Is it possible to realize the Moebius strip as a linear group orbit?

Question

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.

Answer 1

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.

Answer 2

Here is another solution, using the special Euclidean group $\operatorname{SE}(2) := \operatorname{SO}(2) \ltimes \mathbb{R}^2$ instead of Robert Bryant's solution which uses $\operatorname{SO}(2,1)$.

Let $\operatorname{SE}(2)$ act on $\mathbb{R}^2$ in the usual way. Let $V$ be the vector space of (inhomogenous) polynomials of degree $\leq 2$ on $\mathbb{R}^2$, so $\operatorname{SE}(2)$ acts on $V$. Then the orbit of the polynomial $x^2$ is in bijection with the set of lines. (Namely, the zero locus of each such polynomial is a line, and, given a line $\ell$, the function $d(\ell,(x,y))^2$ is a quadratic polynomial in $(x,y)$.) So the orbit of $x^2$ is the space of lines in $\mathbb{R}^2$, which is a Mobius strip.

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Here is the way I would think about this. Let $M$ be a smooth manifold and let $G$ be a group acting on $M$. Let $W$ be any finite dimensional subspace of $C^{\infty}(M)$. Then sending $x \in M$ to the "evaluation at $x$" gives a smooth map $M \to W^{\vee}$. If $W$ is $G$-invariant, then $W^{\vee}$ inherits a $G$-action and the map is $G$-equivariant. Unless we are very unlucky, the map is an embedding.

So I would start by thinking about a group $G$ acting on the Mobius strip $M$, take some function $f \in C^{\infty}(M)$ and see if the $G$-orbit of $f$ spans a finite dimensional vector space. That is how I found the above example, thinking about functions on the space of lines like "slope" and "distance from the origin", until I discovered that "square of distance to the origin" worked.