are all $G/H$ embeddable in a linear representation? 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 $G/H$.
 A: 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.
