Maybe related : Why are Lie groups automatically analytic manifolds?
I'm seraching for a precise statement. I propose two theorems.
Theorem 1
Let $G$ be a smooth manifold (the transitions functions are $C^{\infty}$ as maps between open subsets of $R^n$). We suppose that $G$ is a group for which the mulliplication and the inverse are smooth.
Then the transition functions, the multiplication and the inverse are analytic.
Theorem 2
Let $G$ be a smooth manifold (the transitions functions are $C^{\infty}$ as maps between open subsets of $R^n$). We suppose that $G$ is a group for which the mulliplication and the inverse are smooth.
Then $G$ accepts an analytic atlas for which the multiplication and the inverse are analytic.
Moreover every two such atlas are analytically diffeomorphic.
My guesses:
- Theorem 1 is wrong because the atlas may contain smooth charts which are not analytic.
- Theorem 2 is true.
QUESTIONS:
- is theorem 1 wrong ?
- is theorem 2 true ?
- can I have a link (online; I do not have access to books) to a precise statement and a proof ?
My aim is to be sure that I can always suppose every Lie groups to be analytic (transition charts, multiplication, inverse) without loss of generality.
Best Answer
Consider the following charts: around $e$ it's $X \mapsto \exp X$, from a neighborhood of $0$ in $\frak{g}$ to a neighborhood of $e$ in $G$. Around a $g\in G$ it's $X \mapsto g \exp X$. This give an analytic atlas in which multiplication and inverse is analytic. An argument for this: say we have the chart aroung a $g= \exp X_0$ close to $e$, $X \mapsto \exp X_0 \exp X$. To see the point $\exp X_0 \exp X$ in the chart around $e$ we need to write the equality $$\exp X_0 \exp X = \exp X_1$$ Now, $X_1$ can be expressed in terms of $X_0$ and $X$ using the B-C-H formula.