Prove the sphere is orientable

Question

Is there an easy way to show that the sphere $$\mathbb S^n = \{ x\in \mathbb R^{n+1} : \|x\| =1\}$$ is orientable other then using stereograohic projection? I am preferably looking for something derived from a basic theorem in elementary geometry with respect to the unit normal.

Answer 1

Yes, any level set $X=\{f=0\}$ of a smooth function $f\colon U \to \mathbb{R}$ with $\nabla f \ne 0$ on $\{f=0\}$ is an orientable manifold ( a submanifold of $U$). Here $U$ is an open subset of $\mathbb{R}^n$, or, more generally, an orientable manifold.

To get the orientation: Consider the gradient field $\text{grad} f$ (we need a Riemannian structure on $U$, the standard one if $U$ is a subset of a numeric space). At each point $x$ in $\{f=0\}$ we have $\text{grad}f(x) \perp T_x(X)$. Choose the orientation on $T_x(X)$ as follows: a basis in $(e_1, \ldots , e_{n-1})$ of $T_x(X)$ is positively oriented if $(e_1, \ldots, e_n, \text{grad} f(x))$ is positively oriented in $T_x(U)$.

Answer 2

Here's an orientable atlas with three charts obtained using ordinary spherical coordinates.

Use spherical coordinates $0<\phi<\pi$, $0<\theta<2\pi$, where $x=\sin\phi \cos\theta$, $y=\sin\phi \sin\theta$, $z=\cos\phi$ as a coordinate chart for the complement of the prime meridian (where $\theta=0$).

Use the exact same spherical coordinate formulas but with domains $0<\phi<\pi$, $-\pi<\theta<\pi$ as a coordinate chart for the complement of the meridian $\theta=\pi$.

So far this gives charts that cover all but the north pole $(0,0,1)$ and the south pole $(0,0,-1)$.

Finally, to cover the two poles, change your Euclidean coordinates to $x'=y$, $y'=z$, $z'=x$, and rewrite the spherical coordinate formulas as $x'=\sin\phi \cos\theta$, $y'=\sin\phi \sin\theta$, $z' = \cos\phi$.