Math-1716 · mathematics · Probability

Math-1716

Self-avoiding loops on nonamenable transitive graphs

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Statement

Definitions and mathematical statement

Let $G$ be an infinite connected locally finite vertex-transitive simple graph. For finite nonempty $A\subset V(G)$, let $\partial A$ be the vertices outside $A$ adjacent to it, and suppose $h(G)=\inf_A|\partial A|/|A|>0$.

Fix a root $o$. Let $c_n$ count sequences $(v_0,\ldots,v_n)$ with $v_0=o$, adjacent consecutive vertices, and all vertices distinct. These are rooted self-avoiding walks of length $n$. Their connective constant is \[ \mu(G)=\lim_{n\to\infty}c_n^{1/n}, \] whose existence follows from submultiplicativity and transitivity. Let $\ell_n$ count rooted oriented simple cycles $(v_0,\ldots,v_n)$ with $v_0=v_n=o$, $n\ge3$, and $v_0,\ldots,v_{n-1}$ distinct. Define the loop growth rate by \[ \mu_{\mathrm{loops}}(G)=\limsup_{n\to\infty}\ell_n^{1/n}, \] with $0^{1/n}=0$. Prove that $\mu_{\mathrm{loops}}(G)<\mu(G)$.

The limsup is necessary to make the definition valid without restrictions on cycle lengths: bipartite graphs have no odd cycles, and trees have no cycles at all. Reversing the orientation or counting each cycle without a distinguished orientation changes counts by at most a constant and therefore leaves the exponential rate unchanged.

The assertion is a strict comparison of exponential growth, not simply the observation that a walk has more possible endpoints than a loop. The desired gap may depend on $G$; no uniform positive gap over all nonamenable transitive graphs is specified. Both rates use the same unweighted graph and the same edge-length convention.

Short English statement

On every nonamenable transitive graph, do self-avoiding loops grow exponentially more slowly than self-avoiding walks?

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