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.

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    $\begingroup$ Your previous related question: mathoverflow.net/questions/410275/… $\endgroup$ Jan 21, 2022 at 23:40
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    $\begingroup$ This is almost the same as the previous question Sam Hopkins links to. The Moebius band can be thought of as the space of affine $1$-dimensional subspaces of $\mathbb R^2$, i.e. the space of "infinite lines in the Euclidean plane". From this point of view, the group of Euclidean symmetries acts transitively. So yes, it's a group orbit. $\endgroup$ Jan 22, 2022 at 0:19
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    $\begingroup$ @RyanBudney Although Moebius band is certainly an orbit for a linear group, $ SE_2 $, that does not make it a linear group orbit (bad terminology :( I got it from mathoverflow.net/questions/206618/…). To see the distinction, note that Klein bottle is the orbit of a group action of $ SE_2 $ but Klein bottle cannot be a linear group orbit since its fundamental group does not have a finite commutator subgroup (see same linked question). The distinction: orbit of generic smooth group action vs orbit of linear group action on a vector space $\endgroup$ Jan 22, 2022 at 0:56
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    $\begingroup$ The first sentence of my question defines exactly what a linear group orbit is but it was a huge oversight on my part not to say that that was what I was defining. My apologies. I will edit it now adding the preamble "Is the Moebius strip a linear group orbit? In other words:" And yes $ V $ is a vector space that is why the question said "a vector $ v \in V $ " $\endgroup$ Jan 22, 2022 at 1:25
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    $\begingroup$ @DavidESpeyer The same way a point on the Moebius strip is an affine line in the plane a point on the Klein bottle is a set of all the lines parallel to some affine line in the plane and an integer distance apart. In particular, the same way that the subgroup $ <V,\tau> $ above is the stabilizer of the y axis the subgroup $ <V,\tau,b> $ is the stabilizer of the set of all vertical lines with integer x intercept (here $ b $ just translates by 1 in the x direction). More details in my post math.stackexchange.com/questions/4316503/… $\endgroup$ Jan 22, 2022 at 15:27

2 Answers 2


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.

  • $\begingroup$ Let $D$ be one of the components of a hyperboloid of $2$-sheets in your same $\mathbb{R}^3$. Then we can think of $D$ as the hyperbolic plane, and we can identify $H/\langle \pm 1 \rangle$ as the space of lines in $D$. So one way to describe your construction is "the symmetry group of the hyperbolic plane acts on the set of lines in the hyperbolic plane". $\endgroup$ Jan 22, 2022 at 14:22
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    $\begingroup$ @DavidESpeyer: Yes, but the whole point is to represent the space of lines as a group orbit of a point in a representation of the group, not just to identify it as a homogeneous space. $\endgroup$ Jan 22, 2022 at 14:49
  • $\begingroup$ Of course! On my december 8th question "Moebius strip Riemannian homogeneous?" (its not) Ian Agol commented that it is Lorentzian homogeneous as space of lines in the hyperbolic plane. That seemed important but I wasn't smart enough to put it all together. This is a beautiful flowing answer thanks so much. I wonder if this extends to a general fact that every pseudo Riemannian homogeneous manifold is a linear group orbit? At least every Riemannian homogeneous manifold is a linear group orbit basically bc isometry group is compact then Mostow-Palais realizes every orbit for some orthogonal rep $\endgroup$ Jan 22, 2022 at 16:10
  • $\begingroup$ @IanGershonTeixeira's referenced question Is the Moebius strip Riemannian homogeneous? and @‍IanAgol's comment. $\endgroup$
    – LSpice
    Jan 22, 2022 at 17:22
  • $\begingroup$ @IanGershonTeixeira: Maybe you will be interested in this question: In David Speyer's example, $\mathrm{SE}(2)$ preserves a foliation on the Möbius strip $M$ by lines but no (pseudo-)Riemannian metric, while, in my example, $\mathrm{O}(2,1)$ preserves a pseudo-Riemannian metric on $M$ but no foliation. These are both 3-dimensional groups, but $M$ can be written 'even more' homogeneously as $P/H$ where $P\subset \mathrm{GL}(3,\mathbb{R})$ has dimension 6 and $H$ has dimension $4$. Is there a $P$-representation $\rho:P\to\mathrm{GL}(V)$ and a $v\in V$ whose $P$-stabilizer is $H$? $\endgroup$ Jan 23, 2022 at 12:07

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.

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.

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    $\begingroup$ You might say, if the function $f$ is a polynomial and the Lie group acts as projective or affine transformations, as most that I can think of typically do, then clearly the degree of $f$ is preserved, so an orbit inside the finite dimensional vector space of polynomials of that degree. $\endgroup$
    – Ben McKay
    Jan 22, 2022 at 15:09
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    $\begingroup$ In your construction, $V$ has dimension $6$, but that can be improved by noticing that your group preserves the 1-dimensional subspace of constant functions on the plane, so one can reduce to the $5$-dimensional quotient and still get a representation in which the orbit of $x^2$ is a Möbius strip. Interestingly, this latter representation is reducible (unlike the $5$-dimensional representation that appears at the end in my construction), as it contains (the image of) the linear functions, but the orbit of $x^2$ in this reduced space is just a circle. $\endgroup$ Jan 22, 2022 at 15:18
  • $\begingroup$ Wow this is also super interesting and I really appreciate you walking me through your thought process that is always the best part! Thank you for introducing the perspective of finite dimensional vector spaces of smooth functions ( side note: If I am not mistaken slope is not a well defined smooth function because of vertical lines?). I especially love @BenMcKay perspective about polynomials and how linear $ x \to ax+by $ and affine $ x \to ax+by+c $ transformations don't increase degree so naturally lead to finite dimensional representations on space of polynomials of bounded degree. $\endgroup$ Jan 22, 2022 at 18:57

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