There is a following exercise in my text:
Let $S^n$ be $n-$ dim sphere in $R^{n+1}$ with inclusion function $i:S^{n}\to R^{n+1}$. Let
$$\omega=\sum_{i=1}^{n+1}(-1)^{i-1} x_i dx_1 \wedge… dx_{i-1}\wedge dx_{i+1}\wedge … \wedge dx_{n+1}.$$Prove that $i^*\omega \in \Omega^n(S^n)$ is Riemannian volume form on $S^n$.
I treied to manually compute this expression and the one which uses definition of Riemannian volume form when they act on some vectors in $T_xS^n$ but things gets complicated when $n$ is large and involves sum of matrix determinants which I don't know how to resolve. I managed to prove the result for small values of $n$.
How would you go with the general case?
Best Answer
It's not so hard once you show that the volume form of a submanifold $N^{n-1}\subset M^n$ of codimension 1 is given by $\mathrm d vol_N(x) = (\iota_Z \mathrm{d}vol_{M})(x)$ for $x\in N$, where $Z$ is a normal vector field to $N$, $\mathrm d vol_M$ is the Riemannian volume form of $M$ and $\iota_Z \omega$ denotes the interior product.
In your case $Z(x) = x$ gives you a normal vector field when restricted to $N = S^{n-1}$ and for $M = \mathbb R^n$ we have $\mathrm d vol_{\mathbb R^n} = dx^1 \wedge \dots \wedge dx^n$. So
\begin{align} \mathrm d vol_{S^{n-1}}(x) &= (\iota_Z dx^1 \wedge \dots \wedge dx^n)(x) \\ &= \sum_{i=1}^{n-1} (-1)^{i-1} x_i \; dx^1 \wedge \dots \wedge dx^{i-1}\wedge dx^{i+1}\wedge \dots\wedge dx^{n} \end{align}
for $x\in S^{n-1}$.