AI 中文总结
本文研究线图为补图的线性超图，定义余维数秩并证明其非负性，在有限网完备化假设下给出有界亏格的显式分类，并利用Bruck与Metsch定理得到全局结构推论。
AI 中文摘要
设 $N(H)$ 为有限线性 $s$-一致超图的边-顶点关联矩阵，其线图是补图，并定义 \\[ R_s(H)=(s+1)\operatorname{rank}_{\mathbb R}N(H)-s|E(H)|. \\] 不等式 $R_s(H)\ge0$ 成立，等号成立当且仅当每个行相关分量都是 $s$ 阶仿射平面。在相应的有限网完备化假设下，具有有界 $R_s$ 的连通系统允许显式分类。若 $F$ 连通，$R_s(F)=d\le s-1$，完备化假设在亏格 $d$ 下成立，且 $s>(d-1)^2$，则要么 $N(F)$ 具有满行秩且 $|E(F)|=d$，要么 $F$ 位于唯一的 $s$ 阶仿射平面中。在后一种情况下，对于某个 $0\le h\le d$，系统包含 $s+1-h$ 个平行类中的所有线，并恰好包含来自其余 $h$ 个类的 $d-h$ 条额外线。反之，每个这样的系统都有亏格 $d$。Bruck 定理给出了每个固定 $d$ 和所有足够大 $s$ 的分类，而 Metsch 完备化定理覆盖了 $d=o(s^{1/3})$ 的情况。作为全局推论，如果在完备化范围内总亏格至多为 $D$，则删除至多 $D$ 条边后留下有限网和仿射平面的顶点不相交并集。相关的列亏格和行杠杆也被显式确定。
英文摘要
Let $N(H)$ be the edge--vertex incidence matrix of a finite linear $s$-uniform hypergraph whose line graph is a cograph, and define \[ R_s(H)=(s+1)\operatorname{rank}_{\mathbb R}N(H)-s|E(H)|. \] The inequality $R_s(H)\ge0$ holds, with equality characterized by affine planes of order $s$ on every row-dependent component. Under the corresponding finite-net completion hypothesis, connected systems with bounded $R_s$ admit an explicit classification. If $F$ is connected, $R_s(F)=d\le s-1$, the completion hypothesis holds through deficiency $d$, and $s>(d-1)^2$, then either $N(F)$ has full row rank and $|E(F)|=d$, or $F$ lies in a unique affine plane of order $s$. In the latter case, for some $0\le h\le d$, the system contains every line in $s+1-h$ parallel classes and exactly $d-h$ additional lines from the remaining $h$ classes. Conversely, each such system has defect $d$. Bruck's theorem gives the classification for every fixed $d$ and all sufficiently large $s$, while Metsch's completion theorem covers $d=o(s^{1/3})$. As a global consequence, if the total defect is at most $D$ in the completion range, deleting at most $D$ edges leaves a vertex-disjoint union of finite nets and affine planes. The associated column defect and row leverage are determined explicitly as well.
Comments15 pages
