Many pentagons in triple systems
Dhruv Mubayi, Jozsef Solymosi
TL;DR
This work resolves a supersaturation question for odd cycles in linear 3-graphs by proving that dense linear triple systems contain many pentagons: for $n>10$ and $m>100\,n^{3/2}$, a linear $n$-vertex triple system with $m$ edges has at least $m^6/n^7$ copies of $C_5$. The authors develop a general framework using shadow graphs to extend to longer odd cycles, obtaining a bound of $\Omega\big(m^{3k}/n^{4k-1}\big)$ copies of $C_{2k+1}$ for each $k\ge2$, with $\,m\gg n^{2-1/(3k)}$. They connect these combinatorial results to geometry by showing that sets with many triangles similar to a fixed triangle yield harmonic-point configurations, and they demonstrate near-optimal exponent behavior via explicit constructions (e.g., a pentagon-free, linear 3-graph with many edges) and a Ruzsa-type geometric construction giving many triangles with disjoint harmonic points. Finally, they derive a density-removal-type consequence: if a graph is $\varepsilon$-far from triangle-free, then it contains at least $c\,\varepsilon^{3\ell} n^{2\ell+1}$ copies of $C_{2\ell+1}$ for every fixed $\ell\ge2$, linking extremal and geometric themes for odd cycles in graphs and 3-graphs.
Abstract
We prove that every $n$ vertex linear triple system with $m$ edges has at least $m^6/n^7$ copies of a pentagon, provided $m>100 \, n^{3/2}$. This provides the first nontrivial bound for a question posed by Jiang and Yepremyan. More generally, for each $ \ell \ge 2$, we prove that there is a constant $c$ such that if an $n$-vertex graph is $\varepsilon$-far from being triangle-free, with $\varepsilon \gg n^{-1/3\ell}$, then it has at least $c \, \varepsilon^{3\ell} n^{2\ell+1}$ copies of $C_{2\ell+1}$. This improves the previous best bound of $c \, \varepsilon^{4\ell+2} n^{2\ell+1}$ due to Gishboliner, Shapira and Wigderson. Our result also yields some geometric theorems, including the following. For $n$ large, every $n$-point set in the plane with at least $60\, n^{11/6}$ triangles similar to a given triangle $T$, contains two triangles sharing a special point, called the harmonic point. In the other direction, we give a construction showing that the exponent $11/6\approx 1.83$ cannot be reduced to anything smaller than $\log_3 6 \approx 1.726$.
