Quotient of line bundle by compact Lie group (after Guillemin-Sternberg) I am trying to understand an argument in Guillemin and Sternberg's paper Geometric Quantization and Multiplicities of Group Representations (Inventiones, 1982). The argument (Proof of Theorem 3.2) seems to be based on the following fact:
Lemma (I think). Let $G$ be a compact connected Lie group acting smoothly and freely on a smooth manifold $M$ and let $L$ be a $G$-equivariant complex line bundle on $M$. Then there is a complex line bundle $L_G$ on $M/G$ such that $\pi^*L_G = L$, where $\pi : M \to M/G$ is the quotient map.
They claim this result (without proof) in a more specific setting (where $M$ is the zero fibre of a moment map and $L$ is the pullback of a prequantum line bundle), but I don't see why it should hold.
More specifically, they define $L_G$ by the sheaf of $G$-invariant sections of $L$, but I don't see why it is locally free of rank 1. It is easy to see that the lemma is equivalent to the following claim:
Claim. For every $p \in M$ there is a $G$-invariant neighbourhood $U$ of $p$ in $M$ together with a non-vanishing $G$-invariant section $s : U \to L$.
 A: Yes, the claim is true. It's a special case of a more general fact, that, in quite some generality, equivariant vector bundles are equivariantly locally trivial.
In your case, given $p$, there is a "slice" through $p$, a set $S\subset M$ containing $p$ such that $G\times S \to M$ is one-to-one and a homeomorphism onto an open subspace. By intersecting with a small enough nonequivariant neighborhood, we can assume that $L$ is trivial on $S$, and the action of $G$ then gives an equivariant trivialization of $L$ on $U = GS$. For a line bundle, an equivariant trivialization is equivalent to having a non-vanishing $G$-invariant section (assuming that the isotropy group, in this case the trivial group, acts trivially on the fiber at $p$).
Slices exist pretty generally, as shown by Palais. See, for example, the writeup at the nLab.
A: This is a question of descent, and a proper environment to address it is the theory of sheaves (here, sheaves of categories). Complex line bundles form a sheaf of categories (stack) over the site of smooth manifolds.
This means that one can make complex line bundles descend along surjective submersions. In your case, this is the projection map
$$M \to M/G$$
which is a surjective submersion under the conditions on the actions that you assume. (More generally, any Lie groups works if the action is assumed to be proper.)
In order to achieve descent, one has to equip the complex line bundle with a "descent structure", which is in your case this is exactly a $G$-equivariant structure.
This pov splits the proof of your lemma into a statement about line bundles that has nothing to do with Lie group actions (line bundles form a stack), and another statement about Lie group actions that has nothing to do with line bundles (the quotient map is a surjective submersion).
From a yet more abstract perspective, the projection
$$M//G \to M/G$$
from the "quotient stack" to its "course moduli space" is an equivalence of stacks (or Lie groupoids) under the assumed conditions on the action. Since complex line bundles form a stack, this implies an equivalence of categories
$$LBun(M//G) \cong LBun(M/G).$$
Moreover, the classical definition of a $G$-equivariant structure gives an equivalence of categories
$$LBun(M)^G \cong LBun(M//G).$$
