The geometric girth of a distance-regular graph having certain thin irreducible modules for the Terwilliger algebra

Abstract

Let $\Gamma$ be a distance-regular graph of diameter $D$. Suppose that $\Gamma$ does not have any induced subgraph isomorphic to $K_{2,1,1}$. In this case, the length of a shortest reduced circuit in $\Gamma$ is called the geometric girth $\mathrm{gg}\left(\Gamma \right)$ of $\Gamma$. Except for ordinary polygons, all known examples have a property that $\mathrm{gg}\left(\Gamma \right)\le 12$ in general, and $\mathrm{gg}\left(\Gamma \right)\le 8$ if $a_1\ne 0$. Is there an absolute constant bound on the geometric girth of a distance-regular graph with valency at least three? This is one of the main problems in the field of distance-regular graphs. P. Terwilliger defined an algebra $\mathcal{T}=\mathcal{T}\left(x\right)$ with respect to a base vertex $x$, which is called a subconstituent algebra or a Terwilliger algebra. The investigation of irreducible $\mathcal{T}$-modules and their thin property proved to be a very important tool to study structures of distance-regular graphs. B. Collins proved that if every irreducible $\mathcal{T}$-module is thin then $\mathrm{gg}\left(\Gamma \right)$ is at most 8, and if $\mathrm{gg}\left(\Gamma \right)=8$, then $a_1=0$ and $\Gamma$ is a generalized octagon. In this paper, we prove the same result under an assumption that every irreducible $\mathcal{T}$-module of endpoint at most 3 is thin.