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

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.

• Your previous related question: mathoverflow.net/questions/410275/… Jan 21 at 23:40
• 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. Jan 22 at 0:19
• @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 Jan 22 at 0:56
• 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 $v \in V$ " Jan 22 at 1:25
• @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/… Jan 22 at 15:27

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.

• 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". Jan 22 at 14:22
• @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. Jan 22 at 14:49
• 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 Jan 22 at 16:10
• @IanGershonTeixeira's referenced question Is the Moebius strip Riemannian homogeneous? and @‍IanAgol's comment. Jan 22 at 17:22
• @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 $H$? Jan 23 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.

• 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. Jan 22 at 15:09
• 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. Jan 22 at 15:18
• 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. Jan 22 at 18:57