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Setting: We have a connected Lie group $G$ and a smooth map $G \to GL(V)$, where $V$ is a finite-dimensional vector space. Are the orbits of $G$ on $V$ embedded submanifolds? More precisely, if one has $v \in V$, and $H \subset G$ is the stabilizer of $V$, then $H$ is a Lie subgroup of $G$ and there is an injective immersion $G/H \rightarrow V$ whose image is the orbit of $v$. (This is true generally for a smooth action). Is this map always an embedding?

I know that the orbits need not be closed submanifolds of $V$.

If this is not true, are there straightforward conditions under which one can guarantee that it is true? Certainly $G$ compact is enough, but this is too strong. $G$ reductive, perhaps?

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  • $\begingroup$ rotations fix the origin..... $\endgroup$ – Will Jagy Oct 18 '14 at 0:23
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    $\begingroup$ Yeah, so the origin is one orbit, which is an embedded 0-manifold. $\endgroup$ – Alex Zorn Oct 18 '14 at 0:30
  • $\begingroup$ Alright, was not sure you would view it that way. Trying to picture something about points in $\mathbb R^2$ other than the origin moving along logarithmic spirals $\endgroup$ – Will Jagy Oct 18 '14 at 0:38
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    $\begingroup$ Can't you just take a lie subgroup G of GL acting on itself by multiplication from the left, where G is a (non-embedded) Lie subgroup? Then V is the space of all matrices of dimension $n$. $\endgroup$ – PhoemueX Oct 18 '14 at 5:09
  • $\begingroup$ Ah. So for example you could have $\mathbb{R}$ acting on $\mathbb{R}^3$ as simultaneous rotations about the x and z axis in such a way that the orbits are nonperiodic trajectories on the sphere. $\endgroup$ – Alex Zorn Oct 18 '14 at 15:22

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