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The concept of orientation has already appeared and needs to be defined formally. Let's dig in

In $\mathbb{R}^n$ we have the canonical basis $\{e_1,e_2,...,e_n\}$ with its dual basis $\{\varphi_1,\varphi_2,...,\varphi_n\}$. The $n$-form $\varphi_1\wedge\varphi_2\wedge...\wedge\varphi_n$ is a generator of $\bigwedge^n(\mathbb{R} ^n)$; $\varphi_1\wedge\varphi_2\wedge...\wedge\varphi_n(v_1,v_2,...,v_n)=\text{Det}(v_{ij})$, which is the volume of the parallelepiped made with the vectors $v_i$. Now, to really obtain the volume, which is always positive, we should take the absolute value of the previous determinant: there are basis $\{v_1,v_2,...,v_n\}$ of $\mathbb{R}^n$ with $\text{Det}(v_{ij})>0$ and basis with $\text{Det}(v_{ij})<0$ (if $\{v_1,v_2,...,v_n\}$ are linearly dependent, the form vanishes). Taking absolute values is not very nice, so instead we will restrict ourselves to basis with positive determinant. Now the form $\varphi_1\wedge\varphi_2\wedge...\wedge\varphi_n$ or any multiple gives indeed a value proportional to the volume, but only with these positive-determinant-basis. We might as well have taken the negative-determinant-basis; the crucial point is to always evaluate on the same type of basis, for the form to truly represent something proportional to the volume. So in this spirit we state the following definition:

Let $E$ be a vector space of dimension $n$ and $\mathcal{B}=\{B=\{v_1,v_2,...,v_n\}| \text{\(B\) is basis of \(E\)}\}$ the set of all basis of $E$. Let define the following relation of equivalence in $\mathcal{B}$: $B\sim C$ if the coordinate matrix of expressing the elements of $B$ in the basis $C$ has positive determinant. $\sim$ divides $\mathcal{B}$ into two equivalence sets. An orientation in $E$ is the choice of one of these two classes, whose basis will be called positively oriented

This way, once an orientation in $E$ has been chosen, a $n$-form is truly a measure volume which is coherent when comparing any two positively oriented basis. The order of the vectors in the basis is crucial: permuting two vectors switches the orientation

Let $M$ be a manifold of dimension $n$. An orientation in $M$ consists in choosing an orientation in each tangent space $T_p M$ which is locally constant in the following sense: for each point there exists a chart $\varphi:U\longrightarrow M$ such that in each point of $\varphi(U)$ the chosen orientation matches the class of $\left\{\dfrac{\partial}{\partial x_1},...,\dfrac{\partial}{\partial x_n}\right\}$. Whenever $M$ admits an orientation, it is said to be orientable; with a chosen orientation in $M$ a chart $\varphi:U\longrightarrow M$ is said to be positively oriented if the induced basis $\left\{\dfrac{\partial}{\partial x_1},...,\dfrac{\partial}{\partial x_n}\right\}$ are positively oriented

Now we make sense of the ways in which we previously tried to depict orientation. The direction of the curve shows, in each tangent line, the semiline containing the positively oriented vectors, since in a one-dimensional real vector space the basis are sets of one nonzero vector, and the two orientations match the two semilines divided by the origin. And in the surfaces, the colors denoted the turning direction to go from the first vector of a positively oriented basis to the second: blue counterclockwise and yellow for clockwise. But for higher dimensions, the right concept is to divide the basis into positive and negative in each point of the manifold