Zonal and quadratic sphere moments for the polar surface measure on $S^2$

Define $X(\theta ,\phi ):=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$ and $D:=[0,\pi ]\times [0,2\pi ]$. Then $D$ is a compact Jordan parameter region, being a closed bounded rectangle and the closure of its nonempty connected interior $D^{\circ }=(0,\pi )\times (0,2\pi )$, and $X$ is $C^1$ on the open neighbourhood ${\mathbb{R}}^2$ of $D$ because its components are finite sums and products of the $C^1$ functions $\sin$ and $\cos$ [F5, F8]. Its partial derivatives are $X_{\theta }=(\cos \theta \cos \phi ,\cos \theta \sin \phi ,-\sin \theta )$ and $X_{\phi }=(-\sin \theta \sin \phi ,\sin \theta \cos \phi ,0)$, and the coordinate formula for the cross product gives $X_{\theta }\times X_{\phi }=({\sin }^2\theta \cos \phi ,{\sin }^2\theta \sin \phi ,\sin \theta \cos \theta )=\sin \theta X(\theta ,\phi ),$ since ${\cos }^2\phi +{\sin }^2\phi =1$; also $\mathrm{\mid }X(\theta ,\phi ){\mathrm{\mid }}^2={\sin }^2\theta +{\cos }^2\theta =1$ [F5, F7]. Hence $X_{\theta }\times X_{\phi }=0$ exactly when $\sin \theta =0$; on the interior $\theta \in (0,\pi )$ one has $\sin \theta >0$ by strict monotonicity and the values $\sin 0=\sin \pi =0$ [F5], so the parameter cross product is nonzero on $D^{\circ }$.

The chart covers the sphere and is injective in its interior. If $n\in S^2$, the intermediate value theorem between the endpoint values $\cos 0=1$, $\cos \pi =-1$, and strict monotonicity of cosine on $[0,\pi ]$ gives a unique $\theta \in [0,\pi ]$ with $\cos \theta =n_3$, so $\sin \theta =(n_1^2+n_2^2{)}^{1\mathrm{/}2}$ because $\sin \theta \ge 0$ on $[0,\pi ]$ and ${\sin }^2\theta =1-{\cos }^2\theta$ [F5]. If $\sin \theta =0$ then $n=\pm e_3=X(0,0)$ or $X(\pi ,0)$; otherwise put $c:=n_1\mathrm{/}\sin \theta$ and $s:=n_2\mathrm{/}\sin \theta$, so $c^2+s^2=1$, and choose $\phi \in [0,\pi ]$ with $\cos \phi =c$ when $s\ge 0$ and $\phi \in [\pi ,2\pi ]$ with $\cos \phi =c$ when $s<0$, possible because cosine has range $[-1,1]$ and is strictly monotone on each interval; then $\sin \phi$ has the sign of $s$ there and ${\sin }^2\phi =1-c^2=s^2$, so $\sin \phi =s$ and $n=X(\theta ,\phi )$. Thus $X[D]=S^2$. Conversely, if $X(\theta ,\phi )=X({\theta }^{\mathrm{\prime }},{\phi }^{\mathrm{\prime }})$ with $(\theta ,\phi )\in D^{\circ }$ and $({\theta }^{\mathrm{\prime }},{\phi }^{\mathrm{\prime }})\in D$, then $\cos \theta =\cos {\theta }^{\mathrm{\prime }}$ with both angles in $[0,\pi ]$, so $\theta ={\theta }^{\mathrm{\prime }}$ by strict monotonicity; then $\sin \theta >0$ and the first two coordinates give $\cos \phi =\cos {\phi }^{\mathrm{\prime }}$ and $\sin \phi =\sin {\phi }^{\mathrm{\prime }}$, hence $\cos (\phi -{\phi }^{\mathrm{\prime }})=\cos \phi \cos {\phi }^{\mathrm{\prime }}+\sin \phi \sin {\phi }^{\mathrm{\prime }}={\cos }^2\phi +{\sin }^2\phi =1$ [F5]. Writing $x:=\phi -{\phi }^{\mathrm{\prime }}\in (-2\pi ,2\pi )$ and using $\cos x=1-2{\sin }^2(x\mathrm{/}2)$, we get $\sin (x\mathrm{/}2)=0$, so $x\mathrm{/}2=m\pi$ for some integer $m$ and $x=2m\pi$; since $\mathrm{\mid }x\mathrm{\mid }<2\pi$ this forces $x=0$, that is $\phi ={\phi }^{\mathrm{\prime }}$ [F5]. Therefore no interior parameter point shares its image with a distinct point of $D$, and $(D,X)$ is a regular parametrized surface patch with image $S^2$.

Reduction to the chart, including its seam and poles. Let $q:S^2\to \mathbb{R}$ be continuous. Put $H:=\{y\in {\mathbb{R}}^3:y_2=0,y_1\ge 0\}$ and $Y(r,\theta ,\phi ):=rX(\theta ,\phi )$ on $U:=(0,1)\times D^{\circ }$. By step 2.1, $Y$ is a bijection from $U$ onto the open set $V:=B_1(0)\setminus H$. Its derivative has columns $X,r{X}_{\theta },r{X}_{\phi }$, so $\det DY=r^{2}X\cdot (X_{\theta }\times X_{\phi })=r^2\sin \theta >0$ by step 1.1. The local inverse theorem [F13] makes its bijective inverse $C^1$ locally at every point, hence $Y$ is a $C^1$ diffeomorphism. The omitted set $H\cap B_1$ is contained in the degenerate box $[-1,1]\times \{0\}\times [-1,1]$ and is Lebesgue null by [F13]. The bounded Borel function $q(y\mathrm{/}\mathrm{\mid }y\mathrm{\mid })$ on $B_1\setminus \{0\}$, assigned zero at $0$, is integrable by [F13]. Applying the polar formula of Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma to its positive and negative parts gives ${\int }_{B_1}q(y\mathrm{/}\mathrm{\mid }y\mathrm{\mid })dy=\frac{1}{3}{\int }_{S^2}qd\sigma$. Change of variables by $Y$ and Fubini [F13] give the same integral as ${\int }_0^1{r}^{2}dr{\int }_{D^{\circ }}q(X(\theta ,\phi ))\sin \theta d\theta d\phi$. The parameter boundary is a finite union of degenerate boxes and therefore null; the integrand extends continuously to $D$, and its Lebesgue integral equals its Riemann integral by [F13]. Cancelling ${\int }_0^1{r}^{2}dr=1\mathrm{/}3$ therefore proves ${\int }_{S^2}qd\sigma ={\int }_0^{2\pi }{\int }_0^{\pi }q(X(\theta ,\phi ))\sin \theta d\theta d\phi \mathrm{.}$ This supplies the measure identification despite the seam and pole degeneracies; it does not apply an injective regular atlas chart across those degeneracies. In particular, for $q=f\circ n_3$, $n_3\circ X=\cos \theta$ gives $I:={\int }_{S^2}f(n_3)d\sigma ={\int }_0^{2\pi }{\int }_0^{\pi }f(\cos \theta )\sin \theta d\theta d\phi$.

Conclusion of (a). The function $g(\theta ,\phi ):=f(\cos \theta )\sin \theta$ is continuous on the rectangle $D$ [F5, F8], hence Riemann integrable, and its sections in $\phi$ are continuous; [F10] therefore gives $I={\int }_0^{\pi }({\int }_0^{2\pi }f(\cos \theta )\sin \theta d\phi )d\theta =2\pi {\int }_0^{\pi }f(\cos \theta )\sin \theta d\theta$. Let $P(u):={\int }_{-1}^{u}f(v)dv$ on $[-1,1]$; by [F6] $P^{\mathrm{\prime }}=f$, and the chain rule [F6] with ${\cos }^{\mathrm{\prime }}=-\sin$ [F5] gives $(P\circ \cos {)}^{\mathrm{\prime }}(\theta )=-f(\cos \theta )\sin \theta$ on $[0,\pi ]$. Since this derivative is continuous, the second fundamental theorem [F6] applies and ${\int }_0^{\pi }f(\cos \theta )\sin \theta d\theta =-[P(\cos \theta ){]}_0^{\pi }=P(1)-P(-1)={\int }_{-1}^{1}f(u)du,$ because $\cos 0=1$ and $\cos \pi =-1$ [F5]. Combining with step 3.1 gives ${\int }_{S^2}f(n_3)d\sigma =2\pi {\int }_{-1}^{1}f(u)du$, which is (a); all integrands here are continuous on the compact sets in question, so the displayed integrals are finite.

Conclusion of (b). Let ${\eta }_1(n_1,n_2,n_3):=(-n_1,n_2,n_3)$ and ${\eta }_2(n_1,n_2,n_3):=(n_1,-n_2,n_3)$. Each is an orthogonal bijection of ${\mathbb{R}}^3$ with ${\eta }_i[S^2]=S^2$, so each preserves $\sigma$ [F2]; by the substitution identity of [F11], applied to the bounded Borel function $g_i(n):=n_{i}f(n_3)$ and the fact that $g_i\circ {\eta }_i=-g_i$, we get ${\int }_{S^2}{n}_{i}f(n_3)d\sigma =-{\int }_{S^2}{n}_{i}f(n_3)d\sigma$, whence ${\int }_{S^2}{n}_{i}f(n_3)d\sigma =0$ for $i=1,2$ (finiteness of both integrals is [F9]). For the third component, (a) applied to the continuous function $u\mapsto uf(u)$ gives ${\int }_{S^2}{n}_{3}f(n_3)d\sigma =2\pi {\int }_{-1}^{1}uf(u)du$. Since $n\cdot a=a_1{n}_1+a_2{n}_2+a_3{n}_3$, linearity of the $\sigma$-integral [F9] gives ${\int }_{S^2}(n\cdot a)f(n_3)d\sigma =2\pi a_3{\int }_{-1}^{1}uf(u)du,$ which is (b).

Conclusion of (c). Expanding the square, $(n\cdot a{)}^2={\sum }_{i,j=1}^3{a}_i{a}_j{n}_i{n}_j$, and by linearity [F9] the integral is the finite sum ${\sum }_{i,j}{a}_i{a}_j{\int }_{S^2}{n}_i{n}_{j}f(n_3)d\sigma$. For $i\mathrm{\ne }j$ the function $n_i{n}_{j}f(n_3)$ changes sign under ${\eta }_i$ (or ${\eta }_j$), so the substitution identity of [F11] gives ${\int }_{S^2}{n}_i{n}_{j}f(n_3)d\sigma =0$. For the diagonal terms, the swap ${\eta }_3(n_1,n_2,n_3):=(n_2,n_1,n_3)$ is orthogonal and preserves $\sigma$ [F2], so the substitution identity gives ${\int }_{S^2}{n}_1^{2}f(n_3)d\sigma ={\int }_{S^2}{n}_2^{2}f(n_3)d\sigma$; with $n_1^2+n_2^2=1-n_3^2$ [F5] and linearity, $2{\int }_{S^2}{n}_1^{2}f(n_3)d\sigma ={\int }_{S^2}(1-n_3^2)f(n_3)d\sigma =2\pi {\int }_{-1}^1(1-u^2)f(u)du$ by (a) applied to the continuous function $u\mapsto (1-u^2)f(u)$; and (a) applied to $u\mapsto u^{2}f(u)$ gives ${\int }_{S^2}{n}_3^{2}f(n_3)d\sigma =2\pi {\int }_{-1}^1{u}^{2}f(u)du$. Substituting into the expansion yields ${\int }_{S^2}(n\cdot a{)}^{2}f(n_3)d\sigma =2\pi a_3^2{\int }_{-1}^1{u}^{2}f(u)du+\pi (a_1^2+a_2^2){\int }_{-1}^1(1-u^2)f(u)du,$ which is (c).

Conclusion of (d) and scope (e). Taking $f=1$ in (a) gives $\sigma (S^2)=2\pi {\int }_{-1}^{1}du=4\pi$; independently [F2] gives $\sigma (S^2)=3\mathrm{\mid }{B}_1\mathrm{\mid }=3V_3(1)$, and [F12] evaluates $V_3(1)={\pi }^{3\mathrm{/}2}\mathrm{/}\mathrm{\Gamma }(5\mathrm{/}2)$ with $\mathrm{\Gamma }(5\mathrm{/}2)=(3\mathrm{/}2)(1\mathrm{/}2)\mathrm{\Gamma }(1\mathrm{/}2)=3\sqrt{\pi }\mathrm{/}4$, so $V_3(1)=4\pi \mathrm{/}3$ and the two routes agree. Taking $f=1$ in (b) gives ${\int }_{S^2}(n\cdot a)d\sigma =0$ for every $a$, hence ${\int }_{S^2}nd\sigma =0$, and taking $f=1$ in (c) gives ${\int }_{S^2}(n\cdot a{)}^{2}d\sigma =2\pi a_3^2\cdot \frac{2}{3}+\pi (a_1^2+a_2^2)\cdot \frac{4}{3}=\frac{4\pi }{3}\mathrm{\mid }a{\mathrm{\mid }}^2$; choosing $a=e_i$ and $a=e_i+e_j$ for $i\mathrm{\ne }j$ reads off ${\int }_{S^2}{n}_i^{2}d\sigma =4\pi \mathrm{/}3$ and ${\int }_{S^2}{n}_i{n}_{j}d\sigma =0$ for $i\mathrm{\ne }j$, that is ${\int }_{S^2}{n}_i{n}_{j}d\sigma =(4\pi \mathrm{/}3){\delta }_{ij}$. Finally the norm identity of [F7] gives $\mathrm{\mid }n\times a{\mathrm{\mid }}^2=\mathrm{\mid }n{\mathrm{\mid }}^2\mathrm{\mid }a{\mathrm{\mid }}^2-(n\cdot a{)}^2=\mathrm{\mid }a{\mathrm{\mid }}^2-(n\cdot a{)}^2$ for $n\in S^2$, so by linearity [F9] and the value just computed ${\int }_{S^2}\mathrm{\mid }n\times a{\mathrm{\mid }}^{2}d\sigma =4\pi \mathrm{\mid }a{\mathrm{\mid }}^2-\frac{4\pi }{3}\mathrm{\mid }a{\mathrm{\mid }}^2=\frac{8\pi }{3}\mathrm{\mid }a{\mathrm{\mid }}^2$. This proves all identities of (d); every integrand is continuous on the compact $S^2$ and $\sigma$ is a finite Borel measure by [F1], so the integrals are absolutely convergent, $f$ need not be nonnegative, and no physical interpretation is asserted, which is (e).