# The Moebius Band

• Jan 9th 2014, 06:53 AM
HAL9000
The Moebius Band
I've got a conceptual question (that depends on definitions involved being unambiguous).

Is the Moebius strip as we tend to imagine it in $R^3$ a surface? (If so I would like to see the FORM of the intervals the parameter set)
Is it a compact surface if we imagine it with the edge curve? We imagine a compact set of $R^3$…but which definition of a compact surface, if any, is appropriate here?

Note that the Jordan Curve theorem allows to conclude that a compact surface in $R^3$ is orientable.
And remember that the Moebius band is non-orientable, although compact (!), as a 2-dim (quotient) manifold of a closed rectangle.

The question is essentially this: How does one not become a victim of a misconception?
• Jan 10th 2014, 06:29 AM
MINOANMAN
Re: The Moebius Band
• Jan 10th 2014, 07:19 AM
HallsofIvy
Re: The Moebius Band
You say the Moebius strip is compact. If you mean by that the actual bounded strip that one cuts and pastes from paper and holds in ones hand, then, yes, of course it exists in 3 dimensions. You are holding it in your hand and you are in 3 dimensions, aren't you?

But if you want to make a full surface of it by extending the edges infinitely then it cannot be embedded in 3 dimensions but then it is no longer compact.
• Jan 20th 2014, 07:20 AM
HAL9000
Re: The Moebius Band
$$F: D \rightarrow \bb{R}^3,\,F(\varphi,t)=c(\varphi)+tv(\varphi)$$ with
$$c(\varphi)=(\cos\varphi,\sin\varphi,0), \, v(\varphi)=\left(\cos\frac{\varphi}{2}\cos \varphi,\cos\frac{\varphi}{2}\sin \varphi,\sin\frac{\varphi}{2}\right)$$
If we take $$D=(0,2\pi)\times(-\frac{1}{2},\frac{1}{2})$$ then we conform with the definition of surface theory. But are we allowed to take $$D=[0,2\pi]\times(-\frac{1}{2},\frac{1}{2})$$ and speak of the image of this set under F as of a surface? Are we allowed to take even $$D=[0,2\pi]\times[-\frac{1}{2},\frac{1}{2}]$$ and speak of a compact surface?