TITLE: Prove the sphere is orientable
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: One way to do it is finding a volume form for $\mathbb{S}^n$.
Obviously, using the unit normal assumes the embedding $\iota:\mathbb{S}^n\hookrightarrow \mathbb{R}^{n+1}$.
The idea is to take the usual volume form in $\mathbb{R}^{n+1}$, namely $\eta:=dx_1\wedge...\wedge dx_{n+1}$, contract it with respect to the radial field $X=\sum_{i=1}^{n+1}x_i\,\frac{\partial}{\partial x_i}$ and then restrict it to $\mathbb{S}^n$ (notice that $X|_{\mathbb{S}^n}$ is the unit normal).
Formally, we define $\omega:=\iota^*(i_X\eta)$, which is a $n$-form in $\mathbb{S}^n$. This is in fact a volume form, because if $\{v_1,...,v_n\}$ is a basis for $T_p\mathbb{S}^n$, then $\{X_p,v_1,...,v_n\}$ is a basis for $\mathbb{R}^{n+1}$ (since $T_p\mathbb{S}^n=\text{span}(X_p)^\perp$), so:
$$\omega(v_1,...,v_n):=\eta(X_p,v_1,...,v_n)\neq 0\,\,_\blacksquare$$