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Mobius band

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Theorem. The Mobuis band is a surface with boundary.

What happens if we cut the Mobius band along this line? The result is a thinner band with a double twist.

Image:double twist.jpg

Since this band can be cut, twisted $360$ degrees, and then glued back together along the same edge - exactly as before! - this double twist band is homeomorphic to the cylinder.

Another way to see this is shown in the diagram below:

Image:cutting Mobius band.jpg

Problem. What happens if we cut the Mobius band along the middle line?

The Mobuis band is a rectangle with the left edge identified with the twisted right edge. This diagram will help to solve this problem.

Suppose we cut it with $n-1$ cuts, then the rectangle has $n$ bands. Let's call them $B_1, ..., B_n$. Suppose their left edges are $x_1, ..., x_n$, from top to bottom. Then their right edges are $x_n, ..., x_1$, from top to bottom. More precisely, these are $-x_n, ..., -x_1$ indicating the twisted edges.

Now $B_1$ and $B_n$ share edges, both in fact:

$x_1 : B_1 : -x_n$, and $x_n : B_n : -x_1$.

We need to make sure that the edges match. Because of that $B_n$ has to be flipped:

$$-x_n : -B_n : x_1.$$

Now attach them:

$$x_1 : B_1 : -x_n : - B_n : x_1.$$

The ends match! Therefore, this is a cylinder.

In a identical fashion, $B_2$ and $B_{n-1}$ are glued into another cylinder, and so are $B_3$ and $B_2$, etc.

Let $k = \frac{n}{2}$; then if $n$ is even, then $k$ is an integer. In this case the process will use all bands that we cut. The last one made of $B_k$ and $B_{k+1}$. The total is $k$ cylinders.

If $n$ is odd, then there will be one band left after we have made $k=\frac{n-1}{2}$ cylinders. The last one is:

$$x_k : B_k : -x_k.$$

This is another Mobius band.

$\square$

Exercise. What happens if we cut the Mobius band by three, $n$ cuts along the middle line?

It is obvious that the middle line is homeomorphic to the circle. Its boundary however appears to be made of two pieces, just like that of the cylinder.

Theorem. The boundary of the Mobius band is homeomorphic to the circle.

Proof. You travel along the "lower edge" from $A$ to $B$, and next you are at $B$ on the "upper" edge, the you travel along it to $A$. You come back where you started. QED

Exercise. Compute the homology of Möbius band. Compute the homology of Möbius band relative its edge.

Exercise. Show that the middle line is a deformation retract of the Möbius band.

Exercise. Compute the homology map of the composition of the inclusion of the boundary of the Möbius band and its retraction to the middle line.