Constructing a non-vanishing $n$-form on $S^n$

Question

How can I go about constructing a smooth non-vanishing $n$-form on $S^n$ using the stereographical coordinates?

Answer 1

Following the comment above:

There are many ways to construct volume forms on a sphere.

the below are two different non-vanishing $n$-forms on $S^n$:

• the $n$-form on $S^n$ pulled back using the stereographic projection from the Riemannian volume form on $\mathbb{R}^n$. Its expression is long, so I won't write it out in coordinates.

• Riemannian volume form on $S^n$ induced by the Euclidean metric in $\mathbb{R}^{n+1}$. From the post Riemannian Volume Form of $S^n$ this is given by $$ i_{{\bf r}}(dx_{1}\wedge \dots \wedge dx_{n+1})=\sum_{i=1}^{n+1} (-1)^{i-1}x_{i}dx_{1}\wedge\dots \widehat{dx_{i}} \dots \wedge dx_{n+1}.$$We want to get rid of $dx_{n+1}$. Substituting $dx_{n+1}=\sum_{i=1}^n-\frac{x_i}{x_{n+1}}dx_i$ $$=\sum_{i=1}^{n} (-1)^{n-i}(-1)^{i-1}x_{i}\left(-\frac{x_i}{x_{n+1}}\right)dx_{1}\wedge\dots \wedge dx_n\\+(-1)^nx_{n+1}dx_1\wedge\dots\wedge dx_n\\ =(-1)^{n-1}\left(\frac{-\sum_{i=1}^{n} x_i^2}{x_{n+1}}\right)dx_{1}\wedge\dots \wedge dx_n\\+(-1)^nx_{n+1}dx_1\wedge\dots\wedge dx_n\\ =(-1)^n\left(x_{n+1}+\frac{\sum_{i=1}^{n} x_i^2}{x_{n+1}}\right)dx_{1}\wedge\dots \wedge dx_n\\ =(-1)^n\frac{\sum_{i=1}^{n+1} x_i^2}{x_{n+1}}dx_{1}\wedge\dots \wedge dx_n\\ =(-1)^n\frac1{x_{n+1}}dx_{1}\wedge\dots \wedge dx_n $$ The charts $\{x_i>0\}$ and $\{x_i<0\}$ for $i=1,\dots,n+1$ cover $S^n$.
  On the chart $\{x_{n+1}>0\}$ with local coordinates $(x_1,\dots,x_n)$ define$$\omega=\frac{1}{x_{n+1}} \mathrm{d} x_1 \wedge \mathrm{d} x_2 \wedge \ldots \wedge \mathrm{d} x_n$$ On the chart $\{x_n>0\}$ with local coordinates $(x_{n+1},x_1,\dots,x_{n-1})$ we have $\frac{\partial x_n}{\partial x_{n+1}}=-\frac{x_{n+1}}{x_n}$, so $$\omega=\frac{1}{x_{n+1}} \mathrm{d} x_1 \wedge \ldots \wedge \mathrm{d} x_{n-1}\wedge(-\frac{x_{n+1}}{x_n}\mathrm{d} x_{n+1})=-\frac1{x_n}\mathrm{d} x_1 \wedge \ldots \wedge \mathrm{d} x_{n-1}\wedge\mathrm{d} x_{n+1}$$ In the above $\frac{\partial x_n}{\partial x_{n+1}}=-\frac{x_{n+1}}{x_n}$ is derived by $\frac\partial{\partial x_{n+1}}(x_n^2)=\frac\partial{\partial x_{n+1}}(1-x_1^2-\dots-x_{n-1}^2-x_{n+1}^2)=\frac\partial{\partial x_{n+1}}(-x_{n+1}^2)$ so $2x_n\frac{\partial x_n}{\partial x_{n+1}}=-2x_{n+1}$.
  You can do this on all pairs of charts to get that $\omega$ is actually well-defined but that it's nowhere vanishing is obvious once you have that it's well-defined.

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The link above used interior product, here is another way to show that volume form is $\omega=x_{n+1}^{-1} \mathrm{d} x_1 \wedge \cdots \wedge \mathrm{d} x_n$, using the induced Riemannian metric on the sphere $\mathcal{S}^{n} \subset \mathbb{R}^{n+1}$.

We have $$ x_1^2+x_2^2+\cdots+x_{n+1}^2=1 $$ and so $$ x_{n+1} \mathrm{d} x_{n+1}=-\sum_{i=1}^n x_i \mathrm{d} x_i $$ Hence $$ g=\sum_{i=1}^{n+1}\mathrm{d} x_i^2=\sum_{i=1}^n \mathrm{d} x_i^2+x_{n+1}^{-2}\left(\sum_{i=1}^n x_i \mathrm{~d} x_i\right)^2 $$ so $g(v,v)=\|v\|^2+x_1^{-2}\left(\sum_{i=1}^n x_i^2\right)\langle e_1,v\rangle$ for $v$ in the tangent space, where $\langle \cdot,\cdot\rangle$ is Euclidean inner product and $$ e_1=\left(x_1, \ldots, x_n\right) / \sqrt{\sum_{i=1}^n x_i^2} $$ Extend the unit vector $e_1$ to an orthonormal basis $e_1, \ldots, e_n$ wrt $\langle\cdot,\cdot\rangle$ of the tangent space, then $$g(e_1,e_1)= \|e_1\|^2+x_{n+1}^{-2}\left(\sum_{i=1}^n x_i^2\right)\langle e_1,e_1\rangle=1+x_{n+1}^{-2}\left(\sum_{i=1}^n x_i^2\right) $$ $$\forall 2\le i\le n:\quad g(e_i,e_i)=\|e_i\|^2+\langle e_1,e_i\rangle=1+0=1$$ then the symmetric matrix $g$ becomes $$\operatorname{diag}\left( 1+x_{n+1}^{-2}\left(\sum_{i=1}^n x_i^2\right), 1,1, \ldots, 1\right)\\=\operatorname{diag}\left(x_{n+1}^{-2}, 1,1, \ldots, 1\right) $$ Since changing orthonormal basis vectors doesn't change its determinant, $$\det g=\det\left(\operatorname{diag}\left(x_{n+1}^{-2}, 1,1, \ldots, 1\right)\right)=x_{n+1}^{-2} $$ The volume form is therefore $$ \sqrt{\det g} \mathrm{~d} x_2 \wedge \cdots \wedge \mathrm{d} x_n=x_{n+1}^{-1} \mathrm{~d} x_2 \wedge \cdots \wedge \mathrm{d} x_n $$

Answer 2

Here's a pretty simple way to construct a non-vanishing $n$-form on $S^n$ using stereographic projection but without using the Riemannian metric on $S^n$:

$\newcommand\R{\mathbb{R}}$Let $(e_1,\dots, e_n)$ be the standard basis of $\R^n$. Let $\pi_N: S^n\backslash\{ e_n\} \rightarrow \R^n$ be stereographic projection from $e_n$, and $\pi_S: S^n\backslash\{ -e_n\} \rightarrow \R^n$ be stereographic projection from $-e_n$.

Let $$ \Omega = f(x)\,dx^1\wedge\cdots\wedge dx^n $$ be a compactly supported $n$-form on $\R^n$ such that $f \ge 0$ and $f = 1$ on the unit ball. Then $$ \Theta = \pi_N^*\Omega + \pi_S^*\Omega $$ is a nonvanishing $n$-form on $S^n$.

If you prefer defining $f$ by an explicit formula, $$ f(x) = e^{-|x|^2} $$ suffices.