The element of Volume of the Sphere and two formulas

Question

Let $S^{n-1}$ be the unit sphere with the inner product $<.,.>$ that inherits from $\mathbb{R}^n$ and $V\in S^{n-1}$. Let $\{e_ i \}_{ i=1}^n $ be an orthonormal frame and let $<V,e_i>=v_i$ . Let $\mathrm{d}\sigma$ be an element of volume on $S^{n-1}$.

1. For $i\not =j\ $, $\int_{S^{n-1}}v_iv_j \ \mathrm{d}\sigma(V)=0$ , since the integrand is odd with respect to reflection about the coordinate hyperplane $\{v_i =0\}$ .

I do not understand this reason. How can I find out it?

2. $ \ \int_{S^{n-1}}v_i^2 \ \mathrm{d}\sigma(V)= \ \int_{S^{n-1}}v_j^2 \ \mathrm{d}\sigma(V)$ . Why?

Answer 1

Good to see you know that problem lies in tangent space at a single point. Let's recall one fact.

If $\phi:M_1\rightarrow M_2$ is a diffeomorphism between two manifolds and $\omega$ is a top degree form on $M_2,$ then $$\int_{M_2}\omega=\int_{M_1}\phi^*(\omega).$$

To answer the first problem we let $\phi:S^{n-1}\rightarrow S^{n-1}$ to be a reflection by $\{V_i=0\},$ which is given by the formula $$\phi(V)=V-2v_i e_i.$$ It is linear, hence it's derivative $\phi_*$ is equal to $\phi$ and for volume form $d\sigma$ on $S^{n-1}$ we have $$\phi^*(d\sigma)(X_2,\dots,X_n)=(d\sigma)(X_2,\dots,X_n)-2(d\sigma)((X_2)_ie_i,\dots,(X_n)_ie_i)=(d\sigma)(X_2,\dots,X_n).$$ We see that $\phi^*(d\sigma)=d\sigma.$ Now we can use the mentioned fact. $$\int_{S^{n-1}}v_iv_jd\sigma(V)=\int_{S^{n-1}}\phi^*(v_iv_jd\sigma(V))=\int_{S^{n-1}}(\pi_i\pi_j)(\phi(V)) d\sigma(V)=\int_{S^{n-1}}(v_i-2v_i)(v_j-0)d\sigma(V)=-\int_{S^{n-1}}v_iv_jd\sigma(V).$$ Hence $$\int_{S^{n-1}}v_iv_jd\sigma(V)=0.$$ To proof second problem we use similar trick, but instead of refflection we use rotation. We just proof that $$\int_{S^{n-1}}v_i^2d\sigma=\int_{S^{n-1}}v_j^2d\sigma.$$ But first recall another fact.

If $\omega$ is a volume form on $M$ and $S$ is a submanifold of codimention 1 with specified outward normal vector field $N$ then the volume form in $S$ is equal to $i_N\omega,$ where $i_N$ is an interior product.

Set $\phi:S^{n-1}\rightarrow S^{n-1}$ to be a rotation that sends $e_j$ to $e_i.$ Rotation is linear. Hence $\phi^*(d\sigma)(X_2,\dots,X_n)=i_N\omega(\phi X_2,\dots,\phi X_n)=\omega(\phi N,\phi X_2,\dots,\phi X_n).$ Last equality holds, because pullback "pushes" points. So for every $x\in S^{n-1}$ vector $N$ is taken at $N_{\phi(x)}=\phi N.$ Returning to $\omega(\phi N,\phi X_2,\dots,\phi X_n),$ we have $$\omega(\phi N,\phi X_2,\dots,\phi X_n)=\det(\phi)\omega(N,X_2,\dots,X_n)=i_N\omega(X_2,\dots,X_n)=d\sigma(X_2,\dots,X_n).$$ Again using first fact we have that $$\int_{S^{n-1}}v_i^2 d\sigma(V)=\int_{S^{n-1}}\phi^*(v_i^2 d\sigma(V))=\int_{S^{n-1}}\pi_i^2(\phi(V)) d\sigma(V)=\int_{S^{n-1}}v_j^2 d\sigma(V).$$