One n Remains to Settle the Tree Conjecture

In the famous network creation game of Fabrikant et al. a set of agents play a game to build a connected graph. The $n$ agents form the vertex set $V$ of the graph and each vertex $v\in V$ buys a set $E_v$ of edges inducing a graph $G=(V,\bigcup\limits_{v\in V} E_v)$. The private objective of each vertex is to minimize the sum of its building cost (the cost of the edges it buys) plus its connection cost (the total distance from itself to every other vertex). Given a cost of $\alpha$ for each individual edge, a long-standing conjecture, called the tree conjecture, states that if $\alpha > n$ then every Nash equilibrium graph in the game is a spanning tree. After a plethora of work, it is known that the conjecture holds for any $\alpha>3n-3$. In this paper we prove the tree conjecture holds for $\alpha>2n$. This reduces by half the open range for $\alpha$ with only $[n, 2n)$ remaining in order to settle the conjecture.