Gallai-Colorings of Triples and 2-Factors of B [ subscript 3 ]

A coloring of the edges of the -uniform complete hypergraph is a -coloring if there is no rainbow simplex; that is, every set of vertices contains two edges of the same color. The notion extends -colorings which are often called Gallai-colorings and originates from a seminal paper of Gallai. One well-known property of -colorings is that at least one color class has a spanning tree. J. Lehel and the senior author observed that this property does not hold for -colorings and proposed to study , the size of the largest monochromatic component which can be found in every -coloring of , the complete -uniform hypergraph. The previous remark says that and in this note, we address the case . We prove that and this determines for . We also prove that by excluding certain 2-factors from the middle layer of the Boolean lattice on seven elements.


Introduction
Studying edge colorings of complete graphs without rainbow triangles (no triangles colored with distinct colors) originates from a famous paper of Gallai [1]; we will refer to such colorings as Gallai-colorings or  2 -colorings.A well-known property of  2 -colorings is that the edges of some color class span a connected subgraph containing all vertices.A natural extension of the concept is the   -coloring, of    , the complete -uniform hypergraph on  vertices.In a   -coloring the requirement is that no   +1 is colored with +1 distinct colors; that is, there is no rainbow simplex.For  ≥ 3,   -colorings of  3   do not necessarily contain spanning connected color classes, but they must contain large ones.To define how large, let   () be the size of the largest monochromatic component which can be found in every   -coloring of    .In terms of this function,  2 () = , and in this note we address  3 ().
Proof.The upper bound (an unpublished note of the senior author and J. Lehel) follows from taking first the  3 -coloring of  3  5 with five colors, where the vertex set is {1, 2, 3, 4, 5}, and the edges ,  + 1,  + 2 and ,  + 2,  + 3 are colored with  (using (mod 5) arithmetic).This coloring can be "blown up" by replacing each vertex  by a set   so that the five sets have sizes that differ by at most one and defining the following straightforward extension of the original 5-coloring.Triples with vertices from three different   are colored according to the color of the triple of their indices.Triples inside   and triples in   ∪  +1 or in   ∪  +2 (with at least one vertex in each) are colored with color .
The lower bound follows from the observation that every 4-set must contain two 3-sets of the same color.Thus the color of some 3-set  is repeated on another triple inside at least 4-sets containing .This means that the monochromatic component containing  has at least (−3)/2+3 vertices.
In fact, for odd  > 3, the proof of the lower bound in Theorem 1 gives a condition for equality in terms of the bipartite graph B  4,3 defined by the inclusions of the 4-and 3-element where (V) is the degree of vertex V.

Corollary 2.
If the lower bound of Theorem 1 is sharp for some odd  ≥ 5 then there is a (2, ( − 3)/2)-factor  in B  4,3 , such that the union of the triples and quadruples in every component of  has at most ( + 3)/2 elements.
Theorem 1 implies that  3 (4) =  3 (5) = 4 and  3 (6) = 5 but gives only  3 (7) ∈ {5, 6}.Notice that, for  = 5, the upper bound construction of Theorem 1 is a (2, 1)-factor in B 5  4,3 required in Corollary 2, consisting of the shifts of the 3vertex component 1234, 123, 134 modulo 5. Is it possible that for some  > 5 there is a (2, (−3)/2)-factor providing a better upper bound for  3 ()?The first case to consider is B 3 = B 7 4,3 , the "middle-layer graph" in the boolean lattice on seven elements.To decide whether  3 (7) = 5, we will determine all possible 2-factors required by Corollary 2. In Lemma 3 we show that these are 2-factors in B 3 whose components are 6-, 8-, or 10-cycles of the following type which we call a C-factor: Conversely, suppose that  3 (7) = 5, and consider the  3coloring of  3  7 with all monochromatic components having at most five vertices.Let  ⊂ B 3 be the bipartite graph defined by selecting two 3-sets of the same color from each 4-element subset of {1, 2, . . ., 7}.The graph  must be 2-regular, otherwise some 3-set, say in color red, would be selected into at least three 4-sets, resulting in a red component of at least six vertices.Thus  is a 2-factor in B 3 , and its components are cycles.But B 3 contains no  4 since two 4-sets can share at most one 3-set.Also, B 3 cannot have a   component with  ≥ 12; otherwise the corresponding coloring of  3  5 has a monochromatic component with more than five vertices, since there are only five 4-sets on five elements.Thus the components of  are elements of { 6 ,  8 ,  10 }.However, these cycles have to form subdivisions of { 3 ,  4 ,  5 } in the Johnson graph  (5,3), such that the union of triples on the edges of a cycle forms different 4-sets.This means that the edges of  ∈ { 3 ,  4 ,  5 } which are to be subdivided must be in different main cliques of  (5,3), where a main clique  4 is the complete subgraph of (5, 3) determined by the triples of {1, 2, 3, 4, 5} − {} for some  ∈ {1, 2, 3, 4, 5}.The edge set of (5, 3) is partitioned into five main cliques, and we can label the edges in each main clique by , such that  consists of edges with different labels.If two edges ,  meet, then the corresponding 3-set is {1, 2, 3, 4, 5}−{, }.For a  3 with edges labeled 3, 4, and 5, the corresponding 3-sets are 123, 124, and 125.One can similarly check for  4 and  5 that the only possibilities are the cycles in the definition of the C-factor.
In Sections 2 and 3 we prove the following.It is worth mentioning that, in the family of 2-factors constructed in [2], all components are cycles with length divisible by 14; thus it does not answer Question 1.
The proof of Theorem 4 is based on properties of graphs shown in Figure 2.They are the graphs on seven vertices having 11, 10, or 9 edges with the property that every set of three vertices contains at least one edge (complements of trianglefree graphs that are close to the extremal Turán graphs).We need to analyze star-partitions  3 + 4 + 5 of  3  7 , where the star   is the hypergraph having  triples containing two fixed vertices, called the base.Note that  3  7 is an interesting object to decompose.For example,  3  7 can be partitioned into 5 "tight cycles" (consecutive triples in a cyclic permutation of ( 3 7 )) but cannot be partitioned into 5 Fano planes.In fact,  3  7 does not contain three edge disjoint Fano planes (Cayley [3]).However,  3  7 can be partitioned into 6 linear hypergraphs (but not into 5) [4].
Decompositions of complete hypergraphs into stars (hyperclaws) have been investigated [5,6], and this is strongly related to our subject.In particular, the following decomposition problem is related to the case  = 9.Let  be a star  4 with vertex set  1 , . . .,  6 and with base  1  2 .Define the mate of  as the 4-uniform hypergraph with the six 4-element subsets { 1 ,  2 ,   ,   } where , ,  ̸ =  runs over the unordered pairs of 3, 4, 5, 6. Question 2. Is it possible to partition  3 9 into stars  4 so that their mates partition  4  9 ?(Star partition without the mate condition is possible [5].) In fact, Question 2 is a very special case of the C-factor problem arising from the question whether  3 (9) > 6.Here C should be the list of components of a (2, 3)-factor of B 9 4,3 that contain triples and quadruples of six elements (out of nine).Such components arise by subdividing the edges of several 3regular subgraphs of  (6,3).Question 2 came about by considering the component of a (2, 3)-factor from the subdivision of a  4 ∈  (6,3).However, other components come from the subdivision of the following graphs:  2 ×  3 , the graph obtained from  2 ×  3 by subdividing two edges not in a triangle and connecting the subdivision points with an edge, the cube, and the Petersen graph-we checked with a computer that this list is exhaustive.For example, the cube in  (6,3) generate components with 8 and 10 triples (and 12 and 15 quadruples) in B 9 4,3 .Thus there are five possible components of a (2, 3)-factor in C, and this probably makes it difficult to decide whether  3 (9) > 6.The type of a C-factor is the vector (, , ) where , , and  denote the number of cycles of C from C 6 , C 8 , and C 10 , respectively.One can easily determine the possible types of a C-factor by determining the nonnegative integer solutions of 35 = 3 + 4 + 5.There are 14 types, and we will eliminate all of them.

Special 2-Factors in B
First we reformulate the existence of a C-factor as a decomposition problem of  3 7 .We define   , the -star, as a 3uniform hypergraph having  edges containing a fixed pair of vertices, called the base of the star.A star partition of  3   is a partition of all edges into stars; in particular  3  = ∑  =1     means that  3   is partitioned into   copies of   , for  = 1, 2, . . ., .Notice that the complements of the four-element sets in the cycles of a C-factor provide a starpartition of  3 7 .On the other hand, the three-element sets in the cycles of a C-factor also give a partition of  3 7 into copies of the hypergraphs  3 ,  4 , and  5 which we call the complementary partition ( 3 is isomorphic to  3 ,  4 has four triples containing a fixed vertex and consecutive pairs on a 4-cycle of pairs, and  5 has five consecutive triples on a 5-cycle of pairs; these are illustrated in Figure 1).These considerations lead to the following.Proposition 6. B 3 has a C-factor of type (, , ) if and only if  3  7 has a star partition  3 +  4 +  5 together with the complementary partition  3 +  4 +  5 .

Decomposing 𝐾 3
7 along Turán Graphs.We associate a weighted graph , the base graph, with a star partition  3 +  4 +  5 on  3 7 as follows.The vertex set of  is {1, 2, . . ., 7}, and the ( +  + ) edges are determined by the pairs of the bases of the stars in the partition.The graph obtained this way has no multiple edges because different stars must have different bases, since two stars have at least six edges.The weight of an edge  is  if  is the base of an   .
When ++ = 11, easy inspection reveals five possibilities for the base graph.Let  1 11 be the disjoint union of  5 and  2 .Then  2 11 ,  3 11 , and  4 11 are disjoint unions of  3 and  4 plus two edges between them, with a common vertex in  4 , a common vertex in  3 , and no common vertex, respectively (see Figure 2).Finally,  5  11 is the disjoint union of two  3 -s plus one edge between them plus four edges from the seventh vertex to the vertices of degree two (complement of  3,3 with a subdivided edge).
Then one has on  3 7 a star partition with base graph  together with a complementary partition with the following parameters.
The following lemma will eliminate six types of vectors.
Lemma 8. Suppose that B 3 has a C-factor of type (, , ) with  > 0. Then the base graph has a 5-cycle.
Since  9 ,  10 , and  2 11 do not contain 5-cycles, we have the following.
Lemma 9. Suppose that B 3 has a C-factor of type (, , ).Then one has on  3 7 a star partition with base graph  together with a complementary partition with the following parameters.

3.2.
One of the edges of the  3 must generate the triple 567; say  = 56. must also generate the triples 156, 256, 356, and 456.But  can generate at most 4 triples, a contradiction.
The base edges  = 34 and  = 56 must generate the triples 234, 345, 346 and 156, 356, 456, respectively.They cannot both generate  3 -s because then the triple 127 would be doubly covered by their complementary  3 -s.
One of , , say  = 34 generates an  4 .Suppose first that  generates an  3 .Then 13 and 14 both generate  4 -s; otherwise 247 or 237 would be doubly covered by their complementary  3 -s.But this is a contradiction, having at least three  4 -s in the star partition.Thus  also generates an  4 .Consequently, 25 and 26 both generate  3 -s and their complementary  3 -s contain the triples 167 and 157, respectively.However, at least one of these triples is also covered by the complementary  4 of the  4 generated by , a contradiction.Consider the base edges 12, 13, 14, 24, and 34.Since each of these generates two triples with 6 and 7, their complementary  3 -s contain the triples 345, 245, 235, 135, and 125, respectively.However, the complementary  4 of 67, which w.l.o.g. can be written as either  8 (5; 1234) or  8 (5; 1243) must contain at least three of these triples, a contradiction.The base edge 56 must generate the triples 156, 256, 356, and the base edge 23 must generate 235, 236, and 237.If both of them generate  3 -s, then the complementary  3 -s would doubly cover 147.Hence at least one of 56 or 23 must generate an  4 .Similarly, if 67, 13 both generate  3 -s, then 245 would be doubly covered by their complementary  3 -s, and if 57, 12 both generate  3 -s, then 346 would be doubly covered.Then we have at least three  4 -s, a contradiction.

Theorem 4 .Corollary 5 .Question 1 .
B 3 has no C-factor.Lemma 3 and Theorems 1 and 4 give the following. 3 (7) = 6.Theorem 4 might be true in a stronger form.Is there a 2-factor in B 3 in which all cycles have length of at most 10?

5 Figure 1 :
Figure 1: Diagram of the stars   with base  (red) and the corresponding hypergraphs   (blue).

3 and Star Decompositions of 𝐾 3 7
We use C 6 , C 8 , and C 10 to denote the set of possible cycles of a C-factor with different parameter choices from {1, 2, . . ., 7} (e.g., |C 10 | = 4!( 7 5 )).Then C = C 6 ∪C 8 ∪C 10 , and a C-factor is a 2-factor of B 3 such that each cycle in it belongs to C.
must be based on .Let (w.l.o.g) 1, 2, 3, 4 be the cyclic ordering of the 4-cycle in the copy of  4 complementary to  1 (the triples in  8 (5; 1234)).(ii)Consider the four base edges of  on the 4-cycle  = 1, 2, 3, 4, 1.We claim that all of them have weight 4. Suppose this is not the case, and  =  is covered by a star  2 in  3 where ,  are consecutive on .Observe that the triples , , 6 and , , 7 must be in  2 .Therefore the triple  = {1, 2, . .., 7}\{, , 6, 7} is covered by the copy of  3 complementary to  2 .However,  = {5,  + 2,  + 2} with (mod 4), addition thus it is also covered by the copy of  4 complementary to  1 , a contradiction.(iii)At this point, we know that the five base edges are of weight 4, and the rest of the edges have weight 3.In particular, 13 and 57 have weight 3, and both uniquely determine their stars in  3 , namely, 135, 136, 137 and 571, 572, 573.But then the triple 246 is doubly covered by their complementary copies of  3 -s, a contradiction, finishing the proof.