Is there a smooth embedding of $S^2$ into some Euclidean space that is equivariant with respect to a linear representation of $PSL(2,\mathbb C)$?
A counterexample to a more general question can be found [here.](https://mathoverflow.net/questions/149867/is-there-an-analogue-of-mostow-palais-equivariant-embedding-theorem-for-noncompa)



An analogue of Mostow-Palais equivariant embedding theorem for the group of conformal automorphisms of the 2-sphere
The answer is 'no'. In fact, a stronger statement is true: If $V$ is a finite dimensional vector space and $G\subset\mathrm{GL}(V)$ is a (connected) non-compact simple Lie group, then the only bounded orbits of $G$ are fixed points. *A fortiori*, the only compact $G$-orbits are fixed points.


To see this, it's enough to prove it for $G = \mathrm{SL}(2,\mathbb{R})$, since every connected non-compact simple Lie group contains a copy of $\mathrm{SL}(2,\mathbb{R})$ (or its quotient $\mathrm{SO}(2,1)$).


Moreover, it is enough to prove the claim when $V$ is an irreducible representation of $\mathrm{SL}(2,\mathbb{R})$, so assume that. Thus, $V = S^d(W)$ where $W\simeq\mathbb{R}^2$ is the standard $2$-dimensional representation of $\mathrm{SL}(2,\mathbb{R})$, with basis elements $x$ and $y$. The diagonal subgroup of $\mathrm{SL}(2,\mathbb{R})$, a copy of $\mathbb{R}^\times$, acts on $x$ and $y$ as $r \cdot x = rx$ and $r\cdot y = r^{-1}y$ for $r\in\mathbb{R}^\times$. An element of $S^d(W)$ is a polynomial in $x$ and $y$, homogeneous of degree $d$, and we have $r\cdot (x^k y^{d-k}) = r^{2k-d} x^k y^{d-k}$. Thus, the only element of $S^d(W)$ that could have a bounded $G$-orbit would be $x^{d/2}y^{d/2}$ (and that exists only when $d$ is even). However, unless $d=0$, the element $(xy)^{d/2}$ (when it exists) has its $G$-orbit containing polynomials that are not multiples of $(xy)^{d/2}$, and hence their $G$-orbits are not bounded, so the $G$-orbit of $(xy)^{d/2}$ is not bounded either.


