What exactly does "outward normal vector" means when talking about an orientation?

Question

I am asked to calculate the integral of a $2$-form on the sphere: $$ \int_{S^2} \frac{x\mathrm{d}y\wedge \mathrm{d}z-y\mathrm{d}x\wedge \mathrm{d}z+z\mathrm{d}x\wedge \mathrm{d}y}{(x^2+y^2+z^2)^{\frac{3}{2}}}$$

where we choose the orientation of $S^2$ with "outward normal vector".

So what does this exactly mean by saying "outward normal vector"? For example, to start with my calculation, I need to find an atlas compatible with the given orientation. How should I justify whether the atlas given by Riemann stereographic projection is compatible with the desired orientation or not?

By the way, how about saying that choosing an anticlockwise orientation of $S^1$?

Any Hints or Solutions for this integral are welcomed. THANKS!

Answer 1

For the sphere we use polar coordinates: $$\tag{1} \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\cos\varphi\,\sin\theta\\\sin\varphi\,\sin\theta\\\cos\theta\end{pmatrix}\,. $$ Then, \begin{align} dx&=-\sin\varphi\,\sin\theta\,d\varphi+\cos\varphi\,\cos\theta\,d\theta\,,\\[2mm] dy&=\cos\varphi\,\sin\theta\,d\varphi+\sin\varphi\,\cos\theta\,d\theta\,,\\[2mm] dz&=-\sin\theta\,d\theta\,, \end{align} and \begin{align} dy\wedge dz&=\cos\varphi\,\sin^2\theta\;d\theta\wedge d\varphi\,,\\[2mm] -dx\wedge dz&=\sin\varphi\,\sin^2\theta\;d\theta\wedge d\varphi\,,\\[2mm] dx\wedge dy&=\sin\theta\,\cos\theta\;d\theta\wedge d\varphi\,. \end{align} This system can be written in vector form as $$ \begin{pmatrix} dy\wedge dz\\-dx\wedge dz\\dx\wedge dy\end{pmatrix} =\underbrace{\begin{pmatrix}\cos\varphi\,\sin\theta\\\sin\varphi\,\sin\theta\\\cos\theta\end{pmatrix}}_{\textstyle(1)}\sin\theta\;d\theta\wedge d\varphi\,. $$ Note that the vector (1) points out of the sphere and has length one so we can identify it as the outward pointing unit normal $\vec n$ to the sphere $S^2\,.$

Recall the definition of positively oriented basis in Ted Shifrin's answer.

In this case: Two orthogonal tangent vectors at the point $p$ on the sphere given by (1) are

\begin{align} \vec v_1&:=\partial_\theta\begin{pmatrix}\cos\varphi\,\sin\theta\\\sin\varphi\,\sin\theta\\\cos\theta\end{pmatrix}= \begin{pmatrix}\cos\varphi\,\cos\theta\\\sin\varphi\,\cos\theta\\-\sin\theta\end{pmatrix}\,,\\[2mm] \vec v_2&:=\partial_\varphi\begin{pmatrix}\cos\varphi\,\sin\theta\\\sin\varphi\,\sin\theta\\\cos\theta\end{pmatrix}=\begin{pmatrix}-\sin\varphi\,\sin\theta\\\cos\varphi\,\sin\theta\\0\end{pmatrix}\,. \end{align} It is easy to check that $$ \vec v_1\times\vec v_2=\begin{pmatrix}\cos\varphi\,\sin\theta\\\sin\varphi\,\sin\theta\\\cos\theta\end{pmatrix}\sin\theta=\vec n\sin\theta\,. $$ Therefore $(\vec n,\vec v_1,\vec v_2)$ is a positively oriented basis for $\mathbb R^3$ and $(\vec v_1,\vec v_2)$ a positively oriented basis for the tangent space of $S^2\,.$

Answer 2

I agree with @Kurt that you should be looking at polar/spherical coordinates and not stereographic projection. This is true in general, but particularly because of the form of your integrand. But I'm going to answer your question because it should help you build some geometric intuition for what's going on here.

You can do a formal computation, but I recommend pictures here. The answer will be that stereographic projection from the north pole is orientation-reversing if we use the standard orientation on the $xy$-plane and stereographic projection from the south pole is orientation-preserving.

Do stereographic projection from the north pole to the $xy$-plane. Think in polar coordinates on the plane: Moving radially outward moves you up a great circle on the sphere; moving counterclockwise around a circle centered at the origin moves you counterclockwise around a latitude circle on the sphere. The cross product of the respective tangent vectors to those curves points inward. Thus, in order to get an outward-pointing normal, we must reverse the order of our two coordinates. Using $(\theta,r)$ as polar coordinates on the plane will give an orientation-preserving parametrization of the sphere when we stereographically project. You can certainly write this all out formally and compute partial derivatives.

In order to address (what I think is) your question about the orientation on $S^1$, here's the general set-up. In general, given a hypersurface in $\Bbb R^{k+1}$, we orient the tangent plane at a point as follows: We say that $(v_1,\dots,v_k)$ is a positively-oriented basis if and only if the ordered basis $(\vec n,v_1,\dots,v_k)$ is a positively-ordered basis for $\Bbb R^{k+1}$. In the case $k=1$, this means that $(\vec n,v_1)$ should be a right-handed basis; in the case $k=2$, this means that $v_1\times v_2$ should point in the direction of (positive) $\vec n$. Rephrasing the latter, if you stand upright along $\vec n$ and look down to the tangent plane, $v_1,v_2$ must be a right-handed basis.