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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.

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1 Answer 1

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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.

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  • $\begingroup$ It means "yes" for the existence part of "theorem 2". Thanks. $\endgroup$ Aug 24, 2021 at 9:46
  • $\begingroup$ @Laurent Claessens: The conclusion is that a $C^{\infty}$ Lie group "is" an analytic Lie group in a very precise sense -- unique also. This can be proved similarly for Lie group of class $C^{3}$ (and more difficult but still true for lower class, up to $C^0$). $\endgroup$
    – orangeskid
    Aug 24, 2021 at 18:21

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