TITLE: Question about orientable manifolds
QUESTION [11 upvotes]: Let $M$ be a connected orientable smooth manifold.
Is it true that $M$ must have only 2 orientations?
If yes, why?

REPLY [12 votes]: I am also trying this question and again like you do, I am only given very basic definitions of orientations using atlas. Let me try first and see if this works, and give me your comments! (By the way I am reading sharpe's book)
Let $M$ be a connected orientable smooth manifold.
We say that an atlas $(U_{i},\phi_{i})$ is oriented if the jacobian of the transformation between two charts at the intersection is positive. (i.e. $\det\mbox{Jac}\phi_{j}\circ\phi_{i}^{-1}>0$)
Let us say that two oriented atlas $(U_{i},\phi_{i})$ and $(V_{j},\psi_{j})$ are equivalent if the union forms an oriented atlas for $M$. 
Claim
Let $(U_{i},\phi_{i})$ and $(V_{j},\psi_{j})$ be oriented atlases such that its union forms an atlas for $M$. If there exists $p\in U_{\alpha}\cap V_{\beta}\subseteq M$ such that the determinant of the Jacobian $\psi_{\beta}\circ\phi_{\alpha}^{-1}$ is positive, then both atlases are equivalent.
Proof
We will use the fact that $M$ is connected! We define two subsets as below
$S_{1}=\{p\in M: \exists U_{i},V_{j}\mbox{ such that }\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})|_{p}>0\}$ 
$S_{2}=\{p\in M: \exists U_{i},V_{j}\mbox{ such that }\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})|_{p}<0\}$ 
Note that we don't have $0$ as a case because $\psi_{j}\circ\phi_{i}^{-1}$ is a diffeomorphism on the intersection.
Now we have $S_{1}\cap S_{2}=\emptyset$ because suppose we have $p\in S_{1}\cap S_{2}$, then $p\in U_{i}\cap V_{j}$ and $p\in U_{k}\cap V_{l}$ such that $\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})>0$ and $\det\mbox{Jac}(\psi_{l}\circ\phi_{k}^{-1})<0$. But we will run into problems because while $\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})>0$, we also have
$\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})=\det\mbox{Jac}(\psi_{k}\circ\psi_{l}^{-1})\det\mbox{Jac}(\psi_{l}\circ\phi_{k}^{-1})\det\mbox{Jac}(\phi_{k}\circ\phi_{i}^{-1})<0$.
On the other hand, $S_{1}\cup S_{2}=M$, because if $p$ lies in $M$, it must go into some $U_{i}$ and $V_{k}$ and so the determinant of the jacobian is either $>0$ or $<0$. 
Now it is easy to see that both $S_{1}$ and $S_{2}$ are open because taking jacobian, and then the determinant are all continuous functions and both $(0,\infty)$ and $(-\infty, 0)$ are open in $\mathbb{R}$, (so the pre-image is open in the intersection $U_{i}\cap V_{j}$ and hence is also open in $M$.
But there exists a point which both atlas agree (or compatible), this means $S_{1}$ is not empty, and so $S_{2}$ is empty because $M$ is connected.
In a same manner, if both oriented atlases disagree on one point, they disagree everywhere.
Attempted proof to your question
Since $M$ is orientable, we have already found one oriented atlas $(U_{i},\phi_{i})$.
We take the same atlas and give a negative orientation. This can be easily done by post composing $\phi_{i}$ with the operation $L:(x_{1},\ldots,x_{n})\mapsto (-x_{1},\ldots, x_{n})$ to get $L\circ\phi_{i}$.
These are the only two orientations by our previous claim because if we choose another oriented atlas $(W_{k},\lambda_{k})$ for $M$, just choose one point $p\in U_{i}\cap W_{k}$. Then because the determinant of the jacobian is either positive or negative, therefore $(W_{k},\lambda_{k})$ must agree with either $(U_{i},\phi_{i})$ or $(U_{i},L\circ\phi_{i})$.
Comments?