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Banach Lie Groups are what you'd expect:

https://www.encyclopediaofmath.org/index.php/Lie_group,_Banach

If $B$ is a Banach algebra then why is $GL(B)$, the set of invertible elements of $B$, a Banach Lie group?

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    Which axiom is the holdup?2011-12-17

1 Answers 1

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This follows from three simple observations:

  1. The subset $\operatorname{GL}(B) \subset B$ is open, so it is a Banach manifold modeled on $B$ itself.

  2. Multiplication is the restriction of a continuous linear map, hence it is analytic.

  3. Inversion is locally given by the Neumann series, hence it is analytic, too.

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    Thanks! This confirms what I thought was going on. (No access to a reference.)2011-12-18
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    You're welcome. Does that mean you need a reference? I can try to dig one up, but I don't know of any off-hand.2011-12-18
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    Your answer was very helpful, already. If you happen to know of a reference for the analogue of the Frobenius integrability theorem for Banach Manifolds, I'd be deeply appreciative. (Serre's book may have this, but I haven't seen a copy...)2011-12-18
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    There's a [two-volume treatise](http://www.amazon.com/dp/038709444X/) by Glöckner and Neeb on infinite-dimensional Lie groups about to appear in the Springer GTM Series. Next, there's [this recent preprint](http://arxiv.org/abs/1012.1950) by Pelletier that should lead to the relevant literature on Banach versions of the Frobenius theorem. Then I'd also have a look at the two gems [Abraham-Marsden-Ratiu](http://www.amazon.com/dp/0387967907/) and [Choquet-Bruhat--Dewitt-Morette](http://books.google.ch/books?id=hUWEXphqLo8C). This should give some leads to follow.2011-12-18
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    Thank you very much for your help, t.b.! These references will be very helpful.2011-12-18