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I would like to understand the stack $[X/G]\to\mathfrak{Sch}$, where $X$ is a scheme and $G$ is an algebraic group acting on it. I found two descriptions of this groupoid, but I am unable to relate them, so any help in this direction would be appreciated.

First description (the most popular). An object in $[X/G]$ is a principal $G$-bundle $\pi:E\to B$ together with a $G$-equivariant map $f:E\to X$. A morphism is the obvious cartesian square of bundles, plus the compatibility between the equivariant morphisms.

Second description. There is an object $\underline x$ of $[X/G]$ for every point $x$ of $X$. And there is a morphism $\underline x'\to\underline x$ exactly when there is some $g\in G$ such that $x'=gx$, or in other words, when Orb$(x)=$ Orb$(x')$.

Question. How to see that these two descriptions are equivalent (that is, they describe the same stack over $\mathfrak{Sch}$)? Also, I do not even see how to define the groupoid fibration over $\mathfrak{Sch}$, in the second case.

Of course, any insight that may help to better understand this object is welcome.

Thank you!

Brenin
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    For the second construction, where it says "point", read "scheme morphism with codomain $X$". In other words, define $[X / G]$ to be the stack associated with the (strict!) functor $\textbf{Sch}^\textrm{op} \to \textbf{Grpd}$ that sends $T$ to the groupoid whose objects are $T$-valued points of $X$ and whose morphisms are $g : x \to x'$ where $g \in G(T)$ and $x' = g \cdot x$. – Zhen Lin Mar 02 '13 at 23:13
  • Dear @Zhen Lin, I tried to identify the fiber $[X/G]_T$ to the groupoid $h_X(T)$ in the sense that you suggest (image of your functor), but I feel like there is more information in $[X/G]$. To $(\pi:E\to T, f:E\to X)$ I make correspond $f:E\to X$ and to a morphism I make correspond the compatibility between the equivariant maps. But 1) this doesn't seem to correspond to an element of $G$ and 2) it seems to me that everything regarding the $G$-bundle has been lost. What am I missing? – Brenin Mar 05 '13 at 09:11
  • I'm not entirely convinced that the first description is literally correct. It certainly should be the case that $[X/G]$ is a classifying stack for some kind of $G$-bundle, with $X \to [X/G]$ being the universal such, but it's not obvious to me exactly what conditions are needed. Also, the associated stack ("stackification") may change the objects and morphisms in the fibres. – Zhen Lin Mar 05 '13 at 09:54

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