In Shigeyuki Morita's Geometry of Differential Forms, orientability is defined in the following way:
If we can assign an orientation to each point on a manifold $M$ in such a way that the orientations as any two sufficiently near points on $M$ are coherent, we say that $M$ is orientable. (Page 48)
However, leading up to this definition, it is never explicitly defined what it means that two points have coherent orientations. The only explanation is in the discussion about surfaces:
When an orientation is specified at a point, the "same" orientation is specified at an arbitrary point in a neighborhood of the point. This is called the coherent orientation. We specify an orientation at a point on a surface, and choose the coherent orientation at each point on a curve starting at the point. If the curve goes back to the starting point, the original orientation may or may not coincide with the orientation propagated along the curve. Now a surface is orientable if the orientation propagated along any curve always comes back to the starting orientation. In this case we can assign an orientation to all points on the surface in such a way that near points have mutually coherent orientations. (Page 46)
Page 47 then involves defining orientations at a point $p\in M$ by choosing a basis on the tangent space $T_{p}M$ along with the standard "right-hand orientation" of $\mathbb{R}^{3}$ example, but never is it said what it means for orientations to be the "same" or coherent between points, or how the orientation of one point specifies one on points in a neighborhood of that point.
The closest thing I could think of would be that the existence of an atlas so that the Jacobians of the transition maps between two local charts is positive, as is the definition in do Carmo's Differential Forms and Applications, Page 50. However, the equivalence of these statements is given as a proposition in Morita's text right after the definition of orientability without proof, which does not allow for me to see how the definition is used.
Any help with providing an explicit definition of orientations between two points being coherent would be greatly appreciated. Maybe I'm just missing something obvious, and if this is the case, I would very much like to be shown where it is.
Best Answer
First understand what we mean by an orientation for $\mathbb{R}^3$. This is tied to the idea of a right-handed set of vectors: we say $\{v_1,v_2, v_3 \}$ is a right-handed set of vectors if $\text{det}(v_1|v_2|v_3)>0$. Geometrically, this means that $v_3$ is on the same side of the plane spanned by $v_1$ and $v_2$ as $v_1 \times v_2$. We are most familar with the case of an orthonormal right-handed frame where we simply insist $v_1 \times v_2 =v_3$. In two dimensions, the condition $\{ v_1,v_2 \}$ be right-handed again is captured by $\text{det}(v_1|v_2)>0$. Geometrically, the condition in two-dimensions means that the second vector is obtained from the first by a Counter-Clockwise rotation. In all cases, an orientation allows us to decide which way is up once all the other ways are fixed. To be slighly more precise, if we fix $n-1$ coordinates in $\mathbb{R}^n$ then an orientation will provide us a framework in which we can decide what the positive direction is for the remaining $n$-th coordinate. This is equivalent to insisting there be an ordered set of vectors $\{v_1,v_2,\dots v_n \}$ for which $\text{det}[v_1|v_2|\cdots|v_{n-1}|v_n]>0$.
Details of the choice: ok, suppose $v_1,\dots , v_{n-1}$ are fixed and you are given $a,b$ vectors in $\mathbb{R}^n$ then $\text{det}[v_1|v_2|\cdots|v_{n-1}|a]>0$ whereas $\text{det}[v_1|v_2|\cdots|v_{n-1}|b]<0$. Then, in terms of the given orientation $\{v_1,v_2,\dots v_n \}$ we see $a$ points in the upward-direction whereas $b$-points in the downward-direction. Of course, this comment is relative to the coordinate system defined by the span of the orientation.
Relation to wedge product: $$ v_1 \wedge v_2 \wedge \cdots \wedge v_n = \text{det}[v_1|v_2|\cdots|v_{n}] e_1 \wedge e_2 \wedge \cdots \wedge e_n. $$ Where $(e_i)_j = \delta_{ij}$ defines the standard basis. The coefficient of the $n$-form will be positive iff the set of vectors $\{v_1,v_2, \dots, v_{n}\}$ shares the same orientation as the standard basis. Moreover, all such related bases form the so-called standard orientation of $\mathbb{R}^n$.
Now, for the case of a manifold it is much the same, however, we have to consider the derivations which fill $T_pM$ at various $p$. The natural orientation given by a coordinate chart $(x^i)$ at $p \in M$ is simply the ordered $n$-tuple (using $\partial_i = \frac{\partial}{\partial x^i}$) $$ \{ \partial_1|_p, \partial_2|_p, \dots, \partial_n|_p \}$$ dual to these $dx^1, dx^2, \dots dx^n$ form the volume form: $$ vol=dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n$$ Given this, we can judge if $v_1,v_2, \dots v_n$ is a coherent orientation to the one which is naturally induced from the coordinate derivations. Simply check: $$ vol(v_1,v_2, \dots , v_n) > 0 ? $$ Now, if there is another coordinate system $(y^j)$ also defined at $p$ then we can ask if $$ \{ \partial/\partial y^1,\partial/\partial y^2, \dots, \partial/\partial y^n \}$$ forms a coherent orientation with the one we already induced from $(x^i)$. The chain rule connects the $x$ and $y$ coordinate derivations and when we feed that to the volume form we find the determinant of the transition function appears. Therefore, two coordinate systems provide coherent orientations if their transition functions have Jacobians with positive determinant.
All of this said, it seems that the phrase you quote:
could apply to what I describe above. But, the larger idea concerns a curve so almost certainly the discussion above about coordinate-induced orientations is not the real answer to the question.
What about a curve? Morita talks about choosing an orientation along the curve. I'm not entirely sure which construction he has in mind. But, here's a possibility: given a curve we can frame the curve using the tangent vector, and the change in the tangent vector etc.. well, in the abstract, how to frame a curve? I leave to your imagination, but, in many cases it can be done. So, there is a natural way to assign $n$-vectors $f_1(t), \dots , f_n(t)$ to a given curve at point $\gamma(t)$. Suppose the curve is closed such that after time $T$ the curve returns to the initial point $p=\gamma(0)=\gamma(T)$. Then, we can compare the orientations of $f_1(0), \dots , f_n(0)$ and $f_1(T), \dots , f_n(T)$ and see if they are compatible (coherent). In particular, feed both to the volume form at $p$ and see that they share the same signed result. If you did this for the tangent, normal frame field for a curve on the Mobius band then you'd find the orientation carried by the frame of the curve was not coherent when you go around the band once.