Showing a Mobuis strip is a surface

Question

A Mobius strip is a surface which is diffeomorphic to the surface $S$ defined below.

Let $α : R → R^{3}$ be defined as $α(t) = (\cos{t},\sin{t}, 0)$. Let $ψ : R × (−1, 1) → R^{3}$

be defined as $ψ(t, s) = 5α(t) + s(\cos(\frac {t}{2})(0, 0, 1) + \sin(\frac {t}{2})α(t))$.

$(5\cos{t},5\sin{t}, 0) + s \sin(\frac {t}{2}) (\cos{t},\sin{t}, 0) + s \cos(\frac {t}{2})(0, 0, 1) $ In this form its allittle easier to view this is cylindrical coordinates.

With some help i have finally been able to visualize this as a circle of radius 5 with a line segment on the end of radius of the circle moving around the entire circle while completing a flip of the line segment with one rotation about the axis. s appears to control the length of this line segment and seems to change the orientation of the line segment? as it transitions through s=0.

Sketch the image $S$ of $ψ$. Mobius band turning counter clockwise when s is negative clockwise when s is positive. appears to be a half turn each way. when s=0 we got a circle of radius 5.

Show that S is a $C^{\infty}$ in $ \mathbb{R^{3}}$

i think i need to do this in a range from 0 to $\pi$ and $ \pi $ to 2 $ \pi $ where the area is s 2.5 $ \pi $ for each half?

Bounty Award: I would like an approach to show its a $C^{\infty}$ surface preferably by using patchs not the IFT any ideas how to set this up is sufficient for bounty.

I am given in the Question that this is diffemorphic to mobius strip. Can i somehow use that and the fact that a mobius strip is a surface to show that this if a surface?

Answer 1

The strategy about splitting the Möbius strip into two pieces and using IFT is the right way to do the problem from the point of view of Pressley's definitions.

However, there is another way to look at this: a surface does not really depend on its embedding into $\mathbb{R}^3$. For example, if I gave you two spheres with the same radii but different centers, you would still think of them as really being the same sphere, just put into $\mathbb{R}^3$ at different places. And you can think of the Klein bottle as being a surface, probably, even though it actually can't be embedded into $\mathbb{R}^3$ (you need one more dimension).

Indeed, if you wanted to construct a sphere, it suffices to give you two disks to be surface patches and tell you how to glue them together by specifying smooth and invertible transition functions. Similarly, to construct a Möbius strip, all I actually have to do is take two copies of $(-1,1) \times (-1,1)$ and specify smooth transition functions on the pieces I want to glue together (picture attached). The embedding into $\mathbb{R}^3$ is extraneous data.