Appendix A: Proofs of Lemmas 4.2 and 4.3
1.1 A.1 Proof of Lemma 4.2
We have
$$\begin{aligned} \mathcal {I}: =&\int _{{\mathop {\mathbb {S}}}^{M+N-1}(\sqrt{M+N})} \frac{|L^{\mathrm{out}} (F\circ R^{-1})|^2}{F\circ R^{-1}} {\mathord {\mathrm {d}}}\sigma \\ =&\int _{{\mathop {\mathbb {S}}}^{M+N-1}(\sqrt{M+N})} \frac{\sum ^{{\mathrm {out}}} |\omega _i( C\nabla _{{\mathord {\mathbf {v}}}}F+D\nabla _{{\mathord {\mathbf {w}}}}F)_j-\omega _j ( SC\nabla _{{\mathord {\mathbf {v}}}}F+D\nabla _{{\mathord {\mathbf {w}}}}F)_i|^2}{F} {\mathord {\mathrm {d}}}\sigma , \end{aligned}$$
where the argument of the functions is of the form \((A^T{\mathord {\mathbf {v}}}+C^T {\mathord {\mathbf {w}}}, B^T{\mathord {\mathbf {v}}}+D^T {\mathord {\mathbf {w}}})\), and \(\sum ^{\mathrm{out}}:=\sum _{M+1\le i<j\le N+M}\). Next, we change variable \({\mathord {\mathbf {v}}}=A{\mathord {\mathbf {v}}}+B{\mathord {\mathbf {w}}}\), \({\mathord {\mathbf {w}}}=C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}}\), we have
$$\begin{aligned} \mathcal {I}&=\int _{{\mathop {\mathbb {S}}}^{M+N-1}(\sqrt{M+N})} {\mathord {\mathrm {d}}}\sigma \\&\quad \frac{\sum ^{\mathrm {out}} |(C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}})_i( C\nabla _{{\mathord {\mathbf {v}}}}F+D\nabla _{{\mathord {\mathbf {w}}}}F)_j-(C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}})_j (C\nabla _{{\mathord {\mathbf {v}}}}F+D\nabla _{{\mathord {\mathbf {w}}}}F)_i|^2}{F}\\&=: \int _{{\mathop {\mathbb {S}}}^{M+N-1}(\sqrt{M+N})} \frac{\Sigma _0}{F} {\mathord {\mathrm {d}}}\sigma \end{aligned}$$
By (28), we have
$$\begin{aligned} \Sigma _0 =&\sum ^{\mathrm {out}} |(C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}})_i( C\nabla _{{\mathord {\mathbf {v}}}}F+\ell D {\mathord {\mathbf {w}}})_j-(C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}})_j (C\nabla _{{\mathord {\mathbf {v}}}}F+\ell D{\mathord {\mathbf {w}}})_i|^2\\ =&\sum ^{\mathrm {out}} |(C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}})_i (C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))_j-(C{\mathord {\mathbf {v}}}+D{\mathord {\mathbf {w}}})_j (C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))_i|^2. \end{aligned}$$
Now we average over the directions of the vector \({\mathord {\mathbf {w}}}\). Recall that the function F does not depend on this direction nor does \(\ell \). Denote such an average by \(\langle \cdot \rangle \). Hence, we get, after averaging over each pair (i, j),
$$\begin{aligned} \langle \Sigma _0\rangle&=\sum ^{\mathrm {out}} |(C{\mathord {\mathbf {v}}})_i(C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))_j-(C{\mathord {\mathbf {v}}})_j C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))_i|^2\\&\quad +\sum ^{\mathrm {out}}\Big \langle |(D{\mathord {\mathbf {w}}})_i(C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))_j-(D{\mathord {\mathbf {w}}})_j(C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))_i|^2 \Big \rangle \\&= |C{\mathord {\mathbf {v}}}|^2 |C \nabla _{{\mathord {\mathbf {v}}}} F |^2-(C{\mathord {\mathbf {v}}}\cdot C \nabla _{{\mathord {\mathbf {v}}}} F)^2\\&\quad +\Big \langle |D{\mathord {\mathbf {w}}}|^2 \Big \rangle |C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}})|^2-\Big \langle ({\mathord {\mathbf {w}}}, D^T C (\nabla _{{\mathord {\mathbf {v}}}} F-\ell {\mathord {\mathbf {v}}}))^2\Big \rangle \end{aligned}$$
where we used that \(\langle \omega _i\rangle =0\). Next, by using that
$$\begin{aligned} \langle \omega _i\, \omega _j\rangle =\frac{|{\mathord {\mathbf {w}}}|^2}{N} \delta _{ij}, \end{aligned}$$
we have
$$\begin{aligned} \langle \Sigma _0\rangle =&|C {\mathord {\mathbf {v}}}|^2 |C \nabla _{\mathord {\mathbf {v}}}F|^2 -(C {\mathord {\mathbf {v}}}\cdot C\nabla _{\mathord {\mathbf {v}}}F)^2 \\&+\frac{|{\mathord {\mathbf {w}}}|^2}{N} \left[ {\mathrm{Tr }} (D^T D) |C(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2 -|D^TC(\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F)|^2 \right] \end{aligned}$$
Finally, just note that due to the fact that F does not depend on the direction of \({\mathord {\mathbf {w}}}\), we have
$$\begin{aligned} \mathcal {I}= \int _{{\mathop {\mathbb {S}}}^{M+N-1}(\sqrt{M+N})} \frac{\langle \Sigma _0 \rangle }{F} {\mathord {\mathrm {d}}}\sigma . \end{aligned}$$
This proves the lemma. \(\square \)
1.2 A.2 Proof of Lemma 4.3
Combining Lemma 4.1 with 4.2, it suffices to show
$$\begin{aligned} |LF|^2-\Sigma _1\le \Sigma _2. \end{aligned}$$
(30)
Using the rotation invariance of F in the second variable, we write
$$\begin{aligned} |LF|^2= & {} \sum _{i<j} |v_i\partial _{v_j} F - v_j \partial _{v_i} F|^2 + \sum _i \sum _j |w_j \partial _{v_i}F - v_i \partial _{w_j} F|^2. \\= & {} |{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - ({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F)^2 + |{\mathord {\mathbf {w}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 + |{\mathord {\mathbf {v}}}|^2 |{\mathord {\mathbf {w}}}|^2 \ell ^2 - 2|{\mathord {\mathbf {w}}}|^2 \ell ({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F) \\= & {} |{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - ({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F)^2 + |{\mathord {\mathbf {w}}}|^2 |\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F|^2 . \end{aligned}$$
It remains, thus, to estimate
$$\begin{aligned}&|{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - ({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F)^2 + |{\mathord {\mathbf {w}}}|^2 |\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F|^2 \nonumber \\&\quad - |C {\mathord {\mathbf {v}}}|^2 |C \nabla _{\mathord {\mathbf {v}}}F|^2 + (C {\mathord {\mathbf {v}}}\cdot C\nabla _{\mathord {\mathbf {v}}}F)^2 \nonumber \\&\quad -\frac{|{\mathord {\mathbf {w}}}|^2}{N} \left[ |C(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2{\mathrm{Tr }} D^TD -|D^TC(\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F)|^2 \right] \end{aligned}$$
(31)
Since R is a rotation, we have that
$$\begin{aligned} \left( \begin{array}{cc} A &{} B \\ C &{} D \end{array} \right) \left( \begin{array}{cc} A^T &{} C^T \\ B^T &{} D^T \end{array} \right) = \left( \begin{array}{cc} I_M &{} 0 \\ 0 &{} I_N \end{array} \right) \end{aligned}$$
and
$$\begin{aligned} \left( \begin{array}{cc} A^T &{} C^T \\ B^T &{} D^T \end{array} \right) \left( \begin{array}{cc} A &{} B \\ C &{} D \end{array} \right) =\left( \begin{array}{cc} I_M &{} 0 \\ 0 &{} I_N \end{array} \right) \ . \end{aligned}$$
Using this we get that
$$\begin{aligned} {\mathrm{Tr}} D^TD = N-{\mathrm{Tr}}B^TB = N-M + {\mathrm{Tr}}A^TA , A^TA+C^TC = I_M \end{aligned}$$
and
$$\begin{aligned} D^TC = -B^TA . \end{aligned}$$
Using these relations (31) becomes
$$\begin{aligned}&|{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - ({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F)^2 + |{\mathord {\mathbf {w}}}|^2 |\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F|^2 \\&\quad - (| {\mathord {\mathbf {v}}}|^2 - |A {\mathord {\mathbf {v}}}|^2)( | \nabla _{\mathord {\mathbf {v}}}F|^2-|A \nabla _{\mathord {\mathbf {v}}}F|^2) + \left[ ({\mathord {\mathbf {v}}}, \nabla _{\mathord {\mathbf {v}}}F) -( A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F)\right] ^2 \\&\quad -\frac{|{\mathord {\mathbf {w}}}|^2}{N} \left[ ( |(\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F)|^2-|A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2)(N-{\mathrm{Tr }}B^TB) -|B^TA(\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F)|^2 \right] \end{aligned}$$
which can be simplified to
$$\begin{aligned}&|{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 +|A{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - |A{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 + (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F)^2 \\&\quad -2({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F) (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F) \\&\quad +|{\mathord {\mathbf {w}}}|^2 |A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2 \\&\quad +\frac{|{\mathord {\mathbf {w}}}|^2}{N} \left[ ( |(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2-|A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2){\mathrm{Tr }}B^TB +|B^TA(\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F)|^2 \right] . \end{aligned}$$
Since \( AA^T+BB^T = I_M\) and \({\mathrm{Tr}} B^TB = {\mathrm{Tr}} BB^T = M - {\mathrm{Tr}} AA^T\), we obtain an upper bound
$$\begin{aligned}&|{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 +|A{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - |A{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 + (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F)^2\\&\quad -2({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F) (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F) \\&\quad +|{\mathord {\mathbf {w}}}|^2 |A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2 \\&\quad +\frac{|{\mathord {\mathbf {w}}}|^2}{N} \left[ ( |(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2-|A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2)M +|A(\ell {\mathord {\mathbf {v}}}- \nabla _{\mathord {\mathbf {v}}}F)|^2 \right] . \end{aligned}$$
which can be simplified to
$$\begin{aligned}&|{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 +|A{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - |A{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 + (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F)^2 \\&\quad -2({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F) (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F) \\&\quad + \left( 1-\frac{M-1}{N}\right) |{\mathord {\mathbf {w}}}|^2 |A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2 + \frac{M}{N} |{\mathord {\mathbf {w}}}|^2 |(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2 . \end{aligned}$$
Regarding the terms quartic in A we do not know how to handle but noting that \(- |A{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 + (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F)^2\le 0\) we get the bound
$$\begin{aligned}&|LF|^2-\Sigma _1\le |{\mathord {\mathbf {v}}}|^2 |A\nabla _{\mathord {\mathbf {v}}}F|^2 +|A{\mathord {\mathbf {v}}}|^2 |\nabla _{\mathord {\mathbf {v}}}F|^2 - 2({\mathord {\mathbf {v}}}\cdot \nabla _{\mathord {\mathbf {v}}}F) (A {\mathord {\mathbf {v}}}\cdot A\nabla _{\mathord {\mathbf {v}}}F) \\&\quad + \left( 1-\frac{M-1}{N}\right) |{\mathord {\mathbf {w}}}|^2 |A(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2 + \frac{M}{N} |{\mathord {\mathbf {w}}}|^2 |(\ell {\mathord {\mathbf {v}}}-\nabla _{\mathord {\mathbf {v}}}F)|^2=\Sigma _2 , \end{aligned}$$
as claimed. \(\square \)
A.3 Proof of Lemma 2.1
One can think about K as the top left entry of the matrix
$$\begin{aligned} \sum _{\alpha _1, \dots , \alpha _k} \lambda _{\alpha _1} \cdots \lambda _{\alpha _k} \int _{[-\pi ,\pi ]^k} \nu (\mathrm {d} \theta _1) \cdots \nu (\mathrm {d} \theta _k) \, \left[ \prod _{l=1}^k r_{\alpha _l}(\theta _l)\right] ^{-1} \begin{pmatrix} I_M &{} 0 \\ 0 &{} 0\end{pmatrix} \left[ \prod _{l=1}^k r_{\alpha _l}(\theta _l)\right] . \end{aligned}$$
The computation hinges on a repeated application of the elementary identity
$$\begin{aligned}&\int _{-\pi }^{\pi } \nu (\mathrm {d}\theta ) \, \begin{pmatrix}\cos (\theta ) &{} -\sin (\theta )\\ \sin (\theta ) &{} \cos (\theta )\end{pmatrix} \begin{pmatrix} m_1 &{} 0\\ 0 &{} m_2\end{pmatrix} \begin{pmatrix} \cos (\theta ) &{} \sin (\theta )\\ -\sin (\theta ) &{} \cos (\theta )\end{pmatrix} \\&\quad = \begin{pmatrix} (1-\tilde{\nu }) m_1+\tilde{\nu } m_2 &{} 0\\ 0 &{} (1-\tilde{\nu }) m_2+\tilde{\nu } m_1 \end{pmatrix} , \end{aligned}$$
where \(\tilde{\nu }=\int \nu (\mathrm {d} \theta ) \, \sin ^2(\theta ).\) For this to be true, we just need (9). We easily check that for the rotations \(r_\alpha (\theta )\)
$$\begin{aligned}&\sum _{\alpha } \lambda _{\alpha } \int _{-\pi }^{\pi } \nu (\mathrm {d} \theta ) \, r_\alpha (\theta )^{-1} \begin{pmatrix} m_1 I_M &{} 0 \\ 0 &{}m_2 I_N\end{pmatrix} r_\alpha (\theta )\nonumber \\&\quad = \frac{1}{\Lambda }\left( \frac{M\lambda _S}{2}+\frac{N\lambda _R}{2}\right) \begin{pmatrix} m_1 I_M &{} 0 \\ 0 &{} m_2 I_N\end{pmatrix} \nonumber \\&\qquad + \frac{\mu }{\Lambda N}\begin{pmatrix} N(M-1)+N((1-\tilde{\nu }) m_1 + \tilde{\nu } m_2)I_M &{} 0 \\ 0 &{} (N-1)M+ M(\tilde{\nu } m_1 +(1-\tilde{\nu })m_2) I_N \end{pmatrix} \nonumber \\&\quad = \begin{pmatrix} m_1 I_M &{} 0 \\ 0 &{} m_2 I_N\end{pmatrix}+ \frac{\mu _\nu }{\Lambda N} \begin{pmatrix} N( m_2-m_1)I_M &{} 0 \\ 0 &{} M(m_1 -m_2)I_N \end{pmatrix} . \end{aligned}$$
(32)
where \(\mu _\nu =\tilde{\nu } \mu \). Denote by \(L(\nu _1,\nu _2)\) the \((N+M)\times (N+M)\) matrix
$$\begin{aligned} L(m_1,m_2)=\begin{pmatrix} m_1 I_{M} &{} 0\\ 0 &{} m_2 I_N\end{pmatrix}, \end{aligned}$$
and set
$$\begin{aligned} \mathcal {P}=I_2 - \frac{\mu _\nu }{\Lambda N}\begin{pmatrix} N &{} -N\\ -M &{} M \end{pmatrix} . \end{aligned}$$
Then, (32) is recast as
$$\begin{aligned} \sum _{\alpha } \lambda _{\alpha } \int _{-\pi }^{\pi } \nu (\mathrm {d} \theta ) \, r_\alpha (\theta )^{-1}L(m_1,m_2)r_\alpha (\theta )= L(m_1',m_2') , \end{aligned}$$
(33)
where
$$\begin{aligned} \begin{pmatrix} m_1' \\ m_2'\end{pmatrix}= \mathcal {P}\begin{pmatrix} m_1 \\ m_2\end{pmatrix} . \end{aligned}$$
By a repeated application of (33), we obtain
$$\begin{aligned}&\sum _{\alpha _1, \dots , \alpha _k} \lambda _{\alpha _1} \cdots \lambda _{\alpha _k} \int _{[-\pi ,\pi ]^k} \nu (\mathrm {d}\theta _1) \, \cdots \nu (\mathrm {d}\theta _k) \, \left[ \prod _{j=1}^k r_{\alpha _j}(\theta _j)\right] ^T L(\underline{m})\left[ \prod _{j=1}^k r_{\alpha _j}(\theta _j)\right] \\&\quad = L(\mathcal {P}^k\underline{m}) . \end{aligned}$$
Thus,
$$\begin{aligned} K =\left( \mathcal {P}^k\, \begin{pmatrix} 1 \\ 0\end{pmatrix}\right) _1 I_M . \end{aligned}$$
It is easy to see that \(\mathcal {P}\) has eigenvalues \(\ell _1=1\) and \(\ell _2=1-\mu _\nu (M+N)/(\Lambda N)\) with eigenvectors \(\underline{m}_1=(1, 1)\) and \(\underline{m}_2=(N,-M)^T/(M+N)\). Consequently,
$$\begin{aligned} \begin{pmatrix} 1 \\ 0\end{pmatrix}=\frac{M}{N+M}\underline{m}_1+\underline{m}_2 , \end{aligned}$$
which yields
$$\begin{aligned} \left( \mathcal {P}^k\, \begin{pmatrix} 1 \\ 0\end{pmatrix}\right) _1=\frac{M}{N+M}+ \frac{N}{M+N}\left( 1-\mu _\nu \frac{M+N}{\Lambda N}\right) ^k . \end{aligned}$$
This proves Lemma 2.1. \(\square \)