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.


1 Answer 1


In your non-orientable surface there must be a simple, closed loop that "switches orientation" (i.e. if you go along the loop, using charts and positive Jacobian to keep your orientation all the way around, then you will have switched orientation by the time you go one full lap). A thin neighborhood of that loop will be 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 $\left(\bigcup_{i<n}U_i\right)\cap U_n$ 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 $\bigcup_{i<n}U_i$. Together these two paths make a loop that switches orientation.

  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Commented Oct 11, 2022 at 16:31
  • $\begingroup$ 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. $\endgroup$ Commented Oct 13, 2022 at 4:36
  • 1
    $\begingroup$ 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. $\endgroup$
    – Arthur
    Commented Oct 13, 2022 at 5:34

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