# Triviality of (equivariant) holomorphic vector bundles

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 $$\mathbb P(1,2)$$.
• $\mathbb C P^n$ is simply connected but has many nontrivial holomorphic vector bundles. Aug 30, 2013 at 11:43
• $\mathbb C^\times$ 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. Aug 30, 2013 at 14:03
• @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 $X$. Aug 31, 2013 at 19:26