# What goes wrong with Stokes theorem if a surface is not orientable?

For the Möbius strip parametrized by $$\sigma(\theta, r) = \begin{pmatrix} (1 + r\sin\frac{\theta}{2})\cos{\theta} \\ (1 + r\sin\frac{\theta}{2})\sin\theta \\ r\cos\frac{\theta}{2} \end{pmatrix},\qquad (\theta, r)\in A = (0,2\pi) \times (-\tfrac{1}{2},\tfrac{1}{2})$$ we get the normal vector $$\nu$$ by doing the cross product $$\nu(\theta,r) = \frac{\partial \sigma}{\partial \theta}\times\frac{\partial \sigma}{\partial r} = \begin{pmatrix} (1+r\sin\tfrac{\theta}{2})\cos\frac{\theta}{2}\cos\theta + \frac{r}{2}\sin\theta \\ (1+r\sin\tfrac{\theta}{2})\cos\frac{\theta}{2}\sin\theta - \frac{r}{2}\cos\theta \\ -(1 + r\sin\frac{\theta}{2})\sin\frac{\theta}{2} \end{pmatrix}$$ which is continuous on $$A$$ but not on $$\overline A$$ because $$\nu(0,0)=(1,0,0)\ne(-1,0,0)=\nu(2\pi,0)$$

For what kind of $$F(x,y,z)=(F^1,F^2,F^3)$$ with $$F^i\in C^1(\Bbb R^3)$$ can there be a problem? I lack some intuition for the necessity of the orientability for Stokes' theorem:

$$\iint\limits_\Sigma \nabla \times F\cdot ds=\int\limits_{\partial\Sigma}F\cdot dl$$

• The problem is not with Stokes, but with defining the integral on a non-orientable manifold. Commented Dec 29, 2018 at 15:34
• This a Möbius strip which has been cut at $\theta = 0$ so it should be orientable. If the domain was $\overline{A}$, it wouldn't be orientable. Commented Dec 29, 2018 at 15:49

Since this has gotten bumped: The problem (as noted in the comments) is not with Stokes' theorem, but defining a flux integral (or the integral of a differential $$2$$-form) on a non-orientable surface at all.

Well, if you cut the strip at $$\theta = 0$$ you get an orientable surface and you lose only a segment of measure zero, which shouldn't affect the integral, so why don't we define the integral that way (for this particular case of the Möbius strip)? Is it because the value will change depending on where we make the cut? That would kind of make sense.

That is, in fact, the issue. For each real $$t$$, the rectangle $$R_{t} = (t, t + 2\pi) \times (-\frac{1}{2}, \frac{1}{2})$$ is the domain of a parametrization of a dense subset of the Möbius band. Let $$dS_{t}$$ denote the oriented area element induced by the parametrization $$\sigma$$ on $$R_{t}$$. Direct calculation (left as an exercise) shows that if $$F$$ is a vector field, the integral $$\iint_{R_{t}} F \cdot dS_{t}$$ generally depends on $$t$$. Particularly, $$R_{2\pi}$$ and $$R_{0}$$ have the same image under $$\sigma$$ but induce opposite unit normal fields, so for every vector field $$F$$ we have $$\iint_{R_{2\pi}} F \cdot dS_{2\pi} = -\iint_{R_{0}} F \cdot dS_{0}.$$ The point is, removing a set of measure zero to get an orientable surface does not give a well-defined integral.

(We can, however, integrate a density—locally the absolute value of a $$2$$-form—over the Möbius strip. For instance, area and mass-per-unit-area do not depend on orientation, and a Möbius strip has both area and mass.)

Edit: Just noticed that your strip does not include the points $$\theta = 0, 2\pi$$ which means that it can be oriented.

However, on the full domain you cannot define the vector surface integral to begin with and so the problem is not with Stoke's Theorem per say.Remember, that by definition a vector surface integral is: $$\iint_{\mathcal{S}} \overrightarrow{F} \cdot \, \mathrm{d}\overrightarrow{S} = \iint_\mathcal{S} \left(\overrightarrow{F} \cdot \hat{n}\right) \mathrm{d}S$$

In other words you need a clear definition of $$\hat{n}$$ throughout $$\mathcal{S}$$. The Möbius strip has only one side and therefore cannot be oriented and that's why you cannot even define one side of equation for the Stoke's Theorem.

To get an intuition for why the Möbius strip cannot be oriented I suggest making one yourself. It's easy --just take a strip of paper and glue the two ends together by introducing a 180 degree twist somewhere in between. Now take a normal vector at a point of your choice anywhere on the surface. Translate this vector along the strip and arrive back at this point. Your normal vector will now be pointing in the reverse direction from when you started at that point.

This shows that you cannot define $$\hat{n}$$ consistently for this surface.

• Well, if you cut the strip at $\theta = 0$ you get an orientable surface and you lose only a segment of measure zero, which souldn't affect the integral, so why don't we define the integral that way (for this particular case of the Möbius strip)? Is it because the value will change depending on where we make the cut? That would kind of make sense Commented Dec 30, 2018 at 7:47
• @JohnCataldo Yes, in this way your vector surface integral can be calculated but how would you orient $\partial \mathcal{S}$ then? Commented Dec 30, 2018 at 11:17
• Does it matter? Don't we usually have to chose an orientation anyway? The result will be the same up to the sign $\pm$ Commented Dec 30, 2018 at 12:31
• I am not too sure. When you define the curve you will have to exclude the start and the end points. The proof of Stoke’s theorem relies on the curve being closed. Commented Dec 30, 2018 at 14:30
• @JohnCataldo. Nonetheless this is a very interesting question. I will try to work out what exactly goes wrong with Stokr’s theorem and revert back to you if I find something. Commented Dec 30, 2018 at 14:32