A “double” Möbius Strip

Question

A few days back, I was taught about equivalence relations, classes and was given some glimpses of quotient topology.

Suppose we are given the space $[0,1]^{2} \subset \Bbb{R}^{2}.$ Then, the quotient space $[0,1]^{2}/\sim$ looks like a cylinder (i.e the equivalence classes look like cylinders where $\sim$ is an equivalence relation) when $\sim$ identifies $p\sim p$ and $(0,p)\sim(1,p) \space\forall\space p $. Similarly, $[0,1]^{2}/\sim$ looks like a Möbius strip when $(0,p)\sim(1,1-p)$ and $p\sim p$.

Now, my question is that - "Is there an equivalence relation that can be explicitly stated such that we get a Möbius-like strip with two twists?" It is clear that this structure is homeomorphic to the cylinder in $\Bbb{R}^{3}$ because when we add another twist to $(0,p)\sim(1,1-p)$, we again get $(0,p)\sim(1,p)$ , but when we look at the edges of the two structures, the cylinder has two disconnected circles while the double Möbius has two linked circles, hence there is no continuous deformation of space so that the cylinder can be converted to this. This also implies that they have different knot groups, which I know nothing of. Please express your views and discuss.

Answer 1

I think I (finally) understood your question. You know how to obtain surfaces (such as Moebius bands) by identifying sides of polygons. What you are asking is for a rigorous description of a "cut-and-paste operation on the Euclidean 3-space $R^3$ which produces the doubly twisted Moebius band in $R^3$."

I will write down an answer but it might take you awhile to understand it. (For the record: This is not the best way to get "doubly twisted Moebius bands": A better way is to used a "framed unknot": Say, a round circle $C$ equipped with a normal vector field that twists once around $C$.)

Here is one way to make this work. Start with $R^3$ and a "vertical round annulus" $A$ contained the $xz$-plane. The boundary of $A$ is the union of two round circles $C_1, C_2$, which I will assume to be concentric with the common center at the origin $(0,0,0)$ and respective radii 1 and 2. Then $C_1$ intersects the $xy$-plane in two points $p_1=(-1,0,0), q_1=(1,0,0)$, while $C_2$ intersects the $xy$-plane also in two points $p_1=(-2,0,0), q_1=(2,0,0)$.

Now, draw a round circle $\alpha$ in the $xy$-plane which separates $p_1,q_1$ from $p_2,q_2$. The precise location of this circle is irrelevant.

Consider the surface $P$ equal to the $xy$-plane and let a homeomorphism $f: P\to P$ be the Dehn twist along the loop $\alpha$. (For the record, there is a left and right Dehn twists, it does not matter for our purpose which one do we use.) This homeomorphism will be identity outside of a small neighborhood of the loop $\alpha$. In particular, it is the identity on the intersection $A\cap P$.

Let $P_\pm$ denote the lower and upper half-spaces in $R^3$ bounded by the plane $P$. I will define a new topological space $X$ obtained by gluing $P_-$ to $P_+$ using the homeomorphism $f$ of their boundaries. The annulus $A$ still sits inside $X$ (since $f$ was the identity at $A\cap P$). One can show that $X$ is homeomorphic to $R^3$ (for instance, because $f: P\to P$ is isotopic to the identity map). However, under the homeomorphic identification $X\to R^3$ the annulus $A$ maps to a "once overtwisted annulus" (doubly twisted Moebius band that you are interested in). Proving that this is true will take some effort and I will not do it. I note only that if instead of gluing using $f$ you perform gluing using its $n$th power, you obtain an $n$-times overtwisted annulus in $R^3$. One can also use a "half-twist" $g$ instead of $f$ ($g$ has the property that $g\circ g=f$: It is a 180 degree rotation inside the circle $\alpha$ in $P$ and swaps $p_1, q_1$ and fixes the midpoint of $p_1q_1$). If you glue $P_+, P_-$ using $g$ you obtain a Moebius band embedded in $R^3$ in a standard way. You can also replace $g$ with its odd powers. This will result in overtwisted Moebius bands in $R^3$.

In order to understand what is going on, I suggest you draw a picture.

Answer 2

This is more or less the same question as Why aren't the "higher twist" Möbius bands distinct line-bundles over $S^1$?

A "Möbius-like strip with two (half) twists" is topologically a cylinder.

In fact you have to distinguish between the cylinder $C = S^1 \times [0,1]$ as a topological space which is obtained as the quotient space described in your question and the embeddings of $C$ into $\mathbb R^3$. You can embed $C$ with an arbitrary number $n$ of (full) twists into $\mathbb R^3$ and for $n \ne n'$ these embeddings cannot be deformed into each other via an isotopy.

Answer 3

Playing with home made Mobius Bands is good fun. You can try cutting them down the middle, or one third the way from the edge. Do the same with Bands of different numbers of twists.

See also this presentation.