7
$\begingroup$

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?

  • 0
    Which axiom is the holdup?2011-12-17

1 Answers 1

8

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.

  • 0
    Thank you very much for your hel$p$, t.b.! These references will be very helpful.2011-12-18