The Ramsey Number of Fano Plane Versus Tight Path

The hypergraph Ramsey number of two $3$-uniform hypergraphs $G$ and $H$, denoted by $R(G,H)$, is the least integer $N$ such that every red-blue edge-coloring of the complete $3$-uniform hypergraph on $N$ vertices contains a red copy of $G$ or a blue copy of $H$. The Fano plane $\mathbb{F}$ is the unique 3-uniform hypergraph with seven edges on seven vertices in which every pair of vertices is contained in a unique edge. There is a simple construction showing that $R(H,\mathbb{F}) \ge 2(v(H)-1) + 1.$ Hypergraphs $H$ for which the equality holds are called $\mathbb{F}$-good. Conlon asked to determine all $H$ that are $\mathbb{F}$-good. In this short paper we make progress on this problem and prove that the tight path of length $n$ is $\mathbb{F}$-good.


The Ramsey number of the Fano plane versus the tight path 1 Introduction
Ramsey theory is one of the most intensively studied topics in combinatorics. Given two k-uniform hypergraphs G and H, we denote by R(G, H) the hypergraph Ramsey number of G and H. That is, R(G, H) is the least integer such that any red-blue edge-coloring of the complete k-uniform hypergraph on that many vertices contains a red G or a blue H as a subhypergraph. The existence of R(G, H) is guaranteed by Ramsey's theorem [15]. Given a bounded degree k-uniform hypergraph H, it is known that the Ramsey number R(H, H) is linear in the number of vertices of H [4,6,7,10,12]. However, estimating or even determining Ramsey numbers precisely is often a difficult problem.
In this short paper we will determine exactly the Ramsey number of the tight path and the Fano plane. This result is the first progress on a question asked at the AIMS workshop on hypergraph Ramsey problems in 2015 by Conlon [16]. A simple construction by Burr [1] shows that provided H is connected and v(G) σ(H), where χ(H) is the chromatic number of H and σ(H) is the size of the smallest color class in any χ(H)-coloring of H. Following Burr and Erdős [1,2], we will say that G is H-good, if (1) holds with equality. The intuition behind this definition was that H-good graphs tend to be poor expanders (see [5,Section 2.5] for further details). Denote by F the Fano plane, i.e. the unique 3-uniform hypergraph with seven edges on seven vertices in which every pair of vertices is contained in a unique edge. Conlon asked which hypergraphs are F-good [16].
In the graph case, there are many exact Ramsey numbers known. Erdős [2] started the systematic study of cliques versus large graphs. Nikiforov and Rousseau [13] gave a new approach to provide exact result for several families of graphs. Recently, the Ramsey number of the cycle and the clique (Keevash, Long and Skokan [11]), and the Ramsey number of the clique and the hypercube (Griffiths, Morris, Fiz Pontiveros, Saxton, Skokan [8]) have been determined. For similar results we refer the interested reader to the excellent recent survey [5]. In the hypergraph case there are only few instances where the Ramsey number is known exactly. Our result is the first Ramsey-goodness-result for hypergraphs.
From now on, we consider only 3-uniform hypergraphs. Let P t n be the tight path on n vertices, i.e. it contains distinct vertices v 1 , v 2 , . . . , v n and edges e 1 , e 2 , . . . , e n−2 where Theorem 1. There exists n 0 ∈ N such that for any n n 0 , we have R(P t n , F) = 2n − 1.
Using the definitions from above, Theorem 1 states that P t n is F-good. The lower bound R(P t n , F) 2n − 1 follows easily by the following folklore construction. Write the vertex set of the complete 3-uniform hypergraph on 2n − 2 vertices K (3) 2n−2 as the disjoint union of two sets A and B with |A| = |B| = n − 1. Color all 3-edges that are fully contained in either A or in B red, and all other edges blue. Observe that as |A|, |B| < n the electronic journal of combinatorics 27(1) (2020), #P1.60 there is no red P t n in this coloring. Since the chromatic number 1 χ(F) = 3, in every copy of F there is one edge that is fully contained in either A or B, hence this coloring cannot contain a blue copy of F either. This establishes that R(P t n , F) > 2n − 2. There is another almost extremal example, which is very different. For simplicity, let n be divisible by 6. Write the vertex set of the complete 3-uniform hypergraph on 2n−3 vertices K (3) 2n−3 as the disjoint union of three sets A,B and C with |A| = |B| = |C| = 2n/3 − 1.
Color red all triples of vertices {x, y, z} such that either x, y ∈ A and z ∈ A ∪ B, or x, y ∈ B and z ∈ B ∪ C, or x, y ∈ C and z ∈ A ∪ C. Color all other triples blue. A short case analysis (see Lemma 4) shows that there is no blue Fano plane F. The longest red tight path is obtained by alternating between taking two vertices from one of the sets and one vertex from another. Such a path can have length at most |A| + |B|/2 + 1 n − 1. The main contribution of our work is to establish the upper bound R(P t n , F) 2n − 1. In the proof of the upper bound, we will build up a picture of what a potentially bad coloring could look like and quickly realize that a bad coloring needs to be close to one of the two previous described colorings. In the remainder of the proof we then rule out these two types separately.
We remark that this proof technique also works for proving that the tight cycle is Fgood. Let C Theorem 2. There exists n 0 ∈ N such that for any n n 0 , we have R(C We choose not to present the proof of Theorem 2 since the proof is almost the same as the proof of Theorem 1 and differs only in some technicalities which do not give the reader more insight into the methods used.
The organization of the paper is as follows. In Section 2 we will show that Theorem 1 is sharp in the sense that the Ramsey number increases when one adds a small number of edges to the tight path. In Section 3 we will give some definitions and basic tools which will be needed for the proof of the upper bound in Theorem 1. The proof itself will be given in Section 4.

Sharpness example
The following example shows that Theorem 1 is best possible in the following sense: Let P ′ be the 3-uniform hypergraph obtained from the tight path P t n by adding three edges We claim that Ramsey number of P ′ and F is bigger than the Ramsey number of P t n and F.
Assume, for simplicity, that n is divisible by 3. Take three sets A, B, C of size 2n/3 each. Color red all triples of vertices {x, y, z} such that either x, y ∈ A and z ∈ A ∪ B, or x, y ∈ B and z ∈ B ∪ C, or x, y ∈ C and z ∈ A ∪ C. All other triples are colored blue. The following two Lemmas check that there is no blue F and no red P t n .
Lemma 4. The coloring described above does not contain a blue F.
Proof. Let F be a blue Fano plane in the previously described coloring. For i, j, k ∈ [7] with i + j + k = 7, we say that a Fano plane is of type (i, j, k) if i vertices come from A, j vertices come from B and k vertices come from C. Let a 1 , . . . , a i be the vertices in A; b 1 , . . . , b j be the vertices in B; and c 1 , . . . , c k the vertices in C. Since the chromatic number of the Fano plane is three, F needs to have one vertex from each set. Thus, F has to be of type (5, 1, 1), (4, 2, 1), (4, 1, 2), (3, 2, 2) or (3, 3, 1) up to rotation. Since every 5-subset of vertices in a Fano plane contains an edge, F cannot be of type (5, 1, 1). If F is of type (4, 2, 1), then b 1 b 2 c 1 forms an edge or there is an edge inside A, because F has chromatic number three. However, by construction of the coloring both of these edge are red. If F is of type (4, 1, 2), then again c 1 c 2 b 1 forms an edge or there is an edge inside A. Since the edge inside A is red, c 1 c 2 b 1 forms an edge. In a Fano plane every pair is in exactly one edge, thus b 1 cannot be in any further blue edge. This contradicts that every vertex is in three edges. Now, let F be of type (3,2,2). There has to be an edge inside {b 1 , b 2 , c 1 , c 2 }. Without loss of generality let this edge be c 1 c 2 b 1 . Now, b 1 can only be in one further edge (one containing b 2 ). This again contradicts that every vertex is in three edges. Finally, let F be of type (3, 3, 1). Again, there has to be an edge inside Without loss of generality let this edge be b 1 b 2 c 1 . Now, c 1 can only be in one further edge (one containing b 3 ), contradicting that c 1 needs to have degree three.
Lemma 5. The previously described coloring does not contain a red P t n .
Proof. Let us assume there is an embedding of a red P ′ . The first three vertices of such an embedding cannot come from different sets A, B and C. Without loss of generality, let A be the set which contains at least two of them. The only way to embed a red copy of P t n is to use all vertices of A and n/3 vertices of B. Since between 2 vertices from B there has to be at least 2 vertices from A, the only way for an embedding of P t n to start is with the first 6 vertices having the following patterns: AABAAB, ABAABA, ABAAAB, BAABAA, BAAABA or BAAAAB. However, regardless of which pattern we use, the resulting red tight path cannot be extended to a red copy of P ′ : one of {v 2 , v 3 , v 6 }, {v 1 , v 2 , v 5 } and {v 1 , v 4 , v 6 } would be of the form BBA and therefore blue.

Preparations
Let the hyperedges of H := K 2n−1 be two-colored with colors red and blue, without a blue F. In the proof we will build up a picture of how this bad coloring could potentially look like over a sequence of Lemmas and eventually rule out its existence entirely. Our starting point in the proof of Theorem 1 will be an upper bound on the off-diagonal hypergraph Ramsey numbers. We choose to use an upper bound from [3], but any weaker bound would suffice.
the electronic journal of combinatorics 27(1) (2020), #P1.60 Theorem 6 (Conlon-Fox-Sudakov [3]). There exists C > 0 so that for every integer s 4 and sufficiently large t, In the proof we will make use of the following definitions.
Definition 7. Given two disjoint sets A, B of vertices in H, we say four vertices a 1 , the graph on W with ab being an edge iff abv is blue in H. Analogously, G red v,W defines the red link graph.
For t ∈ N, we define the complete directed bipartite graph − → K t,t to be the directed graph on vertex set A ∪ B with |A| = |B| = t, A and B disjoint, and the arc set The following theorem is a directed version of the Kövári-Sós-Turán Theorem [14].
The following tools consisting of the next two Lemmas will be used multiple times in the main proof. 2n−1 be two-colored with colors red and blue, without a blue F. Further, let m ∈ N be big enough and A, B, C ⊆ V (H) be disjoint sets such that |A| = |B| = |C| = m. Assume that there are at most 1000 vertex-disjoint red butterflies connecting each pair of the three sets A, B, C.
Then there exists an absolute constant t > 0 such that Proof. Removing at most 4000 vertices from each set, we end up with sets A 1 ⊂ A, A 2 ⊂ B, A 3 ⊂ C so that there are no red butterflies connecting them. Note that if two vertices a ∈ A i and b ∈ A j are not contained in any red butterfly, then either all hyperedges the set of hyperedges {uvy : y ∈ A j } is entirely blue. Note that some edges might be oriented in both ways.
the electronic journal of combinatorics 27(1) (2020), #P1.60 Let t be the bipartite Ramsey number for K 4,4 . That is, t is the least integer such that every two-coloring of the edges of K t,t contains a monochromatic copy of K 4,4 . Note that by Irving [9], we have t 48.
Suppose between A 1 and A 2 , both the left and right density is at least C ′ m −1/t for C ′ being a constant large enough. By Theorem 8 we can find complete directed Since x was an arbitrary vertex in D 1 , every vertex in D 1 has an out-neighbourhood in A 3 of size at most t. Repeating the same argument after replacing D 1 by E 2 , we get that every vertex in E 2 has an out-neighbourhood in A 3 of size at most t. So by removing at most 2t 2 vertices from A 3 we get a set A ′ 3 of the property that all edges from A ′ 3 to D 1 and all edges from By the choice of t, we can find sets W 1 ⊂ D 1 and W 2 ⊂ E 2 of sizes at least four such that without loss of generality (W 1 , W 2 ) forms a directed K 4,4 . Hence we have found a 4-blowup of a transitive triangle. But this is impossible, as letting This proves that in − → G between A 1 and A 2 , in one of the directions the density has to be less than C ′ m −1/t . Repeating this argument for the other two pairs, we get that between any pair of sets from A 1 , A 2 , A 3 , in one of the directions the density has to be less than C ′ m −1/t whereas the density in the other direction has to be at least 1 − C ′ m −1/t . The majority orientation forms a transitive triangle or an oriented 3-cycle. Suppose now that the majority orientation forms a transitive triangle. Pick four vertices from each set at random. Then the probability that the 12 vertices do not form a 4-blowup of the transitive triangle is at most 48C ′ m −t . Therefore, there exists a 4-blowup of a transitive triangle in − → G , giving a blue Fano plane in H. Thus, the majority orientation has to form a 3-cycle.
Without loss of generality let 2n−1 be two-colored with colors red and blue, without a blue F. Further, let m ∈ N be big enough and A, B, C ⊆ V (H) be disjoint sets such that |A| = |B| = |C| = m. Assume that there are at most 1000 vertex-disjoint red butterflies connecting each pair of the three sets A, B, C.
Then there exists an absolute constant t > 0 such that Proof. Applying Lemma 9, we get a positive constant t ′ such that w.l.o.g.
Let t = 3t ′ . For the sake of contradiction, say that | −→ AC| b m 3−1/t . Define 99 100 Then |Y 1 | 4/5m as otherwise | − − → BC| r 2m 3−1/t ′ . Because of the size of Y 1 , G blue v,B has to contain an edge inside Y 1 . Let w 1 w 2 be such an edge. The number of 4-tuples (a, b, c, d) of distinct vertices a, b, c, d ∈ C with ab, cd ∈ E(G blue v,C ) is at least ; ad, bc ∈ E(G blue w 1 ,C ) and ac, bd ∈ E(G blue w 2 ,C ). Thus, the hyperedges vw 1 w 2 , vab, vcd, w 1 ad, w 1 bc, w 2 ac, w 2 bd form a blue Fano plane; a contradiction, therefore we conclude that without a blue F and without a red P t n . Fix such a coloring. Let ε > 0 be a sufficiently small constant and assume that n is sufficiently large. Set m = ε 5 log n log log n .
Observe that m 5 log m ε 5 5 log n, hence we have by Theorem 6 7 ) 2 Cm 5 log m n ε 4 .
Since H contains no blue F, it cannot contain a blue K and we conclude that it contains a red K In the next two Subsections the two cases from Lemma 12 will be handled separately. The strategy is to build a long red tight path using these two or three blocks.
Remark 13. For the proof of Theorem 2 one can use the fact that when a long path say starting in A 1 and ending in A a is found, it is clear that some of the vertices inside the block can be used to close the cycle.

The two paths case
In this case G 1 is the vertex-disjoint union of two paths, i.e. in H we have vertex-disjoint red K m -s {A 1 , . . . , A a , B 1 , . . . , B b } and a set J of junk vertices with |J| n ε 4 . For every i, j there are at least 1000 red butterflies between A i and A i+1 and also between B j and B j+1 . Slightly abusing notation, let P 1 = ∪ i A i and P 2 = ∪ j B j . Note that if |P i | n for some i ∈ {1, 2} then we can embed the tight path P t n into P i just by walking through each blob and jumping from blob to blob by using the hyperedges from the red butterflies. We know that Definition 14. A red triple triangle between A i and B j is a set of vertices w, x, y, z ∈ A i and v ∈ B j (or w, x, y, z ∈ B j and v ∈ A i ) so that wxv, xyv, yzv is red in H.
Observe that when we have a red triple triangle between A i and B j we can find a red tight path of length m + 1 by swallowing one additional vertex from B j using the red triple triangle. If there is no red triple triangle between two blobs then there also have to be few red hyperedges between the blobs.

Lemma 15. If there is no red triple triangle between
Proof. Pick any vertex v ∈ B j and consider its red link graph in A i . If v is not in a red triple triangle, then the red link graph does not contain a path of length 3, hence the number of edges in this link graph is at most 10m. So the number of red hyperedges between B j and A i , assuming that there are no red triple triangles, is at most 20m 2 . edges from the matching have to go in the same direction. In particular since |P i | n−n ε 4 for i = 1, 2 this implies |M | 2n ε 4 . Put all blobs covered by M into J and get a new rubbish set J ′ . We will have |J ′ | |J| + 2|M |m n ε 3 vertices. The subgraph of G 2 on the blobs which have not been removed spans an independent set. Let A be the set of vertices in P 1 which have not been removed and let B be the vertices in P 2 which have not been removed. The following argument shows that for three different blobs For contradiction, assume there are more than that many blue hyperedges. Take a blob B i which has not been removed from {B 1 , . . . , B b }. By Lemma 15 Picking at random one vertex each of A ′ 1 , A ′ 2 and A ′ 3 , and 4 vertices from B i , these vertices do not form a blue Fano plane with probability at most 1 − 400m −1 + 6 · 50m −1 , thus, there has to exist a blue Fano plane. We conclude Therefore there are at most 400m 2 (n/m) 3 + m 3 (n/m) 2 500n 3 /m blue hyperedges inside A. Similarly, this holds for B. is more than 2ε 100 n 4 . We will now walk along the red hyperedges step by step adding in each step 5 vertices to the tight path. Let v 1 , v 2 , . . . be the special vertices in B. Let (a 1 , b 1 ) ∈ Good(v 1 ) and (c 1 , d 1 ) be an open tuple such that (c 1 , d 1 ) is reachable from (a 1 , b 1 ) via v 1 . We begin the walk with a 1 , b 1 , v 1 , c 1 , d 1 . Now take a look at step i. Assume we already have defined a 1 , b 1 , v 1 , c 1 , d 1 , a 2 , . . . , a i−1 the electronic journal of combinatorics 27(1) (2020), #P1.60 We can remove all special vertices from A and B and add them to the junk set J ′ . So we obtain A ′ , B ′ , J ′′ so that for each v ∈ A ′ the red link graph in B ′ has at most 1 5 ε 5 n 2 edges, for each w ∈ B ′ the red link graph of w in A ′ has at most 1 5 ε 5 n 2 edges and Proof. Let V (H) = A ′ ∪ B ′ ∪ J ′′ be the decomposition from Lemma 19. We actually will prove that if a rubbish vertex v ∈ J ′′ has e(G blue v,A ′ ) εn 2 , then it cannot have a blue hyperedge into B ′ . Indeed suppose there are a, b ∈ B ′ with abv blue. Then both a and b are part of at least |A ′ | 2 − 1 5 ε 5 n 2 blue hyperedges with the other two vertices being in A ′ by the final statement in Lemma 19. Therefore there are at least |A ′ | 2 − 2 5 ε 5 n 2 pairs (c, d) with c, d ∈ A ′ and cda, cdb blue, call this set of edges S. Now suppose v leads at least εn 2 blue hyperedges into A ′ , i.e. e(G blue v,A ′ ) εn 2 and hence G blue v,A ′ contains a path P of length ε 2 n. The restriction of S onto the vertex set of P = {p 1 , p 2 , . . .} contains at least (1 − ε) |P | 2 edges and hence contains four vertices p i , p i+1 , p j , p j+1 with p i p j , p i p j+1 , p i+1 p j , p i+1 p j+1 in S. Then ap i p j+1 , avb, ap j p i+1 , p j+1 bp i+1 , p j+1 vp j , p i vp i+1 , p i p j b form a blue Fano plane. Hence we can split up the junk set J ′′ = J 1 ∪ J 2 such that for all v ∈ J 1 e(G blue v,A ′ ) εn 2 and all v ∈ J 2 e(G blue v,B ′ ) εn 2 .
Lemma 21. In the setting of Lemma 19, we can find a red P t n in H.
the electronic journal of combinatorics 27(1) (2020), #P1.60 Proof. Let J ′′ = J 1 ∪ J 2 be the decomposition of the junk vertices from Lemma 20. Set A * = A ′ ∪ J 1 and B * = B ′ ∪ J 2 . Then either |A * | n or |B * | n. W.l.o.g. |A * | n. Now one can find a red tight path of length n inside A * . Let v 1 , v 2 , . . . be the vertices from J 1 . Call a tuple (a, b), a, b ∈ A ′ good* for v ∈ J 1 if the red link graph of v in A ′ contains at least 9 10 n 2 tuples (c, d), c, d ∈ A ′ such that abv, bvc, vcd are red in H. Since the blue link graph of v contains at most εn 2 edges, for each v ∈ J 1 the number of good* tuples is at least 9 10 n 2 . Now start with an arbitrary good* tuple (a 1 , b 1 ) for v 1 . The start of the walk is a 1 , b 1 , v 1 . Now assume we already have chosen a 1 , b 1 , v 1 red. This is possible for all i εn, because 9 10 n 2 − 1 10 Enlarge the path by a i b i v i . After all vertices from J 1 are used, just walk through A ′ until all vertices in A ′ are used. This is possible, because all hyperedges inside A ′ are red. Thus, we find a red tight path of length |A * | n. . . , C c } be the three paths in G 1 . There cannot be an edge between blobs of different paths, otherwise we can reduce this case to the two path case from Section 4.2 in the following way. Without loss of generality, assume that there is an edge between A j and B k . Split up each blob from A and B into two blobs of equal size (if m is odd one vertex ends up in J) in such a way that blobs coming from consecutive blobs still have at least 100 disjoint red butterflies between them and such that there are also still at least 100 disjoint red butterflies between the blobs coming from A j and B k . Now, when one constructs the graph of all new blobs, where two blobs are adjacent with each other when they have at least 100 disjoint red butterflies between them, then this graph can be decomposed into two paths. We already handled this case in Subsection 4.2. Therefore we can assume that there are no edges between blobs of different paths in G 1 .

The three paths case
Using Lemma 10 it follows that w.l.o.g.
for all i, j, k. Let P 1 = ∪A i , P 2 = ∪B i and P 3 = ∪C i . Clearly, |P 1 |, |P 2 |, |P 3 | < n as otherwise one could find a red tight path of length n just by going through a blob and then jumping to the next by using a red butterfly and so on.
Definition 22. For X, Y ⊂ V (H) and 0 α 1, denote G(X, Y, α) the graph with vertex set X, and ab is an edge iff the number of red hyperedges abc with c from Y is at least α|Y |.
Lemma 23. There exists a constant C ′ such that the electronic journal of combinatorics 27(1) (2020), #P1.60 Proof. Let A i ∈ A, B i ∈ B. We will show that one can find a red tight path of length at least 3m/2 − 4000m 1−1/t just using A i and B i starting with two vertices from A i , ending with two vertices from A i , not using two vertices being part of a butterfly to A i−1 and not using two other vertices being part of a butterfly to A i+1 . Consider the graph G(A i , B i , 0.99). The number of vertices in this graph with degree at most 0.9m is at most 2000m 1−1/t as otherwise | −−→ B i A i | r > 2000m 1−1/t · 0.1m · 0.01m · 0.5 = m 3−1/t . Let A ′ ⊆ A i be the set of all vertices of degree at least 0.9m not containing the two vertices being part of a butterfly to A i−1 and not containing the two vertices being part of a butterfly to A i+1 . Then |A ′ | m − 2000m 1−1/t − 4, G(A ′ , B i , 0.99) has minimum degree at least 0.8m and thus there exists a Hamiltonian path v 1 , v 2 , . . . , v A ′ in this graph. After every second vertex in this path we will now add a vertex from B i to find a long red tight path in A i ∪ B i . Assume we already found the tight red path v 1 , v 2 , w 1 , v 3 , v 4 , w 2 , . . . , v 2j−1 , v 2j . Then we can pick a vertex w j ∈ B i which has not been used yet and such that v 2j−1 v 2j w j , v 2j w j , v 2j+1 , w j , v 2j+1 v 2j+2 are red for j < |A ′ |/2. This is possible because m − 0.01m − 0.01m − 0.01m − j > 0. Thus, we can find a red tight path of length at least If |P 1 | |P 2 |, then we can find a tight red path of length at least 3/2|P 1 | − 5000nm −1/t by the following argument. Jump from blob to blob in A using the vertices from the butterflies and always absorbing the vertices from the index corresponding blob in B.
When we are done with all blobs in A we stop and have found a red tight path of length at least 3 2 m − 4000m 1−1/t |P 1 | m If |P 1 | |P 2 |, then we can find a tight red path of length at least (|P 1 | + |P 2 |/2) − 5000nm −1/t by the following argument. Jump from blob to blob in A using the vertices from the butterflies and always absorbing the vertices from the index-corresponding blob in B. When we are done with all blobs in B we go back to A and walk through the remaining blobs from A using the butterflies. So we can find a tight red path of length at least 3 2 m − 4000m 1−1/t |P 2 | m + |P 1 | − |P 2 | |P 1 | + |P 2 | 2 − 5000nm −1/t .