# analytic structure on lie groups

I need a reference for a result I have heard only very vaguely "A lie group (smooth) has a compatible analytic manifold structure". (Would even appreciate a concise way to refer to the result..)

I gather from a discussion on a related question that "There is an amazing theorem of Morrey and Grauert that says that not only does every (paracompact) smooth manifold have a real analytic structure, the real analytic structure is unique.". I assume however that the aforementioned more specific statement was earlier known and easier to prove..

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These statements are entirely different: first involves functoriality, 2nd concerns objects up to isomorphism. The forgetful functor from real-an. Lie groups to smooth Lie groups is an equivalence of categories, whereas the functor from (paracompact) real-an. manifolds to (paracompact) smooth manifolds is merely essentially surjective on objects (Morrey-Grauert) and bijective between isomorphism classes. It is much weaker to say that smoothly isomorphic real-an. mnflds are analytically isom. than to say that some smooth isom. is real-an. (or that all smooth maps are real-an.). –  BCnrd Dec 16 '10 at 1:03
With suitable analytic foundations (including real-analytic Frobenius-integrability!) the entire development of elementary Lie theory works in the real-analytic category verbatim, up to and including the fact that the functor from a (connected and) simply connected Lie group (either smooth or real-analytic) to its Lie algebra is an equivalence of categories. So comparison through the Lie algebras does the job. Can be seen in more direct ways too (e.g., using graph arguments and real-analytic Frobenius-integrability). –  BCnrd Dec 16 '10 at 1:11
was not quite aware that was how the latter statement read...thanx.. –  faquarl Dec 17 '10 at 6:31

$\def\ad{\text{ad}}$ This follows immediately from the Baker-Campbell-Hausdoff formula: For complex $z$ near $1$ we consider the function $$f(z):= \frac{\log(z)}{z-1} = \sum_{n\geq0}\frac{(-1)^n}{n+1}(z-1)^n$$ Then for $X$, $Y$ near $0$ in $\mathfrak g$ we have $\exp X.\exp Y= \exp C(X,Y)$, where $$C(X,Y) = Y + \int_0^1 f(e^{t. \ad X}.e^{ \ad Y}).X\,dt$$ $$= X + Y + \sum_{n\geq1}\frac{(-1)^n}{n+1}\int_0^1\biggl( \sum_{{k,\ell\geq0, k+\ell\geq1}} \frac{t^k}{k!\,\ell!}\; ( \ad X)^k( \ad Y)^\ell\biggr)^nX\;dt$$ $$= X + Y + \sum_{n\geq1}\frac{(-1)^n}{n+1} \sum_{{k_1,\dots,k_n\geq0, \ell_1,\dots,\ell_n\geq0,k_i+\ell_i\geq1}} \frac{(\ad X)^{k_1}(\ad Y)^{\ell_1}\dots (\ad X)^{k_n}(\ad Y)^{\ell_n}} {(k_1+\dots+k_n+1)k_1!\dots k_n!\ell_1!\dots\ell_n!}X$$ $$= X + Y + \tfrac12[X,Y] +\tfrac1{12}([X,[X,Y]]-[Y,[Y,X]]) + \cdots .$$ For a short proof of this formula see 4.29 of here. This series has radius of convergence $\pi$ in each norm on the Lie algebra in which the bracket (as bilinear operator) is bounded by 1. This even works for $C^2$-Lie groups, since $C^2$ suffices to get the Lie bracket at the tangent space of the identity.

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