1. Introduction
Let
$H_1,H_2$
be two
$k$
-uniform hypergraphs (
$k$
-graphs for short). Their Ramsey number
$r(H_1,H_2)$
is defined as the least integer
$N$
such that every red/blue edge-colouring of the complete
$k$
-graph
$K_N^{(k)}$
contains a monochromatic red copy of
$H_1$
or a monochromatic blue copy of
$H_2$
. In this paper, we will be concerned with the off-diagonal regime, where
$H_1=H$
is a fixed
$k$
-graph, and
$H_2=K_n^{(k)}$
is a clique, and we are interested in the asymptotics of
$r(H, K_n^{(k)})$
as
$n \to \infty$
. Note that
$r(H, K_n^{(k)})$
can also be viewed as the least integer
$N$
such that every
$N$
-vertex
$H$
-free
$k$
-graph contains an independent set of size
$n$
.
In the special case of
$k=2$
, it follows from the classical Erdős–Szekeres bound [Reference Erdös and Szekeres12] that
$r(H,K_n^{(2)}) = n^{\Theta _H(1)}$
, i.e. that
$r(H,K_n^{(2)})$
grows polynomially in
$n$
for any fixed
$H$
. However, our knowledge of the correct order of polynomial growth is limited to extremely few graphs
$H$
. For example, while an early result of Erdős [Reference Erdős8] implies that
$r(K_3^{(2)}, K_n^{(2)}) = n^{2+o(1)}$
(with many further developments [Reference Ajtai, Komlós and Szemerédi1–Reference Campos, Jenssen, Michelen and Sahasrabudhe3, Reference Pontiveros, Griffiths and Morris13, Reference Hefty, Horn, King and Pfender15, Reference Kim16, Reference Shearer23] obtaining much more precise asymptotic bounds), it was only extremely recently that Mattheus and Verstraëte [Reference Mattheus and Verstraete17] proved that
$r(K_4^{(2)}, K_n^{(2)}) = n^{3+o(1)}$
. It remains a major open problem to determine the order of polynomial growth of
$r(H,K_n^{(2)})$
for most other graphs
$H$
, for example when
$H$
is a clique of order at least
$5$
, or a cycle of length at least
$4$
.
For hypergraphs of uniformity
$k \geq 3$
, our understanding is even more limited. For a general fixed
$H$
, the best known upper bound on
$r(H,K_n^{(k)})$
follows from the stepping-down technique of Erdős and Rado [Reference Erdős and Rado11], which implies that
\begin{equation} r\big(H,K_n^{(k)}\big) \leq {\textrm {twr}}_{k-1}\big(n^{O_H(1)}\big), \end{equation}
where the tower function is recursively defined by
${\textrm {twr}}_1(x)=x$
and
${\textrm {twr}}_i(x) = 2^{{\textrm {twr}}_{i-1}(x)}$
for
$i \geq 2$
.
Perhaps surprisingly, the upper bound (1.1) turns out to be tight (up to the implicit constant) in many cases, including when
$H$
is a clique of order at least
$k+2$
[Reference Mubayi and Suk19]. The proof of this result, and of many other lower bounds [Reference Conlon, Fox and Gunby4, Reference Conlon, Fox and Sudakov6, Reference Conlon, Fox and Sudakov7, Reference Mubayi and Suk18] on hypergraph Ramsey numbers, uses a variant of the celebrated stepping-up construction of Erdős, Hajnal, and Rado [Reference Erdős, Hajnal and Rado10], which we shortly discuss in more detail.
Although the bound in (1.1) is tight when
$H$
is a clique, it is natural to expect that it is far from tight when
$H$
is “far from being a clique”. For example, a simple supersaturation argument, going back at least to the early work of Erdős and Hajnal [Reference Erdős and Hajnal9], demonstrates that
$r(H,K_n^{(k)})=n^{O_H(1)}$
whenever
$H$
is
$k$
-partite, or, more generally, iterated
$k$
-partite.Footnote
1
It is conjectured in [Reference Conlon, Fox and Gunby5] that this is in fact a complete characterisation:
$H$
is iterated
$k$
-partite if and only if
$r(H,K_n^{(k)}) = n^{O_H(1)}$
.
Towards resolving this conjecture, or more generally towards understanding the behaviour of
$r(H,K_n^{(k)})$
for general
$H$
, it is natural to ask what happens when
$H$
is still “far from clique-like”, but is not necessarily iterated
$k$
-partite. One such class of hypergraphs is the class of linear hypergraphs, in which every pair of edges intersects in at most one vertex. A longstanding folklore conjecture in the field was that in uniformity
$k=3$
, every linear
$3$
-graph
$H$
satisfies
$r(H,K_n^{(3)}) = n^{O_H(1)}$
. However, this conjecture was recently disproved in [Reference Conlon, Fox and Gunby4].
Theorem 1.1 ([Reference Conlon, Fox and Gunby4, Theorem 1.4]). For every
$C\gt 1$
, there exists a linear
$3$
-graph
$H$
such that
$r(H,K_n^{(3)}) \geq 2^{(\log n)^C}$
for all sufficiently large
$n$
.
In particular, this lends some credence to the conjecture of [Reference Conlon, Fox and Gunby5]: since linear hypergraphs need not have polynomial off-diagonal Ramsey numbers, perhaps the only relevant structure is that of being iterated
$k$
-partite.
Our main result is a generalisation of Theorem1.1 to arbitrary uniformities.
Theorem 1.2.
For every
$C\gt 1$
and every
$k \geq 3$
, there exists a linear
$k$
-graph
$H$
such that
\begin{equation*} r\big(H,K_n^{(k)}\big) \geq {\textrm {twr}}_{k-2}\big(2^{(\log n)^C}\big) \end{equation*}
for all sufficiently large
$n$
.
We note that the tower height in Theorem1.2 is nearly best possible, by (1.1). We also remark that, if we let
$K_{n,\ldots ,n}^{(k)}$
denote the complete
$k$
-partite
$k$
-graph with parts of size
$n$
, then a simple supersaturation argument (see [Reference Fox and He14, Proposition 7.2]) demonstrates that
$r(H,K_{n,\ldots ,n}^{(k)}) \leq n^{O_H(1)}$
for any linear
$k$
-graph
$H$
. Thus, Theorem1.2 implies that
$r(H,K_n^{(k)})$
and
$r(H,K_{n,\ldots ,n}^{(k)})$
can be extremely far apart, in that the latter is polynomial and the former grows as a tower of height at least
$k-2$
. Previously no such separation was known for any
$H$
; in particular, if
$H$
is a fixed clique, then
$r(H,K_n^{(k)})$
and
$r(H,K_{n,\ldots ,n}^{(k)})$
are expected to be of roughly the same order.
In order to prove Theorem1.2, we need to both define the linear
$k$
-graph
$H$
and to define a colouring of
$E(K_N^{(k)})$
containing no red copy of
$H$
and no large blue clique. As in many previous works, our colouring is based on the stepping-up construction of Erdős, Hajnal, and Rado [Reference Erdős, Hajnal and Rado10]. In particular, our technique takes as input a linear
$(k-1)$
-graph
$F$
, and outputs a linear
$k$
-graph
$H$
whose off-diagonal Ramsey number is exponential in that of
$F$
, as stated in the following proposition.
Proposition 1.3.
Let
$k \geq 4$
and let
$H$
be a linear
$(k-1)$
-graph. There exists a linear
$k$
-graph
$H'$
such that
\begin{equation*} r\big(H',K_{2n+2k}^{(k)}\big) \gt 2^{r\big(H,K_n^{(k-1)}\big)-1}. \end{equation*}
Note that Proposition1.3 immediately implies Theorem1.2 by induction on
$k$
, with Theorem1.1 serving as the base case.
As discussed above, the colouring used in the proof of Proposition1.3 is based on the stepping-up technique. Although variants of this technique appear frequently in the literature, most papers simply record certain ad hoc properties needed for the particular application. We therefore present in section 2 a slightly more general treatment of the properties of this construction, in the hopes that future researchers can use some of these in a black-box way. We then present the construction of the linear
$k$
-graph
$H$
in section 3, and complete the proof of Proposition1.3, and thus of Theorem1.2.
2. Stepping-up
As mentioned in the introduction, the problem of estimating
$r(H,K_n^{(k)})$
can be equivalently viewed as estimating the minimum independence number of
$N$
-vertex
$H$
-free
$k$
-graph. In the remainder of this paper we use the latter perspective, as our arguments are slightly more natural in this language.
In classical stepping-up constructions, given a
$(k-1)$
-graph
$G$
on
$\{0,\ldots ,N-1\}$
, one constructs a
$k$
-graph
$\widetilde {G}$
whose vertices are all vectors
$\varepsilon =(\varepsilon _{N-1},\varepsilon _{N-2},\ldots , \varepsilon _0)\in \{0,1\}^N$
, ordered lexicographically. One also considers, for any two distinct strings
$\varepsilon ,\varepsilon '$
, the largest index
$\delta =\delta (\varepsilon ,\varepsilon ')$
where they differ. Observe that
$\varepsilon \lt \varepsilon '$
if and only if
$\varepsilon _{\delta }\lt \varepsilon '_{\delta }$
where
$\delta = \delta (\varepsilon , \varepsilon ')$
. Finally, for any
$k$
strings
$\varepsilon ^{(1)}\lt \cdots \lt \varepsilon ^{(k)}$
, we decide whether they form an edge in
$\widetilde {G}$
by some deterministic rule about the tuple
$(\delta _1,\ldots ,\delta _{k-1})$
where
$\delta _i =\delta (\varepsilon ^{(i)},\varepsilon ^{(i+1)})$
. The exact rule depends on the application, though typically if
$\delta _1,\ldots , \delta _{k-1}$
is monotone, then
$\{\varepsilon ^{(1)},\ldots , \varepsilon ^{(k)}\}$
is an edge if and only if
$\{\delta _1,\ldots ,\delta _{k-1}\}$
is an edge in
$G$
.
In this paper, we switch to a new set of notations that represents each vertex in the stepping-up construction with a single number instead of a string. We will also rephrase the conditions on
$\delta _1,\ldots , \delta _{k-1}$
in terms of binary trees, which should allow one to visualise the constructions better.
First of all, suppose that
$a,a'$
are two distinct numbers in
$\{0,1,\ldots , 2^N-1\}$
. If their binary representations are
$\varepsilon ,\varepsilon '$
, then it is clear that
$a\lt a'$
if and only if
$\varepsilon \lt \varepsilon '$
. More importantly,
$\delta (\varepsilon ,\varepsilon ')$
is simply the largest non-negative integer
$\delta$
such that
$\lfloor 2^{-\delta }a\rfloor \neq \lfloor 2^{-\delta }a'\rfloor$
. We generalise this observation to the definition below.
Definition 2.1.
Let
$S$
be a set of at least two positive integers. The top splitting level
$\ell (S)$
of
$S$
is the maximum non-negative integer
$\ell$
such that
$\{\lfloor 2^{-\ell }s\rfloor \, :\, s\in S\}$
has size at least two. The left subset
$S_{\textrm {left}}$
of
$S$
is the set of
$s\in S$
with
$\lfloor 2^{-\ell (S)}s\rfloor$
even, and the right subset
$S_{\textrm {right}}$
of
$S$
is the set of
$s\in S$
with
$\lfloor 2^{-\ell (S)}s\rfloor$
odd.
For example, if
$S=\{5,6,7,8,9\}$
, then
$\ell (S)=3$
,
$S_{\textrm {left}}=\{5,6,7\}$
, and
$S_{\textrm {right}}=\{8,9\}$
. Note that, by the definition of
$\ell (S)$
, for any
$s\in S$
, there exists an integer
$q_s$
and an integer
$r_s$
such that
$s=q_s2^{\ell (S)+1}+r_s$
where
$0\le r_s\lt 2^{\ell (S)+1}$
. Thus
$\lfloor 2^{-\ell (S)}s\rfloor$
is either
$2q_s$
or
$2q_s+1$
, depending on whether
$r_s\le 2^{\ell (S)}$
or not. Hence,
$S_{\textrm {left}}$
,
$S_{\textrm {right}}$
are well-defined and always non-empty.
To get the full list of
$\delta _1,\ldots , \delta _{k-1}$
, we need to iteratively split the subsets into left subsets and right subsets until they all have size
$1$
. If we record how the subsets split using a binary tree, we get the following definition.
Definition 2.2.
For any set of non-negative integers
$S$
, its binary structure
$b(S)$
is a weighted rooted ordered binary tree defined iteratively as follows.
-
(1) If
$\left \lvert S\right \rvert =1$
, then
$b(S)$
is a single root vertex with weight
$1$
. -
(2) If
$\left \lvert S\right \rvert \gt 1$
, let
$b(S)$
be the binary tree so that the left subtree of the root is
$b(S_{\textrm {left}})$
, and the right subtree of the root is
$b(S_{\textrm {right}})$
. The weight of the root is
$\left \lvert S\right \rvert$
.
Note that the weight of any vertex
${v}$
in
$b(S)$
is equal to the number of leaves in the subtree rooted at
${v}$
.
For example, the following figure is the binary structure of
$S=\{5,6,7,8,9\}$
. To compute the
$\delta$
’s, we can simply read the internal nodes from left to right and see which level they split. For example, for
$\varepsilon ^{(1)} = (0,1,0,1)$
,
$\varepsilon ^{(2)}=(0,1,1,0)$
,
$\ldots$
,
$\varepsilon ^{(5)} = (1,0,0,1)$
, we have
$(\delta _1,\delta _2,\delta _3,\delta _4) = (1,0,3,0)$
.
Figure 1. Binary structure of
$S=\{5,6,7,8,9\}$
.
As mentioned earlier, the case where
$\delta _1,\ldots , \delta _{k-1}$
is monotone is usually special. Rephrasing this in terms of binary structures gives the following definition.
Definition 2.3.
The binary structure
$b(S)$
is increasing if all right subtrees of internal nodes are singletons, and it is decreasing if all left subtrees of internal nodes are singletons. The binary structure is monotone if it is increasing or decreasing. If
$S$
is a set of non-negative integers whose binary structure is monotone, let
$L(S)$
be iteratively defined as follows.
-
(1) If
$\left \lvert S\right \rvert =1$
, let
$L(S)$
be the empty set.
-
(2) If
$\left \lvert S\right \rvert \gt 1$
, let
$S'$
be
$S_{\textrm {left}}$
if
$b(S)$
is increasing, and let
$S'$
be
$S_{\textrm {right}}$
if
$b(S)$
is decreasing. Then set
$L(S) {\stackrel {\mbox{ def}}{=}} L(S')\cup \{\ell (S)\}$
.
For example,
$\{5,6,7,8,9\}$
does not have a monotone binary structure (Figure 1), whereas
$\{1,2,4,8,16\}$
is decreasing with
$L(\{1,2,4,8,16\}) = \{0,1,2,3\}$
(Figure Reference Bohman and Keevash2). Note that if
$S$
is monotone, then
$L(S)$
is simply the set of levels of the internal nodes in the binary structure of
$S$
.
Figure 2. Binary structure of
$S=\{1,2,4,8,16\}$
.
Finally, instead of classifying
$k$
-tuples using the relative order of
$\delta _1,\ldots , \delta _{k-1}$
, we classify them based on types of their structures.
Definition 2.4.
A type of binary structure
$T$
is a weighted rooted ordered binary tree such that each leaf vertex has a positive integer weight, and each internal vertex has weight equal to the sum of the weights of its children. Its size is the weight of its root. A set of positive integers
$S$
is of type
$T$
if
$T$
can be obtained from
$b(S)$
by iteratively removing leaves. The type
$T$
is monotone if there is some set of positive integer
$S$
of type
$T$
with
$b(S)$
monotone.
Before we introduce our reformulation of stepping-up, let us first review the classical construction given by Erdős, Hajnal and Rado [Reference Erdős, Hajnal and Rado10]. For simplicity, assume that
$k=4$
, and we are given a
$3$
-graph
$G$
on
$N$
vertices with small independence number and clique number. The stepping-up of
$G$
is a
$4$
-graph on
$2^N$
vertices defined as follows: for any
$\{\varepsilon ^{(1)},\ldots ,\varepsilon ^{(4)}\}$
, include it as an edge if one of the following holds:
-
•
$\delta _1,\delta _2,\delta _3$
is monotone and forms an edge in
$G$
. -
•
$\delta _1\lt \delta _2\gt \delta _3$
.
Using our terminology, the first condition is satisfied for some set
$S$
of size
$4$
precisely when
$b(S)$
is monotone and
$L(S)$
is an edge in
$G$
. With some thought, we can see that the second condition is precisely when
$b(S)$
is of type
$T_{2,2}$
, where
$T_{2,2}$
is the binary tree with weight
$4$
at the root and weight
$2$
at both of its children, both of which are leaves (Figure 3).
Figure 3. Type
$T_{2,2}$
.
Our variant of stepping-up will also include two such conditions, which we now define in turn. We begin with the monotone edges.
Definition 2.5.
Let
$G$
be a
$(k-1)$
-graph on
$\{0,1,\ldots , N-1\}$
. Its left stepping-up is the
$k$
-graph on
$\{0,\ldots , 2^N-1\}$
consisting of all edges
$e$
with
$b(e)$
increasing and
$L(e)\in E(G)$
. Its right stepping-up is the
$k$
-graph on
$\{0,\ldots , 2^N-1\}$
consisting of all edges
$e$
with
$b(e)$
decreasing and
$L(e)\in E(G)$
.
Now we can define the stepping-up that we will use throughout the paper.
Definition 2.6.
For any tuple
$(G_1,G_2,\mathcal{T})$
where
$G_1,G_2$
are two
$(k-1)$
-graphs on
$\{0,1,\ldots , N-1\}$
and
$\mathcal{T}$
is a family of types of binary structure of size
$k$
, its stepping-up is the
$k$
-graph
$G$
on vertex set
$\{0,\ldots , 2^N-1\}$
formed by taking the edge union
\begin{equation*}G {\stackrel {\mbox{ def}}{=}} \widetilde {G_1}\cup \widetilde {G_2}\cup G_{\mathcal{T}}\end{equation*}
where
$\widetilde {G_1}$
is the left stepping-up of
$G_1$
,
$\widetilde {G_2}$
is the right stepping-up of
$G_2$
, and
$G_{\mathcal{T}}$
is the
$k$
-graph with
\begin{equation*}E(G_{\mathcal{T}}) = \left \{e\in \binom {\{0,\ldots , 2^N-1\}}{k}\,:\, e\text { is of type }T\text { for some }T\in \mathcal{T}\right \}\!.\end{equation*}
For example, the classical stepping-up construction discussed above for
$k=4$
is precisely the stepping-up of
$(G,G,\{T_{2,2}\})$
. Note that the complement can also be described as a stepping-up: it is the stepping-up of
$(G^c, G^c, \{T_{1,(2,1)},T_{(1,2),1}\})$
where
$T_{1,(2,1)}$
and
$T_{(1,2),1}$
are drawn in Figure 4 below.
Figure 4. Type
$T_{1,(2,1)}$
and type
$T_{(1,2),1}$
.
Having rephrased the classical construction, we now turn to bounding its independence number, recalling that our ultimate goal is to construct a
$k$
-graph with no large independent sets. To bound the independence number of the stepped-up hypergraph, we define the following auxiliary functions.
Definition 2.7.
For any positive integers
$n_1, n_2$
and any family
$\mathcal{T}$
of types of binary structure, let
$f(n_1,n_2,\mathcal{T})$
be the maximum possible size of set
$S$
of non-negative integers that does not contain any
$n_1$
-set with an increasing binary structure, any
$n_2$
-set with a decreasing binary structure, or any set of type
$T$
for any
$T\in \mathcal{T}$
.
The following statement then follows easily from the definition.
Lemma 2.8.
Let
$G_1,G_2$
be two
$(k-1)$
-graphs on
$\{0,1,\ldots , N-1\}$
, and let
$\mathcal{T}$
be a family of types of binary structure of size
$k$
. Let
$G$
be the stepping-up of
$(G_1,G_2,\mathcal{T})$
. Then
\begin{equation*}\alpha (G)\leq f\left (\alpha (G_1)+2,\alpha (G_2)+2,\mathcal{T}\right ).\end{equation*}
Proof. Let
$S\subseteq \{0,\ldots , 2^N-1\}$
be an independent set in
$G$
. If
$S$
contains an
$(\alpha (G_1)+2)$
-set
$S'$
with an increasing binary structure, then
$\left \lvert L(S')\right \rvert =\alpha (G_1)+1\gt \alpha (G_1)$
, hence
$L(S')$
contains an edge
$e'$
in
$G_1$
. This shows that
$S$
contains a
$k$
-set
$e$
with
$b(e)$
increasing and
$L(e)=e'$
, which is a contradiction. Therefore
$S$
does not contain any
$(\alpha (G_1)+2)$
-set
$S'$
with
$b(S')$
increasing. Similarly,
$S$
does not contain any
$(\alpha (G_2)+2)$
-set
$S'$
with
$b(S')$
decreasing. Lastly, from the way
$G$
is constructed, it is clear that
$S$
does not contain any sets of type
$T$
for any
$T\in \mathcal{T}$
. As a consequence,
$\left \lvert S\right \rvert \leq f\left (\alpha (G_1)+2,\alpha (G_2)+2,\mathcal{T}\right ).$
In order to apply Lemma2.8, we also need to compute
$f(n_1,n_2,\mathcal{T})$
. This can be done by dynamic programming if
$\mathcal{T}$
is explicitly known. Here we carry out the computation for two particularly useful cases.
For the first case, let
$T_{a,b}$
be the type of binary structure where the children of the root are leaves with weight
$a$
on the left and
$b$
on the right. We show that
$f(n_1,n_2,\mathcal{T})$
is linear in
$n_1+n_2$
if
$\mathcal{T}$
contains some type of the form
$T_{a,b}$
.
Lemma 2.9.
Let
$n_1,n_2,k$
be positive integers with
$n_1,n_2\geq 2$
. Let
$\mathcal{T}$
be a family of types of binary structure of size
$k$
containing types of the form
$T_{a,b}$
for some
$a+b=k$
. Furthermore, let
$a_{\min }$
be the minimum possible value for
$a$
, and
$b_{\min }$
be the minimum possible value for
$b$
. Then
\begin{equation*}f\left (n_1,n_2,\mathcal{T}\right )\leq (a_{\min }-1)(n_2-2)+(b_{\min }-1)(n_1-2)+2k.\end{equation*}
Proof. We will proof by induction on
$n_1+n_2$
. The case
$n_1+n_2=4$
holds trivially as
$n_1=n_2=2$
in this case, which forces
$f(n_1,n_2,\mathcal{T})=1$
as any set of size
$2$
has an increasing binary structure. The inductive step is going to be established by the inequality
\begin{equation} f(n_1,n_2,\mathcal{T})\leq \max \left \{a_{\min }-1+f(n_1,n_2-1,\mathcal{T}), b_{\min }-1+f(n_1-1,n_2,\mathcal{T}),2k\right \} \end{equation}
Indeed, assuming the claim holds for
$f(n_1,n_2-1,\mathcal{T})$
and
$f(n_1-1,n_2,\mathcal{T})$
, then we have
\begin{equation*}a_{\min }-1+f(n_1,n_2-1,\mathcal{T}) \leq (a_{\min }-1)(n_2-2)+(b_{\min }-1)(n_1-2)+2k,\end{equation*}
\begin{equation*}b_{\min }-1+f(n_1-1,n_2,\mathcal{T}) \leq (a_{\min }-1)(n_2-2)+(b_{\min }-1)(n_1-2)+2k,\end{equation*}
and
\begin{equation*}2k\leq (a_{\min }-1)(n_2-2)+(b_{\min }-1)(n_1-2)+2k.\end{equation*}
Therefore (2.1) and the inductive hypothesis together imply
\begin{equation*}f(n_1,n_2,\mathcal{T})\leq (a_{\min }-1)(n_2-2)+(b_{\min }-1)(n_1-2)+2k,\end{equation*}
and it remains to prove (2.1).
Suppose that
$S$
is a set with no
$n_1$
-sets with increasing binary structures, no
$n_2$
-sets with decreasing binary structures, and no sets of type
$T$
for any
$T\in \mathcal{T}$
. Note that
$S_{\textrm {right}}$
now cannot contain any
$(n_2-1)$
-set with decreasing binary structure: indeed, if
$S'\subseteq S_{\textrm {right}}$
satisfies
$\left \lvert S'\right \rvert =n_2-1$
and
$b(S')$
is decreasing, then
$\{s\}\cup S'$
also has decreasing structure for any
$s\in S_{\textrm {left}}$
, which is a contradiction. Therefore
\begin{equation*}\left \lvert S_{\textrm {right}}\right \rvert \leq f(n_1,n_2-1,\mathcal{T}).\end{equation*}
Similarly
\begin{equation*}\left \lvert S_{\textrm {left}}\right \rvert \leq f(n_1-1,n_2,\mathcal{T}).\end{equation*}
If
$\left \lvert S_{\textrm {left}}\right \rvert \lt a_{\min }$
or
$\left \lvert S_{\textrm {right}}\right \rvert \lt b_{\min }$
, then (2.1) now follows immediately. Therefore let us now assume
$\left \lvert S_{\textrm {left}}\right \rvert \geq a_{\min }$
and
$\left \lvert S_{\textrm {right}}\right \rvert \geq b_{\min }$
. Since
$S$
does not contain any set of type
$T_{a_{\min },k-a_{\min }}$
or
$T_{k-b_{\min },b_{\min }}$
, we see that
$\left \lvert S_{\textrm {right}}\right \rvert \lt k-a_{\min }$
and
$\left \lvert S_{\textrm {left}}\right \rvert \lt k-b_{\min }$
. This shows that
$\left \lvert S\right \rvert \lt 2k-a_{\min }-b_{\min }\lt 2k$
, as desired.
We will use Lemma2.9 to get a better quantitative dependence when stepping-up. However, if one is only interested in understanding the tower heights of Ramsey numbers, then it suffices to use the much cheaper bound below.
The depth of a binary tree is the number of edges in the longest path from the root to a leaf.
Lemma 2.10.
Let
$k,d$
be two non-negative integers. If
$T$
is a type of binary structure of size
$k$
and depth
$d$
, then
\begin{equation*}f(n_1,n_2,\{T\})\leq k(n_1+n_2)^d.\end{equation*}
Proof. We induct on
$d$
. If
$d=0$
, then any set of non-negative integers of size
$\left \lvert T\right \rvert =k$
is of type
$T$
, which shows that
$f(n_1,n_2,\{T\})\leq k-1\leq k(n_1+n_2)^{d}$
, proving the base case of the induction.
Now suppose that the claim holds for all smaller depths, and we wish to prove it for
$d$
. We now do an internal induction on
$n_1+n_2$
. The base case
$n_1=n_2=1$
is again trivial, since any
$1$
-set has an increasing binary structure, hence
$f(1,1,\{T\}) =0 \leq k(1+1)^d$
. So we may now assume that the claim holds for any positive integers
$n_1',n_2'$
with
$n_1'+n_2'\lt n_1+n_2$
.
Let
$S$
be a set of non-negative integers that avoids
$n_1$
-sets with increasing binary structures,
$n_2$
-sets with decreasing binary structures, and
$k$
-sets of type
$T$
. Let
$T_{\textrm {left}}$
and
$T_{\textrm {right}}$
be the left and right subtrees of the root of
$T$
, respectively. Note that both of these trees have order at most
$k$
and depth at most
$d-1$
. Suppose that
$\left \lvert S\right \rvert \geq 2$
. As in the previous proof, we know that
$S_{\textrm {left}}$
does not contain any
$(n_1-1)$
-sets with increasing binary structures. Similarly,
$S_{\textrm {right}}$
does not contain any
$(n_2-1)$
-sets with decreasing binary structures.
Moreover, it cannot simultaneously hold that
$S_{\textrm {left}}$
contains a set of type
$T_{\textrm {left}}$
, and that
$S_{\textrm {right}}$
contains a set of type
$T_{\textrm {right}}$
, as these would combine to give a set of type
$T$
in
$S$
. If
$S_{\textrm {left}}$
does not contain a set of type
$T_{\textrm {left}}$
, then
\begin{align*} \left \lvert S\right \rvert &= \left \lvert S_{\textrm {left}}\right \rvert +\left \lvert S_{\textrm {right}}\right \rvert \\ &\leq f(n_1-1,n_2,\{T_{\textrm {left}}\})+f(n_1,n_2-1,\{T\})\\ &\leq k(n_1-1+n_2)^{d-1} + k(n_1+n_2-1)^d\\ &\leq k(n_1+n_2)^d, \end{align*}
where we use the inductive hypothesis in the penultimate step and the inequality
$(x-1)^{d-1}+(x-1)^{d} \leq x^d$
(valid for all integers
$x,d \geq 1$
) in the final step. Similarly, if
$S_{\textrm {right}}$
does not contain a set of type
$T_{\textrm {right}}$
, then
\begin{align*} \left \lvert S\right \rvert &= \left \lvert S_{\textrm {left}}\right \rvert +\left \lvert S_{\textrm {right}}\right \rvert \\ &\leq f(n_1-1,n_2,\{T\})+f(n_1,n_2-1,\{T_{\textrm {right}}\})\\ &\leq k(n_1-1+n_2)^d + k(n_1+n_2-1)^{d-1}\\ &\leq k(n_1+n_2)^d. \end{align*}
In either case we get the desired bound, completing the induction.
We briefly remark that from Lemma2.9, it is easy to see that the stepping-up construction for
$k=4$
of Erdős, Hajnal and Rado only grows the independence number at most linearly. To bound the clique number, we can apply Lemma2.10 to see that the independence number of the complement only grows at most quadratically as
$T_{1,(2,1)}$
has depth
$2$
. To show that the clique number grows at most linearly, one would need to compute
$f(n_1,n_2,\{T_{1,(2,1)},T_{(1,2),1}\})$
, and it is not hard to see that once we take
$T_{(1,2),1}$
into consideration, this function does indeed grow linearly in
$n_1$
and
$n_2$
. We remark too that one can similarly rephrase stepping-up for general uniformities using our binary structures framework, but we omit the details as this does not improve on the classical construction.
3. Linear hypergraphs
We now turn to the proof of our technical result, Proposition1.3. The main remaining difficulty is the construction of the linear
$k$
-graph; the following result gives the main property we need of this construction, namely that it is avoided by the stepping-up constructions discussed in the previous section.
Theorem 3.1.
Let
$k\geq 4$
, and let
$H$
be a linear
$(k-1)$
-graph with no isolated vertices. Then there exists a linear
$k$
-graph
$H'$
such that the following holds. Suppose that
$G$
is a
$(k-1)$
-graph that is
$H$
-free, and
$\mathcal{T}$
is a family of types of binary structure of size
$k$
that are not monotone. Then the stepping-up
$\widetilde {G}$
of
$(G,G,\mathcal{T})$
is
$H'$
-free.
We remark that we need
$k\geq 4$
because any type of binary structure of size at most
$3$
is monotone. Given Theorem3.1, we can readily complete the proof of Proposition1.3 and hence of Theorem1.2.
Proof of Theorem 1.3. For any linear
$(k-1)$
-graph
$H$
, let
$H'$
be the linear
$k$
-graph obtained from Theorem3.1. By definition, there exists a
$(k-1)$
-graph
$G$
on
$ \{0,1,\ldots , r(H,K_n^{(k-1)})-2 \}$
that is
$H$
-free with
$\alpha (G)\leq n-1$
. Let
$\mathcal{T}$
be
$\{T_{2,k-2},T_{k-2,2}\}$
. As
$k\geq 4$
, we know that both children have weight at least
$k-2\geq 2$
and so
$T_{2,k-2}$
and
$T_{k-2,2}$
are not monotone.
Let
$\widetilde {G}$
be the stepping-up of
$(G,G,\mathcal{T})$
. Then we know that
$\widetilde {G}$
is
$H'$
-free and
\begin{equation*}\alpha (\widetilde {G})\leq 2\alpha (G)+2k\lt 2n+2k\end{equation*}
by Lemmas 2.8 and 2.9. This shows that
\begin{equation*}r\big(H', K_{2n+2k}^{(k)}\big)\gt \lvert {V(\widetilde {G})}\rvert = 2^{r\big(H,K_n^{(k-1)}\big)-1},\end{equation*}
as desired.
We now turn to the proof of Theorem3.1, which occupies the remainder of this section. We first make the following two definitions, which arise naturally when analysing stepping-up constructions.
Definition 3.2.
A dyadic partition of a set
$A$
is a partition
$A=A_1\sqcup A_2 \sqcup \cdots \sqcup A_t$
with the property that
$|A_i| \le \frac {1}{2} (|A_i| + \cdots + |A_t|)$
for all
$i \lt t$
and
$|A_t| = 1$
.
The following lemma captures some properties we need of our construction of a linear
$k$
-graph.
Lemma 3.3.
Let
$k \geq 3$
, and fix an ordered linear
$k$
-graph
$F^\lt$
. There exists a linear
$k$
-graph
$H'$
with the following property. For every dyadic partition
$V(H')=A_1 \sqcup \ldots \sqcup A_t$
, and for every two-colouring
$\beta \,:\,[t] \to \{L,R\}$
, and for every ordering
$\pi$
of
$V(H')$
, there exists an ordered monochromatic transversal copy of
$F^\lt$
. That is, there is a copy of
$F^\lt$
each of whose vertices lie in a different part
$A_i$
, such that all these
$A_i$
receive the same colour under
$\beta$
, and such that the order of these vertices under
$\pi$
agrees with the fixed ordering of
$F^\lt$
.
We remark that Lemma3.3 works for
$k=3$
. However, our stepping-up construction only works for
$k\geq 4$
, as in the statement of Theorem3.1. We will apply Lemma3.3 with
$F^\lt$
chosen to be a certain
$k$
-graph derived from a given
$(k-1)$
-graph, which we now define.
Definition 3.4.
Let
$H$
be a
$(k-1)$
-graph. Its expansion
$H^+$
is the
$k$
-graph obtained from
$H$
by adding to each
$e \in E(H)$
a new vertex
$v_e$
, which participates in no other edges.
An ordered expansion of
$H$
is any ordering of
$V(H^+)$
in which, for all
$e \in E(H)$
, the vertex
$v_e$
is the smallest vertex of
$e\cup \{v_e\}$
.
For example, we may obtain an ordered expansion by putting all the new vertices
$v_e$
first, followed by all the old vertices of
$H$
. Using Lemma3.3 and the properties of the stepping-up construction discussed above, we are now ready to prove Theorem3.1.
Proof of Theorem 3.1. Let
$F^\lt$
be an ordered expansion of
$H$
, and let
$H'$
be the
$k$
-graph given by Lemma3.3. We will show that
$H'$
satisfies the requirement of Theorem3.1. Suppose for the sake of contradiction that
$\widetilde {G}$
contains a copy of
$H'$
. In this case, we will think of
$V(H')$
as a subset of
$V(\widetilde {G})$
and produce a dyadic partition
$A_1\sqcup \cdots \sqcup A_t$
with a colouring
$\beta \,:\,[t]\to \{L,R\}$
of
$V(H')$
as follows. Starting with
$B_0 = V(H')$
, suppose that we have defined
$B_i$
and
$A_1,\ldots , A_i$
so that
$V(H') = A_1\sqcup \cdots \sqcup A_i\sqcup B_i$
. If
$B_i$
is a singleton, then we set
$t=i+1, A_{i+1}=B_i$
and then terminate. Therefore we will now assume that
$\left \lvert B_i\right \rvert \geq 2$
, and so its left and right subsets are well-defined and non-empty. Let
$(A_{i+1},B_{i+1})$
be a permutation of
$((B_i)_{\textrm {left}}, (B_i)_{\textrm {right}})$
so that
$\left \lvert A_{i+1}\right \rvert \leq \left \lvert B_{i+1}\right \rvert$
. Moreover, we set
$\beta (i+1)=L$
if
$A_{i+1}$
is the left subset of
$B_i$
, and
$\beta (i+1)=R$
otherwise. Finally, we increase
$i$
by
$1$
and continue the process until it terminates. To check that
$A_1\sqcup \cdots \sqcup A_t$
is indeed a dyadic partition of
$V(H')$
, note that the procedure guarantees that it is a partition with
$\left \lvert A_t\right \rvert =1$
. In addition, for all
$1\leq i\lt t$
, we have that
$\left \lvert A_i\right \rvert \leq \frac{1}{2}\left (\left \lvert A_i\right \rvert +\left \lvert B_i\right \rvert \right ) = \frac{1}{2}\left (\left \lvert A_i\right \rvert +\cdots +\left \lvert A_t\right \rvert \right )$
.
Having constructed the dyadic partition and the colouring, we now construct the ordering
$\pi$
of
$V(H')$
. For any
$v\in V(H')$
, let
$i(v)$
be the unique
$i$
such that
$v\in A_i$
. Then for any
$u,v\in V(H')$
, we set
$u\gt _{\pi } v$
if
$i(u)\lt i(v)$
. If
$i(u)=i(v)$
, then we arbitrarily order
$u$
and
$v$
with respect to
$\pi$
.
By Lemma3.3, there is an ordered monochromatic transversal copy of
$F^\lt =H^+$
. We will use this copy to find a copy of
$H$
in
$G$
, which gives a contradiction. First assume that the ordered monochromatic transversal copy of
$H^+$
is monochromatic of color
$R$
. We will again think of
$V(H^+)$
as a subset of
$V(H')$
, which is a subset of
$V(\widetilde {G})$
. By the definition of
$H^+$
, there is an injection
$\varphi \, : \, V(H)\to V(H^+)$
such that for each edge
$e\in E(H)$
, we have that
$\varphi (e)$
is contained in an edge
$e'$
in
$V(H^+)$
and
$\varphi (e)$
does not contain the smallest element
$v_e$
(with respect to
$\pi$
) in
$e'$
. Now we define
$\varphi ^* \, : \, V(H)\to V(G)=\{0,1,\ldots , N-1\}$
as
$\varphi ^*(v) = \ell \left (B_{i(\varphi (v))-1}\right )$
, the top splitting level of
$B_{i(\varphi (v))-1}$
. We will show that this embeds
$H$
into
$G$
.
For each
$e\in H$
, we can order the vertices in
$e$
as
$v_1,\ldots , v_{k-1}$
so that
$i(\varphi (v_1))\gt \cdots \gt i(\varphi (v_{k-1}))$
. (Recall that they are distinct as the copy of
$H'$
is transversal). By the definition of
$\pi$
, we have that
$v_e\lt _{\pi }\varphi (v_1)\lt _{\pi }\cdots \lt _{\pi } \varphi (v_{k-1})$
, which shows that
$i(v_e)\gt i(\varphi (v_1))$
again by the definition of
$\pi$
. Now for every
$j\in [k-1]$
, we will show that the top splitting level of
$\{v_e,\varphi (v_1),\ldots , \varphi (v_j)\}$
is
$\varphi ^*(v_j)$
and its right subset is the singleton
$\{\varphi (v_j)\}$
. Note that by construction, it is clear that
$B_{i(\varphi (v_j))-1}\supseteq \{v_e,\varphi (v_1),\ldots , \varphi (v_j)\}$
. Moreover, as
$\beta (i(\varphi (v_j)))=R$
, the left subset of
$B_{i(\varphi (v_j))-1}$
is
$B_{i(\varphi (v_j))}$
, which contains
$v_e,\varphi (v_1)\ldots , \varphi (v_{j-1})$
as
$i(v_e)\gt i(\varphi (v_1))\gt \cdots \gt i(\varphi (v_{j-1}))\gt i(\varphi (v_j))$
; whereas its right subset is
$A_{i(\varphi (v_j))}$
, which contains
$\varphi (v_j)$
by definition. Therefore the top splitting level of
$\{v_e,\varphi (v_1),\ldots , \varphi (v_j)\}$
is
$\ell (B_{i(\varphi (v_j))-1}) = \varphi ^*(v_j)$
, and its right subset is
$\{\varphi (v_j)\}$
.
This immediately implies that
$\{v_e\}\cup \varphi (e)$
has an increasing binary structure with
$L(\{v_e\}\cup \varphi (e))$
is
$\varphi ^*(e)$
. Note that for any
$T\in \mathcal{T}$
, as
$T$
is not monotone, we know that
$\{v_e\}\cup \varphi (e)$
is not of type
$T$
. We also know that
$\{v_e\}\cup \varphi (e)$
is not decreasing as
$k\geq 3$
. Therefore
$\{v_e\}\cup \varphi (e)$
must be in the left stepping-up of
$G$
, and so
$\varphi ^*(e) = L(\{v_e\}\cup \varphi (e))\in E(G)$
by definition. As a consequence,
$\varphi ^*$
is an embedding from
$H$
to
$G$
, which is a contradiction with the assumption that
$G$
is
$H$
-free. Therefore we have reached a contradiction from assuming that there is a monochromatic ordered transversal copy of
$H^+$
in
$H$
with color
$R$
.
The other case where the ordered transversal copy of
$H^+$
is monochromatic of color
$L$
can be dealt with analogously. We swap the roles of left and right, and the rest of the argument holds verbatim. As we reach a contradiction either way, we see that
$\widetilde {G}$
must be
$H'$
-free.
3.1 Properties of dyadic partitions
All that remains is to prove Lemma3.3, which we now turn to. In this subsection, we record two useful facts we will need about the sizes of various parts in a dyadic partition, and which we use in the next subsection to complete the proof. The first demonstrates that it is not possible to put too many elements of
$A$
into a small collection of the sets
$A_i$
.
Lemma 3.5.
Let
$A=A_1\sqcup A_2 \sqcup \cdots \sqcup A_t$
be a dyadic partition. Then for every
$I\subseteq [t]$
,
\begin{equation*} |A|-\sum _{i\in I}|A_i|\ge \left \lfloor \frac {|A|}{2^{|I|}}\right \rfloor . \end{equation*}
Proof. Let
$I=\{i_1,i_2,\ldots ,i_s\}$
where
$i_1\lt i_2\lt \ldots \lt i_s$
, and let
$b_j=|A|-\sum _{k=1}^j|A_{i_k}|$
. Suppose first that
$i_s\lt t$
. By the definition of a dyadic partition, we have
$b_1\ge |A|/2$
. For
$j\ge 2$
, suppose
$b_{j-1}\ge |A|/2^{j-1}$
. Then
\begin{align*} b_j&=|A|-\sum _{k=1}^{j}|A_{i_k}|=\frac {1}{2}\left (|A|-\sum _{k=1}^{j-1}|A_{i_k}|\right )+\frac {1}{2}\left (|A|-\sum _{k=1}^{j-1}|A_{i_k}|\right )-|A_{i_j}|\\ &\ge \frac {b_{j-1}}{2}+\frac {1}{2}\left (|A|-\sum _{k=1}^{j-1}|A_{i_k}|-\sum _{i=i_j}^t|A_i|\right )\ge \frac {|A|}{2^j}, \end{align*}
where the first inequality uses the fact that
$|A_{i_j}|\le \frac{1}{2} \sum _{i=i_j}^t|A_i|$
, which follows from the definition of a dyadic partition and the fact that
$i_j\lt t$
, and the second inequality uses the inductive hypothesis and the fact that the term in parentheses is non-negative.
This completes the proof in case
$i_s\lt t$
. Additionally, the result is vacuous if
$|A|\lt 2^{|I|}$
. So we may assume that
$i_s=t$
and
$|A|/2^{|I|-1}\gt 2$
. We then apply the argument above to
$I \setminus \{t\}$
to conclude that
\begin{equation*} |A|-\sum _{i\in I}|A_i| = |A|-\sum _{i \in I \setminus \{t\}} |A_i| - |A_t| \geq \frac {|A|}{2^{|I|-1}}-1 \geq \left \lfloor \frac {|A|}{2^{|I|}}\right \rfloor , \end{equation*}
where we use the fact that
$|A_t|=1$
in a dyadic partition, as well as the inequality
$x-1 \geq \lfloor x/2\rfloor$
, valid for all real
$x \geq 1$
.
Our next lemma, which will be crucial in the analysis of the stepping-up construction, is a simple corollary of Lemma3.5.
Lemma 3.6.
Let
$A=A_1\sqcup A_2 \sqcup \cdots \sqcup A_t$
be a dyadic partition. For every
$a\in A$
, let
$i(a)$
be the unique index
$i$
such that
$a\in A_i$
. For every integer
$s\lt t/2$
and every 2-colouring
$\beta \,:\,[t]\rightarrow \{L,R\}$
, there exists
$X\in \{L, R\}$
such that for every
$I\subseteq [t]$
with
$|I|=s-1$
,
\begin{equation*} |\{a\in A\,\big |\,i(a)\not \in I,\beta (i(a))=X\}|\ge \left \lfloor \frac {|A|}{2^{2s-1}}\right \rfloor . \end{equation*}
Proof. Suppose for contradiction that for each
$X\in \{L,R\}$
there exists
$I_X$
with
$|I_X|=s-1$
such that
\begin{equation*} |\{a\in A\,\big |\,i(a)\not \in I_X,\beta (i(a))=X\}|\lt \left \lfloor \frac {|A|}{2^{2s-1}}\right \rfloor . \end{equation*}
Then we have
\begin{equation*} |A|-\sum _{i\in I_L\cup I_R}|A_i|\le \sum _{X\in \{L,R\}}|\{a\in A\,\big |\,i(a)\not \in I_X,\beta (i(a))=X\}|\lt 2 \left \lfloor \frac {|A|}{2^{2s-1}}\right \rfloor \leq \left \lfloor \frac {|A|}{2^{2s-2}}\right \rfloor . \end{equation*}
Note that
$|I_L\cup I_R|\le 2s-2$
, hence by Lemma3.5,
\begin{equation*} |A|-\sum _{i\in I_L\cup I_R}|A_i|\ge \frac {|A|}{2^{|I_L\cup I_R|}}\ge \left \lfloor \frac {|A|}{2^{2s-2}}\right \rfloor . \end{equation*}
This gives a contradiction, and therefore completes the proof.
3.2 Proof of Lemma 3.3
In this section we prove Lemma3.3, thus completing the proof of Theorem1.2.
Let us define an oriented
$s$
-graph to be an
$s$
-graph
$H_0$
, as well as a bijection
$\psi _e:e \to [s]$
for every edge
$e \in E(H_0)$
. Note that in the case
$s=2$
, this precisely recovers the definition of an oriented graph. We say an oriented
$s$
-graph is linear if its underlying unoriented
$s$
-graph is linear.
In the proof of Lemma3.3, we will use the following statement, which is simply the same statement in case
$F^\lt$
is an ordered
$s$
-uniform edge.
Lemma 3.7.
Let
$s \geq 3$
. There exists a linear oriented
$s$
-graph
$H_0$
with the following property. For every dyadic partition
$V(H_0)=A_1 \sqcup \ldots \sqcup A_t$
, and for every two-colouring
$\beta \,:\,[t] \to \{L,R\}$
, and for every ordering
$\pi$
of
$V(H_0)$
, there exists an ordered monochromatic transversal edge. That is, there exists an edge
$e \in E(H_0)$
which goes between
$s$
distinct parts
$A_i$
, all of which receive the same colour under
$\beta$
, such that the ordering given by
$\psi _e$
agrees with the ordering induced by
$\pi$
on
$e$
.
Proof. Let
$n$
be a sufficiently large integer with respect to
$s$
, and let
$c\gt 0$
be a sufficiently small constant. Let
$p = cn^{2-s}$
. Let
$H_0 \sim G_s(n,p)$
be a random
$s$
-graph on
$k$
vertices in which each
$s$
-tuple is made an edge with probability
$p$
, independently over all choices. Additionally, we make
$H_0$
into an oriented
$s$
-graph in a uniformly random way: each edge
$e \in E(H_0)$
selects a uniformly random bijection
$\psi _e\,:\,e \to [s]$
, independently of all other choices. Let
$\mathcal{E}$
be the event that there is an ordered monochromatic transversal edge in
$H_0$
for every dyadic partition of
$V(H_0)$
, every two-colouring of
$[t]$
, and for every ordering of
$V(H_0)$
. We begin by estimating the probability of
$\mathcal{E}$
, and then show that with positive probability we can ensure both that
$\mathcal{E}$
holds and that
$H_0$
is linear.
Let us fix a dyadic partition
$V(H_0)=A_1 \sqcup \ldots \sqcup A_t$
, a colouring
$\beta \,:\,[t] \to \{L,R\}$
, and an ordering
$\pi$
of
$V(H_0)$
. Note that there are at most
$n^n$
dyadic partitions of a set of order
$n$
(because every such partition yields a unique function
$[n] \to [t]$
, and
$t \leq n$
), there are
$2^t \leq 2^n$
choices for
$\beta$
, and there are
$n! \lt n^n$
orderings of
$V(H_0)$
. So in total we need to union-bound over at most
$(2n)^{2n}$
choices for these three data.
For any
$a\in A_1\sqcup \cdots \sqcup A_t$
, let
$i(a)$
be the unique
$i$
such that
$a\in A_i$
. We begin by estimating the number of potential monochromatic transversal edges. This number is exactly
\begin{equation*}T {\stackrel {\mbox{ def}}{=}} \frac {1}{s!}\#\{a_1,\ldots , a_s\in V(H_0)\, : \, i(a_1),\ldots , i(a_s)\text { distinct, and } \beta (i(a_1))=\cdots =\beta (i(a_s))\}.\end{equation*}
We first claim that
$T = \Omega _s(n^s)$
, which we prove as follows. Let
$X\in \{L,R\}$
be such that the conclusion of Lemma 3.6 holds.Footnote
2
Then for any
$i_1,\ldots , i_{s-1}\in [t]$
, there exist at least
$\Omega _s(n)$
choices for
$a_s$
with
$i(a_s)\notin \{i_1,\ldots , i_{s-1}\}$
and
$\beta (i(a_s))=X$
. This shows that for any
$a_1,\ldots , a_m$
with
$i(a_1),\ldots , i(a_m)$
distinct,
$\beta (i(a_1))=\cdots =\beta (i(a_m))=X$
and
$m\lt s$
, there are at least
$\Omega _s(n)$
ways to extend this sequence by
$a_{m+1}$
while maintaining the properties. This shows
$T=\Omega _s(n^s)$
, as claimed.
Now, let
$Z$
be the random variable counting the number of monochromatic transversal edges in
$H_0$
which are correctly ordered, i.e. whose ordering
$\psi _e$
agrees with
$\pi$
. Recall that the random ordering is chosen independently of the other random choices, showing that
$Z$
is distributed as a binomial random variable with parameters
$T$
and
$p/s!$
. Therefore,
\begin{equation*} {\mathbb{P}}(Z=0) = (1-p/s!)^T \leq e^{-pT/s!} \leq e^{-\Omega _s(pn^s)} = e^{-\Omega _s(cn^2)}. \end{equation*}
Applying the union bound over the at most
$(2n)^{2n}$
choices for the dyadic partition as well as
$\beta$
and
$\pi$
, we see that with probability at least
$1-e^{-\Omega _s(cn^2)}$
, we have the claimed property for all possible choices of these three data. In other words,
${\mathbb{P}}({\mathcal{E}}) \geq 1-e^{-\Omega _s(cn^2)}$
.
It remains to ensure that
$H_0$
is linear. For a pair of vertices
$u,v$
, let
${\mathcal{E}}_{u,v}$
be the event that
$u$
and
$v$
are contained in at most one edge of
$H_0$
. By the union bound,
${\mathbb{P}}(\overline {{\mathcal{E}}_{u,v}}) \leq 4^s n^{2s-4} p^2=c^2 4^s$
, as we have at most
$n^{2s-4}$
ways of picking the remaining vertices, at most
$4^s$
ways of partitioning them into two edges, and probability
$p^2$
that both edges appear. Now, we observe that each
${\mathcal{E}}_{u,v}$
is a down-event, so by Harris‘s inequality (or the FKG inequality), we have
\begin{equation*} {\mathbb{P}} \left ( \bigwedge _{\{u,v\} \in \binom {V(G)}2} {\mathcal{E}}_{u,v}\right ) \geq \prod _{u,v} (1-c^2 4^s) = (1-c^2 4^s)^{\binom n2} \geq e^{-O_s(c^2 n^2)}. \end{equation*}
Recall that
${\mathbb{P}}({\mathcal{E}})\geq 1-e^{-\Omega _s(cn^2)}$
. Therefore, for
$c$
sufficiently small in terms of
$s$
, and for
$n$
sufficiently large, we have that the sum of these two probabilities is larger than
$1$
. Hence, with positive probability,
$H_0$
is linear and contains an ordered monochromatic transversal edge for every dyadic partition, every colouring, and every ordering.
Remark. The same proof shows that we can actually take
$H_0$
to have (Berge) girth
$\gt g$
, for any fixed
$g$
. The linear case is simply this statement for
$g=2$
. Indeed, another application of Harris‘s inequality shows that the probability that
$H_0$
does not contain a Berge cycle of length at most
$g$
is also at least
$e^{-O_{s,g}(c^2 n^2)}$
.
We are finally ready to prove Lemma3.3.
Proof of Theorem 3.3. Let
$s=\left \lvert F^\lt \right \rvert$
. Let
$H_0$
be the oriented
$s$
-graph given by Lemma3.7. Obtain a
$k$
-graph
$H'$
by placing a copy of
$F^\lt$
in each
$s$
-uniform edge of
$H_0$
, in an order-respecting way. That is, given an edge
$e \in E(H_0)$
, we insert
$F^\lt$
into
$e$
by mapping the
$i$
th vertex in the ordering of
$F^\lt$
to
$\psi _e^{-1}(i)$
, for all
$1 \leq i \leq s$
. Note that since
$F^\lt$
and
$H_0$
are both linear, so is
$H'$
. Additionally, by Lemma3.7, we know that in any dyadic partition, colouring, and ordering of
$H_0$
, there is a monochromatic correctly-ordered transversal edge. This precisely yields the claimed monochromatic correctly-ordered transversal copy of
$F^\lt$
in
$G$
.
Remark. As above, if we assume that
$F^\lt$
has girth
$\gt g$
, then we can ensure that
$G$
also has girth more than
$g$
, since any short Berge cycle in
$G$
must either stay entirely within one edge of
$H_0$
, or traverse at least
$g$
edges in
$H_0$
, and both of these are ruled out by the assumption that
$F$
and
$H_0$
have girth
$\gt g$
.
4. Concluding remarks
One of the most fascinating special cases of a linear hypergraph is the Fano plane
$F_3$
; this is the seven-vertex
$3$
-graph whose hyperedges are the lines in the projective geometry
${\textrm {PG}}(2,2)$
over
${\mathbb{F}}_2$
. The growth rate of
$r(F_3,K_n^{(3)})$
remains wide open: the best known lower bound is polynomial and the best known upper bound is exponential, and it would be very interesting to narrow the gap. In particular,
$F_3$
is not iterated tripartite, so the conjecture of [Reference Conlon, Fox and Gunby5] predicts that
$r(F_3,K_n^{(3)})$
grows super-polynomially; proving or disproving this is a natural step towards resolving the conjecture.
There is a natural
$4$
-uniform analogue of the Fano plane, namely the 13-vertex linear
$4$
-graph
$F_4$
whose hyperedges are the lines in the projective geometry
${\textrm {PG}}(2,3)$
over
${\mathbb{F}}_3$
.
$F_4$
is again not iterated
$4$
-partite; curiously, however, it is quite straightforward to show that
$r(F_4,K_n^{(4)})$
grows super-polynomially.
Proposition 4.1.
We have
$r(F_4,K_n^{(4)}) \gt 2^{n-2}$
.
Proof. Let
$G_k$
be the
$4$
-graph iteratively definedFootnote
3
as follows:
$G_1$
is a single edge, and
$G_k$
is obtained from the disjoint union of two copies of
$G_{k-1}$
by adding in all edges intersecting each copy in exactly
$2$
vertices. Note that
$\alpha (G_k) = \alpha (G_{k-1})+1$
, since every independent set in
$G_k$
cannot contain two vertices from both copies of
$G_{k-1}$
. Together with the base case
$\alpha (G_1)=3$
, we see that
$\alpha (G_k) = k+2$
for all
$k$
. Since
$G_k$
has
$2^{k+1}$
vertices, the claimed result follows if we can show that
$G_k$
is
$F_4$
-free for all
$k$
.
If
$F_4 \subseteq G_k$
for some
$k$
, then in particular there is a partition of
$V(F_4)$
into two non-empty parts
$A\cup B$
such that every edge is either fully contained in a part, or else intersects each part in exactly two vertices, and it can be checked by computer or minor casework that there is no such partition. For example, the following SageMath code does the job.
This code outputs only
$A = \varnothing$
and
$A=V(F_4)$
, hence there is no such partition into non-empty parts.
Note that Theorem1.2 yields a linear
$4$
-graph
$H$
with
$r(H,K_n^{(4)}) \geq 2^{2^{(\log n)^C}}$
, which is larger than the lower bound guaranteed by Proposition4.1. However, the construction of
$H$
in Theorem1.2 is probabilistic, whereas Proposition4.1 gives a fully explicit linear
$4$
-graph, namely
$F_4$
. We remark that by using this as our base case, one can similarly construct explicit linear
$k$
-graphs
$H$
with
$r(H,K_n^{(k)}) \geq {\textrm {twr}}_{k-2}(\Omega _k(n))$
. Indeed, while the proof of Lemma3.3 is probabilistic, it can be made explicit by using the Ramsey theorem for ordered linear hypergraphs of Nešetřil and Rödl [Reference Nesetril and Rödl20]. One can use this powerful machinery to provide an alternative proof of Lemma3.3; as the proof of the Nešetřil–Rödl theorem is fully deterministic, this would yield an explicit (but extraordinarily large and complicated) linear
$k$
-graph
$H$
with
$r(H,K_n^{(k)}) \geq {\textrm {twr}}_{k-2}(\Omega (n))$
. To get the stronger result that
$H$
can have arbitrarily large girth, one could alternatively use the ordered version [Reference Reiher, Rödl and Schacht22, Theorem3.1] of the recent girth Ramsey theorem of Reiher and Rödl [Reference Reiher and Rödl21], whose proof is again explicit.
Finally, it would be very interesting to improve the lower bound in Theorem1.1 to
$r(H,K_n^{(3)}) \geq 2^{n^c}$
for some absolute constant
$c\gt 0$
, and some linear
$3$
-graph
$H$
. By repeated applications of Proposition1.3, this would immediately imply the existence of a linear
$k$
-graph
$H$
with
$r(H,K_n^{(k)}) \geq {\textrm {twr}}_{k-1}(n^c)$
, matching the upper bound in (1.1) up to the constant.
Acknowledgements
We are grateful to the anonymous referees for several helpful comments which greatly improved the presentation of this paper.
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