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The minimum degree of
minimal Ramsey graphs for cliques

John Bamberg1 1Centre for the Mathematics of Symmetry and Computation, The University of Western Australia, Australia. [email protected] Anurag Bishnoi2 2 Delft Institute of Applied Mathematics, TU Delft, Netherlands. [email protected]  and  Thomas Lesgourgues3 3School of Mathematics and Statistics, UNSW, Australia. [email protected]
Abstract.

We prove that sr(Kk)=O(k5r5/2)s_{r}(K_{k})=O(k^{5}r^{5/2}), where sr(Kk)s_{r}(K_{k}) is the Ramsey parameter introduced by Burr, Erdős and Lovász in 1976, which is defined as the smallest minimum degree of a graph GG such that any rr-colouring of the edges of GG contains a monochromatic KkK_{k}, whereas no proper subgraph of GG has this property. The construction used in our proof relies on a group theoretic model of generalised quadrangles introduced by Kantor in 1980.

Key words and phrases:
Ramsey graphs, generalised quadrangles, Kantor family
2010 Mathematics Subject Classification:
05C55,05D10,51E12
Anurag Bishnoi’s research supported by a Discovery Early Career Award of the Australian Research Council (No. DE190100666)
Thomas Lesgourgues’ research supported by an Australian Government Research Training Program Scholarship and the School of Mathematics and Statistics, UNSW

1. Introduction

A graph GG is called rr-Ramsey for another graph HH, denoted by G(H)rG\rightarrow(H)_{r}, if every rr-colouring of the edges of GG contains a monochromatic copy of HH. Observe that if G(H)rG\rightarrow(H)_{r}, then every graph containing GG as a subgraph is also rr-Ramsey for HH. Some very interesting questions arise when we study graphs GG which are minimal with respect to G(H)rG\rightarrow(H)_{r}, that is, G(H)rG\rightarrow(H)_{r} but there is no proper subgraph GG^{\prime} of GG such that G(H)rG^{\prime}\rightarrow(H)_{r}. We call such graphs rr-Ramsey minimal for HH and we denote the set of all rr-Ramsey minimal graphs for HH by r(H)\mathcal{M}_{r}(H). The classical result of Ramsey [21] implies that for any finite graph HH and positive integer rr, there exists a graph GG that is rr-Ramsey for HH, that is, r(H)\mathcal{M}_{r}(H) is non-empty.

Some of the central problems in graph Ramsey theory are concerned with the case where HH is a clique KkK_{k}. For example, the most well studied parameter is the Ramsey number Rr(k)R_{r}(k), that denotes the smallest number of vertices of any graph in r(Kk)\mathcal{M}_{r}(K_{k}). The classical work of Erdős [8] and Erdős and Szekeres [9] shows that 2k/2R2(k)22k2^{k/2}\leqslant R_{2}(k)\leqslant 2^{2k}. While these bounds have been improved since then, most recently by Sah [23] (also see [24] and [5]), the constants in the exponent have stayed the same. We refer the reader to the survey of Conlon, Fox and Sudakov [6] for more on this and other graph Ramsey problems.

Several other questions on r(H)\mathcal{M}_{r}(H) have also been explored; for example, the well studied size-Ramsey number R^r(H)\hat{R}_{r}(H) which is the minimum number of edges of a graph in r(H)\mathcal{M}_{r}(H). We refer the reader to [1, 3, 18, 22] for various results on minimal Ramsey problems. In this paper, we will be interested in the smallest minimum degree of an rr-Ramsey minimal graph, which is defined by

sr(H)minGr(H)δ(G),s_{r}(H)\coloneqq\min_{G\in\mathcal{M}_{r}(H)}\delta(G),

for a finite graph HH and positive integer rr, where δ(G)\delta(G) denotes the minimum degree of GG. Trivially, we have sr(H)Rr(H)1s_{r}(H)\leqslant R_{r}(H)-1, since the complete graph on Rr(H)R_{r}(H) vertices is rr-Ramsey for HH and has minimum degree Rr(H)1R_{r}(H)-1. The study of this parameter was initiated by Burr, Erdős and Lovász [2] in 1976. They were able to show the rather surprising exact result, s2(Kk)=(k1)2s_{2}(K_{k})=(k-1)^{2}, which is far away from the trivial exponential bound of s2(Kk)Rr(k)1s_{2}(K_{k})\leqslant R_{r}(k)-1. The behaviour of this function is still not so well understood for r>2r>2 colours. Fox et al. [10] determined this function asymptotically for every fixed kk up-to a polylogarithmic factor, and for k=3k=3 their result was further improved by Guo and Warnke [12] who managed to obtain matching logarithmic factors.

Theorem 1.1 (Fox, Grinshpun, Liebenau, Person, Szabó).
  1. (i)

    There exist constants c,C>0c,C>0 such that for all r2r\geqslant 2, we have

    cr2lnrsr(K3)Cr2ln2r.cr^{2}\ln r\leqslant s_{r}(K_{3})\leqslant Cr^{2}\ln^{2}r.
  2. (ii)

    For all k4k\geqslant 4 there exist constants ck,Ck>0c_{k},C_{k}>0 such that for all r3r\geqslant 3, we have

    ckr2lnrlnlnrsr(Kk)Ckr2(lnr)8(k1)2.c_{k}r^{2}\frac{\ln r}{\ln{\ln r}}\leqslant s_{r}(K_{k})\leqslant C_{k}r^{2}(\ln r)^{8(k-1)^{2}}.
Theorem 1.2 (Guo, Warnke).

sr(K3)=Θ(r2lnr)s_{r}(K_{3})=\Theta(r^{2}\ln r).

The constant in the upper bound of Theorem 1.1(ii) is rather large (Ckk2e4k2ln2C_{k}\sim k^{2}e^{4k^{2}\ln 2}), and in particular not polynomial in kk. To remedy this, they proved the following general upper bound which is polynomial in both kk and rr.

Theorem 1.3 (Fox, Grinshpun, Liebenau, Person, Szabó).

For all k,r3k,r\geqslant 3, sr(Kk)8(k1)6r3s_{r}(K_{k})\leqslant 8(k-1)^{6}r^{3}.

For a fixed rr and kk\rightarrow\infty, Hàn, Rödl and Szabó [13] determined this function up-to polylogarithmic factors by proving the following.

Theorem 1.4 (Hàn, Rödl, Szabó).

There exists a constant k0k_{0} such that for every k>k0k>k_{0} and r<k2r<k^{2}

sr(Kk)803(rlnr)3(klnk)2.s_{r}(K_{k})\leqslant 80^{3}(r\ln r)^{3}(k\ln k)^{2}.

We prove the following general upper bound that improves Theorem 1.3, and thus provides the best known upper bound on sr(Kk)s_{r}(K_{k}) outside the special ranges covered by Theorem 1.1 and 1.4.

Theorem 1.5.

There exists an absolute constant CC such that for all r2,k3r\geqslant 2,k\geqslant 3, sr(Kk)C(k1)5r5/2s_{r}(K_{k})\leqslant C(k-1)^{5}r^{5/2}.

Our proof uses the equivalence between sr(Kk)s_{r}(K_{k}) and another extremal function, called the rr-colour kk-clique packing number [10], defined as follows. Let Pr(k)P_{r}(k) denote the minimum nn for which there exist Kk+1K_{k+1}-free pairwise edge disjoint graphs G1,,GrG_{1},\dots,G_{r} on a common vertex set VV of size nn such that for any rr-colouring of VV, there exists an ii such that GiG_{i} contains a copy of KkK_{k} all of whose vertices are coloured in the iith colour.

Lemma 1.6 (see [10, Theorem 1.5]).

For all integers r,k2r,k\geqslant 2 we have sr(Kk+1)=Pr(k)s_{r}(K_{k+1})=P_{r}(k).

Our graphs GiG_{i} in the packing would come from certain point-line geometries known as generalised quadrangles that we define in the next section. In Section 3, we show that any packing of ‘triangle-free’ point-line geometries implies an upper bound on Pr(k)P_{r}(k), assuming certain conditions on the parameters of the geometry. In Section, 4 we give a packing of certain subgeometries of the so-called Hermitian generalised quadrangles using a group theoretic model given by Kantor in the 1980’s [15], and deduce that this packing implies our main result.

2. Background

A (finite) generalised quadrangle 𝒬\mathcal{Q} of order (s,t)(s,t) is an incidence structure of points 𝒫\mathcal{P}, lines \mathcal{L}, together with a symmetric point-line incidence relation satisfying the following axioms:

  1. (i)

    Each point lies on t+1t+1 lines (t1t\geqslant 1) and two distinct points are incident with at most one line.

  2. (ii)

    Each line lies on s+1s+1 points (s1s\geqslant 1) and two distinct lines are incident with at most one point.

  3. (iii)

    If PP is a point and \ell is a line not incident with PP, then there is a unique point on \ell collinear with PP.

Notice that the third axiom above ensures that there are no triangles (i.e., three distinct lines pairwise meeting in three distinct points) in 𝒬\mathcal{Q}. The standard reference on finite generalised quadrangles is the book by Payne and Thas [20]. The collinearity graph of a generalised quadrangle is the graph on the set of points with two points adjacent when they are both incident with a common line. A collineation θ\theta of 𝒬\mathcal{Q}, that is, an automorphism of its collinearity graph, is an elation about the point PP if it is either the identity collineation, or it fixes each line incident with PP and fixes no point not collinear with PP. If there is a group EE of elations of 𝒬\mathcal{Q} about the point PP such that EE acts regularly on the points not collinear with PP, then we say that 𝒬\mathcal{Q} is an elation generalised quadrangle with elation group EE and base point PP. Necessarily, EE has order s2ts^{2}t, as there are s2ts^{2}t points not collinear to a given point in any generalised quadrangle.

Now suppose we have a finite group EE of order s2ts^{2}t where s,t>1s,t>1. A Kantor family of EE is a set 𝒜:={Ai:i=0,,t}\mathcal{A}:=\{A_{i}\colon i=0,\ldots,t\} of subgroups of order ss, and a set 𝒜:={Ai:i=0,,t}\mathcal{A}^{*}:=\{A_{i}^{*}\colon i=0,\ldots,t\} of subgroups of order stst, such that the following are satisfied:

  1. (K0)

    AiAiA_{i}\leqslant A_{i}^{*} for all i{0,,t}i\in\{0,\ldots,t\};

  2. (K1)

    AiAj={1}A_{i}\cap A_{j}^{*}=\{1\} whenever iji\neq j;

  3. (K2)

    AiAjAk={1}A_{i}A_{j}\cap A_{k}=\{1\} whenever i,j,ki,j,k are distinct.

Due to a theorem of Kantor (c.f., [15, Theorem A.3.1]), a Kantor family as described above, gives rise to an elation generalised quadrangle of order (s,t)(s,t), which we briefly describe in Table 1.

Points Lines
elements gg of EE the right cosets AigA_{i}g
right cosets AigA_{i}^{*}g symbols [Ai][A_{i}]
a symbol \infty.

Incidence: gg \sim AigA_{i}g AihA_{i}^{*}h \sim AigA_{i}g, where AigAihA_{i}g\subseteq A_{i}^{*}h AihA_{i}^{*}h \sim [Ai][A_{i}] \infty \sim [Ai][A_{i}]

Table 1. The points and lines of the elation generalised quadrangle arising from a Kantor family (n.b., Ai𝒜A_{i}\in\mathcal{A}, Ai𝒜A_{i}^{*}\in\mathcal{A}^{*}, gEg\in E).

We will simply be needing to use the Kantor family for a well-known family of generalised quadrangles, where the Heisenberg groups appear as the group EE in the description above. We remark that the main property we will need is (K2), since it ensures that lines of the form AigA_{i}g, never form a triangle.For self-containment, we give a proof here. Suppose f,g,hf,g,h are three elements of EE forming the vertices of a triangle. Then there are three elements A,B,C𝒜A,B,C\in\mathcal{A} such that Af=AgAf=Ag, Bg=BhBg=Bh, Ch=CfCh=Cf. Therefore, fg1Afg^{-1}\in A, gh1Bgh^{-1}\in B, fh1Cfh^{-1}\in C, from which it follows that fh1=(fg1)(gh1)ABCfh^{-1}=(fg^{-1})(gh^{-1})\in AB\cap C. Since fhf\neq h, we have ABC{1}AB\cap C\neq\{1\}. So the condition ABC={1}AB\cap C=\{1\} given by (K2) ensures that there are no triangles.

In this paper, we will also need two commonly used maps on finite fields: the trace and norm maps. For a finite field, they are the analogues of ‘real part’ and ‘square-modulus’ for the complex numbers. We will not be needing the general theory of traces and norms, just two functions in particular. The relative trace map from 𝔽q2\mathbb{F}_{q^{2}} to 𝔽q\mathbb{F}_{q} is defined by 𝖳𝗋(x)=x+xq\operatorname{\mathsf{Tr}}(x)=x+x^{q}. If ϕ\phi is a field-automorphism of 𝔽q2\mathbb{F}_{q^{2}}, then 𝖳𝗋(ϕ(x))=𝖳𝗋(x)\operatorname{\mathsf{Tr}}\left(\phi(x)\right)=\operatorname{\mathsf{Tr}}(x) for all x𝔽q2x\in\mathbb{F}_{q^{2}}. Also, 𝖳𝗋\operatorname{\mathsf{Tr}} is additive; that is, 𝖳𝗋(x+y)=𝖳𝗋(x)+𝖳𝗋(y)\operatorname{\mathsf{Tr}}(x+y)=\operatorname{\mathsf{Tr}}(x)+\operatorname{\mathsf{Tr}}(y) for 𝔽q2\mathbb{F}_{q^{2}}. The relative norm map 𝔽q2\mathbb{F}_{q^{2}} to 𝔽q\mathbb{F}_{q} is defined by 𝖭(x)=xq+1\mathsf{N}(x)=x^{q+1}. It is also invariant under field-automorphisms, and is surjective. We will make use of the following property: if qq is odd, then every element of 𝔽q\mathbb{F}_{q} is a square in 𝔽q2\mathbb{F}_{q^{2}}. To see this, let t𝔽qt\in\mathbb{F}_{q}. Since 𝖭\mathsf{N} is surjective, there exists x𝔽q2x\in\mathbb{F}_{q^{2}} such that t=xq+1t=x^{q+1}. Since qq is odd, the element y=x(q+1)/2y=x^{(q+1)/2} is well defined, and then t=y2t=y^{2}.

3. Packing Generalised Quadrangles

A partial linear space is a point-line incidence structure with the property that any two distinct points are incident to at most one common line. A triangle-free partial linear space of order (s,t)(s,t) is an incidence structure satisfying Axioms (i) and (ii) of a generalised quadrangle, and (iii) there are no three distinct lines pairwise meeting each other in three distinct points. Clearly, any subgeometry of a generalised quadrangle where the number of points on a line and the number of lines through a point are constants is a triangle-free partial linear space. We now prove the main lemma that will imply Theorem 1.5 once we have the construction outlined in Section 4. Our proof follows the same idea as in Dudek and Rödl [7], and Fox et al. [10].

Lemma 3.1.

Let r,k,s,tr,k,s,t be positive integers. Say there exists a family (i)i=1r(\mathcal{I}_{i})_{i=1}^{r} of triangle-free partial linear spaces of order (s,t)(s,t), on the same point set 𝒫\mathcal{P} and pairwise disjoint line-sets 1,,r\mathcal{L}_{1},\dots,\mathcal{L}_{r}, such that the point-line geometry (𝒫,i=1ri)\left(\mathcal{P},\bigcup_{i=1}^{r}\mathcal{L}_{i}\right) is also a partial linear space. If s3rklnks\geqslant 3rk\ln k and t3k(1+lnr)t\geqslant 3k(1+\ln r), then Pr(k)|𝒫|P_{r}(k)\leqslant\lvert\mathcal{P}\rvert.

Proof.

In order to show that Pr(k)|𝒫|P_{r}(k)\leqslant\lvert\mathcal{P}\rvert, we will exhibit Kk+1K_{k+1}-free pairwise edge disjoint graphs G1,,GrG_{1},\dots,G_{r} on the common vertex set V=𝒫V=\mathcal{P}, such that for any rr-colouring of VV, there exists an ii such that GiG_{i} contains a copy of KkK_{k} all of whose vertices are coloured in the iith colour. We start by recalling the following properties about each partial linear space i\mathcal{I}_{i}, i{1,,r}i\in\{1,\ldots,r\}:

  1. (P1)

    Every point p𝒫p\in\mathcal{P} is incident with t+1t+1 lines of i\mathcal{L}_{i}.

  2. (P2)

    Every line i\ell\in\mathcal{L}_{i} contains s+1s+1 points from 𝒫\mathcal{P}.

  3. (P3)

    Any two points of 𝒫\mathcal{P} lie on at most one line of i\mathcal{L}_{i}.

  4. (P4)

    i\mathcal{I}_{i} is triangle-free.

Furthermore, given that (𝒫,i=1ri)\left(\mathcal{P},\bigcup_{i=1}^{r}\mathcal{L}_{i}\right) is a partial linear space and the line-sets 1,,r\mathcal{L}_{1},\dots,\mathcal{L}_{r} are disjoint,

  1. (P5)

    For any iji\neq j, and any i\ell\in\mathcal{L}_{i}, mjm\in\mathcal{L}_{j}, \ell and mm are incident with at most one common point.

Let i{1,,r}i\in\{1,\ldots,r\}, 1=s+1k\ell_{1}=\left\lfloor\frac{s+1}{k}\right\rfloor and 2=s+1k\ell_{2}=\left\lceil\frac{s+1}{k}\right\rceil. For each line i\ell\in\mathcal{L}_{i}, we select uniformly at random one partition of \ell among all =j=1kLj()\ell=\bigcup_{j=1}^{k}L_{j}^{(\ell)}, where Lj()L_{j}^{(\ell)} denotes the jjth component of the partition, such that for some kk^{\prime}, |L1()|,,|Lk()|=1\lvert L_{1}^{(\ell)}\rvert,\ldots,\lvert L_{k^{\prime}}^{(\ell)}\rvert=\ell_{1} and |Lk+1()|,,|Lk()|=2\lvert L_{k^{\prime}+1}^{(\ell)}\rvert,\ldots,\lvert L_{k}^{(\ell)}\rvert=\ell_{2}. Choices for distinct lines in i\mathcal{L}_{i} are independent.

We define a graph Gi=(V,Ei)G_{i}=(V,E_{i}) on the vertex set V=𝒫V=\mathcal{P} as follows. For every i\ell\in\mathcal{L}_{i}, we include the edges of a complete kk-partite graph between the vertex sets Lj()L_{j}^{(\ell)} for j{1,,k}j\in\{1,\ldots,k\}. Note that the graph GiG_{i} is a collection of Turán graphs on (s+1)(s+1) vertices with kk parts. Each Turán graph comes from one line i\ell\in\mathcal{L}_{i}. By property (P3), any two points are incident with at most one line, therefore the different Turán graphs are edge-disjoint. Furthermore, by property (P4), GiG_{i} is Kk+1K_{k+1}-free. Finally, by property (P5), for any ij{1,,r}i\neq j\in\{1,\ldots,r\}, GiG_{i} and GjG_{j} are edge disjoint.

In order to conclude, we need to show that with positive probability, for any rr-colouring of VV, there exists an ii such that GiG_{i} contains a copy of KkK_{k} all of whose vertices are coloured in the iith colour. Note that given G1,,GrG_{1},\ldots,G_{r} on the vertex set V=𝒫V=\mathcal{P}, in any rr-colouring of VV, at least one of the colours occurs at least |𝒫|/r\lvert\mathcal{P}\rvert/r times. Therefore if for every GiG_{i}, every set of at least |𝒫|/r\lvert\mathcal{P}\rvert/r vertices contains a copy of KkK_{k}, then we get the desired property. The choices of partitions being done independently, to conclude our proof it suffices to show that for each i{1,,r}i\in\{1,\ldots,r\}, with positive probability every set of at least |𝒫|/r\lvert\mathcal{P}\rvert/r vertices contains a copy of KkK_{k} in GiG_{i}.

Fix i{1,,r}i\in\{1,\ldots,r\}. For a subset W𝒫W\subseteq\mathcal{P}, let 𝒜(W)\mathcal{A}(W) denote the event that the induced subgraph Gi[W]G_{i}[W] contains no copy of KkK_{k}. Let U𝒫U\subset\mathcal{P} with |U|=|𝒫|r\lvert U\rvert=\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor. By property (P4), any copy of KkK_{k} can only appear from one line i\ell\in\mathcal{L}_{i}, i.e.

𝒜(U)i𝒜(U).\mathcal{A}(U)\subseteq\bigcap_{\ell\in\mathcal{L}_{i}}\mathcal{A}(U\cap\ell).

All the events 𝒜(U)\mathcal{A}(U\cap\ell) are independent, therefore

(𝒜(U))i(𝒜(U)).\mathbb{P}(\mathcal{A}(U))\leqslant\prod_{\ell\in\mathcal{L}_{i}}\mathbb{P}(\mathcal{A}(U\cap\ell)).

For a given line i\ell\in\mathcal{L}_{i}, let u=|U|u_{\ell}=\lvert U\cap\ell\rvert, and let =j=1kLj()\ell=\bigcup_{j=1}^{k}L_{j}^{(\ell)} be the random partition of \ell. Note that UU\cap\ell contains no copy of KkK_{k} if and only if there exists j{1,,k}j\in\{1,\ldots,k\} such that ULj()=U\cap L_{j}^{(\ell)}=\varnothing. For a fixed j{1,,k}j\in\{1,\ldots,k\},

(ULj()=)=(s+1u|Lj()|)(s+1|Lj()|)(1us+1)|Lj()|exp(1us+1).\mathbb{P}\left(U\cap L_{j}^{(\ell)}=\varnothing\right)=\frac{\binom{s+1-u_{\ell}}{|L_{j}^{(\ell)}|}}{\binom{s+1}{|L_{j}^{(\ell)}|}}\leqslant\left(1-\frac{u_{\ell}}{s+1}\right)^{|L_{j}^{(\ell)}|}\leqslant\exp\left(-\frac{\ell_{1}u_{\ell}}{s+1}\right).

Therefore

(𝒜(U))\displaystyle\mathbb{P}(\mathcal{A}(U)) i(j{1,,k},ULj()=)\displaystyle\leqslant\prod_{\ell\in\mathcal{L}_{i}}\mathbb{P}\left(\exists\ j\in\{1,\ldots,k\},\ U\cap L_{j}^{(\ell)}=\varnothing\right)
k|i|exp(i1us+1).\displaystyle\leqslant k^{\lvert\mathcal{L}_{i}\rvert}\exp\left(-\sum_{\ell\in\mathcal{L}_{i}}\frac{\ell_{1}u_{\ell}}{s+1}\right).

Because every point of UU is incident with t+1t+1 lines from i\mathcal{L}_{i} (by property (P1)), iu=i|U|=(t+1)|U|\sum_{\ell\in\mathcal{L}_{i}}u_{\ell}=\sum_{\ell\in\mathcal{L}_{i}}\lvert U\cap\ell\rvert=(t+1)\lvert U\rvert, and thus

(𝒜(U))k|i|exp(1(t+1)|U|s+1).\mathbb{P}(\mathcal{A}(U))\leqslant k^{\lvert\mathcal{L}_{i}\rvert}\exp\left(-\frac{\ell_{1}(t+1)\lvert U\rvert}{s+1}\right).

Finally,

(U(𝒫|𝒫|r):𝒜(U))\displaystyle\mathbb{P}\left(\exists U\in\binom{\mathcal{P}}{\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor}:\ \mathcal{A}(U)\right) (|𝒫||𝒫|r)k|i|exp(t+1s+11|𝒫|r)\displaystyle\leqslant\binom{\lvert\mathcal{P}\rvert}{\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor}k^{\lvert\mathcal{L}_{i}\rvert}\exp\left(-\frac{t+1}{s+1}\ell_{1}\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor\right)
(re)|𝒫|/rk|i|exp(t+1s+1s+1k|𝒫|r)\displaystyle\leqslant(re)^{\lvert\mathcal{P}\rvert/r}k^{\lvert\mathcal{L}_{i}\rvert}\exp\left(-\frac{t+1}{s+1}\left\lfloor\frac{s+1}{k}\right\rfloor\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor\right)

Given that |𝒫|s3rklnk\lvert\mathcal{P}\rvert\geqslant s\geqslant 3rk\ln k, r2r\geqslant 2 and k3k\geqslant 3, we have

s+1k56s+1k,|𝒫|r56|𝒫|r.\left\lfloor\frac{s+1}{k}\right\rfloor\geqslant\frac{5}{6}\frac{s+1}{k},\qquad\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor\geqslant\frac{5}{6}\frac{\lvert\mathcal{P}\rvert}{r}.

Therefore,

(U(𝒫|𝒫|r):𝒜(U))\displaystyle\mathbb{P}\left(\exists U\in\binom{\mathcal{P}}{\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor}:\ \mathcal{A}(U)\right) (re)|𝒫|/rk|i|exp(t+1s+15(s+1)6k5|𝒫|6r)\displaystyle\leqslant(re)^{\lvert\mathcal{P}\rvert/r}k^{\lvert\mathcal{L}_{i}\rvert}\exp\left(-\frac{t+1}{s+1}\frac{5(s+1)}{6k}\frac{5\lvert\mathcal{P}\rvert}{6r}\right)
exp[|𝒫|(1+lnrr+|i||𝒫|lnk2536t+1rk)].\displaystyle\leqslant\exp\left[\lvert\mathcal{P}\rvert\left(\frac{1+\ln r}{r}+\frac{\lvert\mathcal{L}_{i}\rvert}{\lvert\mathcal{P}\rvert}\ln{k}-\frac{25}{36}\frac{t+1}{rk}\right)\right].

By double counting (using properties (P1) and (P2)) we know that

|i|(s+1)=|𝒫|(t+1),\lvert\mathcal{L}_{i}\rvert(s+1)=\lvert\mathcal{P}\rvert(t+1),

and therefore

(U(𝒫|𝒫|r):𝒜(U))exp[|𝒫|(1+lnrr+t+1s+1lnk2536t+1rk)]\mathbb{P}\left(\exists U\in\binom{\mathcal{P}}{\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor}:\ \mathcal{A}(U)\right)\leqslant\exp\left[\lvert\mathcal{P}\rvert\left(\frac{1+\ln r}{r}+\frac{t+1}{s+1}\ln k-\frac{25}{36}\frac{t+1}{rk}\right)\right] (3.1)

Note that since s3rklnks\geqslant 3rk\ln k we have

2536t+1rk>2(t+1)s+1lnk,\frac{25}{36}\frac{t+1}{rk}>\frac{2(t+1)}{s+1}\ln k,

and since t3k(1+lnr)t\geqslant 3k(1+\ln r) we have

2536t+1rk>2(1+lnr)r.\frac{25}{36}\frac{t+1}{rk}>\frac{2(1+\ln r)}{r}.

Therefore,

(U(𝒫|𝒫|r):𝒜(U))<1.\mathbb{P}\left(\exists U\in\binom{\mathcal{P}}{\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor}:\ \mathcal{A}(U)\right)<1.

Then there exists an instance of GiG_{i} such that every subset of 𝒫\mathcal{P} with at least |𝒫|r\left\lfloor\frac{\lvert\mathcal{P}\rvert}{r}\right\rfloor vertices contains a copy of KkK_{k} in GiG_{i}. ∎

4. The construction

Let qq be a prime power, and denote the finite field of order q2q^{2} by 𝔽q2\mathbb{F}_{q^{2}}. We will use the model of the Hermitian generalised quadrangle H(3,q2)H(3,q^{2}) that appears in [14, Section 3] (see Example 3). For a definition of H(3,q2)H(3,q^{2}) see [20, Chapter 3].

Let EE be the group defined on 𝔽q2×𝔽q×𝔽q2\mathbb{F}_{q^{2}}\times\mathbb{F}_{q}\times\mathbb{F}_{q^{2}} by the following operation:

(a,γ,b)(a,γ,b)=(a+a,γ+γ+𝖳𝗋(bqa),b+b),(a,\gamma,b)\circ(a^{\prime},\gamma^{\prime},b^{\prime})=(a+a^{\prime},\gamma+\gamma^{\prime}+\operatorname{\mathsf{Tr}}(b^{q}a^{\prime}),b+b^{\prime}),

where 𝖳𝗋(x)=x+xq\operatorname{\mathsf{Tr}}(x)=x+x^{q} is the relative trace map from 𝔽q2\mathbb{F}_{q^{2}} to 𝔽q\mathbb{F}_{q}. It turns out that EE is the Heisenberg group of order q5q^{5} with centre of order qq.

We can construct a generalised quadrangle by constructing a Kantor family of EE. Define

A\displaystyle A^{*}_{\infty} ={(0,γ,a):a𝔽q2,γ𝔽q},\displaystyle=\{(0,\gamma,a)\colon a\in\mathbb{F}_{q^{2}},\gamma\in\mathbb{F}_{q}\},
At\displaystyle A^{*}_{t} ={(a,γ,at):a𝔽q2,γ𝔽q},t𝔽q,\displaystyle=\{(a,\gamma,at)\colon a\in\mathbb{F}_{q^{2}},\gamma\in\mathbb{F}_{q}\},\quad t\in\mathbb{F}_{q},
A\displaystyle A_{\infty} ={(0,0,a):a𝔽q2},\displaystyle=\{(0,0,a)\colon a\in\mathbb{F}_{q^{2}}\},
At\displaystyle A_{t} ={(a,aq+1t,at):a𝔽q2},t𝔽q.\displaystyle=\{(a,a^{q+1}t,at)\colon a\in\mathbb{F}_{q^{2}}\},\quad t\in\mathbb{F}_{q}.

Then 𝒜:={A}{At:t𝔽q}\mathcal{A}:=\{A_{\infty}\}\cup\{A_{t}\colon t\in\mathbb{F}_{q}\} and 𝒜:={A}{Ab:b𝔽q}\mathcal{A}^{*}:=\{A^{*}_{\infty}\}\cup\{A^{*}_{b}\colon b\in\mathbb{F}_{q}\} form a Kantor family of EE giving rise to a generalised quadrangle isomorphic to H(3,q2)H(3,q^{2}).111In [14] the dual of this generalised quadrangle is defined, denoted by O(6,q)O^{-}(6,q), but it is well known that H(3,q2)H(3,q^{2}) is isomorphic to the dual of the elliptic generalised quadrangle O(6,q)O^{-}(6,q) (see [20, Chapter 3], where O(6,q)O^{-}(6,q) is denoted by Q(5,q)Q(5,q)).

From now on we will assume that qq is odd. Let κ\kappa be an element of 𝔽q2\mathbb{F}_{q^{2}}. For each λ𝔽q2\lambda\in\mathbb{F}_{q^{2}}, define τλ:EE\tau_{\lambda}:E\rightarrow E as follows. Let

τλ:\displaystyle\tau_{\lambda}\colon (a,0,0)(a,𝖳𝗋(λa+κλaq+12λqa2),λaq),\displaystyle(a,0,0)\mapsto\left(a,\operatorname{\mathsf{Tr}}(\lambda a+\kappa\lambda a^{q}+\tfrac{1}{2}\lambda^{q}a^{2}),\lambda a^{q}\right),
τλ:\displaystyle\tau_{\lambda}\colon (0,γ,b)(0,γ,b).\displaystyle(0,\gamma,b)\mapsto(0,\gamma,b).

By Axiom (K1), we have |A0A|=|A0||A|/|A0A|=s(st)/1=|E||A_{0}A_{\infty}^{*}|=|A_{0}||A_{\infty}^{*}|/|A_{0}\cap A_{\infty}^{*}|=s(st)/1=|E|. So E=A0AE=A_{0}A_{\infty}^{*}. Therefore, we can write every element gEg\in E as g=g0gg=g_{0}g_{\infty}^{*}, with g0A0g_{0}\in A_{0} and gAg_{\infty}^{*}\in A_{\infty}^{*}. Define τλ(g)τλ(g0)τλ(g)\tau_{\lambda}(g)\coloneqq\tau_{\lambda}(g_{0})\circ\tau_{\lambda}(g_{\infty}^{*}).

Lemma 4.1.

For every λ𝔽q2\lambda\in\mathbb{F}_{q^{2}}, τλ\tau_{\lambda} is an automorphism of EE.

Proof.

Let λ𝔽q2\lambda\in\mathbb{F}_{q^{2}}. It suffices to show that τλ\tau_{\lambda} is a homomorphism from A0A_{0} to EE, since τλ\tau_{\lambda} is clearly bijective. Let a1,a2𝔽q2a_{1},a_{2}\in\mathbb{F}_{q^{2}}. Then

τλ(a1+a2,0,0)=(a1+a2,𝖳𝗋(λ(a1+a2)+κλ(a1+a2)q+12λq(a1+a2)2),λ(a1+a2)q)\tau_{\lambda}(a_{1}+a_{2},0,0)=\left(a_{1}+a_{2},\operatorname{\mathsf{Tr}}(\lambda(a_{1}+a_{2})+\kappa\lambda(a_{1}+a_{2})^{q}+\tfrac{1}{2}\lambda^{q}(a_{1}+a_{2})^{2}),\lambda(a_{1}+a_{2})^{q}\right)

Using (x+y)q=xq+yq(x+y)^{q}=x^{q}+y^{q} in 𝔽q2\mathbb{F}_{q^{2}}, we get,

τλ(a1+a2,0,0)=\displaystyle\tau_{\lambda}(a_{1}+a_{2},0,0)= (a1+a2,𝖳𝗋(λa1+κλa1q+λqa12+λa2+κλa2q+λqa22+λqa1a2),λa1q+a2q).\displaystyle\left(a_{1}+a_{2},\operatorname{\mathsf{Tr}}(\lambda a_{1}+\kappa\lambda a_{1}^{q}+\lambda^{q}a_{1}^{2}+\lambda a_{2}+\kappa\lambda a_{2}^{q}+\lambda^{q}a_{2}^{2}+\lambda^{q}a_{1}a_{2}),\lambda a_{1}^{q}+a_{2}^{q}\right).

Using the fact that 𝖳𝗋()\operatorname{\mathsf{Tr}}{(\cdot)} is additive, we obtain,

τλ(a1+a2,0,0)=\displaystyle\tau_{\lambda}(a_{1}+a_{2},0,0)= (a1+a2,𝖳𝗋(λa1+κλa1q+12λqa12)+\displaystyle\big{(}a_{1}+a_{2},\operatorname{\mathsf{Tr}}(\lambda a_{1}+\kappa\lambda a_{1}^{q}+\tfrac{1}{2}\lambda^{q}a_{1}^{2})+
𝖳𝗋(λa2+κλa2q+12λqa22)+𝖳𝗋(λqa1a2),λa1q+λa2q)\displaystyle\operatorname{\mathsf{Tr}}(\lambda a_{2}+\kappa\lambda a_{2}^{q}+\tfrac{1}{2}\lambda^{q}a_{2}^{2})+\operatorname{\mathsf{Tr}}(\lambda^{q}a_{1}a_{2}),\lambda a_{1}^{q}+\lambda a_{2}^{q}\big{)}
=\displaystyle= (a1,𝖳𝗋(λa1+κλa1q+12λqa12),λa1q)(a2,𝖳𝗋(λa2+κλa2q+12λqa22),λa2q)\displaystyle\left(a_{1},\operatorname{\mathsf{Tr}}(\lambda a_{1}+\kappa\lambda a_{1}^{q}+\tfrac{1}{2}\lambda^{q}a_{1}^{2}),\lambda a_{1}^{q}\right)\circ\left(a_{2},\operatorname{\mathsf{Tr}}(\lambda a_{2}+\kappa\lambda a_{2}^{q}+\tfrac{1}{2}\lambda^{q}a_{2}^{2}),\lambda a_{2}^{q}\right)
=\displaystyle= τλ(a1,0,0)τλ(a2,0,0).\displaystyle\tau_{\lambda}(a_{1},0,0)\circ\tau_{\lambda}(a_{2},0,0).

Therefore, τλ\tau_{\lambda} is an automorphism of EE. ∎

Lemma 4.2.

For every odd prime power qq, there exists κ𝔽q2\kappa\in\mathbb{F}_{q^{2}} such that 𝖳𝗋(κa+a2q1)0\operatorname{\mathsf{Tr}}(\kappa a+a^{2q-1})\neq 0 for all nonzero a𝔽q2a\in\mathbb{F}_{q^{2}}.

Proof.

Suppose that 𝖳𝗋(κa+a2q1)=κa+κqaq+a2q1+a2q=0\operatorname{\mathsf{Tr}}(\kappa a+a^{2q-1})=\kappa a+\kappa^{q}a^{q}+a^{2q-1}+a^{2-q}=0 for some non-zero a𝔽q2a\in\mathbb{F}_{q^{2}}. Multiplying by aq2a^{q-2} and defining y=aq1y=-a^{q-1} we get the following cubic equation,

y3κqy2+κy1=0.y^{3}-\kappa^{q}y^{2}+\kappa y-1=0.

Therefore if κ\kappa is such that the cubic is irreducible, then we get a contradiction. We show that such a choice of κ\kappa exists. Let tt be a non-square in 𝔽q\mathbb{F}_{q} and suppose that α\alpha is an element of 𝔽q2\mathbb{F}_{q^{2}} such that α2=t\alpha^{2}=t. By the Hansen-Mullen Irreducibility Conjecture (which is true, see [4, Theorem 2.7]) there exists a monic cubic of the form x3+ux2tx+vx^{3}+ux^{2}-tx+v irreducible in 𝔽q[x]\mathbb{F}_{q}[x] for some u,v𝔽qu,v\in\mathbb{F}_{q}. Let

κ=(tu+3vtu+v+4ttu+vα)q.\kappa=\left(\frac{-tu+3v}{tu+v}+\frac{4t}{tu+v}\alpha\right)^{q}.

By [16, Theorem 3], y3κqy2+κy1y^{3}-\kappa^{q}y^{2}+\kappa y-1 is irreducible in 𝔽q2[y]\mathbb{F}_{q^{2}}[y]. ∎

Theorem 4.3.

Let κ\kappa be an element of 𝔽q2\mathbb{F}_{q^{2}} such that 𝖳𝗋(κa+a2q1)0\operatorname{\mathsf{Tr}}(\kappa a+a^{2q-1})\neq 0 for all non-zero a𝔽q2a\in\mathbb{F}_{q^{2}}. For each λ𝔽q2\lambda\in\mathbb{F}_{q^{2}} and t𝔽qt\in\mathbb{F}_{q}, let

Atλ={(a,aq+1t+𝖳𝗋(λa+κλaq+12λqa2),at+λaq):a𝔽q2},A_{t}^{\lambda}=\{(a,a^{q+1}t+\operatorname{\mathsf{Tr}}(\lambda a+\kappa\lambda a^{q}+\tfrac{1}{2}\lambda^{q}a^{2}),at+\lambda a^{q}):a\in\mathbb{F}_{q^{2}}\},

and let

𝒮:={{Atλ:t𝔽q}:λ𝔽q2}.\mathcal{S}:=\{\{A_{t}^{\lambda}\colon t\in\mathbb{F}_{q}\}\colon\lambda\in\mathbb{F}_{q^{2}}\}.

Then:

  1. (i)

    Every element of 𝒮\mathcal{S} is a set of subgroups and any two cosets from subgroups of different such sets intersect each other in at most one element.

  2. (ii)

    If we let 𝒫\mathcal{P} be the underlying set of EE, then for every λ𝔽q2\lambda\in\mathbb{F}_{q^{2}} the set of lines λ={Atλg:gE,t𝔽q}\mathcal{L}_{\lambda}=\{A_{t}^{\lambda}g:g\in E,t\in\mathbb{F}_{q}\} gives rise to a triangle-free partial linear space (𝒫,λ)(\mathcal{P},\mathcal{L}_{\lambda}) of order (q21,q1)(q^{2}-1,q-1).

Proof.

First, for each λ𝔽q2\lambda\in\mathbb{F}_{q^{2}} and t𝔽qt\in\mathbb{F}_{q}, we have

Atλ=τλ(At).A_{t}^{\lambda}=\tau_{\lambda}(A_{t}).

Since τλ\tau_{\lambda} is an automorphism of the underlying group (as per Lemma 4.1), it follows that {τλ(A)}{τλ(At):t𝔽q}\{\tau_{\lambda}(A_{\infty})\}\cup\{\tau_{\lambda}(A_{t})\colon t\in\mathbb{F}_{q}\} and {τλ(A)}{τλ(Ab):b𝔽q}\{\tau_{\lambda}(A^{*}_{\infty})\}\cup\{\tau_{\lambda}(A^{*}_{b})\colon b\in\mathbb{F}_{q}\} also form a Kantor family, and give rise to an isomorphic generalised quadrangle. Therefore, by taking cosets of the subgroups AtλA_{t}^{\lambda}, t𝔽qt\in\mathbb{F}_{q}, as lines we get a subgeometry of this generalised quadrangle which is a triangle-free partial linear space of order (q21,q1)(q^{2}-1,q-1). We now prove the first part.

First we show that At1λ1At2λ2={(0,0,0)}A_{t_{1}}^{\lambda_{1}}\cap A_{t_{2}}^{\lambda_{2}}=\{(0,0,0)\} whenever λ1λ2\lambda_{1}\neq\lambda_{2}. An element of At1λ1At2λ2A_{t_{1}}^{\lambda_{1}}\cap A_{t_{2}}^{\lambda_{2}} is of the form (a,aq+1t1+𝖳𝗋(λ1a+κλ1aq+12λ1qa2),at1+λ1aq)(a,a^{q+1}t_{1}+\operatorname{\mathsf{Tr}}(\lambda_{1}a+\kappa\lambda_{1}a^{q}+\tfrac{1}{2}\lambda_{1}^{q}a^{2}),at_{1}+\lambda_{1}a^{q}) for some a𝔽q2a\in\mathbb{F}_{q^{2}}, but it is also (b,bq+1t2+𝖳𝗋(λ2b+κλ2bq+12λ2qb2),bt2+λ2bq)(b,b^{q+1}t_{2}+\operatorname{\mathsf{Tr}}(\lambda_{2}b+\kappa\lambda_{2}b^{q}+\tfrac{1}{2}\lambda_{2}^{q}b^{2}),bt_{2}+\lambda_{2}b^{q}) for some b𝔽q2b\in\mathbb{F}_{q^{2}}. Therefore, a=ba=b and hence

aq+1(t1t2)\displaystyle a^{q+1}(t_{1}-t_{2}) =𝖳𝗋((λ2λ1)(a+κaq+12a2q))\displaystyle=\operatorname{\mathsf{Tr}}\left((\lambda_{2}-\lambda_{1})(a+\kappa a^{q}+\tfrac{1}{2}a^{2q})\right)
a(t1t2)\displaystyle a(t_{1}-t_{2}) =(λ2λ1)aq.\displaystyle=(\lambda_{2}-\lambda_{1})a^{q}.

Now aq+1(t1t2)=aqa(t1t2)a^{q+1}(t_{1}-t_{2})=a^{q}a(t_{1}-t_{2}) and so

𝖳𝗋((λ2λ1)(a+κaq+12a2q))\displaystyle\operatorname{\mathsf{Tr}}\left((\lambda_{2}-\lambda_{1})(a+\kappa a^{q}+\tfrac{1}{2}a^{2q})\right) =a2q(λ2λ1).\displaystyle=a^{2q}(\lambda_{2}-\lambda_{1}).

Expanding this equation gives us

(λ2λ1)(a+κaq12a2q)+(λ2λ1)q(a+κaq+12a2q)q=0.(\lambda_{2}-\lambda_{1})(a+\kappa a^{q}-\tfrac{1}{2}a^{2q})+(\lambda_{2}-\lambda_{1})^{q}(a+\kappa a^{q}+\tfrac{1}{2}a^{2q})^{q}=0.

Suppose, by way of contradiction, that a0a\neq 0. Since λ2λ1\lambda_{2}\neq\lambda_{1}, we can rewrite the equation as

(λ2λ1)q1=a+κaq12a2q(a+κaq+12a2q)q.(\lambda_{2}-\lambda_{1})^{q-1}=-\frac{a+\kappa a^{q}-\tfrac{1}{2}a^{2q}}{(a+\kappa a^{q}+\tfrac{1}{2}a^{2q})^{q}}.

Let 𝖭:𝔽q2𝔽q\mathsf{N}\colon\mathbb{F}_{q^{2}}\rightarrow\mathbb{F}_{q} be the relative norm function, defined by 𝖭(x)=xq+1\mathsf{N}(x)=x^{q+1}. The left-hand side has norm 1 and hence

𝖭((aq+κqa12a2))=𝖭(aq+κqa+12a2).\mathsf{N}(-(a^{q}+\kappa^{q}a-\tfrac{1}{2}a^{2}))=\mathsf{N}(a^{q}+\kappa^{q}a+\tfrac{1}{2}a^{2}).

Now 𝖭(1)=1\mathsf{N}(-1)=1 and we can factor out 𝖭(a)\mathsf{N}(a), so

𝖭(aq1+κq+12a)𝖭(aq1+κq12a)=0.\mathsf{N}(a^{q-1}+\kappa^{q}+\tfrac{1}{2}a)-\mathsf{N}(a^{q-1}+\kappa^{q}-\tfrac{1}{2}a)=0.

Note that by definition of the relative norm map,

𝖭(x+y+z)\displaystyle\mathsf{N}(x+y+z) =(x+y+z)(x+y+z)q\displaystyle=(x+y+z)(x+y+z)^{q}
=(x+y+z)(xq+yq+zq)\displaystyle=(x+y+z)(x^{q}+y^{q}+z^{q})
=xxq+yyq+zzq+xyq+yxq+xzq+zxq+yzq+zyq\displaystyle=xx^{q}+yy^{q}+zz^{q}+xy^{q}+yx^{q}+xz^{q}+zx^{q}+yz^{q}+zy^{q}

Using that in 𝔽q2\mathbb{F}_{q^{2}}, yxq=(yqx)qyx^{q}=(y^{q}x)^{q}, we obtain

𝖭(x+y+z)=𝖭(x)+𝖭(y)+𝖭(z)+𝖳𝗋(xyq+xzq+yqz).\mathsf{N}(x+y+z)=\mathsf{N}(x)+\mathsf{N}(y)+\mathsf{N}(z)+\operatorname{\mathsf{Tr}}{(xy^{q}+xz^{q}+y^{q}z)}.

Therefore,

0=\displaystyle 0= (𝖭(aq1)+𝖭(κq)+𝖭(12a)+𝖳𝗋(aq1κ+12a2q1+12κa))\displaystyle\left(\mathsf{N}(a^{q-1})+\mathsf{N}(\kappa^{q})+\mathsf{N}(\tfrac{1}{2}a)+\operatorname{\mathsf{Tr}}(a^{q-1}\kappa+\tfrac{1}{2}a^{2q-1}+\tfrac{1}{2}\kappa a)\right)
(𝖭(aq1)+𝖭(κq)+𝖭(12a)+𝖳𝗋(aq1κ12a2q112κa))\displaystyle-\left(\mathsf{N}(a^{q-1})+\mathsf{N}(\kappa^{q})+\mathsf{N}(\tfrac{1}{2}a)+\operatorname{\mathsf{Tr}}(a^{q-1}\kappa-\tfrac{1}{2}a^{2q-1}-\tfrac{1}{2}\kappa a)\right)
=\displaystyle= 𝖳𝗋(κa+a2q1),\displaystyle\operatorname{\mathsf{Tr}}(\kappa a+a^{2q-1}),

a contradiction. So a=0a=0 and At1λ1At2λ2={(0,0,0)}A_{t_{1}}^{\lambda_{1}}\cap A_{t_{2}}^{\lambda_{2}}=\{(0,0,0)\}.

Now for any two subgroups H,KH,K of a group, the intersection of two cosets of HH and KK is either empty, or a coset of HKH\cap K, which proves our claim. ∎

Corollary 4.4.

There exists an absolute constant CC such that for all r2,k3r\geqslant 2,k\geqslant 3, we have sr(Kk)C(k1)5r5/2s_{r}(K_{k})\leqslant C(k-1)^{5}r^{5/2}.

Proof.

Let r2r\geqslant 2, k3k\geqslant 3, c=10+9ln232c=\frac{10+9\ln{2}}{3\sqrt{2}} and let qq be the smallest prime such that qckrq\geqslant ck\sqrt{r}. By Lemma 4.2 and Theorem 4.3(ii), there exists a family of rq2r\leqslant q^{2} triangle-free partial linear spaces of order (q21,q1)(q^{2}-1,q-1), on the same point set 𝒫\mathcal{P} and pairwise disjoint line-sets 1,,r\mathcal{L}_{1},\dots,\mathcal{L}_{r}, and by Theorem 4.3(i), the point-line geometry (𝒫,i=1ri)\left(\mathcal{P},\bigcup_{i=1}^{r}\mathcal{L}_{i}\right) is also a partial linear space. Note that q213rklnkq^{2}-1\geqslant 3rk\ln k and q13k(1+lnr)q-1\geqslant 3k(1+\ln r). Combining Lemmas 1.6 and 3.1, sr(Kk+1)=Pr(k)|𝒫|s_{r}(K_{k+1})=P_{r}(k)\leqslant\lvert\mathcal{P}\rvert. By Bertrand’s postulate, q2ckrq\leqslant 2ck\sqrt{r}, and using |𝒫|=q5\lvert\mathcal{P}\rvert=q^{5} yields the desired bound, with C=(2c)5=26,282C=\lceil(2c)^{5}\rceil=26,\!282. Note that, in light of Conjecture 5.2, we did not try to further optimise this constant. ∎

5. Concluding remarks

While generalised quadrangles have been used extensively in extremal combinatorics, and particularly Ramsey theory (e.g. [7, 11, 17, 19, 25]), our result appears to be the first instance in Ramsey theory where the group theoretic structure of these geometries is exploited. We are hopeful that Kantor’s model of generalised quadrangles will lead to new results in other Ramsey problems as well.

In the probabilistic argument of Section 3, note that if we use s+1=q2s+1=q^{2} and t+1=qt+1=q, then from equation (3.1) it follows that we can solve the following quadratic inequality in qq to ensure that the probability is <1<1:

25361rkq21+lnrrqlnk>0.\frac{25}{36}\frac{1}{rk}q^{2}-\frac{1+\ln r}{r}q-\ln k>0.

One can check that this inequality is satisfied for all q3625k(1+lnr)+65rklnkq\geqslant\frac{36}{25}k(1+\ln r)+\frac{6}{5}\sqrt{rk\ln k}. Using that for any a,b>0a,b>0, (a+b)524(a5+b5)(a+b)^{5}\leqslant 2^{4}(a^{5}+b^{5}), we obtained the following more refined upper bound.

Theorem 5.1.

For all r2,k2r\geqslant 2,k\geqslant 2,

sr(Kk)212[(k1)5ln5r+(k1)5/2r5/2ln5/2(k1)].s_{r}(K_{k})\leqslant 2^{12}\left[(k-1)^{5}\ln^{5}r+(k-1)^{5/2}r^{5/2}\ln^{5/2}(k-1)\right].

For further improvements to our upper bound we should perhaps explore triangle-free partial linear spaces that do not arise from generalised quadrangles. Moreover, if we could make the probabilistic argument of Section 3 deterministic, then this could also lead to an improvement in the bound. We would like to make the following conjecture.

Conjecture 5.2.

For all r2,k2r\geqslant 2,k\geqslant 2

sr(Kk)Ck2r2f(lnk,lnr)s_{r}(K_{k})\leqslant Ck^{2}r^{2}f(\ln k,\ln r)

for some constant C>0C>0 and a constant degree polynomial function ff.

Acknowledgements

We are grateful to Anita Liebenau for helpful discussions. We also thank Simona Boyadzhiyska for pointing out a significant gap in our first draft that was fixed later. We also want to thank the anonymous referee for a thorough proofreading and many helpful remarks.

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