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Put $X := \mathbb A^{n+1}\!-\lbrace0\rbrace$. Let $G=\mathbb C^*$ act on $X$ with (positive) weights $w_0,\dots,w_n$. The quotient stack $[X/G]$ is called the weighted projective stack.

Each vector bundle on $[X/G]$ corresponds to a $G$-equivariant vector bundle $E$ on $X$, and vice versa. So I want to understand the latter (because it seems to be easier than the former).

Question: is every complex $G$-equivariant vector bundle $E$ on $X$ of the form $X \times V$ for any representation $V$ of $G$?

Notes:

  1. This question comes from this paper by Prof. Edidin. In §4.2 he says that the statement of my question holds when $n=1$. Because I can not understand whether the dimension $n$ is important or not, I wrote my question with a general settings.

  2. If we forget the $G$-action, then $E$ is trivial. (Read the comments of my question at SE for details.)

  3. If a holomorphic structure of the bundle $E$ is important, then please assume it.

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up vote 5 down vote accepted

No, that is not true. The easiest counterexample is when $n$ equals $2$ and the weights are all $1$, i.e., the quotient stack is actually the scheme $\mathbb{P}^2$. For the tangent sheaf on $\mathbb{P}^2$, the pullback sheaf on $\mathbb{A}^3\setminus \{0\}$ is not a trivial locally free sheaf (even without considering the $G$-equivariance). If it were a trivial locally free sheaf, then by Hartog's theorem / the S2 property, the pushforward of this sheaf to all of $\mathbb{A}^3$ would also be a locally free sheaf. However, it is straightforward to compute that the pushforward is not locally free: the "fiber" at the origin has rank $3$ instead of $2$.

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