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!
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.)
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.
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 $O(\phi)=1$.
I will use (elementary) results from Lee's Introduction to Smooth Manifolds to streamline the argument.
Let's define the (compact) Mobius band intrinsically (i.e., not via any particular embedding in $\mathbb R^3$) by $M=\mathbb R\times[0,1]/\sim$ where $(x,y)\sim (x+2n,y)$ and $(x,y)\sim(x+2n+1,1-y)$ ($n\in\mathbb N$). Then $M$ has a unique structure which makes the canonical projection map $$ q:\mathbb R\times [0,1]\to M=\mathbb R^2/\sim,(x,y)\mapsto [x,y], $$ a local diffeomorphism. Charts in terms of $q$ are then given by $q\vert_U^{-1}$ where $U\subset\mathbb R\times[0,1]$ is an open neighbourhood that doesn't contain distinct points that are identified under the equivalence relation (see also Example 10.3 and Problem 10-1(a) in Lee for the very similar Möbius bundle).
Suppose $M$ is oriented. By Exercise 15.4 in Lee we know that every local frame with connected domain is either positive oriented or negatively oriented. So let's describe certain coordinate frames. To this end, we take specific charts $\phi=q\vert_{(0,1)\times [0,1]}^{-1}$ and $\psi=q\vert_{(1/2,3/2)\times [0,1]}^{-1}$.
Now consider the coordinate frame $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ corresponding to $\phi$. This frame is either positively or negatively oriented (by Exercise 15.4). Likewise, we know that the frame $\frac{\partial}{\partial\tilde x},\frac{\partial}{\partial\tilde y}$ corresponding to $\psi$ is either positively or negatively oriented.
By Proposition 14.6 in Lee, we know that the collection of all oriented smooth charts is a consistently oriented atlas for $M$. It follows that the transition matrix between $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial\tilde x},\frac{\partial}{\partial\tilde y}$ needs to have either positive determinant at each point in $q((0,1/2)\cup (1/2,1)\times [0,1])$ or negative determinant at each point in $q((0,1/2)\cup (1/2,1)\times [0,1])$. Note that the transition map $$\psi\circ\phi^{-1}:(0,1/2)\cup (1/2,1)\times[0,1]\to (1/2,1)\cup(1,3/2)\times[0,1] $$ is the identity on $(1/2,1)\times [0,1]$, hence the Jacobian of this map has determinant equal to $1$, which is positive, and therefore it must be positive on $(0,1/2)\cup (1/2,1)\times [0,1]$. However, on $(0,1/2)\times [0,1]$ the transition map $\psi\circ\phi^{-1}$ is given by $(x,y)\mapsto (x+1,1-y)$, whose Jacobian has negative determinant. We've arrived at a contradiction, and therefore $M$ is not orientable.
