analytic vector bundles

Question

Let $E$ be a real analytic vector bundle on an analytic manifold $M$. Assume that $E$, as a smooth vector bundle, is a trivial bundle.

Is $E$ a trivial analytic vector bundle?

I need to the answer to this question for the following question:

Analytic version of the Cartan lemma

Answer 1

Here's just a sketch, maybe someone else can fill in the details. Most of the steps will require the use of the nontrivial fact that $C^\omega(M)$ is dense in $C^\infty(M)$. Let's assume $M$ is compact.

Step 1: Construct a real-analytic embedding of vector bundles $E\to M\times\mathbb R^N$ for some finite $N$. This gives a real-analytic classifying map $M\to\operatorname{Gr}_n(\mathbb R^N)$.

Step 2: Increase $N$ to $N'$ so that the composition $\gamma:M\to\operatorname{Gr}_n(\mathbb R^N)\to\operatorname{Gr}_n(\mathbb R^{N'})$ is smoothly null-homotopic.

Step 3: Approximate this smooth null-homotopy of $\gamma$ with a real-analytic homotopy from $\gamma$ to a map $\gamma':M\to\operatorname{Gr}_n(\mathbb R^{N'})$ which is contained in an $\epsilon$-ball over which the universal bundle is real-analytically trivial.

I suspect this argument also works for $M$ non-compact (taking $N=\infty$) though I haven't checked carefully.

Answer 2

Here's an easier strategy:

Consider the real-analytic vector bundle $\operatorname{Hom}(\mathbb R^n,E)$ where $\mathbb R^n$ denotes the trivial bundle of rank $n=\operatorname{rk}E$. By assumption, it has a smooth section which is a fiberwise isomorphism. Approximate this smooth section by a real-analytic section (this is where we use a hard theorem to tell us such an approximation exists). Now the property of being a fiberwise isomorphism is an open condition, so the real-analytic section is also a fiberwise isomorphism, and thus gives the desired real-analytic trivialization.