# Is there a way to prove algebraically that a Möbius strip is non-orientable?

I am doing my HL Maths coursework on non-orientability of surfaces and am trying to prove whether a möbius strip is orientable or not (of course it isn't) Is there a way to prove algebraically that a mobious strip is non-orientable via vectors and normal vectors? And if so how? Or am I approaching this in the wrong way? (please avoid using ''math jargon'' as much as possible because I am not well versed in topological concepts and terminology yet, but I'm slowly learning)

• – Vasya Mar 26 '18 at 15:47
• @Vasya: that looks to be over the OP's head based on the last sentence of the post. – Cheerful Parsnip Mar 26 '18 at 15:48
• Perhaps your first step should be to define a Mobius strip algebraically. – S. McGrew Mar 26 '18 at 17:27

## 1 Answer

Essentially, you just need a parity argument. A Moebius strip can be seen as a quotient of a square in $\mathbb{R}^2$, where the "left" and "right" sides are identified but glued together by firstly reversing the orientation of one of them. To help you visualizing this, in the following diagram points labelled by the same letter are the same point:

This object is orientable iff there is a continuous, non-vanishing vector field orthogonal to $\mathbb{R}^2$ at each point of the above square with identified sides. If that happens, at each sub-square the vector field is either pointing toward my face (we may say it is positive) or entering in my screen (we may say it is negative). Given a sub-square and its sign (i.e. the sign of the vector field defined over the sub-square), we may assign an orientation to the sides of such sub-square: counter-clockwise for positive squares, clockwise for negative squares. But if the left-$CD$-sub-square is positively oriented the right-$CD$-sub-square is negatively oriented, so there are adjacent sub-squares with opposite signs and an undefined orientation of the common edge, contradiction.

Ergo a Moebius strip is non-orientable.