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$\DeclareMathOperator{\Spec}{Spec}$ $\DeclareMathOperator{\Sym}{Sym}$ $\newcommand{\func}{\mathcal{O}}$ $\newcommand{\M}{\mathcal{M}}$ It is well known that the notions of locally free $\func_X$-modules on a scheme $X$ and vector bundles on $X$ ($X$-schemes $E$ that are locally $\mathbb{A}_X^n$) are equivalent : given vector bundle, one takes the sheaf of sections, and given a locally free $\func_X$-module $\M$, the total space of the bundle is $\Spec_X(\Sym^\bullet\M^\vee)$.

Given a group sheme $G$ and a $G$-torsor over $X$, is there a simple procedure that gives the total space of the torsor ?

A few clarifications about the question :

  • the general answer seems to be "no", according to this answer
  • the same link refers to Milne's Etale cohomology, section III.4, where a partial answer is given. However it is not what I would call a "simple procedure" : I am looking for something more similar to taking the relative spectrum of the symmetric algebra for a vector bundle.
  • however I have no hope that directly taking the relative spectrum of a clever $\func_X$-algebra will work : this is a procedure that "takes closed subschemes", while often a torsor will look like an "open subscheme" (see the following example).

My question is in fact motivated by this particular case : given a locally free $\func_X$-module $\M$, we have the $GL$-torsor of trivializations of $\M$, that is the sheaf whose sections over $U\subset X$ is the set of bases of $\M(U)$ as $\func_X(U)$-module. The question becomes :

Is there a simple description of the total space of the torsor of trivializations of a vector bundle $\M$ ?

(The point of this, ultimately, is that it will be a final object in the category of couples $(Y,f,\phi)$, where $f\colon Y\to Y$ is a morphism and $\phi$ is an $\func_Y$-isomorphism $f^*\M\simeq\func_Y^n$)

I think this procedure works :

Call $E\to X$ the total space of the bundle $\M$, and form the product $F=E\times_X\ldots\times_XE$ with $n$ terms. By construction, locally over a sufficiently small open $U\subset X$, it is isomorphic to $\mathbb{A}^{n^2}\times U$ (with compatibility between the $U$). We can then take the open subset of points corresponding to an invertible matrix, and this gives the desired space.

What I am interested in is to find a "simpler" way, i.e. using existing general constructions, without needing to manipulating the details like I just did. Admittedly the construction above is not hard in anyway, it just doesn't generalize well.

Last, I feel that this all carries verbatim to rigid analytic geometry (in the sense of Tate), eg the relative spectrum is done by Brian Conrad in Relative ampleness in rigid geometry §2.2. But since I am not overly familiar with this setting (especially Grothendieck topologies), I still ask :

What about the setting of rigid analytic geometry ?

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  • $\begingroup$ This is more suitable for mathoverflow and you may transfer it there. I think Brian Conrad may answer most of your questions. $\endgroup$ – Lao-tzu Jan 12 at 19:09

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