Crowns in linear $3$-graphs

A \textit{linear $3$-graph}, $H = (V, E)$, is a set, $V$, of vertices together with a set, $E$, of $3$-element subsets of $V$, called edges, so that any two distinct edges intersect in at most one vertex. The linear Tur\'an number, ${\rm ex}(n,F)$, is the maximum number of edges in a linear $3$-graph $H$ with $n$ vertices containing no copy of $F$. We focus here on the \textit{crown}, $C$, which consists of three pairwise disjoint edges (jewels) and a fourth edge (base) which intersects all of the jewels. Our main result is that every linear $3$-graph with minimum degree at least $4$ contains a crown. This is not true if $4$ is replaced by $3$. In fact the known bounds of the Tur\'an number are $$6 \left\lfloor{\frac{n - 3}{4}}\right\rfloor \leq {\rm ex}(n, C) \leq 2n,$$ and in the construction providing the lower bound all but three vertices have degree $3$. We conjecture that ${\rm ex}(n, C) \sim \frac{3n}{2}$ but even if this were known it would not imply our main result. Our second result is a step towards a possible proof of ${\rm ex}(n,C) \leq \frac{3n}{2}$ (i.e., determining it within a constant error). We show that a minimal counterexample to this statement must contain certain configurations with $9$ edges and we conjecture that all of them lead to contradiction.