Manifold #16. The Möbius Strip매니폴드 #16. 뫼비우스 띠

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers Lesson 16: The Möbius strip. In the previous post we glued two sides of a square in the same direction to make a cylinder. This time we glue one side flipped and check whether the result is a manifold.

Starting from an endless strip

This time we start from a strip that goes on forever instead of a finite sheet.

It carries the subspace topology of the plane. The horizontal direction $s$ runs on without end, and the vertical direction $t$ has width 2. For each integer $k$ we define a map of this strip to itself.

For even $k$ it only shifts sideways by $2\pi k$. For odd $k$ it shifts and also flips up and down. Each $f_k$ is continuous and its inverse $f_{-k}$ is continuous, so it is a homeomorphism. Also $f_k \circ f_l = f_{k+l}$ holds. (Flipping an even number of times gives back the original.)

The equivalence relation with a flip

For two points $(s, t)$ and $(s', t')$ we write $(s, t) \sim (s', t')$ when there is an integer $k$ with $(s', t') = f_k(s, t)$. Reflexivity comes from $k = 0$, symmetry from $f_{-k}$, and transitivity from $f_k \circ f_l = f_{k+l}$, so it is an equivalence relation. Let $M = S / \!\sim$ be the quotient space and $P : S \to M$ the canonical projection, and give $M$ the final topology. This $M$ is the Möbius strip.

Read over one period, the relation says this. Keep only the piece with $s \in [0, 2\pi]$, and the point $(0, t)$ at the left end is glued to the point $(2\pi, -t)$ at the right end. Gluing equal heights gave the cylinder, and gluing with the height flipped gives the Möbius strip. In the toy below, switch the way of gluing and watch whether the two marked points become one point.

Choose Straight and push Glue to the end and you get a cylinder. Choose Flipped and the strip joins with a half twist. Change the marked height $t$ to see how the two points meet. With the flipped gluing, the point at height $t$ on the left meets the point at height $-t$ on the right. After gluing to the end, also check whether the blue and green faces continue into each other or stay separate. Blue and green mark the two sides of the paper.

Q. How was the glued shape drawn in three dimensions?
A. The toy is a picture I built that changes continuously from the flat strip to the glued strip. The center line is rolled into a circle and the width direction is twisted by a half turn so the two ends meet. The lecture does not derive this three-dimensional formula; it is a picture for seeing the gluing that the definition of the quotient space describes. During the curling the length is not kept fixed as with real paper, and stretches to fit the picture. (This explanation is added to help read the toy and was not covered in the lecture.)

The projection is an open map

A quotient map is in general not open, but this projection is. Take an open set $U \subset S$. By the definition of the quotient topology, $P(U)$ is open when the preimage $P^{-1}(P(U))$ is open. This preimage is the set of all points equivalent to a point of $U$, and written out it is

Each $f_k$ is a homeomorphism, so $f_k(U)$ is open, and a union of open sets is open. Hence $P(U)$ is open. For the same reason the projection is an open map for the cylinder too.

Two charts

To be a manifold, every point needs a neighborhood homeomorphic to an open set of the plane. We cover the strip with two pieces.

(In the lecture the widths of the two pieces look slightly different from the period, so here I fixed the width at $2\pi$.)

The projection $P$ restricted to $V_1$ is one-to-one. Two points of $V_1$ can be equivalent only if their $s$ coordinates differ by $2\pi k$ for some $k \neq 0$, and that cannot happen because $V_1$ is narrower than $2\pi$ and omits its endpoints. $P$ is continuous and open, so $P|_{V_1}$ is a homeomorphism from $V_1$ onto $P(V_1)$. The same holds for $V_2$. The inverses of these two homeomorphisms,

are the charts. Let us also check that the two pieces cover all of $M$. Shifting by a multiple of $2\pi$, the $s$ coordinate of any point lands in $(-\pi, \pi)$ or in $(0, 2\pi)$. The point $s = 0$ is handled by $V_1$, and $s = \pi$ by $V_2$.

The transition map is smooth

Let us see how the coordinates change where the two pieces overlap. Start from the coordinates $(s, t)$ on the $V_1$ side.

• If $0 < s < \pi$, the point is already in $V_2$. Then $\psi_2 \circ \psi_1^{-1}(s, t) = (s, t)$.
• If $-\pi < s < 0$, we must use $f_1$ to enter $V_2$. Then $\psi_2 \circ \psi_1^{-1}(s, t) = (s + 2\pi, -t)$.

With the segment $s = 0$ removed, the overlap splits into two parts (the same happens for the cylinder), and on each part the coordinate change is only a shift or a flip, which is smooth. So these two pieces form a smooth atlas, and $M$ is a smooth two-dimensional manifold.

Q. If we put coordinates on the cylinder the same way, what are the transition maps?
A. Since equal heights are glued, on both parts it is $(s, t) \mapsto (s, t)$ or $(s + 2\pi, t)$. Having no flipped part at all is the difference between the cylinder and the Möbius strip.

The flip seen through the Jacobian

Write down the Jacobian matrices of the transition maps. On the first part it is the identity, with determinant $+1$. On the second part it is $\mathrm{diag}(1, -1)$, with determinant $-1$. For an oriented manifold (an orientable manifold) we need an atlas whose transition maps all have positive determinant. With these two pieces the overlap splits into two parts with opposite signs, and reversing one chart still cannot make them both positive. This is the background for calling the Möbius strip a non-orientable surface. (A proof that no atlas works does not finish with these two pieces alone, and the lecture did not go that far.)

As a picture: going once around the strip turns the front into the back. If you carry a small clockwise loop once around, it comes back counterclockwise. In the toy, choose Flipped and glue to the end, and the note says the blue and green faces continue into each other, so there is only one face. For the cylinder the same note says there are two faces.

Hausdorff and second countable

The definition of a manifold needs, besides charts, the Hausdorff condition and second countability. The lecture did not check these two. Here is a short look. (This paragraph is added and was not covered in the lecture.)

• Second countable: send a countable base of $S$ (for example open rectangles with rational corners) through $P$, and we get a countable base of $M$. This works because $P$ is open and surjective.
• Hausdorff: take two different classes $[x], [y]$. We can choose small neighborhoods $U, V$ of $x$ and $y$ with $f_k(U) \cap V = \varnothing$ for all $k$. For large $|k|$ the map $f_k$ sends points far away, so they do not meet, and for the finitely many remaining $k$ we have $f_k(x) \neq y$, so shrinking the neighborhoods is enough. Then $P(U)$ and $P(V)$ are disjoint open sets.

Neighbors of this shape

The same method builds other shapes. Glue one pair of sides of a square in the same direction and the other pair flipped, and you get the Klein bottle. Identify opposite points $x$ and $-x$ of the sphere and you get the projective plane. The Klein bottle and the projective plane are also non-orientable manifolds. (This paragraph is added and was not covered in the lecture.)

That completes the way to build manifolds from quotient spaces. The next lecture continues from here.