A NOTE ON PERFECT MATCHINGS IN UNIFORM HYPERGRAPHS

We determine the exact minimum `-degree threshold for perfect matchings in k-uniform hypergraphs when the corresponding threshold for perfect fractional matchings is significantly less than 1 2 ( n k−` ) . This extends our previous results that determine the minimum `-degree thresholds for perfect matchings in k-uniform hypergraphs for all ` ≥ k/2 and provides two new (exact) thresholds: (k, `) = (5, 2) and (7, 3).

A connection between m (k, n) and the minimum -degree that forces a perfect fractional matching was discovered in [1]. Let H be a k-uniform hypergraph on n vertices. A fractional matching in H is a function w : E(H) → [0, 1] such that for each v ∈ V (H) we have that e v w(e) ≤ 1. Then e∈E(H) w(e) is the size of w. If the size of the largest fractional matching w in H is n/k then we say that w is a perfect fractional matching. Given k, ∈ N such that ≤ k − 1, define c * k, to be the smallest number c such that every k-uniform hypergraph H on n vertices with δ (H) ≥ (c+o(1)) n− k− contains a perfect fractional matching. It is easy to see that the hypergraph H * (n, k) defined earlier contains no perfect fractional matching. Thus c * k, ≥ 1 − k−1 k k− . Alon et al. [1,Theorem 1.1] showed that for fixed k, , as n ∈ kN tends to infinity, Furthermore, in [1] the authors conjectured that c * k, = 1− k−1 k k− and confirmed this for ≥ k−4.
Let us highlight the ideas behind the proof of Theorem 2. It is informative to first recall the proof of (1), which is an application of the absorbing method. The authors of [1] first applied a lemma of Hàn, Person and Schacht [5,Lemma 2.4], which states that every k-uniform hypergraph either contains the aforementioned M or looks like a hypergraph in H ext (n, k). If H contains M , then we proceed as in [1] (except that we apply a lemma from [11] when converting a fractional matching to an integer matching); if H looks like a hypergraph in H ext (n, k) and δ (H) ≥ δ(n, k, ) + 1, then we obtain a perfect matching by applying [19,Theorem 4.1].
Theorem 5 is the main contribution of this note -it is stronger than two absorbing theorems in our previous papers [19,Theorem 3.1] and [20,Theorem 3.1], in which we assume that ≥ k/2. The proof of Theorem 5 is actually shorter than those of the two previous absorbing theorems because 1) we use a different absorbing structure which allows us to apply a lemma from [13]; 2) when proving Lemma 10, we avoid using auxiliary hypergraphs and obtain the structure of H by considering the neighborhoods of the vertices of H directly.
Notation: Given a set X and r ∈ N, we write X r for the set of all r-element subsets of X. Let H be a k-uniform hypergraph. We When |A| = n/2 and |B| = n/2 , we define B n,k := B n,k (A, B) and B n,k := B n,k (A, B).
We will often write 0 < a 1 a 2 a 3 to mean that we can choose the constants a 1 , a 2 , a 3 from right to left. More precisely, there are increasing functions f and g such that, given a 3 , whenever we choose some a 2 ≤ f (a 3 ) and a 1 ≤ g(a 2 ), all calculations needed in our proof are valid. Hierarchies with more constants are defined in the obvious way.

Proof of Theorem 2
The lower bound for m (k, n) in Theorem 2 follows from the definitions of δ(n, k, ) and c * k, immediately. The following (more general) result provides the desired upper bound for m (k, n). Theorem 4. Given any θ > 0, k, , ∈ N where 1 ≤ , ≤ k − 1 there is an n 0 ∈ N such that the following holds. Let n ≥ n 0 where k divides n. If H is a k-uniform hypergraph on n vertices with then H contains a perfect matching.
The proof of Theorem 4 splits into extremal and non-extremal cases, the former case being when H looks like an element of H ext (n, k). To make this precise we introduce more notation. Let ε > 0. Suppose that H and H are k-uniform hypergraphs on n vertices. We say that H is ε-close to H if H becomes a copy of H after adding and deleting at most εn k edges. More precisely, Our proof of the non-extremal case uses the absorbing method. Given a k-uniform hypergraph and H[S ∪ Q] contain perfect matchings. In this case, if the matching covering S is M , we also say M absorbs Q.
Our main result, Theorem 5, extends [20,Theorem 3.1]. It ensures that if H is as in Theorem 4 then H contains a small absorbing matching or H is close to one of B n,k and B n,k . We postpone its proof to the next subsection.
Theorem 5. Given any ε > 0 and integer k ≥ 2, there exist 0 < α, ξ < ε and n 0 ∈ N such that the following holds. Suppose that H is a k-uniform hypergraph on n ≥ n 0 vertices. If The next result from [19] ensures a perfect matching when our hypergraph H is close to one of the extremal hypergraphs B n,k and B n,k . Theorem 6. [19, Theorem 4.1] Given 1 ≤ ≤ k − 1, there exist ε > 0 and n 0 ∈ N such that the following holds. Suppose that H is a k-uniform hypergraph on n ≥ n 0 vertices such that n is divisible by k. If δ (H) > δ(n, k, ) and H is ε-close to B n,k or B n,k , then H contains a perfect matching.
The final tool required for the proof of Theorem 4 is a weaker version of Lemma 5.6 in [11]. Lemma 7. [11] Let k ≥ 2 and 1 ≤ ≤ k − 1 be integers, and let ε > 0. Suppose that for some b, c ∈ (0, 1) and some n 0 ∈ N, every k-uniform hypergraph H on n ≥ n 0 vertices with δ (H) ≥ cn k− has a fractional matching of size (b + ε)n. Then there exists an n 0 ∈ N such that any k-uniform hypergraph H on n ≥ n 0 vertices with δ (H) ≥ (c + ε)n k− contains a matching of size at least bn.
Proof of Theorem 4. Choose ε > 0 from Theorem 6. We may additionally assume that ε θ, 1/k. Let 0 < α, ξ < ε be as in Theorem 5. Let n be sufficiently large and divisible by k. Assume that H is a k-uniform hypergraph on n vertices satisfying (2) where the last inequality follows since ξ θ, 1/k. Let c := c * /(k − )! + ξ 2 . By the definition of c * , for sufficiently largeñ, every k-uniform hypergraph F onñ vertices with δ (F ) ≥ cñ k− contains a perfect fractional matching. Applying Lemma 7 with ξ 2 /k and (1 − ξ 2 )/k playing the roles of ε and b respectively, we conclude that H contains a matching M of size at least (1 − ξ 2 )n 1 /k. Let W be the uncovered vertices of H . Then |W | ≤ ξ 2 n. We finally absorb W using the absorbing property of M .
2.1. Proof of Theorem 5. The proof of Theorem 5 follows from the following three lemmas. Lemma 8 is a special case of [13, Lemma 1.1] and gives a sufficient condition for a hypergraph H to contain a small matching that absorbs any much smaller set of vertices from H.
Lemma 9. Let k ∈ N and 0 < γ γ 1/k. There exists an n 0 ∈ N such that the following holds. Let H = (V, E) be a k-uniform hypergraph on n ≥ n 0 vertices. Suppose that for every x, y ∈ V at least one of the following conditions holds.
Lemma 10. Let k ∈ N and 0 < α γ ε, 1/k. Then there exists an n 0 ∈ N such that the following holds. Let H = (V, E) be a k-uniform hypergraph on n ≥ n 0 vertices such that Then H is ε-close to B n,k or B n,k .
Proof of Lemma 9. Note that (i) and (ii) imply that δ 1 (H) ≥ γn k−1 and so e(H) ≥ γn k /k. Consider any x, y ∈ V . First assume that (i) holds. Fix any X ⊆ V where |X | = k − 1 and X ∪ {x}, X ∪ {y} ∈ E. By (i) there are at least γn k−1 choices for X . Next choose some X ⊆ V \(X ∪{x, y}) such that |X | = k and X ∈ E. There are at least γn k /k−(k+1) n k−1 ≥ γn k /(2k) choices for X . Set X := X ∪ X . Note that both H[X ∪ {x}] and H[X ∪ {y}] contain perfect matchings. Further, since there are at least γn k−1 choices for X , at least γn k /(2k) choices for X and each (2k − 1)-set may be counted at most 2k−1 k−1 times, there are at least choices for X (as γ γ 1/k), as desired. Now suppose that (ii) holds. Fix any z ∈ V such that |N H (x) ∩ N H (z)| ≥ γn k−1 and |N H (y) ∩ N H (z)| ≥ γn k−1 . There are at least γn choices for z. Next fix some X ∈ N H (x) ∩ N H (z) that is disjoint from y. There are at least γn k−1 − n k−2 ≥ γn k−1 /2 choices for X . Finally, fix some X ∈ N H (y) ∩ N H (z) so that X is disjoint from X ∪ {x}. There are at least γn k−1 − k n k−2 ≥ γn k−1 /2 choices for X . Set X := X ∪ X ∪ {z}. So |X| = 2k − 1 and both H[X ∪ {x}] and H[X ∪ {y}] contain perfect matchings. Further, there are at least choices for X (as γ γ 1/k), as desired.
The rest of this section is devoted to the proof of Lemma 10. We draw on ideas used in the proof of Lemma 5.4 in [20]. We need two results from [20]. The first one implies that if any two vertices in a hypergraph have roughly the same neighborhood, then the hypergraph is near complete or empty.
Lemma 11. [20, Lemma 2.2] Given any k ∈ N and ρ > 0 there exists an n 0 ∈ N such that the following holds. Let F = (V, E) be a k-uniform hypergraph on n ≥ n 0 vertices with edge Proof of Lemma 10. Define Then by Lemma 10 (i), x 0 ∈ X and y 0 ∈ Y . Let V 0 := V \ (X ∪ Y ). We have |V 0 | ≤ γn by Lemma 10 (ii). Roughly speaking, our goal is to show that |X| ≈ |Y | ≈ n/2 and H ≈ B n,k (X, Y ) or H ≈ B n,k (X, Y ). We first provide several properties of X and Y , for example, Claim 13. The following conditions hold. ( Consequently, we have d H (x), d H (y) ≤ (1/2 + α) n−1 k−1 + γn k−1 . To see (ii), suppose that there exists v ∈ X ∩ Y . Then by (3), which implies that |N H (v)| ≤ 4γn k−1 , contradicting the minimum degree condition of H. To see (iii), consider x ∈ X. By the definition of X and the minimum degree condition of H, The same bound holds for |N H (x) \ N H (x )|. Hence Analogously we can derive that |N H (y) N H (y )| < 5γn k−1 for any y, y ∈ Y . To see (iv), consider x ∈ X and y ∈ Y . By (4), we have which proves the first assertion of (iv). By the minimum degree condition and (5), we have

It follows that
which proves the second assertion of (iv).
Since |V 0 | ≤ γn and X ∩ Y = ∅ we have |X| ≥ (1 − γ)n/2 or |Y | ≥ (1 − γ)n/2. Without loss of generality we may assume that |X| ≥ (1 − γ)n/2 ≥ n/3. Let 0 < γ 0 < 1/2 such that for any v, v ∈ X (Claim 13 (iii)), there are two possible cases: In the rest of the proof we assume that one of the two cases holds. Once we have obtained more information we will prove that |X| and |Y | are close to n/2. At present we require the following weaker lower bounds on |X| and |Y |.
The bound on |X| follows since |X| ≥ (1−γ)n/2. Since |X|+|Y |+|V 0 | = n and |V 0 | ≤ γn, to prove the bound on |Y |, it suffices to show that |X| ≤ ( 1 2 + 2γ 0 ) Together with the minimum degree condition, this gives In either case we have (1 − γ 0 ) |X|−1 Given two disjoint subsets A, B ⊂ V and two integers i, j ≥ 0, we call an (i + j)-set S ⊆ V an A i B j -set if |S ∩ A| = i and |S ∩ B| = j, and let A i B j denote the family of all A i B j -sets. Let Proof. We prove both cases by induction on i. In Case 1 there exists a vertex Fix a vertex y ∈ Y . By Claim 13 (iv), |N H (x 1 ) ∩ N H (y)| ≤ 4γn k−1 . Thus at least (1 − γ 0 ) |X|−1 k−1 − 4γn k−1 X k−1 -sets are neighbors of y. By Claim 14, |X|, |Y | ≥ c 0 n. Then for any 0 ≤ i ≤ k − 1, Together with the definition of γ 1 , we conclude that at least -sets are non-neighbors of y. This confirms (ii) for i = 1.
Since γ ε, 1/k, we have that γ k , η ε. Therefore, which implies that H is ε-close to B n,k (Y , X ). Analogously we can show that H is ε-close to B n,k (Y , X ) in Case 2. This completes the proof of Lemma 10.
3. An application of Lemma 9 The following simple application of Lemma 9 implies that the minimum -degree condition that forces a perfect fractional matching also forces a perfect matching in a k-uniform hypergraph H, if we additionally assume that H has a small number of vertices of large degree.
Theorem 17. Given any 0 < ε ≤ δ and k, ∈ N where < k, there is an n 0 ∈ N such that the following holds. Let H be a k-uniform hypergraph on n ≥ n 0 vertices where k divides n. Suppose that δ 1 (H) ≥ δ n−1 k−1 and δ (H) ≥ (c * k, + ε) n− k− . If there are at least εn vertices x ∈ V (H) so that d H (x) ≥ (1 − δ + ε) n−1 k−1 then H contains a perfect matching. Sketch proof. It is easy to see that H satisfies Lemma 9 (ii) (where we choose 0 < γ ε) and so by Lemma  and so by Lemma 7, H contains a matching covering all but a very small set of vertices. After absorbing the uncovered vertices by M , we obtain a perfect matching in H.