There are some surfaces \(S\) for which it is not possible to choose a continuous orientation map \(\hN: S\rightarrow\bbbr^3\text{.}\) Such surfaces are said to be non-orientable. The most famous non-orientable surface is the Möbius strip, which you can construct as follows. Take a rectangular strip of paper.
Lay it flat and then introduce a half twist so that the arrow on the right hand end points upwards, rather than downwards. Then glue the two ends of the strip together, with the two arrows coinciding. That’s the Möbius strip.
Let’s parametrize it. Think of the strip of paper that we used to construct it as consisting of a backbone (the horizontal black line in the figure below) with a bunch of ribs (like the thick blue line in the figure) emanating from it.
When we glue the two ends of the strip together, the black line forms a circle. If the strip has length \(\ell\text{,}\) the circle will have circumference \(\ell\) and hence radius \(\frac{\ell}{2\pi}\text{.}\) We’ll parametrize it as the circle
\begin{gather*}
\frac{\ell}{2\pi}\hr(\theta)
\qquad\text{where } \hr(\theta) = \cos(\theta)\,\hi + \sin(\theta)\,\hj
\end{gather*}
This circle is in the \(xy\)-plane. It is the black circle in the figure below. (The figure only shows the part of the circle in the first octant, i.e. with \(x,y,z\ge 0\text{.}\))
Now we’ll add in the blue ribs. We’ll put the blue rib, that is attached to the backbone at \(\frac{\ell}{2\pi}\hr(\theta)\text{,}\) in the plane that contains the vectors \(\hr(\theta)\) and \(\hk\text{.}\)
A side view of the plane that contains the vectors \(\hr(\theta)\) and \(\hk\) is sketched in the figure below.
To put the half twist into the strip of paper, we want the blue rib to rotate about the backbone by \(180^\circ\text{,}\) i.e. \(\pi\) radians, as \(\theta\) runs from \(0\) to \(2\pi\text{.}\) That will be the case if we pick the angle \(\varphi\) in the figure to be \(\frac{\theta}{2}\text{.}\) The vector that is running along the blue rib in the figure is
\begin{equation*}
\vu(v,\theta,\varphi)=v\cos(\varphi)\,\hr(\theta) + v\sin(\varphi)\,\hk
\end{equation*}
where the length, \(v\text{,}\) of the vector is a parameter. If the width of our original strip of paper is \(w\text{,}\) then as the parameter \(v\) runs from \(-\frac{w}{2}\) to \(+\frac{w}{2}\text{,}\) the tip of the vector \(\vu(v,\theta,\varphi)\) runs over the entire blue rib. So, choosing \(\varphi=\frac{\theta}{2}\text{,}\) our parametrization of the Möbius strip is
\begin{align*}
\vr(\theta,v)
&= \frac{\ell}{2\pi}\hr(\theta)
+ \vu\left(v,\theta,\frac{\theta}{2}\right)\\
&= \frac{\ell}{2\pi}\hr(\theta)
+ v\cos\left(\frac{\theta}{2}\right) \hr(\theta)
+ v\sin\left(\frac{\theta}{2}\right) \hk\\
& 0\le\theta \lt 2\pi,\ -\frac{w}{2}\le v\le \frac{w}{2}
\end{align*}
where \(\hr(\theta) = \cos(\theta)\,\hi + \sin(\theta)\,\hj\text{.}\)
Now that we have parametrized the Möbius strip, let’s return to the question of orientability. Recall, from Definition
3.5.1, that, if the Möbius strip were orientable, there would exist a continuous function
\(\hN\) which assigns to each point
\(\vr\) of the strip a unit normal vector
\(\hN(\vr)\) at
\(\vr\text{.}\) First, we’ll find the normal vectors to the surface using
3.3.1. The partial derivatives
\begin{align*}
\frac{\partial\vr}{\partial\theta}(\theta,v)
&=\frac{\ell}{2\pi}\hr'(\theta)
+ v\cos\left(\frac{\theta}{2}\right) \hr'(\theta)
- \frac{v}{2}\sin\left(\frac{\theta}{2}\right) \hr(\theta)
+ \frac{v}{2}\cos\left(\frac{\theta}{2}\right) \hk\\
\frac{\partial\vr}{\partial v}(\theta,v)
&=\cos\left(\frac{\theta}{2}\right) \hr(\theta)
+ \sin\left(\frac{\theta}{2}\right) \hk
\end{align*}
are relatively messy, so let’s just consider the case \(v=0\) (i.e. find the normal vectors on the backbone). Then
\begin{align*}
\frac{\partial\vr}{\partial\theta}(\theta,0)
&=\frac{\ell}{2\pi}\hr'(\theta)\\
\frac{\partial\vr}{\partial v}(\theta,0)
&=\cos\left(\frac{\theta}{2}\right) \hr(\theta)
+ \sin\left(\frac{\theta}{2}\right) \hk
\end{align*}
Since
\begin{align*}
\hr'(\theta)\times \hr(\theta)
&=\big(-\sin(\theta)\,\hi + \cos(\theta)\,\hj\big)\times
\big(\cos(\theta)\,\hi + \sin(\theta)\,\hj\big)
=-\hk\\
\hr'(\theta)\times \hk
&=\big(-\sin(\theta)\,\hi + \cos(\theta)\,\hj\big)\times \hk
=\hr(\theta)
\end{align*}
we have
\begin{gather*}
\frac{\partial\vr}{\partial\theta}(\theta,0) \times
\frac{\partial\vr}{\partial v}(\theta,0)
=-\frac{\ell}{2\pi}\Big[\cos\left(\frac{\theta}{2}\right)\,\hk
-\sin\left(\frac{\theta}{2}\right)\,\hr(\theta) \Big]
\end{gather*}
As \(\hk\) and \(\hr(\theta)\) are mutually perpendicular unit vectors, \(\cos\big(\frac{\theta}{2}\big)\,\hk
-\sin\big(\frac{\theta}{2}\big)\,\hr(\theta)\) has length one, and the two unit normal vectors to the Möbius strip at \(\vr(\theta,0)\) are
\begin{equation*}
\pm \Big[\cos\left(\frac{\theta}{2}\right)\,\hk
-\sin\left(\frac{\theta}{2}\right)\,\hr(\theta) \Big]
\end{equation*}
So, for each \(\theta\text{,}\) \(\hN\big(\vr(\theta,0)\big)\) must be either
\begin{equation*}
\cos\left(\frac{\theta}{2}\right)\,\hk
-\sin\big(\frac{\theta}{2}\big)\,\hr(\theta)
\qquad\text{or}\qquad
-\Big[\cos\left(\frac{\theta}{2}\right)\,\hk
-\sin\left(\frac{\theta}{2}\right)\,\hr(\theta) \Big]
\end{equation*}
Imagine walking along the Möbius strip. The normal vector \(\hN\big(\vr(\theta,v)\big)\) is our body when we are at \(\vr(\theta,v)\) — our feet are at the tail of the vector \(\hN\big(\vr(\theta,v)\big)\) and our head is at the arrow of \(\hN\big(\vr(\theta,v)\big)\text{.}\) We start walking at \(\vr(0,0)=\frac{\ell}{2\pi}\hi\text{.}\) Our body, \(\hN\big(\frac{\ell}{2\pi}\hi\big)=\hN\big(\vr(0,0)\big)\) has to be one of \(\pm \big(\cos(0)\,\hk-\sin(0)\,\hr(0) \big)=\pm\hk\text{.}\) Let’s suppose that \(\hN\big(\vr(0,0)\big)=+\hk\text{.}\) (We start upright.) Now we start walking along the backbone of the Möbius strip, increasing \(\theta\text{.}\) Because \(\hN\big(\vr(\theta,0)\big)\) has to be continuous, \(\hN\big(\vr(\theta,0)\big)\) has to be \(+\big(\cos\big(\frac{\theta}{2}\big)\,\hk
-\sin\big(\frac{\theta}{2}\big)\,\hr(\theta) \big)\text{.}\) We keep increasing \(\theta\text{.}\) By continuity, \(\hN\big(\vr(\theta,0)\big)\) has to be \(+\big(\cos\big(\frac{\theta}{2}\big)\,\hk
-\sin\big(\frac{\theta}{2}\big)\,\hr(\theta) \big)\) for bigger and bigger \(\theta\text{.}\) Eventually we get to \(\theta=2\pi\text{,}\) i.e. to
\begin{equation*}
\vr(2\pi,0)= \frac{\ell}{2\pi}\hr(2\pi) = \frac{\ell}{2\pi}\hi
=\frac{\ell}{2\pi}\hr(0)=\vr(0,0)
\end{equation*}
We are back to our starting point. Continuity has forced
\begin{equation*}
\hN\big(\vr(2\pi,0)\big)
=\hN\big(\vr(\theta,0)\big)\Big|_{\theta=2\pi}
=+\Big[\cos\left(\frac{\theta}{2}\right)\,\hk
-\sin\left(\frac{\theta}{2}\right)\,\hr(\theta) \Big]
\Big|_{\theta=2\pi}
=-\hk
\end{equation*}
So we have arrived back upside down. That’s a problem — \(\hN\big(\vr(2\pi,0)\big)
=\hN\big(\frac{\ell}{2\pi}\hi\big)\) and we have already defined \(\hN\big(\frac{\ell}{2\pi}\hi\big)=+\hk\text{,}\) not \(-\hk\text{.}\) So the Möbius strip is not orientable. The interested reader should look up M. C. Escher’s Möbius Strip II (Red Ants).
