Forest expansion of two-body partition functions for sparse interaction graphs

TL;DR

It is shown that for sparse graphs, the partition function above a certain temperature Tc can be approximated by a graph polynomial expansion over forests of the interaction graph, and in the high-temperature regime, T can be obtained via a maximal spanning tree algorithm on a (given) weighted graph.

Abstract

We study tree approximations to classical two-body partition functions on sparse and loopy graphs via the Brydges-Kennedy-Abdessalam-Rivasseau forest expansion. We show that for sparse graphs (with large cycles), the partition function above a certain temperature $T^*$ can be approximated by a graph polynomial expansion over forests of the interaction graph. Within this "forest phase", we show that the approximation can be written in terms of a reference tree $\mathcal T$ on the interaction graph, with corrections due to cycles. From this point of view, this implies that high-temperature models are easy to solve on sparse graphs, as one can evaluate the partition function using belief propagation. We also show that there exists a high- and low-temperature regime, in which $\mathcal T$ can be obtained via a maximal spanning tree algorithm on a (given) weighted graph. We study the algebra of these corrections and provide first- and second-order approximation to the tree Ansatz, and give explicit examples for the first-order approximation.

Figures (8)

• Figure 1: Forest expansion for $K_3$, the complete graph on 3-nodes.
• Figure 2: Temperature hierarchy derived "sparse" graphs: the "cycle perturbative" regime is hidden in a "forest" phase.
• Figure 3: This is an example of a cycle and the closure variable involved.
• Figure 4: Example of a cut in the case of a multi-cycle forest element.
• Figure 5: Two examples for cycle dense and cycle sparse graphs. In (a) we see that each cycle is only finitely connected to many other cycles via connected links. In (b) we have that each cycle is connected to all the other $K$ cycles via a single edge, independently from the length of the cycle. In this case, if these cycles are of length $L_1\cdots L_K$, then the total number of edges of the graph is $E=\sum_{i=1}^K L_i-K+1$, and if the cycles are all equal, $E=K(L-1)+1$. Since these are all cycles, we have also $N=K(L-2)+2$. Thus $E-N=K-1$ and $K=\frac{N-2}{L-2}$, which corresponds to $\eta=1$ in eqn. (\ref{['eq:eta']}).