Extending a $G$-torsor/vector bundle over a codimension 2 locus

Question

I want to prove the following statement: Let $X$ be a scheme over a field $k$ locally of finite presentation, and let $S$ be a closed subset of $X$ of codimension $\geq 2$. If $\mathcal{E}$ is a locally free sheaf of rank $n$ on $X-S$, then one can extend $\mathcal{E}$ to $X$ uniquely.

I think the following argument should be able to show the statement: $\mathcal{E}$ is described by a cocycle $(g_{ij})$ where $g=g_{ij}$ is an element of $\mathrm{GL_n}(\Gamma(U,\mathcal{O}_{X-S}))$. Here $U$ is an open of $X-S$ that is a restriction of an affine open $\mathrm{Spec}(B)$ of $X$, i.e., $\mathrm{Spec}(B)|_{X-S}=U$. At this point, if I assume $X$ is also normal, then I can use Algebraic Hartog to deduce $\mathrm{Spec}(B)=\Gamma(U,\mathcal{O}_{X-S})$. Therefore, the same cocycle $(g_{ij})$ defines the unique extension to $X$.

However, in this argument, I still need to assume $X$ is normal. How can I remove this condition?

I also want to ask a more general question. For a group scheme $G$ over $X$ (with some further mild assumptions on $X$ if necessary), when can we still extend a $G$-torsor on $X-S$ to $X$ uniquely?

Answer 1

As Mohan points out in the comments, your actual question is doomed to have a positive answer. To help conceptualize this, it's useful to set up some terminology.

Let us fix $X$ to be a locally Noetherian integral scheme. For a $\mathcal{O}_X$-module $\mathcal{F}$ on $X$, we define $\mathcal{F}^\ast$ to be the sheaf-theoretic dual: $\mathcal{Hom}(\mathcal{F},\mathcal{O}_X)$.

Definition: A coherent sheaf $\mathcal{F}$ on $X$ is called reflexive if the natural double-duality map $\mathcal{F}\to \mathcal{F}^{\ast\ast}$ is an isomorphism.

Denote by $\mathbf{Rflx}(X)$ the category of reflexive coherent sheaves on $X$, and $\mathbf{Vect}(X)$ the full subcategory of vector bundles on $X$.

Let us further call an open subset $U\subseteq X$ large if $Z:=X-U$ has the property that $\mathrm{depth}(\mathcal{O}_{X,z})$ is at least $2$ for all points $z$ in $Z$. If $X$ is normal (or Cohen--Macaulay) then this is equivalent to the fact that $Z$ is of codimension at least $2$ in $X$.

Proposition 1 (see Tag 0EBJ): Suppose that $U\subseteq X$ is large. Then, the pair
$$j_\ast\colon \mathbf{Rflx}(U)\to\mathbf{Rflx}(X)\qquad j^\ast\colon \mathbf{Rflx}(X)\to \mathbf{Rflx}(U)$$ are well-defined quasi-inverses.

The slogan for this is simple: reflexive sheaves don't change under the passage to reflexive subsets.

It also suggests a simple reality.

Corollary: If $U\subseteq X$ is large, and $\mathcal{E}$ is a vector bundle on $U$, then it extends to a vector bundle on $X$ if and only if $j_\ast\mathcal{E}$ is a vector bundle.

Proof: Indeed, if $\mathcal{V}$ is a vector bundle on $X$ extending $U$, then

$$j^\ast\mathcal{V}\cong \mathcal{E}\cong j^\ast j_\ast\mathcal{E}.$$

As $j_\ast\mathcal{E}$ and $\mathcal{V}$ are reflexive sheaves on $X$, we see by the proposition that $\mathcal{V}\cong j_\ast\mathcal{E}$, and so the claim follows. $\blacksquare$

With this in mind, we can inspect the example mentioned by Mohan.

Example: Let

$$M=\left(k[x,y,z]e_1\oplus k[x,y,z]e_2\oplus k[x,y,z]e_3\right)/(xe_1+ye_2+ze_3),$$

and set $\mathcal{F}=\widetilde{M}$. We claim that $\mathcal{F}$ is a reflexive module which is not a vector bundle, but $\mathcal{F}|_U$ is a vector bundle, where $U=\mathbb{A}^3_k-\{(0,0,0)\}$. This will show by the above corollary that $\mathcal{F}|_U$ is a vector bundle on $U$ which does not extend to $\mathbb{A}^3_k$.

Let us first show that $\mathcal{F}$ is not a vector bundle, but $\mathcal{F}|_U$ is. It suffices to assume that $k$ is algebraically closed. Then, recall that as $\mathbb{A}^3_k$ and $U$ are reduced Noetherian schemes, and $\mathcal{F}$ is coherent, that it suffices to show that the function

$$r\colon \mathbb{A}^3_k(k)\to\mathbb{Z},\qquad x\mapsto \dim_{k(x)}(\mathcal{F}\otimes k(x)),$$

is not constant, but is constant on $U(k)$. Indeed, this follows from Tag 0FWG, together with the fact that $r$ is upper semi-continuous (see Mohan's answer here), and that the $\mathbb{A}^3(k)$ is very dense in $\mathbb{A}^3_k$ (see [GW, Proposition 3.3.5]), and similarly for $U$. But, this is clear as if $x=(a,b,c)\in k^3$ then $\mathcal{F}\otimes k(x)$ is $$(ke_1\oplus ke_2\oplus ke_3)/(ae_1+be_2+ce_3),$$

which evidently has dimension $2$ unless $(a,b,c)=(0,0,0)$.

Finally, to show that $\mathcal{F}$ is reflexive, we see by Tag 0AY5 and the fact that $\mathcal{F}|_U$ is a vector bundle (and so reflexive), it suffices to show that the stalk $\mathcal{F}_{(x,y,z)}$ has depth at least $2$. But, one can explicitly check that $(x,y)$ is a regular sequence for this module. $\blacksquare$

---

A similar story exists for $G$-torsors, where $G$ is a smooth group $X$-scheme, at least when $X$ is normal (but this is most likely just an assumption of convenience). The operative notion being that of a reflexive pseudo-torsor for $G$. This is developed in the joint paper [IKY] of mine, specifically in §A.6. Of course, when $G=\mathrm{GL}_{n,X}$ this recovers the theory of reflexive modules discussed above.

The only extra noteworthy result that doesn't have an analogue in the theory of reflexive modules is the following.

Proposition 2 (cf. [IKY, Proposition A.25]): Let $\rho\colon\mathcal{H}\hookrightarrow\mathcal{G}$ be a closed embedding of reductive $X$-group schemes. Then, if $\mathcal{Q}$ is a reductive pseudo-torsor for $\mathcal{H}$, then it is a $\mathcal{H}$-torsor if and only if $\rho_\ast\mathcal{Q}$ is a $\mathcal{G}$-torsor.

In practice one applies this for a faithful representation $\rho\colon \mathcal{G}\hookrightarrow \mathrm{GL}(\Lambda_0)_X$. If one thinks of $\mathcal{Q}$ as an association of reflexive $\mathcal{O}_X$-modules to representations $\Lambda$ of $\mathcal{G}$ (e.g., see [IKY, Proposition A.24 and Proposition A.25]) then this says that it suffices to check that the reflexive $\mathcal{O}_X$-module $\mathcal{Q}(\Lambda_0)$ is a vector bundle.

The reason I mention this is two-fold:

• it's surprisingly useful (it plays a non-trivial role in the proof of the main results of [IKY]),

• its proof looks more similar to that which you attempted to carry out in your post (cf. [IKY, Remark A.27]).

EDIT: Let me just add the observation made by ASadMathematician above, which is that your question does admit a positive answer if $X$ is regular and dimension at most $2$. This follows easily from our above discussion (namely Proposition 1) and the following (classical) fact.

Proposition 3 (see Tag 0B3N): Suppose that $X$ is a locally Noetherian connected regular scheme of dimension at most $2$. Then, the fully faithful embedding $\mathbf{Vect}(X)\to\mathbf{Rflx}(X)$ is an equivalence (i.e., every reflexive sheaf is a vector bundle).

References

[GW] Görtz, U. and Wedhorn, T., 2010. Algebraic Geometry I: Schemes. Vieweg+ Teubner.

[IKY] https://arxiv.org/pdf/2310.08472.pdf