Does the Mobius Strip have an homogenous embedding?

Question

So in this question I’m trying to do two things at once. 1. Define what a “homogenous embedding” is by describing what it is like and then furthermore ask if such an embedding exists for the mobius strip.

We start with the first part. A manifold is homogenous, loosely, if every point is like every other point in a “rigid” sense ie there are isometries of the manifold sending one point to the other. Examples of this are the following:

1. The real line, or any line in $\mathbb{R}^n$. any point can be mapped to any other point by shifting the line. So all points are kind of the same.

2. The perimeter of a circle embedded in $\mathbb{R}^2$, any point can be mapped to any other point by rotation so again all points are kind of the same.

3. The surface of a sphere (which naturally embeds in $\mathbb{R}^3$ or higher, any point can clearly be rotated to any other point.

4. The surface of an infinite cylinder embedded in $\mathbb{R}^3$. We can translate or rotate any point to any other (notice with a finite cylinder there is a clear difference between boundary points and interior points so this really HAS to be infinite and can be viewed as $S_1 \times R$ where the homogenous group actions from those factors gives rise to the homogenous group action for the whole space)

5. A torus in $\mathbb{R}^4$ this one is a little tricky to see. In $\mathbb{R}^3$ the torus isn’t homogenous because most obvious the inner perimeter has a different length than the outer perimeter. In $\mathbb{R}^4$ it becomes possible to draw a torus so that all perimeters are identical in length and so we can really, say all points are identical with a natural action of $S_1 \times S_1$ acting on them.

So now we get to the meat of the question. What about the mobius strip? Is there a way to embed an infinite mobius strip in $\mathbb{R}^4$ or higher so that it doesn’t self intersect and really is homogenous? IE every point is basically identical, meaning for any pair of points there is a function from the manifold to itself that preserves all relative distances and maps one point to the other?

It’s not clear to me if this is possible but I do feel optimistic.

Answer 1

Yes, there is an embedding, although as pointed out in the comments of the question and in response to this answer, the homogeneous property fails due to non-orientability. This infinite Möbius strip is often referred to as the nontrivial line bundle over the circle and can be thought of abstractly as the space $$ X = ([0, 1] \times \mathbb{R})/{\sim}, $$ where the equivalence classes are singleton points except at the boundary, where $(0, t) \sim (1, -t)$ for all $t \in \mathbb{R}$.

How to embed this space isometrically in some $\mathbb{R}^N$ though? Just as there's an isometric embedding of the torus $S^1 \times S^1 \hookrightarrow \mathbb{R}^4$ built out of the embedding $S^1 \hookrightarrow \mathbb{R}^2$ (see the Clifford torus embedding), we can do something similar for $X$. Here's a parametrization: $$ \begin{cases} x = \cos s \\ y = \sin s \\ z = t \cos \frac{1}{2}s \\ w = t \sin \frac{1}{2}s \end{cases} \qquad s \in [0, 2\pi], \quad t \in \mathbb{R} $$ This describes a line in the $zw$-plane of slope $\tan \frac{1}{2}s$ "over" the point $(\cos s, \sin s)$ in the $xy$-plane. Notice that the line is parametrized in opposite directions at $s=0$ vs $s=2\pi$: $(z, w) = (t, 0)$ vs $(z, w) = (-t, 0)$, respectively.

As for your rigid homogeneity, any translation in the parameters $(s, t)$, where $s$ is considered modulo $2\pi$, i.e. translation by an element of the Lie group $S^1 \times \mathbb{R}$, will map a point in your space to another point with identical local geometry.

Answer 2

I think you're confounding "homogeneous" and something about embeddings.

The cylinder with the product metric is (metrically) homogeneous because for any point $(\alpha, C)$, the map $$ (\theta, s) \mapsto ((\theta + \alpha) \bmod 2 \pi, s + C) $$ sends the point $(0, 0)$ to $(\alpha, C)$ in a metric-preserving way.

The embedded cylinder in $3$-space, $$ \{(\cos t, \sin t, u) \mid 0 \le t < 2\pi, u \in \mathbb R \} $$ happens to be a lot like that abstract manifold, and there is a linear isometry taking any point (like, say $P = (1, 0, 0)$) to any other point $(\cos t, \sin t, u)$, where $t$ and $u$ are fixed, and taking the whole cylinder to itself; the isometry in this case is $(x, y, z) \mapsto (\cos(a + t), \sin(a + t), z+u)$m where $a$ is any angle whose cosine and sine are $x$ and $y$ respectively.

There might be some other isometric embedding of the abstract cylinder that does not admit any linear isometry of 3-space taking each point to each other point (e.g. if you did an isometric embedding of the circle as an eccentric ellipse rather than a as a round circle).

So the first question I think you need to ask is "is the infinite mobius strip with the "product" metric in fact homogeneous in the first sense?"

Then there's the followup: is there an isometric embedding $f$ of $M$ in some $\mathbb R^k$ with the property that for any $P, Q \in F(M)$, there's a linear isometry $T_{PQ}$ of $\mathbb R^k$ such that

1. $T(P) = Q$
2. $T(f(M)) = f(M)$ as sets

My own offhand guess about the first question is that the answer is "no" -- (1) the "centerline" of the Mobius band is special and (2) if you want to take a (non-centerline) point $A$ on some ray to a different (non-centerline) point $B$ on that ray "stretching all rays by the same amount" doesn't work. So the right question is probably something like this:

"For certain pairs $P, Q$ of points on the mobius strip $M$, there's an isometry $S_{PQ}$ of $M$ taking $P$ to $Q$. Is there an embedding $f$ of $M$ in $\mathbb R^k$, for some $k$, with the property that for every such pair, there's a linear isometry $T_{PQ}$ of $\mathbb R^k$ such that $$ T_{PQ}(f(A)) = f(S_{PQ}(A)) $$ for all $A \in M$?"

Is that in fact what you want? If so, then I have a little bit of an answer: Writing $M$ as $[0, 2\pi] \times \mathbb R / \sim$, where "$\sim$" is the smallest equivalence relation for which $(0, t) \sim (2\pi, -t)$, I suspect that the isometries of $M$ look like (a) "rotation in the $S^1$ direction, with flipping as you pass $2\pi$," and (b) reflection ($t \mapsto -t$) in the $\mathbb R$ direction, and combinations of these.

In that case, the desired embedding in 4-space is a lot like the flat torus. You send the point $(\alpha, t)$ to $$ (\cos \alpha, \sin \alpha, t\cos \frac{\alpha}{2}, t\sin \frac{\alpha}{2}) $$ and the "rotational" isometries are realized by multiplication by matrices like $$ \begin{pmatrix} \cos \theta & -\sin \theta & 0 & 0 \\ \sin \theta & -\cos \theta & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ while the "reflection" isometry is represented by multiplication by $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} $$

Answer 3

The accepted answer is simply wrong, there are no such things as "translations in the parameters (,), where is considered modulo 2." I will prove below that the open Moebius band does not admit any homogeneous Riemannian metrics. However, Moebius band does admit flat metrics, which are (obviously) locally homogeneous, i.e. for any two points $p, q\in M$ there are neighborhoods $U, V$ of, respectively, $p, q$, and an isometry $U\to V$.

Let's first convince ourselves that "translations" do not exist for any values of the parameter $t\ne 0$ and $s=0$. The supposed formula for such a translation is $$ T: (x,y)\mapsto (x, y+t). $$ However, $(0,y)$ is equivalent to $(1,-y)$ and we get $T(1,-y)=(1, -y+t)$. But for $T$ to be well-defined as a map of the Moebius band, you have to have that $T(2\pi,-y)=(2\pi, -y+t)$ is equivalent to $(0, y+t)$, implying that $-y+t= -(y+t)$, i.e. $t=0$.

Now, let's proceed with a general proof. Suppose that $g$ is a homogeneous Riemannian metric on the Moebius band $M$. The actual definition of homogeneity simply means that the isometry group of the Riemannian manifold $(M,g)$ acts transitively on $M$, i.e. for any two points $p, q\in M$ there exists an isometry $f: (M,g)\to (M,g)$ sending $p$ to $q$ (you will find this definition in pretty much any textbook on Riemannian geometry). It is known that every homogeneous Riemannian manifold is complete (see for instance, this question). Since $M$ is noncompact, completeness of $(M,g)$ implies that it has infinite diameter (this follows for instance from the Hopf-Rinow theorem). Now, let $C$ denote the central circle of the Moebius band. More precisely, in the notation of the accepted answer, this is the projection to $M$ of the interval $$ \{ [x, 0]: x\in [0,1]\}. $$ The circle $C$ has the property that for every self-diffeomorphism $f: M\to M$, $f(C)\cap C\ne \emptyset$ (same is true for self-homeomorphisms). You need to know a bit of differential or algebraic topology to verify this. (Briefly, the mod 2 fundamental class of $C$ generates $H_1(M, Z_2)\cong Z_2$ and the self-intersection of this class is nonzero.)

Now, suppose that $g$ is a homogeneous Riemannian metric on $M$ and $d$ is the corresponding distance function on $M$. Let $D$ denote the diameter of $C$ in $(M,d)$, i.e. $$ D=\max_{p, q\in C} d(p, q). $$ For every isometry $f$ of $(M, g)$, the diameter of $f(C)$ is also $D$. Now, using the fact that $M$ has infinite diameter, find a point $q\in M$ such that $d(q, C)\ge 2D$, where $$d(q,C)=\max\{ d(q,p): p\in C\}.$$ Let $f$ be an isometry of $(M,g)$ such that $q\in f(C)$ (such an isometry exists by the homogeneity assumption). Since $f(C)\cap C\ne \emptyset$, let $p$ denote a point of this intersection. Then $$2D\le d(q, C)\le d(q,p)\le D.$$ A contradiction.

Answer 4

Although, as Moishe Kohan has pointed out, the open Moebius strip has no homogeneous Riemannian metric, there is a metric tensor $g$ for the open Moebius strip $M$ giving it a transitive group of isometries. However, this metric tensor is indefinite, and doesn't come from an embedding of the strip into ${\Bbb R}^n$. (Added: That is, the construction in this post doesn't construct it from an embedding. As Moishe Kohan has pointed out in the comments, you could find an isometric embedding of $(M, g)$ into some pseudo-Euclidean space ${\Bbb R}^{p,q}$. Given the poster's question, though, it seems better to find some metric on $M$ so that it can be embedded into ${\Bbb R}^{p,q}$ and so that the group of isometries of ${\Bbb R}^{p,q}$ induces a transitive group of isometries on $M$. I don't know whether that's possible for this metric or for any other.)

To see this, coordinatize the Moebius strip by $t$, $u\in \Bbb R$, where we identify the points $(t, u)$ and $(t + \pi, -u)$, coordinatize ${\Bbb R}^3$ by $x$, $y$ and $z$, and give ${\Bbb R}^3$ the metric $$ ds^2=-dx^2+dy^2+dz^2. $$

We also identify $(x,y,z)\sim (-x,-y,-z)$ in ${\Bbb R}^3$. Then we can embed the Moebius strip into ${\Bbb R}^3/\sim$ via $$ (t,u)\mapsto (\sinh u, \cosh u \cos t, \cosh u \sin t) $$ and give it the induced metric. The image of the strip is the quotient of the one-sheeted hyperboloid $$ H=\{(x,y,z)\mid -x^2+y^2+z^2=1\} $$ by $\sim$. Any matrix $N\in O_{1,2}({\Bbb R})$ (meaning any 3 by 3 real matrix $N$ with $$N^T\left(\begin{array}{ccc}-1 & \ & \ \\ \ & 1 & \ \\ \ & \ & 1 \end{array}\right)N=\left(\begin{array}{ccc}-1 & \ & \ \\ \ & 1 & \ \\ \ & \ & 1 \end{array}\right)),$$ will preserve the metric in ${\Bbb R}^3$ and send $H$ into itself; also, $v\sim w$ implies that $Nv\sim Nw$. So, any such $N$ induces an isometry on the Moebius strip. Then, the isometry group of the strip is transitive since any point on $H$ can be sent to a point $(x, y, 0)$ on $H$ via some rotation around the $x$ axis, $$ \left(\begin{array}{ccc}1 & \ & \ \\ \ & \cos \theta & \sin \theta\ \\ \ & -\sin \theta & \cos \theta \end{array}\right)\in O_{1,2}({\Bbb R}), $$ and any point $(x, y, 0)$ on $H$ can be sent to $(0, 1, 0)$ via the matrix $$ \left(\begin{array}{ccc}y & -x & \ \\ -x & y & \ \\ \ & \ & 1 \end{array}\right)\in O_{1,2}({\Bbb R}). $$