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I know that the Möbius band is a nonorientable surface. However, the following exercise seems to contradict this.

A Möbius band can be constructed as a ruled surface by

$x(u,v)=\beta(u)+v\gamma(u)$, where $-1/3<v<1/3$

$\beta(u)=(\cos u, \sin u, 0)$ and

$\gamma(u)=(\cos [u/2]\cos u, \cos [u/2]\sin u,\sin[u/2])$.

Then the mapping $x$ from an open set of $\mathbf{R}^2$ to the band is regular and one to one. Its unit normal vector field can be got by the cross porduct of partial deravatives of $x$. This seems to contradict nonorientability of the Möbius band, because it has a unit normal vector field.

Where am I mistaken?

azimut
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noot
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2 Answers2

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In order to get a parametrization defined on an open subset of $\mathbf{R}^{2}$ you need to restrict your angular parameter $u$ to the open interval $(0, 2\pi)$. But the image of $x(u,v)$ in this case is not precisely the Möbius band, but the band minus one vertical line where it should "close". If you take $u$ belonging to the interval $[0, 2 \pi]$, then you cover the band, but you also obtain a normal field that is not globally continuous. Indeed, doing the calculations for the central line of your band (that is, to the curve $v = 0$), you may check that $N(0,0) = - N(2\pi, 0)$, where $N$ is a normal field obtained by taking the cross product of the coordinate derivatives $x_{u}$ and $x_{v}$.

ulilaka
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The normal vector field has to be differentiable, in this case it's not even continuous.

Michael Hoppe
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