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I've read somewhere (but cannot locate the source) that the following classification holds: simply connected homogeneous 3-manifolds are either isometric to $S^2 \times \mathbb{R}$ or to a metric Lie group.

I'm looking for a reference with a proof of the above statement. In particular, I'm curious if there is a proof of the compact case that does not rely on the Poincaré conjecture (since, if I'm not mistaken, $S^3$ is the only simply connected homogeneous closed 3-manifold, and is a Lie group).

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    $\begingroup$ A proof can follow by classifying Lie groups $G$ of dimension $\le 6$ with a compact subgroup of codimension 3 (in the compact case, $G$ itself being compact which makes the classification quite straightforward) This doesn't require the Poincaré conjecture at all. $\endgroup$
    – YCor
    Sep 3, 2022 at 8:52
  • $\begingroup$ Your title only asks that the $3$-manifold be simply-connected and homogeneous, but then you talk about 'isometric to' in your desired conclusion instead of 'diffeomorphic to'. Did you mean to ask about classifying simply-connected homogeneous Riemannian $3$-manifolds up to isometry? $\endgroup$
    – Robert Bryant
    Sep 3, 2022 at 12:37
  • $\begingroup$ @RobertBryant I guess I forgot to include the Riemannian part. I suppose if we drop that requirement, and say instead that they are either diffeomorphic to $S^2 \times \mathbb{R}$ or to a Lie group, the classification still holds? $\endgroup$
    – Paul Cusson
    Sep 3, 2022 at 16:49
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    $\begingroup$ @PaulCusson: Yes, but the list of diffeomorphism types of connected, homogeneous, simply-connected $3$-manifolds has a much simpler statement: There are three examples: $\mathbb{R}^3$, $S^2\times\mathbb{R}$, and $S^3$. $\endgroup$
    – Robert Bryant
    Sep 3, 2022 at 20:39
  • $\begingroup$ See (in Russian) "О трехмерных однородных пространствах" ("On three-dimensional homogeneous spaces") mathnet.ru/links/d34f0956c834ac21cc1b8db76cb82e88/smj3925.pdf $\endgroup$
    – Vladimir47
    Oct 13, 2023 at 13:15

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