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Write $\overline{\rho }={\sum }_i{p}_i{\rho }_i$. The Holevo quantity iswhere $S$ is Von Neumann entropy, with a consistent logarithm base. Attach an orthogonal classical label register $X$ to form the classical-quantum stateThe entropy of an orthogonal quantum mixture givesHere $H(p)$ is the Shannon entropy of the classical label. Hence $I(X:B)=\chi (\mathcal{E})$, where $I$ denotes quantum mutual information.

Let $\Phi :B\to C$ be the proposed quantum channel. Using its Kraus representation, define an linear isometry $V={\sum }_{\ell }{A}_{\ell }\otimes |\ell {⟩}_E$, since $V^{†}V=I$. This is a Stinespring representation. Apply it to $B$, obtaining a state ${\eta }_{XCE}$; tracing out $E$ applies $\Phi$ to every ensemble member. An linear isometry preserves nonzero eigenvalues and therefore Von Neumann entropy, soTheir difference isThe Strong subadditivity of Von Neumann entropy states that this expression is nonnegative for every tripartite state. ConsequentlyThus the Holevo quantity under a quantum channel decreases because discarding the dilation environment cannot improve correlations with the classical label. The probabilities remain unchanged throughout.

Using base-two logarithms, define the quantum conditional entropy byThe Subadditivity of Von Neumann entropy, $S(AB)\le S(A)+S(B)$, immediately gives $S(A\mid B)\le S(A)$.

For the lower bound, purify ${\rho }_{AB}$ by an auxiliary system $C$. Complementary subsystems of a pure state have the same nonzero reduced-state eigenvalues, by the Schmidt decomposition. Hence $S(BC)=S(A)$ and $S(C)=S(AB)$. Applying Subadditivity of Von Neumann entropy to $BC$ givesRearranging yields $S(A\mid B)\ge -S(A)$. This is the relevant side of the Araki–Lieb inequality, here derived using purification.

Finally the maximum entropy of a quantum state gives $S(A)\le {\log }_2\dim {\mathcal{H}}_A$. For example, this follows directly fromby nonnegativity of quantum relative entropy. Combining the two bounds provesBoth signs can be attained: a maximally mixed $A$ independent of $B$ has the positive extreme, while a maximally entangled state with a sufficiently large $B$ has the negative extreme.

A density operator is Hermitian, positive semidefinite and has trace one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:The remaining basis vectors are traceless by orthogonality to $I/\sqrt{N}$. The Generalized Bloch representation is consequentlyUsing orthonormality to expand the squared Hilbert-Schmidt norm gives the purity bound for generalized Bloch vectors:If $p_1,\dots ,p_N$ are its nonnegative eigenvalues, then ${\sum }_j{p}_j=1$ andIt follows thatEquality requires every pairwise product $p_j{p}_k$ to vanish, so exactly one eigenvalue equals one and all others are zero. Conversely, a rank-one projector has precisely that spectrum and attains equality. Thus equality holds exactly for a pure state. This normalization is different from the commonly used unit-radius qubit Bloch vector.

For $N=2$, take the normalized Pauli operators as the traceless basis. The matrix associated with a vector $s$ is $\rho =I/2+s\cdot \sigma$, where these basis matrices are the usual Pauli matrices divided by $\sqrt{2}$. Their multiplication relations giveAt radius $1/\sqrt{2}$, these eigenvalues are one and zero. Every sphere point therefore gives a positive rank-one projector. Any other orthonormal Hermitian basis merely changes the real coordinates by an orthogonal transformation, so the result is basis independent.

For every $N>2$, choose a rank-one projector $P$ and consider the Hermitian trace-one matrixIts eigenvalues are $2/N-1$ once and $2/N$ with multiplicity $N-1$. They sum to one and satisfyBy orthonormality its traceless coordinate vector has squared length $1-1/N$, so it is on the specified sphere. But $2/N-1<0$, so $A$ is not a density operator. Thus for $N>2$, the pure states occupy a proper subset of the sphere; positivity excludes some of its points. In fact, this example is the antipodal Bloch vector to that of $P$.

The usual assertion presumes $N\ge 2$. If the trivial dimension $N=1$ is admitted, the radius-zero sphere consists of the single coordinate point, also corresponding to the single pure state, so that dimension is an additional trivial exception.

Density operator controllability cannot defeat unitary invariants: it grants access to the unitary orbit of a density operator, whose members have the same eigenvalues, rather than to every density operator. In the Hilbert-Schmidt inner product orthonormal normalization of this question,Quantum Hamiltonian evolution preserves this norm, so the target is not reachable from the stated initial state. Equivalently, their values of purity of a density operator are $1/2+1/4=3/4$ and $1/2+1/2=1$: the initial state is a mixed state and the target a pure state. Dissipation, an appropriate measurement protocol or an auxiliary system with additional resources would be needed to change the spectrum; freely controlling the closed-system Quantum Hamiltonian alone does not suffice.

Subtract a steady state to obtain $e(t)=s(t)-s_{\ast }$ with $\dot{e}=Ae$, hence $e(t)=e^{At}e(0)$. The steady state attracts every initial condition exactly whenIn other words, $A$ must be a Hurwitz stable matrix. To see sufficiency without assuming diagonalizability, each block in its Jordan normal form contributes terms of the form $t^k{e}^{\mu t}$ with finite $k$. Negative real part makes every such term tend to zero, so the whole matrix exponential tends to zero.

For necessity, an eigenmode with positive real part grows, and a mode with zero real part persists or oscillates instead of decaying. For a complex eigenvalue of a real matrix, take the real and imaginary parts of its eigenvector to obtain real nondecaying solutions. Thus any eigenvalue in the closed right half-plane prevents convergence of all deviations. Physical density operators span the trace-one affine space, so convergence from all physical initial states also forces decay on every traceless direction.

This gives global asymptotic stability for the affine linear system and uniqueness of its steady state. A merely negative-semidefinite symmetric part proves only nonincrease of a particular norm, not this necessary-and-sufficient condition.

With $q=2\lambda +1$, the reduced coefficient matrix and its symmetric part areIts diagonal entries are both strictly negative exactly when $q\ne 0$, equivalently $\boxed{\lambda \ne -1/2}$. The coherent driving cancels from the symmetric part because it is a rotation generator.

For a steady state $s_{\ast }$ put $e=s-s_{\ast }$. Differentiating the squared Euclidean distance givesFor $q\ne 0$ the derivative is strictly negative whenever $e\ne 0$. Integrating the inequality proves the symmetric-part contraction criterion here explicitly:Thus every nonstationary initial state approaches the unique steady state, with strictly decreasing distance; the steady state itself has constant zero distance. If the decoupled $y$ coordinate is retained, it contributes $-e_y^2$ to the squared-distance derivative and also decays, so the conclusion holds for the full qubit state.

For the target-dependent settings in (e), $q=-\cos {\theta }_d$. Every nonequatorial target is therefore globally attractive. At an equatorial target, $q=\alpha =0$: $x(t)=x(0)$, $z(t)=z(0)e^{-t/2}$, and $y(t)=y(0)e^{-t/2}$. A general state approaches $(x(0),0,0)$ rather than the specified pure target $x=\pm 1$. This proves the equatorial exception, not merely failure of a sufficient estimate.

The parameter dependence matters: $q=0$ alone does not exclude attractivity for every possible driving strength. If $\alpha \ne 0$, the reduced matrix has negative trace and positive determinant $4{\alpha }^2$, so both eigenvalues have negative real parts even though its symmetric part is only semidefinite. The nonattractive case asserted for the equatorial target also uses its prescribed $\alpha =0$.

Take the vectors in the quantum state ensemble to be normalized, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The corresponding density matrix isEach rank-one projector is a self-adjoint operator, so ${\rho }^{†}=\rho$. For every $|v⟩$ in the Hilbert space,proving that $\rho$ is positive semidefinite. The trace is $\mathrm{Tr}\rho ={\sum }_i{p}_i⟨{\psi }_i|{\psi }_i⟩=1$.

Conversely, diagonalize a self-adjoint operator $\rho$ in an orthonormal eigenbasis, writing $\rho ={\sum }_j{\lambda }_j|e_j⟩⟨e_j|$. Its positive semidefiniteness implies ${\lambda }_j=⟨e_j|\rho |e_j⟩\ge 0$, and its unit trace gives ${\sum }_j{\lambda }_j=1$. Thus $\{{\lambda }_j,|e_j⟩\}$ is a quantum state ensemble with precisely this density matrix. Terms with zero eigenvalue may be omitted. This argument includes rank-one pure states; an ensemble representation need not describe a genuinely mixed state.

The stated condition is the definition of an extreme point in the convex set of density matrices. Suppose first that $\rho =|\psi ⟩⟨\psi |$ with $\Vert \psi \Vert =1$, and consider any convex combination $\rho ={\sum }_i{a}_i{\rho }_i$ with all $a_i>0$. For every $|w⟩$ orthogonal to $|\psi ⟩$,Every term is nonnegative, so $⟨w|{\rho }_i|w⟩=0$ for each $i$. To see that this implies ${\rho }_{i}w=0$, expand ${\rho }_i={\sum }_j{\mu }_j|u_j⟩⟨u_j|$, ${\mu }_j\ge 0$. The equality ${\sum }_j{\mu }_j|⟨u_j|w⟩{|}^2=0$ implies ${\mu }_j⟨u_j|w⟩=0$ for every $j$, which kills ${\rho }_{i}w$. Thus each density matrix ${\rho }_i$ vanishes on ${\psi }^{\perp }$. By self-adjointness its range lies in the line $\mathbb{C}\psi$, so ${\rho }_i=c_i|\psi ⟩⟨\psi |$. Its unit trace forces $c_i=1$, establishing the required purity.

Conversely, suppose $\rho$ is not rank one. In its spectral decomposition, at least two eigenvalues are positive. Select a normalized eigenvector $|e_1⟩$ with positive eigenvalue ${\lambda }_1$; then $0<{\lambda }_1<1$. DefineThe remaining eigenvalues show that $\sigma$ is a density matrix. The decomposition $\rho ={\lambda }_1|e_1⟩⟨e_1|+(1-{\lambda }_1)\sigma$ is a convex combination of two different density matrices, neither equal to $\rho$. This violates the given purity condition. Hence the extreme points of the density-operator state space obey

The EPR criterion of reality says that if a physical quantity can be predicted with certainty without disturbing the system, then it represents an element of physical reality. For separated particles, the EPR locality assumption regards choices of measurements on the other particles as nondisturbing. The perfect correlations below therefore imply predetermined local values for both $X={\sigma }_x$ and $Y={\sigma }_y$, independent of which distant measurements are chosen.

Put $|0⟩=|\uparrow ⟩$ and $|1⟩=|\downarrow ⟩$. The Pauli matrices satisfyFor three particles, each operator with one $X$ and two $Y$ takes $|000⟩$ to $-|111⟩$ and $|111⟩$ to $-|000⟩$, whereas $X\otimes X\otimes X$ interchanges the two strings without a minus sign. Thus the GHZ state in the question has the certain outcomesEach local $X$ or $Y$ value can be predicted from measurements at the other two sites in one of these contexts. If the EPR criterion of reality and locality assign values $x_j,y_j\in \{\pm 1\}$, the first three equalities requireMultiplying these ordinary numbers and using $y_j^2=1$ yields $x_1{x}_2{x}_3=1$, contradicting the certain quantum outcome $-1$. This derives the GHZ theorem explicitly; the multiplication is of hypothetical local values, not of mutually incompatible local observables.

For general $N$, let $O_j=X_j{⨂}_{k\ne j}{Y}_k$, with the tensor factors ordered by their site labels, and let $X_N=X^{\otimes N}$. Directly,For the difference of the two strings to be an eigenstate, these phases must agree, which happens exactly when $N-1$ is even. For odd $N$ the common phase is $(-1{)}^{(N-1)/2}$, and interchanging the strings negates their difference. ConsequentlyFor even $N$, $X_N$ still has eigenvalue $-1$, but every $O_j$ takes the difference state to a phase times the sum state, which is orthogonal to it. Thus it is not an eigenstate of any $O_j$; those operators each have outcomes $\pm 1$ with equal probabilities. Among the stated $N\ge 4$, simultaneous eigenstates therefore occur precisely at $N=5,7,9,\dots$.

For odd $N$, certainty of each $O_j$ again permits remote prediction of both local $X$ and $Y$. Writing ${\lambda }_N=(-1{)}^{(N+1)/2}$, the local hidden-variable theory would require $x_j{\prod }_{k\ne j}{y}_k={\lambda }_N$ for all $j$. Their product ishence ${\prod }_j{x}_j={\lambda }_N$, since $N-1$ is even and $N$ is odd. Quantum theory requires the same product to be $-1$. The contradiction occurs when ${\lambda }_N=+1$, giving the infinite familyFor $N\equiv 1(\mathrm{mod}4)$, these particular certain product constraints are consistent; for example $x_j=-1$, $y_j=+1$ satisfies them. This distinguishes the GHZ contradiction for N congruent to 3 modulo 4 from the larger set of odd $N$ for which simultaneous eigenstates exist.

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

The axisymmetric magnetic mode spans a one-dimensional real representation of axial rotations, so its signed amplitude is a real variable $B$. An equatorial dipole has two real horizontal components. Rotation through an angle $\vartheta$ rotates that pair; encoding the components by one complex amplitude gives $A\mapsto e^{i\vartheta }A$, and rotation through $\pi$ gives $A\mapsto -A$. The velocity mode has the same rotational representation, so write it as $V\mapsto e^{i\vartheta }V$. Magnetic reversal acts asThus the two relevant symmetries are axial rotation and magnetic reversal. Reflection symmetry has not been assumed, so complex coupling coefficients are permitted.

The equations for $B,A$ must be linear in the magnetic variables at fixed velocity. Combining this with rotational equivariance, their general local form iswhere $f$ is real valued and $g,h,j,\ell$ are complex valued. These functions can be understood through their local Taylor expansions; the symmetry argument does not require a convergent infinite series. Reality of $\dot{B}$ explains its conjugate pairing.

For the velocity equation, a monomial $B^p{A}^q(A^{\ast }{)}^r{V}^s(V^{\ast }{)}^t$ must have the same rotational weight as $V$ and be even under magnetic reversal. Its selection rules areA formal sum of all such monomials with complex coefficients is the most general local velocity evolution. At degree one there is $V$; at degree two there is $AB$; at degree three there are $V|V{|}^2,V|A{|}^2,V^{\ast }{A}^2,V{B}^2$. Expanding the magnetic equations similarly gives the equivariant three-mode dynamo normal formThe ${\mu }_i,{\omega }_i,e$ are real, and the remaining coefficients are complex. The minus signs on cubic coefficients are a convention, not an assumption about their signs. Terms such as $B^3$ and $A|A{|}^2$ are forbidden in the magnetic equations by magnetic linearity, even though symmetry alone would allow some of them.

For the normalization of nonzero dynamo coupling coefficients, assume $c_1{c}_3\ne 0$. Make the variable changesThe transformed coefficients are ${\tilde{c}}_1=qv{e}^{i({\chi }_V-{\chi }_A)}{c}_1=1$ and ${\tilde{c}}_3=(q/v)e^{i({\chi }_A-{\chi }_V)}{c}_3$, of modulus one. ThusThe remaining phase is genuine; the two complex coefficients cannot both generally be made positive real. A zero coupling cannot be normalized to one, so the printed normalization presupposes this generic nonzero-coupling case.

Now impose the specified simplification, in the normalized variables, and write $A=r{e}^{i\theta }$, $V=s{e}^{i\varphi }$ with $r,s>0$ and signed real $B$. Put $\psi =\varphi -\theta$. Separating real and imaginary parts givesConstant moduli require the first three right sides to vanish. Since $rs\ne 0$, the first equation also fixes $\cos \psi$, so a continuously evolving phase difference must be constant. Hence the last right side must vanish as well. Both phases then advance at one common frequency, giving a relative equilibrium. These four algebraic conditions and the displayed common-frequency equations govern all the requested nonzero constant-modulus solutions, without needing to solve them.

The pure-velocity branch isThis presumes $d_4\ne 0$; the usual saturated branch has ${\mu }_3,d_4>0$. Near it, set $A=e^{i({\omega }_{3}t+{\varphi }_0)}(u+iv)$, $B=b$, and define $\Delta ={\omega }_2-{\omega }_3$ and $\kappa ={\mu }_1-e{s}_0^2$. Magnetic feedback in the velocity equation is quadratic, so the first magnetic linearization isA branch of the stated relative equilibria has a zero growth rate in this rotating frame. Its determinant therefore gives the magnetic onset on a rotating velocity branch conditionFor ${\mu }_2^2+{\Delta }^2\ne 0$, the zero-mode relation is $u+iv=-c_2{s}_{0}b/({\mu }_2+i\Delta )$, so a nonzero $b$ gives a nonzero $A$ when $c_2\ne 0$. With a transverse eigenvalue crossing and nondegenerate nonlinear saturation this is the generic magnetic bifurcation surface; magnetic reversal pairs its two signs. Degenerate cases require separate treatment. For example, if ${\mu }_2=\Delta =0$ and $c_2\ne 0$, determinant zero is automatic but the stationary $u$ equation forces $b=0$, so that zero mode does not establish the requested branch with both magnetic amplitudes nonzero.

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.