Towards Lehel's conjecture for 4-uniform tight cycles

A $k$-uniform tight cycle is a $k$-uniform hypergraph with a cyclic ordering of its vertices such that its edges are all the sets of size $k$ formed by $k$ consecutive vertices in the ordering. We prove that every red-blue edge-coloured $K_n^{(4)}$ contains a red and a blue tight cycle that are vertex-disjoint and together cover $n-o(n)$ vertices. Moreover, we prove that every red-blue edge-coloured $K_n^{(5)}$ contains four monochromatic tight cycles that are vertex-disjoint and together cover $n-o(n)$ vertices.


Introduction
An r-edge-colouring of a graph (or hypergraph) is a colouring of its edges with r colours.A monochromatic subgraph of an r-edge-coloured graph is one in which all the edges have the same colour.
Lehel conjectured that every 2-edge-colouring of the complete graph on n vertices admits a partition of the vertex set into two monochromatic cycles of distinct colours, where the empty set, a single vertex and a single edge are considered to be degenerate cycles.This conjecture was proved for large n by Luczak, Rödl and Szemerédi [16] using Szemerédi's Regularity Lemma.Allen [1] improved the bound on n by giving a different proof.Finally Bessy and Thomassé [3] proved Lehel's conjecture for all n ≥ 1.
Similar problems have also been considered for colourings with a general number of colours.In particular, a lot of attention has been given to the problem of determining the number of monochromatic cycles that are needed to partition an r-edge-coloured complete graph.Erdős, Gyárfás and Pyber [6] proved that every r-edge-coloured complete graph can be partitioned into O(r 2 log r) monochromatic cycles and conjectured that r monochromatic cycles would suffice.Their result was improved by Gyárfás, Ruszinkó, Sárközy and Szemerédi [11] who showed that O(r log r) monochromatic cycles are enough.However, Pokrovskiy [17] disproved the conjecture and proposed a weaker version of the conjecture that each r-edge-coloured complete graph contains r monochromatic vertex-disjoint cycles Proposition 1.1.For all k ≥ 3 and m ≥ k+1, there exists a 2-edge-colouring of that does not admit a partition into two tight cycles of distinct colours.
It is natural to ask whether we can cover almost all vertices of a 2-edge-coloured complete k-graph with two vertex-disjoint monochromatic tight cycles of distinct colours.The case when k = 3 is affirmed in [5].Here, we show that this is true when k = 4. Theorem 1.2.For every ε > 0, there exists an integer n 1 such that, for all n ≥ n 1 , every 2-edge-coloured complete 4-graph on n vertices contains two vertex-disjoint monochromatic tight cycles of distinct colours covering all but at most εn of the vertices.
When k = 5, we prove a weaker result that four monochromatic tight cycles are sufficient to cover almost all vertices.
Theorem 1.3.For every ε > 0, there exists an integer n 1 such that, for all n ≥ n 1 , every 2-edge-coloured complete 5-graph on n vertices contains four vertex-disjoint monochromatic tight cycles covering all but at most εn of the vertices.
To prove Theorems 1.2 and 1.3, we use the connected matching method that has often been credited to Luczak [15].We now present a sketch proof for Theorem 1.2.Consider a 2-edge-coloured complete 4-graph K (4) n on n vertices.We start by applying the Hypergraph Regularity Lemma to the 2-edge-coloured complete 4-graph K (4) n .More precisely the Regular Slice Lemma of Allen, Böttcher, Cooley and Mycroft [2], see Lemma 4.3.We obtain a 2-edge-coloured reduced graph R that is almost complete.A monochromatic matching in a k-graph is a set of vertex-disjoint edges of the same colour.We say that it is tightly connected if, for any two edges f and f ′ , there exists a sequence of edges e 1 , . . ., e t of the same colour such that e 1 = f , e t = f ′ and |e i ∩ e i+1 | = k − 1 for all i ∈ [t − 1].Using Corollary 4.12, it suffices to find two vertex-disjoint monochromatic tightly connected matchings of distinct colours in the reduced graph R. The main challenge is to identify the 'tightly connected components' (see Section 2 for the formal definition) in which we will find the matchings.To do so, we introduce the concept of 'blueprint', which is a 2-edge-coloured 2-graph with the same vertex set as R. The key property is that connected components in the blueprint correspond to tightly connected components in R.
We conclude the introduction by outlining the structure of the paper.In Section 2, we introduce some basic notation and definitions.In Section 3, we prove Proposition 1.1.In Section 4, we introduce the statements about hypergraph regularity and prove the crucial Corollary 4.12 that allows us to reduce our problem of finding cycles in the complete graph to one about finding tightly connected matchings in the reduced graph.In Section 5, we give the definition of blueprint and setup some useful results.In Sections 6 and 7, we prove Theorems 1.2 and 1.3, respectively.Finally, we make some concluding remarks in Section 8.

Preliminaries
If we say that a statement holds for 0 < a ≪ b ≤ 1, then we mean that there exists a non-decreasing function f : (0, 1] → (0, 1] such that the statement holds for all a, b ∈ (0, 1] with a ≤ f (b).Similar expressions with more variables are defined analogously.If 1/n appears in one of these expressions, then we implicitly assume that n is a positive integer.
Throughout this paper, any 2-edge-colouring uses the colours red and blue.Let H be a 2-edge-coloured k-graph.We denote by H red (and H blue ) the subgraph of H on V (H) induced by the red (and blue) edges of H. Two edges f and f ′ in H are tightly connected if there exists a sequence of edges e 1 , . . ., e t such that e 1 = f , e t = f ′ and |e i ∩ e i+1 | = k − 1 for all i ∈ [t − 1].A subgraph H ′ of H is tightly connected if every pair of edges in H ′ is tightly connected in H.A maximal tightly connected subgraph of H is called a tight component of H.Note that a tight component is a subgraph rather than a vertex subset as in the traditional graph case.A red tight component and a red tightly connected matching are a tight component and a tightly connected matching in H red , respectively.We define these terms similarly for blue.
Let H be a k-graph and S, W ⊆ V (H).We denote by H − W the k-graph with V (H − W ) = V (H) \ W and E(H − W ) = {e ∈ E(H) : e ∩ W = ∅}.We call H − W the k-graph obtained from H by deleting W .Further we let Let F be a k-graph or a set of k-element sets.We denote by H − F the subgraph of H obtained by deleting the edges in F .We define N H (S, W ) to be the set {e ∈ For the 1-st shadow of H, we also simply write ∂H instead of ∂ 1 H.For µ, α > 0, we say that a k-graph H on n vertices is (µ, α)-dense if, for each i . It follows that f and f ′ are tightly connected.
The following proposition shows that any k-graph that has all but a small fraction of the possible edges contains a (1 − ε, α)-dense subgraph.The proof was inspired by the proof of Lemma 8.8 in [12].A different generalisation of this lemma can also be found as Lemma 2.3 in [14].
This implies This proves the first part of the claim.We now prove the second part of the claim.Fix i ∈ [k − 1].We proceed by induction on j − i.For j = i + 1, the statement holds by the definition of A i .Now let S ∈ V (H) i and j ≥ i + 2 be such that We have and thus Hence by the induction hypothesis we have S ∈ A i .
For each j ∈ [k − 1], let F j be the set of edges e ∈ H for which there exists some By the above, it suffices to show that X ∈ B i ∪ S i .Let e ∈ H ′ with X ⊆ e.Since e ∈ F i , we have X ∈ A i and thus X ∈ B i .It remains for us to show that X ∈ S i .Assume the contrary that X is contained in more that

Extremal example
In this section, we prove Proposition 1.1, that is, we prove that, for k ≥ 3, there exist arbitrarily large 2-edge-coloured complete k-graphs that do not admit a partition into two tight cycles of distinct colours.
A k-uniform tight path is a k-graph obtained by deleting a vertex from a tight cycle.First we need the following proposition.Proposition 3.1.Let k ≥ 3, let P and C be a k-uniform tight path and tight cycle, respectively.We have the following.
Proof.We first prove (i).Let M be a matching of maximum size in P .Since each edge of P contains at least 2 vertices of Y , Now we prove (ii).Since |e ∩ Y | ≥ 2 and |e ∩ X| ≤ k − 2 for each edge e ∈ C, we have We are now ready to give our extremal example.Note that the case k = 3 of the extremal example is already given in [7].Recall that, in a k-graph, we consider a single edge and any set of fewer than k vertices to be degenerate cycles.n − z has the following 3 monochromatic tight components: Note that B 1 and B 2 are blue and R is red.Suppose for a contradiction that K n can be partitioned into a red tight cycle C R and a blue tight cycle C B .
First assume z ∈ V (C R ).Since all the red edges containing z are in a red tight component disjoint from R, we have This implies that m ≤ 2, a contradiction.Hence, we may assume that z ∈ Hence, V (P B ) ∩ Y = ∅ and |V (P B )| ≥ (n − 1) − km ≥ k.We must have P B ⊆ B 2 .By Proposition 3.1(i), we have that Thus k+1 that has a blue edge containing z and at least two vertices of Y , a contradiction.

Hypergraph regularity
In this section, we give the formulation of hypergraph regularity that we use, following closely the presentation of Allen, Böttcher, Cooley and Mycroft [2].A hypergraph H is an ordered pair (V (H), E(H)), where E(H) ⊆ 2 V (H) .Again, we identify the hypergraph H with its edge set E(H).A subgraph and f ⊆ e, then f ∈ H.A kcomplex is a complex with only edges of size at most k.We denote by H (i) the spanning subgraph of H containing only the edges of size i.Let P be a partition of For P ′ = {V i 1 , . . ., V ir } ⊆ P, we define the subgraph of H induced by P ′ , denoted by H[P ′ ] or H[V i 1 , . . ., V ir ], to be the subgraph of H[ P ′ ] containing only the edges that are P ′ -partite.The hypergraph H is P-partite if all of its edges are P-partite.In this case we call the parts of P the vertex classes of H.We say that H is s-partite if it is P-partite for some partition P of V (H) into s parts.Let H be a P-partite hypergraph.If X is a k-set of vertex classes of H, then we write H X for the k-partite subgraph of H (k) induced by X, whose vertex classes are the elements of X.Moreover, we denote by Let i ≥ 2, and let P i be a partition of a vertex set V into i parts.Let H i and H i−1 be a P i -partite i-graph and a P i -partite (i − 1)-graph on a common vertex set V , respectively.We say that a P i -partite i-set in V is supported on H i−1 if it induces a copy of the complete (i − 1)-graph K (i−1) i on i vertices in H i−1 .We denote by K i (H i−1 ) the P i -partite i-graph on V whose edges are all P i -partite i-sets contained in V which are supported on H i−1 .Now we define the density of H i with respect to H i−1 to be Finally, given an i-graph G whose vertex set contains that of H i−1 , we say that G is (d i , ε, r)-regular with respect to H i−1 if the i-partite subgraph of G induced by the vertex classes of H i−1 is (d i , ε, r)-regular with respect to H i−1 .We refer to the density of this i-partite subgraph of G with respect to H i−1 as the relative density of G with respect to H i−1 .Now let s ≥ k ≥ 3 and let H be an s-partite k-complex on vertex classes V 1 , . . ., V s .For any set A ⊆ [s], we write V A for i∈A V i .Note that, if e ∈ H (i) for some 2 ≤ i ≤ k, then the vertices of e induce a copy of 1) .Therefore, for any set A ∈ [s]  i , the density d( i , the induced subgraph (a) J is P-partite for some P which partitions V (J ) into t parts, where t 0 ≤ t ≤ t 1 , of equal size.We refer to P as the ground partition of J , and to the parts of P as the clusters of J .
(b) There exists a density vector d = ( For any k-set X of clusters of J , we denote by ĴX the k-partite and call ĴX a polyad.Given a (t 0 , t 1 , ε)-equitable (k − 1)-complex J and a k-graph G on V (J ), we say that G is (ε k , r)-regular with respect to a k-set X of clusters of J if there exists some d such that G is (d, ε k , r)-regular with respect to the polyad ĴX .Moreover, we write d * G,J (X) for the relative density of G with respect to ĴX ; we may drop either subscript if it is clear from context.
We can now give the crucial definition of a regular slice.
)-regular with respect to all but at most ε k t k of the k-sets of clusters of J , where t is the number of clusters of J .
If we specify the density vector d and the number of clusters t of an equitable complex or a regular slice, then it is not necessary to specify t 0 and t 1 (since the only role of these is to bound d and t).In this situation we write that Given a regular slice J for a k-graph G, we define the d-reduced k-graph R J d (G) as follows.
Definition 4.2 (The d-reduced k-graph).Let k ≥ 3. Let G be a k-graph and let J be a (t 0 , t 1 , ε, ε k , r)-regular slice for G.Then, for d > 0, we define the d-reduced k-graph R J d (G) to be the k-graph whose vertices are the clusters of J and whose edges are all k-sets X of clusters of J such that G is (ε k , r)-regular with respect to X and d * (X) ≥ d.
We now state the version of the Regular Slice Lemma that we need, which is a special case of [2, Lemma 10].Lemma 10]).Let k ≥ 3.For all positive integers t 0 and s, positive ε k and all functions r : N → N and ε : N → (0, 1], there are integers t 1 and n 0 such that the following holds for all n ≥ n 0 which are divisible by t 1 !. Let K be a 2-edge-coloured complete k-graph on n vertices.Then there exists a (k − 1)-complex J on V (K) which is a (t 0 , t 1 , ε(t 1 ), ε k , r(t 1 ))-regular slice for both K red and K blue .
Given a 2-edge-coloured complete k-graph H we want to apply the Regular Slice Lemma to H red and H blue .The following lemma shows that in this setting the union of the corresponding reduced graphs r)-regular slice for both K red and K blue there are at least (1 − 2ε k ) t k k-sets X of clusters of J such that both K red and K blue are (ε k , r)-regular with respect to X. Let X be such a k-set.Since K red and K blue are complements of each other, we have d Let H be a k-graph.A fractional matching in H is a function ω : E(H) → [0, 1] such that for all v ∈ V (H), ω(v) := e∈H:v∈e ω(e) ≤ 1.The weight of the fractional matching is defined to be e∈H ω(e).A fractional matching is tightly connected if the subgraph induced by the edges e with ω(e) > 0 is tightly connected in H.The following result from [2] converts a tightly connected fractional matching in the reduced graph into a tight cycle in the original graph.Lemma 4.5 ([2, Lemma 13]).Let k, r, n 0 , t be positive integers, and let ψ, ε, ε k , d k , . . ., d 2 be positive constants such that 1/d i ∈ N for each 2 ≤ i ≤ k − 1, and such that 1/n 0 ≪ 1/t, Then the following holds for all integers n ≥ n 0 .Let G be a k-graph on n vertices, and J be a (•, •, ε, ε k , r)-regular slice for G with t clusters and density vector (d k−1 , . . ., d 2 ).Suppose that R J d k (G) contains a tightly connected fractional matching with weight µ.Then G contains a tight cycle of length ℓ for every ℓ ≤ (1 − ψ)kµn/t that is divisible by k.
We use the following fact, lemma and proposition to prove Lemma 4.10 which is a stronger version of Lemma 4.5 that allows us to control the location of the tight cycle.
For any r, s ∈ N and , and let G be an s-partite k-complex The following proposition shows that a refinement of a regular slice is also a regular slice.
Proof.We construct J from J as follows.Let the ground partition of J be {V i,j : i ∈ Starting with the edges of J we iteratively add additional edges at random as follows.For each 2 ≤ i ≤ k − 1, beginning with i = 2, we add each i-edge that contains two vertices that are in vertex classes with the same first index and is supported on the (i − 1)-edges independently with probability d i .
We now show that with high probability J is the desired (k − 1)-complex.Note that it suffices to show that with high probability J is (d, √ ε, √ ε, 1)-regular. Let , let B i be the event that If the r j are distinct, then the claim holds by Lemma 4.7 with G = J [V A ] and α = 1/N.
If not all the r j are distinct, then K i ( Thus for each subgraph Q of Since there are at most 2 (im) i−1 choices for Q, the claim follows by a union bound.
Note that if J is not (d, √ ε, √ ε, 1)-regular, then there exists some i ∈ [k − 1] and i such that B i,A holds.Further by choosing i minimal we can ensure that B i−1 holds.Thus, by a union bound and Claim 4.9, we have The following lemma is a strengthening of Lemma 4.5.We believe the constant β and the corresponding condition could be removed if one were to go through the proof of Lemma 4.5 to prove a stronger result.
We first explain the main ideas of the proof.We would like to find a regular slice for G ′ = G[ i∈[t] W i ] so that we can then apply Lemma 4.5 to G ′ .The issue is that not all vertex classes in G ′ have the same size.To get around this we take a refinement of the original partition and use Proposition 4.8 to find a new regular slice with that ground partition.The reduced graph for this new regular slice will be a blow up of the original reduced graph.We can find a corresponding tightly connected matching in this new reduced graph.Then we simply apply Lemma 4.5.
By Proposition 4.8, there exists a (•, •, ε 1/4 )-equitable (k − 1)-complex J * with density vector (d k−1 , . . ., d 2 ) and ground partition {V i,j : i Let G be the subgraph of G[ W ] obtained by removing all edges contained in k-tuples of density less than d k and in irregular k-tuples.We show that J is a regular slice for G. Let X be a set of k clusters of J .If the k clusters in X are all contained in distinct clusters of J that form a regular k-tuple of density at least d k , then let Y denote the k-set of these clusters.Note that (G 1) .Note that, for all other k-sets of clusters X, the k-partite subgraph of G induced by the clusters in X is empty.For these k-sets of clusters, G is (0, √ ε k , r)-regular with respect to the polyad ( Consider the tightly connected fractional matching ϕ on R J d k (G) with weight µ.We construct a tightly connected matching on R as follows.For each e ∈ R J d k (G), we will pick a matching M e in R of size ϕ(e) = ⌊(1 − 3ε ′ )ϕ(e)N⌋.Note that, for each i ∈ [t], For each vertex In the second inequality above we used the fact that since ϕ is a fractional matching with weight µ and all edges have weight at least β, there are at most µ/β edges of positive weight.Since R is a blow-up of R J d k (G), M is tightly connected.We conclude by applying Lemma 4.5 with k, r, n, t, For the next result, we need the following definition.Proof.We prove the first statement.The other two statements can be proved similarly (where for the third statement we additionally make use of the fact that Lemma 4.10 also allows us to control the length of the resulting cycle).Without loss of generality assume that η ≤ 1/3.Let )t for all t ≥ t 0 .We choose functions ε(•) and r(•) where ε(•) approaches zero sufficiently quickly and r(•) increases sufficiently quickly such that for any integer t * ≥ t 0 and d 2 , . . ., d k−1 ≥ 1/t * we may apply Lemma 4.10 with ε(t * ) and r(t * ) playing the roles of ε and r, respectively.We apply Lemma 4.3 to obtain n 0 and t 1 .Let ε = ε(t 1 ) and r = r(t 1 ).Let n 1 ≥ n 0 be large enough such that for all n ≥ n 1 and d 2 , . . ., d k−1 ≥ 1/t 1 we may apply Lemma 4.10.Let n 2 = n 1 + t 1 !.We show that the theorem holds for all n ≥ n 2 .Let K be a 2-edge-coloured complete k-graph on n vertices.Let n ≤ n be the largest integer such that t 1 !divides n.Let K be a complete subgraph of K on n vertices.Note that n ≥ n 1 .By Lemma 4.3, there exists a (t 0 , t 1 , ε, ε k , r)-regular slice J for both K red and K blue .Let t be the number of clusters of J and let (d k−1 , . . ., d 2 ) be the density vector of t such that all edges with non-zero weight have weight at least β and lie in s monochromatic tight components K 1 , . . ., K s of H.For each j ∈ [s], we define a fractional matching ϕ j in H by setting ϕ j (e) = ϕ(e) if e ∈ K i and ϕ(e) = 0 otherwise.For each j ∈ [s], let µ j be the weight of ϕ j .It follows that j∈[s] µ j = µ.
Let V 1 , . . ., V t be the clusters of J .For each i ∈ [t] and j ∈ [s], we define For each i ∈ [t], let V i,1 , . . ., V i,s be disjoint subsets of V i such that |V i,j | = ⌈w i,j n/t⌉.By Lemma 4.10, there exist tight cycles C 1 , . . ., C s in K such that, for all j ∈ [s], V i,j and C j has the same colour as K j .Hence C 1 , . . ., C s are vertex-disjoint and together cover

Blueprints
Let H be a 2-edge-coloured k-graph.We define what we call a blueprint for H which is an auxiliary graph that can be used as a guide when finding connected matchings in H.A form of the notion of a blueprint for k = 3 already appeared in [13].We say that e induces H(e) and write R(e) or B(e) instead of H(e) if e is red or blue, respectively.We simply say that G is a blueprint, when H and ε are clear from context.For S ∈ V (H) k−3 , all the red (blue) edges of a blueprint containing S induce the same red (blue) tight component, so we call that component the red (blue) tight component induced by S. Note that any subgraph of a blueprint is also a blueprint.The main aim of this section is to prove the following lemma that establishes the existence of blueprints for 2-edge-coloured (1 − ε, α)-dense graphs.
We need a few simple preliminary results to prove Lemma 5.3.For a 2-graph G, we denote by δ(G) the minimum degree of G. First we show that any 2-edge-coloured 2-graph with large minimum degree contains a large monochromatic connected subgraph.This proposition is implied by [8, Lemma 1.5] but we include a proof for completeness.Proof.Let F ′ be an induced subgraph of F of maximum order that contains a spanning monochromatic component.Assume without loss of generality that F ′ contains a spanning red component.Let S = V (F ′ ) and S = V (F ) \ V (F ′ ).Since δ(F ) ≥ (1 − β)n, we have that |S| ≥ (1 − β)n/2.Suppose, for a contradiction, that |S| < (1 − β)n.Note that all edges between S and S are blue.If δ(F ) − |S| + 1 > S /2, then each pair of vertices in S has a common neighbour in S and so there is a blue component strictly containing S which contradicts the maximality of F ′ .Therefore But now every pair of vertices in S has a common neighbour in S, since S = |V (F )|−|S| ≤ 2βn and so Thus S ∪ N F (S) is spanned by a blue component.But since we have a contradiction.It is easy to see that δ(F ′ ) ≥ (1 − 2β)n.
Then there exists a subgraph F ′ of F of order at least Proof.By Proposition 5.5, there exists a subgraph F * of F with δ(F * ) ≥ (1 − 3 √ ε)n.We conclude by applying Proposition 5.4 with F = F * and β = 3 √ ε.

Proof of Lemma 5.3
Now we show that for any (1 − ε, α)-dense 2-edge-coloured graph we can find a dense blueprint.
Proof of Lemma 5.3.
and In the next claim we show that, for each S ∈ V (H) k−3 , almost all edges in F of the same colour containing S induce the same monochromatic tight component in H. Let D be the directed graph with vertex set N red F (S) and edge set Note that, for y 1 y 2 ∈ E(D), there exists an edge in R(S ∪y 2 ) containing S ∪y 1 y 2 .So if y 1 y 2 is a double edge (that is, Hence the number of double edges in D is at least Thus there exists a vertex y 0 ∈ N red F (S) that is incident to at least N red Note that Observe that every edge of D * is an edge of F .Every edge in For each y ∈ X B * , there exists By averaging, there exists an x ∈ X R * such that For each y ∈ Y x , since H is (1 − ε, α)-dense, there are at least |X| − εn vertices x ′ ∈ X such that S ∪ xx ′ yy ′ ∈ H. Thus, by averaging, there exists a vertex x ′ ∈ X and a set The following lemma shows that if we have a vertex set T ∈ V (G) k−3 such that d red G (T ) and d blue G (T ) are both large, then T is contained a lot of sets in ∂R ∩ ∂B, where R and B are the red and blue tight components induced by the red and blue edges incident to T , respectively.Lemma 5.9.
Then there exists a vertex y ∈ S blue such that, for , then we may assume that m blue = m red = ⌈δn⌉ by deleting vertices in S blue and S red if necessary.Let D be the bipartite directed graph with vertex classes S blue and S red such that, for each y ∈ S blue and x ∈ S red , we have Thus the number of double edges in D is at least m blue m red − 3 √ εn(m blue + m red ).For each y ∈ S blue , let Γ y = {x ∈ S red : xy, yx ∈ D}.Hence there is some vertex y ∈ S blue such that Note that if xy, yx ∈ D with x ∈ S red and y ∈ S blue , then T ∪ xy ∈ ∂R(T ∪ x) ∩ ∂B(T ∪ y).Hence Γ y ⊆ Γ red y and thus the lemma follows.
Roughly speaking, in the next lemma we consider the following situation.Let R be a red tight component in H, G be a blueprint for H and R G ⊆ G red be such that H(R G ) ⊆ R. We pick a maximal matching in R + G and let U be the remaining vertices of H not in this matching, so R + G [U] is empty.Then the lemma implies that the number of monochromatic tight components in U is less than what we would expect.In particular, if k = 4, then the edges in G[U] induce only two monochromatic tight components in H.For simplicity we assume k = 4 and S = ∅.It is easy to see that an analogous argument works in the general case.Thus for the rest of the proof, we omit the subscript S.
, then we are done by setting J blue = ∅ as Thus |X| ≥ δ 1/2 n.Let D be the digraph with vertex set X such that, for each x ∈ X, for some y ∈ N blue K (x)}.We now bound δ + (D) as follows.If d red K (x, X) ≥ δn, then by applying Lemma 5.9 (with x, N blue G (x, U), N red G (x, X), δ playing the roles of T, S blue , S red , δ), we deduce that On the other hand, if Let F be the graph with vertex set X in which xx ′ forms an edge if and only if it forms a double edge in D. Note that |F | ≥ (1 − 4δ 1/2 ) |X| 2 .By Proposition 5.5, there exists a subgraph We now show that B(x 1 z 1 ) = B(x 2 z 2 ) for all x 1 z 1 , x 2 z 2 ∈ J blue .Since F * is connected and d J blue (x) > 0 for all x ∈ V (F * ), it suffices to consider the case when 6 Monochromatic connected matchings in K (4) n In this section, we prove that every almost complete red-blue edge-coloured 4-graph H contains a red and a blue tightly connected matching that are vertex-disjoint and together cover almost all vertices of H.
Let H be a 2-edge-coloured (1 − ε, α)-dense 4graph on n vertices.Then H contains two vertex-disjoint monochromatic tightly connected matchings of distinct colours such that their union covers all but at most 3ηn of the vertices of H.
To prove Lemma 6.1 we first need the following lemma which chooses the initial tight components in H in which we find our tightly connected matchings.
Let H be a 2-edge-coloured (1 − ε, α)-dense 4-graph on n vertices.Suppose that H does not contain two vertex-disjoint monochromatic tightly connected matchings of distinct colours such that their union covers all but at most 3ηn of the vertices of H.Then, there exists )n and a matching M 0 in R ∪ B such that the following holds, where W 0 = V (G) \ V (M 0 ).
Proof.By Lemma 5.3, there exists a 3 √ ε-blueprint G 0 for H with V (G 0 ) = V (H) and By Corollary 5.6, there exists a subgraph G 1 of G 0 of order at least (1 − 6 √ α)n that contains a spanning monochromatic component We assume without loss of generality that G 1 contains a spanning red component.Since G 1 is a blueprint, all the red edges in G 1 induce the same red tight component R in Let W = V (G) \ V (M).Next, we show that (i) and (ii) hold but with M, W instead of M 0 , W 0 .Note that M blue = ∅, so (ii) holds by our construction.Since G red ⊆ G red 1 and G red 1 is connected and a blueprint, R(e) = R for all e ∈ G red .Note that for all e ∈ G blue [V (M blue ) ∪ W ]. Hence (i) holds.We now add vertex-disjoint edges of (G red ) + [W ] ∪ (G blue ) + [W ] to M and call the resulting matching M 0 .We deduce that M 0 satisfies (i)-(iii).
We now prove Lemma 6.1.
Proof of Lemma 6.1.Suppose the contrary that H does not contain two vertex-disjoint monochromatic tightly connected matchings of distinct colours such that their union covers all but at most 3ηn of the vertices of H.We call this the initial assumption.Apply Lemma 6.2 and obtain a red tight component R, )n and a matching M 0 in R ∪ B satisfying Lemma 6.2(i)-(iii).We now fix G, R and B. We use the following notation for the rest of the proof.For a matching M in R ∪ B, we set Note Note that (i ′ ) and (ii ′ ) are weaker statements of those in Lemma 6.2(i) and (ii), so M 0 ∈ M. Let M ′ be the set of M ∈ M also satisfying Observe that M 0 ∈ M ′ , so M ′ is nonempty.Let γ = 10α 1/30 .We now show that, for all M ∈ M, W red and W blue partition W , and moreover one of them is small.We now prove Lemma 7.1.The proof works by first finding a maximal matching in R ∪ B, where R and B are the components given by Lemma 7.2, and then finding maximal connected matchings in the remaining vertices.
Proof of Lemma 7.1.Assume, for a contradiction, that such matchings do not exist.We call this the initial assumption.Apply Lemma 7.2 and obtain V red , V blue , G, R 3 , R, B 3 , B and let V * = V red ∪ V blue .Since there are only few vertices in V (H) \ V * we ignore these vertices from the start and construct our matchings in H[V * ].
We begin by choosing a matching M ⊆ and |U| ≥ ηn by the initial assumption.Let U red = U ∩V red and U blue = U ∩V blue .The following claim shows that if U red and U blue are both large, then G[U] must contain many edges in R 3 or many edges in B 3 .
Proof of Claim.Define a bipartite graph K 0 with vertex classes U red and U blue such that x ∈ U red and y ∈ U blue are joined by an edge if and only if xy ∈ ∂R 3 ∩ ∂B 3 .Recall that d ∂R 3 (x) ≥ (1 − α 1/75 )n and d ∂B 3 (y) ≥ (1 − α 1/75 )n for all x ∈ U red and y ∈ U blue .Hence Hence it suffices to show that almost all edges of K 0 are of the same colour.Indeed, if we have that at least U red U blue − 3α 1/154 n 2 edges of K 0 are red, then we have We show that each edge xy ∈ K 0 is coloured either red or blue.It suffices to show that either d R 3 (xy, U) < α 1/76 n or d B 3 (xy, U) < α 1/76 n.Indeed if d R 3 (xy, U), d B 3 (xy, U) ≥ α 1/76 n, then by Lemma 5.9, there exists u, u , a contradiction to the maximality of M.Moreover, by Lemma 5.8, we have that min{d red K 0 (u), d blue K 0 (u)} ≤ α 1/76 n for all u ∈ U.
Let K 1 be the graph obtained from K 0 by, for each u ∈ U, deleting all red edges incident to u if d red K (u) ≤ α 1/76 n and all blue edges incident to u if d blue K (u) ≤ α 1/76 n.Note that |K 1 | ≥ U red U blue − α 1/77 n 2 and that, in K 1 , each vertex is incident to only edges of one colour.It is not too hard to see that by deleting at most 2α 1/154 n 2 additional edges, we can obtain a subgraph K 2 of K 1 for which each vertex has degree 0 or large degree.More precisely, for all u ∈ U red , Since each vertex is incident to only edges of one colour and any two vertices in U red that have non-zero degree have a common neighbour this implies that all edges in K 2 are of the same colour.Since |K 2 | ≥ U red U blue − 3α 1/154 n 2 , this concludes the proof.
The following claim shows that there is a red tight component R * and a blue tight component B * of H such that almost all the edges in G[U] induce one of these components.
By Lemma 5.9, the fact that F is (1 − γ 1/5 , γ 1/5 )-dense and the fact that H is (1 − ε, α)dense, we have, for u ′ ∈ Γ L (u), .By an easy greedy argument, there exists a matching M ′ in H blue that covers all but at most ηn of the vertices in W .The matching M ′ ∪ M ⋄ covers all but at most 3ηn of the vertices of H.This contradicts the initial assumption.

Concluding Remarks
For k ≥ 3, let f (k) be the minimum integer m such that, for all large 2-edge-coloured complete k-graphs, there exists m vertex-disjoint monochromatic tight cycles covering almost all vertices.Note that f (k) is well defined by [4] but the bound is very large.It is easy to see that f (k) ≥ 2 for all k ≥ 3. Indeed, consider the k-graph H = K (k) (A, B) given in Example 5.2 with |A| = 3k−1 3k n.Note that H[A] is a red tight component.Moreover, note that any tight cycle contained in a monochromatic tight component other than H[A] covers at most about a third of the vertices of H and any tight cycle in H[A] leaves all n 3k vertices in B uncovered.Hence no monochromatic tight cycle covers almost all vertices in H.We have f (3) = 2 by [5].Theorems 1.2 and 1.3 imply f (4) = 2 and f (5) ≤ 4, respectively.In general, we believe that f (k) = 2 for all k.However, we believe that new ideas may be needed as indicated by again considering the k-graph H = K (k) (A, B) with |A| = 3k−1 3k n (as above).If H contains two vertex-disjoint monochromatic tight cycles of distinct colour covering almost all vertices, then one of the two cycles must lie entirely in the red tight component H[A].However, this tight component is not induced by any edge in the blueprint of H (which is K (k−2) (A, B) with colours swapped).Thus we ask the weaker question of whether one can bound f (k) by some suitable function of k.

Proof of Proposition 1 . 1 .
Let k ≥ 3, m ≥ k + 1 and n = k(m + 1) + 1.Let X, Y and {z} be three disjoint vertex sets of K (k) n of sizes (k − 1)m + k − 2, m + 2 and 1, respectively.We colour an edge e in K (k) n red if z ∈ e and |e ∩ Y | ≥ 2 or z ∈ e and |e ∩ Y | = 1.Otherwise we colour it blue.Note that K (k) 1) and thus x B = 2k − 3 and y B = 3.Let P B = v 1 . . .v 2k .Either the edge v 1 . . .v k or the edge v k+1 . . .v 2k contains at most one vertex of Y , a contradiction to P B ⊆ B 2 .Thus we may assume y R ≥ m and since y R ≤ m by (3.1), we have y R = m.By (3.3), we have x R = (k − 1)m and thus x B = k − 2 and y B = 2. Hence, C B is a copy of K (k)

Definition 4 . 11 .
Let µ s k (β, ε, n) be the largest µ such that every 2-edge-coloured (1 − ε, ε)-dense k-graph on n vertices contains a fractional matching with weight µ such that all edges with non-zero weight have weight at least β and lie in s monochromatic tight components.Let µ s k (β) = lim inf ε→0 lim inf n→∞ µ s k (β, ε, n)/n.Similarly, let µ * k (β, ε, n) be the largest µ such that every 2-edge-coloured (1 − ε, ε)-dense k-graph on n vertices contains a fractional matching with weight µ such that all edges with non-zero weight have weight at least β and lie in one red and one blue tight component.Let µ * k (β) = lim inf ε→0 lim inf n→∞ µ * k (β, ε, n)/n.The following is the crucial result that reduces finding cycles in the original graph to finding tightly connected matchings in the reduced graph.

Definition 5 . 1 .
Let ε > 0, k ≥ 3 and let H be a 2-edge-coloured k-graph on n vertices.We say that a 2-edge-coloured(k − 2)-graph G with V (G) ⊆ V (H) is an ε-blueprint for H, if(BP1) for every edge e ∈ G, there exists a monochromatic tight component H(e) in H such that H(e) has the same colour as e and d ∂H(e) (e) ≥ (1 − ε)n and (BP2) for e, e ′ ∈ G of the same colour with |e ∩ e ′ | = k − 3, we have H(e) = H(e ′ ).

Example 5 . 2 .
Let k ≥ 3 and let n be a positive integer.Let A and B be disjoint vertex sets with |A ∪ B| = n.Let K (k) (A, B) be the 2-edge-coloured complete k-graph with vertex set A ∪ B where an edge e is red if and only if |e ∩ A| is even (and blue otherwise).Let H be K (k) (A, B) and let G be K (k−2) (A, B) with colours reversed.If ε ≥ k−2 n , then G is an ε-blueprint for H. Indeed, for an edge e ∈ G we can set H(e) = {f ∈ H : |f ∩ A| = |e ∩ A| + 1}.

F (S) − 6 √
εn double edges.Let Γ red (S) = {y 0 } ∪ {y ∈ N red F (S) : yy 0 , y 0 y ∈ E(D)}.Note that Γ red (S) ≥ N red F (S) − 6 √ εn and R(S ∪ y) = R(S ∪ y 0 ) for all y ∈ Γ red (S).Consider the multi-(k − 2)-graph D * with for H. Consider any e, e ′ ∈ G red * with |e ∩ e ′ | = k − 3. Let S = e ∩ e ′ , y = e ′ \ S and y ′ = e \ S. Since e, e ′ ∈ G red * , we have y, y ′ ∈ Γ red (S) and so R(e) = R(S ∪ y . The same statements hold when the colours are reversed.Proof.Let m blue = |S blue | and m red = |S red |.If δ < ε 1/9 2 and H(S ∪ e) = H(S ∪ e ′ ) for all e, e ′ ∈ J S of the same colour.In particular, if k = 4, then the edges in J induce only one red and one blue tight component in H.The same statement holds when the colours are reversed.Proof.Set J red S = G red S [U].Note that for e, e ′ ∈ J red S , we have e, e ′ ∈ (R G ) S and thus H(S ∪ e) = H(S ∪ e ′ ) = R since H(R G ) ⊆ R. Therefore to prove the lemma, it suffices to prove that there exists J blue S ⊆ G blue S [U] such that J red S + J blue S ≥ |G S [U]| − 7δ 1/4 n 2 and H(S ∪ e) = H(S ∪ e ′ ) for all e, e ′ ∈ J blue S .
that |W | ≥ ηn by the initial assumption.Without loss of generality, |W blue (M 0 )| ≤ |W red (M 0 )|.We define M be the set of matchings M in R ∪ B such that (i ′ ) R(e) = R and B(e ′ ) = B for all edges e ∈ G red [W ] and e

Claim 7 . 4 .
Let γ = α 1/1110 .There exists a red tight component R * and a blue tight component B * of H such that

3 and(
iii) R * = R or B * = B.Proof of Claim.First we show that, for each u ∈ U, there existsJ u ⊆ G u [U], where G u is the link graph of G at u, such that |J u | ≥ |G u [U]| − α 1/14n 2 and R(e ∪ u) = R(e ′ ∪ u) for e, e ′ ∈ J red u and B(e ∪ u) = B(e ′ ∪ u) for e, e ′ ∈ J blue u .To show this fix u ∈ U. Without loss of generality assume that u ∈ U red .By Lemma 7.2,
.1)We now colour each edge e of F as follows.Note that the link graph H e is a 2-graph.We induce a 2-edge-colouring on H e by colouring the 2-edge f ∈ H e with the colour of the k-edge e ∪ f ∈ H.By Corollary 5.6, there exists a monochromatic component in H e of order at least (1 − 3 √ ε)n.Let K e be such a component chosen arbitrarily.We colour the edge e according to the colour of K e .If e is red in F , then we define R(e) ⊆ H to be the red tight component containing all the edges e ∪ f where f ∈ K e .If e is blue in F , then we define B(e) analogously.
By (iii ′ ), we have that if xyz ∈ T and w ∈ N H (xyz, W * red ), then wxyz ∈ H blue .For xyz ∈ T , let B(xyz) be the maximal blue tight component containing all the edges xyzw, where w ∈ N H (xyz, W * red ).We say that xyz generates the blue tight component B(xyz).It suffices to show that all xyz ∈ T generate the same blue tight component.First we show that triples that share two vertices generate the same blue tight component.Note that, for xyz 1 , xyz 2 ∈ T , we have d H (xyz 1 , W red .Since the edges wxyz 1 and wxyz 2 are blue, it follows that B(xyz 1 ) = B(xyz 2 ).induce the same red tight component R G of G. Let V red = J red .Since R G is a red tight component of G all its edges induce the same red tight component R of H. Define V blue and B analogously.Two edges f and f ′ in H are loosely connected if there exists a sequence of edges e 1 , . . ., e t such that e 1 = f , e t = f ′ and |e i ∩ e i+1 | ≥ 1 for all i ∈ [t − 1].A subgraph H ′ of H is loosely connected if every pair of edges in H ′ is loosely connected.A maximal loosely connected subgraph of H is called a loose component of H.