Monomial integrals on the sphere and orthonormality on the distinguished torus

Apply [F2] to the indicator of the open, hence Borel, unit ball. For $0<r<1$ every $r\omega$ lies in ${\mathbb{B}}^m$, while for $r>1$ none does, so ${\lambda }_{2m}({\mathbb{B}}^m)={\int }_0^1{r}^{2m-1}dr\sigma (S^{2m-1})=\frac{\sigma (S^{2m-1})}{2m}\mathrm{.}$ By [F4], $\sigma (S^{2m-1})=2m{\pi }^m\mathrm{/}m!=2{\pi }^m\mathrm{/}(m-1)!$, which is finite and positive, so ${\sigma }_1$ is well defined.

Fix $j<m$ and a unit complex number $\eta \in \mathbb{C}$ with $\mathrm{\mid }\eta \mathrm{\mid }=1$, and let $R_{j,\eta }$ multiply the $j$th complex coordinate by $\eta$ and fix the others. It maps the unit sphere to itself; its real matrix is the identity except for the $j$th block in [F9], whose determinant is $\mathrm{\mid }\eta {\mathrm{\mid }}^2=1$. Since $R_{j,\eta }$ and its inverse are continuous, both $R_{j,\eta }E$ and $R_{j,\eta }^{-1}E$ are Borel for Borel $E\subseteq S^{2m-1}$ by [F12]. The conical set in [F3] is carried onto the cone for $R_{j,\eta }E$, so [F10] gives $\sigma (R_{j,\eta }E)=\sigma (E)$; applying this equality to $R_{j,\eta }^{-1}E$ gives $\sigma (R_{j,\eta }^{-1}E)=\sigma (E)$. Thus the continuous map is measure preserving, and [F11] makes the integrals of the bounded monomial products in [F12] invariant.

For a fixed $\alpha$, set $M_{\alpha }:={\int }_{S^{2m-1}}\mathrm{\mid }{\omega }^{\alpha }{\mathrm{\mid }}^{2}d\sigma$. The integrand $\mathrm{\mid }{z}^{\alpha }{\mathrm{\mid }}^2{\mathbf{1}}_{{\mathbb{B}}^m}$ is nonnegative Borel by [F12] and [F14]; since $\mathrm{\mid }(r\omega {)}^{\alpha }{\mathrm{\mid }}^2=r^{2\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }{\omega }^{\alpha }{\mathrm{\mid }}^2$, [F2] and [F4] give $\frac{{\pi }^m\alpha !}{(m+\mathrm{\mid }\alpha \mathrm{\mid })!}={\int }_0^1{r}^{2m-1+2\mathrm{\mid }\alpha \mathrm{\mid }}dr{M}_{\alpha }=\frac{M_{\alpha }}{2(m+\mathrm{\mid }\alpha \mathrm{\mid })}\mathrm{.}$ Hence $M_{\alpha }=2{\pi }^m\alpha !\mathrm{/}(m-1+\mathrm{\mid }\alpha \mathrm{\mid })!$, and division by the total mass in step 1.1 gives the normalized diagonal moment.

Suppose $\alpha \mathrm{\ne }\beta$ and choose $j<m$ with $a:={\alpha }_j\mathrm{\ne }b:={\beta }_j$. Let $d:=\mathrm{\mid }a-b\mathrm{\mid }>0$ and $\eta :=e^{i\pi \mathrm{/}d}$, so [F7] gives $\mathrm{\mid }\eta \mathrm{\mid }=1$ and ${\eta }^d=-1$. By the recursive powers in [F6], induction and commutativity give ${\eta }^a{\bar{\eta }}^b=-1$: if $a>b$, factor $(\eta \bar{\eta }{)}^b{\eta }^{a-b}={\eta }^d$; if $b>a$, factor $(\eta \bar{\eta }{)}^a{\bar{\eta }}^{b-a}=\overline{{\eta }^d}$. Under $R_{j,\eta }$ the integrand ${\zeta }^{\alpha }\overline{{\zeta }^{\beta }}$ is multiplied by this scalar. Integral invariance from step 1.2 therefore makes its integral equal to its negative, so it is zero. The integrand is integrable by [F12] and finiteness in step 1.1.

On ${\mathbb{T}}^m$, the normalized product Haar measure is invariant under translation of any one coordinate by [F13]. For $\alpha \mathrm{\ne }\beta$, choose $j,a,b,d,\eta$ as in step 2.2 and translate that coordinate by the torus element represented by $1\mathrm{/}(2d)$, which multiplies it by $\eta =e^{i\pi \mathrm{/}d}$. The integrand is multiplied by ${\eta }^a{\bar{\eta }}^b=-1$ by the same power calculation, so integral invariance [F11] makes the integral zero. If $\alpha =\beta$, the integrand is identically one and [F5] gives total measure one; hence the integral is one.

Step 2.1 gives the unnormalized and normalized sphere diagonal moments; step 2.2 gives the off-diagonal sphere moments; step 3.1 gives all distinguished-torus moments. Together these are exactly the three displayed formulas.