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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 $H_*$.

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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

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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.

On the other hand, in the separable case there are some results showing that we can avoid such anomalies.

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