# Associating a principal bundle to a torsor

in arXiv:math/0212266, Moerdijk defines a torsor to be a sheaf $\mathcal{S}$ on $X$ with a freely transitive left-action of a sheaf of groups $\mathcal{G}$, such that $X =\bigcup \{ U \in \mathbf{Open}(X), \mathcal{S}(U) \neq \emptyset\}$. I was thinking if one can associate (functorially) a (locally trivial) $G$-principal bundle for some group or, maybe, groupoid $G$ (or, even a family of groupoids) from a given torsor.

For a $G$-principal bundle, one can always make the reverse way, by picking the sheaf of sections and the cocycles as the sheaf of group. However, the fibers of the étale space are discrete, so I don´t know exactly how to recover $G$ and its topology. Is there any known result like this? How is this functor?

• A $G$ principale bundle on $X$ is exactly the same as a $\mathcal{G}$-torsor for $\mathcal{G}$ the sheaf of groups on $X$ locally constant equal to $G$. Basically, in this context, torsor is a synonym for principal bundle. – Simon Henry May 7 '14 at 22:38
• @SimonHenry I agree with you if $G$ is discrete, because you can associate the etale space, however for the general case I don't know how you are making such bundle. – user40276 May 7 '14 at 23:10
• Yes I mean't for $G$ a discrete group. I understand your question now... – Simon Henry May 8 '14 at 8:51

So let $G$ be a topological group, $X$ a topological space, and let $\mathcal{G}$ be the sheaf of local functions from $X$ to $G$ (which is a sheaf of group over $X$).

Let $T$ be a (locally trivial) principal $G$-bundle over $X$, then you can easily check that the sheaf of (local) section of $T$ is a $\mathcal{G}$-torsor : being a torsor is a purely locale property and this is trivial on open subsets of $X$ on which $T$ is trivial.

Conversely, if you have a a $\mathcal{G}$-torsor $\mathcal{T}$, then you can construct a locally trivial $G$ principal bundle in the following way:

Let $U_i$ be a covering of $X$ such that for each $i$ on has a section $t_i$ of $\mathcal{T}$ on $U_i$. then on $U_i \wedge U_j$ there is a unique function $\gamma_{i,j}$ from $U_i \wedge U_j$ to $G$ such that $\gamma_{i,j} t_i =t_j$. this give you a $G$ cocycle which will allow to reconstruct a locally trivial bundle and one easily check that this two operations induce an equivalence between principal bundles and torsors.

• Thanks, I've tried this too, but $t_i$ is a section of $\mathcal{S}$ (in my notation), so its values are defined on the étale space. But the étale space has discrete fibers, so, if $G$ is not discrete, we have a problem. Now, seeing better I think I can solve this problem: just pick $\coprod \mathcal{G} (U_{\alpha})/ \{ (\alpha, x, g) \cong (\beta, x, g_{\alpha \beta} g) \}$, then the fibers are not discrete, but the bundle seems to be dependent of the cocyles without the equivalence modulo coboundary... – user40276 May 8 '14 at 12:27
• Oh, I cannot edit it anymore, but I meant the coproduct over $\mathcal{G}_x$ the stalks. – user40276 May 8 '14 at 12:34
• Well in the construction I explain it is easy to see that a different choice of $t_i$ give an equivalent cocycle modulo coboundary: simply because $t'_i = \gamma_i t_i$ and the $\gamma_i$ are a coboundary between the two cocycles. – Simon Henry May 8 '14 at 12:54
• Oh, you're right it does not depend – user40276 May 8 '14 at 12:57
• Yes, I agree, but for a principal bundle the fibers must be the topological group in question, so I think doing what I said above would be the right way or I missing something that you said. – user40276 May 8 '14 at 12:59