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Write the translation group action as $(A,B,C)\mapsto (e^{i{\theta }_1}A,e^{i{\theta }_2}B,e^{-i({\theta }_1+{\theta }_2)}C)$. Modulo lattice translations, the stabilizer subgroup of $(h,h,h)$, with nonzero real $h$, is the sixfold rotation group. Its fixed-point subspace of a group action is the real line $(a,a,a)$. The stabilizer subgroup of $(r,0,0)$, with nonzero real $r$, consists of the circle ${\theta }_1=0$ together with a half-turn; its fixed-point subspace of a group action is the real line $(a,0,0)$. The equivariant branching lemma therefore supplies generic branches of both axial types.

For a steady-state equivariant dynamical system, an axial subgroup has a one-dimensional real fixed-point subspace of a group action. A complex axial subgroup for a Hopf problem has a one-dimensional complex fixed-point subspace under the spatial group combined with the common phase action. Distinct conjugacy classes may yield distinct types of bifurcating branches.

The normalizer of an axial isotropy subgroup preserves its one-dimensional fixed-point subspace of a group action. If its quotient by the isotropy contains an element acting as minus one, the reduced scalar vector field is odd, forcing a pitchfork bifurcation rather than a generic quadratic branch. For odd dihedral groups the reflection normalizer has no such quotient; an even-degree anisotropic term can distinguish the two half-axes at higher order.

// Target: algebra.bigb

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 a vector $x$, its isotropy group is the stabilizer subgroupFor a subgroup $H\le G$, its fixed-point subspace of a group action isIt is a linear subspace, the intersection of the kernels of $R_G(h)-I$. The subgroup fixes a particular vector pointwise; the fixed-point subspace of a group action consists of all vectors it fixes, and need not be one-dimensional.

Suppose $f$ is sufficiently smooth and equivariant, $f(0,\mu )=0$, and the group representation on the critical space is absolutely irreducible. Its linearization is then $D_{x}f(0,\mu )=a(\mu )I$. Assume $a(0)=0$ and $a^{\prime }(0)\ne 0$. The equivariant branching lemma states that each isotropy subgroup $H$ with $\dim \mathrm{Fix}(H)=1$ has a local nontrivial branch of equilibria with that isotropy, apart from the common origin. Conjugate isotropy subgroups give symmetry-related branches, so branch types are counted up to conjugacy.

To see the scalar mechanism, write the fixed-point subspace of a group action as $\mathbb{R}e$ and restrict $f(se,\mu )=eh(s,\mu )$. Equivariance makes the fixed-point subspace of a group action invariant. Since $h(0,\mu )=0$, write $h=sk(s,\mu )$; $k(0,0)=0$ and $k_{\mu }(0,0)=a^{\prime }(0)\ne 0$. The implicit function theorem solves $k(s,\mu (s))=0$ near zero. For small nonzero $s$, the stabilizer subgroup is exactly that of $e$. A normalizer element reversing $e$ makes $h$ odd and $\mu (s)$ even; the generic branch is then a pitchfork bifurcation. The lemma ensures existence, not dynamical stability, and does not exclude further branches of smaller isotropy.

Choose three representative vectors, with nonzero coordinate scale suppressed. Their stabilizer subgroups and fixed-point subspaces of a group action areHere ${\kappa }_x{r}_z$ exchanges the first two coordinates and ${\kappa }_y{r}_x$ exchanges the last two. On the first line the stabilizer subgroup is every signed permutation of the two zero coordinates; fixing it forces those coordinates to vanish. On the second line it exchanges the equal coordinates and independently reflects the zero coordinate, forcing $x=y,z=0$. On the third line all coordinate permutations fix the vector, forcing $x=y=z$.

The three subgroups have different orders and hence cannot be conjugate. Their group orbits contain respectively six face-axis, twelve edge-axis and eight vertex-axis directions. Each fixed-point subspace of a group action is one-dimensional, so the equivariant branching lemma guarantees three distinct symmetry types of bifurcating equilibria. The axial isotropy types of the full cube symmetry group are precisely these: a vector with several distinct nonzero absolute coordinate values has a separate freely variable coordinate block for each such value in the fixed-point subspace of a group action of its stabilizer subgroup, giving dimension greater than one.

Let a finite or compact group $G$ act linearly on a real space $V$, and let $\dot{x}=F(x,\mu )$ be a smooth equivariant dynamical system: $F(gx,\mu )=gF(x,\mu )$. Assume $F(0,\mu )=0$. The representation is an absolutely irreducible real representation when its complexification is irreducible; equivalently, every real linear operator commuting with $G$ is a scalar multiple of the identity. Consequently the linearization at the trivial branch is $D_{x}F(0,\mu )=\alpha (\mu )I$. A generic steady-state crossing has $\alpha (0)=0$ and ${\alpha }^{\prime }(0)\ne 0$, with stable noncritical directions removed by a centre manifold if this is a reduction of a larger system.

The isotropy group of a vector $x$ is ${\Sigma }_x=\{g\in G:gx=x\}$. For a subgroup $\Sigma$, its fixed-point subspace of a group action is $\mathrm{Fix}(\Sigma )=\{x:\sigma x=x\text{ for every }\sigma \in \Sigma \}$. Its normaliser isThe normaliser preserves $\mathrm{Fix}(\Sigma )$, and $N_G(\Sigma )/\Sigma$ describes its residual symmetry. A steady-state axial subgroup is an actual isotropy subgroup with ${\dim }_{\mathbb{R}}\mathrm{Fix}(\Sigma )=1$.

The equivariant branching lemma states that, under the smoothness, equivariance, absolute irreducibility, and transverse scalar-eigenvalue crossing hypotheses above, each conjugacy class of axial isotropy subgroup gives a branch of nonzero equilibria in its fixed-point line through $(0,0)$. The generic leading nonlinear coefficient determines which parameter side the branch occupies and whether it is transcritical-like or pitchfork-like; the theorem does not assert that these branches are stable or exhaust every possible branch. The usual nondegeneracy condition excludes a vanishing leading nonlinear coefficient. Conjugate subgroups produce symmetry-related copies of a branch, rather than new isotropy types.

To explain the dynamical role of the fixed-point subspaces of a group action, if $x$ is fixed by $\Sigma$, then $\sigma F(x,\mu )=F(\sigma x,\mu )=F(x,\mu )$ for every $\sigma \in \Sigma$. Thus the vector field is tangent to that subspace, and uniqueness makes it flow invariant. On a fixed-point line one solves a scalar equilibrium equation and determines radial stability. Stability in the full space additionally requires all transverse eigenvalues to be stable. Higher-dimensional fixed-point spaces support their own restricted dynamics and may contain symmetry-constrained connecting orbits.

Finally, the normalizer action determines axial branch parity: if $N_G(\Sigma )/\Sigma$ contains an element acting as $x\mapsto -x$ on the fixed line, its scalar vector field must be odd, so a generic cubic term gives a pitchfork bifurcation. If that residual sign reversal is absent, a quadratic term is permitted and can give a transcritical-like branch. This is why fixed-space dimension alone does not determine either the branch type or its full stability.

For the Equivariant branching lemma, assume a generic steady-state crossing $DF(0,\mu )=\mu I$ after parameter normalization, and nonzero leading nonlinear coefficient on each axial line. The lemma gives an equilibrium branch for each axial subgroup with a one-dimensional real fixed-point subspace of a group action.

For $H_{\mathrm{ax}}=⟨{\kappa }_y,{\kappa }_z⟩$, $\mathrm{Fix}(H_{\mathrm{ax}})=\{(r,0,0)\}$. Conjugation by $\rho$ supplies the other two coordinate axes. For $H_{\mathrm{diag}}=⟨\rho ⟩$, $\mathrm{Fix}(H_{\mathrm{diag}})=\{(r,r,r)\}$. Sign reflections supply all the equal-magnitude diagonals. These isotropy types are not conjugate: their subgroup orders are $4$ and $3$. Thus there are two distinct primary axial branch types: coordinate-axis states and equal-magnitude diagonal states.

The global sign reversal belongs to the group and makes the restricted scalar equations odd, so generically each has the form $\dot{r}=r(\mu +\gamma r^2+\cdots )$, with $\gamma \ne 0$. Its two nonzero solutions satisfy $r^2=-\mu /\gamma +\cdots$ on the appropriate parameter side. Symmetry-related solutions belong to the same branch type; their existence does not imply stability or that both types lie on the same side of the crossing.