4 JIE HAN, CHUANYUN ZANG, AND YI ZHAO
balanced and both H[T ] and H[S ∪T ] contain perfect matchings. These two definitions work very differently
– they are needed for the following two different absorbing lemmas.
Our first absorbing lemma is an analog of [23, Fact 2.3] for k-partite k-graphs.
Lemma 2.1 (Absorbing lemma I). Let 1/n α 1/k. Suppose H is a k-partite k-graph with parts of
size n such that δ
[k]\{i}
(H) ≥ n for i ∈ [3], then there exists a matching M
0
in H of size at most
√
αn such
that for every balanced 2k-set S of H, the number of S-absorbing edges in M
0
is at least αn.
Our second absorbing lemma deals with the case when only two partite minimum codegrees are large and
their sum is not significantly smaller than n.
Lemma 2.2 (Absorbing Lemma II). Let 1/n α 1/t 1/k. Suppose H is a k-partite k-graph
with parts of size n and a
i
:= δ
[k]\{i}
(H) for each i ∈ [k]. If
P
k
i=1
a
i
≥ (1 − )n, a
1
≥ a
2
≥ n and a
j
< n
for j ≥ 3, then one of the following holds.
(i) H is 2k
2
-D-extremal.
(ii) There exists a family F
0
of disjoint tk-sets such that |F
0
| ≤
√
αn, each F ∈ F
0
spans a matching of
size t and for every crossing k-set S of H, the number of S-perfect-absorbing sets in F
0
is at least
αn.
We first prove the following proposition, which is a standard application of Chernoff’s bound. We will
apply it in both proofs of Lemmas 2.1 and 2.2 for randomly sampling the absorbing sets.
Proposition 2.3. Let 1/n λ, 1/k, 1/i
0
. Let V be a vertex set with k parts each of size n, and let F
1
, . . . , F
t
be families of balanced i
0
k-sets on V such that |F
i
| ≥ λn
i
0
k
for i ∈ [t] and t ≤ n
2k
. Then there exists a family
F
0
⊆
S
i∈[t]
F
i
of disjoint balanced i
0
k-sets on V such that |F
0
| ≤ λn/(4i
0
k) and |F
i
∩F
0
| ≥ λ
2
n/(32i
0
k) for
each i ∈ [t].
Proof. We build F
0
by standard probabilistic arguments. Choose a collection F of balanced i
0
k-sets in H by
selecting each balanced i
0
k-set on V independently and randomly with probability p = /(2n
i
0
k−1
), where
= λ/(4i
0
k). Since t ≤ n
2k
, Chernoff’s bound implies that, with probability 1 −o(1), the family F satisfies
the following properties:
|F| ≤ 2p
n
i
0
k
≤ 2n
i
0
k
p = n and |F
i
∩ F| ≥
p
2
· λn
i
0
k
=
1
4
λn for any i ∈ [t].
Furthermore, the expected number of intersecting pairs of members in F is at most
p
2
n
i
0
k
· i
0
k · n
i
0
k−1
=
2
i
0
kn/4.
By Markov’s inequality, F contains at most
2
i
0
kn/2 intersecting pairs of i
0
k-sets with probability at least
1/2.
Let F
0
⊂ F be the subfamily obtained by deleting one i
0
k-set from each intersecting pair and removing
all i
0
k-sets that do not belong to any F
i
, i ∈ [t]. Therefore, |F
0
| ≤ |F| ≤ n and for each i ∈ [t], we have
|F
i
∩ F
0
| ≥ |F
i
∩ F| −
1
2
2
i
0
kn ≥
1
4
λn −
1
2
2
i
0
kn =
λ
2
32i
0
k
n
and we are done.
Now we prove our first absorbing lemma.
Proof of Lemma 2.1. We claim that for every balanced 2k-set S, there are at least
3
n
k
/2 S-absorbing
edges. Since there are at most n
2k
balanced 2k-sets, the existence of the desired matching follows from
Proposition 2.3.
Indeed, assume that {w, v} := S ∩ V
3
and u ∈ S ∩ V
2
. We obtain S-absorbing edges e = {v
1
, v
2
, . . . , v
k
}
as follows. First, for each j ∈ [4, k], we choose arbitrary v
i
∈ V
j
\ S – there are n − 2 choices for each v
j
.
Having selected {v
4
, v
5
, . . . , v
k
}, we select a neighbor of {u, v, v
4
, . . . , v
k
} as v
1
. Next, we choose a neighbor
of S
0
as v
2
, where S
0
is an arbitrary crossing (k − 1)-subset of S \ V
2
that contains w. Finally, we choose a
neighbor of {v
1
, v
2
, v
4
, . . . , v
k
} as v
3
. There are at least n −2 choices for v
j
for j = 1, 2, 3. Hence, there are
at least
(n − 2)
k−3
(n − 2)
3
≥
1
2
3
n
k
>
√
32kαn
k