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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?

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What do you mean, “compute both sides”? – PseudoNeo Aug 2 '12 at 9:24
BTW,… is hugely relevant for your question. Eric's answer there translates quite directly to your case. – PseudoNeo Aug 2 '12 at 9:55
up vote 7 down vote accepted

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 $x\in S^{n-1}$.

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Beat me to it. :) – ZulfiqarIII Aug 2 '12 at 13:57

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