Endomorphisms of quadratic forms graph in characteristic two

Abstract

Let q be a power of 2, and let ${\mathbb {F}}_q$ be the finite field with q elements. The quadratic forms graph, denoted by Q(n, q) where $n\ge 2$, has all quadratic forms on ${\mathbb {F}}_q^n$ as vertices and two vertices f and g are adjacent if $\mathrm{rk}(f-g) = 1$ or 2. A graph G is called a pseudo-core if every endomorphism of G is either an automorphism or a colouring. A graph G is a core if every endomorphism of G is an automorphism. We prove that Q(n, q) is a pseudo-core and Q(2m, q) is a core. Moreover, we gave the smallest eigenvalue of Q(n, q).

1 Introduction

Throughout, let ${\mathbb {F}}_q$ be the finite field with q elements where q is a power of a prime. Write ${\mathbb {F}}_q^*={\mathbb {F}}_q{\setminus } \{0\}$. Let ${\mathbb {F}}_q^n$ be the vector space whose elements are n-dimensional row vectors over ${\mathbb {F}}_q$. A quadratic form f on ${\mathbb {F}}_q^n$ is a map $f: {\mathbb {F}}_q^n\rightarrow {\mathbb {F}}_q$ such that, for every $u,v\in {\mathbb {F}}_q^n$ and $a,b\in {\mathbb {F}}_q$, $f(au+bv)=a^2f(u)+b^2f(v)+abB_f(u,v)$ for some bilinear form $B_f(u,v)=:B_f$ on ${\mathbb {F}}_q^n$. Note that $B_f$ is alternating if q is even and symmetric if q is odd. We denote by $\mathrm{Rad}(B_f)$ and $\mathrm{Rad}(f)$ the radicals of $B_f$ and f, i.e.

$$\begin{aligned} \mathrm{Rad}(B_f)= & {} \left\{ u\in {\mathbb {F}}_q^n: B_f(u,v)=0, \quad \forall \, v\in {\mathbb {F}}_q^n\right\} , \end{aligned}$$

(1.1)

$$\begin{aligned} \mathrm{Rad}(f)= & {} \left\{ u\in {\mathbb {F}}_q^n: f(u)=0, B_f(u,v)=0,\quad \forall \, v\in {\mathbb {F}}_q^n\right\} . \end{aligned}$$

(1.2)

The rank of a quadratic form f on ${\mathbb {F}}_q^n$, denoted by $\mathrm{rk}(f)$, is the dimension of the quotient space ${\mathbb {F}}_q^n/\mathrm{Rad}(f)$, i.e. $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$. There is a one-to-one correspondence between quadratic forms on ${\mathbb {F}}_q^n$ and homogeneous polynomials of degree 2 in n variables over ${\mathbb {F}}_q$.

All graphs are simple [4] and finite. For a graph $\varGamma$, we denote by $V(\varGamma )$ and $\mathrm{diam}(\varGamma )$ the vertex set and the diameter of $\varGamma$, respectively. Let $d_\varGamma (x, y)$ (d(x, y) for short) be the distance between vertices x and y in $\varGamma$. We write $x \sim y$ if $x,y\in V(\varGamma )$ are adjacent. Let |X| be the cardinal number of a set X.

The quadratic forms graph, denoted by Q(n, q) where $n\ge 2$, has all quadratic forms on ${\mathbb {F}}_q^n$ as vertices and two vertices f and g are adjacent if $\mathrm{rk}(f-g) = 1$ or 2. We have (cf. [1])

$$\begin{aligned} |V(Q(n, q))|=q^{n(n+1)/2}. \end{aligned}$$

When $n=2$, Q(n, q) is a complete graph. The automorphism group of Q(n, q) is well known (cf. [1, 16, 17]). For any $f_0\in V(Q(n,q))$ and any invertible $n\times n$ matrix P over ${\mathbb {F}}_q$, maps $\varphi (f)=f-f_0$ and $\phi (f(x))=f(xP)$ for all $f=f(x)\in V(Q(n,q))$ are two automorphisms of Q(n, q). Thus, Q(n, q) is vertex-transitive, but Q(n, q) is not distance-transitive when $n>3$ (cf. [1, Proposition 9.6.4].

For a graph $\varGamma$ and $x\in V(\varGamma )$, let $\varGamma _i(x)$ denote the set of vertices of $\varGamma$ which are distance i from x, $i=1, \ldots , d,$ where $d=\mathrm{diam}(\varGamma )$. $\varGamma _1(x)$ is called the neighbourhood of x. A connected graph $\varGamma$ with diameter d is distance-regular if, for any vertices x, y with $d(x,y)=i$, there are $b_i$ and $c_i$ such that they depend only on i and independent of the vertices x, y and

$$\begin{aligned} |\varGamma _{i+1}(x)\cap \varGamma _1(y)| = b_i, \quad |\varGamma _{i-1}(x)\cap \varGamma _1(y)| = c_i, \quad i=0,1, \ldots , d. \end{aligned}$$

The sequence $\{b_0, b_1, \ldots ,b_{d-1}; c_1, c_2, \ldots , c_d\}$ is called the intersection array of $\varGamma$. In a distance-regular graph $\varGamma$, we have $a_i:=|\varGamma _{i}(x)\cap \varGamma _1(y)|=b_0-b_i-c_i$ where $d(x,y)=i$. For a rational number x, let [x] be the largest integer not larger than x.

Lemma 1.1

(see [1, Theorem 9.6.3]) When $n\ge 3$, Q(n, q) is a distance-regular graph with diameter $d=[(n+1)/2]$, and intersection array is given by

$$\begin{aligned} b_j= & {} q^{4j}(q^{n-2j+1}-1)(q^{n-2j}-1)/(q^2-1), \nonumber \\ c_j= & {} q^{2j-2}(q^{2j}-1)/(q^2-1), \quad 0\le j\le d. \end{aligned}$$

(1.3)

Recall that a clique of a graph $\varGamma$ is a complete subgraph of $\varGamma$, and a clique ${\mathcal {C}}$ is maximal if there is no clique of $\varGamma$ which properly contains ${\mathcal {C}}$ as a subset. A maximum clique of a graph is a clique which has maximum cardinality. For convenience, we regard that a clique and its vertex set are the same. Let $\omega (\varGamma )$ be the clique number of a graph $\varGamma$ which is the number of vertices in a maximum clique.

Let G and H be two graphs. A homomorphism from G to H is a map $\varphi : V(G)\rightarrow V(H)$ such that $\varphi (x)\sim \varphi (y)$ provided $x\sim y$. An isomorphism from G to H is a bijection $\varphi : V(G)\rightarrow V(H)$ such that $x\sim y$ $\Leftrightarrow$ $\varphi (x)\sim \varphi (y)$. We write $G\cong H$ if graphs G and H are isomorphic. A homomorphism (resp. isomorphism) from G to itself is called an endomorphism (resp. automorphism) of G. Let $\mathrm{End}(G)$ (resp. $\mathrm{Aut}(G)$) be the set of all endomorphisms (resp. automorphisms) of G. $\mathrm{End}(G)$ is monoid and $\mathrm{Aut}(G)$ is a group. A homomorphism $\varphi$ from G to the complete graph $K_r$ on r vertices is called an r-colouring of G (colouring for short). The chromatic number of G, denoted by $\chi (G)$, is the least value k for which G can be k-coloured. It is easy to see that $\varphi \in \mathrm{End}(G)$ is a colouring if and only if $\varphi \left( V(G)\right)$ is a maximum clique of G.

A graph $\varGamma$ is called a core if $\mathrm{End}(\varGamma )=\mathrm{Aut}(\varGamma )$. A subgraph H of a graph $\varGamma$ is called a core of $\varGamma$ if H is a core and there exists a homomorphism from $\varGamma$ to H. Every graph has a core, which is an induced subgraph and is unique up to isomorphism [4, Lemma 6.2.2].

A graph $\varGamma$ is called a pseudo-core if every endomorphism of $\varGamma$ is either an automorphism or a colouring. Every core is a pseudo-core. A graph is called a core-complete graph if it is a core or its each core is complete [5]. Every pseudo-core is core-complete but not vice versa. For more information, see [5, 7, 13, 14].

The research of graph homomorphisms is an important subject in the algebraic graph theory [4, 7, 8]. If a graph $\varGamma$ is a pseudo-core and $\mathrm{Aut}(\varGamma )$ was known, then the structure of $\mathrm{End}(\varGamma )$ is clear. Thus, it is an interesting question to distinguish whether a graph is a pseudo-core. Recently, we knew that some important graphs are pseudo-cores, for example bilinear forms graphs [12], alternating forms graphs [14], Grassmann graphs [13], twisted Grassmann graphs [15], primitive strongly regular graphs [19], etc.

In this paper, we discuss endomorphisms of Q(n, q). Our main results are summarized as follows:

• If q is even, then Q(n, q) is a pseudo-core (see Theorem 4.8).

• If q is even, then Q(2m, q) is a core (see Theorem 5.6).

• If q is a power of a prime and $d=[(n+1)/2]$ ($n\ge 3$), then the smallest eigenvalue of Q(n, q) is $\theta _d=-\,(q^{2d}-1)/(q^2-1)$ (see Theorem 5.2 ).

2 Matrix representation and rank for Q(n, q), q even

From now on, we assume that q is a power of 2 unless specified otherwise. Thus, the characteristic of ${\mathbb {F}}_q$ is 2 and ${\mathbb {F}}_q=\left\{ x^2: x\in {\mathbb {F}}_q\right\}$. Note that in ${\mathbb {F}}_q$ the q is even if and only if q is a power of 2.

Let $\in {\mathbb {F}}_q^{m\times n}$ denote the set of $m\times n$ matrices over ${\mathbb {F}}_q$ and $\mathrm{GL}_n({\mathbb {F}}_q)$ the set of $n\times n$ invertible matrices over ${\mathbb {F}}_q$. Write $^tA$ as the transpose matrix of a matrix A. Denote by $I_r$ (I for short) the $r\times r$ identity matrix and $\mathrm{diag}(A_1,\ldots , A_k)$ a block diagonal matrix where $A_1,\ldots , A_k$ are square matrices. Let $E^{n\times n}_{ij}$ ($E_{ij}$ for short) denote the $n\times n$ matrix whose (i, j)-entry is 1 and all other entries are 0’s. We also write $E_{ij}=E_{i,j}$ if i or j is a formula. Denote by $0_{n}$ (0 for short) the $n\times n$ zero matrix.

A matrix $A=(a_{ij})\in \mathbb {F}_q^{n\times n}$ is called alternate, if $a_{ij}=-\,a_{ij}$ for all $i\ne j$ and $a_{ii}=0$ for all $i=1,\ldots ,n$. Let ${\mathcal {K}}_n({\mathbb {F}}_q)$ (${\mathcal {K}}_n$ for short) be the set of $n\times n$ alternate matrices over ${\mathbb {F}}_q$. Denote by ${\mathcal {T}}_n({\mathbb {F}}_q)$ (${\mathcal {T}}_n$ for short) the set of $n\times n$ upper triangular matrices over ${\mathbb {F}}_q$. Every quadratic form f over ${\mathbb {F}}_q$ can be uniquely written as $f(x)=\sum _{i\le j}a_{ij}x_ix_j=xA\,^tx$, where $A=(a_{ij})\in {\mathcal {T}}_n$ and $x=(x_1,\ldots , x_n)$. Thus, there is a bijective map from V(Q(n, q)) to ${\mathcal {T}}_n$.

Let $B=(b_{ij})\in {\mathbb {F}}_q^{n\times n}$. Then, $f=xB\,^tx$ is a quadratic form over ${\mathbb {F}}_q$. Let $T_B=(t_{ij})\in {\mathcal {T}}_n$ where $t_{ii}=b_{ii}$ ($1\le i\le n$) and $t_{ij}=b_{ij}+b_{ji}$ ($1\le i<j\le n$). Clearly, there is a unique alternate matrix $K_B\in {\mathcal {K}}_n({\mathbb {F}}_q)$ such that $T_B=B+K_B$ and $f=xB\,^tx=xT_B\,^tx$ since $xK_B\,^tx=0$ for all $x\in {\mathbb {F}}_q^n$. Therefore, $B+{\mathcal {K}}_n=T_B+{\mathcal {K}}_n$. For any $B\in {\mathbb {F}}_q^{n\times n}$, we let

$$\begin{aligned} \left[ B \right] =T_B+{\mathcal {K}}_n=B+{\mathcal {K}}_n. \end{aligned}$$

(2.1)

The $\left[ B \right]$ is called the matrix representation of the quadratic form $f=xB\,^tx$. When $B\in {\mathcal {T}}_n$, we have $T_B=B$, and hence, we can assume that $\left[ B \right] =B$ since B and $\left[ B \right]$ represent the same quadratic form.

For a subset ${\mathcal {S}}$ of ${\mathbb {F}}_q^{n\times n}$, we let $\left[ {\mathcal {S}}\right] =\left\{ [X]: X\in {\mathcal {S}}\right\}$ and $\left[ {\mathcal {S}}+A\right] =\left\{ [X+A]: X\in {\mathcal {S}}\right\}$ where $A\in {\mathbb {F}}_q^{n\times n}$. The $\left[ {\mathcal {S}}\right]$ is called the matrix representation of the subset $\left\{ xA\,^tx: A\in {\mathcal {S}}, x\in {\mathbb {F}}_q^n \right\}$ of V(Q(n, q)). If ${\mathcal {S}}\subseteq {\mathcal {T}}_n$, then ${\mathcal {S}}$ and $\left[ {\mathcal {S}}\right]$ represent the same subset of V(Q(n, q)), and hence, we can assume that ${\mathcal {S}}=\left[ {\mathcal {S}}\right]$. For any subset ${\mathcal {S}}\subseteq {\mathbb {F}}_q^{n\times n}$ and $P\in \mathrm{GL}_n({\mathbb {F}}_q)$, we let

$$\begin{aligned} \left[ P{\mathcal {S}}\,^tP \right] =\left\{ [PX\,^tP]: X\in {\mathcal {S}}\right\} =:P\left[ {\mathcal {S}}\right] \,^tP. \end{aligned}$$

(2.2)

By (2.1) it is clear that $\left[ P{\mathcal {T}}_n\,^tP \right] =P\left[ {\mathcal {T}}_n \right] \,^tP=\left[ {\mathcal {T}}_n \right] =\left[ {\mathbb {F}}_q^{n\times n}\right]$.

Let $A,B\in {\mathbb {F}}_q^{n\times n}$. Then, there are $K_A, K_B\in {\mathcal {K}}_n$ such that $T_A=A+K_A, T_B=B+K_B\in {\mathcal {T}}_n$ and $T_{A+ B}=T_A+T_B\in {\mathcal {K}}_n$. It follows from (2.1) that

$$\begin{aligned} \left[ A+ B\right] =\left[ A\right] + \left[ B\right] , \quad \hbox {if}\, A,B\in {\mathbb {F}}_q^{n\times n}. \end{aligned}$$

(2.3)

For a quadratic form $f=xA\,^tx$ where $A\in {\mathbb {F}}_q^{n\times n}$, by (1.1) and (1.2), we have that $\mathrm{Rad}(\lambda f)=\mathrm{Rad}(f)$ for all $\lambda \in {\mathbb {F}}_q^*$ and

$$\begin{aligned} \mathrm{Rad}(f)=\left\{ u\in {\mathbb {F}}_q^n: uA\,^tu=0, uA\,^tv+vA\,^tu=0, \quad \forall \, v\in {\mathbb {F}}_q^n \right\} . \end{aligned}$$

(2.4)

For $A\in {\mathbb {F}}_q^{n\times n}$, we let $\mathrm{rk}(\left[ A \right] )=\mathrm{rk}(f)$ and $\mathrm{Rad}([A])=\mathrm{Rad}(f)$, where $f=xA\,^tx$ is the quadratic form on A. The $\mathrm{rk}(\left[ A \right] )$ (resp. $\mathrm{Rad}([A])$) is also called quadratic form rank (resp. quadratic form radical) of A. They are simply written as

$$\begin{aligned} \mathrm{rk}(A):=\mathrm{rk}(\left[ A \right] ), \quad \mathrm{Rad}(A):=\mathrm{Rad}([A]). \end{aligned}$$

Note that $\mathrm{rk}(A)\ne \mathrm{rank}(A)$ (the usual rank of matrix A) in general. By $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$ and (2.4), it is easy to see that

$$\begin{aligned} \mathrm{rk}(A )= & {} \mathrm{rk}(PA\,^tP)\hbox {, where } A\in {\mathbb {F}}_q^{n\times n}, P\in \mathrm{GL}_n({\mathbb {F}}_q); \end{aligned}$$

(2.5)

$$\begin{aligned} \mathrm{rk}(\lambda A)= & {} \mathrm{rk}(A)\hbox {, where }\lambda \in {\mathbb {F}}_q^*. \end{aligned}$$

(2.6)

By [1, Lemma 9.6.1], we have

$$\begin{aligned} \mathrm{rk}(A+B)\le \mathrm{rk}(A)+\mathrm{rk}(B). \end{aligned}$$

(2.7)

Two matrices $X, Y\in {\mathbb {F}}_q^{n\times n}$ are said to be congruent mod ${\mathcal {K}}_n$, if $X+Y\in {\mathcal {K}}_n$, and denoted by $X\equiv Y$ (mod ${\mathcal {K}}_n$) or simply, $X\equiv Y$. Two matrices $A, B\in {\mathbb {F}}_q^{n\times n}$ are said to be cogredient, if there is $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $A\equiv PB\,^tP$ (mod ${\mathcal {K}}_n$). Thus, ${\mathbb {F}}_q^{n\times n}$ is partitioned into mutually disjoint classes of cogredient matrices. By (2.1), it is clear that

$$\begin{aligned}{}[A]=[B]\hbox {, if }A,B\in {\mathbb {F}}_q^{n\times n}\quad \hbox {and}\quad A\equiv B (\hbox {mod}\, {\mathcal {K}}_n). \end{aligned}$$

(2.8)

A matrix $A\in {\mathbb {F}}_q^{n\times n}$ is called definite, if $xA\,^tx=0$ for all $x\in {\mathbb {F}}_q^{n}$ implies that $x=0$. It is easy to see that a matrix cogredient to a definite matrix (resp. an alternate matrix) is also definite (resp. alternate). Let

$$\begin{aligned} N=\left\{ x^2+x:x\in {\mathbb {F}}_q \right\} . \end{aligned}$$

Then, N is an additive subgroup of ${\mathbb {F}}_q$ of index 2 and $N^2=N$ (see [21, Lemma 11.15]). From now on, we choose a fixed element $\alpha \in {\mathbb {F}}_q$ but $\alpha \notin N$.

Lemma 2.1

(Wan [21, Lemma 11.16] or [20, Lemma 1.32]) In ${\mathbb {F}}_q^{2\times 2}$, the matrix $\left( \begin{array}{c@{\quad }c} a &{} 1 \\ 0 &{} b \\ \end{array} \right)$ is cogredient to either $\left( \begin{array}{c@{\quad }c} 0 &{} 1 \\ 0 &{} 0 \\ \end{array} \right)$ or $\left( \begin{array}{c@{\quad }c} \alpha &{} 1 \\ 0 &{} \alpha \\ \end{array} \right)$ according to $ab\in N$ or $ab\notin N$, respectively. Moreover, the first one of the above two matrices is not definite and the second one is definite.

Using (2.4), it is easy to prove that $\mathrm{rk}\left( \begin{array}{c@{\quad }c} 0 &{} 1 \\ 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c} \alpha &{} 1 \\ 0 &{} \alpha \\ \end{array} \right) =2$. Applying Lemma 2.1 and (2.6)–(2.8), we have

$$\begin{aligned} \mathrm{rk}\left( \begin{array}{c@{\quad }c} a &{} c \\ 0 &{} b \\ \end{array} \right) =2,\quad \hbox {where}\quad a,b\in {\mathbb {F}}_q\quad \hbox {and}\quad c\in {\mathbb {F}}_q^*. \end{aligned}$$

(2.9)

By (2.4), it is easy to verify that

$$\begin{aligned}&\mathrm{rk}\left( E_{11}\right) =1, \end{aligned}$$

(2.10)

$$\begin{aligned}&\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0_\upsilon &{} I_\upsilon &{}0\\ 0 &{} 0_\upsilon &{}0 \\ 0 &{} 0 &{}0_{n-2\upsilon } \\ \end{array} \right) =2\upsilon , \quad \hbox { where }1\le \upsilon \le n/2. \end{aligned}$$

(2.11)

By [21, Theorem 11.21] (or [20, Theorem 1.35]) and (2.4)–(2.11), we have the following lemma.

Lemma 2.2

Every $A\in {\mathbb {F}}_q^{n\times n}$ is cogredient to a matrix of one and only one of the following forms

$$\begin{aligned} {\left( \begin{array}{c@{\quad }c@{\quad }c} 0_\upsilon &{} I_\upsilon &{} 0 \\ 0 &{} 0_\upsilon &{} 0 \\ 0 &{} 0 &{} 0_{n-2\upsilon }\\ \end{array} \right) , \quad \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0_\upsilon &{} I_\upsilon &{} 0 &{} 0 \\ 0 &{} 0_\upsilon &{} 0&{} 0 \\ 0 &{} 0&{} 1 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0_{n-2\upsilon -1} \\ \end{array} \right) , \quad \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0_{\upsilon -1} &{} I_{\upsilon -1} &{} 0 &{} 0 &{}0\\ 0&{} 0_{\upsilon -1} &{} 0&{} 0 &{}0\\ 0 &{} 0&{} \alpha &{} 1&{}0 \\ 0 &{} 0 &{} 0 &{} \alpha &{}0 \\ 0 &{} 0 &{} 0 &{} 0 &{}0_{n-2\upsilon } \\ \end{array} \right) .} \end{aligned}$$

Moreover, if A is cogredient to the first one (resp. second one, the third one) of the above three matrices, then $\mathrm{rk}(A)=2\upsilon$ (resp. $\mathrm{rk}(A)=2\upsilon +1$, $\mathrm{rk}(A)=2\upsilon$), respectively.

Corollary 2.3

Let $0\ne A\in {\mathbb {F}}_q^{n\times n}$. Then, $\mathrm{rk}(A)=1$ if and only if there exists $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $[A]=\left[ PE_{11}\,^tP \right]$, and if and only if $[A]=\left[ D\right]$ where $D=\mathrm{diag}(a_1, \ldots , a_n)\ne 0$.

Proof

By Lemma 2.2, (2.5) and (2.8), it is clear that $\mathrm{rk}(A)=1$ if and only if there exists $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $[A]=\left[ PE_{11}\,^tP \right]$. If $[A]=\left[ PE_{11}\,^tP \right]$, then $PE_{11}\,^tP$ being a symmetric matrix implies that $0\ne T_{PE_{11}\,^tP}=\mathrm{diag}(a_1, \ldots , a_n)=:D$, and hence, $[A]=\left[ D\right]$. Conversely, if $[A]=\left[ D\right]$ where $D=\mathrm{diag}(a_1, \ldots , a_n)\ne 0$, then by (2.4) and $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$, we have $\mathrm{rk}(A)=1$. Thus, $\mathrm{rk}(A)=1$ if and only if $[A]=\left[ D\right]$ where $D=\mathrm{diag}(a_1, \ldots , a_n)\ne 0$. $\square$

3 Maximal cliques of Q(n, q), q even

Hemmeter and Woldar [9,10,11] studied the maximal cliques of Q(n, q). For any fixed vertex $f_0$ of Q(n, q), the type one clique of Q(n, q) is defined to be

$$\begin{aligned} C(f_0)=\left\{ f\in V(Q(n, q)): \mathrm{rk}(f-f_0)\le 1\right\} . \end{aligned}$$

(3.1)

The definition of the type two clique of Q(n, q) is complex in [9,10,11]; here we omit this definition. Applying the matrix representation for Q(n, q), we will give another definition of the type two clique below. For any fixed $(n-2)$-dimensional subspace U of ${\mathbb {F}}^n_q$ and a vertex $f_0$ of Q(n, q), the cubic clique of Q(n, q) is defined to be

$$\begin{aligned} C(U, f_0)=\left\{ f\in V(Q(n, q)): \mathrm{Rad}(f-f_0)\supseteq U\right\} . \end{aligned}$$

(3.2)

Every cubic clique have size $q^3$ (cf. [9,10,11]). A cubic clique also is called a type three clique. Cliques of types one and two are also called grand cliques of Q(n, q). Every grand clique of Q(n, q) has size $q^n$ (cf. [9,10,11]).

Lemma 3.1

(Hemmeter [9, Theorem 24]) Let ${\mathcal {M}}$ be a maximal clique of Q(n, q) where q is even. Then, ${\mathcal {M}}$ is grand or cubic, or $|{\mathcal {M}}|\le q^2+q+2$.

Since every cubic clique of Q(n, q) has size $q^3$, from Lemma 3.1 we have that

$$\begin{aligned} \omega (Q(n,q))=q^n, \quad \hbox {if }n\ge 3. \end{aligned}$$

(3.3)

Corollary 3.2

When $n\ge 4$, ${\mathcal {M}}$ is a maximum clique of Q(n, q) if and only if ${\mathcal {M}}$ is a grand clique of Q(n, q). Moreover, when $q\ge 4$ and $n\ge 3$, ${\mathcal {C}}$ is a maximal clique of Q(n, q) if and only if ${\mathcal {C}}$ is either a grand clique or a cubic clique.

Proof

Let ${\mathcal {M}}$ be a maximum clique of Q(n, q). Then, $|{\mathcal {M}}|= q^n$. When $n\ge 4$, by Lemma 3.1 we have that ${\mathcal {M}}$ is a grand clique. Conversely, when $n\ge 4$, Lemma 3.1 and (3.3) imply that every grand clique of Q(n, q) is a maximum clique.

When $q\ge 4$ and $n\ge 3$, we have $q^3>q^2+q+2$; thus, Lemma 3.1 implies that ${\mathcal {C}}$ is a maximal clique of Q(n, q) if and only if ${\mathcal {C}}$ is either a grand clique or a cubic clique. $\square$

Lemma 3.3

(Hemmeter [9, Theorem 22, Corollary 23]) In Q(n, q), the followings hold:

• (i) C(0) is the only type one clique containing 0.

• (ii) Let $C_1$ and $C_2$ be distinct type two cliques, both containing 0. Then, $\left| C_1\cap C_2 \right| =q$.

• (iii) Let $C_1$ be a type two clique containing 0. Then, $\left| C_1\cap C(0)\right| =q$.

• (iv) If $C_1$ and $C_2$ are grand cliques, then $\left| C_1\cap C_2 \right| =q$ or 0.

By Lemma 3.3 (iv) and Corollary 3.2, we have the following corollary.

Corollary 3.4

Let ${\mathcal {C}}$ and ${\mathcal {C}}'$ be maximum cliques of Q(n, q) where $n\ge 4$. If $\left| {\mathcal {C}}\cap {\mathcal {C}}'\right| >q$, then ${\mathcal {C}}={\mathcal {C}}'$.

When q is even and $n\ge 2$, let $\varGamma ^{(n)}({\mathbb {F}}_q)$ ($\varGamma ^{(n)}$ for short) be the graph that the vertex set is $\left[ {\mathcal {T}}_n \right]$ and two vertices [A], [B] are adjacent if $1\le \mathrm{rk}([A-B])\le 2$. By the matrix representation of Q(n, q), the graph $\varGamma ^{(n)}$ is isomorphic to Q(n, q), and $\varGamma ^{(n)}$ is called the matrix representation graph of Q(n, q). Since $\varGamma ^{(n)}\cong Q(n,q)$, $\varGamma ^{(n)}$ is also called a quadratic forms graph. Note that the use of matrix representation can better express our proof. We will study $\varGamma ^{(n)}$.

Let ${\mathcal {D}}_n({\mathbb {F}}_q)$ (${\mathcal {D}}_n$ for short) be the set of $n\times n$ diagonal matrices over ${\mathbb {F}}_q$. For any $A\in \mathbb {F}_q^{n\times n}$ where $n\ge 3$, by Corollary 2.3 and (3.3), $\left[ {\mathcal {D}}_n+A \right]$ is a maximum clique of $\varGamma ^{(n)}$. The $\left[ {\mathcal {D}}_n+A \right]$ is called a type one clique of $\varGamma ^{(n)}$. If $C(f_0)$ is a type one clique of Q(n, q), then (3.1) and Corollary 2.3 imply that the matrix representation of $C(f_0)$ is $\left[ {\mathcal {D}}_n+A \right]$ where $f_0=xA\,^tx$. Therefore, ${\mathcal {C}}$ is a type one clique in $\varGamma ^{(n)}$ if and only if ${\mathcal {C}}$ is the matrix representation of a type one clique in Q(n, q).

Clearly, $\left[ {\mathcal {D}}_n+A \right] =\left[ {\mathcal {D}}_n+A_1\right]$ where $A_1$ is a strictly upper triangular matrix. Thus, $\left[ {\mathcal {D}}_n+A \right] \ne \left[ {\mathcal {D}}_n+B\right]$ if and only if $\left[ {\mathcal {D}}_n+A \right] \cap \left[ {\mathcal {D}}_n+B\right] =\emptyset$. On the other hand, there are $q^{n(n+1)/2}$ type one cliques in $\varGamma ^{(n)}$. Let $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ and $X\in {\mathcal {D}}_n$. Since $PX\,^tP$ is symmetric, $T_{PX\,^tP}\in {\mathcal {D}}_n$. Thus, for any $P\in \mathrm{GL}_n({\mathbb {F}}_q)$, (2.1) implies that $\left[ {\mathcal {D}}_n\right] =\left[ P{\mathcal {D}}_n\,^tP \right]$. Thus, for any $A\in {\mathbb {F}}_q^{n\times n}$ we have that

$$\begin{aligned} \left[ {\mathcal {D}}_n+A\right] =\left[ P{\mathcal {D}}_n\,^tP +A\right] . \end{aligned}$$

(3.4)

In ${\mathcal {T}}_n$, we let

$$\begin{aligned} {\mathcal {L}}_1= & {} \left\{ \sum _{j=1}^nx_jE_{1j}: x_j\in {\mathbb {F}}_q,\quad 1\le j\le n \right\} ,\\ {\mathcal {L}}_i= & {} \left\{ \sum _{k=1}^{i-1}x_kE_{ki}+\sum _{k=i}^nx_kE_{ik}: x_k\in {\mathbb {F}}_q, 1\le k\le n \right\} , \quad i=2, \ldots , n. \end{aligned}$$

Clearly, $\mathrm{rk}([X])\le 2$ for all $[X]\in \left[ {\mathcal {L}}_1\right]$. Thus, $\left[ {\mathcal {L}}_1\right]$ is clique of $\varGamma ^{(n)}$ and $\left| {\mathcal {L}}_1\right| =q^n$. By (3.3), $\left[ {\mathcal {L}}_1\right]$ is a maximum clique of $\varGamma ^{(n)}$ when $n\ge 3$. It is clear that ${\mathcal {L}}_i$ ($1\le i\le n$) is a linear space over ${\mathbb {F}}_q$. For $2\le i\le n$, there is an $n\times n$ permutation matrix Q such that $\left[ {\mathcal {L}}_i\right] =\left[ Q{\mathcal {L}}_1 \,^tQ\right]$.

For any $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ and $A\in {\mathbb {F}}_q^{n\times n}$, the set $\left[ P{\mathcal {L}}_1\,^tP+A\right]$ is called a type two clique of $\varGamma ^{(n)}$. Recall that $\varGamma ^{(n)}\cong Q(n, q)$. By Corollary 3.2, there are only two types of maximum cliques in Q(n, q) if $n\ge 4$. Thus, when $n\ge 4$, there are only two types of maximum cliques in $\varGamma ^{(n)}$; they are the type one clique and the type two clique. Clearly, any type two (resp. type one) clique in $\varGamma ^{(n)}$ is not of type one (resp. type two) in $\varGamma ^{(n)}$. Therefore, when $n\ge 4$, ${\mathcal {C}}$ is a type two (resp. type one) clique in $\varGamma ^{(n)}$ if and only if ${\mathcal {C}}$ is the matrix representation of a type two (resp. type one) clique in Q(n, q).

Let $\mathrm{diag}\left( {\mathcal {T}}_2, 0\right) =\left\{ \mathrm{diag}(X, 0): X\in {\mathcal {T}}_2 \right\}$ be a subset of ${\mathbb {F}}_q^{n\times n}$. Clearly, $\left[ \mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \right]$ is a clique of $\varGamma ^{(n)}$ with size $q^3$. For any $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ and $A\in {\mathbb {F}}_q^{n\times n}$, the clique $\left[ P\mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \,^tP+A\right]$ is called a cubic clique of $\varGamma ^{(n)}$.

Proposition 3.5

Assume that $C(U,f_0)$ is a cubic clique of Q(n, q). Then, the matrix representation of $C(U,f_0)$ is a cubic clique in $\varGamma ^{(n)}$.

Proof

By (3.2), $C(U,f_0)=\left\{ f\in V(Q(n, q)): \mathrm{Rad}(f-f_0)\supseteq U\right\}$, where U is a fixed $(n-2)$-dimensional subspace of ${\mathbb {F}}^n_q$ and $f_0$ is a fixed vertex of Q(n, q). Without loss of generality, we assume that $f_0=0$. There is a $P\in \mathrm{GL}_{n}({\mathbb {F}}_q)$ such that $U=\left\{ (0, y)P: y\in {\mathbb {F}}_q^{n-2}\right\}$. Write $W:=\left\{ (0, y): y\in {\mathbb {F}}_q^{n-2}\right\}$. Then, $W=UP^{-1}$.

Let vertex $f=xA\,^tx\in {\mathcal {M}}$, where $A\in {\mathcal {T}}_n$ and $x=(x_1,\ldots , x_n)$. Put $Y:= PA\,^tP=\left( \begin{array}{c@{\quad }c} Y_{11} &{} Y_{12} \\ Y_{21} &{} Y_{22} \\ \end{array} \right)$ where $Y_{11}\in {\mathbb {F}}_q^{2\times 2}$. Then, $[Y]=\left[ T_Y \right] =T_Y$, where $T_Y=\left( \begin{array}{c@{\quad }c} Y_{11}' &{} Y_{12}' \\ 0 &{} Y_{22}' \\ \end{array} \right)$, $Y_{11}'\in {\mathcal {T}}_2$ and $Y_{22}'\in {\mathcal {T}}_{n-2}$. Clearly, $W\subseteq \mathrm{Rad}(xY\,^tx)=\mathrm{Rad}(xT_Y\,^tx)$. By (2.4), we have $uT_Y\,^tu=0$, $uT_Y\,^tv+vT_Y\,^tu=0$, $\forall \ u\in W$, $\forall \ v\in {\mathbb {F}}^n$. It follows that $Y_{22}'=0$ and $Y_{12}'=0$, and hence, $T_Y=\left( \begin{array}{c@{\quad }c} Y_{11}' &{} 0 \\ 0 &{} 0\\ \end{array} \right)$. Therefore, $[A]=\left[ P^{-1}Y\,^tP^{-1} \right] =\left[ P^{-1}T_Y\,^tP^{-1} \right]$. Consequently, the matrix representation of $C(U,f_0)$ is the cubic clique $\left[ P^{-1}\mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \,^tP^{-1}\right]$ in $\varGamma ^{(n)}$. $\square$

Lemma 3.6

Let ${\mathcal {L}}$ and ${\mathcal {L}}'$ be two distinct type two cliques in $\varGamma ^{(n)}$ with ${\mathcal {L}}\cap {\mathcal {L}}'\ne \emptyset$. Then, for any $[A]\in {\mathcal {L}}\cap {\mathcal {L}}'$, there is $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that

$$\begin{aligned} {\mathcal {L}}=\left[ P{\mathcal {L}}_1\,^tP+A\right] \quad \hbox {and}\quad {\mathcal {L}}'=\left[ P{\mathcal {L}}_2\,^tP+A\right] . \end{aligned}$$

Proof

Let $[A]\in {\mathcal {L}}\cap {\mathcal {L}}'$. Note that ${\mathcal {L}}_1$ is a linear space over ${\mathbb {F}}_q$. By the definition of type two clique, there are matrices $P_1, Q_1\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that ${\mathcal {L}}=\left[ P_1{\mathcal {L}}_1\,^tP_1+A\right]$ and ${\mathcal {L}}'=\left[ Q_1{\mathcal {L}}_1\,^tQ_1+A\right]$. Note that the map $[X]\longmapsto [X-A]$ and the map $[X]\longmapsto [P_1^{-1}X\,^tP_1^{-1}]$ are two automorphisms of $\varGamma ^{(n)}$. Without loss of generality, we assume that $P_1=I_n$ and $A=0$. Hence, ${\mathcal {L}}=\left[ {\mathcal {L}}_1\right]$ and ${\mathcal {L}}'=\left[ Q_1{\mathcal {L}}_1\,^tQ_1\right]$. Write $Q_1=(q_{ij})$.

Let $X=Q_1\left( \begin{array}{c} x \\ 0 \\ \end{array} \right) \,^tQ_1\in Q_1{\mathcal {L}}_1\,^tQ_1$ where $0\ne x\in {\mathbb {F}}_q^n$. Put $y=x\,^tQ_1\ne 0$. Then, $X=\left( \begin{array}{c} q_{11}y \\ q_{21}y\\ \vdots \\ q_{n1}y \\ \end{array}\right)$. If $\left( \begin{array}{c} q_{21}\\ \vdots \\ q_{n1} \\ \end{array} \right) =0$, then $Q_1=\left( \begin{array}{c@{\quad }c} q_{11} &{} *\\ 0 &{} *\\ \end{array} \right)$; thus, ${\mathcal {L}}'=\left[ Q_1{\mathcal {L}}_1\,^tQ_1\right] =\left[ {\mathcal {L}}_1\right]$, a contradiction. Thus, $\left( \begin{array}{c} q_{21}\\ \vdots \\ q_{n1} \\ \end{array} \right) \ne 0$. Using elementary transformations of matrix, there is $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $X=P\left( \begin{array}{c} 0 \\ y'\\ 0\\ \vdots \\ 0 \\ \end{array} \right) \,^tP$ with $T_X\in P{\mathcal {L}}_2\,^tP$ and ${\mathcal {L}}_1=P{\mathcal {L}}_1\,^tP$. It follows that ${\mathcal {L}}=\left[ P{\mathcal {L}}_1\,^tP\right]$ and ${\mathcal {L}}'=\left[ P{\mathcal {L}}_2\,^tP\right]$. $\square$

Corollary 3.7

If ${\mathcal {L}}, {\mathcal {L}}'$ is two distinct type two cliques in $\varGamma ^{(n)}$ with ${\mathcal {L}}\cap {\mathcal {L}}'\ne \emptyset$, then $\left| {\mathcal {L}}\cap {\mathcal {L}}' \right| =q$.

Lemma 3.8

In $\varGamma ^{(n)}$, there exist exactly two type two cliques which contain 0 and $aE_{ij}$ in common, where $i<j$ and $a\in {\mathbb {F}}_q^*$. More precisely, they are $\left[ {\mathcal {L}}_i\right]$ and $[{\mathcal {L}}_j ]$.

Proof

Clearly, $\left[ {\mathcal {L}}_i\right]$ and $[{\mathcal {L}}_j ]$ are type two cliques which contain 0 and $aE_{ij}$ ($i<j$) in common, where $a\in {\mathbb {F}}_q^*$. Suppose that ${\mathcal {L}}$ is a type two clique containing 0 and $aE_{ij}$. By cogredient transformations, without loss of generality we assume that $E_{ij}=E_{12}$. There is $P=(p_{ij})\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that

$$\begin{aligned} {\mathcal {L}}=\left[ P{\mathcal {L}}_1\,^tP\right] =\left\{ \left[ \left( \begin{array}{c} p_{11}y \\ p_{21}y\\ \vdots \\ p_{n1}y \\ \end{array}\right) \right] : y=x\,^tP\in {\mathbb {F}}_q^n \right\} . \end{aligned}$$

(3.5)

If $p_{21}=\cdots =p_{n1}=0$, then ${\mathcal {L}}=\left[ {\mathcal {L}}_1\right]$. Now, we assume that $^t(p_{21},\ldots ,p_{n1})\ne 0$. We show that $^t(p_{31},\ldots ,p_{n1})= 0$ as follows. Otherwise, if $^t(p_{31},\ldots ,p_{n1})\ne 0$, then there is $Q_1\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that ${\mathcal {L}}=\left[ Q_1{\mathcal {L}}_3\,^tQ_1\right]$ and $\left[ aE_{12}\right] =\left[ Q_1aE_{12}\,^tQ_1\right]$, a contradiction to $\left[ aE_{12}\right] \in {\mathcal {L}}$. Thus, $^t(p_{31},\ldots ,p_{n1})= 0$, and hence, $p_{21}\ne 0$. Since $[aE_{12}]\in {\mathcal {L}}$ and (3.5), there exists $y_0=(a_1, a_2, \ldots , a_n)\ne 0$ such that

$$\begin{aligned} \left[ aE_{12}\right] =\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} p_{11}a_1 &{} p_{11}a_2+p_{21}a_1 &{} p_{11}a_3&{} \cdots &{} p_{11}a_n \\ 0 &{} p_{21}a_2 &{} p_{21}a_3 &{} \cdots &{} p_{21}a_n \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \end{array} \right) . \end{aligned}$$

It follows that $p_{11}=0$, which implies that ${\mathcal {L}}=\left[ {\mathcal {L}}_2\right]$. Therefore, there are only two type two cliques which contain 0 and $aE_{12}$ in common. $\square$

Lemma 3.9

Let ${\mathcal {L}}, {\mathcal {L}}', {\mathcal {L}}''$ be three distinct type two cliques in $\varGamma ^{(n)}$ with ${\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {L}}''\ne \emptyset$. Then, $\left| {\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {L}}'' \right| =1$.

Proof

Let $[B]\in {\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {L}}''$. Note that the map $[X]\longmapsto [X-B]$ is an automorphism of $\varGamma ^{(n)}$. Without loss of generality, we assume that $0\in {\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {L}}''$. By Lemma 3.6, there is $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that ${\mathcal {L}}=\left[ P{\mathcal {L}}_1\,^tP\right]$ and ${\mathcal {L}}'=\left[ P{\mathcal {L}}_2\,^tP\right]$. Thus, ${\mathcal {L}}\cap {\mathcal {L}}'=\left\{ PxE_{12}\,^tP: x\in {\mathbb {F}}_q \right\}$. For any $x\in {\mathbb {F}}_q^*$, Lemma 3.8 implies that $PxE_{12}\,^tP\notin {\mathcal {L}}''$ since ${\mathcal {L}}''\ne {\mathcal {L}}$ and ${\mathcal {L}}''\ne {\mathcal {L}}'$. Therefore, ${\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {L}}''=\{0\}$, and hence, $\left| {\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {L}}'' \right| =1$. $\square$

Lemma 3.10

Let $\varphi \in \mathrm{End}(\varGamma ^{(n)})$ where $n\ge 4$. Suppose that there are two distinct type two cliques ${\mathcal {L}}$ and ${\mathcal {L}}'$ in $\varGamma ^{(n)}$ such that ${\mathcal {L}}\cap {\mathcal {L}}'\ne \emptyset$ and $\varphi ({\mathcal {L}})=\varphi ({\mathcal {L}}')=:{\mathcal {C}}$. If ${\mathcal {M}}$ is a type two clique in $\varGamma ^{(n)}$ with ${\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {M}}\ne \emptyset$, then $\varphi ({\mathcal {M}})={\mathcal {C}}$.

Proof

Clearly, ${\mathcal {C}}$ is a maximum clique of $\varGamma ^{(n)}$. Let ${\mathcal {M}}$ be a type two clique in $\varGamma ^{(n)}$ with ${\mathcal {L}}\cap {\mathcal {L}}'\cap {\mathcal {M}}\ne \emptyset$. Without loss of generality, we assume that ${\mathcal {M}}\ne {\mathcal {L}}$ and ${\mathcal {M}}\ne {\mathcal {L}}'$. By Corollary 3.7 and Lemma 3.9, we obtain that $\left| {\mathcal {M}}\cap ({\mathcal {L}}\cup {\mathcal {L}}') \right| =\left| {\mathcal {M}}\cap {\mathcal {L}}\right| +\left| {\mathcal {M}}\cap {\mathcal {L}}'\right| -\left| {\mathcal {M}}\cap {\mathcal {L}}\cap {\mathcal {L}}'\right| =2q-1>q.$ Since $\varphi ({\mathcal {M}}\cap ({\mathcal {L}}\cup {\mathcal {L}}'))\subseteq \varphi ({\mathcal {M}})\cap \varphi ({\mathcal {L}}\cup {\mathcal {L}}')=\varphi ({\mathcal {M}})\cap {\mathcal {C}}$ and the restriction mapping $\varphi \mid _{{\mathcal {M}}}$ is injective, we have $\left| \varphi ({\mathcal {M}})\cap {\mathcal {C}}\right| >q$. Since $\varphi ({\mathcal {M}})$ is a maximum clique, it follows from Corollary 3.4 that $\varphi ({\mathcal {M}})={\mathcal {C}}$. $\square$

Lemma 3.11

Suppose that $n\ge 4$, ${\mathcal {M}}$ is a maximum clique and ${\mathcal {C}}$ is a cubic clique in $\varGamma ^{(n)}$. Then, $\left| {\mathcal {M}}\cap {\mathcal {C}}\right| \le q^2$.

Proof

Similar to the proof of Lemma 3.9, without loss of generality we assume that $0\in {\mathcal {M}}\cap {\mathcal {C}}$. Let ${\mathcal {C}}=\left[ P\mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \,^tP\right]$ where $P\in \mathrm{GL}_n({\mathbb {F}}_q)$. Recall that $\varGamma ^{(n)}\cong Q(n, q)$. Corollary 3.2 implies that ${\mathcal {M}}$ is either a type one clique or a type two clique in $\varGamma ^{(n)}$. Thus, ${\mathcal {M}}=\left[ {\mathcal {D}}_n\right]$ or ${\mathcal {M}}=\left[ Q{\mathcal {L}}_1\,^tQ\right]$ where $Q\in \mathrm{GL}_n({\mathbb {F}}_q)$. Note that the map $[X]\longmapsto [P^{-1}X\,^tP^{-1}]$ is an automorphism of $\varGamma ^{(n)}$. Without loss of generality, we assume that $P=I_n$. Hence, ${\mathcal {C}}=\left[ \mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \right]$.

Case 1 Suppose that ${\mathcal {M}}=\left[ {\mathcal {D}}_n \right]$. Then, it is clear that $\left| {\mathcal {C}}\cap \left[ {\mathcal {D}}_n\right] \right| =q^2$.

Case 2 Suppose that ${\mathcal {M}}=\left[ Q{\mathcal {L}}_1\,^tQ\right]$. Write $Q=(q_{ij})$. Then,

$$\begin{aligned} {\mathcal {M}}=\left[ Q{\mathcal {L}}_1\,^tQ\right] =\left\{ \left[ \left( \begin{array}{c} q_{11}y \\ q_{21}y\\ \vdots \\ q_{n1}y \\ \end{array}\right) \right] : y=x\,^tQ\in {\mathbb {F}}_q^n \right\} . \end{aligned}$$

Assume that $^t(q_{31}, q_{41},\ldots , q_{n1})\ne 0$. Applying appropriate elementary transformations of matrix, there is $Q_1\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that ${\mathcal {C}}=\left[ \mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \right] =\left[ Q_1\mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \,^tQ_1\right]$, and every element in ${\mathcal {M}}$ is of the form $\left[ Q_1\left( \begin{array}{c@{\quad }c} 0_2 &{} Y_1 \\ 0 &{} Y_2 \\ \end{array} \right) \,^tQ_1\right]$ where $Y_2\in {\mathcal {T}}_{n-2}$. It follows that ${\mathcal {M}}\cap {\mathcal {C}}=\{0\}$ and $\left| {\mathcal {M}}\cap {\mathcal {C}}\right| =1$.

Assume that $^t(q_{31}, q_{41},\ldots , q_{n1})=0$. Write $y=(y_1,\ldots ,y_n)$. Then, ${\mathcal {M}}$ can be written as

$$\begin{aligned} {\mathcal {M}}=\left\{ \left[ \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} q_{11}y_1 &{} q_{11}y_2+q_{21}y_1 &{} q_{11}y_3&{} \cdots &{} q_{11}y_n \\ 0 &{} q_{21}y_2 &{} q_{21}y_3 &{} \cdots &{} q_{21}y_n \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \end{array} \right) \right] : y_i\in {\mathbb {F}}_q \right\} . \end{aligned}$$

Since $^t(q_{11},q_{21})\ne 0$, we obtain that $\left| {\mathcal {M}}\cap {\mathcal {C}}\right| = q^2$. $\square$

Lemma 3.12

Suppose that $n\ge 4$, $q=2$ and ${\mathcal {M}}$ is a maximum clique of $\varGamma ^{(n)}=\varGamma ^{(n)}({\mathbb {F}}_2)$ with $0\in {\mathcal {M}}$. Assume that $A\in {\mathbb {F}}_2^{n\times n}$ with $\mathrm{rk}(A)\le 2$, and [A] is adjacent to five vertices in ${\mathcal {M}}$. Then, $[A]\in {\mathcal {M}}$.

Proof

By Corollary 3.2, ${\mathcal {M}}$ is a type one clique or a type two clique of $\varGamma ^{(n)}$. Since $0\in {\mathcal {M}}$, ${\mathcal {M}}=\left[ {\mathcal {D}}_n\right]$ or ${\mathcal {M}}=\left[ P{\mathcal {L}}_1\,^tP\right]$ where $P\in \mathrm{GL}_n({\mathbb {F}}_2)$.

Case 1 ${\mathcal {M}}=\left[ {\mathcal {D}}_n\right]$. If $\mathrm{rk}(A)\le 1$, then it is clear that $[A]\in {\mathcal {M}}$. Now we assume that $\mathrm{rk}(A)=2$. By Lemma 2.2, there is $P_1\in \mathrm{GL}_n({\mathbb {F}}_2)$ such that either $[A]=\left[ P_1E_{12}\,^tP_1 \right]$ or $[A]=\left[ P_1(E_{11}+E_{12}+E_{22})\,^tP_1 \right]$. Since $\left[ {\mathcal {D}}_n\right] =\left[ P_1{\mathcal {D}}_n\,^tP_1\right]$, without loss of generality we assume that $P_1=I_n$. Since [A] is adjacent to five vertices in ${\mathcal {M}}$, there is $B=\mathrm{diag}(b_1, b_2, \ldots , b_n)$ with $(b_3, \ldots , b_n)\ne 0$ such that $[A]\sim [B]$. Hence, $[A-B]\sim 0$. By (2.4) and $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$, it is easy to see that $\mathrm{rk}(A-B)=3$, a contradiction. Thus, $\mathrm{rk}(A)\le 1$, and hence, $[A]\in {\mathcal {M}}$.

Case 2 ${\mathcal {M}}=\left[ P{\mathcal {L}}_1\,^tP\right]$. Without loss of generality we assume that $P=I_n$, and hence, ${\mathcal {M}}=\left[ {\mathcal {L}}_1\right] ={\mathcal {L}}_1$. Note that ${\mathcal {L}}_1$ is a linear space over ${\mathbb {F}}_2$. Let $[A]=\left( \begin{array}{c@{\quad }c} a_{1} &{} \beta \\ 0 &{} A_{2} \\ \end{array} \right)$ where $A_{2}\in {\mathcal {T}}_{n-1}$ and $\beta =(a_{2}, a_{3}, \ldots , a_{n})$. When $A_{2}=0$, $[A]\in {\mathcal {L}}_1$.

Suppose that $A_{2}\ne 0$. Then, $1\le \mathrm{rk}(A_{2})\le 2$. We show a contradiction as follows. Write $E_{ij}'=E_{ij}^{(n-1)\times (n-1)}$, $1\le i,j\le n-1$. By Lemma 2.2, applying appropriate cogredient transformations, without loss of generality we can assume that $A_{2}=E_{11}'$, or $A_{2}=E_{12}'$, or $A_{2}=E_{11}'+E_{12}'+E_{22}'$. It is easy to see that $(a_{4}, \ldots , a_{n})=0$ in $\beta$. Otherwise, (2.4) and $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$ imply that $\mathrm{rk}(A)\ge 3$, a contradiction. Thus, $A=a_{1}E_{11}+a_{2}E_{12}+a_{3}E_{13}+E_{22}$, or $A=a_{1}E_{11}+a_{2}E_{12}+a_{3}E_{13}+E_{23}$, or $A=a_{1}E_{11}+a_{2}E_{12}+a_{3}E_{13}+E_{22}+E_{23}+E_{33}$. Let

$$\begin{aligned} {\mathcal {L}}_1'=\left\{ x_{1}E_{11}+x_{2}E_{12}+x_{3}E_{13}: x_1,x_2,x_3\in {\mathbb {F}}_2\right\} . \end{aligned}$$

Since $A=[A]$ is adjacent to five vertices in ${\mathcal {L}}_1$, similar to the discussion above, A must be adjacent to five vertices in ${\mathcal {L}}_1'$. Note that ${\mathcal {L}}_1'$ is a linear space over ${\mathbb {F}}_2$. If $A=a_{1}E_{11}+a_{2}E_{12}+a_{3}E_{13}+E_{22}$, then there are five vertices in ${\mathcal {L}}_1'+E_{22}$ such that they are adjacent to 0. If $A=a_{1}E_{11}+a_{2}E_{12}+a_{3}E_{13}+E_{23}$, then there are five vertices in ${\mathcal {L}}_1'+E_{23}$ such that they are adjacent to 0. If $A=a_{1}E_{11}+a_{2}E_{12}+a_{3}E_{13}+E_{22}+E_{23}+E_{33}$, then there are five vertices in ${\mathcal {L}}_1'+E_{22}+E_{23}+E_{33}$ such that they are adjacent to 0.

Using (2.4) and $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$, we have that

$$\begin{aligned} \mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 0 &{} 1 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 1 &{} 1 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 1&{} 0 &{} 1 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 1&{} 1 &{} 1 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =3, \nonumber \\ \end{aligned}$$

(3.6)

$$\begin{aligned} \mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 1&{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 1 &{} 1 \\ 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 1&{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 1&{} 0 &{} 1 \\ 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) =3, \nonumber \\ \end{aligned}$$

(3.7)

$$\begin{aligned} \mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 1&{} 0 &{} 0 \\ 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 1 &{} 0 \\ 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 0 &{} 1 \\ 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) =\mathrm{rk}\left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 1 &{} 1 \\ 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) =3.\nonumber \\ \end{aligned}$$

(3.8)

Note that $\left| {\mathcal {L}}_1'+E_{22}\right| =\left| {\mathcal {L}}_1'+E_{23}\right| =\left| {\mathcal {L}}_1'+E_{22}+E_{23}+E_{33}\right| =8$. By (3.6), there are no five vertices in ${\mathcal {L}}_1'+E_{22}$ such that they are adjacent to 0. Similarly, by (3.7) (resp. (3.8)), there are no five vertices in ${\mathcal {L}}_1'+E_{23}$ (resp. ${\mathcal {L}}_1'+E_{22}+E_{23}+E_{33}$) such that they are adjacent to 0. Thus, we have a contradiction. $\square$

4 Colouring and pseudo-core

In this section, we will discuss colouring and pseudo-core for $\varGamma ^{(n)}$.

Theorem 4.1

Let q be even, and let $\varphi$ be an endomorphism of $\varGamma ^{(n)}$ where $n\ge 4$. Further, assume that there are two distinct type two cliques ${\mathcal {L}}$ and ${\mathcal {L}}'$ in $\varGamma ^{(n)}$, such that ${\mathcal {L}}\cap {\mathcal {L}}'\ne \emptyset$ and $\varphi ({\mathcal {L}})=\varphi ({\mathcal {L}}')$. Then, $\varphi$ is a colouring.

Proof

Step 1 By Lemma 3.6, there are matrices $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ and $A\in {\mathbb {F}}_q^{n\times n}$ such that ${\mathcal {L}}=\left[ P{\mathcal {L}}_1\,^tP+A\right]$ and ${\mathcal {L}}'=\left[ P{\mathcal {L}}_2\,^tP+A\right]$. By (2.5), the map $[X]\longmapsto \varphi \left( \left[ PX\,^tP+A\right] \right)$ is an endomorphism of $\varGamma ^{(n)}$. Replacing $\varphi$ by the endomorphism $[X]\longmapsto \varphi \left( \left[ PX\,^tP+A\right] \right)$, without loss of generality we may assume that ${\mathcal {L}}=\left[ {\mathcal {L}}_1\right]$, ${\mathcal {L}}'=\left[ {\mathcal {L}}_2\right]$ and

$$\begin{aligned} \varphi \left( \left[ {\mathcal {L}}_1\right] \right) =\varphi \left( \left[ {\mathcal {L}}_2\right] \right) =:{\mathcal {C}}. \end{aligned}$$

(4.1)

Since $\left[ {\mathcal {L}}_1\right]$ and $\left[ {\mathcal {L}}_2\right]$ are maximum cliques of $\varGamma ^{(n)}$, ${\mathcal {C}}$ is also a maximum clique and $|{\mathcal {C}}|=q^n=\omega (\varGamma ^{(n)})$. Recall that the matrix representation for Q(n, q). By Corollary 3.2, ${\mathcal {C}}$ is either a type one clique or a type two clique.

Let $1\le k\le n$ and let

$$\begin{aligned} {\mathcal {R}}_k=\left\{ [X]: X\in {\mathbb {F}}_q^{n\times n},\mathrm{rk}(X)\le k \right\} . \end{aligned}$$

We will prove that $\varphi ({\mathcal {R}}_k)={\mathcal {C}}$.

Let $B\in {\mathbb {F}}_q^{n\times n}$ with $\mathrm{rk}(B)=1$. By Corollary 2.3, there exists $P_1\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $[B]=\left[ P_1E_{11}\,^tP_1 \right]$. Put ${\mathcal {M}}_1=\left[ P_1{\mathcal {L}}_1\,^tP_1\right]$. Then, ${\mathcal {M}}_1$ is a type two clique and $[B]\in {\mathcal {M}}_1$. Since $0\in \left[ {\mathcal {L}}_1\right] \cap \left[ {\mathcal {L}}_2\right] \cap {\mathcal {M}}_1$, Lemma 3.10 and (4.1) imply that $\varphi ({\mathcal {M}}_1)={\mathcal {C}}$. It follows that $\varphi ([B])\in {\mathcal {C}}$ for any $B\in {\mathbb {F}}_q^{n\times n}$ with $\mathrm{rk}(B)=1$. From (4.1), we have $\varphi (0)\in {\mathcal {C}}$. Thus, we have proved that $\varphi ({\mathcal {R}}_1)\subseteq {\mathcal {C}}$. Using Corollary 2.3 again, ${\mathcal {R}}_1=\left[ {\mathcal {D}}_n\right]$ is a maximum clique of $\varGamma ^{(n)}$ since $n\ge 4$. Thus, we obtain

$$\begin{aligned} \varphi ({\mathcal {R}}_1)=\varphi \left( \left[ {\mathcal {D}}_n\right] \right) ={\mathcal {C}}. \end{aligned}$$

(4.2)

Let $G\in {\mathbb {F}}_q^{n\times n}$ with $\mathrm{rk}(G)=2$. By Lemma 2.2, there exists $P_2\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that either $[G]=\left[ P_2E_{12}\,^tP_2\right]$ or $[G]=\left[ P_2(\alpha E_{11}+E_{12}+\alpha E_{22})\,^tP_2\right]$, where $\alpha \in {\mathbb {F}}_q$ but $\alpha \notin \left\{ x^2+x:x\in {\mathbb {F}}_q \right\}$ is fixed. Let ${\mathcal {C}}'=\left[ P_2\mathrm{diag}\left( {\mathcal {T}}_2, 0\right) \,^tP_2\right]$ be a cubic clique of $\varGamma ^{(n)}$. Then, $[G]\in {\mathcal {C}}'$. Set ${\mathcal {M}}=\left[ P_2{\mathcal {L}}_1\,^tP_2\right]$ and ${\mathcal {M}}'=\left[ P_2{\mathcal {L}}_2\,^tP_2\right]$. Then, ${\mathcal {M}}$ and ${\mathcal {M}}'$ are two distinct type two cliques. Clearly, $\left| {\mathcal {C}}'\cap {\mathcal {M}}\right| =\left| {\mathcal {C}}'\cap {\mathcal {M}}'\right| =q^2$ and $\left| {\mathcal {C}}'\cap {\mathcal {M}}\cap {\mathcal {M}}'\right| =q$. Since $0\in \left[ {\mathcal {L}}_1\right] \cap \left[ {\mathcal {L}}_2\right] \cap {\mathcal {M}}$ and $0\in \left[ {\mathcal {L}}_1\right] \cap \left[ {\mathcal {L}}_2\right] \cap {\mathcal {M}}'$, Lemma 3.10 and (4.1) imply that

$$\begin{aligned} \varphi ({\mathcal {M}})=\varphi ({\mathcal {M}}')={\mathcal {C}}. \end{aligned}$$

(4.3)

We have that $\left| {\mathcal {C}}'\cap ({\mathcal {M}}\cup {\mathcal {M}}')\right| =\left| {\mathcal {C}}'\cap {\mathcal {M}}\right| +\left| {\mathcal {C}}'\cap {\mathcal {M}}'\right| -\left| {\mathcal {C}}'\cap {\mathcal {M}}\cap {\mathcal {M}}'\right| =2q^2-q$. Since $\varphi ({\mathcal {C}}'\cap ({\mathcal {M}}\cup {\mathcal {M}}'))\subseteq \varphi ({\mathcal {C}}')\cap \varphi ({\mathcal {M}}\cup {\mathcal {M}}')=\varphi ({\mathcal {C}}')\cap {\mathcal {C}}$ and the restriction mapping $\varphi \mid _{{\mathcal {C}}'}$ is injective, we obtain that

$$\begin{aligned} \left| \varphi ({\mathcal {C}}')\cap {\mathcal {C}}\right| \ge 2q^2-q. \end{aligned}$$

(4.4)

If $\varphi ({\mathcal {C}}')$ is a cubic clique of $\varGamma ^{(n)}$, then Lemma 3.11 implies that $\left| \varphi ({\mathcal {C}}')\cap {\mathcal {C}}\right| \le q^2$, a contradiction to (4.4). Thus, $\varphi ({\mathcal {C}}')$ is not a cubic clique. When $q=2$, Lemma 3.12 and (4.4) imply that $\varphi ({\mathcal {C}}')\subseteq {\mathcal {C}}$. When $q\ge 4$, Corollary 3.2 implies that $\varphi ({\mathcal {C}}')$ is not a maximal clique of $\varGamma ^{(n)}$, and hence, $\varphi ({\mathcal {C}})$ is contained in a maximum clique. By Corollary 3.4 and (4.4), we must have $\varphi ({\mathcal {C}}')\subseteq {\mathcal {C}}$ if $q\ge 4$. Therefore, we always have $\varphi ({\mathcal {C}}')\subseteq {\mathcal {C}}$. Since $[G]\in {\mathcal {C}}'$, $\varphi ([G])\in {\mathcal {C}}$. By the arbitrariness of G and (4.2), we have proved that

$$\begin{aligned} \varphi ({\mathcal {R}}_2)={\mathcal {C}}. \end{aligned}$$

(4.5)

Step 2 Suppose that $\varphi ({\mathcal {R}}_{k-1})={\mathcal {C}}$ and $\varphi ({\mathcal {R}}_{k-2})={\mathcal {C}}$, where $k\ge 3$. We prove $\varphi ({\mathcal {R}}_{k})={\mathcal {C}}$ as follows. Let

$$\begin{aligned} Y_1= & {} \sum _{i=1}^\upsilon E_{i, \upsilon +i}+E_{2\upsilon +1, 2\upsilon +1} \quad (\hbox {where }1\le \upsilon \le (n-1)/2);\\ Y_2= & {} \sum _{i=1}^\upsilon E_{i, \upsilon +i} \quad (\hbox {where }1\le \upsilon \le n/2);\\ Y_3= & {} \sum _{i=1}^{\upsilon -1} E_{i, \upsilon -1+i}+\alpha E_{2\upsilon -1, 2\upsilon -1}+E_{2\upsilon -1, 2\upsilon }+\alpha E_{2\upsilon , 2\upsilon } \quad (\hbox {where }1\le \upsilon \le n/2). \end{aligned}$$

Note that $Y_3:=\alpha E_{11}+\alpha E_{22}+E_{12}$ if $\upsilon =1$.

Let $Y\in {\mathbb {F}}_q^{n\times n}$ with $\mathrm{rk}(Y)=k$. By Lemma 2.2, if $k=2\upsilon +1$, then there exists $P_3\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $Y\equiv P_3Y_1\,^tP_3$ (mod ${\mathcal {K}}_n$). If $k=2\upsilon$, then there exists $P_4\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that either $Y\equiv P_4Y_2\,^tP_4$ or $Y\equiv P_4Y_3\,^tP_4$ (mod ${\mathcal {K}}_n$).

Case 1 Suppose that $k=2\upsilon +1$ and $Y\equiv P_3Y_1\,^tP_3$ (mod ${\mathcal {K}}_n$). Let

$$\begin{aligned} {\mathcal {L}}_i'=\left[ P_3({\mathcal {L}}_i+Y_1-E_{kk})\,^tP_3\right] =\left[ P_3({\mathcal {L}}_i+Z)\,^tP_3\right] , i=1, \upsilon +1, \end{aligned}$$

where $Z=\sum _{i=2}^\upsilon E_{i, \upsilon +i}$. Then, both ${\mathcal {L}}_1'$ and ${\mathcal {L}}_{\upsilon +1}'$ are type two cliques with ${\mathcal {L}}_1'\cap {\mathcal {L}}_{\upsilon +1}'\ne \emptyset$. Using (2.4) and (2.7), it is easy to see that $\mathrm{rk}(Z)\le 2\upsilon -2$. Hence, (2.5) and (2.7) imply that $\mathrm{rk}(X)\le 2\upsilon$ for all $[X]\in {\mathcal {L}}_i'$, $i=1, \upsilon +1$. Thus, ${\mathcal {L}}_i'\subseteq {\mathcal {R}}_{2\upsilon }={\mathcal {R}}_{k-1}$, $i=1,\upsilon +1$. It follows from the induction hypothesis that $\varphi ({\mathcal {L}}_1')=\varphi ({\mathcal {L}}_{\upsilon +1}')={\mathcal {C}}$. Put ${\mathcal {M}}=\left[ P_3\left( {\mathcal {L}}_{k}+Y_1-E_{kk}\right) \,^tP_3\right]$. Then, ${\mathcal {M}}$ is a type two clique and $P_3(Y_1-E_{kk})\,^tP_3\in {\mathcal {L}}_1'\cap {\mathcal {L}}_{\upsilon +1}'\cap {\mathcal {M}}$. By Lemma 3.10, we get $\varphi ({\mathcal {M}})={\mathcal {C}}$. Since $\left[ P_3Y_1\,^tP_3\right] \in {\mathcal {M}}$, we obtain $\varphi \left( \left[ P_3Y_1\,^tP_3\right] \right) \in {\mathcal {C}}$. By $Y\equiv P_3Y_1\,^tP_3$ and (2.8), we have $\varphi ([Y])\in {\mathcal {C}}$.

Case 2 Suppose that $k=2\upsilon$ and $Y\equiv P_4Y_2\,^tP_4$ (mod ${\mathcal {K}}_n$). Let

$$\begin{aligned} {\mathcal {L}}_i'=\left[ P_4({\mathcal {L}}_i\,^tP_4+Y_2-E_{\upsilon , 2\upsilon })\,^tP_4\right] =\left[ P_4({\mathcal {L}}_i\,^tP_4+Z)\,^tP_4\right] , \quad i=1, \upsilon +1, \end{aligned}$$

where $Z=\sum _{i=2}^{\upsilon -1} E_{i, \upsilon +i}$. Then, both ${\mathcal {L}}_1'$ and ${\mathcal {L}}_{\upsilon +1}'$ are type two cliques with ${\mathcal {L}}_1'\cap {\mathcal {L}}_{\upsilon +1}'\ne \emptyset$. By (2.4) and (2.7), one has $\mathrm{rk}(Z)\le 2\upsilon -4$. Thus, (2.5) and (2.7) imply that $\mathrm{rk}(X)\le 2\upsilon -2$ for all $[X]\in {\mathcal {L}}_i'$, $i=1, \upsilon +1$. Thus, ${\mathcal {L}}_i'\subseteq {\mathcal {R}}_{k-2}$, $i=1, \upsilon +1.$ By the induction hypothesis, we have $\varphi ({\mathcal {L}}_1')=\varphi ({\mathcal {L}}_{\upsilon +1}')={\mathcal {C}}$. Set ${\mathcal {M}}=\left[ P_4({\mathcal {L}}_{\upsilon }+Y_2- E_{\upsilon , 2\upsilon })\,^tP_4\right]$. Then, ${\mathcal {M}}$ is a type two clique with ${\mathcal {L}}_1'\cap {\mathcal {L}}_{\upsilon +1}'\cap {\mathcal {M}} \ne \emptyset$. It follows from Lemma 3.10 that $\varphi ({\mathcal {M}})={\mathcal {C}}$. Since $\left[ P_4Y_2\,^tP_4\right] \in {\mathcal {M}}$, we get $\varphi \left( \left[ P_4Y_2\,^tP_4\right] \right) \in {\mathcal {C}}$. By $Y\equiv P_4Y_2\,^tP_4$ and (2.8), we obtain $\varphi ([Y])\in {\mathcal {C}}$.

Case 3 Suppose that $k=2\upsilon$ and $Y\equiv P_4Y_3\,^tP_4$ (mod ${\mathcal {K}}_n$). Let

$$\begin{aligned} {\mathcal {L}}_i'= & {} \left[ P_4\left( {\mathcal {L}}_i+Y_3-\alpha E_{2\upsilon -1, 2\upsilon -1}-E_{2\upsilon -1, 2\upsilon }\right) \,^tP_4\right] \\= & {} \left[ P_4\left( {\mathcal {L}}_i+Z\right) \,^tP_4\right] , \quad i=1, \upsilon , \end{aligned}$$

where $Z=\sum _{i=2}^{\upsilon -1} E_{i, \upsilon -1+i}+\alpha E_{2\upsilon , 2\upsilon }$. Then, both $\mathcal {L}_1'$ and $\mathcal {L}_\upsilon '$ are type two cliques with $\mathcal {L}_1'\cap \mathcal {L}_{\upsilon } '\ne \emptyset$. Using (2.4) and (2.7), it is easy to see that $\mathrm{rk}(Z)\le 2\upsilon -3$. Therefore, by (2.5) and (2.7), we have that $\mathrm{rk}(X)\le 2\upsilon -1$ for all $[X]\in {\mathcal {L}}_i'$, $i=1, \upsilon$. Consequently, ${\mathcal {L}}_i'\subseteq {\mathcal {R}}_{k-1}$, $i=1,\upsilon$. By the induction hypothesis, we get $\varphi ({\mathcal {L}}_1')=\varphi ({\mathcal {L}}_\upsilon ')={\mathcal {C}}$. Put ${\mathcal {M}}=\left[ P_4({\mathcal {L}}_{2\upsilon -1}+Y_3-\alpha E_{2\upsilon -1, 2\upsilon -1}-E_{2\upsilon -1, 2\upsilon })\,^tP_4\right]$. Then, ${\mathcal {M}}$ is a type two clique in $\varGamma ^{(n)}$ with ${\mathcal {L}}_1'\cap {\mathcal {L}}_\upsilon '\cap {\mathcal {M}} \ne \emptyset$. Using Lemma 3.10, we obtain $\varphi ({\mathcal {M}})={\mathcal {C}}$. Since $\left[ P_4Y_3\,^tP_4\right] \in {\mathcal {M}}$, we have $\varphi \left( \left[ P_4Y_3\,^tP_4\right] \right) \in {\mathcal {C}}$. From $Y\equiv P_4Y_3\,^tP_4$ and (2.8), we obtain $\varphi ([Y])\in {\mathcal {C}}$.

Combining Cases 1–3, we always have $\varphi ([Y])\in {\mathcal {C}}$ for any $Y\in {\mathbb {F}}_q^{n\times n}$ with $\mathrm{rk}(Y)=k$. It follows from the induction hypothesis that $\varphi ({\mathcal {R}}_{k})={\mathcal {C}}$. Taking $k=n$, we obtain $\varphi \left( \left[ {\mathcal {T}}_n \right] \right) =\varphi ({\mathcal {R}}_{n})={\mathcal {C}}$. Thus, $\varphi$ is a colouring. $\square$

Lemma 4.2

Let $n\ge 4$ and let $\varphi \in \mathrm{End}(\varGamma ^{(n)})$ which is not a colouring. Suppose that ${\mathcal {C}}$ is a type one clique of $\varGamma ^{(n)}$ and ${\mathcal {L}}$ is a type two clique of in $\varGamma ^{(n)}$ with ${\mathcal {C}}\cap {\mathcal {L}}\ne \emptyset$. Then, $\varphi ({\mathcal {C}})\ne \varphi ({\mathcal {L}})$.

Proof

Let $[A]\in {\mathcal {C}}\cap {\mathcal {L}}$. Then, there exists $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that either ${\mathcal {L}}=\left[ P{\mathcal {L}}_1\,^tP+A\right]$ and ${\mathcal {C}}=\left[ {\mathcal {D}}_n+A\right] =\left[ P{\mathcal {D}}_n\,^tP+A\right]$ by (3.4) and (2.2). Put ${\mathcal {L}}'=\left[ P{\mathcal {L}}_2\,^tP+A\right]$. Then, ${\mathcal {L}}'$ is a type two clique, $\left| {\mathcal {L}}'\cap {\mathcal {C}}\right| =\left| {\mathcal {L}}'\cap {\mathcal {L}}\right| =q$ and ${\mathcal {L}}'\cap {\mathcal {L}}\cap {\mathcal {C}}=\{A\}$. It follows that

$$\begin{aligned} \left| {\mathcal {L}}'\cap ({\mathcal {C}}\cup {\mathcal {L}})\right| =\left| {\mathcal {L}}'\cap {\mathcal {C}}\right| +\left| {\mathcal {L}}'\cap {\mathcal {L}}\right| -\left| {\mathcal {L}}'\cap {\mathcal {L}}\cap {\mathcal {C}}\right| =2q-1>q. \end{aligned}$$

(4.6)

Clearly, $\varphi ({\mathcal {C}})$, $\varphi ({\mathcal {L}})$ and $\varphi ({\mathcal {L}}')$ are three maximum cliques of $\varGamma ^{(n)}$.

Suppose that $\varphi ({\mathcal {C}})=\varphi ({\mathcal {L}})=:{\mathcal {M}}$. Then, $\varphi \left( {\mathcal {L}}'\cap ({\mathcal {C}}\cup {\mathcal {L}}) \right) \subseteq \varphi ({\mathcal {L}}')\cap {\mathcal {M}}$; thus, (4.6) implies that $\left| \varphi ({\mathcal {L}}')\cap {\mathcal {M}}\right| >q$. It follows from Corollary 3.4 that $\varphi ({\mathcal {L}}')={\mathcal {M}}=\varphi ({\mathcal {L}})$. By Theorem 4.1, $\varphi$ is a colouring, a contradiction to the conditions. Therefore, we must have $\varphi ({\mathcal {C}})\ne \varphi ({\mathcal {L}})$. $\square$

Lemma 4.3

Let $n\ge 4$ and let $\varphi \in \mathrm{End}(\varGamma ^{(n)})$ which is not a colouring and $\varphi (0)=0$. If $A\in {\mathbb {F}}_q^{n\times n}$ and $\mathrm{rk}(A)=3$, then $\varphi ([A])\ne 0$.

Proof

By Lemma 2.2, there exists $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $[A]=\left[ P(E_{12}+E_{33})\,^tP\right]$. Let ${\mathcal {L}}=\left[ P{\mathcal {L}}_3\,^tP\right]$ and ${\mathcal {L}}'=\left[ P({\mathcal {L}}_1+E_{33})\,^tP\right]$. Then, ${\mathcal {L}}$, ${\mathcal {L}}'$ are type two cliques with $[A]\in {\mathcal {L}}'$, and

$$\begin{aligned} {\mathcal {L}}\cap {\mathcal {L}}'=\left\{ \left[ P(xE_{13}+E_{33})\,^tP\right] : x\in {\mathbb {F}}_q\right\} . \end{aligned}$$

(4.7)

Since $\varphi$ is not a colouring, Theorem 4.1 implies that $\varphi ({\mathcal {L}})\ne \varphi ({\mathcal {L}}')$, and hence, $\varphi ({\mathcal {L}})$ and $\varphi ({\mathcal {L}}')$ are two distinct maximum cliques. By Lemma 3.3 (iv), we have $\left| {\mathcal {L}}\cap {\mathcal {L}}' \right| =q=\left| \varphi ({\mathcal {L}})\cap \varphi ({\mathcal {L}}')\right|$. Note that the restriction mapping $\varphi \mid _{{\mathcal {L}}}$ is injective. We obtain that $\varphi \left( {\mathcal {L}}\cap {\mathcal {L}}'\right) =\varphi ({\mathcal {L}})\cap \varphi ({\mathcal {L}}')$. For any $[X]\in {\mathcal {L}}\cap {\mathcal {L}}'$, (4.7) implies that $[X]\sim 0$; thus, $\varphi ([X])\sim 0$. Consequently, $0\notin \varphi ({\mathcal {L}})\cap \varphi ({\mathcal {L}}')$. Since $0\in \varphi ({\mathcal {L}})$ and $\varphi ([A])\in \varphi ({\mathcal {L}}')$, we have $\varphi ([A])\ne 0$. Otherwise, $0\in \varphi ({\mathcal {L}})\cap \varphi ({\mathcal {L}}')$, a contradiction. $\square$

Lemma 4.4

Let $n\ge 4$ and let $\varphi \in \mathrm{End}(\varGamma ^{(n)})$ which is not a colouring and $\varphi (0)=0$. If $A\in {\mathbb {F}}_q^{n\times n}$ and $\mathrm{rk}(A)=3$, then $\mathrm{rk}\left( \varphi ([A])\right) \ge 3$.

Proof

Since $\left[ {\mathcal {D}}_n\right]$ is a maximum clique of $\varGamma ^{(n)}$, Corollary 3.2 implies either $\varphi \left( \left[ {\mathcal {D}}_n\right] \right)$ is either a type one clique or a type two clique. By $\varphi (0)=0$, we have either $\varphi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right]$ or $\varphi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ P{\mathcal {L}}_1\,^tP\right]$ where $P\in \mathrm{GL}_n({\mathbb {F}}_q)$.

Note that $\left[ {\mathcal {D}}_n\right]$ is the unique type one clique containing 0. Since $\varphi$ is not a colouring and $\varphi (0)=0$, Theorem 4.1 implies that $\varphi$ maps distinct type two cliques containing 0 to distinct maximum cliques. Assume that $\varGamma ^{(n)}$ has and only has $s+1$ maximum cliques containing 0. By Corollary 3.2, $\varGamma ^{(n)}$ has and only has s type two cliques containing 0. Let ${\mathcal {C}}_1, \ldots , {\mathcal {C}}_s$ are s distinct type two cliques containing 0. Since $\varphi$ is not a colouring and $\varphi (0)=0$, Theorem 4.1 and Lemma 4.2 imply that $\varphi ({\mathcal {C}}_1), \ldots , \varphi ({\mathcal {C}}_s), \varphi \left( \left[ {\mathcal {D}}_n\right] \right)$ are $s+1$ distinct maximum cliques containing 0. Then, we have

$$\begin{aligned} \left\{ \varphi ({\mathcal {C}}_1), \ldots , \varphi ({\mathcal {C}}_s), \varphi \left( \left[ {\mathcal {D}}_n\right] \right) \right\} = \left\{ {\mathcal {C}}_1, \ldots , {\mathcal {C}}_s, \left[ {\mathcal {D}}_n\right] \right\} . \end{aligned}$$

(4.8)

By (4.8), if $\varphi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right]$, then $\varphi$ maps any type two clique containing 0 to a type two clique containing 0, and hence, $\varphi \left( \left[ {\mathcal {L}}_1\right] \right) =\left[ Q_1{\mathcal {L}}_1\,^tQ_1\right]$ where $Q_1\in \mathrm{GL}_n({\mathbb {F}}_q)$. If $\varphi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ P{\mathcal {L}}_1\,^tP\right]$, then there is a type two clique ${\mathcal {C}}_k=\left[ P_k{\mathcal {L}}_1\,^tP_k\right]$ where $P_k\in \mathrm{GL}_n({\mathbb {F}}_q)$, such that $\varphi \left( \left[ P_k{\mathcal {L}}_1\,^tP_k\right] \right) =\left[ {\mathcal {D}}_n\right]$. From (3.4), we have $\left[ {\mathcal {D}}_n\right] =\left[ P_k{\mathcal {D}}_n\,^tP_k\right] =\left[ Q_1{\mathcal {D}}_n\,^tQ_1\right]$.

When $\varphi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right]$ with $\varphi \left( \left[ {\mathcal {L}}_1\right] \right) =\left[ Q_1{\mathcal {L}}_1\,^tQ_1\right]$, we let

$$\begin{aligned} \psi ([X])=Q_1^{-1}\varphi \left( \left[ X\right] \right) \,^tQ_1^{-1}, \quad [X]\in [{\mathcal {T}}_n]. \end{aligned}$$

When $\varphi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ P{\mathcal {L}}_1\,^tP\right]$ with $\varphi \left( \left[ P_k{\mathcal {L}}_1\,^tP_k\right] \right) =\left[ {\mathcal {D}}_n\right]$, we let

$$\begin{aligned} \psi ([X])=P^{-1}\varphi \left( \left[ P_kX\,^tP_k\right] \right) \,^tP^{-1}, \quad [X]\in [{\mathcal {T}}_n]. \end{aligned}$$

Then, $\psi \in \mathrm{End}(\varGamma ^{(n)})$, $\psi$ is not a colouring, and $\psi (0)=0$. Moreover, we have either $\psi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right]$ with $\psi \left( \left[ {\mathcal {L}}_1\right] \right) =\left[ {\mathcal {L}}_1\right]$, or $\psi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {L}}_1\right]$ with $\psi \left( \left[ {\mathcal {L}}_1\right] \right) =\left[ {\mathcal {D}}_n\right]$. Let $\phi ([X])=\psi ^2([X])$ for all $[X]\in [{\mathcal {T}}_n]$. Then, $\phi \in \mathrm{End}(\varGamma ^{(n)})$ and $\phi (0)=0$. Moreover, we always have

$$\begin{aligned} \phi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right] \quad \hbox {and} \quad \phi \left( \left[ {\mathcal {L}}_1\right] \right) =\left[ {\mathcal {L}}_1\right] . \end{aligned}$$

(4.9)

By (4.9), it is clear that $\phi$ is not a colouring.

Let $A\in {\mathbb {F}}_q^{n\times n}$ with $\mathrm{rk}(A)=3$. By Lemma 2.2, there is $Q_2\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that

$$\begin{aligned}{}[A]=\left[ Q_2(E_{12}+E_{33})\,^tQ_2 \right] . \end{aligned}$$

(4.10)

Recall that $\left[ {\mathcal {D}}_n\right]$ is the unique type one clique containing 0. By (4.9), Corollary 3.2 and Lemma 4.2, $\phi$ maps any type two clique containing 0 to a type two clique containing 0. Let

$$\begin{aligned} {\mathcal {L}}=\left[ Q_2({\mathcal {L}}_1+E_{33})\,^tQ_2 \right] , \quad {\mathcal {L}}'=\left[ Q_2({\mathcal {L}}_2+E_{33})\,^tQ_2 \right] . \end{aligned}$$

Then, ${\mathcal {L}}$ and ${\mathcal {L}}'$ are type two cliques containing 0 and $[A]\in {\mathcal {L}}\cap {\mathcal {L}}'$. Write $B=Q_2E_{33}\,^tQ_2$. Then, $[B]\in \left[ {\mathcal {D}}_n\right]$, and $\phi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right]$ implies that $\phi ([B])\in \left[ {\mathcal {D}}_n\right]$.

Suppose that both $\phi ({\mathcal {L}})$ and $\phi ({\mathcal {L}}')$ are type one cliques in $\varGamma ^{(n)}$. Then, $\phi ({\mathcal {L}})=\left[ {\mathcal {D}}_n+\phi ([B])\right] =\left[ {\mathcal {D}}_n\right]$ and similarly $\phi ({\mathcal {L}}')=\left[ {\mathcal {D}}_n\right]$. It follows from Theorem 4.1 that $\phi$ is a colouring, a contradiction. Therefore, $\phi ({\mathcal {L}})$ or $\phi ({\mathcal {L}}')$ is a type two clique in $\varGamma ^{(n)}$. Without loss of generality, we assume that $\phi ({\mathcal {L}})$ is a type two clique. Then, there exists $Q_3\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that $\phi ({\mathcal {L}})=\left[ Q_3{\mathcal {L}}_1\,^tQ_3 +\phi ([B])\right]$. Since $\phi ([B])\in \left[ {\mathcal {D}}_n\right]$, $Q_3^{-1}\phi ([B])\,^tQ_3^{-1}\in \left[ {\mathcal {D}}_n\right]$ by (2.2). Hence, we can let $Q_3^{-1}\phi ([B])\,^tQ_3^{-1}=\left[ \mathrm{diag}(b_1, \ldots , b_n)\right]$. Thus,

$$\begin{aligned} \phi ({\mathcal {L}})=\phi \left( \left[ Q_2({\mathcal {L}}_1+E_{33})\,^tQ_2 \right] \right) =\left[ Q_3\left( {\mathcal {L}}_1+\mathrm{diag}(0,b_2, \ldots , b_n)\right) \,^tQ_3 \right] . \end{aligned}$$

By (2.4)–(2.5) and $\mathrm{rk}(f)=n-\mathrm{dim}(\mathrm{Rad}(f))$, it is easy to see that $\mathrm{rk}(X)=1$ or 3 for all $[X]\in {\mathcal {L}}$. Thus, Lemma 4.3 implies that $\phi ([X])\ne 0$ for all $[X]\in {\mathcal {L}}$. Consequently, $0\notin \phi ({\mathcal {L}})$. It follows that $(b_2, \ldots ,b_n)\ne 0$. Otherwise, $(b_2, \ldots ,b_n)=0$ implies that $0\in \phi ({\mathcal {L}})$, a contradiction. Using (2.4) and (2.5) again, we get that $\mathrm{rk}\left( \phi ([X])\right) =1$ or 3 for all $[X]\in {\mathcal {L}}$.

By $\phi \left( \left[ {\mathcal {D}}_n\right] \right) =\left[ {\mathcal {D}}_n\right]$ and $\left| {\mathcal {L}}\cap \left[ {\mathcal {D}}_n\right] \right| =\left| \phi ({\mathcal {L}})\cap \left[ {\mathcal {D}}_n\right] \right| =q$, it is easy to prove that

$$\begin{aligned} \phi \left( {\mathcal {L}}\cap \left[ {\mathcal {D}}_n\right] \right) =\phi ({\mathcal {L}})\cap \left[ {\mathcal {D}}_n\right] . \end{aligned}$$

Since the restriction mapping $\phi \mid _{{\mathcal {L}}}$ is injective, it follows from Corollary 2.3 that $\mathrm{rk}(X)=1$ if and only if $\mathrm{rk}\left( \phi ([X])\right) =1$ for all $[X]\in {\mathcal {L}}$. In other words, $\mathrm{rk}(X)=3$ if and only if $\mathrm{rk}\left( \phi ([X])\right) =3$ for all $[X]\in {\mathcal {L}}$. By $[A]\in {\mathcal {L}}$ and $\mathrm{rk}(A)=3$, we obtain that $\mathrm{rk}\left( \phi ([A])\right) =3$. Recalling that $\phi ([X])=\psi ^2([X])$, we must have $\mathrm{rk}\left( \psi ([A])\right) \ge 3$. Otherwise, $\mathrm{rk}\left( \psi ([A])\right) \le 2$ implies that $\psi ([A])\sim 0$ or $\psi ([A])=0$, and hence, $\mathrm{rk}\left( \phi ([A])\right) \le 2$ a contradiction.

Let $A'=P_k^{-1}A\,^tP_k^{-1}$. Then, $\mathrm{rk}(A')=3$. Similarly, we have that $\mathrm{rk}\left( \psi ([A'])\right) \ge 3$.

Recall that the definition of $\psi ([X])$. If $\psi ([X])=Q_1^{-1}\varphi \left( \left[ X\right] \right) \,^tQ_1^{-1}$, then $\mathrm{rk}\left( \psi ([A])\right) \ge 3$ and (2.5) imply that $\mathrm{rk}\left( \varphi ([A])\right) \ge 3$. If $\psi ([X])=P^{-1}\varphi \left( \left[ P_kX\,^tP_k\right] \right) \,^tP^{-1}$, then $\mathrm{rk}\left( \psi ([A'])\right) \ge 3$ and (2.5) imply that $\mathrm{rk}\left( \varphi ([A])\right) \ge 3$. Therefore, we always have $\mathrm{rk}\left( \varphi ([A])\right) \ge 3$. $\square$

Recall that a graph on n vertices is a strongly regular graph with parameters (n, k, a, c) if it is k-regular, every pair of adjacent vertices has a common neighbours, and every pair of distinct nonadjacent vertices has c common neighbours. A strongly regular graph $\varGamma$ is called primitive if both $\varGamma$ and its complement are connected ( [4, 6]). A strongly regular graph with parameters (n, k, a, c) is primitive if and only if $0<c<k$ (cf. [6, p. 178]). Note that a distance-regular graph with diameter $d=2$ is a connected strongly regular graph.

Lemma 4.5

(Roberson [19, Corollary 4.2]) Every primitive strongly regular graph is a pseudo-core.

Corollary 4.6

Assume that $2\le n\le 4$ and q is a power of a prime. Then, Q(n, q) is a pseudo-core.

Proof

Since Q(2, q) is a clique, Q(2, q) is a core, and hence, it is a pseudo-core. By Lemma 1.1, Q(3, q) is a connected strongly regular graph with parameters (n, k, a, c), where $k=(q^2+1)(q^3-1)$ and $c= q^2(q^2+1)$. Since $0<c<k$, Q(3, q) is a primitive strongly regular graph. Similarly, Q(4, q) is a primitive strongly regular graph parameter $(n',k',a',c')$, where $k'=(q^2+1)(q^5-1)$ and $c'= q^2(q^2+1)$. By Lemma 4.5, both Q(3, q) and Q(4, q) are pseudo-cores. $\square$

For a graph $\varGamma$, an endomorphism $\varphi \in \mathrm{End}(\varGamma )$ is called local injective, if for any $x\in V(\varGamma )$, the restriction of $\varphi$ to the neighbourhood $\varGamma _1(x)$ is injective.

Lemma 4.7

(Ne$\check{\mathrm{s}}$et$\check{\mathrm{r}}$il [18, Lemma 1]) Let G be a connected graph. If $\varphi \in \mathrm{End}(G)$ is local injective, then $\varphi \in \mathrm{Aut}(G)$.

Theorem 4.8

Let q be even. Then, every quadratic forms graph Q(n, q) is a pseudo-core.

Proof

Let $\varGamma =\varGamma ^{(n)}$. By $\varGamma \cong Q(n,q)$, we only need to prove that $\varGamma$ is a pseudo-core. By Corollary 4.6, both $\varGamma ^{(2)}$ and $\varGamma ^{(3)}$ are pseudo-cores. From now on, we assume $n\ge 4$. Let $\varphi \in \mathrm{End}(\varGamma )$ such that $\varphi$ is not a colouring. Since $\varGamma$ is vertex-transitive, by Lemma 4.7 we only need to prove that the restriction of $\varphi$ to the neighbourhood $\varGamma _1(0)$ is injective.

Let $[A], [B]\in \varGamma _1(0)$ with $[A]\ne [B]$. Then, $1\le \mathrm{rk}(A), \mathrm{rk}(B)\le 2$. Define $\psi ([X])=\varphi ([X]+[B])-\varphi ([B])$ for all $[X]\in \left[ {\mathcal {T}}_{n}\right]$. Then, $\psi \in \mathrm{End}(\varGamma )$, $\psi (0)=0$ and $\psi$ is not a colouring. Let $0\ne X\in {\mathbb {F}}_q^{n\times n}$. If $\mathrm{rk}(X)=1$ or 2, then $\psi ([X])\sim 0$, and hence, $\psi ([X])\ne 0$. If $\mathrm{rk}(X)=3$, from Lemma 4.4 we have $\mathrm{rk}\left( \psi ([X])\right) \ge 3$. Now, we assume that $\mathrm{rk}(X)=4$. By Lemma 2.2, there is $P\in \mathrm{GL}_n({\mathbb {F}}_q)$ such that either $X\equiv P\left( \begin{array}{c@{\quad }c@{\quad }c} 0_2&{} I_2 &{} 0 \\ 0 &{} 0_2 &{} 0 \\ 0 &{} 0 &{} 0_{n-4}\\ \end{array} \right) \,^tP$ or $X\equiv P\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0 &{} 1 &{} 0 &{} 0 &{}0\\ 0&{} 0 &{} 0&{} 0 &{}0\\ 0 &{} 0&{} \alpha &{} 1&{}0 \\ 0 &{} 0 &{} 0 &{} \alpha &{}0 \\ 0 &{} 0 &{} 0 &{} 0 &{}0_{n-4} \\ \end{array} \right) \,^tP$ (mod ${\mathcal {K}}_n$). Clearly, there is $B\in {\mathbb {F}}_q^{n\times n}$ such that $\mathrm{rk}(B)=3$ and $[X]\sim [B]$. Since $\psi ([X])\sim \psi ([B])$ and $\mathrm{rk}\left( \psi ([B])\right) \ge 3$ by Lemma 4.4, we must have $\psi ([X])\ne 0$. Therefore, we have $\psi ([X])\ne 0$ if $1\le \mathrm{rk}(X)\le 4$. Since $1\le \mathrm{rk}([A-B])\le 4$ and $[A-B]=[A]-[B]$, we obtain that $\varphi ([A])-\varphi ([B])=\psi ([A-B])\ne 0$ for all $[A], [B]\in \varGamma _1(0)$ with $[A]\ne [B]$. Hence, the restriction of $\varphi$ to the neighbourhood $\varGamma _1(0)$ is injective. $\square$

5 Core and the smallest eigenvalue of Q(n, q)

The smallest eigenvalue of a graph G plays an important role in the estimation of chromatic number of G. Unfortunately, so far we don’t know the eigenvalues of Q(n, q). In this section, we calculate the smallest eigenvalue of Q(n, q).

The number of sign changes in a sequence $x_0, \ldots , x_n$ of nonzero real numbers is the number of indices i such that $x_{i-1}x_i<0$. The number of sign changes in a sequence of real numbers, where some of the terms are equal to zero, is the number of sign changes of the sequence obtained by removing the zero terms.

Let G be a graph with $\left| V(G)\right| =n+1$. Denote n eigenvalues of G by $\theta _0(G)\ge \theta _1(G)\ge \cdots \ge \theta _n(G)$. The $\theta _0(G)$ (resp. $\theta _n(G)$) is called the maximum eigenvalue (resp. smallest eigenvalue) of G. We also write $\theta _{max}(G)=\theta _{0}(G)$ and $\theta _{min}(G)=\theta _{n}(G)$.

Lemma 5.1

(cf. [1, pp. 128–129], [3, Lemma 4.66]) Suppose that $\varGamma$ is a distance-regular graph with diameter d and the intersection array given by $\{b_0, b_1, \ldots ,b_{d-1}; c_1, c_2, \ldots , c_d\}$, and the eigenvalues of $\varGamma$ are $\theta _0>\theta _1>\cdots >\theta _d$, in decreasing order. Let

$$\begin{aligned} L=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0 &{} b_0 &{} &{} &{} &{} \\ c_1 &{} a_1 &{} b_1 &{} &{} &{} \\ &{} c_2 &{} a_2 &{} b_2 &{} &{} \\ &{} &{} \ddots &{} \ddots &{}\ddots &{} \\ &{} &{} &{} c_{d-1} &{}a_{d-1} &{}b_{d-1} \\ &{} &{} &{} &{} c_d &{} a_d \\ \end{array} \right) , \end{aligned}$$

(5.1)

where $a_i=b_0-b_i-c_i$, $i=1, \ldots , d$. Then, the followings hold:

• (i) The eigenvalues of $\varGamma$ are precisely the eigenvalues of L.

• (ii) The eigenvector with eigenvalue $\theta _i$ has exactly i sign changes, $i=1,\ldots , d$.

Theorem 5.2

Let q be a power of a prime, $n\ge 3$ and $d=[(n+1)/2]$. Then, the smallest eigenvalue of Q(n, q) is $\theta _d=-\,(q^{2d}-1)/(q^2-1)$.

Proof

By Lemma 1.1, Q(n, q) is a distance-regular graph with diameter d. Assume that tridiagonal matrix L is of the form (5.1), where

$$\begin{aligned} b_j= & {} q^{4j}\left( q^{n-2j+1}-1\right) \left( q^{n-2j}-1\right) \big /\left( q^2-1\right) ,\\ c_j= & {} q^{2j-2}\left( q^{2j}-1\right) \big /\left( q^2-1\right) ,\\ a_j= & {} b_0-b_j-c_j=\left( q^{n+2j+1}+q^{n+2j}-q^{n+1}-q^{n}\right. \\&-\,\left. q^{4j}-q^{4j-2}+q^{2j-2}+1\right) \big /\left( q^2-1\right) , \end{aligned}$$

$0\le j\le d$. Note that $2d=n$ if n is even and $2d=n+1$ if d is odd.

Let $x_d=1$,

$$\begin{aligned} x_i=(-\,1)^{d-i}\prod _{j=1}^{d-i}\left( q^{2j+(-\,1)^n}-1\right) , \quad i=0, 1,\ldots , d-1, \end{aligned}$$

(5.2)

and $X=\,^t(x_0,x_1, \ldots , x_d)$. Put $\theta = -\sum _{i=0}^{d-1}q^{2i}=-\,(q^{2d}-1)/(q^2-1)$. We prove $LX=\theta X$ as follows. It is easy to see that

$$\begin{aligned} b_0x_{1}=\theta x_0, \quad c_dx_{d-1}+a_dx_d=\theta x_d. \end{aligned}$$

(5.3)

Since $x_{i-1}/x_i=1-q^{2d-2i+2+(-\,1)^n}$ and $x_{i+1}/x_i=1/(1-q^{2d-2i+(-\,1)^n})$, we have that

$$\begin{aligned} c_ix_{i-1}/x_i+a_i+b_ix_{i+1}/x_i=-\,\left( q^{2d}-1\right) /(q^2-1)=\theta , \quad i=1,\ldots , d-1. \end{aligned}$$

Thus,

$$\begin{aligned} c_ix_{i-1}+a_ix_i+b_ix_{i+1}=\theta x_i, \quad i=1,\ldots , d-1. \end{aligned}$$

(5.4)

By (5.3) and (5.4), we obtain $LX=\theta X$. Therefore, by Lemma 5.1(i), $\theta$ is an eigenvalue of Q(n, q). Since the eigenvector X has exactly d sign changes, Lemma 5.1(ii) implies that $\theta$ is the smallest eigenvalue of Q(n, q), i.e. $\theta =\theta _d$. $\square$

Lemma 5.3

(Godsil [5, p. 268]) A graph G has a complete core if and only if $\chi (G)=\omega (G)$.

Lemma 5.4

Let G be a core-complete graph which is not a clique. Then, G is a core if and only if $\chi (G)> \omega (G)$.

Proof

Since G is core-complete, we have either G is a core or every core of G is complete. Assume that G is a core. Then, we must have $\chi (G)> \omega (G)$. Otherwise, $\chi (G)=\omega (G)$ and there is a homomorphism $\varphi$ from G to the complete graph on $\chi (G)$ vertices. Since G is a core, $\varphi \in \mathrm{Aut}(G)$, and hence, $\varphi (V(G))=V(G)$. It follows that $|V(G)|=\chi (G)=\omega (G)$, and hence, G is a clique, a contradiction.

Conversely, assume that $\chi (G)> \omega (G)$. By Lemma 5.3, every core of G is not complete. Since G is a core-complete graph, one has that G is a core. $\square$

Lemma 5.5

(Hoffman, cf. [2, Theorem 3.6.2]) If a graph G is not edgeless, then

$$\begin{aligned} \chi (G)\ge 1- \theta _{max}(G)/\theta _{min}(G). \end{aligned}$$

Theorem 5.6

Let q be even. Then, every quadratic forms graph Q(2m, q) is a core.

Proof

Since Q(2, q) is a clique, it is a core. Now, we assume that $m\ge 2$. Recall that every pseudo-core is core-complete (cf. [5]). By Theorem 4.8 and Lemma 5.4, we only need to prove that $\chi (Q(2m,q))> \omega (Q(2m,q)$. By Theorem 5.2, the smallest eigenvalue of Q(2m, q) is $\theta _m= -\,(q^{2m}-1)/(q^2-1)$. By Lemma 1.1, it is well known that the maximum eigenvalue of Q(2m, q) is $b_0=(q^{2m+1}-1)(q^{2m}-1)/(q^2-1)$ (cf. [1, p. 129]). By Lemma 5.5 and (3.3), we obtain that $\chi (Q(2m,q))\ge 1- b_0/\theta _{m}=q^{2m+1} > q^{2m}=\omega (Q(2m,q)$. $\square$