# Varieties satisfying the extension of vector bundles property

We know if we have a regular variety $$X$$ with $$U$$ an open sub-scheme such that $$codim(X\setminus U)\geq 2$$, then any reflexive sheaf has a unique extension from $$U$$ to $$X$$. My question is when a vector bundle on $$U$$ extends to a vector bundle on $$X$$? More precisely I have two types of questions:

1. What type of restrictions, the assumption that $$X$$ is variety such that every vector bundle on all $$U$$'s ($$codim(X\setminus U)\geq 2$$) extends to a vector bundle on $$X$$, imposes on $$X$$. (What are all these $$X$$'s).

2. What type of restrictions, the assumption that $$X$$ is variety such that every vector bundle on some $$U$$ ($$codim(X\setminus U)\geq 2$$) extends to a vector bundle on $$X$$, imposes on $$X$$ and $$X\setminus U$$?

• As soon as you have a reflexive sheaf $\mathscr F$ on $X$ that is not a vector bundle, taking $U$ the locus where $\mathscr F$ is locally free gives a counterexample to 1. This probably means that 1 is true if and only if $\dim X \leq 2$. Nov 28, 2020 at 20:41

Here is the criterion:

Lemma. Let $$X$$ be a regular variety, and let $$U \subsetneq X$$ be a nonempty open subset such that $$\operatorname{codim}(X - U) \geq 2$$. Then the following are equivalent:

1. Every vector bundle on $$U$$ extends to a vector bundle on $$X$$;
2. Every reflexive sheaf on $$X$$ that is locally free on $$U$$ is locally free;
3. $$\dim(X) \leq 2$$.

Proof. 1 $$\Rightarrow$$ 2: if $$\mathscr F$$ is reflexive and $$\mathscr F|_U$$ is locally free, then $$\mathscr F|_U$$ extends to a vector bundle $$\mathscr E$$ on $$X$$. Since extensions of reflexive sheaves are unique [Tag 0EBJ], we conclude that $$\mathscr F = \mathscr E$$, i.e. $$\mathscr F$$ is locally free.

2 $$\Rightarrow$$ 1: if $$\mathscr F$$ is locally free on $$U$$, then its unique reflexive extension $$j_*\mathscr F$$ is locally free by assumption.

3 $$\Rightarrow$$ 2: see [Tag 0B3N].

2 $$\Rightarrow$$ 3: pick a closed point $$x \in X - U$$. Since $$X$$ is regular, there exists a locally free sheaf $$\mathscr E$$ and a surjection $$\mathscr E \twoheadrightarrow \mathcal I_x$$ [Tag 0F8A]. Its kernel $$\mathscr F$$ is reflexive [Tag 0EBG], and we get an exact sequence $$0 \to \mathscr F \to \mathscr E \to \mathcal O_X \to \mathcal O_x \to 0.\tag{1}\label{1}$$ Restricting to $$U$$ gives a short exact sequence $$0 \to \mathscr F|_U \to \mathscr E|_U \to \mathcal O_U \to 0,$$ showing that $$\mathscr F|_U$$ is locally free. By assumption, this implies that $$\mathscr F$$ is locally free. Then \eqref{1} shows that $$\mathscr Ext^i(\mathcal O_x,\mathscr G) = 0$$ for any $$i > 2$$ and any coherent sheaf $$\mathscr G$$, so in particular $$\operatorname{depth}(\mathcal O_{X,x}) \leq 2$$. Since $$X$$ is regular, this means $$\dim_x X \leq 2$$. Since $$X$$ is integral and of finite type over a field, it is equidimensional, so $$\dim_x X = \dim X$$ since $$x$$ is closed. $$\square$$

• Splendid, bravo! Jan 8, 2021 at 16:28