Robust Repulsion for Growing Crowns in Linear Hypergraphs
Put $q=r-1$, $t=k-1$, and $D=tq+1$. For an edge $e$ of a linear $C^r_{1,k}$-free $r$-uniform hypergraph, define \[ \delta_H(e)=\sum_{v\in e}\frac{1}{d_H(v)}-\frac{r}{D}. \] The defect satisfies $\delta_H(e)\ge 0$. At equality, every vertex of $e$ has degree $D$, and the petal trace at $e$ is a disjoint union of $t$ affine planes of order $q$. Equivalently, restoring the base line gives $t$ projective planes of order $q$ with common line $e$. For growing crowns, the equality structure is stable in the following sense. If $q_j\to\infty$, $2\le t_j\le q_j$, and $e_j$ is an edge of a finite linear $C_{1,t_j+1}^{q_j+1}$-free hypergraph satisfying \[ \frac{t_j^2}{q_j}\to0, \qquad t_j^3\delta_{H_j}(e_j)\to0, \] then, for every fixed $0<\theta<1$, \[ \frac{ |\{f\ne e_j:f\cap e_j\ne\varnothing,\ \delta_{H_j}(f)\ge\theta/t_j^2\}| }{(q_j+1)(t_jq_j)} \to1. \] Thus an edge close to equality is adjacent almost entirely to edges with defect of order at least $t_j^{-2}$. A uniform form gives absolute constants $Q_0,\varepsilon_0,c_0>0$ such that, whenever $q\ge Q_0t^2$ and $\delta_H(e)<\varepsilon_0/t^3$, at least $\tfrac12(q+1)tq$ neighbors of $e$ have defect at least $1/(100t^2)$. Consequently, \[ c^{\mathrm{lin}}_{q+1,t+1}\le t-\frac{c_0}{t}, \] where $c^{\mathrm{lin}}_{r,k}$ denotes the asymptotic linear Tur\'an coefficient for $C^r_{1,k}$.