Example. A nonuniform law on sixteen cycle configurations [fdip-0029]
Example. A nonuniform law on sixteen cycle configurations [fdip-0029]
Use the 4 vertices \(V=\{0,1,2,3\}\) of the alternating cycle, the alphabet \(S=\{0,1\}\), and retained vertices \(J=\{0,2\}\). There are 16 configurations \(\Omega =S^V\) and 4 retained configurations \(\Lambda =S^J\). The map \(\rho (\sigma )=\sigma |_J\) is surjective: assign zero at vertices \(1,3\) to extend any retained configuration. Each fiber has 4 elements. The pullback identifies retained observables with those depending only on the values at vertices \(0,2\).
Choose the following real weights, independently of the edge energies: \[p(\sigma )=\begin {cases} 1/12,&\sigma _0=\sigma _2,\\ 1/24,&\sigma _0\ne \sigma _2. \end {cases}\] There are 8 configurations of each kind, so \(\sum _\sigma p(\sigma )=8/12+8/24=1\). By the finite-state criterion, this defines a real algebraic state. Summing the 4 weights in each fiber gives \[q(0,0)=q(1,1)=1/3,\qquad q(0,1)=q(1,0)=1/6.\] For \(F(\eta )=\eta _0\eta _2\), only \(\eta =(1,1)\) contributes, and hence \[\omega _q(F)=1/3=4\cdot (1/12)=\omega _p(F\circ \rho ).\] This is an explicit instance of the finite fiber-sum identity.
The law here is a chosen finite probability distribution. It is not obtained from the cycle's edge weights or its precision matrix, and the carrier \(S^V\) is distinct from the real configuration space \(\mathbb R^V\) used by the Gaussian law. The algebra convention remains [stacks2017stacks, Section 22.2], Tag 09JF; the positivity convention is [waldmann2004states, Definition 3.9, p. 11]. If the vertex set is empty, \(S^\varnothing \) has one outcome, even for empty \(S\). An empty alphabet with nonempty vertex set instead gives no outcomes and no real state. Thus zero vertices and an empty configuration set are different cases.