Let $G$ be a closed subgroup of $O(n)$ such that $\mathbb R^n/G$ is isometric to $\mathbb R^{n-2} \times \mathbb R_+$. Can we have a classification of $G$ up to conjugation?
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3$\begingroup$ Probably up to conjugation is more interesting. Nevertheless $G$ should have 1-dimensional orbits and this should be quite restrictive. $\endgroup$YCor– YCor2021-05-30 07:58:57 +00:00Commented May 30, 2021 at 7:58
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The generic orbits are 1-dimensional.
The action might have fixed points that correspond to the boundary of the space of orbits $\mathbb{R}^{n-2}\times\mathbb{R}_+$, or these orbits are 1-dimensional as well.
In the first case, it is a subaction of $O(2)$ acting on the first factor of $\mathbb{R}^2\oplus\mathbb{R}^{n-2}$.
In the second case, the action is an extension of $\mathbb{R}$-action by parallel translations.
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$\begingroup$ Oops, since it is a subgroup of $O(n)$, you only have the first case. $\endgroup$Anton Petrunin– Anton Petrunin2021-05-30 20:23:11 +00:00Commented May 30, 2021 at 20:23
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$\begingroup$ There are plenty of other possible actions (e.g., $SO(2)$ acting diagonally on $R^2\oplus R^2$), in many possible ways. How do you discard them? $\endgroup$YCor– YCor2021-05-31 13:03:24 +00:00Commented May 31, 2021 at 13:03
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$\begingroup$ @YCor I only claim that it is a subgroup of $O(2)$ that acts on the first factor of $\mathbb{R}^2\oplus\mathbb{R}^{n-2}$. Since the orbit is 1-dimensional, it has to be either $SO(2)$ or $O(2)$. $\endgroup$Anton Petrunin– Anton Petrunin2021-05-31 21:49:57 +00:00Commented May 31, 2021 at 21:49
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$\begingroup$ Why not $O(2)$ or $SO(2)$ $\times$ a finite group? $\endgroup$YCor– YCor2021-05-31 21:52:21 +00:00Commented May 31, 2021 at 21:52
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$\begingroup$ @YCor if a group acting nontrivially on the $\mathbb{R}^{n-2}$-factor, then the quotient cannot have an isometric copy of $\mathbb{R}^{n-2}$. $\endgroup$Anton Petrunin– Anton Petrunin2021-05-31 21:56:33 +00:00Commented May 31, 2021 at 21:56