# Orientability on Manifolds

So I'm having some issues with the definition of orientability. I'll take the case of $\mathbb R$ for now. I know there's supposed to be only 2 orientations on a smooth, connected manifold, and for $\mathbb R$ I'm guessing these arise from the frames $x \rightarrow \frac{\partial}{\partial x}$ and $x \rightarrow -\frac{\partial}{\partial x}$. However what about the frame $x \rightarrow \frac{\partial}{\partial x}$ for $x \geq 0$ and $x \rightarrow -\frac{\partial}{\partial x}$ $x \leq 0$? Or for that matter literally any choice of $\frac{\partial}{\partial x}$ or $-\frac{\partial}{\partial x}$ for whatever $x$? These all give frames because they are clearly global sections and since the dimension is 1 they are bases for each tangent space. Then since the frame is global there is no problem with 2 sections in the frame disagreeing on their orientation as there is just 1 section and we can just define the pointwise orientation on $\mathbb R$ to be the one given by this random assignment.

• You at least have a typo, as your rule assigns $0$ two distinct vectors. Next, would you please write down your definition of orientability/orientation? I think you don't have the 'standard' definitions... – peter a g Dec 3 '17 at 14:40
• @peter_a_g Ah yeah my bad I meant one of them strict inequailty. Our definition of a manifold being orientable is if it admits a continuous orientation, which means that every point in the manifold is in the domain of positively oriented local frame. And a local frame is oriented if for all points in the domain of the frame, the orientation from the basis you get by evaluating the frame at the point is the same as the chosen orientation on the tangent space. – Fromage Dec 3 '17 at 14:53

Your assignment is not continuous around $0$, so it does not define a section. You could try to patch it by mapping $x$ to $x\frac{\partial}{\partial x}$, but then this does not yield a frame at $0$.