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