Möbius strip

If a smooth Möbius strip in three-space is a rectangular one – that is, created from identifying two opposite sides of a geometrical rectangle with bending but not stretching the surface – then such an embedding is known to be possible if the aspect ratio of the rectangle is greater than ${\displaystyle {\sqrt {3}}}$, with the shorter sides identified. (For a smaller aspect ratio, it is not known whether a smooth embedding is possible.) As the aspect ratio decreases toward ${\displaystyle {\sqrt {3}}}$, any such embedding seems to approach a shape that can be thought of as a strip of three equilateral triangles, folded on top of one another to occupy an equilateral triangle.

If the Möbius strip in three-space is only once continuously differentiable (class C¹), however, then the theorem of Nash-Kuiper shows that no lower bound exists.

A method of making a Möbius strip from a rectangular strip too wide to simply twist and join (e.g., a rectangle only one unit long and one unit wide) is to first fold the wide direction back and forth using an even number of folds—an "accordion fold"—so that the folded strip becomes narrow enough that it can be twisted and joined, much as a single long-enough strip can be joined.[10] With two folds, for example, a 1 × 1 strip would become a 1 × ⅓ folded strip whose cross section is in the shape of an 'N' and would remain an 'N' after a half-twist. This folded strip, three times as long as it is wide, would be long enough to then join at the ends. This method works in principle, but becomes impractical after sufficiently many folds, if paper is used. Using normal paper, this construction can be folded flat, with all the layers of the paper in a single plane, but mathematically, whether this is possible without stretching the surface of the rectangle is not clear.[11]

To turn a rectangle into a Möbius strip, join the edges labelled A so that the directions of the arrows match.

Topologically, the Möbius strip can be defined as the square ${\displaystyle [0,1]\times [0,1]}$ with its top and bottom sides identified by the relation ${\displaystyle (x,0)\sim (1-x,1)}$ for ${\displaystyle 0\leq x\leq 1}$, as in the diagram.

A less used presentation of the Möbius strip is as the topological quotient of a torus.[12] A torus can be constructed as the square ${\displaystyle [0,1]\times [0,1]}$ with the edges identified as ${\displaystyle (0,y)\sim (1,y)}$ (glue left to right) and ${\displaystyle (x,0)\sim (x,1)}$ (glue bottom to top). If one then also identified (x, y) ~ (y, x), then one obtains the Möbius strip. The diagonal of the square (the points (x, x) where both coordinates agree) becomes the boundary of the Möbius strip, and carries an orbifold structure, which geometrically corresponds to "reflection" – geodesics (straight lines) in the Möbius strip reflect off the edge back into the strip. Notationally, this is written as T²/S₂ – the 2-torus quotiented by the group action of the symmetric group on two letters (switching coordinates), and it can be thought of as the configuration space of two unordered points on the circle, possibly the same (the edge corresponds to the points being the same), with the torus corresponding to two ordered points on the circle.

The Möbius strip is a two-dimensional compact manifold (i.e. a surface) with boundary. It is a standard example of a surface that is not orientable. In fact, the Möbius strip is the epitome of the topological phenomenon of nonorientability. This is because two-dimensional shapes (surfaces) are the lowest-dimensional shapes for which nonorientability is possible and the Möbius strip is the only surface that is topologically a subspace of every nonorientable surface. As a result, any surface is nonorientable if and only if it contains a Möbius band as a subspace.

The Möbius strip is also a standard example used to illustrate the mathematical concept of a fiber bundle. Specifically, it is a nontrivial bundle over the circle S¹ with its fiber equal to the unit interval, I = [0, 1]. Looking only at the edge of the Möbius strip gives a nontrivial two point (or Z₂) bundle over S¹.

A simple construction of the Möbius strip that can be used to portray it in computer graphics or modeling packages is:

• Take a rectangular strip. Rotate it around a fixed point not in its plane. At every step, also rotate the strip along a line in its plane (the line that divides the strip in two) and perpendicular to the main orbital radius. The surface generated on one complete revolution is the Möbius strip.
• Take a Möbius strip and cut it along the middle of the strip. This forms a new strip, which is a rectangle joined by rotating one end a whole turn. By cutting it down the middle again, this forms two interlocking whole-turn strips.

The open Möbius band is formed by deleting the boundary of the standard Möbius band. It is constructed from the set S = { (x, y) ∈ R² : 0 ≤ x ≤ 1 and 0 < y < 1 } by identifying (glueing) the points (0, y) and (1, 1 − y) for all 0 < y < 1.

It may be constructed as a surface of constant positive, negative, or zero (Gaussian) curvature. In the cases of negative and zero curvature, the Möbius band can be constructed as a (geodesically) complete surface, which means that all geodesics ("straight lines" on the surface) may be extended indefinitely in either direction.

Constant negative curvature: Like the plane and the open cylinder, the open Möbius band admits not only a complete metric of constant curvature 0, but also a complete metric of constant negative curvature, say −1. One way to see this is to begin with the upper half plane (Poincaré) model of the hyperbolic plane ℍ, namely ℍ = { (x, y) ∈ ℝ² | y > 0} with the Riemannian metric given by (dx² + dy²) / y². The orientation-preserving isometries of this metric are all the maps f : ℍ → ℍ of the form f(z) := (az + b) / (cz + d), where a, b, c, d are real numbers satisfying ad − bc = 1. Here z is a complex number with Im(z) > 0, and we have identified ℍ with {z ∈ ℂ | Im(z) > 0} endowed with the Riemannian metric that was mentioned. Then one orientation-reversing isometry g of ℍ is given by g(z) := −z, where z denotes the complex conjugate of z. These facts imply that the mapping h : ℍ → ℍ given by h(z) := −2⋅z is an orientation-reversing isometry of ℍ that generates an infinite cyclic group G of isometries. (It can be expressed as h(z) = (√2i z + 0) / (0z − I/√2), and its square is the isometry h(h(z)) := 4⋅z, which can be expressed as (2z + 0) / (0z + ​¹⁄₂).) The quotient ℍ / G of the action of this group can easily be seen to be topologically a Möbius band. But it is also easy to verify that it is complete and non-compact, with constant negative curvature equal to −1.

The group of isometries of this Möbius band is 1-dimensional and is isomorphic to the special orthogonal group SO(2).

(Constant) zero curvature: This may also be constructed as a complete surface, by starting with portion of the plane R² defined by 0 ≤ y ≤ 1 and identifying (x, 0) with (−x, 1) for all x in R (the reals). The resulting metric makes the open Möbius band into a (geodesically) complete flat surface (i.e., having Gaussian curvature equal to 0 everywhere). This is the only metric on the Möbius band, up to uniform scaling, that is both flat and complete.

The group of isometries of this Möbius band is 1-dimensional and is isomorphic to the orthogonal group SO(2).

Constant positive curvature: A Möbius band of constant positive curvature cannot be complete, since it is known that the only complete surfaces of constant positive curvature are the sphere and the projective plane. The projective plane P² of constant curvature +1 may be constructed as the quotient of the unit sphere S² in R³ by the antipodal map A: S² → S², defined by A(x, y, z) = (−x, −y, −z). The open Möbius band is homeomorphic to the once-punctured projective plane, that is, P² with any one point removed. This may be thought of as the closest that a Möbius band of constant positive curvature can get to being a complete surface: just one point away.

The group of isometries of this Möbius band is also 1-dimensional and isomorphic to the orthogonal group O(2).

The space of unoriented lines in the plane is diffeomorphic to the open Möbius band.[13] To see why, let L(θ) denote the line through the origin at an angle θ to the positive x-axis. For each L(θ) there is the family P(θ) of all lines in the plane that are perpendicular to L(θ). Topologically, the family P(θ) is just a line (because each line in P(θ) intersects the line L(θ) in just one point). In this way, as θ increases in the range 0° ≤ θ < 180°, the line L(θ) represents a line's worth of distinct lines in the plane. But when θ reaches 180°, L(180°) is identical to L(0), and so the families P(0°) and P(180°) of perpendicular lines are also identical families. The line L(0°), however, has returned to itself as L(180°) pointed in the opposite direction. Every line in the plane corresponds to exactly one line in some family P(θ), for exactly one θ, for 0° ≤ θ < 180°, and P(180°) is identical to P(0°) but returns pointed in the opposite direction. This ensures that the space of all lines in the plane – the union of all the L(θ) for 0° ≤ θ ≤ 180° – is an open Möbius band.

The group of bijective linear transformations GL(2, R) of the plane to itself (real 2 × 2 matrices with non-zero determinant) naturally induces bijections of the space of lines in the plane to itself, which form a group of self-homeomorphisms of the space of lines. Hence the same group forms a group of self-homeomorphisms of the Möbius band described in the previous paragraph. But there is no metric on the space of lines in the plane that is invariant under the action of this group of homeomorphisms. In this sense, the space of lines in the plane has no natural metric on it.

This means that the Möbius band possesses a natural 4-dimensional Lie group of self-homeomorphisms, given by GL(2, R), but this high degree of symmetry cannot be exhibited as the group of isometries of any metric.

The edge, or boundary, of a Möbius strip is homeomorphic (topologically equivalent) to a circle. Under the usual embeddings of the strip in Euclidean space, as above, the boundary is not a true circle. However, it is possible to embed a Möbius strip in three dimensions so that the boundary is a perfect circle lying in some plane. For example, see Figures 307, 308, and 309 of "Geometry and the imagination".[14]

A much more geometric embedding begins with a minimal Klein bottle immersed in the 3-sphere, as discovered by Blaine Lawson. We then take half of this Klein bottle to get a Möbius band embedded in the 3-sphere (the unit sphere in 4-space). The result is sometimes called the "Sudanese Möbius Band",[15] where "sudanese" refers not to the country Sudan but to the names of two topologists, Sue Goodman and Daniel Asimov. Applying stereographic projection to the Sudanese band places it in three-dimensional space, as can be seen below – a version due to George Francis can be found here.

From Lawson's minimal Klein bottle we derive an embedding of the band into the 3-sphere S³, regarded as a subset of C², which is geometrically the same as R⁴. We map angles η, φ to complex numbers z₁, z₂ via

${\displaystyle z_{1}=\sin \eta \,e^{i\varphi }}$

${\displaystyle z_{2}=\cos \eta \,e^{i\varphi /2}.}$

Here the parameter η runs from 0 to π and φ runs from 0 to 2π. Since |z₁|² + |z₂|² = 1, the embedded surface lies entirely in S³. The boundary of the strip is given by |z₂| = 1 (corresponding to η = 0, π), which is clearly a circle on the 3-sphere.

To obtain an embedding of the Möbius strip in R³ one maps S³ to R³ via a stereographic projection. The projection point can be any point on S³ that does not lie on the embedded Möbius strip (this rules out all the usual projection points). One possible choice is ${\displaystyle \left\{1/{\sqrt {2}},i/{\sqrt {2}}\right\}}$. Stereographic projections map circles to circles and preserves the circular boundary of the strip. The result is a smooth embedding of the Möbius strip into R³ with a circular edge and no self-intersections.

The Sudanese Möbius band in the three-sphere S³ is geometrically a fibre bundle over a great circle, whose fibres are great semicircles. The most symmetrical image of a stereographic projection of this band into R³ is obtained by using a projection point that lies on that great circle that runs through the midpoint of each of the semicircles. Each choice of such a projection point results in an image that is congruent to any other. But because such a projection point lies on the Möbius band itself, two aspects of the image are significantly different from the case (illustrated above) where the point is not on the band: 1) the image in R³ is not the full Möbius band, but rather the band with one point removed (from its centerline); and 2) the image is unbounded – and as it gets increasingly far from the origin of R³, it increasingly approximates a plane. Yet this version of the stereographic image has a group of 4 symmetries in R³ (it is isomorphic to the Klein 4-group), as compared with the bounded version illustrated above having its group of symmetries the unique group of order 2. (If all symmetries and not just orientation-preserving isometries of R³ are allowed, the numbers of symmetries in each case doubles.)

But the most geometrically symmetrical version of all is the original Sudanese Möbius band in the three-sphere S³, where its full group of symmetries is isomorphic to the Lie group O(2). Having an infinite cardinality (that of the continuum), this is far larger than the symmetry group of any possible embedding of the Möbius band in R³.

Using projective geometry, an open Möbius band can be described as the set of solutions to a polynomial equation. Adding a polynomial inequality results in a closed Möbius band. These relate Möbius bands to the geometry of line bundles and the operation of blowing up in algebraic geometry.

The real projective line ${\displaystyle \mathbf {RP} ^{1}}$ is the set ${\displaystyle \mathbf {R} ^{2}\setminus \{(0,0)\}}$ modulo scaling. That is, a point in ${\displaystyle \mathbf {RP} ^{1}}$ is an equivalence class of the form

${\displaystyle [A:B]=\{(\lambda A,\lambda B):\lambda \in \mathbf {R} \setminus \{0\}\}.}$

Every equivalence class ${\displaystyle [A:B]}$ with ${\displaystyle B\neq 0}$ has a unique representative whose second coordinate is 1, namely ${\displaystyle (A/B,1)}$. These points form a copy of the Euclidean line ${\displaystyle \mathbf {R} ^{1}}$. However, the equivalence class of ${\displaystyle [1:0]}$ has no such representative. This extra point behaves like an unsigned infinity, making ${\displaystyle \mathbf {RP} ^{1}}$ topologically the same as the circle ${\displaystyle S^{1}}$. The advantage of ${\displaystyle \mathbf {RP} ^{1}}$ over the circle is that some geometric objects have simpler equations in terms of A and B. This is the case for the Möbius band.

A realization of an open Möbius band is given by the set

${\displaystyle M=\{((x,y),[A:B])\in \mathbf {R} ^{2}\times \mathbf {RP} ^{1}:Ax=By\}.}$

If we delete the line ${\displaystyle x=0}$ from M (or in fact any line), then the resulting subset can be embedded in Euclidean space ${\displaystyle \mathbf {R} ^{3}}$. Deleting this line gives the set

${\displaystyle {\begin{aligned}M'&=\{((x,y),[A:B])\in \mathbf {R} ^{2}\times \mathbf {RP} ^{1}:Ax=By,\ B\neq 0\}\\&=\{(x,y,m)\in \mathbf {R} ^{3}:mx=y\},\end{aligned}}}$

where m corresponds to ${\displaystyle A/B}$.

There is a realization of the closed Möbius band as a similar set, but with an additional inequality to create a boundary:

${\displaystyle N=\{((x,y),[A:B])\in \mathbf {R} ^{2}\times \mathbf {RP} ^{1}:Ax=By,\ x^{2}+y^{2}\leq 1\}.}$

The boundary of N is the set of all points with ${\displaystyle x^{2}+y^{2}=1}$. The geometry of N is very similar to that of M, so we will focus on M in what follows.

The geometry of M can be described in terms of lines through the origin. Every line through the origin in ${\displaystyle \mathbf {R} ^{2}}$ is the solution set of an equation ${\displaystyle ax+by=0}$. The solution set does not change when ${\displaystyle (a,b)}$ is rescaled, so the line only depends on the equivalence class ${\displaystyle [A:B]}$. That is, the lines through the origin are parametrized by ${\displaystyle \mathbf {RP} ^{1}}$. Furthermore, every point ${\displaystyle (x,y)}$ in ${\displaystyle \mathbf {R} ^{2}}$, except for ${\displaystyle (0,0)}$, lies on a unique line through the origin, specifically, the line defined by ${\displaystyle [-y:x]}$. The point ${\displaystyle (x,y)=(0,0)}$, however, lies on every line through the origin. For this point, the equation ${\displaystyle Ax+By=0}$ degenerates to ${\displaystyle 0=0}$. This is always true, so every ${\displaystyle [A:B]}$ is a solution. Consequently the set M may be described as the disjoint union of the set of lines through the origin. This is the same as the union of the lines through the origin, except that it contains one copy of the origin for each line. These additional copies of the origin are a copy of ${\displaystyle \mathbf {RP} ^{1}}$ and constitute the center circle of the Möbius band. The lines themselves describe the ruling of the Möbius band. This point of view on M exhibits it both as the total space of the tautological line bundle ${\displaystyle {\mathcal {O}}(-1)}$ on ${\displaystyle \mathbf {RP} ^{1}}$ as well as the blow-up of the origin in ${\displaystyle \mathbf {R} ^{2}}$.

To see the half-twist in M, begin with the point ${\displaystyle (1,0)}$ in ${\displaystyle \mathbf {R} ^{2}}$. This corresponds to a unique point of M, namely ${\displaystyle P=((1,0),[0:1])}$. Draw the counterclockwise half circle to produce a path on M given by ${\displaystyle \gamma (t)=((\cos(2\pi t),\sin(2\pi t)),[-\sin(2\pi t),\cos(2\pi t)])}$. The path stops at ${\displaystyle t=1/2}$, where it gives the point ${\displaystyle Q=((-1,0),[0:1])}$. Except for P and Q, every point in the path lies on a different line through the origin. Therefore ${\displaystyle \gamma (t)}$ travels once around the center circle of M. However, while P and Q lie in the same line of the ruling, they are on opposite sides of the origin. This change in sign is the algebraic manifestation of the half-twist.

Ancient Roman mosaic depicting a Möbius strip

The Möbius strip principle has been used as a method of creating the illusion of magic. The trick, known as the Afghan bands, was very popular in the first half of the twentieth century. Many versions of this trick exist and have been performed by famous illusionists such as Harry Blackstone Sr. and Thomas Nelson Downs.[30][31]

The universal recycling symbol (♲) design has its three arrows forming a Möbius loop. According to its designer Gary Anderson, "the figure was designed as a Mobius strip to symbolize continuity within a finite entity".[32]