I've been thinking about vector bundles recently and came up against the following problem. I'm hoping someone can help me with the intution here.

Background: In this question, we're working over a base scheme (or variety, if you like) $X$. All sheaves I work with are sheaves of $\mathcal O_X$-modules.

Let $\cal E$ and $\cal E'$ be locally free sheaves on $\mathcal O_X$ and suppose that $\cal E'\subseteq\cal E$. Say that $\mathcal E'$ is a saturated subsheaf of $\mathcal E$ if the quotient sheaf $\mathcal E/\mathcal E'$ is again a locally free sheaf.

Question: If the locally free sheaves $\mathcal E'$ and $\mathcal E''$ are saturated subsheaves of the locally free sheaf $\mathcal E$, then is their intersection $\mathcal E'\cap\mathcal E''\subseteq\mathcal E$ locally free? Assuming $\mathcal E'\cap\mathcal E''$ is locally free, is it a saturated subsheaf of $\mathcal E$?

(For both questions, give either a proof or counterexample.)

Thoughts: By taking stalks, you can reduce this question to a problem in commutative algebra. Here $X$ would be a local ring and the locally free sheaves would just be free modules. This reduction might help in the proof, but it hasn't given me much intuition.

If $X$ is a point or a curve then any subsheaf of a locally free sheaf is locally free, meaning that if the first assertion fails, it can only fail in dimension $>2$. This obstruction has made it hard for me to think of examples.

It's tempting to come up with examples just by intersecting the total spaces of the sheaves; then we would be in the situation of this Math Overflow question. Unfortunately, intersection does not commute with formation of the total space of a vector bundle, so I don't know how to use geometric intuition about vector bundles for this problem.

Terminological Warning: Mohan pointed out that my usage of the word "saturated" is nonstandard: the usual definition states that a subsheaf is saturated if its cokernel is torsion-free, which is weaker than locally free. However, I'm not going to go back through my question and replace all instances of "saturated".

Sasha proposed that we say $\mathcal E'\to\mathcal E$ is a fiberwise monomorphism if its cokernel is locally free. This terminology makes sense, and I should have used it instead of "saturated".

  • $\begingroup$ There is already a notion of saturated, where the quotient is only assumed to be torsion-free, not locally free, so may be you should use another word. $\endgroup$
    – Mohan
    Oct 16, 2019 at 20:08
  • $\begingroup$ Ah, I see; thank you for the terminological clarification. $\endgroup$
    – user134824
    Oct 16, 2019 at 21:18
  • $\begingroup$ I would say $\mathcal{E}' \to \mathcal{E}$ is a fiberwise monomorphism. $\endgroup$
    – Sasha
    Oct 17, 2019 at 4:21

1 Answer 1


Under your assumptions the sheaf $\mathcal{E}' \cap \mathcal{E}''$ is necessarily reflexive, bot not necessarily locally free, as the next example shows.

Let $X = \mathbb{P}^3$, $\mathcal{E} = \mathcal{O}^{\oplus 4}$, with $$ \mathcal{E}' = \Omega^1(1) := \mathrm{Ker}(\mathcal{O}^{\oplus 4} \stackrel{(x_0,x_1,x_2,x_3)}\longrightarrow \mathcal{O}(1)), $$ and $$ \mathcal{E}'' = \mathcal{O}^{\oplus 3} := \mathrm{Ker}(\mathcal{O}^{\oplus 4} \stackrel{(1,0,0,0)}\longrightarrow \mathcal{O}(1)). $$ Then $$ \mathcal{E}' \cap \mathcal{E}'' \cong \mathrm{Ker}(\mathcal{O}^{\oplus 3} \stackrel{(x_1,x_2,x_3)}\longrightarrow \mathcal{O}(1)), $$ which is a standard example of a reflexive, but not locally free sheaf.


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