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 '19 at 20:08
  • $\begingroup$ Ah, I see; thank you for the terminological clarification. $\endgroup$ – user134824 Oct 16 '19 at 21:18
  • $\begingroup$ I would say $\mathcal{E}' \to \mathcal{E}$ is a fiberwise monomorphism. $\endgroup$ – Sasha Oct 17 '19 at 4:21

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.


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.