Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $x\in S^{n-1}$.

share|cite|improve this answer
Beat me to it. :) – ZulfiqarIII Aug 2 '12 at 13:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.