Transverse projector and isotropic angular contraction

For any $v\in {\mathbb{C}}^3$, put $w={\sum }_a⟨v,{\epsilon }_a⟩{\epsilon }_a$. Orthonormality gives $w\in n^{\perp }$ and $⟨v-w,{\epsilon }_b⟩=0$ for both $b$; since the two vectors span $n^{\perp }$, $v-w$ is orthogonal to that space. Thus $w$ is its orthogonal projection by [F1]. Also $v-⟨v,n⟩n\in n^{\perp }$ and $⟨v,n⟩n$ is orthogonal to $n^{\perp }$, so uniqueness gives ${\sum }_a⟨v,{\epsilon }_a⟩{\epsilon }_a=v-⟨v,n⟩n$. Taking $v$ to be the $j$th standard basis vector and comparing $i$th coordinates proves the first identity.

Every one-variable integrand used below is a polynomial in $\sin t$ and $\cos t$, so is differentiable by [F4] and [F6], continuous by [F5], and bounded and Riemann integrable on the relevant closed interval by [F5]. Under the assumed Countable Choice, [F5] identifies its Riemann integral with its Lebesgue integral. For any such integrand $f$ on $[a,b]$, [F7] makes $\{a,b\}$ null. Both $f$ and $f{\mathbf{1}}_{(a,b)}$ are bounded measurable functions on this finite measure space; if $\mathrm{\mid }f\mathrm{\mid }\le M$, [F8] gives $\int \mathrm{\mid }f{\mathbf{1}}_{(a,b)}\mathrm{\mid }\le M(b-a)<\mathrm{\infty }$, and likewise for $f$. They agree almost everywhere, so [F7] identifies the closed-interval integral with the open-interval Lebesgue integral. All one-dimensional integrals below may therefore be evaluated by the Darboux FTC in [F4] and used on the open angular intervals.

The addition formulas in [F4] at $\pi +\pi$, using [F6], give $\sin (2\pi )=0$ and $\cos (2\pi )=1$; [F4] gives the values at zero. The derivatives of $t$ and of a constant are $1$ and $0$, directly from their difference quotients. The product rules in [F6] and ${\cos }^{2}t+{\sin }^{2}t=1$ in [F4] show that primitives for ${\cos }^{2}t$, ${\sin }^{2}t$, $\sin t$, $\cos t$, $\sin t\cos t$, and $1$ are, respectively, $(t+\sin t\cos t)\mathrm{/}2$, $(t-\sin t\cos t)\mathrm{/}2$, $-\cos t$, $\sin t$, ${\sin }^{2}t\mathrm{/}2$, and $t$. These primitives are differentiable on all of $\mathbb{R}$, so their restrictions have the one-sided endpoint derivatives required by [F4]. Applying [F4] and step 1.2 on $[0,2\pi ]$ gives ${\int }_0^{2\pi }{\cos }^2\phi d\phi ={\int }_0^{2\pi }{\sin }^2\phi d\phi =\pi$ and ${\int }_0^{2\pi }1d\phi =2\pi$, whereas the integrals of $\sin \phi$, $\cos \phi$ and $\sin \phi \cos \phi$ are zero.

The product rules in [F6] give $(-\cos t+{\cos }^{3}t\mathrm{/}3{)}^{\mathrm{\prime }}=\sin t-{\cos }^{2}t\sin t={\sin }^{3}t$ by [F4], and $(-{\cos }^{3}t\mathrm{/}3{)}^{\mathrm{\prime }}={\cos }^{2}t\sin t$. The endpoint values $\cos 0=1$ and $\cos \pi =-1$ from [F4] and [F6], together with the FTC and step 1.2, therefore give ${\int }_0^{\pi }{\sin }^3\theta d\theta =4\mathrm{/}3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2\mathrm{/}3$.

The sphere functions $n_i{n}_j$ are bounded by $1$ and continuous: for $n,m\in S^2$, $\mathrm{\mid }{n}_i{n}_j-m_i{m}_j\mathrm{\mid }\le \mathrm{\mid }{n}_i-m_i\mathrm{\mid }+\mathrm{\mid }{n}_j-m_j\mathrm{\mid }\le 2\mathrm{\mid }n-m\mathrm{\mid }$. They are Borel measurable by [F8] and have absolute integral at most $\sigma (S^2)=4\pi$ by [F2] and [F8]. Their angular integrands $g(\theta ,\phi )=n_i(\mathrm{\Phi }(\theta ,\phi ))n_j(\mathrm{\Phi }(\theta ,\phi ))\sin \theta$ are continuous and bounded by $1$ on $R=(0,\pi )\times (0,2\pi )$, whose planar Lebesgue measure is $2{\pi }^2$ by [F7]. Hence they are Borel and ${\lambda }_2$-integrable by [F8]. By [F9] with $m=n=1$, the Borel sigma-algebra on this open rectangle is the product of its one-dimensional Borel sigma-algebras, and planar Lebesgue measure there agrees with the product of the one-dimensional Lebesgue measures restricted to Borel sets. Thus the integrals of $g$ and $\mathrm{\mid }g\mathrm{\mid }$ agree for these measures, and $g$ is also integrable for the product measure. Each angular interval has finite measure by [F7], so the two factor spaces are sigma-finite, as required by [F3]. The sphere formula [F2] and Fubini [F3] now give the diagonal moments as $(4\mathrm{/}3)\pi$, $(4\mathrm{/}3)\pi$, and $(2\mathrm{/}3)(2\pi )$ by steps 2.1 and 2.2. For the $12$, $13$ and $23$ off-diagonal moments, the respective $\phi$ factors are $\cos \phi \sin \phi$, $\cos \phi$ and $\sin \phi$, whose integrals vanish by step 2.1; the reversed indices have the same products. Thus ${\int }_{S^2}{n}_i{n}_{j}d\sigma =(4\pi \mathrm{/}3){\delta }_{ij}$. Linearity in [F8] and the total mass in [F2] give ${\int }_{S^2}({\delta }_{ij}-n_i{n}_j)d\sigma =(8\pi \mathrm{/}3){\delta }_{ij}$, completing the second identity.