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Use dimensionless interaction couplings in the exponent of the Boltzmann weight. In the following formulas $K,L$ denote the physical interaction energies divided by $k_{B}T$ if that factor has not already been absorbed into the displayed Hamiltonian. This convention is required for temperature-independent numerical fixed-point coordinates.

Split the original square lattice into retained sites $E$ with even $x+y$ and eliminated sites $O$ with odd $x+y$, where coordinates are in units of $a$. Every nearest-neighbour bond joins $E$ to $O$, whereas a diagonal next-nearest-neighbour bond stays within one sublattice. Denote retained Ising spins by $s_e$ and eliminated Ising spins by ${\eta }_o$. For fixed retained Ising spins defineThe blocked Boltzmann weight in the checkerboard decimation of the square-lattice Ising model isUse the independent uniform product measure on Ising spins on $O$ to perform a cumulant expansion. Its one-spin mean vanishes, and $⟨{\eta }_o{\eta }_{o^{\prime }}{⟩}_0={\delta }_{o{o}^{\prime }}$. Consequently the first-order $K$ term, the first-order eliminated-sublattice $L$ term, and the mixed $KL$ term vanish. To the requested order,Here the truncation keeps $K^2$ and the linear term in $L$; equivalently count $L$ as order $K^2$. At higher orders the interaction family will generally not close.

Each eliminated site has four retained neighbours, soIts six distinct neighbour pairs comprise four corner pairs at distance $\sqrt{2}a$ and two opposite pairs at distance $2a$. A given retained diagonal pair shares two eliminated centres; a given opposite pair shares only one. Thus the generated pair couplings are $2K^2$ for diagonal pairs and $K^2$ for axial pairs separated by two original spacings. The original $L$ interaction directly contributes to each retained diagonal pair. No single-spin term survives spin inversion symmetry, and no three-spin or four-spin term is present in this second cumulant. This proves closure on exactly two nonconstant interactions at the retained order.

The primitive vectors of the retained square lattice are $a(1,1)$ and $a(1,-1)$. Rotate the axes by $45$ degrees and rescale lengths by $1/\sqrt{2}$; retained diagonal pairs become nearest neighbours, and the axial distance-$2a$ pairs become next-nearest neighbours. With $b=\sqrt{2}$, the leading checkerboard Ising decimation recursion is thereforeThe constant part of the weight is also determined: $4(K^2/2)(N/2)=N{K}^2$, so the original partition function equalsto the same order in the logarithm of the weights. That normalization changes free energy but not the retained-spin interaction operators. If physical energy couplings are kept instead, at fixed ${\beta }_T$ the relations read $K_{\mathrm{energy}}^{\prime }=L_{\mathrm{energy}}+2{\beta }_T{K}_{\mathrm{energy}}^2$ and $L_{\mathrm{energy}}^{\prime }={\beta }_T{K}_{\mathrm{energy}}^2$.

The fixed-point equations are $L_{\ast }=K_{\ast }^2$ and $K_{\ast }=3K_{\ast }^2$. Their only finite real solutions areThe origin is the high-temperature fixed point; its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, with both multipliers zero. The nontrivial candidate critical point is $(1/3,1/9)$. Linearize the discrete renormalization-group transformation there:The characteristic polynomial is ${\lambda }^2-(4/3)\lambda -2/3$, givingEigenvectors may be chosen as $(1,2/(3{\lambda }_{\pm }){)}^T$. The positive eigenvalue greater than one is the relevant direction of a fixed point; the other is irrelevant because its magnitude is less than one. Its negative sign only alternates the sign of the perturbation between steps. For a discrete map the magnitude relative to one decides relevance, not the sign rule used for continuous beta-function eigenvalues.

Let $t$ be the relevant thermal scaling coordinate, so $t^{\prime }={\lambda }_{+}t$ to first order. A correlation length measured in restored lattice units satisfies $\xi (t^{\prime })=\xi (t)/b$. If $\xi (t)\sim |t{|}^{-\nu }$, then ${\lambda }_{+}^{-\nu }=b^{-1}$. Hence the thermal exponent from a discrete renormalization map isThis is the value predicted by the stated low-order real-space renormalization group, whose nontrivial fixed point is evaluated beyond the strictly infinitesimal-coupling limit. It is an approximation, not an exact solution of the full square-lattice Ising model.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

For brevity write $b_1=b_{3/2}^{(1)}(\alpha )$, $b_2=b_{3/2}^{(2)}(\alpha )$, and $K=G{M}_p\alpha /(4a_p)$. The quadratic disturbing function is $\mathcal{R}=K[\frac{1}{2}{b}_1{e}^2-b_2{e}_{p}e\cos ({\varpi }_p-\varpi )]$. Differentiation in Lagrange planetary equations givesThe separate longitude of pericentre equation is singular at $e=0$, because a circular orbit has no distinguished periapsis. The complex eccentricity, $z=e{e}^{i\varpi }$, is regular there. Using $z_p=e_p{e}^{i{\varpi }_p}$ and $\dot{z}=e^{i\varpi }(\dot{e}+ie\dot{\varpi })$ givesSince $n{a}^2=\sqrt{G{M}_{\star }a}$ and $n_p=\sqrt{G{M}_{\star }/a_p^3}$,The approximations are planar, small-orbital eccentricity, first order in the planet-to-star mass ratio, and secular averaging away from important mean-motion resonances. The semimajor axis values remain constant at this order.

For a planet with fixed orbit, set $g=A{b}_1$ and $z_f=(b_2/b_1)z_p$. Then $\dot{z}=ig(z-z_f)$, so $z-z_f=(z(0)-z_f)e^{igt}$. With $z(0)=0$,On the Argand diagram, the complex eccentricity traces a circle centred on the forced eccentricity $z_f$, of radius $|z_f|$. For an interior test particle, $0<\alpha <1$ and $g>0$, so the motion is counterclockwise, starting at the origin. The rotating radius $z-z_f$ is the free eccentricity. The orbital eccentricity ranges from zero to $2|z_f|$, attaining its maximum at half a secular period. If $z_p$ is placed on the positive real axis, the initial tangent points downward, which is consistent with counterclockwise motion from the leftmost point of this circle.

Now let the planetary mass grow as $M_p(t)=M_{p}t/t_p$ for $0\le t\le t_p$, while the planet's orbital elements and the test particle's semimajor axis remain fixed. Both the precession and forcing coefficients acquire the same factor $t/t_p$. Their ratio, and hence $z_f$, remain constant. ThusandThis is the secular phase integral for proportional mass growth. The Argand diagram circle and the free eccentricity amplitude are unchanged; only the rate of traversal changes. The angular speed about $z_f$ starts at zero and rises linearly to $g$. The early displacement is quadratic in time rather than linear. Since $t_p=2\pi /g$, $\Phi (t_p)=\pi$, soA planet already at its final mass would instead return the particle to $z=0$ at $t_p$. After growth stops, the circle continues at rate $g$, with $z(t)=z_f[1+e^{ig(t-t_p)}]$ for $t\ge t_p$. Slow mass growth does not damp the free eccentricity here: the fixed centre of the circle and the conservative equation preserve its amplitude.

For several planets with fixed masses, Laplace-Lagrange secular theory gives $\dot{\mathbf{z}}=i\mathbf{Az}$ for the planets andfor a massless particle, with real secular coupling coefficients ${\nu }_j$. Resolve planetary complex eccentricities into secular eigenmodes, $z_j(t)={\sum }_k{E}_{jk}{e}^{i{g}_{k}t}$. Away from secular resonances, the test-particle solution has the formIts Argand diagram trajectory is a superposition of rotations: generally a quasiperiodic curve with beats, rather than a circle with one fixed centre. Even if the planetary orbits are artificially held fixed, their combined forcing shifts the centre to the sum of the individual forced eccentricities. At $g_k=g_0$, the resonant contribution is proportional to $t{e}^{i{g}_{0}t}$ in linear theory; its growing amplitude eventually invalidates the small-orbital eccentricity approximation.

If one planetary mass changes, the Laplace-Lagrange secular matrix, its eigenvalues, its eigenvectors, and the particle's forced response generally change with time. Planetary modes need not remain independent. Frequencies drift and a secular resonance can sweep across the particle, changing or exciting its free eccentricity. Very slow evolution away from resonances permits an approximate adiabatic mode description; rapid evolution or a resonance passage can mix modes. The simple substitution of a phase integral works only in the special proportional-growth case above, where the entire particle equation is multiplied by one scalar and its forced centre stays fixed. It does not apply to a general many-planet system.

During the positive half-period, write $Y=(A,B{)}^T$ and $\dot{Y}=K_{+}Y$, with $K_{+}=\left(\begin{array}{cc}0& D^2\\ i& 0\end{array}\right)$. Then $K_{+}^2=i{D}^{2}I={\sigma }^{2}I$. Its two eigenvalues are $\pm \sigma$, and corresponding eigenvectors are $(\sigma ,i{)}^T$ and $(-\sigma ,i{)}^T$. Since $D>0$, they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, $e^{T{K}_{+}}=I\cosh (\sigma T)+(K_{+}/\sigma )\sinh (\sigma T)$. Because $D^2/\sigma =-i\sigma$, the fundamental matrix of a linear differential equation givesThe lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_{-}=\left(\begin{array}{cc}0& -D^2\\ i& 0\end{array}\right)$ and $K_{-}^2=({\sigma }^{\ast }{)}^{2}I$. The analogous matrix exponential has the stated $M^{-}$, replacing $\sigma$ by its complex conjugate ${\sigma }^{\ast }=-i\sigma$. Both matrices satisfywith ${\sigma }^{\ast }$ in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.

At each complete period the state is multiplied by the monodromy matrix of a periodic linear system $N=M^{-}{M}^{+}$, so $Y(2nT)=N^{n}Y(0)$. The determinant is therefore $\det N=1$. Let $c_{\pm }=\cosh ({\sigma }^{(\ast )}T)$ and $s_{\pm }=\sinh ({\sigma }^{(\ast )}T)$. Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt{2}DT$, the hyperbolic-function identity for $|\cosh (a+ia){|}^2$ consequently yieldsFor $x>0$, $F>2$: differentiating gives $F^{\prime }=\sinh x-\sin x>0$, since this derivative has derivative $\cosh x-\cos x>0$ and vanishes initially. The characteristic polynomial is ${\lambda }^2-F\lambda +1$. Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate $-\gamma$; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.

As $x\to \infty$, $F=\frac{1}{2}{e}^x[1+O(e^{-x})]$, $\Lambda =F[1+O(F^{-2})]$, and $\log \Lambda =x-\log 2+o(1)$. ThereforeIn particular, at $T=1$ the rate is asymptotic to $(\sqrt{2}D-\log 2)/2$, as required. The growth calculation uses the product $M^{-}{M}^{+}$, rather than averaging the two generators: the two generators do not commute.

Use an orthonormal eigenbasis of the Hermitian operator $A$ and set $b_j=⟨{\alpha }_j|B|{\alpha }_j⟩$. Since $B$ is a positive contraction,For $0<r<d$, the missing weight among the first $r$ entries equals the weight in the remaining entries:The ordered eigenvalues satisfy ${\lambda }_j\ge {\lambda }_{r-1}$ for $j<r$ and ${\lambda }_j\le {\lambda }_{r-1}$ for $j\ge r$. ThereforeThis argument does not require the eigenvalues of $A$ to be positive. For $r=0$, positivity and zero trace give $B=0$; for $r=d$, the same argument applied to $I-B$ gives $B=I$. Thus in every caseThe orthogonal projection onto the leading $r$ eigenvectors attains equality. This is the Hermitian effect variational principle, extending the Ky Fan maximum principle to all positive contractions with the prescribed trace.

Expand the square root and quotient in the logarithmic-height plume equations:The linear stability analysis givesIn particular, the second row contains $-2m{M}^{\prime }$, since the reciprocal of $\hat{M}$ contributes another $-M^{\prime }$.

For the growing mode of power-law plume similarity, the characteristic polynomial isWithin the physical interval $q,m>0$ and $f<0$, $P(0)<0$ while $P(\lambda )\to +\infty$ as $\lambda \to +\infty$. There is a positive real eigenvalue, so the similarity equilibrium is unstable with increasing height. Perturbations behave as $e^{\lambda \xi }\propto z^{\lambda }$; the result is not stability in physical time. Thus this exact positive-buoyancy similarity trajectory is not a generic attracting far field.

At the upper endpoint,The limiting polynomial is $\lambda (\lambda +1)(\lambda +10/3)$, reproducing the three stated eigenvalues. The positive eigenvalue tends to zero, with $\lambda \sim -2f/3$. The two negative modes decay as $z^{-1}$ and $z^{-10/3}$; the remaining mode is neutral at the limiting linear system. Its eigenvector is $(1,2,3)$ in relative-flux coordinates.

The neutral buoyancy-flux mode of a pure plume has a simple interpretation: a constant surviving buoyancy flux fixes the amplitudes of a pure plume, with $Q\propto F^{1/3}{z}^{5/3}$ and $M\propto F^{2/3}{z}^{4/3}$. A small fractional change in $F$ produces changes in $Q,M$ in the ratio $1:2:3$. It is retained rather than damped out.

There is a qualification at the exact endpoint. For fixed nonzero $K$ and $\beta =-8/3$, taking $Q\propto z^{5/3}$ gives $F_z\propto -z^{-1}$, not zero. The flux coefficients above also diverge as $f\to 0$. Thus the constant-flux powers are the formal limiting scaling, not an exact constant-buoyancy solution in a nonzero endpoint stratification; cumulative ambient effects can require logarithmic corrections.

For the linear evolution from part (iii), maximize the energy ratio at each time. The optimal energy amplification of a linear system isThus the optimizing initial state is a right singular vector, not generally an eigenvector of $L$.

There is a convenient exact way to find the maximizing time. Let $x_0$ be a unit optimizing initial vector and $y=M(t)x_0$. The singular value decomposition gives $M^{T}y=G_{\mathrm{opt}}{x}_0$. For $t>0$, the top singular value is simple. At an interior time maximum,Since $LM=ML$, this also equals $G_{\mathrm{opt}}{x}_0^{T}L{x}_0$. Thus both initial and final orientations are neutral energy-growth directions. The positive entries of $M^{T}M$ let the top right singular vector be chosen with positive components. Its orientation decreases through the growth sector, so it must start at ${\rho }^{+}$ and end at ${\rho }^{-}$.

Put $q=a/d=e^{-3t/(4R)}$ and $K=4R/3$. The solution givesSetting this equal to ${\rho }^{-}$ yieldsTherefore the exact optimal time for $R>1$ isHere and below $\log$ is the natural logarithm. This is the unique positive stationary time: $q$ decreases monotonically from one to zero, and the neutral-direction condition determines one $q_{\ast }\in (0,1)$. Also $G_{\mathrm{opt}}(0)=1$, $G_{\mathrm{opt}}(\infty )=0$, and $R>1$ admits initial energy growth, so this stationary point is the global maximum. For $R\le 1$, the instantaneous energy-growth criterion for a linear system gives no amplification, and the maximum is one at time zero.

The first partial derivatives areBoth vanish at $(\lambda ,\lambda )$, so every positive diagonal point is a stationary point. The Hessian matrix there isIts eigenvectors $(1,1)$ and $(1,-1)$ have respective eigenvaluesThe zero eigenvalue reflects the entire line of nonisolated stationary points: $f(\lambda ,\lambda )=2$. In fact $f-2=(x-y{)}^2[1/(xy)-1/a^2]$, showing a non-strict local minimum when $\lambda <|a|$, a non-strict local maximum when $\lambda >|a|$, and values on both sides of two near $\lambda =|a|$. At that last point the whole Hessian matrix vanishes, so its signs alone cannot classify the point.

First exclude the short patterns $(a,a+1,a)$ and $(a,a-1,a)$ in the joint spectrum. In the first pattern, the local spectral rules for Young–Jucys–Murphy elements make $s_r$ act by $+1$ and $s_{r+1}$ by $-1$ on the same simultaneous eigenvector. The two sides of the braid relation in a Coxeter group then act by opposite signs. The second pattern gives the same contradiction with signs reversed.

Now induct on the length of a spectral vector. A prefix of length $i-1$ is spectral for $S_{i-1}$: decompose the restricted group representation and retain a nonzero component of the simultaneous eigenvector. If $a_i$ differed by neither $+1$ nor $-1$ from every earlier entry, move it left using the allowed adjacent interchanges. Encountering an equal entry would contradict the distinctness of consecutive eigenvalues. Otherwise it reaches the first position, forcing $a_i=0$. For $i>1$ the original first entry was also zero, so an equal entry would indeed have been encountered. ThusThis argument uses only the stated local rules. In particular, iterating the result from $a_1=0$ also shows that all coordinates are integers.

The first Young–Jucys–Murphy element is the empty sum $X_1=0$. Applying it to the nonzero simultaneous eigenvector defining a point of the joint spectrum gives $0=a_{1}v$. HenceFor the remaining spectral arguments use the following local facts, without assuming the restriction branching rule for a symmetric group: consecutive eigenvalues are distinct; a difference $a_{r+1}-a_r=\pm 1$ makes $s_r=(rr+1)$ act on the corresponding line as $\pm 1$; every other adjacent interchange gives another point of the joint spectrum in the same irreducible representation. These are the local spectral rules for Young–Jucys–Murphy elements. They follow from $s_r{X}_r{s}_r+s_r=X_{r+1}$ and the one- or two-dimensional analysis of the two neighboring eigenspaces. They are the appropriate local results used below.

The most plausible cause is strong dependence between the latent variables and covariance hyperparameters in their joint posterior distribution. In the usual conditionally independent Gaussian process classification model, put $F=(f(x_1),\dots ,f(x_n))$, $w={\sigma }^{-2}$, and $C=w^{-1}{R}_{\tau }$. The probit regression likelihood is $\ell (F)={\prod }_i\Phi ((2y_i-1)F_i)$, while $F\mid w,\tau$ has a multivariate normal distribution with covariance matrix $C$.

For distinct inputs and positive length scales, $R_{\tau }$ is positive definite. Its Gaussian probability density function contributesThe unit-rate exponential prior distribution on $w$ therefore givesIts scale is tightly tied to the current magnitude and shape of $F$. Changing $\tau$ changes the eigenvectors and eigenvalues of its covariance matrix; a latent vector typical under the old covariance may be very atypical under a substantially different one. Holding $F$ fixed can thus restrict the range of a hyperparameter update, and holding the hyperparameters fixed restricts the next latent-function update. Even exact sampling of each full conditional distribution can move slowly along a narrow ridge of the joint posterior distribution.

Binary observations provide limited information about the absolute size of large correctly signed latent values. This can leave substantial scale uncertainty and strengthen the dependence. Long correlation lengths can also make the covariance matrix nearly singular and the latent coordinates highly dependent. These are plausible explanations for a large mixing time, not proofs that every data set causes slow convergence. A blocked Gibbs sampler removes dependence within its chosen blocks, but not dependence between them. Repeated identical inputs should be represented by one shared latent value; otherwise the nominal Gaussian matrix is singular and the displayed inverse formula must be replaced by a representation on its support.

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

The fixed-point equation for $u$ factors asThe cubic derivative is $3u^2+2u+3>0$ for every real $u$; its values at zero and one have opposite signs. Thus the possible first coordinates in the stated domain are exactly $u=0,u_{\ast },1$.

For finite $v$, the second fixed-point equation is $v[v^3-(1+u^2{)}^2]=0$. The renormalization-group fixed points are consequentlyThere are six finite points and three boundary points at infinity. Infinity is interpreted in a compactified domain, not as an ordinary real number. The reciprocal coordinate at infinite coupling $w=1/v$ transforms as $w^{\prime }=(1+u^2{)}^2{w}^4$, making $w=0$ a well-defined boundary fixed point.

At any finite fixed point $(\tilde{u},\tilde{v})$, let $\delta u=u-\tilde{u}$, $\delta v=v-\tilde{v}$. The Jacobian matrix linearization isThe only point with both coordinates strictly interior is $\tilde{u}=u_{\ast }$, $\tilde{v}=(1+u_{\ast }^2{)}^{2/3}$. Using its fixed-point relations, the matrix becomesIts eigenvalues are $\rho$ and $4$. In particular, $u_{\ast }<1/\sqrt{3}$ because the increasing cubic is already positive there, so $\rho >1$ (numerically about $1.68$). The corresponding eigenvectors can be chosen as $(1,c/(\rho -4){)}^T$ and $(0,1{)}^T$. Both discrete multipliers exceed one, henceBoth perturbations are relevant directions of a fixed point under repeated coarse-graining, rather than one stable and one unstable direction. A length-rescaling factor was not specified, so these multipliers should not be assigned numerical critical scaling exponents without additional information. At $a=0,1$, the $u$ multiplier is zero; at $v=0$ or $w=0$ the other multiplier is also zero. Thus the four corner points attract locally within the domain, the two positive finite boundary points at $a=0,1$ are saddles, and the points $(u_{\ast },0),(u_{\ast },\infty )$ have one repulsive and one attractive direction.