# Integral subgroup theorem for Banach-Lie groups

Let $G$ be a Banach-Lie group with Lie algebra $\mathfrak g$ and $\mathfrak h$ a closed subalgebra. Using the exponential map and the Baker-Campbell-Hausdorff-formula one constructs a local Lie group $\mathcal L_H$ corresponding to $\mathfrak h$. Then the subgroup $H = \langle \mathcal L_H\rangle$ of $G$ naturally inherits a topology and smooth structure (one imposes that $\mathcal L_H\to H$ is an open smooth embedding) which makes it into a Lie group and the inclusion $H \to G$ a continuous homomorphism and a smooth immersion. In finite dimensions, this is the concept of analytic subgroup of Chevalley, and in infinite dimensions can be found in B. Maissen, Lie-Gruppen mit Banachr\"aumen als Parameterr\"aume (German), Acta Math.\ \textbf{108} (1962) 229–270. Note that I'm avoiding on purpose use of the Frobenius theorem which in infinite dimensions requires that distributions not only be closed but also complemented.

Since $H$ is locally arcwise connected, it is clear that its topology is stronger that the locally arcwise connected topology (say denoted by $H_*$) constructed from the induced topology from $G$.

(More explicitly, a basis for the topology in $H_*$ is constructed by taking the arcwise connected components of the elements of a basis for the induced topology.)

My question is that I think both topologies coincide but so far I can prove it only in the finite-dimensional case (namely, I use the fact that a continuous bijective homomorphism of a locally compact, second countable group into a locally arcwise connected group must be a homeomorphism). Does anyone have an argument to see that the identity map $H \to H_*$ is a homeomorphism in general? Of course, it suffices to see that $\mathcal L_H$ is a neighborhood of the identity in $H_*$.

• No clue, doc, wabbits don't do Functional Analysis, but if you take an arc from $1$ to $x$ in a neighbourhood in $H_*$-topology, can you not approximate it by nice ones, i.e. piecewise smooth arcs, whose smooth components are cosets of 1-parameter subgroups? If yes, you should be done as the nice arcs have tangent vectors and sit in $<L_H>$... – Bugs Bunny May 18 '11 at 15:46

I have been told by Karl-H. Neeb of the following neat counterexample due to K. H. Hofmann and S. Morris. Consider the Banach-Lie group $\mathfrak g=\ell^1(\mathbf R,\mathbf R)\times\mathbf R$ with the addition and take the closed discrete subgroup $D$ generated by the elements $(n,\sum_{x\in\mathbf R}n(x)x)$ where $n:\mathbf R\to\mathbf Z$ is an almost null function. Then the quotient $G=\mathfrak g/D$ is a Banach-Lie group with Lie algebra $\mathfrak g$. Consider the subalgebra $\mathfrak h = \ell^1(\mathbf R,\mathbf R)$ and the analytic subgroup $H=\exp_G \mathfrak h$ of $G$ it generates. Since $\mathfrak h+D=\mathfrak g$, we have that $H=G$ as a set. Hence the Lie topology on $H$ is much finer than the locally arcwise connected topology constructed from the induced topology.