsto bb' \otimes m$ and not $b' \otimes bm$. The latter map will give you $I \subset Ann(\phi^*(E))$. The important point is that the map $q$ is not injective or surjective. Thanks for the details. It helped me understand the problem better. Jul 5, 2021 at 11:30