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Let $G$ be a connected Lie group and $H$ a normal connected subgroup. I'm looking for examples where $G\to G/H$ has a section as topological spaces, but not as Lie groups.

My motivation here is to find a continuous analogue of extensions of discrete groups, where splittings always exist on the level of sets, but only exist as groups when the extension is a semidirect product.

I've looked at exact sequences of Lie groups, such as: $$1\to SU(n) \to U(n) \to U(1) \to 1$$

But in this case, the sequence is split so $U(n) \cong SU(n)\rtimes U(1)$. I've also looked at: $$1\to SU(2) \to SO(4) \to SO(3) \to 1$$

But I believe there is still a splitting here as well. (I couldn't find a reference but this seems to be the case)

(Side question: does a topological splitting imply that $G\to G/H$ is a trivial fibration, so that $G$ is homeomorphic to $H\times G/H$?)

Ideally, I'd like to see an example where $H$ is abelian. I'd also like to consider situations where $H$ isn't normal so $G/H$ is just a homogeneous space and not a group, but I'm not completely sure how to phrase the question in that case. Cohomological criteria are welcome and appreciated.

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    $\begingroup$ This might not be of interest to you since it's usually not normal, but if $H$ is maximal compact, then the coset space is Euclidean, and there is a section inducing a diffeomorphism between $H \times G/H$ and $G$. $\endgroup$
    – jdc
    Commented Nov 13, 2023 at 15:40
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    $\begingroup$ For your second sequence, in general if $G/Z(G)$ acts on $G$ via the inner automorphisms then $G \rtimes G/Z(G)$ is isomorphic to a central product $G \times_{Z(G)} G$, so the central product splits. I admit that I am surprised by this. $\endgroup$
    – Dave Benson
    Commented Nov 13, 2023 at 15:40
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    $\begingroup$ $G$ simply connected nilpotent not abelian, $H=[G,G]$: then $G\to G/[G,G]$ never splits (even as abstract groups). But topologically this splits (this is topologically conjugate, through exponentials, to a surjective linear map). $\endgroup$
    – YCor
    Commented Nov 13, 2023 at 15:47
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    $\begingroup$ (For a very concrete example of @YCor's suggestion, let $G$ be the Heisenberg group, with abelianisation $\Big(\begin{smallmatrix}1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1\end{smallmatrix}\Big) \mapsto (a,b)$.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 13, 2023 at 16:47
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    $\begingroup$ For the side question, let $s\colon G/H\to G$ be a continuous (or smooth) section, then we have the isomorphism $G/H\times H \to G$ mapping $(x, h)\to s(x)\cdot h$. $\endgroup$
    – Piotr Achinger
    Commented Nov 13, 2023 at 21:04

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A summary of some of the comments:

@YCor proposed a collection of examples given by $G\to [G, G]$ where $G$ is simply-connected nilpotent non-abelian.

Specifically, if $H$ is the Heisenberg group, we have an exact sequence that splits topologically but not as groups:

$$0\to \mathbb{R}\to H\to\mathbb{R}^2\to 0$$

@PiotrAchinger gave an answer to my parenthetical question: a continuous section $s$ does imply that the fibration is trivial, since we can identify $(x,h)$ with $s(x)\cdot h$.

Finally, @KonradWaldorf pointed out that the continuous group cohomology $H^2$ classifies extensions of the kind I'm interested in, at least in the central case, since a continuous 2-cocycle gives rise to a topological section.

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