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Take the coupling parameters to be real. The dipole and quadrupole parity in a mean-field dynamo subspaces are invariant: substitution of $A_2=A_1$, $B_2=-B_1$ makes the second-hemisphere equations equal to the first-hemisphere equations with these same signs. Substitution of $A_2=-A_1$, $B_2=B_1$ likewise preserves the other parity. In either subspace put $A=A_1$, $B=B_1$. The reduced equations and characteristic equations areA marginal normal mode has $\lambda =i\omega$. Equating real and imaginary parts givesConsequently the marginal surfaces areTo identify which side actually grows, write $z=ab$ in $(\lambda +1{)}^2=z$. The largest real part of the two eigenvalues is $-1+\sqrt{(|z|+\mathrm{Re}z)/2}$. It is negative exactly when $\mathrm{Re}z+[\mathrm{Im}z{]}^2/4<1$: squaring $|z|<2-\mathrm{Re}z$ proves this, and the latter condition ensures the right-hand side is positive. Thus dipole modes decay for $F_d<1$ and quadrupole modes decay for $F_q<1$, with growth for the corresponding reversed inequality.

For explicit coupled-hemisphere dynamo parity thresholds, suppose $\Omega \ne 0$ and defineThe dipole stability interval is $\delta -L<{\alpha }_0<\delta +L$, and the quadrupole stability interval is $-\delta -L<{\alpha }_0<-\delta +L$. A comparison of first onset from ${\alpha }_0=0$ assumes both are initially stable, namelyIncrease $|{\alpha }_0|$ along a specified sign $s=\mathrm{sgn}{\alpha }_0$. The first thresholds for the two parities areTherefore the quadrupole reaches onset first if $s({\Omega }^2{ϵ}_1+2{ϵ}_2)>0$; the dipole reaches onset first if it is negative; the thresholds coincide if it vanishes. For positive ${\alpha }_0$ and positive coupling parameters this selects the quadrupole, provided the initial stability condition holds. Reversing the sign of ${\alpha }_0$ exchanges the preference. If both signs are freely allowed, each parity attains the same minimum threshold $|\Omega |(L-|\delta |)$ on opposite sign rays; a unique preference cannot be inferred from the absolute product alone. If the initial stability condition fails, the zero-alpha state is already unstable or marginal and there is no first onset from a stable origin. When $\Omega =0$, the formulas for $F_d,F_q$ still apply, but $|{\alpha }_0\Omega |$ is identically zero and is not a useful onset parameter.