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Use a real physical field, so the three positive wavevectors are accompanied by their complex-conjugate modes. Modulo lattice translations, the effective symmetry group is $G={\mathbb{T}}^2⋊C_6$. Its translation group action isA $120^{\circ }$ rotation cyclically permutes the amplitudes; a half-turn $\kappa$ acts by simultaneous complex conjugation. Together these generate the sixfold rotation group. These formulas specify the group action independently of the choice of orientation of the three wavevectors.

For a nonzero real hexagon $(h,h,h)$, its stabilizer subgroup is the rotation group $C_6$. A translation fixes it only when all three phase factors equal one. Cyclic permutation forces a fixed vector to have equal components, and the half-turn forces those components to be real. Therefore $\mathrm{Fix}(C_6)=\{(a,a,a):a\in \mathbb{R}\}$. For a roll $(r,0,0)$ with nonzero real $r$, the stabilizer subgroup consists of the translations $T_{0,\theta }$ and the half-turn, with $\kappa T_{0,\theta }\kappa =T_{0,-\theta }$. It is $S^1⋊C_2$, isomorphic to $O(2)$ as an abstract group even though spatial reflections are absent. The circle eliminates $B,C$ from its fixed-point subspace of a group action, and the half-turn makes $A$ real. ThusIf unreduced physical translations are used, the full lattice kernel is included in both stabilizer subgroups. The real representation on the six critical components is absolutely irreducible: translations distinguish the three Fourier pairs, rotations permute them, and the half-turn rules out a complex scalar in the commutant. The equivariant branching lemma consequently supplies generic steady branches of both axial types at a transverse crossing of the common linear eigenvalue. It guarantees existence, not stability or exhaustiveness of the possible branches.

Translation equivariance severely restricts the monomials in $\dot{A}$. Through cubic order they are $A$, $\overline{B}\overline{C}$, $|A{|}^{2}A$, $|B{|}^{2}A$ and $|C{|}^{2}A$: each has the same translation weight as $A$, using ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. The half-turn forces their coefficients to be real; the $120^{\circ }$ rotation gives the two cyclic counterpart equations. A spatial reflection would interchange $B,C$ and require equal cross-couplings, but it is not a symmetry here. After a real amplitude rescaling makes the generic nonzero quadratic coefficient one, the rotational hexagon amplitude equations arewith the other two equations obtained by cyclic permutation. All four coefficients are real; $\delta$ records the absence of spatial reflection. The normalization assumes the resonant quadratic coefficient is nonzero; its vanishing would be an additional degeneracy.

For the roll, $r^2=\mu /{\nu }_1$. The radial eigenvalue is $-2\mu$ and the phase of the active mode is a neutral translation. The transverse variables $(b,\overline{c})$ have matrixIts two real eigenvalues, each occurring twice in the full real linearization, are $-({\nu }_2-{\nu }_1)r^2\pm \sqrt{{\delta }^2{r}^4+r^2}$. Define $d={\nu }_2-{\nu }_1$ and $D_r=d^2-{\delta }^2$. Strict attraction modulo translation is therefore equivalent toIndeed the larger transverse eigenvalue is negative precisely when $d{r}^2>\sqrt{{\delta }^2{r}^4+r^2}$, and squaring is legitimate only after requiring $d>0$. If ${\delta }^2>d^2$, that inequality is impossible. Equality ${\delta }^2=d^2$ is also insufficient because of the additional positive term $r^2$. Any nonzero roll that exists with ${\nu }_1<0$ already has a positive radial eigenvalue; the degenerate case ${\nu }_1=0$ has no ordinary isolated nonzero roll branch.

For hexagons put $N={\nu }_1+2{\nu }_2$. A nonzero real amplitude $h$ satisfies $\mu =N{h}^2-h$, or $h=(1\pm \sqrt{1+4N\mu })/(2N)$ when $N\ne 0$. Separate real and imaginary disturbances. The real Jacobian matrix is cyclic, with diagonal $-h-2{\nu }_1{h}^2$ and off-diagonal entries $h-2({\nu }_2\pm \delta )h^2$. Its eigenvalues areThe imaginary Jacobian matrix has every entry equal to $-h$, giving eigenvalues $-3h,0,0$. The two zero modes translate the hexagonal pattern. Thus the general strict stability conditions for any nonzero root $h$ areThese conditions state orbital stability transverse to translations, not asymptotic convergence to one prescribed spatial phase. Equalities require a nonlinear bifurcation calculation.

When ${\nu }_2>{\nu }_1>0$, both $d,N$ are positive. The upper positive hexagon branch is radially stable after its saddle-node bifurcation at $\mu =-1/(4N)$, $h=1/(2N)$; it remains stable until $h=1/d$. At this second threshold the pair ${\lambda }_{\pm }$ crosses the imaginary axis with nonzero frequency if $\delta \ne 0$. The crossing is transverse, since $dh/d\mu =1/(2Nh-1)$ and the derivative of $-2h+2d{h}^2$ with respect to $h$ is two there. Hence, subject to the usual nonzero cubic coefficient, this is a Hopf bifurcation:For $\delta =0$ the same threshold has two zero real eigenvalues, so it is not a Hopf bifurcation.

A possible phase portrait when neither steady branch is stable has an attracting periodic oscillation surrounding the unstable hexagon, with the three amplitudes waxing and waning cyclically. This is realized, for example, by ${\nu }_1=1$, ${\nu }_2=2$, $\delta =2$ just above ${\mu }_H=4$. Rolls are unstable for every positive $\mu$ because ${\delta }^2=4>d^2=1$. At the hexagon Hopf point $h=1$, the real spectrum is $-9,\pm 4i\sqrt{3}$. Evaluation of the cubic Hopf normal form with a unit-norm critical eigenvector gives $\mathrm{Re}{G}_{21}=-64/27<0$, so this example has a supercritical Hopf bifurcation and an attracting small limit cycle in the real invariant subspace. The figure shows such a cycle at $\mu =4.2$ and an approaching trajectory; the positive octant is forward invariant because an amplitude's derivative at zero is the product of the other two. This is a possible portrait, rather than a claim that every parameter set with unstable rolls and hexagons has the same attractor. Stability asserted for the periodic orbit here is within the stipulated real subspace.

For the rotational hexagon amplitude equations, put $d={\nu }_2-{\nu }_1$ and $N={\nu }_1+2{\nu }_2$. A roll with $r^2=\mu /{\nu }_1$ has transverse eigenvalues $-d{r}^2\pm \sqrt{{\delta }^2{r}^4+r^2}$, each twice, radial eigenvalue $-2\mu$, and one neutral translation. A hexagon with $\mu =N{h}^2-h$ has real-amplitude eigenvalues $h-2N{h}^2$ and $-2h+2d{h}^2\pm 2i\sqrt{3}\delta h^2$; its imaginary-amplitude eigenvalues are $-3h,0,0$. These follow by diagonalizing the cyclic Jacobian matrix. Stability is understood modulo translations. If ${\nu }_2>{\nu }_1>0$ and $\delta \ne 0$, the upper positive hexagon branch reaches a Hopf bifurcation at $h=1/d$, $\mu =(2{\nu }_1+{\nu }_2)/d^2$.