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The six critical Fourier modes are at $\pm {\mathbf{k}}_1,\pm {\mathbf{k}}_2,\pm {\mathbf{k}}_3$, with ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. Reality of the pattern pairs each positive-wavevector coefficient with the conjugate coefficient at the negative wavevector. The leading critical part of the extended centre manifold parametrization is consequentlyThe displayed six-mode expression is its leading-order projection; nonlinear harmonics on the full centre manifold need not vanish identically.

A translation $\mathbf{d}$ acts by $A_i\mapsto e^{i{\mathbf{k}}_i\cdot \mathbf{d}}{A}_i$. Each monomial in the equation for $A_i$ must therefore carry wavevector ${\mathbf{k}}_i$. Since $-{\mathbf{k}}_{i+1}-{\mathbf{k}}_{i+2}={\mathbf{k}}_i$, the allowed quadratic term is $A_{i+1}^{\ast }{A}_{i+2}^{\ast }$. At cubic order the allowed terms are $A_i|A_i{|}^2,A_i|A_{i+1}{|}^2,A_i|A_{i+2}{|}^2$. Rotation through $60$ degrees acts, with one choice of orientation, asIts square cyclically permutes the amplitudes, equating the coefficients across the three equations. Its cube conjugates every amplitude; equivariance under this half-turn makes all coefficients real. It does not interchange the two cross-coupling coefficients. A reflection would do that and force $b=c$, but that symmetry is expressly absent. Thus the chiral hexagonal amplitude equations areat the retained order, with cyclic indices and real coefficients. The missing opening modulus bar in the last cubic term of the printed formula must be supplied; phase equivariance requires $|A_{i+2}{|}^2$.

Write $A_i=r_i{e}^{i{\varphi }_i}$, with all three amplitudes nonzero, and let $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. Separating real and imaginary parts givesHence the phase locking of a resonant hexagonal triad obeysFor $\alpha >0$, $\Phi =0$ is phase stable and $\Phi =\pi$ is phase unstable. Two independent translation coordinates remove two phases; since the wavevectors sum to zero, the remaining phase is exactly $\Phi$. Thus a stable fully active phase-locked pattern can be represented by three positive real amplitudes after choosing the spatial origin. The statement is modulo translations, not a restriction on every representative. Boundary states must also be retained: the zero solution is stable for $\mu <0$, and rolls have two zero amplitudes. These are the necessary qualifications to a literal claim that all stable solutions are strictly positive.

In the real nonnegative subspace, a roll isor a cyclic permutation. For hexagons let $s=a+b+c$, which is positive under the assumptions. The equal-amplitude branches solve $\mu +\alpha r-s{r}^2=0$, givingwith only positive roots retained. The upper branch exists for $\mu \ge -{\alpha }^2/(4s)$; the lower branch is positive for $-{\alpha }^2/(4s)\le \mu <0$. They meet at a fold.

For a purported rectangle $(p,p,q)$ with $p,q>0$, subtract its first two equilibrium equations after dividing each by its own amplitude. The quadratic terms agree, and the difference is $(c-b)(p^2-q^2)=0$. Since $(b-a)(c-a)<0$ implies $b\ne c$, it forces $p=q$. A state with exactly two nonzero amplitudes is also impossible because the quadratic term drives the zero third amplitude. Thus there are no nontrivial rectangles.

In fact no other unequal positive equilibria are possible under the straddling cross-couplings. Here is the uniqueness of positive hexagon equilibria with straddling cross-couplings argument. Put $x_i=r_i^2$, $p=r_1{r}_2{r}_3>0$. The equilibrium equations implyConsider $b>a>c$, write $B=b-a>0$, $C=c-a<0$, and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two givesThe left side is positive unless all three are equal, while the right side is nonpositive. If $x_2\le x_3$, subtract equations two and three instead:Now the left side is negative unless all are equal and the right side is nonnegative. Both cases force equality. For $c>a>b$, reverse the cyclic orientation and exchange $b,c$. Thus the only nonzero equilibria in this subspace are the stated rolls and hexagons.

At a roll the active radial eigenvalue is $-2\mu$. The two inactive amplitudes have real-part linearizationIts determinant is ${\mu }^2(a-c)(a-b)/a^2-{\alpha }^2\mu /a<0$. It therefore has a positive eigenvalue: the rolls are unstable for every $\mu >0$, which is stronger than the large-$\mu$ conclusion. In the complex system there is also a neutral translation phase, and the imaginary transverse block has the same determinant.

At a positive hexagon the real-amplitude Jacobian is circulant, with diagonal $d=-2a{r}^2-\alpha r$ and off-diagonals $p=\alpha r-2b{r}^2$, $q=\alpha r-2c{r}^2$. Its uniform eigenvector $(1,1,1)$ has eigenvalueThe other two eigenvectors lie in the zero-sum subspace; using the two nontrivial cube roots of unity givesFor sufficiently large $\mu$, $r\sim \sqrt{\mu /s}$, and the positive term $(b+c-2a)r^2$ dominates the quadratic-coupling term $2\alpha r$. Thus the hexagons are unstable at large $\mu$, while the uniform radial direction on the upper branch is stable. The two translation phases are neutral and the total phase has eigenvalue $-3\alpha r$, so the amplitude instability is genuine.

More precisely, the upper branch is amplitude stable between its fold and the chiral hexagon Hopf thresholdAt this threshold the conjugate pair crosses with nonzero frequency because $b\ne c$, permitting a Hopf bifurcation. The linear calculation locates the crossing; the nonlinear normal form decides the stability and direction of its periodic branch.

The dynamics are bounded in the retained model. For $S=r_1^2+r_2^2+r_3^2$ in the positive phase-locked subspace,Since $r_1{r}_2{r}_3\le (S/3{)}^{3/2}$, the negative quadratic-in-$S$ term dominates at large amplitude. For large positive $\mu$, the origin, rolls and hexagons are all unstable, so generic bounded dynamics cannot end at an asymptotically stable steady pattern. The chiral difference $b-c$ favors successive changes in dominant roll orientation, in the spirit of the Küppers–Lortz instability. Oscillatory competition or periodic switching is the natural behavior near the Hopf transition; linear stability alone does not establish a unique global limit cycle. With $\alpha =0$, coordinate planes are invariant and heteroclinic roll switching is possible. For $\alpha >0$ a missing third mode is regenerated when two others are present, so that exact boundary-plane heteroclinic picture is broken. Finally, sufficiently large $\mu$ is a conclusion of the cubic truncation; far from onset, omitted higher-order terms can change the physical pattern dynamics.