On the structure of self-complementary graphs

TL;DR

This paper provides a new structural proof for Gibbs' theorem on self-complementary graphs, introduces conjectures about their partitions, and verifies the conjecture for specific antimorphism cases.

Contribution

It offers a novel proof of Gibbs' theorem that reveals more structural details and proposes a conjecture about the partitions in self-complementary graphs.

Findings

Every self-complementary graph on 4k vertices has k disjoint paths of 4 vertices.

The new proof provides additional structural insights.

The conjecture holds for graphs with antimorphisms as products of two circular permutations, one of length 4.

Abstract

A \emph{self-complementary} graph is a graph isomorphic to its complement. An isomorphism between $G$ and its complement, viewed as a permutation of $V(G)$, is then called an \emph{antimorphism}. A \emph{skew partition} of $G$ is a partition of $V(G)$ into 4 sets $A,B,C,D$ such that there is no edge between $A,B$ and every possible edge between $C,D$. A \emph{symmetric partition} of $G$ is a partition of $V(G)$ into 4 sets $A,B,C,D$ such that there is no edge between $A, D$, no edge between $B, C$, every possible edge between $A,B$ and every possible edge between $C,D$. We give a new proof of a theorem of Gibbs saying that every self-complementary graph on $4k$ vertices has $k$ disjoint paths on 4 vertices as induced subgraph. This new proof gives more structural information than the original one. We conjecture that every self-complementary graph on $4k$ vertices either has an induced cycle on 5 vertices, or a skew partition, or a symmetric partition. The new proof of Gibb's theorem yields a proof of the conjecture for the self-complementary graphs that have an antimorphism that is the product of a two circular permutations, one of them of length 4.