Is the Moebius strip a linear group orbit? In other words:
Does there exists a Lie group $ G $, a representation $ \pi: G \to GL(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 $ 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
$$
SE_2= \left \{ \
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1
\end{bmatrix} : a^2+b^2=1 \right \}
$$
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
$$
SE_2 \mathbin{/} \langle V, \tau \rangle
$$
is the Moebius strip. Indeed if we consider the model of the Moebius strip as the manifold of affine lines in the plane then the subgroup $ <V,\tau> $ is exactly the stabilizer of the $ y $-axis.
Best Answer
The answer is yes.
Take $ O_{2,1} $ and act on the second symmetric power of the standard rep $ \mathbb{R}^3 $ and the orbit of $ e_1 \otimes e_1 $ works where $ e_1,e_2,e_3 $ is standard basis.
Or take natural $ E_2 $ action of affine transformations on 6d space of all polynomials in $ x,y $ of degree 2 or less. Then the orbit of $ x^2 $ works.
Both of these answers are copied directly from
https://mathoverflow.net/questions/414402/is-it-possible-to-realize-the-moebius-strip-as-a-linear-group-orbit