Matricial characterization of tournaments with maximum number of diamonds

TL;DR

This paper provides a complete matrix-based characterization of n-tournaments with the maximum number of diamonds for certain sizes, assuming skew-conference matrices, and offers bounds for other sizes.

Contribution

It introduces a matricial framework to characterize maximum diamond tournaments for specific sizes, extending understanding of their structure.

Findings

Complete characterization for n ≡ 0 mod 4 and n ≡ 3 mod 4

Upper bounds and matricial descriptions for n ≡ 2 mod 4

Assumption of skew-conference matrices

Abstract

A diamond is a $4$-tournament which consists of a vertex dominating or dominated by a $3$-cycle. Assuming the existence of skew-conference matrices, we give a complete characterization of $n$-tournaments with the maximum number of diamonds when $n\equiv0\pmod{4}$ and $n\equiv3\pmod{4}$. For $n\equiv2\pmod{4}$, we obtain an upper bound on the number of diamonds in an $n$-tournament and we give a matricial characterization of tournaments achieving this bound.

Equations

$${\sigma }_h=(-1{)}^h\sum (\text{all }h\times h\text{ principal minors})$$

$$P_M(x)=x^l(x^2+{\lambda }_1^2)(x^2+{\lambda }_2^2)\cdots (x^2+{\lambda }_k^2)$$

$$P_M(x)=(x^2+{\lambda }_1^2)(x^2+{\lambda }_2^2)\cdots (x^2+{\lambda }_{n/2}^2)$$

$$P_M(x)=x(x^2+{\lambda }_1^2)(x^2+{\lambda }_2^2)\cdots (x^2+{\lambda }_{(n-1)/2}^2)$$

$${\delta }_T=\frac{1}{96}{n}^2(n-1)(n-2)-\frac{1}{16}\sum _{i<j}{m}_{ij}^2$$

$$P_S(x)=\{\begin{array}{ll}(x^2+{\lambda }_1^2)(x^2+{\lambda }_2^2)\cdots (x^2+{\lambda }_m^2)& \text{ if }n\text{ is even }\\ x(x^2+{\lambda }_1^2)(x^2+{\lambda }_2^2)\cdots (x^2+{\lambda }_m^2)& \text{if }n\text{ is odd}\end{array}$$

$$P_{S^2}(x)=\{\begin{array}{ll}(x^2+2{\lambda }_1^{2}x+{\lambda }_1^4)\cdots (x^2+2{\lambda }_m^{2}x+{\lambda }_m^4)& \text{ if }n\text{ is even }\\ x(x^2+2{\lambda }_1^{2}x+{\lambda }_1^4)\cdots (x^2+2{\lambda }_m^{2}x+{\lambda }_m^4)& \text{if }n\text{ is odd}\end{array}$$

$${\alpha }_2=\sum _i{\lambda }_i^2$$

$${\alpha }_4=\sum _{i<j}{\lambda }_i^2{\lambda }_j^2=\frac{1}{2}((\sum _i{\lambda }_i^2{)}^2-\sum _i{\lambda }_i^4)$$

$${\beta }_2=4\sum _{i<j}{\lambda }_i^2{\lambda }_j^2+\sum _i{\lambda }_i^4$$

$${\beta }_2=\sum _{i<j}((n-1{)}^2-m_{ij}^2)=\frac{n(n-1{)}^3}{2}-\sum _{i<j}{m}_{ij}^2$$

$${\alpha }_4=8{\delta }_T+\left(\genfrac{}{}{0ex}{}{n}{4}\right)$$

$${\delta }_T$$

$${\delta }_T={\delta }_{T-v}+c_3(T-v)$$

$$c_3(T)=\frac{1}{24}n(n-1)(n-2)+\frac{1}{4}\sum _{i<j}{m}_{ij}$$

$$d_T^{+}(i)+d_T^{-}(i)=n-1$$

$$\sum _i{d}_T^{+}(i)=\sum _i{d}_T^{-}(i)=\frac{n(n-1)}{2}$$

$$c_3(T)=\left(\genfrac{}{}{0ex}{}{n}{3}\right)-\sum _i\left(\genfrac{}{}{0ex}{}{d_T^{+}(i)}{2}\right)$$

$$m_{ij}=2{\gamma }_{ij}-n+2$$

$$d_T^{+}(i,j)-d_T^{-}(i,j)=d_T^{+}(i)-d_T^{-}(j)$$

$${\gamma }_{ij}=n-2-(d_T^{+}(i,j)+d_T^{-}(i,j))$$

$${\gamma }_{ij}=2n-3-(d_T^{+}(i)+d_T^{+}(j)+2d_T^{-}(i,j))$$

$$\sum _{i<j}{\gamma }_{ij}=\sum _k{d}_T^{+}(k)d_T^{-}(k)$$

$$\sum _{i<j}{\gamma }_{ij}=(n-2)\sum _k{d}_T^{+}(k)-2\left(\genfrac{}{}{0ex}{}{d_T^{+}(k)}{2}\right)$$

$$c_3(T)=\frac{1}{2}\sum _{i<j}{\gamma }_{ij}-\frac{1}{12}\left(n-1\right)\left(n-2\right)n$$

$${\delta }_T\le \frac{1}{96}{n}^2(n-1)(n-2)$$

$${\delta }_T\le \frac{1}{96}n(n-1)(n-3)(n+1)$$

$$\displaystyle\begin{array}[]{lll}\delta_{T}&=&\frac{1}{96}n^{2}(n-1)(n-2)-\frac{1}{16}\sum_{i<j}{m_{ij}^{2}}\\ &\leq&\frac{1}{96}n^{2}(n-1)(n-2)-\frac{1}{16}n(n-1)/{2}\\ &=&\frac{1}{96}n(n-1)(n-3)(n+1)\par\end{array}$$

$$m_{ij}^{{}^{\prime }}=1$$

$$c_3(T^{{}^{\prime }})=\frac{1}{24}n(n-1)(n+1)$$

Full text

1 Introduction

One of the most important problems in Extremal Combinatorics is to determine the largest or the smallest possible number of copies of a given object in a finite combinatorial structure. We address this problem in the case of tournaments. Throughout this paper, we mean by an $n$-tournament, a tournament with $n$ vertices. It is easy to see that, up to isomorphy, there are four distinct $4$-tournaments. The two that contain a single $3$-cycle are called diamonds [3, 4, 8]. They consist of a vertex dominating or dominated by a $3$-cycle. The class of tournaments without diamonds was characterized by Moon [15]. These tournaments appear in the literature under the names local orders [5],* locally transitive tournaments* [12] or vortex-free tournaments [11]. For $n\geq 9$, Bouchaala [3] proved that the number $\delta_{T}$ of diamonds in an $n$-tournament is either [math], $n-3$, $2n-8$ or at least $2n-6$. In another side, motivated by geometric considerations, Leader and Tan [13] proved that $\delta_{T}$ is at most $\frac{1}{4}\binom{n}{4}+O(n^{3})$. Moreover, by a probabilistic method, they showed that there is an $n$-tournament with at least $\frac{1}{4}\binom{n}{4}$ diamonds. To find the Turán density of a particular $4$-uniform hypergraphs, Baber [10] associates with each tournament $T=(V,A)$, the $4$-uniform hypergraph $\mathcal{H}_{T}$ on $V$ whose hyperedges correspond to subsets of $V$ which induce a diamond in $T$. In this hypergraph, every $5$-subset contains either [math] or $2$ hyperedges. Recently, using a combinatorial argument due to de Caen [6], Gunderson and Semeraro [10] proved that an $r$-uniform hypergraph in which every $(r+1)$-subset contains at most $2$ hyperedges has at most $\frac{n}{r^{2}}\binom{n}{r-1}$ hyperedges, in particular, an $n$-tournament contains at most $\frac{n}{16}\binom{n}{3}$ diamonds. Moreover, using Paley tournaments, they showed that this bound is reached if $n=q+1$ for some prime power $q\equiv 3\pmod{4}$.

In this paper, we study the tournaments with the maximum number of diamonds. Our work is closely related to the existence of D-optimal designs. More precisely, assuming the existence of skew-conference matrices, we give a complete characterization of $n$-tournaments with the maximum number of diamonds when $n\equiv 0\pmod{4}$ and $n\equiv 3\pmod{4}$. For $n\equiv 2\pmod{4}$, we obtain an upper bound on the number $\delta_{T}$ in an $n$-tournament. Moreover, we give a matricial characterization of tournaments achieving this bound.

2 Number of diamonds and $3$-cycles in tournaments

Throughout this paper, all matrices are from the set $\{0,\pm 1\}$, unless otherwise noted. The identity matrix of order $n$ is denoted by $I_{n}$ and the all ones matrix is denoted by $J_{n}$. The absolute value $\left|M\right|$, of a matrix $M$, is obtained from $M$ by replacing each entry of $M$ by its absolute value. Two matrices $A$ and $B$ are $\{\pm 1\}$-diagonally similar if $B=DAD$ for some $\pm 1$-diagonal matrix.

Let $T$ be an $n$-tournament. With respect to a labelling, the adjacency matrix of $T$ is the $n\times n$ matrix $A=(a_{ij})$ in which $a_{ij}$ is $1$ if $i$ dominates $j$ and [math] otherwise. The Seidel adjacency matrix of $T$ is $S=A-A^{t}$ where $A^{t}$ is the transpose of $A$. Remark that, with respect to different labellings, the Seidel adjacency matrices of a tournament are permutationally similar.

Remark 1**.**

The determinant of the Seidel adjacency matrix of a $4$-tournament is $9$ if it is a diamond and $1$ otherwise.

The following lemma is crucial in our study.

Lemma 2**.**

Let $T$ be an $n$-tournament and let $S$ be its Seidel adjacency matrix. Then the sum of all $4\times 4$ principal minors of $S$ is $8\cdot\delta_{T}+\binom{n}{4}$.

Proof.

The number of $4\times 4$ principal minors of $S$ is $\binom{n}{4}$. It follows from Remark 1 that the sum of all $4\times 4$ principal minors of $S$ is $9\cdot\delta_{T}+(\binom{n}{4}-\delta_{T})=$ $8\cdot\delta_{T}+\binom{n}{4}$. ∎

Let $M$ be an $n\times n$ complex matrix and let $P_{M}(x):=\det(xI-M)=x^{n}+\sigma_{1}x^{n-1}+\cdots+\sigma_{n-1}x+\sigma_{n}$ be its characteristic polynomial, then

[TABLE]

When $M$ is a real skew-symmetric matrix, its nonzero eigenvalues are all purely imaginary and come in complex conjugate pairs $\pm i\lambda_{1},\ldots,\pm i\lambda_{k}$, where $\lambda_{1},\ldots,\lambda_{k}$ are real positive numbers. Equivalently, the characteristic polynomial of $M$ has the form

[TABLE]

where $l+2k=n$.

Assume now that $M$ is skew-symmetric and all its off-diagonal entries are from the set $\{-1,1\}$. Such matrix is sometimes known as a skew-symmetric Seidel matrix. By using [14, Proposition 1], $\det(S)=0$ if and only if $n$ is odd. Then, if $n$ is even, $l=0$ and

[TABLE]

If $n$ is odd, then by using [14, Proposition 1] again, any $(n-1)\times(n-1)$-principal minor is nonzero and thus, the multiplicity of the eigenvalue [math] is $1$. It follows that

[TABLE]

A useful formula of the number of diamonds is given in the following proposition.

Proposition 3**.**

Let $T$ be an $n$-tournament and let $S$ be its Seidel adjacency matrix. Then,

[TABLE]

where $m_{ij}$ are the entries of $S^{2}$.

Proof.

Let $m$ be the integer part of $\frac{n}{2}$ and $\pm i\lambda_{1},\ldots,\pm i\lambda_{m}$ the nonzero eigenvalues of $S$. As we have seen above

[TABLE]

The nonzero eigenvalues of $S^{2}$ are $-\lambda_{1}^{2},\ldots,-\lambda_{m}^{2}$, each of them appears two times. Hence, we can write $P_{S^{2}}(x)$ in the following form.

[TABLE]

Let $P_{S}(x):=x^{n}+\alpha_{1}x^{n-1}+\ldots+\alpha_{n}$ and let $P_{S^{2}}(x):=x^{n}+\beta_{1}x^{n-1}+\ldots+\beta_{n}$. By expanding expressions (2) and (3), we get

[TABLE]

[TABLE]

and

[TABLE]

It follows that $\beta_{2}=2\alpha_{4}+\alpha_{2}^{2}$. By Equality (1), we have $\alpha_{2}=\frac{n(n-1)}{2}$, and hence $\alpha_{4}=\frac{1}{2}\beta_{2}-\frac{1}{2}\left(\frac{n(n-1)}{2}\right)^{2}$.

Since $S^{2}$ is symmetric and all its diagonal entries are $1-n$, by Equality (1), we have

[TABLE]

Applying Lemma (2) and Equality (1) again, we get

[TABLE]

It follows that

[TABLE]

∎

Let $T=(V,A)$ be a tournament, the switching of $T$, according to a subset $X$ of $V$, consists of reversing all the arcs between $X$ and $V\backslash X$, we denote the resulting tournament by $T_{X}$. We say that two tournaments $T$ and $T^{{}^{\prime}}$ on a vertex set $V$ are switching equivalent, if there exists $X\subset V$ such that $T_{X}=T^{{}^{\prime}}$.

Remark 4**.**

It is well-known that two tournaments are switching equivalent iff their Seidel adjacency matrices are $\{\pm 1\}$-diagonally similar [16]. Since similarity by a $\pm 1$ diagonal matrix preserves the principal minors, by Remark 1, switching equivalent tournaments have the same diamonds.

Let $v$ be a vertex of a tournament $T=(V,A)$, the out-neighbourhood $N^{+}_{T}(v)$ of $v$ is the set of all vertices of $T$ dominated by $v$. The in-neighbourhood $N^{-}_{T}(v)$ of $v$ is the set of all vertices of $T$ which dominate $v$. In the switching $T_{N^{-}(v)}$ of $T$ according to $N^{-}_{T}(v)$, the vertex $v$ dominates $V\setminus\{v\}$. Hence, by Remark 4, to study the number of diamonds, we can assume that there is a vertex $v$ dominating $V\setminus\{v\}$. We obtain then the following lemma connecting the number of diamonds and the number of $3$-cycles.

Lemma 5**.**

Let $T=(V,A)$ be a tournament containing a vertex $v$ that dominates $V\setminus\{v\}$. Then,

[TABLE]

where $c_{3}(T-v)$ is the number of $3$-cycles in $T-v$.

Similarly to $\delta_{T}$, the number $c_{3}(T)$ of $3$-cycles in $T$ can also be expressed in terms of the entries of $S^{2}$.

Proposition 6**.**

Let $T$ be an $n$-tournament and let $S$ be its Seidel adjacency matrix. Then,

[TABLE]

where $m_{ij}$ are the entries of $S^{2}$.

Before proving this proposition, we need some notions and basic results about tournaments. For more details, the reader is referred to [15].

Let $T$ be an $n$-tournament. Without loss of generality, we can assume that the vertex set of $T$ is $\{1,\ldots,n\}$. The out-degree $d^{+}_{T}(i)$ (resp. in-degree $d^{-}_{T}(i)$) of a vertex $i$ is $|N^{+}_{T}(i)|$ (resp. $|N^{-}_{T}(i)|$). The out-degree $d^{+}_{T}(i,j)$ of $(i,j)$ (resp. in-degree $d^{-}_{T}(i,j)$ of $(i,j)$) is $|N^{+}_{T}(i)\cap N^{+}_{T}(j)|$ (resp. $|N^{-}_{T}(i)\cap N^{-}_{T}(j)|$).

The tournament $T$ is regular, if there is a constant $k$ such that $d^{+}_{T}(i)=k$ for every $i\in\{1,\ldots,n\}$; it is doubly regular if there is a constant $h$ such that $d^{+}_{T}(i,j)=h$ for every $i\neq j\in\{1,\ldots,n\}$. A doubly regular tournament is also regular. If $T$ is regular, then $n$ is odd and $d^{+}_{T}(i)=\frac{n-1}{2}$ for $i\in\{1,\ldots,n\}$. If $T$ is doubly regular, then $n\equiv 3\pmod{4}$ and $d^{+}_{T}(i,j)=d^{-}_{T}(i,j)=\frac{n-3}{4}$ for $i\neq j\in\{1,\ldots,n\}$.

Recall the well-known equalities

[TABLE]

[TABLE]

[TABLE]

Remark 7**.**

It follows from Equality (6) that $c_{3}(T)\leq\frac{1}{24}(n^{3}-n)$. Moreover equality holds iff $n$ is odd and $T$ is regular.

For $i\neq j\in\{1,\ldots,n\}$, let $\gamma_{ij}:=|N^{+}_{T}(i)\cap N^{-}_{T}(j)|+|N^{-}_{T}(i)\cap N^{+}_{T}(j)|$. Then, we have

[TABLE]

[TABLE]

[TABLE]

Combining Equalities (8), (9) and (4), we get

[TABLE]

By double-counting principle, we obtain

[TABLE]

Using Equalities (4) and (11), we get

[TABLE]

It follows from Equalities (5) and (6) that

[TABLE]

Equality (7) allows to complete the proof of Proposition 5.

3 Characterization of $n$-tournaments with maximum number of diamonds for $n\equiv 0,2,3\pmod{4}$

Throughout this section, $T$ denotes an $n$-tournament and $S$ denotes its Seidel adjacency matrix.

It follows from Proposition 3 that

[TABLE]

Equality holds if and only if $m_{ij}=0$ for every $i\neq j$. Since $m_{ii}=1-n$, then $\delta_{T}=\frac{1}{96}n^{2}(n-1)(n-2)$ if and only if $S^{2}=(1-n)I_{n}$, or equivalently $SS^{t}=(n-1)I_{n}$. Skew-symmetric Seidel matrices that satisfy this equality are called skew-conference matrices, and exist only if $n$ is divisible by $4$.

Example 8**.**

For any prime power $q\equiv 3\pmod{4}$, the Paley tournament $T(q)$ is the tournament whose vertices are elements of $\mathbb{F}_{q}$ where the vertex $x$ dominates the vertex $y$ iff $y-x$ is a square in $\mathbb{F}_{q}$. Let $T^{*}(q)$ be the tournament on $n=q+1$ vertices obtained by adding to $T(q)$ a new vertex which dominates all vertices of $T(q)$. It is well-known that the Seidel adjacency matrix of $T^{*}(q)$ is a skew conference matrix. Then, the number of diamonds in $T^{*}(q)$ is $\frac{1}{96}n^{2}(n-1)(n-2)$.

For $n$ odd, we obtain the following refinement of Equality (12).

Proposition 9**.**

If $n$ is odd, then

[TABLE]

Moreover, equality holds if and only if $|S^{2}+nI_{n}|=J_{n}$.

Proof.

Let $S^{2}:=(m_{ij})$. Since $n$ is odd, by Equality (7), $m_{ij}$ is also odd. Hence $\sum_{i<j}{m_{ij}^{2}}\geq n(n-1)/{2}$.

By Proposition 3,

[TABLE]

Equality holds if and only if $|m_{ij}|=1$ for $i\neq j$, or equivalently $|S^{2}+nI_{n}|=J_{n}$, because $m_{ii}=1-n$. ∎

Theorem 10**.**

If $n\equiv 3\pmod{4}$, then the following assertions are equivalent

$T$* has $\frac{1}{96}n(n-1)(n-3)(n+1)$ diamonds.* 2. 2.

There exists a diagonal ${\pm 1}$-matrix D such that $DS^{2}D+nI_{n}=J_{n}$. 3. 3.

$T$* is switching equivalent to a doubly regular tournament.*

Proof.

$1.\Leftrightarrow 2.$ Assume that $T$ has $\frac{1}{96}n(n-1)(n-3)(n+1)$ diamonds. By Proposition 9, we have $m_{ij}=\pm 1$ for every $i\neq j$. Using Equalities (7) and (10), we get:

1. i.

If $d^{+}_{T}(i)\equiv d^{+}_{T}(j)\pmod{2}$, then $\gamma_{ij}\equiv 1\pmod{2}$, and $m_{ij}\equiv 1\pmod{4}$. Hence $m_{ij}=1$. 2. ii.

If $d^{+}_{T}(i)\equiv d^{+}_{T}(j)+1\pmod{2}$, then $\gamma_{ij}\equiv 0\pmod{2}$, and $m_{ij}\equiv-1\pmod{4}$. Hence $m_{ij}=-1$.

Let $D=diag(\epsilon_{1},\epsilon_{2},\ldots,\epsilon_{n})$ be the diagonal matrix such that $\epsilon_{i}=1$ if $d^{+}_{T}(i)$ is even and $\epsilon_{i}=-1$ otherwise. It is easy to check that $\epsilon_{i}m_{ij}\epsilon_{j}=1$ for all $i\neq j$. Then, $DS^{2}D+nI_{n}=J_{n}$.

The converse is trivial.

$2.\Leftrightarrow 3.$

Suppose that there exists a diagonal ${\pm 1}$-matrix D such that $DS^{2}D+nI_{n}=J_{n}$. The tournament $T^{{}^{\prime}}$ whose Seidel adjacency matrix is $DSD$ is switching equivalent to $T$. We will prove that $T^{{}^{\prime}}$ is doubly regular.

Let $DS^{2}D:=(m_{ij}^{{}^{\prime}})$. Thus, for $i\neq j\in\{1,\ldots,n\}$, we have

[TABLE]

By Proposition 6, we get

[TABLE]

It follows from Remark 7 that the tournament $T^{{}^{\prime}}$ is regular.

By Identities (7) and (10), the in-degree of each pair in $T^{{}^{\prime}}$ is $(n-3)/4$. Hence, by definition, $T^{{}^{\prime}}$ is doubly regular.

Conversely, assume that $T$ is switching equivalent to a doubly regular tournament $T^{{}^{\prime}}$. The out-degree and the in-degree of each pair $(i,j)$ in $T^{{}^{\prime}}$ is $(n-3)/4$. Let ${S^{{}^{\prime}}}$ be the Seidel adjacency matrix of $T^{{}^{\prime}}$ and let ${S^{{}^{\prime}}}^{2}:=(m^{{}^{\prime}}_{ij})$. By Equalities (7) and (10), we get $m^{{}^{\prime}}_{ij}=1$ for $i\neq j\in\{1,\ldots,n\}$ and hence ${S^{{}^{\prime}}}^{2}+nI_{n}=J_{n}$. Since $T$ and $T^{{}^{\prime}}$ are switching equivalent, $S^{{}^{\prime}}=DSD$ for some $\{\pm 1\}$-diagonal matrix $D$ and then $DS^{2}D+nI_{n}=J_{n}$.

∎

Theorem 11**.**

If $n\equiv 2\pmod{4}$, then

[TABLE]

Moreover, equality holds if and only if

[TABLE]

for some permutation matrix $P$.

Proof.

We label the vertices of $T$ so that the first $c$ vertices have an even out-degree and the remaining $n-c$ vertices have an odd out-degree. With respect to this labelling, the Seidel adjacency matrix of $T$ is $PSP^{t}$ where $P$ is a permutation matrix. Let $PS^{2}P^{t}:=(m^{{}^{\prime}}_{ij})$. Let $i\neq j\in\{1,\ldots,n\}$ such that $d^{+}_{T}(i)\equiv d^{+}_{T}(j)\pmod{2}$. By Equality (10), $\gamma_{ij}\equiv 1\pmod{2}$, and by Equality (7), $m^{{}^{\prime}}_{ij}\equiv 2\pmod{4}$. Therefore, ${m^{{}^{\prime}}_{ij}}^{2}\geq 4$.

By Proposition 3, we have

[TABLE]

Equality holds iff $c=\frac{n}{2}$, $m^{{}^{\prime}}_{ij}=\pm 2$ if $1\leq i<j\leq c$ or $c+1\leq i<j\leq n$ and $m^{{}^{\prime}}_{ij}=0$ otherwise. ∎

We give two classes of tournaments that satisfy the conditions of Theorem 11.

Recall that an EW matrix $B$ of order $n\equiv 2\pmod{4}$ is a $(\pm 1)$-matrix verifying $BB^{t}=B^{t}B=\begin{pmatrix}M&0\\ 0&M\end{pmatrix}$ where $M=(n-2)I_{n/2}+2J_{n/2}$. Ehlich [7] and Wojtas [17] independently proved that EW matrices have the maximum determinant among $\pm 1$-matrices of order $n\equiv 2\pmod{4}$. EW matrices exist only if $2n-2$ is the sum of two squares. An EW matrix is said to be of skew type if $B+B^{t}=2I_{n}$. Such matrix exist only if $2n-3$ is a square, hence, there are no EW matrices of skew type with order $n=10,18,22,30,34,38,46,50$.

Consider the matrix $B-I_{n}$, where $B$ is an EW-matrix of skew type. Clearly, this matrix is skew-symmetric, moreover, it has the maximum determinant among skew-symmetric Seidel matrices of order $n\equiv 2\pmod{4}$ [2]. By simple computation, we have

[TABLE]

Hence, by Theorem 11, if $S+I_{n}$ is an EW matrix, then $T$ has $\frac{1}{96}n(n-3)(n-2)(n+2)$ diamonds. Moreover, by  [9, Lemma 3.3], the characteristic polynomial of $S$ is $P_{S}(x)=(x^{2}+8k+1)(x^{2}+4k-1)^{2k}$ where $n=4k+2$. 2. 2.

Let $T$ be a doubly regular tournament on $n+1=4k+3$ vertices and let $T^{{}^{\prime}}$ be the tournament obtained by removing any vertex $v$ of $T$. It is easy to see that for every vertices $i,j$ of $T^{{}^{\prime}}$, we have

[TABLE]

Let $S$ be the Seidel adjacency matrix of $T^{{}^{\prime}}$. Using Identities (7) and (9), we find that up to permutation

[TABLE]

Hence, again by Theorem 11, the tournament $T^{{}^{\prime}}$ has $\frac{1}{96}n(n-3)(n-2)(n+2)$ diamonds. Moreover, by [9, Lemma 4.2.iii], $P_{S}(x)=(x^{2}+1)(x^{2}+4k+3)^{2k}$.

Remark 12**.**

Let $T$ be a tournament with $n=4k+2$ vertices and let $S$ be its Seidel adjacency matrix. It follows from [9, Lemmata 3.3 and 3.7] that the characteristic polynomial of $S$ is $P_{S}(x)=(x^{2}+8k+1)(x^{2}+4k-1)^{2k}$ iff there is a $\pm 1$-diagonal matrix D such that $DSD$ is a skew-symmetric EW matrix.

Remark 13**.**

Let $T$ be a tournament and let $S$ be its Seidel adjacency matrix. It follows from Lemma 2 and Equality (1) that if $P_{S}(x)=(x^{2}+8k+1)(x^{2}+4k-1)^{2k}$ or $P_{S}(x)=(x^{2}+1)(x^{2}+4k+3)^{2k}$, then $T$ has the maximum number of diamonds.

Up to switching, there are two 6-tournaments with the maximum number of diamonds $6$, one of them is obtained by removing a vertex from a doubly regular tournament, and the other consists of two $3$-cycles one dominating the other, its Seidel adjacency matrix is a skew symmetric EW matrix.

As for $10$-tournaments, using SageMath, we found two switching classes of tournaments with the maximum number of diamonds $70$. The characteristic polynomial of the tournaments in the first class is $(x^{2}+1)(x^{2}+11)^{4}$, we identified one as a tournament obtained by removing a vertex from a doubly regular tournament. The characteristic polynomial of the tournaments in the second class is $(x^{2}+1)(x^{4}+18x^{2}+61)^{2}$. The tournament with the following Seidel adjacency matrix belongs in the second class.

[TABLE]

Curiously, $S$ has the maximum determinant among $\pm 1$ matrices, but $S+I_{n}$ is not an EW matrix. This leads to the following questions.

Question 14**.**

Let $S$ be a skew Seidel matrix with the maximum determinant, does its corresponding tournament have the maximum number of diamonds ?

The answer to this question is positive in the following two cases:

There exists a skew conference matrix of order $n\equiv 0\pmod{4}$. 2. 2.

There exists a skew EW matrix of order $n\equiv 2\pmod{4}$.

Question 15**.**

Let $n=4k+2$. Is there an infinite family of $n$-tournaments with $\frac{1}{96}n(n-3)(n-2)(n+2)$ diamonds, such that their Seidel adjacency matrices has a characteristic polynomial that is neither $(x^{2}+8k+1)(x^{2}+4k-1)^{2k}$ nor $(x^{2}+1)(x^{2}+4k+3)^{2k}$.

4 The case of $n\equiv 1\pmod{4}$

We start with following lemma.

Lemma 16**.**

Let $T=(V,A)$ be a tournament on $n+1\equiv{2}\pmod{4}$ vertices with $\frac{1}{96}(n+1)(n-2)(n-1)(n+3)$ diamonds. Then

[TABLE]

for every $v\in V$.

Proof.

Let $T=(V,A)$ be a tournament on $n+1=4k+2$ vertices containing $\frac{1}{96}(n+1)(n-2)(n-1)(n+3)$ diamonds. Fix a vertex of $v$ and let $T^{{}^{\prime}}$ be the tournament switching equivalent to $T$ in which $v$ dominates the other vertices.

By adapting the proof of [1, Lemma 2.1], we show easily that the the score vector of $T^{{}^{\prime}}-v$ is $2k$, $2k+1$ and $2k-1$, each appearing $2k+1$ , $k$ and $k$ times, respectively.

By Equality (6), the number of $3$-cycles in $T^{{}^{\prime}}-v$ is

[TABLE]

That is,

[TABLE]

Since $v$ dominates every vertex in $V\setminus\{v\}$ in the tournament $T^{{}^{\prime}}$, by Lemma 5

[TABLE]

Hence,

[TABLE]

It follows that

[TABLE]

because $T^{{}^{\prime}}-v$ is switching equivalent to $T-v$. ∎

The previous lemma leads to the following conjecture.

Conjecture 17**.**

Let $T=(V,A)$ be a tournament on $n\equiv 1\pmod{4}$ vertices. Then, the number of diamonds in $T$ is at most

[TABLE]

The following lemma gives another way to obtain $n$-tournaments with $\frac{1}{96}{\left(n+3\right)}{\left(n-1\right)}{\left(n-2\right)}{\left(n-3\right)}$ diamonds.

Lemma 18**.**

Let $T=(V,A)$ be a tournament on $n-1$ vertices such that its Seidel adjacency matrix is a skew-conference matrix . Let $T^{{}^{\prime}}$ be a tournament obtained from $T$ by adding a vertex that dominates all vertices in $V$, then

[TABLE]

Proof.

Let $T^{{}^{\prime}}$ be a tournament obtained from $T$ by adding a vertex $v$ that dominates all vertices in $V$, the number of diamonds in the tournament $T^{{}^{\prime}}$ is

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Let $S$ be the Seidel adjacency matrix of $T$. By Propositions 3 and 6, we have

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Since the non-diagonal entries of $S^{2}$ are equal to zero, then

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