Math-1717 · mathematics · Probability

Math-1717

Locality of connective constants

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Statement

Definitions and mathematical statement

Let $(G_n,o_n)$ and $(G,o)$ be infinite connected locally finite vertex-transitive graphs with distinguished roots. Say that $G_n$ converges locally to $G$ if, for every integer radius $R\ge0$, the induced rooted balls $(B_{G_n}(o_n,R),o_n)$ and $(B_G(o,R),o)$ are isomorphic for all sufficiently large $n$. The threshold index may depend on $R$.

For such a graph $H$ with root $v$, let $c_m(H)$ be the number of nearest-neighbor paths $(v_0,\ldots,v_m)$ with $v_0=v$ and all $m+1$ vertices distinct. Its connective constant is \[ \mu(H)=\lim_{m\to\infty}c_m(H)^{1/m} =\inf_{m\ge1}c_m(H)^{1/m}. \] The limit exists because vertex transitivity gives $c_{m+k}(H)\le c_m(H)c_k(H)$; its value does not depend on the root.

Prove that local convergence implies $\mu(G_n)\to\mu(G)$. No common group presentation, compatible generating set, or graph height function is included in the hypothesis. Local convergence already forces the degrees to agree eventually, so a separate eventual degree bound is unnecessary.

For each fixed $m$, local convergence identifies the walks of length $m$ and gives $c_m(G_n)=c_m(G)$ eventually. This supplies the upper bound $\limsup_n\mu(G_n)\le\mu(G)$. The substantive remaining demand is the reverse lower bound for the asymptotic growth rate, which involves arbitrarily long walks and is not decided by one fixed neighborhood.

Short English statement

Does agreement of increasingly large neighborhoods force convergence of self-avoiding-walk connective constants?

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