Let $\omega$ the (n-1) form on $\mathbb{R}^n$ $$\omega=\sum_{j=1}^{n}(-1)^{j-1}x_{j}dx_{1}\wedge\cdots\wedge \hat{dx_{j}}\wedge\cdots dx_{n}$$ Show that the restriction of $\omega$ to $S^{n-1}$ in precisely the volume for this sphere.

What I did was: $\omega$ never vanish on the sphere, because, defining $\sigma\in \Omega^{n-1}(S)$ for $$\sigma_{p}(v_{1},...,v_{n-1})=det(p,v_{1},...,v_{n-1})$$ and $i:S^{n-1}\rightarrow \mathbb{R}^{n}$ the inclusion function, then $\omega=i^{\ast}(\sigma)$ then $\omega\not=0$ and therfore is a volume form.

  1. How proof that $\omega$ is the volume form?

The first thing that comes to mind is show that $\int_{S^{n-1}}\omega=Vol(S^{n-1})$ but I have serious problems with the definition, I think that is to much.

  1. How see that $\omega$ is invariant on $\mathbb{R}^{n}$ under action of $O(n)$
  • 1
    $\begingroup$ Do you want to show that $\omega$ is a $(n-1)$-volume form on $\mathbb{R}^n$ or that $i^* \omega$ is a volume form on $\mathbb{S}^n$ ? $\endgroup$
    – aGer
    May 15 '15 at 23:48
  • 1
    $\begingroup$ What is your definition of the volume form? $\endgroup$
    – user99914
    May 16 '15 at 3:31
  • $\begingroup$ I want to prove that the restriction of $\omega$ to $S^{n-1}$ is the volume form, $i.e.$ $\omega=\sqrt{det(g_{ij})}dy_{1}\wedge...\wedge dy_{n-1}$ where coodinates $y_{i}$ comes to chart from an orienting atlas $\endgroup$
    – Donyarley
    May 16 '15 at 7:40
  • $\begingroup$ If so, I think you have done so. By writing $\sigma$ as $\det$, it shows that if you plug in $n-1$ orthornormal basis, then it's one. $\endgroup$
    – user99914
    May 16 '15 at 19:18
  • $\begingroup$ Everybody say it, but I can't see it. $\endgroup$
    – Donyarley
    May 16 '15 at 23:46


The formula for the volume form of a hypersurface $i:S^{n-1}\rightarrow \mathbb{R}^{n}$ is

$$ds=i^{*}(\iota_\nu dV),$$ where $dV$ is a given volume form on the $\mathbb{R}^{n}$ and $\nu\in T\mathbb{R}^{n}$ is a smooth unit normal field to the surface; with interior product (contraction) and the pull-back (restriction) operations. Intuition: $dV$ measures volumes of n-vectors, "feeding" it a unit vector makes it measure "areas" orthogonal to that unit vector.

Example, $S^2$:

We use $\nu=(x,y,z)$; the linearity of $\iota$ gives

$$ ds=x\ \iota_{\partial_{x}}dV+y\ \iota_{\partial_{y}}dV+z\ \iota_{\partial_{z}}dV. $$

Now remember that $dV=dx\wedge dy\wedge dz=dy\wedge dz\wedge dx=dz\wedge dx\wedge dy$; so contracting $dV$ with $\partial_{x^{i}}$ just removes the $dx^{i}$. Thus $$ ds=x\ dy\wedge dz+y\ dz\wedge dx+z\ dx\wedge dy. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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