Remembering that in our coordinate system ω = (sin θ, 0 , cos θ ) and ω ′ = (cos θ ′ sin θ ′ , sin φ ′ sin θ ′ , cos θ ′ ), as required. (Here the inner two integrals alone give π $^{3}$/ 2 which can be found straight- forwardly). There are three further integrals required for the bifurcation analysis. I 3 = ∫ d ω [ P$_{2}$ (cos θ )] 2 = 4 π/ 5 trivially. Let us turn our attention to with coefficients where the double factorial is defined by Therefore the inner integral of (A7) becomes We now utilise the addition formula for Legendre polynomials, 25 P$_{2}$$_{n}$ (cos γ ) = P$_{2}$$_{n}$ (cos θ ) P$_{2}$$_{n}$ (cos θ $^{′}$) + 2 2 n ∑ m =1 (2 n − m )! (2 n + m )! × P m 2 $_{n}$(cos θ ) P m 2 $_{n}$(cos θ $^{′}$) cos m ( φ − φ $^{′}$) . (A12) ∫ d ω $^{′}$sin γP$_{2}$ (cos θ $^{′}$) = ∫ d ω ′ ∞ ∑ n =0 c$_{2}$$_{n}$P$_{2}$$_{n}$ (cos γ ) P$_{2}$ (cos θ $^{′}$) = ∫ 2 π 0 dφ ′ ∫ π 0 dθ $^{′}$sin θ ′ ∞ ∑ n =0 c$_{2}$$_{n}$P$_{2}$$_{n}$ (cos γ ) P$_{2}$ (cos θ $^{′}$) . (A11) n !! ≡     $^{n}$ · ( n − 2) · · · 5 · 3 · 1 n > 0 odd n · ( n − 2) · · · 6 · 4 · 2 n > 0 even 1 n = − 1 , 0. (A10) c$_{2}$$_{n}$ = − π (4 n + 1)(2 n − 3)!!(2 n − 1)!! 2 n $^{+1}$( n + 1)! , (A9) I 4 = ∫ d ω ∫ d ω $^{′}$sin γ [ P$_{2}$ (cos θ ) P$_{2}$ (cos θ $^{′}$)] . (A7) We expand sin γ as sin γ = ∞ ∑ n =0 c$_{2}$$_{n}$P$_{2}$$_{n}$ (cos γ ) (A8) I 2 = ∫ d ω ′′ ∫ d ω ∫ d ω ′ ∣ ∣    sin θ 0 cos θ    ·    sin φ ′ sin θ ′ − cos θ ′ sin θ ′ 0    ∣ ∣ = ∫ d ω ′′ ∫ d ω ∫ d ω $^{′}$| sin θ sin θ $^{′}$sin φ $^{′}$| = ∫ d ω ′′ · 2 π 3 = 8 π 4 (A6)