On MSE this got 5 upvotes but no answers not even a comment so I figured it was time to cross-post it on MO:


Is the Moebius strip a linear group orbit? In other words:


**Does there exists a Lie group $ G $ a representation $ \pi: G \to \operatorname{Aut}(V) $ and a vector $ v \in V $ such that the orbit
$$
\mathcal{O}\_v=\{ \pi(g)v: g\in G \} 
$$
is diffeomorphic to the Moebius strip?**


My thoughts so far:


The only two obstructions I know for being a linear group orbit is that the manifold (1) must be smooth homogeneous (shown below for the the group $ \operatorname{SE}\_2 $) and (2) must be a vector bundle over a compact Riemannian homogeneous manifold (here the base is the circle $ S^1 $).


The Moebius strip is homogeneous for the special Euclidean group of the plane
$$
\operatorname{SE}\_2= \left \{ \
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1 
\end{bmatrix} : a^2+b^2=1 \right \}.
$$
There is a connected group $ V $ of translations up each vertical line
$$
V= \left \{ \ 
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & y \\
0 & 0 & 1 
\end{bmatrix} : y \in \mathbb{R} \right \}.
$$
Now if we include the rotation by 180 degrees
$$
\tau:=\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 
\end{bmatrix}
$$
then $ \langle V, \tau \rangle$ has two connected components and
$$
\operatorname{SE}\_2/\langle V, \tau \rangle
$$
is the Moebius strip.



Is it possible to realize the Moebius strip as a linear group orbit?
Yes. Here is one way: Consider standard $\mathbb{R}^3$ endowed with the Lorentzian quadratic form $Q = x^2+y^2-z^2$, and let $G\simeq\mathrm{O}(2,1)\subset\mathrm{GL}(3,\mathbb{R})$ be the symmetry group of $Q$. Then $G$ preserves the hyperboloid $H$ of $1$-sheet given by the level set $Q=1$, which is diffeomorphic to a cylinder. Consider the quotient of $H$ by $\mathbb{Z}\_2$ defined by identifying $v\in H\subset\mathbb{R}^3$ with $-v$. This abstract quotient is a smooth Möbius strip.


This quotient can be identified as a linear group orbit as follows: Let $V = S^2(\mathbb{R}^3)\simeq \mathbb{R}^6$ and consider the smooth mapping $\sigma:\mathbb{R}^3\to V$ given by $\sigma(v) = v^2$ for $v\in\mathbb{R}^3$. Then $\sigma$ is a $2$-to-$1$ immersion except at the origin. The action of $G$ on $\mathbb{R}^3$ extends equivariantly to a representation $\rho:G\to \mathrm{Aut}(V)$ such that $\rho(g)(v^2) = \rho\bigl(\sigma(v)\bigr)=\sigma(g v)= (gv)^2$. It follows that $\sigma(H)\subset S^2(\mathbb{R}^3)\simeq\mathbb{R}^6$, which is a Möbius strip, is a linear group orbit under the representation $\rho$.


Note that the representation of $G$ on $S^2(\mathbb{R}^3)\simeq\mathbb{R}^6$ is actually reducible as the direct sum of a trivial $\mathbb{R}$ and an irreducible $\mathbb{R}^5$. Projecting everything into the $\mathbb{R}^5$ factor, one obtains a representation of $G$ on $\mathbb{R}^5$ that has a Möbius strip as a $G$-orbit.


