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Take X to be a scheme over a (say algebraic closed) field k and G a group scheme act on X. I want to get more convenient knowledge about G-invariant vector bundles over X.

I'm inspired by this example: $X=A^n$ is the affine space and $G=G_m$ is the multiplication group which acts on X through obvious scaler multiplication. Then to know a vector bundle over X invariant under G is equivalent to know a vector bundle over the projective space $P^{n-1}$.

Of course I do not expect in general there exists a scheme Y playing the role as $P^{n-1}$ in the above example, but under what conditions about the action of G can we get such a scheme? More weakly, if we allow Y to be an object in some bigger category (algebraic space, stack, presheaf...), can we get similar results which adapt to more general actions?

edit: i should have used the term G-equivariant...sorry

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    $\begingroup$ I think the claim about $A^n$ is false: vector bundles on $P^{n-1}$ correspond to $G_m$-equivariant vector bundles on $A^n \setminus \{0 \}$, and these don't have to extend to vector bundles on $A^n$. The tangent bundle of $P^{n-1}$ is a counterexample. $\endgroup$
    – Piotr Achinger
    Commented Sep 29, 2014 at 7:30
  • $\begingroup$ "$G$-invariant" per se doesn't make much sense: it would mean $g^* E = E$ for every $g\in G$, but there is a choice involved! This leads to the notion of linearization of a vector bundle $E$: it's an isomorphism $f:\pi^* E \to \mu^* E$, where $\pi, \mu:G\times X\to X$ are the projection resp. the action, satisfying a certain "cocycle condition". The pair $(E, f)$ is called a $G$-equivariant bundle. This $f$ may be non-unique, as the example of $E=\mathcal{O}_X$ on $X=\mathbb{A}^1$ with the usual $\mathbb{G}_m$-action shows (here one has $\mathbb{Z}$ worth of possible choices). $\endgroup$
    – Piotr Achinger
    Commented Sep 29, 2014 at 7:34
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    $\begingroup$ You probably mean $G$-equivariant vector bundles. Then your $Y$ will just be the quotient stack $[X/G]$. $\endgroup$
    – abx
    Commented Sep 29, 2014 at 7:37
  • $\begingroup$ yes i mean G-equivariant...sorry $\endgroup$
    – KylinChen
    Commented Sep 29, 2014 at 7:44
  • $\begingroup$ Any choice of $\mathbb{G}_m$-torsor gives you the sort of object you want. See the last section in Vistoli's notes on descent (on the arXiv). $\endgroup$
    – S. Carnahan ♦
    Commented Sep 29, 2014 at 23:47

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