Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $G$?


  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.

share|improve this question

1 Answer 1

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 $2$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.