无3-簇的子空间族
$3$-cluster-free families of subspaces
AI总结：
该研究定义了向量空间中的3-簇与覆盖三元组，证明不含覆盖三元组的k维子空间族最大规模等于最大星型族规模，且k<n/2时星型族是唯一极值结构，推导出3-簇的对应结论，给出集合系相关定理的向量空间类比。
摘要：
有限域上n维向量空间V中的三个k维子空间A、B、C，若满足A∩B∩C={0_V}且dim(A+B+C)≤2k，则被称为一个3-簇。我们将一类特殊的3-簇称为覆盖三元组，即子空间A、B、C满足A=(A∩B)⊕(A∩C)。我们证明，当2≤k≤n/2时，不含覆盖三元组的k维子空间族的最大规模，与最大星型族（所有子空间都包含一个指定非零向量的子空间族）的规模相同。此外我们还证明，若k<n/2，则星型族是唯一达到该最大规模的族。这一结果可推导出3-簇的对应结论，给出了Mubayi关于集合系的定理在向量空间中的类比。
英文摘要：
Three $k$-dimensional subspaces $A$, $B$, and $C$ of an $n$-dimensional vector space $V$ over a finite field are called a $3$-cluster if $A \cap B \cap C = \{\mathbf{0}_V\}$ and yet $\dim(A+B+C) \leq 2k$. A special kind of $3$-cluster, which we call a covering triple, consists of subspaces $A,B,C$ such that $A = (A \cap B )\oplus (A \cap C)$. We prove that, for $2 \leq k \le n/2$, the largest size of a covering triple-free family of $k$-dimensional subspaces is the same as the size of the largest such star (a family of subspaces all containing a designated non-zero vector). Moreover, we show that if $k < n/2$, then stars are the only families achieving this largest size. This in turn implies the same result for $3$-clusters, which gives the vector space-analogue of a theorem of Mubayi for set systems.
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