Eigenbundle decomposition

Question

Let $G$ be a finite cyclic group and $X$ a smooth manifold equipped with a trivial $G$-action.

It is known that we can decompose every $G$-equivariant vector bundle with respect to the action:

Every smooth $G$-equivariant vector bundle $V \to X$ can be decomposed as the Whitney sum $V=\bigoplus_{\chi}V_\chi$ of $G$-equivariant vector bundles.

Here $\chi$ runs the character group $X(G):=\mathrm{Hom}(G,\mathbb C^\ast)$ and the $G$-action on $V_\chi$ is given by $g \cdot v = \chi(g)v$ ($g \in G$, $v \in V_\chi$).

I understand the case when $X$ is a point. This is just the eigenspace decomposition.

Question: Why do the eigenspaces vary smoothly so that $V_\chi$ gives a smooth subbundle of $V$?

Notes:

1. I do not mention above whether the vector bundle is real or complex. Maybe we should assume it to be complex.
2. If you know a reference containing the proof, please let me know.

Answer 1

Not a full answer, but as I thought about this for a while, and managed to form something resembling a coherent picture of what's happening, I scribble down the reason, why I think this should be true.

Let $V_x$ be the fiber of $x\in X$, and $V_{x,\chi}\subset V_x$ be the subspace of those vectors $v$ that are acted on according to the character $\chi$, i.e. $v\in V_{x,\chi}$, iff $g\cdot v=\chi(g) v$ for all $g\in G$.

The smoothness of the $G$-action surely means that the inner products $$m_\chi(x):=\langle \chi, V_x\rangle_G=\dim V_{x,\chi}$$ between $\chi$ and the representation of $G$ on the fiber $V_x$ are smooth functions of $x$. Presumably $X$ is connected, so this means that $m_\chi(x)$ is a constant, so at least the dimension of $V_{x,\chi}$ is independent from $x$.

But we can say more. The group algebra $\Bbb{C}G$ acts linearly on the fibers, and as the individual elements of $G$ act smoothly on $V$, so do the elements of the group algebra. Consider the idempotent $$ e_\chi=\frac1{|G|}\sum_{g\in G}\overline{\chi(g)}g\in \Bbb{C}G. $$ At all the fibers $V_x$, we have $V_{x,\chi}=e_\chi(V_x)$. Because this projection $e_\chi$ depends smoothly on $x$. This is because if we denote by $\pi_x$ the representation of $G$ on the fiber $V_x$, then we get a smooth projection $V\to V_\chi$ as $$ e_\chi(x)=\frac1{|G|}\sum_{g\in G}\overline{\chi(g)}\pi_x(g). $$ In place of the image of $e_\chi(x)$ we also get $V_\chi$ as the kernel of $I-e_\chi(x)$.

I think that the claim follows. I'm too ignorant about manifolds/ fiber bundles to call upon a suitable theorem at this point :-(

Answer 2

This is clear on the level of a local trivialisation. If $U$ is any open subset of $X$ such that the pre-image of $U$ under $V\rightarrow X$ is isomorphic to $U\times \mathbb{C}^k$, then the $G$-action on this pre-image is just a $G$-representation on $\mathbb{C}^k$, so locally above $U$ you get the required decomposition. Since the open $U$ overlap and the trivialisations at the overlapping open sets are all compatible, the decomposition into eigenspaces fits together to give you a smooth bundle on the whole manifold.