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I'm in particular interested in understanding Grothendieck's argument for this in SGA 1 (page 232 in http://arxiv.org/pdf/math/0206203v2.pdf)

Let $G$ be $\text{GL}_n$ over a scheme $S$ for some integer $n$, and let $P/S$ be a principal $G$-bundle. Then we know that there is an fpqc morphism $S'\rightarrow S$ such that $P' := P\times_S S'$ is isomorphic to $G' := G\times_S S'$ over $S'$.

He claims in his proof that the statement follows from noting that $G(T) = \text{Aut}(\mathcal{O}_T^n)$ (for any $S$-scheme $T$), and that fpqc morphisms are morphisms of effective descent in the category of locally free $\mathcal{O}_T$-modules of rank $n$.

I guess his argument must rely on something special about automorphism group schemes which I'm missing. I admit I haven't read the entirety of his prior chapters on descent. Everything I know about descent pretty much comes from reading the stacks project.

I'd like to know how Grothendieck envisaged his argument would go, though I'd also appreciate any other argument proving this fact or relevant references.

thanks

  • will
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1 Answer 1

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I would argue using the associated fiber bundle construction in Exp. XI just after corollary 4.3 (a few pages back from your spot). For any $GL_n$-torsor $P$, you may use the action of $GL_n$ on $\mathbb{G}_a^n$ to construct a canonical vector bundle $E = P \times^{GL_n} \mathbb{G}_{a,S}^n$, and there is a canonical locally free sheaf $\mathcal{F}$ such that $E = \mathbb{V}(\mathcal{F}) = \operatorname{Spec}_S \operatorname{Sym}_{\mathcal{O}_S}(\mathcal{F})$. The structures $E$ and $\mathcal{F}$ can be constructed on $S$ by descent from the corresponding trivial objects on $S'$. Furthermore, the locally free sheaf $\mathcal{F}$ is Zariski-locally trivial.

By unwinding the definition of $E$, we see that there is a canonical isomorphism $P \cong \underline{\operatorname{Isom}}_{S-vect}(\mathbb{G}_{a,S}^n, E)$. Indeed, sections of $P$ are in canonical bijection with trivializations of $E$. Using the inverse of the $\mathbb{V}$ equivalence, we have an isomorphism $P \cong \underline{\operatorname{Isom}}_{\mathcal{O}_S-mod}(\mathcal{O}^{\oplus n}_S, \mathcal{F})$. Thus, a Zariski-local trivialization of $\mathcal{F}$ induces a Zariski-local trivialization of $P$.

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  • $\begingroup$ By "canonical locally free sheaf", do you mean the vector bundle $(P'\times_S \underline{Spec }\text{Sym}({O}^n))/G$? ($G = \text{GL}_n$) I'm sorry but I'm relatively new to this stuff. Can you explain why/how $P'$ is identified with (the vector bundle associated to) $\underline{Hom}_{S'}(\mathcal{O}_{S'}^n,E')$. $P'$ is just $\text{Aut}(\mathcal{O}_{S'}^n)$, and it seems like there are more homomorphisms $\mathcal{O}_{S'}^n\rightarrow E'$ than automorphisms of $\mathcal{O}_{S'}^n$... $\endgroup$
    – Will Chen
    May 24, 2014 at 22:48
  • $\begingroup$ Sorry for the confusion. I wrote Homs when I should have written Isoms. $\endgroup$
    – S. Carnahan
    May 25, 2014 at 9:03
  • $\begingroup$ Actually, reading this again, I'm wondering why your $E$ has to be a Zariski-locally trivial vector bundle. All the definitions of vector bundles in algebraic geometry define them as being zariski locally trivial, but I can only see that $E$ is a fpqc-locally trivial principal $\mathbb{G}_{a,S}^n$-bundle. Are all such bundles Zariski-locally trivial? How can I see this? $\endgroup$
    – Will Chen
    May 26, 2014 at 22:36
  • $\begingroup$ I just realized that Grothendieck addresses this question in the same proposition, though again I don't understand his proof. $\endgroup$
    – Will Chen
    May 26, 2014 at 22:57
  • $\begingroup$ @oxeimon I find it easier to argue using the locally free sheaf instead of the vector bundle. You can pass between them using EGA2 chapter 1. $\endgroup$
    – S. Carnahan
    May 26, 2014 at 23:15

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