# Why is the Möbius strip not orientable?

I am trying to understand the notion of an orientable manifold.
Let M be a smooth n-manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1}))> 0$ (where defined). My question is:
Using this definition of orientation, how can one prove that the Möbius strip is not orientable?

Thank you!

• It's intuitively "obvious" but I wouldn't know how to "rigorously" prove the thing, so +1 for that. – J. M. is a poor mathematician Dec 27 '10 at 1:21
• This isn't the argument you were looking for, but you may find the following notes interesting: math.uwaterloo.ca/~karigian/teaching/calculus/moebius.pdf – WWright Dec 27 '10 at 1:31
• I remember seeing an explicit construction of two charts that cover all the Möbius strip and have jacobian determinant negative, in do Carmo's book "Differential geometry of curves and surfaces" (at least in the portuguese edition). I think he proposes as an exercise to show that a surface of this kind is non-orientable. – Ronaldo Dec 27 '10 at 2:43
• @WWright: actually, the notes you link to do give a proof of the nonorientability of the Mobius strip: they show that Stokes' theorem is not valid on the Mobius strip, whereas it is valid on smooth orientable manifolds. (One might of course prefer a more direct argument.) – Pete L. Clark Dec 27 '10 at 4:46
• @Pete L. Clark - When I read the original question, I'd imagined that the difficulty was connecting the various definitions of orientation and this is why I said it "isn't the argument you were looking for." Maybe I assumed incorrectly, but in my own experience, I've found rectifying the various definitions of orientation to be the main roadblock. – WWright Dec 27 '10 at 16:16

If you had an orientation, you'd be able to define at each point $p$ a unit vector $n_p$ normal to the strip at $p$, in a way that the map $p\mapsto n_p$ is continuous. Moreover, this map is completely determined once you fix the value of $n_p$ for some specific $p$. (You have two possibilities, this uses a tangent plane at $p$, which is definable using a $(U_\alpha,\phi_\alpha)$ that covers $p$.)

The point is that the positivity condition you wrote gives you that the normal at any $p'$ is independent of the specific $(U_{\alpha'},\phi_{\alpha'})$ you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of $n_p$ changes sign when you return to $p$, which of course is a contradiction.

(This is just a formalization of the intuitive argument.)

• So the normal vector takes the place of the pencil one usually uses to demonstrate the nonorientability... awesome. :) – J. M. is a poor mathematician Dec 27 '10 at 1:49
• +1: this is the answer I was about to give. One key point explained in more detail: an orientation induces a map from the surface $S$ into the unit normal bundle of $S$, any fiber of which has just two points. On a sufficiently small connected neigborhood $U_p$ of a point $p$, the unit normal bundle trivializes and we are looking at a continuous map $U_p \rightarrow \{ \pm 1 \}$: hence it is constant. Therefore the entire map from $S$ to the unit normal bundle is determined by its value at a single point. Now following the normal vector around the central circle gives a contradiction. – Pete L. Clark Dec 27 '10 at 1:57
• It might help the OP to leave an even more detailed answer, explaining how for a hypersurface $S$ embedded in Euclidean space, an orientation on $S$ induces an orientation on the unit normal bundle of $S$.... – Pete L. Clark Dec 27 '10 at 2:04
• The point being that there are ten million (or so!) slightly different ways to define orientations, many of which look plausibly equivalent but for which a beginning student may have difficulty proving the equivalence. Certainly this was a problem for me when I was first learning this subject: I had on one hand the geometric intuition and on the other the formal definitions, but it was not so easy to combine the two (clap?). I don't have the time to do this just now, but perhaps later if no one else beats me to it. – Pete L. Clark Dec 27 '10 at 2:05

A simple answer that IMO is easy to justify using your definition of orientation goes like this.

Given any manifold $M$ and a point $p \in M$ there is a homomorphism $O : \pi_1(M,p) \to \mathbb Z_2$ and the idea is this: if $\phi : [0,1] \to M$ is a path such that $\phi(0)=\phi(1)=p$, given any basis for the tangent space to $M$ at $p$, $T_pM$ you can parallel transport that basis along the path, and you'll get a second basis for the tangent space at $\phi(1)=p$, $T_pM$. And you can ask, is the change-of-basis map from your 1st to your 2nd basis for $T_pM$ orientation-preserving -- i.e. is the determinant of that linear transformation positive? If it is, define $O(\phi)=0$, if the determinant is negative, define $O(\phi)=1$.

Fact: the path-component of $p$ in the manifold $M$ is orientable if and only if $O$ is the zero function, $O=0$. You prove it by cutting your path $\phi$ into small segments and comparing orientations within charts -- the key analytical step is the intermediate value theorem, using that determinant is a continuous function of matrices.

Of course, in this discussion "parallel transport" assumes a Riemann metric but you don't really need a Riemann metric for this argument to work. The parallel transport of vectors along a path $\phi$ simply means continuously-varying vectors such that the vector corresponding to $t \in [0,1]$ is always tangent to the manifold, i.e. elements of $T_{\phi(t)} M$. And of course if you're transporting $n$ vectors you demand that these $n$ vectors always make basis for $T_{\phi(t)}M$.

And in the case of the Moebius band, given any concrete model of the Moebius band you transport a basis along any path that goes once around the band and $O(\phi)=1$.

Let $M:=\{(x,y)|x\in\mathbb R, -1<y<1\}$ be an infinite strip and choose an $L>0$. The equivalence relation $(x+L,-y)\sim(x,y)$ defines a Möbius strip $\hat M$. Let $\pi: M \to \hat M$ be the projection map. The Möbius strip $\hat M$ inherits the differentiable structure from ${\mathbb R}^2$. We have to prove that $\hat M$ does not admit an atlas of the described kind which is compatible with the differentiable structure on $\hat M$. Assume that there is such an atlas $(U_\alpha,\phi_\alpha)_{\alpha\in I}$. We then define a function $\sigma:{\mathbb R}\to\{-1,1\}$ as follows: For given $x\in{\mathbb R}$ the point $\pi(x,0)$ is in $\hat M$, so there is an $\alpha\in I$ with $\pi(x,0)\in U_\alpha$. The map $f:=\phi_\alpha^{-1}\circ\pi$ is a diffeomorphism in a neighbourhood $V$ of $(x,0)$. Put $\sigma(x):=\mathrm{sgn}\thinspace J_f(x,0)$, where $J_f$ denotes the Jacobian of $f$. One easily checks that $\sigma(\cdot)$ is well defined and is locally constant, whence it is constant on ${\mathbb R}$. On the other hand we have $f(x+L,y)\equiv f(x,-y)$ in $V$ which implies $\sigma(L)=-\sigma(0)$ -- a contradiction.