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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 $O(n)$
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    $\begingroup$ 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
    Commented May 15, 2015 at 23:48
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    $\begingroup$ What is your definition of the volume form? $\endgroup$
    – user99914
    Commented May 16, 2015 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
    Commented May 16, 2015 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
    Commented May 16, 2015 at 19:18
  • $\begingroup$ Everybody say it, but I can't see it. $\endgroup$
    – Donyarley
    Commented May 16, 2015 at 23:46

1 Answer 1

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Hint:

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 $$ ds=x\ dy\wedge dz+y\ dz\wedge dx+z\ dx\wedge dy. $$

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