A result on spanning trees with bounded total excess

TL;DR

This work addresses when a connected graph $G$ contains a spanning tree with bounded total excess $\mathrm{te}(T,k)$. It advances a spectral-radius approach by proving that if $\rho(G)$ is at least the spectral radius of a particular join graph $K_1\vee(K_{n-k-b-1}\cup(k+b)K_1)$, then $G$ possesses a spanning tree with $\mathrm{te}(T,k)\le b$, for suitable $k$ and $b$ (with $k\ge\max\{5,b+3\}$ and $(b,k)\neq(2,5)$). The result generalizes known spanning $k$-tree conditions (corresponding to $b=0$) and strengthens order bounds relative to prior theorems by using equitable-partition spectral techniques and subgraph monotonicity. The work broadens the application of spectral radius methods in extremal graph theory, offering a concrete criterion to guarantee bounded total excess in spanning trees.

Abstract

Let $G$ be a connected graph and $T$ a spanning tree of $G$. Let $ρ(G)$ denote the adjacency spectral radius of $G$. The $k$-excess of a vertex $v$ in $T$ is defined as $\max\{0,d_T(v)-k\}$. The total $k$-excess $\mbox{te}(T,k)$ is defined by $\mbox{te}(T,k)=\sum\limits_{v\in V(T)}{\max\{0,d_T(v)-k\}}$. A tree $T$ is said to be a $k$-tree if $d_T(v)\leq k$ for any $v\in V(T)$, that is to say, the maximum degree of a $k$-tree is at most $k$. In fact, $T$ is a spanning $k$-tree if and only if $\mbox{te}(T,k)=0$. This paper studies a generalization of spanning $k$-trees using a concept called total $k$-excess and proposes a lower bound for $ρ(G)$ in a connected graph $G$ to ensure that $G$ contains a spanning tree $T$ with $\mbox{te}(T,k)\leq b$, where $k$ and $b$ are two nonnegative integers with $k\geq\max\{5,b+3\}$ and $(b,k)\neq(2,5)$.