Edge‐transitive homogeneous factorizations of complete uniform hypergraphs

Abstract

For a finite set V and a positive integer k with k≤n:=|V|, letting Vfalse{kfalse} be the set of all k‐subsets of V, the pair scriptKnk:=false(V,Vfalse{kfalse}false) is called the complete k‐hypergraph on V, while each k‐subset of V is called an edge. A factorization of the complete k‐hypergraph Knk of index s≥2, simply a (k,s)‐factorization of order n, is a partition {E1,E2,…,Es} of the edges into s disjoint subsets such that each k‐hypergraph (V,Ei), called a factor, is a spanning subhypergraph of Knk. Such a factorization is homogeneous if there exist two transitive subgroups G and M of the symmetric group of degree n such that G induces a transitive action on the set {E1,E2,…,Es} and M lies in the kernel of this action. In this article, we give a classification of homogeneous factorizations of Knk that admit a group acting transitively on the edges of Knk. It is shown that, for 6≤2k≤n and s≥2, there exists an edge‐transitive homogeneous (k,s)‐factorization of order n if and only if (n,k,s) is one of (32, 3, 5), (32, 3, 31), (33, 4, 5), ()2d,3,false(2d−1false)false(2d−1−1false)3, and (q+1,3,2), where d≥3 and q is a prime power with q≡1(mod4).

On graph norms for complex‐valued functions

“…Chen and Lu [3] obtained an analogous result for 𝑟-graphs , 𝑟 ⩾ 3. In fact, the vertex-transitivity condition in Theorem 7.4 and the isolated-vertices-free condition in Theorem 7.5 are not necessary.…”

“…As every vertex is incident to 𝑘 − 1 pairs in the complete graph, exactly half of which are in . The case 𝑟 = 3 is proved in [3]; see Remark 2 and Table 2 therein Some of the set-inclusion graphs do have transitive colourings. For example, one may check that 𝐼(6, 4, 1) has one.…”

“…The edge‐transitive self‐complementary $$r$$‐graphs has been studied, for example in [3, 19, 22], and their complete classification has been obtained under certain additional conditions. The following result by Zhang [22] is about graphs.…”

“…As every vertex is incident to $$k-1$$ pairs in the complete graph, exactly half of which are in $$\mathcal {G}$$. The case $$r=3$$ is proved in [3]; see Remark 2 and Table 2 therein.$$\Box$$…”

On graph norms for complex-valued functions

“…Indeed, for r = 2, G is regular. As every vertex is incident to k − 1 pairs in the complete graph, exactly half of which are in G. The case r = 3 is proved in [3]; see Remark 2 and Table 2 therein Some of the set-inclusion graphs do have transitive colourings. For example, one may check that I(6, 4, 1) has one.…”