I'm supposed to prove that the Möbius strip is not orientable by using this lemma:

Lemma: Let $S\subset\mathbb{R}^3$ be a surface such that $S=S_1\cup S_2$, where $S_1$, $S_2$ are connected, orientable surfaces, whose intersection $S_1\cap S_2$ has exactely two connected components $A$ and $B$. If there is an orientation $N_1$ for $S_1$ and $N_2$ for $S_2$ such that $N_1=N_2$ in $A$ and $N_1\neq N_2$ in $B$, then $S$ is not orientable.

I'm using the following parametrization for the Möbius strip:

$$X(u, v)=\left((2-v\sin\frac{u}{2})\sin u, (2-v\sin\frac{u}{2})\cos u,v\cos\frac{u}{2}\right)$$

for $u\in[0,2\pi]$ and $u\in(-1,1)$

I've defined $S_1:=X\left((\frac{\pi}{2}, \frac{3\pi}{2})\times(-1,1)\right)$ and $S_2:=X\left((0,\frac{3\pi}{4})\cup(\frac{5\pi}{4}, 2\pi)\times(-1,1)\right)$, so that $S_1\cap S_2=X\left((\frac{\pi}{2}, \frac{3\pi}{4})\times(-1,1)\right)\cup X\left((\frac{5\pi}{4}, \frac{3\pi}{2})\times(-1,1)\right)$ (one of the components is $A$, the other one $B$).

My question is: how do I prove that the induced orientations are the same in $A$ and opposites in $B$, since for both $S_1$ and $S_2$ we are using the exact same formula $X(u,v)$? How does the minus sign actually appear?


If my computation is correct, a vector $\vec{n}_{u,v}$ normal to the strip would be given by the following formula $$ \vec{n}_{u,v}= \begin{pmatrix} \frac{1}{2}cos\left(\frac{u}{2}\right)cos^2(u)-\frac{v}{8}sin\left(\frac{u}{2}\right)cos\left(\frac{u}{2}\right)cos^2(u)-\frac{v}{4}cos^2\left(\frac{u}{2}\right)sin(u)cos(u)\\ \frac{1}{2}cos\left(\frac{u}{2}\right)sin^2(u)-\frac{v}{8}sin\left(\frac{u}{2}\right)cos\left(\frac{u}{2}\right)sin^2(u)-\frac{v}{4}cos^2\left(\frac{u}{2}\right)sin(u)cos(u) \\ \frac{v}{8}cos\left(\frac{u}{2}\right)sin\left(\frac{u}{2}\right) \end{pmatrix} $$ For the sake of simplicity we can assume that the vector is "centered" along the strip, meaning $v=0$, such that $$ \vec{n}_{u,0}= \begin{pmatrix} \frac{1}{2}cos\left(\frac{u}{2}\right)cos^2(u) \\ \frac{1}{2}cos\left(\frac{u}{2}\right)sin^2(u) \\ 0 \end{pmatrix} $$ The "minus" sign you are talking about should appear when you evaluate this vector for $u=0$ and $u=2 \pi$ (informally, you make the vector move along the path parameterized by $u$ such that it makes a whole loop and comes back to its starting point). Indeed we have that $$ \vec{n}_{0,0}= \begin{pmatrix} 1/2\\ 0 \\ 0 \end{pmatrix} $$ and $$ \vec{n}_{2\pi,0}= \begin{pmatrix} -1/2\\ 0 \\ 0 \end{pmatrix} $$

I hope this can help (I would've made this a comment but I don't have enough privilege yet !)

EDIT : I used a different parametrization (this one) so your calculations may yield slightly different results, but the argument is still the same.

  • $\begingroup$ In deed, that shows that the strip is not orientable. But I was actually looking for some argument which uses the specific lemma above $\endgroup$ – rmdmc89 Apr 12 '17 at 23:19

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