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Take three Ising spins $s_i\in \{-1,1\}$, with an antiferromagnetic interaction of energy scale $J$ on each edge of a triangle. An individually satisfied bond has $s_i{s}_j=-1$. But multiplying all three bond products gives $(s_1{s}_2)(s_2{s}_3)(s_3{s}_1)=1$, whereas three satisfied bonds would give $-1$. At least one bond must therefore remain unsatisfied. This is an exactly finite example of geometrical frustration.

There are six ground states with one unsatisfied bond and ground-state energy $-J$. The two all-aligned microstates have energy $3J$. No interaction is switched off in a ground state: the system minimizes their sum by accepting one local disappointment. Allowing continuous three-component spins instead of Ising spins changes this optimization problem and permits the $120$-degree arrangement described under geometrical frustration.

Writing $\beta =1/(k_{B}T)$ for inverse temperature, the exact partition function and mean internal energy areAt $T>0$ these expressions are smooth: a finite triangle has no thermodynamic phase transition. In zero magnetic field, spin inversion symmetry also gives $⟨s_i⟩=0$ at every site, even as the Boltzmann distribution concentrates on the six frustrated ground states.

An odd periodic nearest-neighbour Ising model with antiferromagnetic coupling has unavoidable geometrical frustration. For $k$ parallel bonds, $H=-NJ+2Jk$, while ${\prod }_j{S}_j{S}_{j+1}=1$ forces $N-k$ to be even. Thus odd $N$ requires at least one parallel bond. A ground state has exactly one, with $N$ possible locations and two global spin inversion symmetry choices, proving the displayed ground-state degeneracy and ground-state energy. The zero-temperature canonical ensemble distributes that defect uniformly. A directed arc of $n$ bonds contains it with probability $n/N$, giving two-point correlation function $(-1{)}^n(1-2n/N)$. The finite antiferromagnetic Ising-ring correlation gives the corresponding positive-temperature result.

For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:Thus the Helmholtz free energy per site has the expansionDifferentiate at fixed $J$, remembering that $K=J/(k_{B}T)$. The entropy per site isThe nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy $J[st+(s+t)\tau ]$ has minimum energy $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have energy $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at $-J$ by requiring $t=-s^{\prime }$, where $s^{\prime }$ is the next cell's first base spin.

For fixed $s$ and $s^{\prime }$, the number of allowed apex choices is therefore two when $s=s^{\prime }$ and one when $s=-s^{\prime }$. The zero-temperature counting transfer matrix on successive first base spins isThe eigenvalues of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the thermodynamic limit gives $k_B\ln 3/3$, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and $s\to k_B\ln 2$ per site.

Let $k$ count parallel bonds. Then $N-k$ bonds are antiparallel and the energy isThe periodic boundary conditions imposeHence $N-k$ must be even. For odd $N$, $k$ is odd, so its minimum is one: one bond must be parallel. All other bonds alternate. Choosing that bond in $N$ ways and then choosing either value of $S_1$ determines every Ising spin, givingThis is an odd antiferromagnetic Ising ring: geometrical frustration forces one defect, whose location is unrestricted. The ground-state degeneracy includes both its possible locations and global spin inversion symmetry. The formula also includes the self-bond case $N=1$ under the stated Hamiltonian convention.

For $J>0$ each bond minimizes its energy when its two Ising spins are opposite, giving bond energy $-J$. An even ring allows every bond to do this simultaneously, including the closing bond required by the periodic boundary conditions. Its two ground states areThe two configurations are related by spin inversion symmetry. There is no geometrical frustration.

For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:Thus the Helmholtz free energy per site has the expansionDifferentiate at fixed $J$, remembering that $K=J/(k_{B}T)$. The entropy per site isThe nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy $J[st+(s+t)\tau ]$ has minimum energy $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have energy $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at $-J$ by requiring $t=-s^{\prime }$, where $s^{\prime }$ is the next cell's first base spin.

For fixed $s$ and $s^{\prime }$, the number of allowed apex choices is therefore two when $s=s^{\prime }$ and one when $s=-s^{\prime }$. The zero-temperature counting transfer matrix on successive first base spins isThe eigenvalues of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the thermodynamic limit gives $k_B\ln 3/3$, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and $s\to k_B\ln 2$ per site.

Choose two Ising spins with Hamiltonian $H=-J{s}_1{s}_2$, where $J$ independently takes $J_0$ or $2J_0$ with equal prior probabilities and $J_0>0$. This original finite example has no geometrical frustration and no phase transition. Its four spin assignments give $Z_J=4\cosh (\beta J)$. ThereforeThe concave function $\log$ puts $F_{\mathrm{a}}\le F_{\mathrm{q}}$. For this two-value case the claim is just $(Z_1+Z_2)/2\ge \sqrt{Z_1{Z}_2}$, obtained by squaring $\sqrt{Z_1}-\sqrt{Z_2}$. Thus no formal continuation of replicas is needed to see the distinction.

In the annealed disorder ensemble, the stronger bond has probability $Z_{2J_0}/(Z_{J_0}+Z_{2J_0})$, approaching one at low temperature. A quenched disorder collection instead retains one half at each coupling. As $T\to 0$, $F_{\mathrm{q}}\to -3J_0/2$ and $F_{\mathrm{a}}\to -2J_0$: averaging samples and allowing the material variables to equilibrate describe different physics.