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For $m\ge 3$, put $\eta ={\zeta }_m+{\zeta }_m^{-1}$. Since $\eta$ is real and ${\zeta }_m$ is nonreal but satisfies $T^2-\eta T+1=0$, the cyclotomic field $\mathbb{Q}({\zeta }_m)$ has degree two over $\mathbb{Q}(\eta )$. Complex conjugation fixes $\eta$ and is nontrivial. The Fundamental theorem of Galois theory therefore identifies its fixed field with $\mathbb{Q}(\eta )$. That fixed field is exactly $\mathbb{Q}({\zeta }_m)\cap \mathbb{R}$, proving maximality among real subfields.

Here $⟨+y|=(⟨0|-i⟨1|)/\sqrt{2}$. Keeping that complex conjugation, the joint amplitude isHence $\mathrm{Pr}(A=+,B=+y)=0$. The cancellation is destructive quantum interference between the two source components, not the absence of either local outcome. Each local outcome individually has probability $1/2$, since each reduced density matrix is $I/2$. Omitting the conjugation in the second bra would turn the cancellation into reinforcement and give the wrong answer.

In an orthonormal computational basis, the squared norm is the sum of the squared moduli of the coefficients:Thus the vector is normalized and represents a pure state of a qubit. The imaginary part contributes positively to the squared modulus; squaring the coefficient without complex conjugation would give the wrong norm.

Alice uses conjugate-basis steering of a Bell pair. Write the desired orthonormal basis for Bob as $|\psi ⟩=a|0⟩+b|1⟩$ and $|{\psi }^{\perp }⟩=c|0⟩+d|1⟩$. Entrywise complex conjugation in the computational basis gives Alice's orthonormal basisFor either $|u⟩$ in Bob's chosen orthonormal basis, the corresponding unnormalized conditional quantum state isIts squared norm is $1/2$. Thus Alice measures in the complex-conjugate basis, obtaining the required Bob pure states with probability $1/2$ each. Averaging over Alice's undisclosed outcome leaves Bob's density matrix equal to $I/2$, consistent with the no-communication theorem.

The element $\zeta$ satisfies $T^2-\eta T+1=0$, so $[\mathbb{Q}(\zeta ):\mathbb{Q}(\eta )]\le 2$. Since $\eta =\zeta +\overline{\zeta }$ is real, $\mathbb{Q}(\eta )\subseteq \mathbb{R}$. A primitive root of order $m\ge 3$ is nonreal: the only real roots of unity are $\pm 1$, of orders at most two. Hence $\zeta \notin \mathbb{Q}(\eta )$ andEquivalently, nontrivial complex conjugation fixes precisely the maximal real subfield of a cyclotomic field in this degree-two extension.

Let $S=s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Entrywise complex conjugation interchanges the diagonal entries of $r$ and fixes $s$, soSince $r,s$ generate the group, these identities hold for every element. Thus $S$ is an invertible intertwiner, provingThe character of a tensor product representation is the product of the factor characters. Hence ${\chi }_{R_2\otimes R_2}=(4,1,0)$. Its character inner products with the three rows of solution give multiplicities $(1,1,1)$, soFor explicit Clebsch-Gordan decomposition of group representations bases, let $e_{+},e_{-}$ be the coordinate basis. The symmetric tensor $e_{+}\otimes e_{-}+e_{-}\otimes e_{+}$ spans $1$; the antisymmetric tensor $e_{+}\otimes e_{-}-e_{-}\otimes e_{+}$ spans $\epsilon$; and the ordered pair $e_{-}\otimes e_{-},e_{+}\otimes e_{+}$ carries $R_2$, since $r$ acts by $\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s$ exchanges them. In particular $S^2{R}_2=1\oplus R_2$ and ${\Lambda }^2{R}_2=\epsilon$.

Multiplying by the sign representation preserves the character of $R_2$, because its character is zero on all three reflections. Thus $\epsilon \otimes R_2\cong R_2$, and associativity givesThereforeThe dimensions give $8=1+1+3\cdot 2$. Its two one-dimensional summands can also be seen directly: $e_{+}^{\otimes 3}+e_{-}^{\otimes 3}$ and $e_{+}^{\otimes 3}-e_{-}^{\otimes 3}$ transform as $1$ and $\epsilon$, respectively.

Here the star means entrywise complex conjugation, not Hermitian conjugation. Put $s=g_3$ and $r=g_1{g}_3$. Since ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$,The generators are $g_1=rs$, $g_2=r^{2}s$ and $g_3=s$. Moving every occurrence of $s$ to the right using $sr=r^{-1}s$ reduces every word to $r^k$ or $r^{k}s$ with $k=0,1,2$. The three diagonal matrices are distinct, as are the three off-diagonal matrices, soFor an explicit group isomorphism, send $r$ to $(123)$ and $s$ to $(12)$ in the symmetric group $S_3$, composing permutations from right to left. These permutations satisfy the same relations and generate $S_3$. The normal forms above then give an isomorphism; in particular $g_1\mapsto (13)$, $g_2\mapsto (23)$ and $g_3\mapsto (12)$.

Conjugation by $s$ exchanges $r$ with $r^2$, and conjugation by $r$ cycles the three reflections. The conjugacy classes are thereforeTheir sizes are $1,2,3$, respectively. They correspond to the identity, three-cycles and transpositions in $S_3$.

All gamma matrices in this basis are real, and the Lorentz transformation $S(\theta )$ is real. Therefore complex conjugation commutes with the connected Spinor representation of the Lorentz group. With a suitable phase convention, charge conjugation is simply ${\psi }^c={\psi }^{\ast }$. The Majorana spinor condition ${\psi }^c=\psi$ becomesThe reality constraint is preserved by Lorentz transformations because $(S{\psi }_M{)}^{\ast }=S{\psi }_M$. For fermion fields, real components mean self-conjugate Grassmann variables or self-adjoint field components, rather than ordinary commuting real numbers.

Moreover, $\Gamma ={\sigma }_3$ is real, so charge conjugation commutes with the chiral projectors. The Majorana spinor and Weyl spinor conditions can therefore be imposed simultaneously. The Majorana-Weyl spinors areThe independent real component counts areThese are component counts before imposing an equation of motion. In particular the simultaneous reality and chirality constraints are compatible in this spacetime dimension, unlike the familiar four-dimensional case.

On the complex vector space $\mathrm{CF}(G)$ of class functions, define the linear operatorThis is a complex-linear map and satisfies $T^2=I$. It is well-defined on class functions, because inversion of conjugacy classes takes an entire conjugacy class $K$ to another conjugacy class $K^{-1}$.

Use first the basis consisting of the indicator functions $1_K$ of the conjugacy classes. The linear operator $T$ permutes this basis by $1_K\mapsto 1_{K^{-1}}$. A permutation matrix has trace equal to the number of fixed basis vectors, soThe right-hand side counts the real conjugacy classes.

By character orthogonality and completeness, the irreducible characters are another basis of $\mathrm{CF}(G)$. Part (a) gives $T\chi =\overline{\chi }$. The pointwise complex conjugate of an irreducible character is again an irreducible character: conjugate the representing matrices, and conjugate any hypothetical proper invariant subspace back to contradict irreducibility. Thus $T$ also permutes this basis, andHere $\chi =\overline{\chi }$ means exactly that $\chi$ is a real-valued character. Equality of the trace in these two bases provesThis is the real character and real conjugacy class counting theorem. The linear operator used in the proof is $f\mapsto f\circ \mathrm{inv}$; on the irreducible character basis it agrees with pointwise complex conjugation, although pointwise complex conjugation itself is not a complex-linear map on all class functions.

The quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus

The dual space is the vector space of all linear functionals on $V$:Addition and scalar multiplication are defined pointwise. The dual map sends an output measurement back along $T$:The function composition of these linear maps is linear, and $T^{\prime }$ is linear in $\phi$. This is the algebraic dual map, so its definition involves no complex conjugation or inner product.

For the normal operator $T$, the adjoint operator identity and $T^{\ast }T=T{T}^{\ast }$ giveFor the last equality, $⟨T{T}^{\ast }v,v⟩=⟨T^{\ast }v,T^{\ast }v⟩$ and complex conjugation allow reversal of the two arguments. Taking the nonnegative square roots provesIn particular the two kernels coincide.

Use the convention that the Hermitian inner product is conjugate-linear in its first argument and linear in its second. In orthonormal coordinates, $⟨v,w⟩={\sum }_i\overline{v_i}{w}_i$. It satisfies conjugate symmetry, sesquilinearity and positive definiteness. Expanding in the orthonormal basis givesApplying the Hermitian inner product and $⟨e_i,e_j⟩={\delta }_{ij}$ yieldsThis identifies exactly where complex conjugation enters; replacing the first factor by $⟨e_i,v⟩$ would generally be wrong under this convention.

Parametrize the positively oriented circle by $z=r{e}^{i\theta }$. The Cauchy derivative formula at the origin becomesTaking the complex conjugate of the Cauchy integral theorem and using $d\bar{z}=-ir{e}^{-i\theta }d\theta$ givesAdd this zero integral to the derivative integral. Since $f+\bar{f}=2\mathrm{Re}f=2u$, recovering an analytic derivative from the boundary real part yieldsThe complex conjugation on the integrand and the reversed sign in $d\bar{z}$ are both required; the circle's orientation itself has not been reversed.

The degree-five row of the wavevector selection rule for equivariant monomials in part (a) has two vectors other than $(1,0,0,0)$. They correspond exactly toThe vector $(1,0,0,0)$ gives only regular fifth-order terms $A_1|A_j{|}^2|A_k{|}^2$. Omitting those as required, and applying the same two square-symmetry generators, the truncated normal form of a dynamical system isHere $e,f$ are real because rotation by $\pi$ acts by complex conjugation. These are equations for steady patterns, so all four complex right-hand sides must vanish; constancy of the two invariant angles alone would only establish a relative equilibrium and could allow translation drift.

Take a nonzero equal-magnitude state, $|A_j|=R>0$, and set $z=R^2$, $S=c_1+c_2+c_3+c_4$, $\rho =\sigma +Sz$. Dividing each equation by its nonzero amplitude gives, in order,Taking imaginary parts, adding and subtracting pairs, yieldsThus, unless $e=f=0$, each invariant phase is zero or $\pi$ modulo $2\pi$. Equality of the real parts further givesThis identifies the equal-amplitude states of an eight-mode square pattern completely. For generic $e\ne f$, the two nonzero phase types, with their amplitude equations, areRepresentatives are respectively $(A_1,A_2,A_3,A_4)=R(1,1,1,1)$ and $R(-1,1,-1,1)$. Spatial translations change individual phases but leave the indicated phase pair fixed. Reflection acts on the invariant angles as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_1,{\chi }_2)$, and interchange of $x,y$ as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_2,-{\chi }_1)$. Thus the two rows are not equivalent under square symmetry and translations. Every positive root of the corresponding polynomial gives an amplitude branch of that type; if there is no positive root, that type is absent at those parameter values.

There are two coefficient degeneracies to include. When $e=f\ne 0$, the mixed pairs $(0,\pi )$ and $(\pi ,0)$ are allowed and are equivalent under square symmetry. They form one additional phase type, with amplitude equation $\boxed{\sigma +Sz=0}$; $iR(1,1,1,1)$ is a representative. When $e=f=0$, all invariant phases are unrestricted, and the same cubic amplitude equation gives a continuous family modulo translations and the finite square-group action. Even if $e+f=0$ but $e\ne f$, only the original two phase types remain: the equality of their amplitude equations does not remove the phase constraints. Finally, the all-zero state is always a separate steady state and has no defined polar phases.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

The hypotheses make $K$ a CM-field, with nontrivial automorphism $c$ over the totally real number field $k$. In every complex embedding of $K$, this automorphism acts as complex conjugation. Write $E_K={\mathcal{O}}_K^{\times }$, $E_k={\mathcal{O}}_k^{\times }$ and ${\mu }_K$ for the finite cyclic group of roots of unity in $K$.

For any $u\in E_K$, the element $u/c(u)$ is an algebraic integer, indeed a unit, and every conjugate has absolute value one. The Kronecker theorem on algebraic integers in the unit disk makes it a root of unity. For clarity, the essential proof is finite: the conjugates of all positive powers have modulus one, so the integer coefficients of their monic minimal polynomials are bounded by fixed binomial coefficients. There are only finitely many such polynomials and roots; two powers coincide, forcing a root of unity.

Thusis a homomorphism. Its kernel is exactly $E_k$, since a conjugation-fixed element lies in $k$ and both it and its inverse are integral there. For $\zeta \in {\mu }_K$, $c(\zeta )={\zeta }^{-1}$ and $\varphi (\zeta )={\zeta }^2$. Consequently $\varphi$ induces an injectionTo check injectivity explicitly, if $\varphi (u)={\zeta }^2$, then $\varphi (u/\zeta )=1$, so $u/\zeta \in E_k$. The cyclic group ${\mu }_K$ has even order because it contains $-1$, and its quotient by squares has order two. Therefore the unit index of a CM-field satisfies

Complex conjugation fixes $0,1,\infty$, but sends $i$ to $-i$, so it is not the identity. The preceding classification of fixed points of a Möbius transformation says that a nonidentity Möbius transformation has at most two fixed points. Since complex conjugation has at least three, it cannot be a Möbius transformation.

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.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.