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Say $G$ is a connected Lie group. The orbits in any linear representation are of the form $G/H$, where $H$ is the stabilizer of any element of the orbit. Which $G/H$ appear this way?

edit: It seems the compact case works whenever $G/H$ is nice enough.

Actually it works very generally. There is a theorem of Mostow ( https://www.jstor.org/stable/1970055 ) which proves if $G$ is a compact Lie group acting on a separable finite-dimensional space $E$ with finitely many orbit types then $E$ embedds equivariantly into a finite dimensional representation.

Still, I wonder if there's a simple construction for embedding $G/H$.

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    $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$
    – Shaun
    Apr 26 at 19:54
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    $\begingroup$ Do you mean continuous finite dimensional complex representations, or what? $\endgroup$
    – reuns
    Apr 26 at 20:15
  • $\begingroup$ @Shaun I think it is an interesting question, and sufficiently precise as stated. The context would be a distraction to the mathematicians in the audience, and my thoughts have so far lead nowhere and would be irrelevant to include. $\endgroup$
    – wzzx
    Apr 26 at 20:43
  • $\begingroup$ @reuns Yes let's take real representations, $G \to GL(N,\mathbb{R})$ (which includes complex ones). $\endgroup$
    – wzzx
    Apr 26 at 20:45
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    $\begingroup$ I do not really understand the question: Are you asking for an easy proof of Mostow's theorem? Are you asking for a complete classification of pairs $(G,H)$ for which such a representation exists? $\endgroup$ Apr 27 at 0:45

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I can prove it by expanding on the $L^2(G/H,\mathbb{R})$ idea, so long as $G$ is compact, $H<G$ is closed (in this case $G/H$ is a smooth manifold), and $H$ has finitely many fixed points. Probably there's a more elegant argument but this is what I came up with.

Let me choose a basepoint $\bar e \in G/H$ with stabilizer $H$. Take any smooth function $f \in L^2(G/H,\mathbb{R})$ which attains its unique maximum at $\bar e$. By averaging $f$ over the $H$ action we can suppose $f$ is $H$-symmetric. By construction, the stabilizer of $f$ in $L^2(G/H,\mathbb{R})$ is $H$, so its orbit is $G/H$. This proves the infinite dimensional version, now I want to reduce to a finite dimensional representation.

The strategy will be to look for $f$ among the harmonic functions (ie. those with $\nabla^2 f = \lambda f$ using a $G$-invariant metric), since these give a nice decomposition of $L^2(G/H,\mathbb{R})$ into finite dimensional representations for each allowed $\lambda$. We will find $G/H$ appears as an orbit inside a finite dimensional subrep of $L^2(G/H,\mathbb{R})$.

So let $f'$ be a harmonic with eigenvalue $\lambda > 0$. In particular $f'$ is not constant. We can use the $G$ action to choose $f'$ so that $\bar e$ is a (global) maximum. Now, since $\nabla^2$ is $G$-invariant one can show that $f$ averaged over $H$ is also harmonic with eigenvalue $\lambda$, so we can suppose $f'$ is $H$ invariant.

However, we are not quite done because there may be other global maxima at the other fixed points of $H$. I think that this can be avoided by choosing $\lambda$ appropriately. Instead, as a hack I will try to find finitely many $H$-symmetric harmonics to add to $f'$ to get one with a unique maximum at $\bar e$, which is not an eigenfunction perhaps but only finitely many $\lambda$ appear, so it is a vector in a finite dimensional subrep.

The harmonic functions are a basis for arbitrary $L^2$ functions and the averaging argument shows that $H$-symmetric harmonics are a basis for $H$-symmetric $L^2$ functions. In particular the function $f$ constructed previously is a sum of $H$-symmetric harmonics, which cannot each take all the same values on the $H$ fixed points. If there are finitely many fixed points, some finite combination of these functions plus $f'$ will have a unique maximum at $\bar e$, and therefore have stabilizer $H$. Phew.

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