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arXiv:2409.15140v1 [math.CO] 23 Sep 2024
Bisection Width, Discrepancy, and Eigenvalues of Hypergraphs
Eero R¨aty∗
, Istv´an Tomon†
Abstract
A celebrated result of Alon from 1993 states that any d-regular graph on nvertices (where
d=O(n1/9)) has a bisection with at most dn
2(1
2−Ω( 1
√d)) edges, and this is optimal. Recently,
this result was greatly extended by R¨aty, Sudakov, and Tomon. We build on the ideas of the
latter, and use a semidefinite programming inspired approach to prove the following variant for
hypergraphs: every r-uniform d-regular hypergraph on nvertices (where d≪n1/2) has a bisection
of size at most dn
r1−1
2r−1−c
√d,
for some c=c(r)>0. This bound is the best possible up to the precise value of c. Moreover, a
bisection achieving this bound can be found by a polynomial-time randomized algorithm.
The minimum bisection is closely related to discrepancy. We also prove sharp bounds on the
discrepancy and so called positive discrepancy of hypergraphs, extending results of Bollob´as and
Scott. Furthermore, we discuss implications about Alon-Boppana type bounds. We show that
if His an r-uniform d-regular hypergraph, then certain notions of second largest eigenvalue λ2
associated with the adjacency tensor satisfy λ2≥Ωr(√d), improving results of Li and Mohar.
1 Introduction
1.1 Bisection width
An equipartition of a finite set is a partition into two parts, whose sizes differ by at most one. A
bisection of a graph Gis an equipartition of its vertex set, together with all the edges containing
one vertex in each part, and the size of a bisection is the number of its edges. The bisection width
or minimum bisection of a graph Gis the minimum size of a bisection, and it is denoted by bw(G).
Due to its importance in theoretical computer science, the algorithmic aspect of the bisection width
problem has been studied extensively. For the background and recent advances in the area, see e.g.
[21,22,26,27,31]
A fundamental result in this topic is due to Alon [1] (see also [2]), which states that every
d-regular graph on nvertices has bisection width at most nd
4−Ω(√dn) whenever d=On1/9. On
the other hand, Bollob´as [4] proved that the bisection width of almost all d-regular graph is at least
nd
4−n√dln 2
2, thus matching the bound of Alon [1] up to a constant factor in the error term. The
problem of improving these bounds for random d-regular graphs has been a line of active research in
the interface of combinatorics and theoretical computer science, for example due to its connections
with finding internal partitions. Due to the difficulty of the problem, a specific point of focus has
been on small constant values of d, see e.g. the works [7,11,12,29,33].
∗Ume˚a University, e-mail :eero.raty@umu.se. Supported by a postdoctoral grant from the Osk. Huttunen
Foundation.
†Ume˚a University, e-mail :istvan.tomon@umu.se. Supported by the Swedish Research Council grant 2023-03375.
1
Recently, R¨aty, Sudakov, and Tomon [36] substantially extended the result of Alon [1] in the
range d≤(1
2−ε)n. They proved that nd
4−Ω(√dn) remains an upper bound for the bisection width
if d=On2/3, thus random graphs minimize this quantity asymptotically. However, surprisingly, in
case n2/3≪d < 1
2−εn, the situation completely changes. When n2/3≤d≤n3/4, the maximum
of the bisection width among d-regular graphs is nd
4−Θn2/d, and random graphs are beaten by
certain families of strongly-regular graphs. The range n3/4≤d≤1
2−εnis even more mysterious,
we refer the interested reader to [36] for further details.
The bisection width also extends to hypergraphs naturally as follows. The bisection of a
hypergraph His an equipartition of its vertex set, together with all the edges that contain at
least one vertex in each part. The bisection width of a hypergraph His the minimum number of
edges in a bisection, and we again denote it by bw(H). The study of hypergraph bisection width has
attracted attention from the algorithmic point of view [15,25,35]. A particular topic of interest has
been the hypergraph-generalisation of the s-tcut problem. Here the aim is to find a minimum cut
(not necessarily bisection) of Hthat has two predetermined vertices sand tplaced on the different
sides of the cut [8].
In this paper, our goal is to study the minimum bisection problem for hypergraphs from an
extremal point of view, in particular to generalize the aforementioned results of Alon [1], and R¨aty,
Sudakov, and Tomon [36]. We remark that the case of 3-uniform hypergraphs can be reduced to
multigraphs, as the size of a bisection in a 3-uniform hypergraph is equal to half the size of the
corresponding bisection of the underlying multi-graph (i.e., the graph in which each edge is included
as many times as it appears in a hyperedge). However, no similar reduction is possible if the
uniformity is at least 4.
Let us consider the minimum bisection of an r-uniform hypergraph Hwith nvertices and average
degree d. The expected size of a random bisection in His
e(H)·1−1
2r−1+ Θr1
n,
so this is always a trivial upper bound for bw(H). On the other hand, if His the random hypergraph
in which every edge is included independently with probability p=d/n−1
r−1≤1/2, then the average
degree of His ≈dand
bw(H)≥e(H)·1−1
2r−1−O(√dn)
with high probability. See the next subsection for a detailed argument. As one of our main results, we
establish an upper bound for the bisection width of d-regular hypergraphs that matches the previous
lower bound, given dis not too large with respect to n.
Theorem 1.1. Let r≥2be an integer, then there exists c, ε > 0such that the following holds. Let
Hbe an r-uniform d-regular hypergraph on nvertices, where d≤εn1/2. Then
bw(H)≤e(H)·1−1
2r−1−c√dn.
Here, we remark that the same result is true if instead of regularity, we only assume that Hhas
average degree dand maximum degree O(d). Also, a bisection achieving this bound can be found
with a polynomial-time randomized algorithm. However, the result no longer holds without some
restriction on the maximum degree: if Gis the complete bipartite graph with vertex classes of size
2
d/2 and (n−d/2), then Ghas average degree d(1 −on(1)), but bw(G)≥1
2e(G). Furthermore, a
similar result no longer holds if d≫n2(which also assumes r≥4). Indeed, if His the random
hypergraph in which every edge is included independently with probability p=d/n−1
r−1≤1/2, then
bw(H) = e(H)·1−1
2r−1+ Θr(d). See the next section for further discussion.
Finally, we remark that Theorem 1.1 naturally extends to hypergraphs that are not necessarily
uniform. If His such a hypergraph, then a random bisection has size at least Pe∈E(H)1−21−|e|.
We show that if the maximum size of an edge is bounded by r, there is a bisection with significantly
less edges.
Theorem 1.2. Let r≥2be an integer, then there exists c, ε > 0such that the following holds.
Let Hbe a hypergraph on nvertices with medges such that each edge is of size at most r, and the
maximum degree is ∆, where ∆2≤εmn1/2. Then
bw(H)≤
X
e∈E(H)
1−21−|e|
−cm
√∆.
Note that this theorem immediately implies Theorem 1.1 by having ∆ = d.
1.2 Discrepancy
Let Hbe an r-uniform hypergraph with nvertices and edge density p=|E(H)|
(n
r). Given U⊂V(H),
define the discrepancy of Uas
disc(U) = e(U)−p|U|
r.
Then disc(U) measures how much Udeviates from its expected size. The discrepancy of His defined
as the maximum absolute discrepancy over all subsets of vertices, that is,
disc(H) = max
U⊂V(H)|disc(U)|.
This notion of discrepancy was introduced by Erd˝os, Goldbach, Pach and Spencer [13] in the 80’s,
extending earlier notions studied by Erd˝os and Spencer [14]. In [13], it is proved that if Gis a graph
on nvertices and its edge density psatisfies 1
n≤p≤1
2, then disc(G) = Ω(p1/2n3/2), and equality
is a achieved by the Erd˝os-R´enyi random graph Gn,p . When p < 1/n, it is not difficult to show
that the right lower bound is Ω(pn2). Noting that the discrepancy of a hypergraph is equal to the
discrepancy of its complement, these provide sharp bounds in case 1
2≤p≤1 as well. Bollob´as and
Scott [5] extended this result to r-uniform hypergraphs in case pis not too small. More precisely,
they proved that if 1
n≤p≤1
2, and His an nvertex r-uniform hypergraph of edge density p, then
disc(H) = Ωr(p1/2nr+1
2). Again, equality is achieved by random hypergraphs. Here, we prove that
the same bound holds for the whole range of interest n−(r−1) ≪p≤1
2
Theorem 1.3. Let Hbe an r-uniform hypergraph on nvertices of average degree d, where 1≤d≤
1
2n−1
r−1. Then
disc(H) = Ωr(√dn).
If d < 1, it is easy to argue that the minimum discrepancy is Ωr(dn) = Ωr(pnr). Indeed, in this
case Hcontains an independent set Uof size Ω(n), and |disc(U)|=p|U|
r= Ωr(pnr).
3
While the discrepancy of a hypergraph measures the maximum absolute deviation of the size of
an induced subhypergraph compared to its expected size, it is also natural to consider whether this
deviation is positive or negative. The positive discrepancy of His defined as
disc+(H) = max
U⊂V(H)disc(U),
and the negative discrepancy of His
disc−(H) = max
U⊂V(H)−disc(U).
Note that one trivially has disc+(G) = disc−(Gc) and disc(G) = max(disc+(G),disc−(G)).
We observe that the positive discrepancy and the bisection width of a hypergraph are closely
connected in a sense that small bisection width implies large positive discrepancy.
Lemma 1.4. Let Hbe an r-uniform hypergraph on nvertices with average degree d. Let s(H)be
such that
bw(H) = nd
r1−21−r−s(H).
Then
disc+(H)≥s(H)
2.
Proof. Let V(H) = X∪Ybe an equipartition such that the size of the corresponding bisection is
bw(H). Note that for all n≥rwe have
⌊n
2⌋
r+⌈n
2⌉
r≤21−rn
r.
Hence, it follows that
disc(X) + disc(Y) = e(X) + e(Y)−p⌊n
2⌋
r+⌈n
2⌉
r≥e(X) + e(Y)−21−r·pn
r.
Since e(X) + e(Y) = e(H)−bw(H), we get
disc(X) + disc(Y)≥s(H).
In particular, we conclude that disc+(H)≥s(H)
2.
In the case of graphs, exploring the connection between the bisection width and positive
discrepancy, R¨aty, Sudakov, and Tomon [36] proved sharp bounds on both of these quantities.
However, in the case of hypergraphs, less is known. Bollob´as and Scott [5] proved that if H
is an r-uniform hypergraph with density psatisfying p(1 −p)≥1
n, then disc+(H)·disc−(H) =
Ωr(p(1 −p)nr+1). Unfortunately, this inequality does not provide any information on disc+(H) and
disc−(H) individually, beyond that both are at least Ωr(n).
On the other hand, Bollob´as and Scott [5] proved that for d≥1, the random r-uniform hypergraph
H, where every edge is included independently with probability p=d/n−1
r−1satisfies disc(H) =
O(√dn) with high probability, which implies disc+(H),disc−(H) = O(√dn) as well. Thus, by Lemma
1.4, we also have bw(H)≥(1 −21−r)e(H)−O(√dn), confirming our claim in the previous section.
Our main result concerning the positive discrepancy is the following lower bound for sufficiently
sparse hypergraphs. We prove that whenever the average-degree dis at most n2/3, the positive
discrepancy is bounded below by Ω(√dn), which is sharp by the aforementioned result.
4
Theorem 1.5. Let Hbe an r-uniform hypergraph on nvertices of average degree d < n2/3. Then
disc+(H) = Ωr(d1/2n).
1.3 Eigenvalues
Given a d-regular graph G, let Abe the adjacency matrix of Gand let d=λ1≥ ··· ≥ λnbe the
eigenvalues of A. The Alon-Boppana theorem [34] is one of the central results in spectral graph
theory, stating that the second largest eigenvalue satisfies
λ2≥2√d−1−on(1).
This is known to be tight for infinite values of ddue to the existence of so called Ramanujan graphs
[23]. Furthermore, as a celebrated result of Friedman [18] shows, λ2= 2√d−1 + on(1) for random
d-regular graphs as well. Here, it is good to point out that 2√d−1 also coincides with the spectral
radius of the d-regular infinite tree. This might not be unexpected, as random d-regular graphs are
locally tree-like. However, the exact connection between these two quantities might be more subtle,
as we discuss later in this section.
There has been a plethora of work concerning the spectral theory of hypergraphs under various
frameworks [3,9,10,16,19,24,28,32]. We now briefly discuss the main directions and some of
the highlights of these works. Some authors [3,9] define the adjacency matrix of a hypergraph H
as the matrix Awhose entry Ai,j is the co-degree of the distinct vertices iand j. In contrast, the
rest of the literature considers the so called adjacency tensor. A tensor naturally corresponds to a
multilinear map, so we rather define this map directly instead. We follow the notation of Li and
Mohar [32], which coincides with the notation of other sources up to constant factors depending on
the uniformity.
Definition 1. Given an r-uniform hypergraph Hon vertex set V, its adjacency map τH: (CV)r→R
is the symmetric multilinear function defined as follows: for every x1,...,xr∈CV,
τH(x1,...,xr) = 1
(r−1)! X
v1,...,vr∈V
{v1,...,vr}∈E(H)
x1(v1). . . xr(vr).
We define the normalized adjacency map of Has
σH=τH−r|E(H)|
nrJ,
where Jis the ”all-one” tensor defined as J(x1,...,xr) = Pv1,...,vr∈Vx1(v1). . . xr(vr).
Note that in case Gis a graph with adjacency matrix A, then τG(x, y) = xTAy. Also, if 1is
the all-one vector, then σH(1,...,1) = 0. We now define two sets of quantities that are potential
candidates for the second largest eigenvalue of r-uniform hypergraphs.
Definition 2. For p > 0, the Lp-norm of a vector x∈Cnis defined as
||x||p= (|x(1)|p+···+|x(n)|p)1/p.
5
Given an r-uniform hypergraph H, let
λ(p)
2(H) = sup
x∈RV,||x||p=1
σH(x, . . . , x),
and
µ(p)(H) = sup
||x1||p=1,||x2||p=1,...,||xr||p=1 |σH(x1,...,xr)|.
Note that in case r= 2, λ2(H) = λ(2)
2(H) and µ(2)(H) = max (|λ2(H)|,|λn(H)|). Thus, in a
sense, λ(p)
2is more related to the second largest eigenvalue of H, while µ(p)is related to the second
largest absolute value of the eigenvalues. Clearly, µ(p)(H)≥λ(p)
2(H), but the gap between these
quantities can be arbitrarily large: when Gis the complete balanced bipartite graph on nvertices,
λ2(G) = 0, while µ(2)(G) = n/2. Also, while µ(2) (G)≥Ω(√d) holds for every d-regular graph G
on n≥2dvertices, and this bound is sharp, the minimum of λ(2)
2(G) among d-regular graphs has
a much stranger behavior, see the recent manuscript [36]. In the case of r-uniform hypergraphs for
r≥3, typically µ(2)(H) or µ(r)(H) is studied as the second eigenvalue in the literature, but we
propose λ(p)
2(H) as a stronger alternative.
A hypergraph His said to be k-co-degree regular if for distinct v1,...,vr−1, the number of
edges containing {v1,...,vr−1}is k. Friedman and Wigderson [19] studied the quantity µ(2)(H) for
co-degree regular hypergraphs, motivated by the theory of Cayley hypergraphs. They proved that
if His a 3-uniform k-co-degree regular hypergraph on nvertices, then µ(2)(H)≥Ω(pk(n−k)/n),
and they noted that similar conclusion holds for higher uniformities as well. Lenz and Mubayi
[30] also considered µ(2)(H) in relation to hypergraph quasirandomness. While µ(2)(H) might be a
good measure of the second eigenvalue of dense hypergraphs, it becomes unusable for sparse ones.
Indeed, if His the random 3-uniform hypergraph on nvertices, in which each edge is included with
probability p, then with high probability µ(2)(H) = Θ(√pn) if 1/n ≪p≪1/2, but µ(2)(H) = Θ(1)
if p≪1/n.
Another analogue of the Alon-Boppana theorem for d-regular hypergraphs is proposed by Li
and Mohar [32]. They study the quantity µ(r)(H), and prove that if His an r-uniform d-regular
hypergraph, then
µ(r)(H)≥r
r−1((r−1)(d−1))1/r −on(1).
Here, the quantity r
r−1((r−1)(d−1))1/r coincides with the spectral radius of the infinite d-regular
hypertree by a result of Friedman [17]. Therefore, it is tempting to believe that this is the right
quantity due to its attractive parallel with the graph case. However, as the first part of our next
theorem shows, even λ(r)
2(H) is always at least Ωr(√d) for any uniformity, thus greatly improving
the result of Li and Mohar. We also note that the analogous bound can be achieved for µ(p)(H) in
a slightly wider range of d. This gives an alternative proof of the result of Friedman and Wigderson
[19] for 3-uniform hypergraphs, and also extends the result to regular hypergraphs.
Theorem 1.6. Let Hbe a d-regular r-uniform hypergraph on nvertices, r≥3.
(i) If d≤n2/3, then
λ(r)
2(H)≥c√d
for some c=c(r)>0depending only on r. In general, for any p≥1,
λ(p)
2(H)≥cn1−r/p√d.
6
(ii) If d < n2
4, then
µ(r)(H)≥c√d
for some c=c(r)>0depending only on r. In general, for any p≥1,
µ(p)(H)≥cn1−r/p√d.
(iii) If r≥4and n2
4≤d≤1
2n−1
r−1, then
µ(r)(H)≥cd
n
for some c=c(r)>0depending only on r. In general, for any p≥1,
µ(p)(H)≥cdn−r/p
This result follows almost immediately from our bounds on the discrepancy. Indeed, it turns
out that µ(p)(H) controls the discrepancy of H, while λ(p)
2(H) controls the positive discrepancy. In
particular, we have the following relationship between these quantities.
Lemma 1.7. Let Hbe an r-uniform hypergraph on nvertices of average degree d. Then
nr/p ·λ(p)
2(H)≥rdisc+(H)−Or(d)
and
nr/p ·µ(p)(H)≥rdisc(H)−Or(d).
Proof. Let U⊂V(H), and let ybe the characteristic vector of U, then x=|U|−1/p ·ysatisfies
||x||p= 1. Also,
σH(x, . . . , x) = |U|−r/p ·re(U)−dn
nr· |U|r.(1)
Here, we have the following relationship between the discrepancy and normalized adjacency map:
rdisc(U)− |U|r/pσH(x, . . . , x) = dn ·|U|r−1
nr−1−|U|...(|U| − r+ 1)
n . . . (n−r+ 1) =Or(d).
Therefore, choosing Usatisfying disc(U) = disc+(H), or disc(U) = disc(H), verifies the two desired
inequalities.
Hence, as long as disc+(H) = Ω(√dn)≫d, which is satisfied for d≪n2/3by Theorem 1.5, the
inequality λ(p)
2(H)≥cn1−r/p√dfollows from the previous lemma. Similarly, the second inequality
follows by using Theorem 1.3, and observing that there exists ε > 0 for which rdisc+(H)−Or(d) =
Ωr(√dn) provided that d < εn2.
When εn2≤d≤1
2n−1
r−1, one can show the improved lower bound µ(p)(H) = Ωrn−r/pd. Indeed,
let Ube a uniformly random subset of V(G) of size ⌈n
2⌉, let ydenote the characteristic vector of U,
and let x=|U|−1/p ·y. Then ||x||p= 1, and xalso satisfies equation (1). Since
Ee(U)−dn
rnr·|U|r
nr=p|U|
r−pn
r|U|r
nr≤ −cpnr−1
for some constant c > 0 depending only on r, it follows that there exists a set Ufor which we have
|σ(x, . . . , x)| ≥ Ω(n−r/pd),
which implies the desired bound µ(p)(H) = Ωr(n−r/p d).
7
2 Proof overview
Let us give a short overview of our proofs, in particular the proofs of Theorem 1.1 and Theorem 1.5.
We follow the ideas of [36], which in turn were inspired by the semidefinite programming approach
of Goemans and Williamson [20] on the MaxCut problem.
Let Hbe an r-uniform d-regular hypergraph. We assign certain unit vectors in RV(H)to the
vertices of Hwith the property that if the vertices vand ware both contained in some edge, then
the scalar product of their corresponding vectors is slightly positive, in particular at least Ω(d−1/2).
Then, we chose a random linear half-space, and define X⊂V(H) to be the set of vertices whose
corresponding vectors are contained in this half-space. The vectors are constructed in a manner to
ensure that r-tuples of vertices forming an edge are more likely to be contained in Xthan the average
r-tuple. In particular, we show that the expected discrepancy of Xis Ωr(√dn), proving Theorem 1.5
for regular hypergraphs. To prove the general statement, we argue that large degree vertices can be
either omitted, or already contribute large discrepancy. In order to prove Theorem 1.1, we further
argue that the size of Xmust be close to n/2. Hence, we can add or remove a few vertices to get a
set X′of size ⌊n/2⌋. We show that the bisection given by the partition X′∪(V(H)\X′) is of size
nd
r1−1
2r−1−Ω(√dn) in expectation.
In order to execute this strategy, we first need a bound on the probability that given rvectors
v1,...,vr, they are simultaneously contained in a random linear half-space. This problem is discussed
in the next subsection.
3 A probabilistic geometric lemma
In this section, we consider the following probabilistic problem in geometry, which is the backbone
of our proofs.
Problem. Given rvectors v1,...,vr∈Rn, what is the probability that they are simultaneously
contained in a random linear half-space?
To this end, let wbe a random unit vector in Rn, chosen from the uniform distribution. Define
µ(v1,...,vr) := P(hw, vii ≥ 0 for every i∈[r]) .
In case r= 2, this probability is easy to calculate: µ(v1, v2) = π−α
2π, where αis the angle between
v1and v2. However, for r≥3, we are unable to provide an easy to use formula. Note that when
v1,...,vrare pairwise orthogonal, then µ(v1,...,vr) = 1/2r. In the next lemma, we show that under
some mild assumptions,
µ(v1,...,vr) = 1
2r+ Θr
X
1≤i<j≤rhvi, vji
.
Lemma 3.1. For every r, there exist 0< c1< c2and α > 0such that the following holds. Let
v1,...,vr∈Rnbe unit vectors such that 0≤ hvi, vji ≤ αfor every i, j ∈[r]. Then
µ(v1,...,vr)∈1
2r+ [c1, c2]·X
1≤i<j≤rhvi, vji.
8
Proof. Without loss of generality, we may assume that a=hv1, v2iis maximal among hvi, vji,
1≤i < j ≤r. Then, our goal is to show that µ(v1,...,vr) = 1
2r+ Θr(a) assuming a≤αis
sufficiently small with respect to r.
Let Hbe an r-dimensional linear hyperplane containing v1,...,vr. Note that if w′is the
projection of wonto H, then hvi,wi=hvi,w′iand w′
||w′||2is uniformly distributed on the unit
sphere of Rr. Hence, we may assume that n=r. Furthermore, note that if Ais an isometry of
Rr, then hAx, Ayi=hx, y ifor any x, y ∈Rr, and Awhas the same distribution as w. Hence,
after applying a suitable isometry to the vectors v1,...,vr, we may assume that the matrix with
rows v1,...,vris lower-triangular with non-negative diagonal entries. In other words, vi(i)≥0 and
vi(i+ 1) = vi(i+ 2) = ··· =vi(r) = 0 for i∈[r]. Note that the numbers hvi, vjithen uniquely
determine the rvectors v1,...,vr. Finally, we may assume that wis chosen randomly in the unit
ball of Rrfrom the uniform distribution, instead of the unit sphere.
Recall that a=hv1, v2i. Then v1= (1,0,...,0) and v2= (a, √1−a2,0,...,0). Also, by the
maximality of a,
a≤X
1≤i<j≤rhvi, vji ≤ r2a.
Next, let us bound the entries of vℓfor ℓ∈[r].
Claim 3.2. Let ℓ∈[r]. Then vℓ(ℓ)≥1/2. Also, there exists c=c(r)>0such that if i < ℓ, then
vℓ(i)∈[−ca2, ca].
Proof. Assume that α < 1
18r. We prove the following statement by double induction, first on ℓ, then
on i: for every ℓ≥2, if i∈ {1,...,ℓ−1}, then vℓ(i)∈[−18ra2,3a], and vℓ(ℓ)∈[1/2,1]. As
vℓ(ℓ) = 1−
ℓ−1
X
i=1
vℓ(i)2!1/2
≥1−9ra21/2,
the inequality vℓ(ℓ)≥1/2 follows by noting that a≤αand assuming that our induction hypothesis
holds for i≤ℓ−1.
Note that vℓ(1) = hv1, vℓi ∈ [0, a], so the statement is true for ℓ≥2 and i= 1. Now fix ℓ≥3
and 2 ≤i < ℓ, and assume that our induction hypothesis is true for every pair (ℓ0, i0) with ℓ0< ℓ or
ℓ0=ℓand i0< i. Noting that vi(j) = 0 if j > i, we have
hvi, vℓi=
i
X
k=1
vi(k)vℓ(k).
Hence,
vi(i)vℓ(i) = hvi, vℓi −
i−1
X
k=1
vi(k)vℓ(k).
Here, vi(i)∈[1/2,1], hvi, vℓi ∈ [0, a], and each term vi(k)vℓ(k) in the sum is in [−9a2,9a2]. Hence,
vi(i)vℓ(i)∈[−9ra2, a + 9ra2], and vℓ(i)∈[−18ra2,2a+ 18ra2]. The statement follows as a≤
1/(18r).
Let B1denote the unit ball in Rr, let B+
1⊂B1be the set of vectors with nonnegative coordinates,
and let S⊂B1be the set of vectors wfor which hvi, wi ≥ 0 holds for every i∈[r]. Note that
vol(B+
1)/vol(B1) = 1/2rand µ(v1,...,vr) = vol(S)/vol(B1), where vol(.) denotes the volume.
9
First, we establish an upper bound on µ(v1,...,vr). Let c > 0 be the constant given by the
previous claim, and for i= 1,...,r, let
Ri={z∈[−1,1]r:−2cra ≤z(i)≤0}.
We claim that S⊂B+
1∪R1∪···∪Rr. Indeed, let w∈B1be a vector, and assume that w(ℓ)<−2cra
for some ℓ∈[r]. Then
hw, vℓi=
ℓ
X
i=1
w(i)vℓ(i)≤(ℓ−1)ca +w(ℓ)
2<0
by Claim 3.2, so w6∈ S. But
vol(B+
1∪R1∪ · ·· ∪ Rr)≤vol(B1)
2r+r·(2r−1·(2cra)).
Therefore,
µ(v1,...,vr) = vol(S)
vol(B1)≤1
2r+2rr2ca
vol(B1)≤1
2r+c2X
1≤i<j≤rhvi, vji
with c2=2rr2c
vol(B1).
Next, we establish the lower bound on µ(v1,...,vr). First, let Qbe the set of vectors in Bwhose
every coordinate is at least 2rca2. If w∈Q, then
hvℓ, wi=
ℓ
X
i=1
vℓ(i)w(i)≥ ℓ−1
X
i=1 −ca2!+w(ℓ)vℓ(ℓ)≥0
by Claim 3.2, so Q⊂S. Furthermore, vol(Q)≥vol(B+
1)−2r2ca2. Let
R=1
2r,1
r×h−a
2r,0i×1
2r,1
rr−2
.
Observe that R⊂B1. For every w∈R, we have
hv2, wi=aw(1) + p1−a2·w(2) ≥a
2r−a
2r= 0,
and if ℓ6= 2, then
hvℓ, wi=vℓ(ℓ)w(ℓ) + vℓ(2)w(2) + X
i≤ℓ−1,i6=2
vℓ(i)w(i)≥1
4r−ca2
2r−(ℓ−2) ·ca2
r≥0.
Thus R⊂Sholds as well. But then Q∪R⊂S, and we have
µ(v1,...,vr) = vol(S)
vol(B1)≥vol(Q)
vol(B1)+vol(R)
vol(B1)
≥1
2r−2rca2
vol(B1)+a
(2r)rvol(B1)≥1
2r+c1X
1≤i<j≤rhvi, vji
with suitable c1>0, assuming a≤αis sufficiently small.
10
4 Bisection width
In this section, we prove Theorems 1.1 and 1.2. We prepare the proof with a technical lemma, which
is also the key result in the proof of Theorem 1.5. But first, let us recall the definition of discrepancy
and positive discrepancy.
Definition 3. Let Hbe an r-uniform hypergraph with nvertices and edge density p=|E(H)|
(n
r). Given
U⊂V(H), define the discrepancy of Uas
disc(U) = e(U)−p|U|
r.
Then the positive discrepancy of His defined as
disc+(H) = max
U⊂V(H)disc(U).
Similarly, the negative discrepancy of His
disc−(H) = max
U⊂V(H)−disc(U),
and the discrepancy is
disc(H) = max
U⊂V(H)|disc(U)|= max{disc+(H),disc−(H)}.
Now we are ready to state our key lemma.
Lemma 4.1. Let Hbe an r-uniform hypergraph on nvertices of maximum degree ∆.
(i) If ∆≤n2/3, then for some c=c(r)>0,
disc+(H)≥ce(H)
√∆.
(ii) If e(H)> C ∆2√nfor some sufficiently large C=C(r)>1, then the vertex set of Hcan be
partitioned into two parts, Xand Y, such that for some c′=c′(r)>0,
e(X) + e(Y)−∆· ||X| − |Y|| ≥ e(H)·1
2r−1+c′
√∆.
Proof. Let α=α(r) satisfying 0 < α < min{0.1, α0}be specified later, where α0is the constant
given by Lemma 3.1 as α. We may assume that ∆ and nare sufficiently large with respect to α, so
in turn, with respect to r. Also, let p=e(H)/n
rbe the edge density of H. Let V=V(H), and for
every vertex v∈V, assign the vector xv∈RVas follows: for u∈V,
xv(u) =
1 if v=u,
α
√2r∆if there exists e∈E(H) such that v, u ∈e,
0 otherwise.
11
Note that
1≤ ||xv||2
2≤1 + (r−1) ·∆·α
√2r∆2
≤2.
Let yv=xv/||xv||2be the normalization of xv. Clearly, 0 ≤ hyu, yvifor any u, v ∈V, and if u6=v,
then
hyu, yvi ≤ hxu, xvi ≤ 2α
√2r∆+ (r−1) ·∆·α
√2r∆2
< α.
Furthermore, if uand vboth appear in some edge e, then
hyu, yvi ≥ 1
2hxu, xvi ≥ α
√2r∆.
Let wbe a random unit vector in RV, chosen from the uniform distribution, and define
X={v∈V:hyv,wi ≥ 0}.
First, we calculate the expected discrepancy of Xin H. We have
E(disc(X)) = E(e(X)) −p·E|X|
r.
For each r-element set e={v1,...,vr} ⊂ V, we have P(e⊂X) = µ(yv1,...,yvr). As the vectors
yv1,...,yvrsatisfy the required conditions of Lemma 3.1, we can write that
1
2r+c1X
1≤i<j≤rhyvi, yvji ≤ P(e⊂X)≤1
2r+c2X
1≤i<j≤rhyvi, yvji,
where c1, c2>0 are suitable constants depending only on r. Therefore, if eis an edge of H, then
P(e⊂X)≥1
2r+c1α
√2r∆,
and so
E(e(X)) ≥e(H)1
2r+c1α
√2r∆.(2)
On the other hand, E|X|
ris equal to the expected number of r-element sets contained in X, so
E(disc(X)) ≥e(H)·1
2r+c1α
√2r∆−pX
{v1,...,vr}∈V(r)
1
2r+c2X
1≤i<j≤rhyvi, yvji
.
The terms containing 1
2rcancel, so we get
E(disc(X)) ≥c1αe(H)
√2r∆−c2pX
{v1,...,vr}∈V(r)X
1≤i<j≤rhyvi, yvji.
Here, one can write
X
{v1,...,vr}∈V(r)X
1≤i<j≤rhyvi, yvji=n−2
r−2X
{v,v′}∈V(2) hyv, yv′i=n−2
r−2X
u∈VX
{v,v′}∈V(2)
yv(u)yv′(u).
12
Now fix some u∈V, and let us bound P{v,v′}∈V(2) yv(u)yv′(u), which in particular is bounded by
P{v,v′}∈V(2) xv(u)xv′(u). There are at most (r∆)2pairs {v, v′}such that both {u, v }and {u, v′}are
contained in some edge of H. If v6=uand v′6=u, each such pair contributes α2
2r∆to the second sum.
Also, there are at most 2r∆ pairs {v, v ′}such that v=uor v′=u, and both {u, v}and {u, v′}are
contained in some edge. Each such pair contributes α
√2r∆to the sum. Hence,
X
{v,v′}∈V(2)
yv(u)yv′(u)≤(r∆)2·α2
2r∆+ 2r∆·α
√2r∆≤4r∆α2,
where the last inequality holds by our assumption that ∆ is sufficiently large with respect to α. From
this, we get
X
{v,v′}∈V(2) hyv, yv′i ≤ 4r∆α2n, (3)
and we can write
E(disc(X)) ≥c1αe(H)
√2r∆−c2pn−2
r−2·(4r∆α2n)≥c3αe(H)
√∆−c4α2e(H)∆
n
with suitable c3, c4>0 only depending on r. If ∆ < n2/3, there is a choice for αdepending only on
rsuch that the right-hand-side is at least c3α
2·e(H)
√∆. Hence, we have E(disc(X)) ≥c5e(H)
√∆with some
constant c5>0 depending only on r. Therefore, there is a choice for the random unit vector wsuch
that the resulting set satisfies the required conditions of (i).
Now let us turn to the proof of (ii). Let Y=V\X, then by symmetry, i.e. by noting that
X(w) = Y(−w) with probability 1, we have E(e(X)) = E(e(Y)). Recall that by (2), we have
E(e(X)) = E(e(Y)) ≥e(H)( 1
2r+c1α
√∆). It remains to bound the expectation of ||X|−|Y|| =|2|X|−n|.
By convexity,
(E(|2|X| − n|))2≤E((2|X| − n)2) = E(n2−4|X|n+ 4|X|2).
For every vertex v,P(v∈X) = 1
2, so E(|X|) = n/2. Furthermore, for every pair of distinct vertices
{v, v′}, we have P(v, v′∈X) = µ(yv, yv′)≤1
4+c6hyv, yv′ifor some constant c6>0 by Lemma 3.1
applied with r= 2. Hence,
E(|X|2) = X
v,v′∈V
P(v, v′∈X)≤n
2+ 2 X
{v,v′}∈V1
4+c6hyv, yv′i≤n2
4+n
4+ 8c6r∆α2n,
where the last inequality follows by (3). In conclusion,
E((2|X| − n)2)≤n+ 32c6r∆α2n,
and thus E(||X| − |Y||)≤c7√∆nwith some c7=c7(r)>0.
Putting everything together,
E(e(X) + e(Y)−∆||X| − |Y||)≥e(H)
2r−1+2c1αe(H)
√∆−c7∆3/2√n.
Assuming C > 0 is sufficiently large, the condition e(H)> C∆2√nensures that the right hand side
is at least e(H)·(1
2r−1+c1α
√∆). But then there is a choice for the random unit vector wsuch that the
partition X∪Ysatisfies the requirements of (ii).
13
Proof of Theorem 1.1.Let C, c′be the constants guaranteed by Lemma 4.1, (ii). Setting ε=1
rC ,
the condition d≤εn1/2implies dn
r=e(H)> Cd2√n. Then there exists a partition X∪Yof V(H)
such that
e(X) + e(Y)−d||X| − |Y|| ≥ e(H)1
2r+c′
√d.
Without loss of generality, we may assume that |X| ≤ |Y|. Let Sbe an arbitrary ⌊n/2⌋−|X|element
subset of Y, and let X0=X∪Sand Y0=Y\S. Then X0∪Y0is a bisection of H, and
e(X0) + e(Y0)≥e(X) + e(Y)− |S|d=e(X) + e(Y)−d|Y| − |X|
2≥e(H)1
2r+c′
√d.
Finally,
e(X, Y ) = e(H)−e(X)−e(Y)≤e(H)1−1
2r−c′
√d,
finishing the proof.
Straightforward modifications of Lemma 4.1 imply Theorem 1.2 as well. We omit the details.
5 Positive discrepancy
In this section, we prove Theorem 1.5, which we restate here for the reader’s convenience.
Theorem 5.1. Let Hbe an r-uniform hypergraph on nvertices of average degree d < n2/3. Then
disc+(H) = Ωr(d1/2n).
Note that in case the the maximum degree of His not much larger than its average degree,
Lemma 4.1 immediately implies Theorem 5.1. We show that the general case can be reduced to
this special subcase. It will be useful to define the discrepancy of collections of sets as well. Given
s1,...,sksuch that s1+···+sk=rand disjoint sets U1,...,Uk, define
es1,...,sk(U1,...,Uk) = #{e∈E(H) : |e∩Ui|=sifor i∈[k]}
and
discs1,...,sk(U1,...,Uk) = es1,...,sk(U1,...,Uk)−p|U1|
s1...|Uk|
sk.
Note that if Uand U′are disjoint, then
disc(U∪U′) =
r
X
i=0
disci,r−i(U, U ′).
Finally, let ∂(X) be the set of all edges of Hthat have a vertex in X. Then
|∂(X)|=
r−1
X
i=0
er−i,i(X, X c),
where Xc=V(H)\Xis the complement of X. Next, we prove a lemma which is used to handle
large degree vertices of H.
14
Lemma 5.2. For every r≥2, there exist c1, C > 0such that the following holds. Let Hbe an
r-uniform hypergraph on nvertices of average degree d. Let X⊂V(H)such that the degree of every
vertex in Xis at least Cd. Then
disc+(H)≥c1|∂(X)|.
Proof. Let p=d
(n
r−1)= Θ( d
nr−1) be the density of H. Let s=|Xc|, and let α∈(0,1) be specified
later, depending only on δand r. Furthermore, let b=⌊s/2⌋, and let Ybe a random belement
subset of Xc, chosen from the uniform distribution. We have
disc(X∪Y) =
r
X
i=0
discr−i,i(X, Y).
Here,
E(discr−i,i(X, Y)) =
i−1
Y
j=0
b−j
s−j
·discr−i,i(X, X c).
Writing βifor the coefficient of discr−i,i (X, X c) in the previous line, we thus get
E(disc(X∪Y)) =
r
X
i=0
βidiscr−i,i(X, X c) =
r−1
X
i=0
(βi−βr) discr−i,i(X, X c),
where we used 0 = disc(X∪Xc) = Pr
i=0 discr−i,i(X, X c) in the second equality. Here, the right
hand side can be written as
r−1
X
i=0
(βi−βr)e(X, Xc)−p r−1
X
i=0
(βi−βr)|X|
r−i|Xc|
i!(4)
Next, we use that βi−βr≥(r−i)(βr−1−βr), which easily follows from the fact that βi≥2βi+1 for
every i= 0,...,r−1. Also, |X|
r−i|Xc|
i≤ |X|nr−1for i= 0,...,r−1. Hence, we can lower bound
(4) as
(βr−1−βr) r−1
X
i=0
(r−i)er−i,i(X, X c)!−cd|X|
with a suitable c=c(r)>0. But observe that Pr−1
i=0 (r−i)er−i,i(X, X c) is just the sum of degrees
of the vertices of X, which is at least Cd|X|by our assumption on X. Hence, choosing C > 2c, we
conclude that
E(disc(X∪Y)) ≥βr−1−βr
2 r−1
X
i=0
(r−i)er−i,i(X, X c)!≥βr−1−βr
2|∂(X)|,
so c1=βr−1−βr
2suffices.
Proof of Theorem 5.1.Let ∆ = Cd, where Cis given by Lemma 5.2. Let Xbe the set of vertices of
Hof degree more than ∆, and let H′be the hypergraph we get by removing every edge of Hhaving
a vertex in X. Note that the maximum degree of H′is at most ∆, and e(H′) = e(H)− |∂(X)|.
Hence, in case e(H′)≤e(H)/2, we have |∂(X)| ≥ e(H)/2, which gives disc+(H)≥Ωr(e(H)) by
Lemma 5.2, and we are done. Hence, we may assume that e(H′)≥e(H)/2.
15
By Lemma 4.1,
disc+(H′)≥c2e(H′)
√∆≥c3√dn
with some c2, c3>0 depending only on r.
Now let us bound the discrepancy of H. If |∂(X)| ≥ 1
2disc+(H′), then Lemma 5.2 states that
disc+(H)≥c1|∂(X)| ≥ c1c3
2√dn,
so we are done. Hence, we may assume that |∂(X)| ≤ 1
2disc+(H′). Let U⊂V(H) be such that
disc+(H′) = discH′(U), and let p′=e(H′)/n
rbe the density of H′. Then
discH(U)−discH′(U) = eH(U)−eH′(U)−(p−p′)|U|
r≥ −(p−p′)n
r=−|∂(X)|,
hence
discH(U)≥disc+(H′)− |∂(X)| ≥ 1
2disc+(H′)≥c3
2√dn,
so we are done in this case as well.
6 Discrepancy
In this section, we prove Theorem 1.3. A key preliminary tool in our proof is to observe that the
theorem of Bollob´as and Scott [5] concerning the product of the positive and negative discrepancy
of hypergraphs also extends to multi-hypergraphs. We now state the version of the theorem that is
needed for our purposes. For completeness, the full proof of this theorem is added to the Appendix.
Theorem 6.1 (Multi-hypergraph version of Theorem 14, [5]).Let Hbe an r-uniform
multi-hypergraph of order nwith pn
redges, counted with multiplicities. Suppose that psatisfies
the condition 1
2n≤p≤1−1
2n. Then
disc+(H)·disc−(H)≥Ωp(1 −p)nr+1.
Given disjoint sets Xand Yand an integer i∈ {0,...,r}, recall that we write
disci,r−i(X, Y ) = ei,r−i(X, Y )−p|X|
i|Y|
r−i.
If the host hypergraph His not clear from the context, we highlight it by writing disci,r−i(H;X, Y )
or ei,r−i(H;X, Y ) instead.
Lemma 6.2 (Multi-hypergraph version of Lemma 9, [5]).Let Hbe an r-uniform multi-hypergraph
and let D= disc(H). Then for every pair of disjoint sets Xand Yand i∈[r], we have
|disci,r−i(X, Y )| ≤ r2rD.
Proof. Let p∈[0,1], and let Zbe a random subset of X, each element chosen independently with
probability p. Then
f(p) := E(disc(Z∪Y)) =
r
X
i=0
pidisci,r−i(X, Y ).
Thus, fis a degree rpolynomial, and f(p)∈[−D, D] for every p∈[0,1]. This implies that every
coefficient of fis at most 2rr2rD/r!< r2rDin absolute value, see [6]. This finishes the proof.
16
Proof of Theorem 1.3.By taking complements if necessary, we may assume that the density of H
is at most 1
2. For each t∈ {2,...,r}, define the t-uniform multi-hypergraph Htby setting the
multiplicity of f∈V(t)to be
m(f) = |{e∈E(H) : f⊆e}|.
Note that Hris simply Hitself.
Let dtdenote the average degree of Ht,ptdenote the edge density, and let d=drand p=pr.
Since
ptn
t=e(Ht) = pn
rr
t,
it follows that
pt=n−t
r−tp.
Similarly, it is easy to conclude that we have
dt=r−1
t−1d.
In particular, it is straightforward to verify that for all t∈ {2,...,r},
pt+1
pt
=r−t
n−t.(5)
We start by proving that for each H, at least one of the graphs Hthas a suitable density so that
Theorem 6.1 applies.
Claim 6.3. There exists t∈ {2,...,r}for which we have 1
2n≤pt≤1
2.
Proof. When d=1
2n−1
r−1, we can take t=ras one clearly has pr=1
2. When d= 1, it is easy to
verify that e(H2) = r
2e(H)≥n(r−1)
2. Hence it follows that p2≥r−1
n−1. But equation (5) implies that
p2≥ ··· ≥ prand pt+1 > npt, so taking the smallest index tsatisfying pt≥1
2n, we have pt≤1/2 as
well.
Since Htsatisfies the conditions of Theorem 6.1 and d= Θr(dt), we have disc (Ht) = Ωr(√dn).
Let U⊆V(H) be chosen so that |discHt(U)|= disc (Ht). In order to infer results concerning
disc(H), we rewrite the terms eHt(U) and pt|U|
toccurring in the expression discHt(U). First of all,
observe that
eHt(U) =
r
X
j=tj
tej,r−j(H;U, U c).(6)
By using standard identities for binomial coefficients, we also conclude that
pt|U|
t=pn−t
r−t|U|
t=pj
tr
X
j=t|U|
j|Uc|
r−j.(7)
Combining equations (6) and (7), we conclude that
|discHt(U)|=
r
X
j=tj
tdiscj,r−j(H;U, U c)
= Ωr(√dn).
17
Thus, by the triangle inequality there exists j∈ {t, . . . , r}for which
|discj,r−j(H;U, U c)|= Ωr(√dn).
But then Lemma 6.2 implies that disc(H) = Ωr(√dn),which completes the proof.
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Appendix: Proof of Theorem 6.1
For completeness, we now present the proof of Theorem 6.1. The proof follows the same lines as the
proof in [5], with appropriate trivial modifications added to accommodate the fact we are dealing
with multi-hypergraphs. In the proof, we also need four preliminary results from [5]. The only one
of these that is not identical to its counterpart in [5] is the fourth one, whose proof we include.
Given an edge-weighting won a complete r-uniform hypergraph Hand disjoint sets X1,...,Xt⊆
V(H), we define dk1,...,kt(X1,...,Xt) by setting dk1,...,kt(X1,...,Xt) = Pew(e), where the sum
is taken over all edges esatisfying the condition |e∩Xi|=kifor each 1 ≤i≤t. In case no
edge-weighting is given, we assume w(e) = 1 for every edge.
Lemma 6.4 (Lemma 6, [5]).Let εibe i.i.d. Bernoulli random variables with εi∈ {−1,1}, and let
a= (ai)n
i=1 be a sequence of real numbers. Then
E
n
X
j=1
aiεi≥||a||1
√2n.
Lemma 6.5 (Lemma 10, [5]).Let Hbe a complete r-uniform hypergraph of order nwith
edge-weighting w. Let V(H) = U∪Wbe a random bipartition, with each vertex assigned to one
of the sides independently with probability 1
2. Then
EX
K∈U(r−1)
r−1
|dr−1,1(K, W )| ≥ r2−rX
L∈V(G)r|w(L)|/√2n.
20
Lemma 6.6 (Lemma 11, [5]).Let Hbe a r-uniform hypergraph of order nwith edge-weighting w.
Suppose that α≥1and that there exists disjoint sets X, Y ⊂V(H)with
d1,r−1(X, Y ) + αd(Y) = M≥0.
Then at least one of the following holds
(i) disc+(H) = 2−3r2M/α, or
(ii) disc−(H) = 2−3r2Mα.
Lemma 6.7 (Multi-hypergraph version of Lemma 13, [5]).Let Hbe an r-uniform multi-hypergraph
of order nwith pn
redges with 1
2n≤p≤1−1
2nand nsufficiently large. Let V(H) = X∪Ybe a
random bipartition. Then
EK∈X(r−1) |dr−1,1(K, Y )−p|Y|| = Ωrpp(1 −p)nr−1/2
Proof. As usual, we may assume that p≤1
2. Given K∈V(H)(r−1), we write d(K) for the number
of edges containing K,s(K) for the number of vertices vso that K∪{v}is an edge with multiplicity
at least 1, and we define r(K) = d(K)−p(n−r+ 1). We also write m(e) for the multiplicity of an
r-tuple e∈V(H)(r). Furthermore, for each v∈V(H), we set ρv∈ {0,1}to be the indicator random
variable of the event v∈Y, and we set εv= 2ρv−1∈ {−1,1}.
Consider a fixed set K∈V(H)(r−1), and assume that d(K)>0. As in [5], we deduce that
E|dr−1,1(K, Y \K)−p|Y\K|| =EX
v6∈K
ρv(m(K∪ {v})−p)
=E
1
2X
v6∈K
(m(K∪ {v})−p) + 1
2X
v6∈K
εv(m(K∪ {v})−p)
≥1
2max
|r(K)|,EX
v6∈K
εv(m(K∪ {v})−p)
≥1
4
|r(K)|+EX
v6∈K
εv(m(K∪ {v})−p)
Label the vertices of V(H) with 1,...,n so that K={n−r+ 2,...,n}, and so that m(K∪ {i}) is
positive if and only if i≤s(K). Thus, using that E|X+Y| ≥ E|X|for any random variable Xand
independent Bernoulli random variable Y, we can write
EX
v6∈K
εv(m(K∪ {v})−p)≥E
s(K)
X
i=1
εi(m(K∪ {i})−p)
.
Since m(K∪ {i})−p > 0 for every i≤s(K), it follows that
s(K)
X
i=1 |m(K∪ {i})−p|=d(K)−ps(K)≥(1 −p)d(K).
21
Thus Lemma 6.4 implies that
E
s(K)
X
i=1
εi(m(K∪ {i})−p)≥(1 −p)d(K)
p2s(K)≥(1 −p)rd(K)
2.
Note that this bound remains also true when d(K) = 0. In particular, it follows that for every Kwe
have
E|dr−1,1(K, Y \K)−p|Y\K|| ≥ 1
4|r(K)|+ (1 −p)pd(K)
8.
Hence we conclude that
EK∈X(r−1) |dr−1,1(K, Y )−p|Y|| =X
K∈V(r−1)
P(K⊆X)·E|dr−1,1(K, Y \K)−p|Y\K||
≥2−r−2X
K∈V(r−1) |r(K)|+pd(K)
2!
Our aim is to show that there exists a uniform constant αso that each term in the sum is bounded
below by α√pn when nis sufficiently large. Since p≤1
2and there are n
r−1terms in the sum, this
certainly implies the lemma.
Suppose that n≥2r, and first consider the case when d(K)≥1. Observe that the expression
|d(K)−p(n−r+ 1)|+√d(K)
2as a function of d(K) is clearly increasing when d(K)≥p(n−r+ 1).
It can also be easily shown to be decreasing for d(K)∈1
16 , p(n−r+ 1). Thus whenever d(K)≥1,
we conclude that
|r(K)|+pd(K)
2≥pp(n−r+ 1)
2≥α1√pn
for some constant α1>0.
On the other hand, when d(K) = 0, we have |r(K)|=p(n−r+ 1). Since 1
2n≤p≤1
2, it follows
that p(n−r+ 1) ≥α2√pn for some constant α2>0. Thus we may take α= min(α1, α2), and the
result follows.
We are now ready to prove Theorem 6.1.
Proof of Theorem 6.1.The proof follows exactly the one presented in [5]. To start with, let Fdenote
the complete r-uniform weighted hypergraph on V(H) with edge-weight w(e) = m(e)−p, where
m(e) denotes the multiplicity of the edge e∈H. It is clear that we have w(F) = 0 and disc±(F) =
disc±(H). We may also assume that p≤1
2and that disc−(H)≥disc+(H) = crpp(1 −p)n(r+1)/2/α,
where cris a constant to be chosen later and α≥1.
Define random sets Wr=V(H)⊃Wr−1⊃ ··· ⊃ W1so that for each i≤r−1, Wi+1 =Wi∪Xi+1
is a random bipartition of Wi+1, where each vertex is assigned independently to either of the two
vertex classes with probability 1
2. Define weightings wisuch that for every K∈W(i)
i,
wi(K) = di,1,...,1(K, Xi+1,...,Xr),
with the convention that wr=w. Lemma 6.5 implies that iwe have
EX
K∈W(i)
i
|wi(K)| ≥ (i+ 1)2−i−1X
L∈W(i+1
i+1
|wi+1(L)|/√2n.
22
Note that Lemma 6.7 implies that
EX
K∈W(r−1)
r−1
|wr−1(K)|= Ωrpp(1 −p)nr−1/2.
Combining these two observations, we conclude that
EX
x∈W1|w1(x)|= Ωrpp(1 −p)n(r+1)/2.
Let X+
1={w∈W1:d1,...,1(w, X2,...,Xr)>0}. Note that
Ed1,...,1(X+
1, X2,...,Xr) = Ωrpp(1 −p)n(r+1)/2.(8)
Define V0by setting V0=X+
1∪Sr
i=2 Xi- Given a non-empty set S⊂ {2,...,r}, let VS=Si∈SXi
and
ES=K∪ {x}:x∈X+
1, K ∈Vr−1
S,|K∩Xi|>0∀i∈S.
Note that the sets ESpartition the edges in V0that intersect X+
1in exactly one vertex. We
also write dS=PK∈ESw(K), and observe that d1,r−1(X+
1, VS) = P∅6=T⊂SdTand d{2,...,r}=
d1,...,1(X+
1, X2,...,Xr).
Let S0be minimal with |dS0| ≥ (2k)−k+|S|d{2,...,r}. As in [5], we conclude that
max
S⊂{2,...,r}|d1,r−1(X+
1, VS)| ≥ |d1,r−1X+
1, VS0|
≥ |dS0| − X
∅6=T⊂S0|dT|
= Ωr(d{2,...,r}).
Hence we conclude that there exists S⊂ {2,...,r}with
Ed1,r−1X+
1, VS= Ωrpp(1 −p)n(r+1)/2.
Define X+
S={x∈W1:d1,r−1({x}, VS)>0}. Since Ed1,r−1(W1, VS) = 0 and Ed(VS) = 0, we
conclude that
Ed1,r−1(X+
S, VS) + αd(VS) = βrpp(1 −p)n(r+1)/2,
where βris some constant depending only on r. Set cr=βr·2−3r2−1, and recall that the constant
αis chosen so that
disc+(H) = crpp(1 −p)n(r+1)/2/α.
If α≤1, we are certainly done, as in this case both disc+(H) and disc−(H) are
Ωrpp(1 −p)n(r+1)/2. Otherwise, since disc+(F) = disc+(H)<2−3r2βrpp(1 −p)n(r+1)/2/α,
Lemma 6.6 implies that
disc−(H) = disc−(F)≥2−3r2βrpp(1 −p)n(r+1)/2α.
In particular, it follows that
disc−(H)·disc+(H) = Ωrp(1 −p)nr+1,
which completes the proof.
23
