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Let $G$ the 1-dimensional diagonalisable linear complex analytic group $\mathbb C^*$. We suppose that $G$ acts linearly on $\mathbb C^n$ with positive weights. Set $X=\mathbb C^n -\lbrace 0 \rbrace$. Then every complex $G$-equivariant vector bundle $E$ descends into a complex vector bundle on the (weighted) projective space (or stack) $X/G$.

Question:

Fix a $G$-equivariant complex vector bundle $E$ on $X$. Does there exist a $G$-module $V$ such that $E$ is isomorphic to $X \times V$ as a $G$-equivariant complex vector bundle?

Notes:

  1. This question comes from arXiv:1205.4742. In §4 we use the above statement for a scpecial case where the quotient $X/G$ is $\mathbb P(1,2)$.
  2. I initially asked this question with adjective "holomorphic".
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    $\begingroup$ $\mathbb C P^n$ is simply connected but has many nontrivial holomorphic vector bundles. $\endgroup$
    – Eric O. Korman
    Aug 30, 2013 at 11:43
  • $\begingroup$ Maybe you need to add a (semi)stability assumption for the vector bundle? $\endgroup$
    – Moishe Kohan
    Aug 30, 2013 at 12:27
  • $\begingroup$ @EricO.Korman I see. Maybe I forgot important conditions and I should concentrate on the second question. $\endgroup$
    – H. Shindoh
    Aug 30, 2013 at 13:34
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    $\begingroup$ $\mathbb C^\times$ is homotopic to $S^1$, which has only trivial complex vector bundles (the space of rank $k$ complex vector bundles over $S^n$ is equivalent to homotopy classes of map $S^{n-1} \to GL(k,\mathbb C)$. So in this case you get maps $S^0 \to GL(k,\mathbb C)$ and there is just one homotopy class since $GL(k,\mathbb C)$ is path-connected). So topologically, $E \simeq X\times V$ with $V$ a $G$-module. However, I think there can be non-trivial holomorphic structures on a trivial bundle over $\mathbb C^\times$. I'm not sure about this though. $\endgroup$
    – Eric O. Korman
    Aug 30, 2013 at 14:03
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    $\begingroup$ @H.Shindoh For the result on complex vector bundles over spheres, you can see p. 22 of Hatcher's Vector Bundles and K-theory available here: math.cornell.edu/~hatcher/VBKT/VBpage.html. Actually I guess the action of $G$ on $X \times V$ doesn't have to come from a single action of $G$ on $V$, presumably it could vary with $X$. $\endgroup$
    – Eric O. Korman
    Aug 31, 2013 at 19:26

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