# Inducing orientations on boundary manifolds

Given a $k$-manifold $M$, such that $\partial M$ is a $(k-1)$-manifold, there is a standard way in which $\partial M$ inherits the orientation of $M$. So if $M$ is oriented by the form field $\omega$, then $\omega_\partial=\omega(v_{\text{out}},v_1,\dots,v_{k-1})$ orients $\partial M$, where the $v_1,\dots,v_{k-1}$ are basis vectors for a tangent space on a point of the boundary, and $v_\text{out}$ is an outward pointing vector (see here: http://math.ucsd.edu/~jeggers/math31ch/pieces.pdf, which follows the exposition of Hubbard & Hubbard).

In practice I find this very difficult to interpret. How can one find this $v_\text{out}$ in practice?

For instance, take $M$ to be defined by $z=x^2+y^2$ for $x\ge0$ (technically a manifold with boundary, in fact a half-paraboloid). It's a fact that $\partial M$ is defined by the parabola $z=y^2$. If $M$ is oriented by $dx\wedge dy$, how does this induce an orientation on the boundary (in fact we should obtain a 1-form that induces an orientation on the boundary)?

Note: I find the whole discussion of boundary orientation to be rather clumsy in its exposition. In the case of surfaces inducing orientations on boundary curves, I was originally taught to imagine that the normal vector that induces the surface orientation is me walking on the surface. This normal should walk along the boundary such that the surface lies to its left. OK - but how does this translate to the $v_{\text{out}}$ interpretation, so that we can generalize to other dimensions easily?)

Edit: An ideal answer would merge the intuitive approach with the rigorous approach, if possible. I would grateful to anybody who could help order my (at present confused) understanding!

Let $M$ be a smooth, oriented, $n$-manifold and $S$ be an embedded ($n-1$) hypersurface. Let $\omega$ be an orientation form on $M$, so that $\omega$ is just a nowhere vanishing $n$-form. We would like to find a way of tweaking this $\omega$ to define a nowhere vanishing $n-1$ form on $S$. How do we do this? The key is that since $\omega$ is nowhere vanishing then (without loss of generality) it is always positive; that is, for all $p \in M$, $$\omega_p\left(\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_n}\right|_p\right) >0$$ where the partials form a basis for $T_pM$. If we take $p \in S$, then $T_pS$ will give use $n-1$ vectors to plug into $\omega_p$, so our goal is to find a vector field $v_\text{out}: S \to TM$ such that if $B_p=\left\{\left.\frac{\partial}{\partial x_1}\right|_p, \ldots, \left.\frac{\partial}{\partial x_{n-1}}\right|_p\right\}$ is a basis for $T_pS$ then $\{(v_\text{out})_p\} \cup B_p$ is a basis for $T_pM$. In that case we can define an $n-1$ form $\hat \omega_p$ on $S$ as $$\hat \omega_p\left(\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_{n-1}}\right|_p\right) = \omega_p \left((v_\text{out})_p,\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_{n-1}}\right|_p\right)$$ which will also be positive for every $p \in S$ since $\omega$ is positive for every $p \in S \subseteq M$.
So how do we guarantee that $(v_\text{out})_p$ is always "linearly independent" to $B_p$? We simply demand that $(v_\text{out})_p$ never lies in $T_pS$!
Now if we specifically look at boundaries, our discussion above suggests that we find a vector field $N: \partial M \to TM$ such that $N_p \notin T_p\partial M$ (these are usually called inward-pointing vector fields). The Stokes' orientation is to take the $-N$ and define the $\hat \omega_p$ as above.
The other way of defining induced orientations is to do it in terms of the charts, so that it suffices to look at just the upper half plane $$\mathbb H^n = \{(x_1,\ldots,x_n) \in \mathbb R^n: x_n \geq 0\}.$$ Diffeomorphisms $T: \mathbb H^n \to\mathbb H^n$ with positive Jacobian induce diffeomorphisms $\partial T: \partial \mathbb H^n \to \partial \mathbb H^n$ with positive Jacobian (this is a straightforward but messy exercise with determinants). The orientation form on $\mathbb H^n$ is the standard form $\omega = dx_1 \wedge \cdots \wedge dx_n$, and the vector $-\frac{\partial}{\partial x_n}$ is outward pointing along the boundary. Furthermore, $$\omega\left(-\frac{\partial}{\partial x_n}, \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_{n-1}} \right) = (-1)^n \omega\left(\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2}, \ldots, \frac{\partial}{\partial x_{n}} \right)$$ so this suggests that we should take $\hat \omega = (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}$ to be the induced orientation on $\partial \mathbb H^n$. We then define the induced boundary on $\partial M$ to be $$[\partial M] = \phi^*[\partial \mathbb H^n]$$ where the square brackets are the equivalence class of orientations and $\phi: M \to \mathbb H^n$ is a coordinate chart of $M$.
In your example, let's use $x$ and $y$ as coordinates for $M$. The outward vector on the boundary should be tangent to M. A suitable choice is $-\partial_x$, considered as a coordinate basis vector on $M$. The minus sign is there because $M$ is on the postive $x$ side, so the outward vector should point to negative $x$ values. The induced orientation on the boundary is therefore $-\partial_x\cdot(dx\wedge dy)= -dy$. So the orientation on the boundary runs from positive $y$ to negative $y$.