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The zero-field ferromagnetic Curie–Weiss model is an Ising model on $n$ spins with an equal interaction between every pair. For $s_i\in \{-1,1\}$ and $\beta >0$, its normalized probability mass function is $Z_{n,\beta }^{-1}\exp [\beta ({\sum }_i{s}_i{)}^2/(2n)]$. The partition function $Z_{n,\beta }$ sums these weights over all configurations. The square includes constant diagonal terms; these affect only normalization. Spin inversion symmetry makes every spin's expected value zero, while grouping configurations by their spin magnetization givesHere $m=-1+2k/n$, $0\le k\le n$.

Use the positive exponent in the PDF. The TeX introduces a minus sign in the numerator while retaining the positive-sign normalizing constant, which would not define the stated probability distribution. The Curie–Weiss model assigns equal weights to $s$ and $-s$, since $({\sum }_j{s}_j{)}^2$ is unchanged by simultaneous reversal. This spin inversion symmetry pairs every configuration with one having the opposite value of $S_i$, so

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.

For the spin inversion symmetry parameterization $W=c\left(\begin{array}{ccc}1& x& y\\ x& z& x\\ y& x& 1\end{array}\right)$ with positive entries, spin decimation gives the same form with $x^{\prime }=x(1+y+z)/(1+x^2+y^2)$, $y^{\prime }=(x^2+2y)/(1+x^2+y^2)$ and $z^{\prime }=(z^2+2x^2)/(1+x^2+y^2)$. Multiply $W$ by itself and divide by its $(+,+)$ entry to prove these ratios. The leftover scalar is an additive free-energy coupling; dropping it leaves normalized expectations unchanged but loses the full free energy.