# What does it mean for a manifold to be oriented?

I'm currently working through Spivak's Calculus on Manifolds. I've got to Stokes' Theorem, which is stated thus (the bold is my emphasis):

Stokes' Theorem

If $M$ is a compact oriented $k$-dimensional manifold-with-boundary and $\omega$ is a $(k-1)$-form on $M$, then $$\int_M d \omega = \int_{\partial M} \omega$$

I've read and understood the proof Spivak gives, but I'm unsure about what oriented means. In applications (such as in fluid mechanics), I've sometimes seen that this condition is ignored, or hand-waved over, but it's obviously key.

I understand what is meant by an orientation (the fact that any non-zero $\omega \in \Lambda (\mathbb{R}^n)$ splits the bases of a linear space into two groups - those with $\omega(v_1,...,v_n) >0$ and those with $\omega(v_1,...,v_n) <0$, and these two groups are orientations for $V$), but I don't see how I would verify that a given manifold-with-boundary is oriented (other than it having something to do with the orientation of the tangent space).

To make things more concrete, I know (i.e. have it on good authority) that the Möbius Band is not oriented. However, I know that the 2-sphere is oriented.

So, in short: Given a manifold-with-boundary, how do we know if it's oriented or not?

• An oriented manifold is one for which there exists a consistent normal at each point on the manifold. Such a consistent normal can be defined on a sphere or plane (orientable) but not on a Möbius strip or Klein bottle (non-orientable). One can test orientability by parallel transport of a defined normal along arbitrary closed paths within the manifold. – David G. Stork Apr 16 '15 at 18:30
• @DavidG.Stork, I don't think we want to mention parallel transport here. Moreover, your answer seems to make sense only when $M$ is a hypersurface in an orientable manifold? – Ted Shifrin Apr 16 '15 at 18:48
• Whoever downvoted this is an arse. – goblin Apr 23 '17 at 7:43

An orientation of a manifold can be defined in various ways. Let's suppose all throughout that $M$ is a connected smooth manifold (possibly with boundary), of dimension $m$. If $\omega, \eta \in \Omega^{m}(M)$ are two non-vanishing top forms on $M$, then one can prove that there exists a non-vanishing $f \in C^{\infty}(M)$ such that $\omega = f \eta$. Since $M$ is connected and $f$ is non-vanishing, it is either positive everywhere or negative everywhere. If $f$ is positive everywhere we say $\omega$ and $\eta$ define the same orientation. In this definition, an orientation is an equivalence class of non-vanishing top forms (where the equivalence relation is "define the same orientation" as defined above). An oriented manifold is then a manifold $M$ along with a choice of orientation, i.e. a particular choice of a non-vanishing top form.
We usually say that $S^2$ is an orientable manifold, meaning it can be given an orientation. You would consider $S^2$ an oriented manifold if you have fix an orientation of it. One usually takes $\omega = x\ dy \wedge dz - y\ dx \wedge dz + z\ dx \wedge dy$ as the orientation for $S^2$.
It is important in Stokes' theorem to understand that if you have an oriented manifold with boundary, it induces an orientation on the boundary in a natural way. The most intuitive way that I know to explain this is the following: suppose $M$ is your connected manifold with boundary and $\omega$ is your choice of non-vanishing top form on $M$. First we fix $X$ a non-vanishing outward-pointing vector field on $\partial M$. By outward-pointing I mean that if we take coordinates $x^1, \ldots, x^m$ near a point in the boundary, and the boundary is defined by the condition $x^m = 0$, then the vector field $X$ has an everywhere negative $\frac{\partial}{\partial x^m}$ component. Such a vector field can always be obtained (construct it locally then patch it together by partitions of unity). Then we define a non-vanishing top form on $\partial M$ given by $(\iota_X \omega)(X_1, \ldots, X_{m - 1}) = \omega(X, X_1, \ldots, X_{m - 1})$. It turns out that this choice of orientation doesn't depend on the choice of $X$, as long as it is outward-pointing (for a proof of this, see Lee's Introduction to Smooth Manifolds, in the chapter about orientations). Hence this orientation on $\partial M$ depends only on the choice of orientation we had in $M$. This is the choice of orientation we always make for the boundary of an oriented manifold.
It is important to note one thing about the induced boundary orientation. Set $\mathbb{H}^{m} = \{(x_1, \ldots, x_m) \in \mathbb{R}^{m}\ |\ x_m = 0\}$. Note that $\partial \mathbb{H}^{m} \cong \mathbb{R}^{m - 1}$. The induced boundary orientation on $\partial \mathbb{H}^{m}$ matches with the natural orientation of $\mathbb{R}^{m - 1}$ when $m$ is even, but is the opposite orientation when $m$ is odd.