Baby Rudin, Chapter 10, Problem 23 (d) - Differential forms.

Question

In problems 21 and 22, Rudin defines the differential forms $\eta=\dfrac{xdy-ydx}{x^2+y^2}$ and $\zeta=\dfrac{x dy \wedge dz+ydz \wedge dx+z dx \wedge dy}{r^3}$ and the reader is asked to prove various statements regarding them (for example, that integrating these forms gives the surface area, and that integrating over homotopic surfaces yields the same result).

In problem 23, we are asked to try to generalize some of the assertions from problems 21,22 for arbitrary $n$, for the differential n-form $\omega_n=\frac{1}{r^n} \sum_{i=1}^n(-1)^{i-1} x_i dx_1 \wedge \ldots \wedge dx_{i-1} \wedge dx_{i+1} \wedge \ldots \wedge dx_n$.

I'd love to hear any well known results about $\omega_n$, and with proofs if possible (I'm new to high-dimensional analysis). I already know it's closed, and not exact.

Thank you.

Answer 1

Elaborating on the comment by Ehsan M. Kermani: $\omega_n$ (with $r=1$) is the volume form on $S^{n-1}$. In other words, for any reasonable function $f:\mathbb S^{n-1}\to\mathbb R$ the integral $\int_{S^{n-1}}f\omega_n$ agrees with the integral of $f$ with respect to the Lebesgue measure on $S^{n-1}$.

Sketch of proof. It suffices to check that $T^*\omega_n = \omega_n$ for every orthogonal linear map $T$, because this demonstrates that $\int_{S^{n-1}}f\omega_n$ is invariant under rotations (and the Lebesgue measure is the only rotationally invariant measure). Instead of computing $T^*\omega_n$ directly, introduce $f(x)=\|x\|^2$ and check that $*df = 2\omega_n$ where * is the Hodge star. Since $f\circ T=f$, it follows that $T^*df=df$, hence (by applying * on both sides) $T^*\omega_n=\omega_n$.