# Why is $GL(B)$ a Banach Lie Group?

Banach Lie Groups are what you'd expect:

http://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? –  Nate Eldredge Dec 17 '11 at 21:59

## 1 Answer

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.) –  Jon Bannon Dec 18 '11 at 0:15
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. –  t.b. Dec 18 '11 at 0:18
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...) –  Jon Bannon Dec 18 '11 at 0:42
There's a two-volume treatise by Glöckner and Neeb on infinite-dimensional Lie groups about to appear in the Springer GTM Series. Next, there's this recent preprint 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 and Choquet-Bruhat--Dewitt-Morette. This should give some leads to follow. –  t.b. Dec 18 '11 at 1:03
Thank you very much for your help, t.b.! These references will be very helpful. –  Jon Bannon Dec 18 '11 at 1:32