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Seymour's Second Neighborhood Conjecture for Subsets of Vertices

Published 20 Aug 2018 in math.CO | (1808.06293v3)

Abstract: Seymour conjectured that every oriented simple graph contains a vertex whose second neighborhood is at least as large as its first. In this note, we put forward a conjecture that we prove is actually equivalent: every oriented simple graph contains a subset of vertices $S$ whose second neighborhood is at least as large as its first. This subset perspective gives some insight into the original conjecture. For example, if there is a counterexample to the second neighborhood conjecture with minimum degree $\delta$, then there exists a counterexample on at most ${\delta + 1 \choose 2}$ vertices. Given a vertex $v$, let $d_1+(v)$ and $d_2+(v)$ be the size of its first and second neighborhoods respectively. A digraph is $m$-free if there is no directed cycle on $m$ or fewer vertices. Let $\lambda_m$ be the largest value such that every $m$-free graph contains a vertex $v$ with $d_2+(v) \geq \lambda_m d_1+(v)$. The second neighborhood conjecture implies $\lambda_m = 1$ for all $m \geq 2$. Liang and Xu provided lower bounds for all $\lambda_m$, and showed that $\lambda_m \to 1$ as $m \to \infty$. We improve on Liang and Xu's bound for $m \geq 3$ using this subset perspective.

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