# Using Möbius band to define orientability of surfaces

Given a surface $$S$$, the usual definition of orientability is that there is an atlas of charts with all transition maps having non-negative Jacobian.

Donaldson, in his book on Riemann surfaces, says a closed surface $${S}$$ is orientable if it does not contain any Möbius band, and non-orientable if it does. How can we show that this is equivalent to our usual definition of non-negative jacobian of transition maps?

One direction is clear: if there is a Möbius band embedded then there has to be a transition map with negative Jacobian, as Möbius band is non-orientable.

Now assume a closed surface is non-orientable. How can we show there is an embedded Möbius band in it?

An attempt: Let $$\mathcal C =\{A\subseteq S: A \text{ is path-connected, open, and orientable}\}$$ ordered by set inclusion. Using Zorn's lemma, pick a maximal element $$B\subseteq S$$. Pick a point $$p\in \partial B$$.

Claim: We can choose a closed path in $$S$$ that lies entirely in $$B$$ except for endpoint which is $$p$$ such that a nbd of the closed path is a Möbius band.

As for why such a loop exists, take a finite cover of charts of your surface (which exists because of compactness). Now, start with one of them $$U_1$$. Next, pick a second chart $$U_2$$ that intersects $$U_1$$. If they have negative Jacobian on the overlap, invert the orientation of $$U_2$$. You have now picked an orientation on $$U_1\cup U_2$$. Continue with $$U_3$$, which intersects $$U_1\cup U_2$$, to get an orientation on $$U_1\cup U_2\cup U_3$$. And so on.
At some step $$n$$, this will necessarily fail, because the surface is non-orientable and our cover is finite. In practice, what will happen is that the overlap $$\left(\bigcup_{i has several connected components, and among these there are two components with opposite Jacobian sign. No matter what orientation you choose for $$U_n$$, one of those two overlaps will be orientation reversing.
Now take a point $$p$$ in one of those two connected components, and a point $$q$$ in the other. Connect them by a path through $$U_n$$ and by a path through $$\bigcup_{i. Together these two paths make a loop that switches orientation.
• This assumes, of course, that the surface is connected, or at the very least that $U_1$ is on a non-orientable connected component of the surface. Commented Oct 11, 2022 at 16:31
• How does one argue that the thickening of the simple closed loop is a Mobius band and not some other non-orientable surface? Attempt: Locally in each $U_i$ the thickening looks like $(a,b)\times (-\delta,\delta)$. We can similarly cover a real Mobius band with nbds $W_i$ and then write a homeomorphism between $U_i\cap$ {thickening} and $W_i \cap$ {Mobius band} which agrees on intersections. Commented Oct 13, 2022 at 4:36
• That looks like a very reasonable argument. The fact that it's a thickening of a circle means it's necessarily a band. That leaves two options (Mobius or open cylinder), and the covering of the band by the $U_i$ makes it non-orientable. Commented Oct 13, 2022 at 5:34