Vector spaces and Grassmann graphs over residue class rings

TL;DR

This paper explores the structure of vector spaces over residue class rings, analyzing subspace interactions and automorphisms of Grassmann graphs over these rings, revealing their combinatorial properties.

Contribution

It introduces a matrix-based approach to study subspaces over residue class rings and characterizes the automorphisms of Grassmann graphs in this setting.

Findings

Grassmann graph $G_{p^s}(n,m)$ is connected and vertex-transitive

Valency, clique number, and maximum cliques are determined

Automorphisms of the Grassmann graph are characterized

Abstract

Let $\mathbb{Z}_{p^s}$ be the residue class ring of integers modulo $p^s$, where $p$ is a prime number and $s$ is a positive integer. Using matrix representation and the inner rank of a matrix, we study the intersection, join, dimension formula and dual subspaces on vector subspaces of $\mathbb{Z}^n_{p^s}$. Based on these results, we investigate the Grassmann graph $G_{p^s}(n,m)$ over $\mathbb{Z}_{p^s}$. $G_{p^s}(n,m)$ is a connected vertex-transitive graph, and we determine its valency, clique number and maximum cliques. Finally, we characterize the automorphisms of $G_{p^s}(n,m)$.

Equations

$$\left|{\mathbb{Z}}_{p^s}\right|=p^s,\left|{\mathbb{Z}}_{p^s}^{\ast }\right|=(p-1)p^{s-1},\left|J_{p^s}\right|=p^{s-1}.$$

$$A=BC$$

$$\rho (A)=\rho (PAQ)$$

$$\rho (AB)\le \min \left\{\rho (A),\rho (B)\right\};$$

$$\rho (A_1,A_2)\ge \max \left\{\rho (A_1),\rho (A_2)\right\};$$

$$\rho\left(\begin{array}[]{cc}A&*\\ &*\\ \end{array}\right)\geq\rho(A).$$

$$\mathrm{rk}(A)=\max \left\{k:{\mathrm{Ann}}_R(I_k(A))=(0)\right\}.$$

$$A=P\left(\begin{array}[]{ccccc}I_{r}&&&&\\ &p^{k_{1}}&&&\\ &&\ddots&&\\ &&&p^{k_{t}}&\\ &&&&0\\ \end{array}\right)Q,$$

$$\rho\left(\begin{array}[]{cc}A&0\\ 0&B\\ \end{array}\right)=\rho(A)+\rho(B)$$

$$\rho (A+C)\le \rho (A)+\rho (C).$$

$$x=t_0+t_{1}p+\cdots +t_{s-1}{p}^{s-1}$$

$$X=X_0+X_{1}p+\cdots +X_{s-1}{p}^{s-1}$$

$$\pi :{\mathbb{Z}}_{p^s}^{m\times n}\to {\mathbb{Z}}_p^{m\times n}$$

$$\pi (X+Y)=\pi (X)+\pi (Y),$$

$$\pi (XQ)=\pi (X)\pi (Q),$$

$$\pi {(}^{t}X)={}^t(\pi (X)).$$

$$\pi (X^{-1})=(\pi (X){)}^{-1}.$$

$$X=P\mathrm{diag}\left(I_r,D,0_{m-r-t,n-r-t}\right)Q,$$

$${\mathcal{M}}_1=\left\{\sum _{j=1}^n{x}_j{E}_{1j}:x_j\in {\mathbb{Z}}_{p^s}\right\}$$

$$\omega \left(\Gamma \right)=p^{sk}.$$

$$\alpha \left(\Gamma ({\mathbb{Z}}_{p^s}^{m\times n})\right)=p^{sn(m-1)}$$

$$A=(0,I_k)U$$

$$B_1=\mathrm{diag}(p^{i_1},\dots ,p^{i_r},0_{m-r,n-k-r})$$

$$B=\left(\begin{array}[]{cccc}I_{t}&0&0&Y_{1}\\ 0&D_{2}&0&Y_{2}\\ 0&0&I_{m-r}&0\\ \end{array}\right),$$

$$U_{3}=\left(\begin{array}[]{cccc}I_{t}&0&0&C_{1}\\ 0&I_{n-m-t}&0&0\\ 0&0&I_{m-r}&0\\ 0&0&0&I_{r}\\ \end{array}\right)\in GL_{n}(\mathbb{Z}_{p^{s}}).$$

$$\small BU_{3}=\left(\begin{array}[]{cccc}I_{t}&0&0&Y_{1}\\ 0&D_{2}&0&Y_{2}\\ 0&0&I_{m-r}&0\\ \end{array}\right)\left(\begin{array}[]{cccc}I_{t}&0&0&C_{1}\\ 0&I_{n-m-t}&0&0\\ 0&0&I_{m-r}&0\\ 0&0&0&I_{r}\\ \end{array}\right)=\left(\begin{array}[]{cccc}I_{t}&0&0&C_{1}+Y_{1}\\ 0&D_{2}&0&Y_{2}\\ 0&0&I_{m-r}&0\\ \end{array}\right).$$

$${\rm dim}(A\vee B)={\rm dim}(A)+{\rm dim}(B)-{\rm dim}(A\cap B)=\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right).$$

$$A=(0,I_{k}),\ \ \ B=\left(\begin{array}[]{cccc}D_{1}&0_{r,n-m-r}&I_{r}&0\\ 0&0&0&I_{m-r}\\ \end{array}\right).$$

$$\dim (A\cap B)\ge m-r,\dim (A\vee B)\le k+r.$$

$$\alpha =(0,\dots ,0,t_1,\dots ,t_r,y_1,\dots ,y_{m-r})$$

Full text

Keywords: residue class ring, subspace, dimension formula, Grassmann graph, maximum clique, automorphism

2010 AMS Classification: 05C25, 15B33, 05C60, 05B25, 51A10

1 Introduction

Throughout, let $R$ be a commutative local ring and $R^{*}$ the set of all units of $R$. For a subset $S$ of $R$, let $S^{m\times n}$ be the set of all $m\times n$ matrices over $S$, and $S^{n}=S^{1\times n}$. Denote by $I_{r}$ ($I$ for short) the $r\times r$ identity matrix. For $A\in R^{m\times n}$ and $B\in R^{n\times m}$, if $AB=I_{m}$, we call that $A$ has a right inverse and $B$ is a right inverse of $A$. Similarly, if $AB=I_{m}$, than $B$ has a left inverse and $A$ is a left inverse of $B$. The cardinality of a set $X$ is denoted by $|X|$.

A $1\times n$ matrix over $R$ is called an $n$-dimensional vector over $R$. For $\alpha_{i}\in R^{n}$, $i=1,\ldots,k$, the vector subset $\{\alpha_{1},\ldots,\alpha_{k}\}$ is called unimodular if the matrix \scriptsize\left(\begin{array}[]{c}\alpha_{1}\\ \vdots\\ \alpha_{k}\\ \end{array}\right) has a right inverse. Let $V\subseteq R^{n}$ be a linear subset (i.e., an $R$-module). A largest unimodular vector set of $V$ is a unimodular vector subset of $V$ which has maximum number of vectors. The dimension of $V$, denoted by ${\rm dim}(V)$, is the number of vectors in a largest unimodular vector set of $V$.

A linear subset $X$ of $R^{n}$ is called a $k$-dimensional vector subspace ($k$-subspace or subspace for short) of $R^{n}$, if $X$ has a basis $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{k}\}$ being unimodular. Every basis of a subspace can be extended to a basis of $R^{n}$ (cf. [19, Corollary I.5]). We define the [math]-subspace to be $\{0\}$.

Let $\mathbb{Z}_{p^{s}}$ denote the residue class ring of integers modulo $p^{s}$, where $p$ is a prime and $s$ is a positive integer. The $\mathbb{Z}_{p^{s}}$ is a Galois ring, a commutative local ring, a finite principal ideal ring (cf. [18, 23]), and a Hermite ring (according to Cohn’s definition [4, 5]). The principal ideal $(p)$ is the unique maximal ideal of $\mathbb{Z}_{p^{s}}$, and denoted by $J_{p^{s}}$. The $J_{p^{s}}$ is also the Jacbson radical [1, 4] of $\mathbb{Z}_{p^{s}}$. When $s=1$, $\mathbb{Z}_{p}$ is a finite field. For any $x\in\mathbb{Z}_{p^{s}}$, $x$ is invertible (i.e., a unit) if and only if $x\notin J_{p^{s}}$. We have (cf. [18, 23]) that

[TABLE]

The residue class ring $\mathbb{Z}_{p^{s}}$ plays an important role in mathematics and information science. However, since there are zero divisors, the properties of the vector spaces over $\mathbb{Z}_{p^{s}}$ are essentially different from the vector spaces over a finite field. For example, the intersection of two vector subspaces over $\mathbb{Z}_{p^{s}}$ may not be a vector subspace (cf. Section 3 below). This brings difficulty to study of geometry and graph theory on $\mathbb{Z}_{p^{s}}$. By using matrix representation and the inner rank [5] of a matrix, we will discuss the basic properties of vector spaces over $\mathbb{Z}_{p^{s}}$. For example, intersection, join, dimension formula and dual subspaces of $\mathbb{Z}_{p^{s}}^{n}$.

Recently, some scholars studied some graphs (for instance, symplectic graphs and bilinear forms graph) on $\mathbb{Z}_{p^{s}}$ or a finite commutative ring (cf. [10, 14, 15, 16, 20]). Note that the Grassmann graph over a finite field plays an important role in geometry [22, 24, 25], graph theory [3] and coding theory [7, 8, 22]. Thus, the study of Grassmann graph over $\mathbb{Z}_{p^{s}}$ has good significance for geometry, combinatorics and coding theory.

We define the Grassmann graph over $\mathbb{Z}_{p^{s}}$ as follows. Suppose $m,n$ are integers with $1\leq m<n$. The Grassmann graph over $\mathbb{Z}_{p^{s}}$, denoted by $G_{p^{s}}(n,m)$, has vertex set the set all $m$-subspaces of $\mathbb{Z}_{p^{s}}^{n}$, and two vertices are adjacent if their intersection is a linear subset of dimension $m-1$. Based on our results on vector subspaces of $\mathbb{Z}_{p^{s}}^{n}$, we will study Grassmann graph over $\mathbb{Z}_{p^{s}}$ and its automorphisms.

The paper is organized as follows. In Section 2, we recall the basic properties of matrices over $\mathbb{Z}_{p^{s}}$ and the bilinear forms graphs over $\mathbb{Z}_{p^{s}}$. In Section 3, we will discuss vector subspaces of $\mathbb{Z}_{p^{s}}^{n}$. For example, intersection, join and dimension formula of two vector subspaces of $\mathbb{Z}_{p^{s}}^{n}$; dual subspaces and arithmetic distances of subspaces. In Section 5, we show that the Grassmann graph $G_{p^{s}}(n,m)$ is a connected vertex-transitive graph, and determine its valency, clique number and maximum cliques. In Section 6, we characterize the automorphisms of $G_{p^{s}}(n,m)$.

2 Matrices and bilinear forms graphs over $\mathbb{Z}_{p^{s}}$

In this section, we recall some basic properties of matrices over $\mathbb{Z}_{p^{s}}$. For instance, matrix factorization, the inner rank and McCoy rank of matrix over $\mathbb{Z}_{p^{s}}$. We also introduce the bilinear forms graph over $\mathbb{Z}_{p^{s}}$.

2.1 Matrices

Let $GL_{n}(R)$ be the set of $n\times n$ invertible matrices over $R$. For $A\in R^{m\times n}$, let ${}^{t}A$ denote the transpose matrix of a matrix $A$ and ${\rm det}(X)$ the determinant [2] of a square matrix $X$ over $R$. Let $0_{m,n}$ ([math] for short) be the $m\times n$ zero matrix. Denote by $E^{m\times n}_{ij}$ ($E_{ij}$ for short) the $m\times n$ matrix whose $(i,j)$-entry is $1$ and all other entries are [math]’s. Let ${\rm diag}(A_{1},\ldots,A_{k})$ denote a block diagonal matrix where $A_{i}$ is an $m_{i}\times n_{i}$ matrix.

For $0\neq A\in R^{m\times n}$, by Cohn’s definition [5], the inner rank $\rho(A)$ of $A$, is the least integer $r$ such that

[TABLE]

Let $\rho(0)=0$. Any factorization such that as (2.1) where $r=\rho(A)$ is called a minimal factorization of $A$. For $A\in R^{m\times n}$, it is clear that $\rho(A)\leq\min\left\{m,n\right\}$ and $\rho(A)=0$ if and only if $A=0$. When $R$ is a field, we have $\rho(A)={\rm rank}(A)$, where ${\rm rank}(A)$ is the usual rank of matrix $A$ over a field.

For matrices over $R$, the followings hold (cf. [5, 4, Section 5.4]):

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Denote by $I_{k}(A)$ the ideal in $R$ generated by all $k\times k$ minors of $A$, $k=1,\ldots,{\rm min}\{m,n\}$. Let ${\rm Ann}_{R}(I_{k}(A))=\left\{x\in R:xI_{k}(A)=0\right\}$ denote the annihilator of $I_{k}(A)$. The McCoy rank of $A$, denoted by ${\rm rk}(A)$, is the following integer:

[TABLE]

We have that ${\rm rk}(A)={\rm rk}(^{t}A)$; ${\rm rk}(A)={\rm rk}(PAQ)$ where $P$ and $Q$ are invertible matrices of the appropriate sizes; and ${\rm rk}(A)=0$ if and only if ${\rm Ann}_{R}(I_{1}(A))\neq(0)$ (cf. [2, 17]).

Since $\mathbb{Z}_{p^{s}}$ is an Hermite ring (according to Cohn’s definition [4]), every matrix over $\mathbb{Z}_{p^{s}}$ which has right inverse is completable (i.e., it can be extended to an invertible matrix over $\mathbb{Z}_{p^{s}}$). Since $\mathbb{Z}_{p^{s}}$ is a commutative local ring, a matrix \scriptsize\left(\begin{array}[]{c}\alpha_{1}\\ \vdots\\ \alpha_{m}\\ \end{array}\right)\in\mathbb{Z}_{p^{s}}^{m\times n} has a right inverse if and only if $\alpha_{1},\ldots,\alpha_{m}$ are $m$ linearly independent vectors over $\mathbb{Z}_{p^{s}}$.

Lemma 2.1

(cf. [21, Chap. II]), or [18, p.327], or [14, Lemma 2.3])*  Let $R=\mathbb{Z}_{p^{s}}$, and let $A\in R^{m\times n}$ be a non-zero matrix. Then there are $P\in GL_{m}(R)$ and $Q\in GL_{n}(R)$ such that*

[TABLE]

where $1\leq k_{1}\leq\cdots\leq k_{t}\leq{\rm max}\{s-1,1\}$. Moreover, the parameters $(r,t,k_{1},\ldots,k_{t})$ are uniquely determined by $A$. In (2.6), $I_{r}$ or ${\rm diag}\left(p^{k_{1}},\ldots,p^{k_{t}}\right)$ may be absent.

Lemma 2.2

(see [14, Lemmas 2.4 and 2.7])*  Let $0\neq A\in\mathbb{Z}_{p^{s}}^{m\times n}$ ($s\geq 2$) be of the form (2.6). Then $r+t$ is the inner rank of $A$, and $r$ is the McCoy rank of $A$.*

Let $A\in\mathbb{Z}_{p^{s}}^{m\times n}$ ($s\geq 2$). By Lemma 2.2, ${\rm rk}(A)\leq\rho(A)$, and ${\rm rk}(A)=0$ if and only if $A\in J_{p^{s}}^{m\times n}$. For matrices $A,B,C$ over $\mathbb{Z}_{p^{s}}$, By Lemmas 2.1 and 2.2, it is easy to see that (cf. [14])

[TABLE]

[TABLE]

Lemma 2.3

(see [2, Corollary 2.21])*  Let $A\in\mathbb{Z}_{p^{s}}^{n\times n}$. Then $A\in GL_{n}(\mathbb{Z}_{p^{s}})$ if and only if $\det(A)\in\mathbb{Z}_{p^{s}}^{*}$.*

Note that if $a\in\mathbb{Z}_{p^{s}}^{*}$ and $b\in J_{p^{s}}$, then $a\pm b\in\mathbb{Z}_{p^{s}}^{*}$. Lemma 2.1 implies the following lemma.

Lemma 2.4

(cf. [14, Lemma 2.10])*  Let $A\in\mathbb{Z}_{p^{s}}^{m\times n}$ where $s\geq 2$ and $n\geq m$. Then ${\rm rk}(A)={\rm rk}(A\pm B)$ for all $B\in J_{p^{s}}^{m\times n}$. Moreover, $A$ has a right inverse if and only if ${\rm rk}(A)=m$.*

Let $T_{p}=\{0,1,\ldots,p-1\}\subseteq\mathbb{Z}_{p^{s}}$. For two distinct elements $a,b\in T_{p}$, we have $a-b\in\mathbb{Z}_{p^{s}}^{*}$. Without loss of generality, we may assume that $T_{p}=\mathbb{Z}_{p}$ in our discussion.

Lemma 2.5

(see [1, Proposition 6.2.2] or [18, p.328])*  Every non-zero element $x$ in $\mathbb{Z}_{p^{s}}$ can be written as $x=up^{t}$ where $u$ is a unit and $0\leq t\leq s-1$. Moreover, the integer $t$ is unique and $u$ is unique modulo the ideal $(p^{s-t})$ of $\mathbb{Z}_{p^{s}}$.*

Lemma 2.6

(cf. [18, p.328])*   Every non-zero element $x$ in $\mathbb{Z}_{p^{s}}$ can be written uniquely as*

[TABLE]

By Lemma 2.6, every matrix $X\in\mathbb{Z}_{p^{s}}^{m\times n}$ can be written uniquely as

[TABLE]

Note that every matrix in $T_{p}^{m\times n}$ can be seen as a matrix in $\mathbb{Z}_{p}^{m\times n}$. We define the natural surjection

[TABLE]

by $\pi(X)=X_{0}$ for all $X\in\mathbb{Z}_{p^{s}}^{m\times n}$ of the form (2.9). Clearly, $\pi(A)=A$ if $A\in\mathbb{Z}_{p}^{m\times n}$. For $X,Y\in\mathbb{Z}_{p^{s}}^{m\times n}$ and $Q\in\mathbb{Z}_{p^{s}}^{n\times k}$, We have

[TABLE]

[TABLE]

[TABLE]

Theorem 2.7

Let $X\in\mathbb{Z}_{p^{s}}^{m\times n}$ $(n\geq m$). Then the following hold:

• (i)

${\rm rk}(X)={\rm rank}(\pi(X))$.

• (ii)

$X$* has a right inverse if and only if $\pi(X)$ has a right inverse.*

• (iii)

When $n=m$, $X$ is invertible if and only if $\pi(X)$ is invertible. Moreover, if $X$ is invertible, then

[TABLE]

*Proof. *Without loss of generality, we assume that $s\geq 2$.

(i).  Let ${\rm rk}(X)=r$. By Lemma 2.1, there are matrices $P\in GL_{m}(\mathbb{Z}_{p^{s}})$ and $Q\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that

[TABLE]

where $D={\rm diag}\left(p^{k_{1}},\ldots,p^{k_{t}}\right)$, $1\leq k_{1}\leq\cdots\leq k_{t}\leq s-1$. Using (2.12), we have that $\pi(P)$ and $\pi(Q)$ are invertible, and $\pi(X)=\pi(P){\rm diag}\left(I_{r},0_{t},0_{m-r-t,n-r-t}\right)\pi(Q)$. Thus ${\rm rank}(\pi(X))=r={\rm rk}(X)$.

(ii).  If $X$ has a right inverse, then (2.12) implies that $\pi(X)$ has a right inverse. Conversely, suppose $\pi(X)$ has a right inverse. Using (i), we get ${\rm rank}(\pi(X))=m={\rm rk}(X)$. If follows from Lemma 2.4 that $X$ has a right inverse.

(iii).  Suppose that $n=m$. Similar to the proof of (ii), $X$ is invertible if and only if $\pi(X)$ is invertible. Now, let $X=X_{0}+X_{1}p+\cdots+X_{s-1}p^{s-1}\in\mathbb{Z}_{p^{s}}^{n\times n}$ be invertible, where $X_{i}\in\mathbb{Z}_{p}^{n\times n}$, $i=0,\ldots,s-1$. Write $X^{-1}=Y_{0}+Y_{1}p+\cdots+Y_{s-1}p^{s-1}\in\mathbb{Z}_{p^{s}}^{n\times n}$, where $Y_{i}\in\mathbb{Z}_{p}^{n\times n}$, $i=0,\ldots,s-1$. Then $X_{0}Y_{0}=I_{n}$, and hence (2.14) holds. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Lemma 2.8

(see [14, Lemma 4.2])*   If $A\in GL_{n}(\mathbb{Z}_{p^{s-1}})$ where $s\geq 2$, then $A\in GL_{n}(\mathbb{Z}_{p^{s}})$.*

Lemma 2.9

(see [14, Lemma 4.3])*   If $A\in\mathbb{Z}_{p^{s-1}}^{m\times n}$ where $s\geq 2$, then both $A$ and $Ap$ can be viewed as matrices in $\mathbb{Z}_{p^{s}}^{n\times n}$ with the same inner rank.*

2.2 Bilinear forms graphs

All graphs in this paper are simple [9] and finite. Suppose $G$ is a graph. Denote the distance between vertices $x$ and $y$ in $G$ by $d_{G}(x,y)$ ($d(x,y)$ for short). Let $V(G)$ be the vertex set of $G$. For $x,y\in V(G)$, we write $x\sim y$ if vertices $x$ and $y$ are adjacent.

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

An independent set of $G$ is a subset of vertices such that no two vertices are adjacent. A largest independent set of $G$ is an independent set of maximum cardinality. The independence number $\alpha(G)$ is the number of vertices in a largest independent set of $G$.

As a natural extension of the bilinear forms graph over a finite field, the bilinear forms graph over $\mathbb{Z}_{p^{s}}$ was defined as follows (cf. [14]). The bilinear forms graph over $\mathbb{Z}_{p^{s}}$, denoted by $\Gamma(\mathbb{Z}_{p^{s}}^{m\times n})$, has the vertex set $\mathbb{Z}_{p^{s}}^{m\times n}$ where $m,n\geq 2$, and two vertices $A$ and $B$ are adjacent if $\rho(A-B)=1$.

In $\mathbb{Z}_{p^{s}}^{m\times n}$, let

[TABLE]

A maximal clique $\mathcal{C}$ of $\mathbb{Z}_{p^{s}}^{m\times n}$ is called of type one (resp. type two), if $\mathcal{C}$ is of the form $\mathcal{C}=P\mathcal{M}_{1}+A$ (resp. $\mathcal{C}=\mathcal{N}_{1}Q+A$), where $P$ and $Q$ are fixed invertible matrices over $\mathbb{Z}_{p^{s}}$ and $A\in\mathbb{Z}_{p^{s}}^{m\times n}$ is fixed.

Lemma 2.10

(see [14, Lemma 3.4])*  Let $\mathcal{C}$ be a maximal clique of $\Gamma(\mathbb{Z}_{p^{s}}^{m\times n})$ ($s\geq 2$) containing the vertex [math]. Then $\mathcal{C}\nsubseteq J_{p^{s}}^{m\times n}$.*

Lemma 2.11

(see [14, Theorem 3.6])*  Let $\Gamma=\Gamma(\mathbb{Z}_{p^{s}}^{m\times n})$ and $k={\rm max}\{m,n\}$. Then*

[TABLE]

Moreover, if $n>m$, then every maximum clique of $\Gamma$ is of type one. If $m=n$, then every maximum clique of $\Gamma$ is of type one or type two.

By [14, Theorem 4.4], if $2\leq m\leq n$, then

[TABLE]

3 Subspaces of $\mathbb{Z}_{p^{s}}^{n}$

In this section, we study vector subspaces of $\mathbb{Z}_{p^{s}}^{n}$. For example, intersection, join and dimension formula of two subspaces of $\mathbb{Z}_{p^{s}}^{n}$. These results have many applications in geometry and combinatorics.

In general, a linear subset $X$ in $R^{n}$ may not be a subspace, although it has a dimension. For example, let $R=\mathbb{Z}_{p^{s}}$, and let $X\subseteq J_{p^{s}}^{n}$ be a linear subset with $X\neq\{0\}$. Then ${\rm dim}(X)=0$ because $p^{s-1}x=0$ for all $x\in X$. Thus, $X$ is not a subspace of $R^{n}$. On the other hand, if $X\subseteq\mathbb{Z}_{p^{s}}^{n}$ is a linear subset with ${\rm dim}(X)=0$, we cannot imply $X=\{0\}$ (i.e., a [math]-subspace).

Let $X=[\alpha_{1},\ldots,\alpha_{k}]$ be a $k$-subspace of $\mathbb{Z}_{p^{s}}^{n}$. Then $X$ has a matrix representation \scriptsize\left(\begin{array}[]{c}\alpha_{1}\\ \vdots\\ \alpha_{k}\\ \end{array}\right)\in\mathbb{Z}_{p^{s}}^{k\times n} (cf. [11, 12]). For simpleness, the matrix representation of $X$ is also written as $X$. If $X$ is a matrix representation of the subspace $X$, then for any invertible matrix $P\in GL_{k}(\mathbb{Z}_{p^{s}})$, $PX$ is also a matrix representation of $X$. Thus, the matrix representation is not unique. However, the subspace $X$ has a unique matrix representation which is the row-reduced echelon form $X=(I_{k},B)Q$, where $Q$ is a permutation matrix.

3.1 Join and dimensional formula of subspaces

In $\mathbb{Z}_{p^{s}}^{n}$, a join of two subspaces $A$ and $B$ is a minimum dimensional subspace containing $A$ and $B$. In general, when $s\geq 2$ the join is not unique. For example, suppose that $s\geq 3$, $\alpha=(1,p,0)$, $\beta=(1,p^{2},0)$, $\gamma=(0,1,p^{s-1})$, $e_{1}=(1,0,0)$ and $e_{2}=(0,1,0)$. Then $[e_{1},e_{2}]$ and $[e_{1},\gamma]$ are two distinct joins of $[\alpha]$ and $[\beta]$. Denoted by $A\vee B$ the set of all joins of subspaces $A$ and $B$ in $\mathbb{Z}_{p^{s}}^{n}$ with the same minimum dimension $\dim(A\vee B)$. We have $A\vee B=B\vee A$. When $B\subseteq A$ are two subspaces, we have $A\vee B=\{A\}$.

Let $X,Y$ be two subspaces of $\mathbb{Z}_{p^{s}}^{n}$. In general, $X\cap Y$ is a linear subset but it may not be a subspace of $\mathbb{Z}_{p^{s}}^{n}$. For example, let \scriptsize X=\left(\begin{array}[]{cccc}1&0&0&0\\ 0&1&0&0\\ \end{array}\right), \scriptsize Y=\left(\begin{array}[]{cccc}1&0&0&0\\ 0&1&p^{s-1}&0\\ \end{array}\right) ($s\geq 2$) are $2$-subspaces of $\mathbb{Z}_{p^{s}}^{4}$. Then $(1,p^{i},0,0)\in X\cap Y$, $i=0,1,\ldots,s-1$. It follows that $X\cap Y$ is not a free $\mathbb{Z}_{p^{s}}$-module, and hence $X\cap Y$ is not a subspace of $\mathbb{Z}_{p^{s}}^{4}$.

Theorem 3.1

Let $n>k\geq m\geq 1$. Suppose $A$ and $B$ are $k$-subspace and $m$-subspace of $\mathbb{Z}_{p^{s}}^{n}$, respectively, and $B\nsubseteq A$. Then there is $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that

[TABLE]

where $D={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}},0_{m-r,n-m-r})$, $1\leq r\leq{\rm min}\{m,n-k\}$ and $0\leq i_{1}\leq\ldots\leq i_{r}\leq{\rm max}\{s-1,1\}$.

*Proof. *Without loss of generality, we assume that $s\geq 2$. By the matrix representation of subspace and Lemma 2.1, there is $U_{1}\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,I_{k})U_{1}$. Without loss of generality, we assume $U_{1}=I_{n}$ and hence $A=(0,I_{k})$. Write $B=(B_{1},B^{\prime}_{2})$ where $B_{1}\in\mathbb{Z}_{p^{s}}^{m\times(n-k)}$ and $B^{\prime}_{2}\in\mathbb{Z}_{p^{s}}^{m\times k}$. When $k>m$, by elementary operations of matrix, we may assume with no loss of generality that $B_{2}^{\prime}=(0_{m,k-m},B_{2})$ where $B_{2}\in\mathbb{Z}_{p^{s}}^{m\times m}$. Thus $B=(B_{1},0_{m,k-m},B_{2})$. Using Lemma 2.1, we may assume with no loss of generality that

[TABLE]

where $1\leq r\leq{\rm min}\{m,n-k\}$ and $0\leq i_{1}\leq\ldots\leq i_{r}\leq s-1$.

Case 1.  $B_{2}$ is invertible.

Let $U_{2}={\rm diag}(I_{n-m},B_{2}^{-1})$. Then $A=(0,I_{k})=(0,I_{k})U_{2}$ and $B=(B_{1},0_{m,k-m},I_{m})U_{2}$. Thus (3.1) holds.

Case 2.  $B_{2}$ is not invertible.

Without loss of generality, we let ${\rm diag}(p^{i_{1}},\ldots,p^{i_{r}})={\rm diag}(I_{t},D_{1})$, where $D_{1}={\rm diag}(p^{i_{t+1}},\ldots,p^{i_{r}})$, $1\leq i_{t+1}\leq\ldots\leq i_{r}\leq s-1$. Write \scriptsize B_{2}=\left(\begin{array}[]{c}B_{21}\\ B_{22}\\ \end{array}\right)\in\mathbb{Z}_{p^{s}}^{m\times m} where $B_{21}\in\mathbb{Z}_{p^{s}}^{r\times m}$ and $B_{22}\in\mathbb{Z}_{p^{s}}^{(m-r)\times m}$. Then $B_{22}$ has a right inverse. Without loss of generality, we may assume that $B_{22}=(I_{m-r},0)$. Then $A=(0,I_{k})$ and

[TABLE]

where $D_{2}$ is an $(r-t)\times(n-m-t)$ diagonal matrix over $J_{p^{s}}$, $Y_{1}\in\mathbb{Z}_{p^{s}}^{t\times r}$ and $Y_{2}\in\mathbb{Z}_{p^{s}}^{(r-t)\times r}$. Since $(0,D_{2},0,Y_{2})$ has a right inverse, it is easy to see that $Y_{2}$ has a right inverse. Hence there exists $C_{1}\in\mathbb{Z}_{p^{s}}^{t\times r}$ such that \scriptsize\left(\begin{array}[]{c}C_{1}+Y_{1}\\ Y_{2}\\ \end{array}\right) is invertible. Put

[TABLE]

Then $A=AU_{3}$ and

[TABLE]

Since \scriptsize B_{2}^{\prime\prime}:=\left(\begin{array}[]{cc}0&C_{1}+Y_{1}\\ 0&Y_{2}\\ I_{m-r}&0\\ \end{array}\right) is invertible, Case 1 implies that (3.1) holds. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Theorem 3.2

(dimensional formula)*  Let $A$ and $B$ be two subspaces of $\mathbb{Z}_{p^{s}}^{n}$. Then*

[TABLE]

*Proof. *Let ${\rm dim}(A)=k$ and ${\rm dim}(B)=m$. Without loss of generality, we assume that $s\geq 2$, $n>k\geq m\geq 1$ and $B\nsubseteq A$. By Theorem 3.1, there is $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,I_{k})U$  and  $B=(D,I_{m})U$, where $D={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}},0_{m-r,n-m-r})$, $1\leq r\leq{\rm min}\{m,n-k\}$ and $0\leq i_{1}\leq\ldots\leq i_{r}\leq s-1$. Put $D_{1}={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}})$. Without loss of generality, we may assume that $U=I_{n}$ and $m\geq 2$. Hence we can assume further that

[TABLE]

In above matrices, some zero elements of matrices may be absent.

Clearly, the $(m-r)$-subspace $(0,0,0,I_{m-r})$ is contained in $A\cap B$. On the other hand, the $(k+r)$-subspace \small\left(\begin{array}[]{ccccc}I_{r}&0_{r,n-k-r}&0\\ 0&0&I_{k}\\ \end{array}\right) contains $A$ and $B$. It follows that

[TABLE]

Let $\alpha\in A\cap B$ be an $n$-dimensional vector. Then there are matrices $T_{1}\in\mathbb{Z}_{p^{s}}^{1\times k}$ and $T_{2}=(T_{21},T_{22})\in\mathbb{Z}_{p^{s}}^{1\times m}$, where $T_{21}\in\mathbb{Z}_{p^{s}}^{1\times r}$ and $T_{22}\in\mathbb{Z}_{p^{s}}^{1\times(m-r)}$, such that $\alpha=T_{1}A=(0,T_{1})$, $\alpha=T_{2}B=(T_{21}D_{1},0,T_{21},T_{22})$. Hence $T_{21}D_{1}=0$ and $\alpha=(0,0,T_{21},T_{22})$. By $T_{21}D_{1}=0$ we get $T_{21}\in J_{p^{s}}^{1\times r}$. Thus,

[TABLE]

where $t_{1},\ldots,t_{r}\in J_{p^{s}}$. It follows that ${\rm dim}(A\cap B)\leq m-r$. Hence, from (3.3) we obtain

[TABLE]

Assume that $d={\rm dim}(A\vee B)$. Then there exists a $d$-subspace $W\in A\vee B$. Since $B\nsubseteq A$, one has $d>k$. Thus \scriptsize W=\left(\begin{array}[]{c}W_{1}\\ A\\ \end{array}\right)=\left(\begin{array}[]{cc}W_{11}&0\\ 0&I_{k}\\ \end{array}\right), where $W_{11}\in\mathbb{Z}_{p^{s}}^{(d-k)\times(n-k)}$ has a right inverse. Since $B\subset W$, the $r$-subspace $B_{1}:=(D_{1},0_{r,n-m-r},I_{r},0)\subset W$. Thus, there is a matrix $P=(P_{1},P_{2})\in\mathbb{Z}_{p^{s}}^{r\times d}$ where $P_{1}\in\mathbb{Z}_{p^{s}}^{r\times(d-k)}$ and $P_{2}\in\mathbb{Z}_{p^{s}}^{r\times k}$, such that $B_{1}=PW$ and hence $(D_{1},0_{r,n-m-r},I_{r},0)=(P_{1}W_{11},P_{2})$. It follows from (2.3) that

[TABLE]

Hence $d\geq k+r$. By (3.3) we obtain

[TABLE]

Clearly, we have \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=\rho\left(\begin{array}[]{ccc}0&0&I_{k}\\ D_{1}&0&0\\ 0&0&0\\ \end{array}\right)=k+r. By (3.4) and (3.5), we get (3.2). $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Remark 3.3

Let $A$ and $B$ be two subspaces of $\mathbb{Z}_{p^{s}}^{n}$. By (3.2), $A\vee B$ is the set of subspaces containing $A$ and $B$ with the same dimension \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right).

3.2 On intersection and join of subspaces

Theorem 3.4

Let $A$ and $B$ be two subspaces of $\mathbb{Z}_{p^{s}}^{n}$ and ${\rm dim}(A)+{\rm dim}(B)\leq n$. Then ${\rm dim}(A\cap B)=0$ if and only if \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=\rho(A)+\rho(B). Moreover, $A\cap B=\{0\}$ if and only if \scriptsize{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=\rho(A)+\rho(B).

*Proof. *Let ${\rm dim}(A)=k=\rho(A)$ and ${\rm dim}(B)=m=\rho(B)$. Without loss of generality, we assume that $s\geq 2$, $k\geq m\geq 1$ and $B\nsubseteq A$. Clearly, \scriptsize{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)\leq\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)\leq k+m.

Suppose \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)<k+m. By Theorem 3.1, there is $U_{1}\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,I_{k})U_{1}$ and $B=(D,I_{m})U_{1}$, where $D={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}},0_{m-r,n-m-r})$, $1\leq r<m$ and $0\leq i_{1}\leq\ldots\leq i_{r}\leq s-1$. Thus $\alpha:=(0,\ldots,0,1)U_{1}\in A\cap B$. Since $\alpha$ is unimodular, ${\rm dim}(A\cap B)\geq 1$. Conversely, if ${\rm dim}(A\cap B)\geq 1$, then it is clear that \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)<k+m. Therefore, ${\rm dim}(A\cap B)\geq 1$ if and only if \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)<k+m. It follows that ${\rm dim}(A\cap B)=0$ if and only if \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+m.

If \scriptsize{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+m, then Theorem 3.1 implies that there is $U_{2}\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,0,I_{k})U_{2}$ and $B=(I_{m},0,I_{m})U_{2}$. Thus, it is easy to see that $A\cap B=\{0\}$.

Now, we assume that $A\cap B=\{0\}$. Then ${\rm dim}(A\cap B)=0$, and hence \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+m because (3.2). Suppose \scriptsize{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+t<k+m. By Theorem 3.1, there is $U_{3}\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,0,I_{k})U_{3}$ and $B=(D_{1},0,I_{m})U_{3}$, where $D_{1}={\rm diag}(I_{t},p^{i_{t+1}},\ldots,p^{i_{m}})$, $1\leq i_{t+1}\leq\ldots\leq i_{m}\leq s-1$. Let $\beta=(0,\ldots,0,p^{s-i_{m}})U_{3}$. Then $\beta\neq 0$. By $\beta=p^{s-i_{m}}(0,\ldots,0,p^{i_{m}},0,\ldots,0,1)U_{3}=p^{s-i_{m}}(0,\ldots,0,1)U_{3}$, we get $\beta\in A\cap B$, a contradiction. Hence we must have \scriptsize{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+m. Thus, we have that $A\cap B=\{0\}$ if and only if \scriptsize{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+m. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Theorem 3.5

Let $A$ and $B$ be two subspaces of $\mathbb{Z}_{p^{s}}^{n}$. Then:

• (i)

$A\cap B$* is a fixed subspace of dimension ${\rm dim}(A\cap B)$ if and only if \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right).*

• (ii)

$A\vee B$* is a fixed subspace of dimension ${\rm dim}(A\vee B)$ if and only if \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right) or ${\rm dim}(A\vee B)=n$.*

*Proof. *Put ${\rm dim}(A)=k$ and ${\rm dim}(B)=m$. Without loss of generality, we assume that $s\geq 2$, $n>k\geq m\geq 1$ and $B\nsubseteq A$. By Theorem 3.1, there is $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,I_{k})U$ and $B=(D,I_{m})U$, where $D={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}},0_{m-r,n-m-r})$, $1\leq r\leq{\rm min}\{m,n-k\}$ and $0\leq i_{1}\leq\ldots\leq i_{r}\leq s-1$. Thus \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=k+r and hence $n\geq k+r$. Without loss of generality, we may assume $U=I_{n}$. By Theorem 3.2, we have ${\rm dim}(A\vee B)=k+r$ and ${\rm dim}(A\cap B)=m-r$.

(i).  Suppose \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right). Then $D={\rm diag}(I_{r},0_{m-r,n-m-r})$, \scriptsize B=\left(\begin{array}[]{cccc}I_{r}&0&I_{r}&0\\ 0&0&0&I_{m-r}\\ \end{array}\right) and $A=(0,I_{k})$. Thus $A\cap B$ contains an $(m-r)$-subspace $A_{1}:=(0,0,0,I_{m-r})$, For any vector $x\in A\cap B$, it is easy to verify $x\in A_{1}$. Therefore, $A\cap B=A_{1}$ is a fixed $(m-r)$-subspace.

Conversely, suppose $A\cap B$ is a fixed $(m-r)$-subspace. We show \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right). Otherwise, \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)>{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right), and hence $0\neq p^{i_{r}}\in J_{p^{s}}$ in the matrix $D$. By row elementary operations of matrix, it is easy to see that the $(m-r)$-subspace \small A_{2}:=\left(\begin{array}[]{cccc}0&p^{s-i_{r}}&1&0\\ 0&0&0&I_{m-r-1}\\ \end{array}\right)\subseteq A\cap B. Since the $(m-r)$-subspace $A_{1}=(0,0,0,I_{m-r})\subseteq A\cap B$ and $A_{1}\neq A_{2}$, we get a contradiction.

(ii).   Suppose \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right). Similarly, \scriptsize B=\left(\begin{array}[]{cccc}I_{r}&0&I_{r}&0\\ 0&0&0&I_{m-r}\\ \end{array}\right) and $A=(0,I_{k})$. Clearly, $A\vee B$ contains a $(k+r)$-subspace \scriptsize B_{1}:=\left(\begin{array}[]{ccc}I_{r}&0&0\\ 0&0&I_{k}\\ \end{array}\right). Since a basis of $A$ can be extended to a basis of any $(k+r)$-subspace $C\in A\vee B$, it is easy to prove that every $(k+r)$-subspace containing $A$ and $B$ must be $B_{1}$. Thus $A\vee B=\{B_{1}\}$ is a fixed $(k+r)$-subspace. If ${\rm dim}(A\vee B)=k+r=n$, then ${\rm dim}(A\vee B)=\mathbb{Z}_{p^{s}}^{n}$ and hence $A\vee B$ is fixed.

Conversely, suppose $A\vee B$ is a fixed $(k+r)$-subspace. If $k+r=n$, then (ii) holds. Now, we assume that $k+r<n$. We assert \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right). Otherwise, \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)>{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right), and hence $0\neq p^{i_{r}}\in J_{p^{s}}$ in the matrix $D$. By elementary row operations of matrix, it is easy to see that the $(k+r)$-subspace \scriptsize B_{2}:=\left(\begin{array}[]{ccccc}I_{r-1}&0&0&0&0\\ 0&1&p^{s-1}&0&0\\ 0&0&0&0&I_{k}\\ \end{array}\right) contains $A$ and $B$. Also, the $(k+r)$-subspace \small B_{1}=\left(\begin{array}[]{ccc}I_{r}&0&0\\ 0&0&I_{k}\\ \end{array}\right) contains $A$ and $B$, a contradiction since $B_{1}\neq B_{2}$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

3.3 Enumeration on subspaces

For any non-negative integers $m,n,q$ with $n\geq m$ and $q\geq 2$, the Gaussian binomial coefficient is

[TABLE]

where ${\ n\ \brack 0}_{q}:=1$. Let ${0\brack m}_{q}:=0$ (if $m>0$). We have ${n\brack m}_{q}={n\brack n-m}_{q}$.

Theorem 3.6

Let $n>m>k\geq 1$. Then:

• (i)

The number of $m$-subspaces of $\mathbb{Z}_{p^{s}}^{n}$ is $p^{(s-1)m(n-m)}{n\brack m}_{p}$.

• (ii)

In $\mathbb{Z}_{p^{s}}^{n}$, the number of $k$-subspaces in a given $m$-subspace is $p^{(s-1)k(m-k)}{m\brack k}_{p}$.

• (iii)

In $\mathbb{Z}_{p^{s}}^{n}$, the number of $m$-subspaces containing a given $k$-subspace is $p^{(s-1)(m-k)(n-m)}{n-k\brack m-k}_{p}$.

*Proof. *(i).  Let $n_{m,s}$ be the number of $m$-subspaces of $\mathbb{Z}_{p^{s}}^{n}$. When $s=1$, it is well-known that $n_{m,1}={n\brack m}_{p}$. Now, we assume $s\geq 2$. Let $\pi:\mathbb{Z}_{p^{s}}^{m\times n}\rightarrow\mathbb{Z}_{p}^{m\times n}$ be the natural surjection (2.10). Suppose $S$ is the set of all $m$-subspaces of $\mathbb{Z}_{p^{s}}^{n}$. By Theorem 2.7, $\pi(S)$ is the set of all $m$-subspaces of $\mathbb{Z}_{p}^{n}$. Thus $|\pi(S)|={n\brack m}_{p}$. Let $A$ be any fixed $m$-subspace in $\pi(S)$. By Theorem 3.1, there is $U_{A}\in GL_{n}(\mathbb{Z}_{p})$ such that $A=(0,I_{m})U_{A}$. It is easy to see that the preimages $\pi^{-1}(A)=\left\{(X,I_{m})U_{A}:X\in J_{p^{s}}^{m\times(n-m)}\right\}$. By $|\pi^{-1}(A)|=p^{(s-1)m(n-m)}$, we obtain $n_{m,s}=p^{(s-1)m(n-m)}{n\brack m}_{p}$.

(ii).  By (i), it is clear.

(iii).  Let $S$ be the set of $m$-subspaces in $\mathbb{Z}_{p^{s}}^{n}$ containing a given $k$-subspace $B$ in $\mathbb{Z}_{p^{s}}^{n}$. By Theorem 3.1, we can assume that $B=(0,I_{k})U$ where $U\in GL_{n}(\mathbb{Z}_{p^{s}})$. Then

[TABLE]

Using (i), we obtain $|S|=p^{(s-1)(m-k)(n-m)}{n-k\brack m-k}_{p}$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

3.4 Dual subspace and arithmetic distance

Now, we study dual subspace and arithmetic distance on vector subspace of $\mathbb{Z}_{p^{s}}^{n}$. The dual subspace is an important tool that can simplify some complex calculations.

Let $P$ be an $m$-subspace of $\mathbb{Z}_{p^{s}}^{n}$, and let

[TABLE]

The $P^{\perp}$ is the set of vectors which are orthogonal to every vector of $P$. Obviously, $P^{\perp}$ is a linear subset of $\mathbb{Z}_{p^{s}}^{n}$. By Theorem 3.1, we can assume that $P=(0,I_{m})U$, where $U\in GL_{n}(\mathbb{Z}_{p^{s}})$. Thus, it is easy to prove that

[TABLE]

Hence $P^{\perp}$ is an $(n-m)$-subspace of $\mathbb{Z}_{p^{s}}^{n}$. The subspace $P^{\perp}$ is called the dual subspace of $P$. We have

[TABLE]

[TABLE]

If $P_{1}\subseteq P_{2}$ are two subspaces of $\mathbb{Z}_{p^{s}}^{n}$, then the definition of $P^{\perp}$ implies that $P_{2}^{\perp}\subseteq P_{1}^{\perp}$. It follows from (3.8) that

[TABLE]

Lemma 3.7

Let $A$ be an $m$-subspace of $\mathbb{Z}_{p^{s}}^{n}$, and let $\pi:\mathbb{Z}_{p^{s}}^{m\times n}\rightarrow\mathbb{Z}_{p}^{m\times n}$ be the natural surjection (2.10). Then

[TABLE]

*Proof. *By Theorem 3.1, there is $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,I_{m})U$. From (3.6) we have $A^{\perp}=(I_{n-m},0)\,^{t}U^{-1}$. Let $U=U_{0}+U_{1}p+\cdots+U_{s-1}p^{s-1}$ and $U^{-1}=Y_{0}+Y_{1}p+\cdots+Y_{s-1}p^{s-1}$, where $U_{i},Y_{i}\in\mathbb{Z}_{p}^{m\times n}$, $i=0,\ldots,s-1$. It is clear that $Y_{0}=U_{0}^{-1}$. Then ${}^{t}U^{-1}=\,^{t}U_{0}^{-1}+\,^{t}Y_{1}p+\cdots+\,^{t}Y_{s-1}p^{s-1}$. Using (2.12), we obtain that $\pi(A)=(0,I_{m})U_{0}$ and $\pi(A^{\perp})=(I_{n-m},0)\,^{t}U_{0}^{-1}$. Thus (3.10) holds. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

For two subspaces $A,B$ of $\mathbb{Z}_{p^{s}}^{n}$, the arithmetic distance between $A$ and $B$, denoted by ${\rm ad}(A,B)$, is defined by

[TABLE]

Clearly, ${\rm ad}(A,B)\geq 0$ and ${\rm ad}(A,B)={\rm ad}(B,A)$. Moreover, ${\rm ad}(A,B)=0$ $\Leftrightarrow$ $A\subseteq B$ or $B\subseteq A$.

Theorem 3.8

Let $A$ and $B$ be two subspaces of $\mathbb{Z}_{p^{s}}^{n}$. Suppose that $C$ is a subspace of $\mathbb{Z}_{p^{s}}^{n}$ with ${\rm dim}(C)={\rm min}\left\{{\rm dim}(A),{\rm dim}(B)\right\}$. Then

[TABLE]

*Proof. *Without loss of generality, we assume that ${\rm dim}(A)={\rm max}\left\{{\rm dim}(A),{\rm dim}(B)\right\}$. Thus ${\rm dim}(B)={\rm min}\left\{{\rm dim}(A),{\rm dim}(B)\right\}$. By the conditions, ${\rm dim}(C)={\rm dim}(B)$. By (3.2) and (3.11), we have that ${\rm ad}(A,B)={\rm dim}(A\vee B)-{\rm dim}(A)={\rm dim}(B)-{\rm dim}(A\cap B)$, ${\rm ad}(A,C)={\rm dim}(A\vee C)-{\rm dim}(A)={\rm dim}(C)-{\rm dim}(A\cap C)$, and ${\rm ad}(C,B)={\rm dim}(C\vee B)-{\rm dim}(B)={\rm dim}(C)-{\rm dim}(C\cap B)$. Thus, the inequality (3.12) is equivalent to the following inequality:

[TABLE]

Write $d_{1}={\rm dim}(A\cap C)$ and $d_{2}={\rm dim}(C\cap B)$. Let $A_{1}$ be a $d_{1}$-subspace in $A\cap C$ and $B_{1}$ a $d_{2}$-subspace in $C\cap B$. Then $A_{1}\cap B_{1}\subseteq A\cap B$ and hence ${\rm dim}(A_{1}\cap B_{1})\leq{\rm dim}(A\cap B)$. Since $A_{1},B_{1}\subseteq C$, ${\rm dim}(C)\geq{\rm dim}(A_{1}\vee B_{1})$. Using (3.2) we get ${\rm dim}(A_{1}\vee B_{1})=d_{1}+d_{2}-{\rm dim}(A_{1}\cap B_{1})$. Thus,

[TABLE]

Therefore, we have proved the inequality (3.13). $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Theorem 3.9

Let $n>m$ and let $A$ and $B$ be two $m$-subspaces of $\mathbb{Z}_{p^{s}}^{n}$. Then

[TABLE]

*Proof. *Assume that ${\rm ad}(A,B)=r$. By (3.11) and Theorem 3.1, without loss of generality, we assume that $A=(0,I_{m})$ and $B=(D,I_{m})$, where $D={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}},0_{m-r,n-m-r})$, $1\leq r\leq{\rm min}\{m,n-m\}$ and $0\leq i_{1}\leq\ldots\leq i_{r}\leq{\rm max}\{s-1,1\}$.

By (3.7), both $A^{\perp}$ and $B^{\perp}$ are $(n-m)$-subspaces of $\mathbb{Z}_{p^{s}}^{n}$. Applying (3.6), $A^{\perp}=(I_{n-m},0)$. Let $B^{\perp}=(X_{1},X_{2})$ where $X_{1}\in\mathbb{Z}_{p^{s}}^{(n-m)\times(n-m)}$ and $X_{2}\in\mathbb{Z}_{p^{s}}^{(n-m)\times m}$. Since $B^{\perp}\cdot\,^{t}B=0$, we have $X_{1}\,^{t}D+X_{2}=0$ and hence $\rho(X_{2})=\rho(-X_{1}\,^{t}D)\leq r$ because (2.3). Thus

[TABLE]

On the other hand, using (3.8) we have similarly ${\rm ad}(A,B)={\rm ad}\left((A^{\perp})^{\perp},(A^{\perp})^{\perp})\right)\leq{\rm ad}(A^{\perp},B^{\perp})$. Therefore (3.14) holds. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

4 Grassmann graphs over $\mathbb{Z}_{p^{s}}$

In this section, we discuss the basic properties of Grassmann graph over $\mathbb{Z}_{p^{s}}$. We will determine the valency of every vertex, the clique number and maximum cliques of the Grassmann graph.

4.1 Valency, clique number and independence number

For any two vertices $A,B$ of the Grassmann graph $G_{p^{s}}(n,m)$, by Theorem 3.2 and (3.11), we have

[TABLE]

By (3.11), for vertices $A,B,C$ of $G_{p^{s}}(n,m)$, \scriptsize{\rm ad}(A,B)=\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)-m is the arithmetic distance between $A$ and $B$ in $G_{p^{s}}(n,m)$. Clearly, ${\rm ad}(A,B)\geq 0$, ${\rm ad}(A,B)=0\Leftrightarrow A=B$; ${\rm ad}(A,B)={\rm ad}(B,A)$; ${\rm ad}(A,B)\leq{\rm ad}(A,C)+{\rm ad}(C,B)$ by Theorem 3.8.

Theorem 4.1

If $A$ and $B$ are two vertices of $G_{p^{s}}(n,m)$, then

[TABLE]

*Proof. *Without loss of generality, we assume that $s\geq 2$. Let $A,B\in V(G_{p^{s}}(n,m))$ and ${\rm ad}(A,B)=r$, where $1\leq r\leq{\rm min}\{m,n-m\}$. By Theorem 3.1, we may assume with no loss of generality that $A=(0,I_{m})$ and $B=(D,I_{m})$, where $D={\rm diag}(p^{i_{1}},\ldots,p^{i_{r}},0_{m-r,n-m-r})$, and $0\leq i_{1}\leq\ldots\leq i_{r}\leq s-1$. Let $A_{j}=(D_{j},I_{m})$ where $D_{j}={\rm diag}(p^{i_{1}},\ldots,p^{i_{j}},0_{m-j,n-m-j})$, $j=1,\ldots,r-1$. Then $A\sim A_{1}\sim\cdots\sim A_{r-1}\sim B$. Hence $d(A,B)\leq{\rm ad}(A,B)$. Applying the triangle inequality, it is easy to prove that $d(A,B)\geq{\rm ad}(A,B)$. Thus (4.2) holds. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

By Theorem 3.6(i), we have

[TABLE]

By Theorem 3.9, (4.1) and (3.6), it is easy to see that $G_{p^{s}}(n,m)$ is isomorphic to $G_{p^{s}}(n,n-m)$. When $m=1$, $G_{p^{s}}(n,1)$ a complete graph. Thus, we always assume that $n\geq 2m\geq 4$ in our discussion on $G_{p^{s}}(n,m)$ unless specified otherwise.

Theorem 4.2

The $G_{p^{s}}(n,m)$ is a connected vertex-transitive graph of the valency

[TABLE]

*Proof. *Let $A=(0,I_{m})$ be a fixed vertex of of $G_{p^{s}}(n,m)$. For any vertex $Y$ of $G_{p^{s}}(n,m)$, Theorem 3.1 implies that $Y=(0,I_{m})U$ where $U\in GL_{n}(\mathbb{Z}_{p^{s}})$. Since the map $X\mapsto XU^{-1}$ is an automorphism of $G_{p^{s}}(n,m)$, $G_{p^{s}}(n,m)$ is vertex-transitive. Thus $G_{p^{s}}(n,m)$ is an $r$-regular graph, where $r$ is the same valency of each vertex. Without loss of generality, we assume $s\geq 2$. By (4.2), it is clear that $G_{p^{s}}(n,m)$ is connected.

Let $X$ be any vertex of $G_{p^{s}}(n,m)$ with $X\sim A$. Write $X=(X_{1},X_{2})$ where $X_{1}\in\mathbb{Z}_{p^{s}}^{m\times(n-m)}$ and $X_{2}\in\mathbb{Z}_{p^{s}}^{m\times m}$. Since \scriptsize\rho\left(\begin{array}[]{c}A\\ X\\ \end{array}\right)=m+1, $\rho(X_{1})=1$. Using Lemma 2.1, $X$ can be written as \scriptsize X=\left(\begin{array}[]{cc}\alpha&\beta\\ 0&X_{22}\\ \end{array}\right) where $0\neq\alpha\in\mathbb{Z}_{p^{s}}^{n-m}$ and $X_{22}\in V(G_{p^{s}}(m,m-1))$. Thus, (4.3) implies that $X_{22}$ has $p^{(s-1)(m-1)}{m\brack 1}_{p}$ different choices. For every fixed choice $X_{22}$, we can assume further that \scriptsize\left(\begin{array}[]{cc}\alpha&\beta\\ 0&X_{22}\\ \end{array}\right)=\left(\begin{array}[]{ccc}x&y&0\\ 0&0&I_{m-1}\\ \end{array}\right), where $x\in\mathbb{Z}_{p^{s}}^{n-m}$, $y\in\mathbb{Z}_{p^{s}}$ and $(x,y)$ is unimodular with $x\neq 0$, and $A=(0,I_{m})$ does not change.

Case 1. $y\in\mathbb{Z}_{p^{s}}^{*}$. We can assume $y=1$. Then for every fixed choice $X_{22}$, $(x,1)$ with $x\neq 0$ has $\left|\mathbb{Z}_{p^{s}}\right|^{n-m}-1=p^{s(n-m)}-1$ different choices.

Case 2.  $y\in J_{p^{s}}$. Then for every fixed choice $X_{22}$, unimodular $(x,y)$ with $x\neq 0$ has

[TABLE]

different choices.

Combination Case 1 with Case 2, for every fixed choice $X_{22}$, unimodular $(x,y)$ with $x\neq 0$ has $p^{(s-1)(n-m)}{n-m\brack 1}_{p}+p^{s(n-m)}-1$ different choices. It follows that vertex $X$ with $X\sim A$ has

[TABLE]

different choices. Thus (4.4) holds. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

For every $(m-1)$-subspace $P$ of $\mathbb{Z}_{p^{s}}^{n}$, let $[P\rangle_{m}$ denote the set of all $m$-subspaces containing $P$, which is called a star. For every $(m+1)$-subspace $Q$ of $\mathbb{Z}_{p^{s}}^{n}$, let $\langle Q]_{m}$ denote the set of all $m$-subspaces of $Q$, which is called a top. Note that Theorem 3.2. It is easy to see that every star or top is a clique of $G_{p^{s}}(n,m)$. In $G_{p^{s}}(n,m)$, by Theorem 3.6, we have

[TABLE]

When $n=2m$, we have $|[P\rangle_{m}|=|\langle Q]_{m}|$. On the other hand, $\left|[P\rangle_{m}\right|<\left|\langle Q]_{m}\right|$ if $n<2m$; and

[TABLE]

Let $P$ and $Q$ be an $m$-subspace and an $(m+1)$-subspace of $\mathbb{Z}_{p^{s}}^{n}$, respectively. By Lemma 3.1, there are $U_{1},U_{2}\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $P=(0,I_{m-1})U_{1}$ and $Q=(0,I_{m+1})U_{2}$. Thus

[TABLE]

[TABLE]

When $n=2m$, we define

[TABLE]

[TABLE]

In $G_{p^{s}}(2m,m)$, using (3.9) and (4.5), it is easy to see that

[TABLE]

Lemma 4.3

In $G_{p^{s}}(n,m)$, every star or top is a maximal clique.

*Proof. *For every star $[P\rangle_{m}$, by (4.7), we have

[TABLE]

where $U_{1}\in GL_{n}(\mathbb{Z}_{p^{s}})$ is fixed. Let $X=(X_{1},X_{2})U_{1}\in V(G_{p^{s}}(n,m))$ and $X\sim Z$ for all $Z\in[P\rangle_{m}$, where $X_{1}\in\mathbb{Z}_{p^{s}}^{m\times(n-m+1)}$ and $X_{2}\in\mathbb{Z}_{p^{s}}^{m\times(m-1)}$. Then

[TABLE]

for all unimodular $\alpha\in\mathbb{Z}_{p^{s}}^{n-m+1}$. Thus $\rho(X_{1})=1$. By Lemma 2.1, \scriptsize X=\left(\begin{array}[]{cc}\alpha_{1}&\alpha_{2}\\ 0&X_{2}^{\prime}\\ \end{array}\right)U_{1} where $X^{\prime}_{2}\in\mathbb{Z}_{p^{s}}^{(m-1)\times(m-1)}$. Clearly, $X^{\prime}_{2}$ is invertible. It follows that $X\in[P\rangle_{m}$, and hence $[P\rangle_{m}$ is a maximal clique.

For every top $\langle Q]_{m}$, by (4.8), we may assume that

[TABLE]

where $U_{2}\in GL_{n}(\mathbb{Z}_{p^{s}})$ is fixed. Let $W\in V(G_{p^{s}}(n,m))$ and $W\sim Z$ for all $Z\in\langle Q]_{m}$. By $W\sim(0,I_{m})U_{2}$, we have similarly that \scriptsize W=\left(\begin{array}[]{ccc}\alpha_{1}&\alpha_{2}&\alpha_{3}\\ 0&0&W_{22}\\ \end{array}\right)U_{2}, where $\alpha_{1}\in\mathbb{Z}_{p^{s}}^{n-m-1}$, $\alpha_{2}\in\mathbb{Z}_{p^{s}}$ and $W_{22}\in\mathbb{Z}_{p^{s}}^{(m-1)\times m}$. Since $W_{22}$ has a right inverse, it is easy to see that there is $Y_{0}\in V(G_{p^{s}}(m+1,m))$ such that \scriptsize\rho\left(\begin{array}[]{c}(0,W_{22})\\ Y_{0}\\ \end{array}\right)=m+1. By $W\sim(0,Y_{0})U_{2}\in\langle Q]_{m}$, we get $\alpha_{1}=0$. Thus $W\in\langle Q]_{m}$, and hence $\langle Q]_{m}$ is a maximal clique. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Theorem 4.4

(i)*  The clique number of $G_{p^{s}}(n,m)$ (when $n\geq 2m$) is*

[TABLE]

• (ii)

Let $\mathcal{M}$ be a maximum clique of $G_{p^{s}}(n,m)$, and let $\pi:\mathbb{Z}_{p^{s}}^{m\times n}\rightarrow\mathbb{Z}_{p}^{m\times n}$ be the natural surjection (2.10). Then when $n>2m$, $\pi(\mathcal{M})$ is a star in $G_{p}(n,m)$. When $n=2m$, $\pi(\mathcal{M})$ is a star or a top in $G_{p}(n,m)$.

*Proof. *Put $G_{p^{s}}=G_{p^{s}}(n,m)$. When $s=1$, this theorem is well-known (cf. [6, 13, 22]). From now on we assume that $s\geq 2$ and $n\geq 2m$. Write $\omega=\omega(G_{p^{s}})$. Let $\pi:\mathbb{Z}_{p^{s}}^{m\times n}\rightarrow\mathbb{Z}_{p}^{m\times n}$ be the natural surjection (2.10).

Let $\mathcal{M}=\{A_{1},\ldots,A_{\omega}\}$ be a maximum clique of $G_{p^{s}}$. Suppose $\{\pi(A_{i_{1}}),\ldots,\pi(A_{i_{k}})\}$ is the set all different elements in $\{\pi(A_{1}),\ldots,\pi(A_{\omega})\}$. By (4.1) and Theorem 2.7(i), $\{\pi(A_{i_{1}}),\ldots,\pi(A_{i_{k}})\}$ is a clique of $G_{p}$. Thus

[TABLE]

Moreover, $\mathcal{M}$ has a partition into $k$ cliques: $\mathcal{M}=M_{1}\cup M_{2}\cup\cdots\cup M_{k}$, where $M_{t}$ is a clique with $\pi(M_{t})=\pi(A_{i_{t}})$, $t=1,\ldots,k$. Thus

[TABLE]

(i).  Put $n_{t}=|M_{t}|$, $t=1,\ldots,k$. Then $M_{t}=\{\pi(A_{i_{t}})+B_{t1},\pi(A_{i_{t}})+B_{t2},\ldots,\pi(A_{i_{t}})+B_{tn_{t}}\}$, where $B_{tj}\in J_{p^{s}}^{m\times n}$, $j=1,\ldots,n_{t}$. By Theorem 3.1, there is $U_{t}\in GL_{n}(\mathbb{Z}_{p})$ such that $\pi(A_{i_{t}})=(0,I_{m})U_{t}$. Note that $I_{m}+B$ is invertible for all $B\in J_{p^{s}}^{m\times m}$. The matrix representation $\pi(A_{i_{t}})+B_{tj}$ can be written as $\pi(A_{i_{t}})+B_{tj}=(C_{tj},I_{k})U_{t}$, where $C_{tj}\in J_{p^{s}}^{m\times(n-m)}$, $j=1,\ldots,n_{t}$. Therefore,

[TABLE]

By (4.1), it is clear that $\left\{C_{t1},C_{t2},\ldots,C_{tn_{t}}\right\}$ is a clique of the bilinear forms graph $\Gamma(\mathbb{Z}_{p^{s}}^{m\times(n-m)})$. Since $C_{tj}\in J_{p^{s}}^{m\times(n-m)}$, from (2.9) we can write $C_{tj}=D_{tj}p$ where $D_{tj}\in\mathbb{Z}_{p^{s-1}}^{m\times(n-m)}$, $j=1,\ldots,n_{t}$. By Lemma 2.9, $\left\{D_{t1},D_{t2},\ldots,D_{tn_{t}}\right\}$ is a clique of $\Gamma(\mathbb{Z}_{p^{s-1}}^{m\times(n-m)})$. By Lemma 2.11, $|M_{t}|=n_{t}\leq\omega\left(\Gamma(\mathbb{Z}_{p^{s-1}}^{m\times(n-m)})\right)=p^{(s-1)(n-m)}$, $t=1,\ldots,k$. Therefore,

[TABLE]

By (4.5) and (4.13), every star is a maximum clique of $G_{p^{s}}$ and (4.12) holds.

(ii).  Using (4.13) and (4.12), it is easy to see that $k={n-m+1\brack 1}_{p}$, and hence $\pi(\mathcal{M})$ is a maximum clique of $G_{p}$ for every maximum clique $\mathcal{M}$ of $G_{p^{s}}$. When $n>2m$ (resp. $n=2m$), it is well-known (cf. [13, 22]) that every maximum clique of $G_{p}$ is a star (resp. a star or a top). Therefore, when $n>2m$, $\pi(\mathcal{M})$ is a star of $G_{p}(n,m)$. When $n=2m$, $\pi(\mathcal{M})$ is a star or a top of $G_{p}(n,m)$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

It is a difficult open problem to compute the independence number of Grassmann graph. We can only give the following estimation of independence number on $G_{p^{s}}(n,m)$.

Theorem 4.5

Let $n\geq 2m$ and let $G_{p^{s}}=G_{p^{s}}(n,m)$. Then

[TABLE]

*Proof. *Without loss of generality, we assume $s\geq 2$. Let $N=\{B_{1},\ldots,B_{h}\}$ be a largest independent set of $G_{p}$, where $h=\alpha(G_{p})$. Applying Theorem 3.1, there is $Q_{t}\in GL_{n}(\mathbb{Z}_{p})$ such that $B_{t}=(0,I_{m})Q_{t}$, $t=1,\ldots,h$. Let $\left\{P_{1},P_{2},\ldots,P_{r}\right\}$ be a largest independent set of the bilinear forms graph $\Gamma(\mathbb{Z}_{p^{s-1}}^{m\times(n-m)})$, where $r=p^{(s-1)(m-1)(n-m)}$ because (2.16). Put

[TABLE]

Then $|N_{t}|=r$. By Lemma 2.9 and (4.1), $N_{t}$ is an independent set of $G_{p^{s}}$, $t=1,\ldots,h$. Note that $(pP_{i},0)Q_{t}$ is a matrix over $J_{p^{s}}$, $i=1,\ldots,r$, $t=1,\ldots,h$. For any $X=(pP_{i},I_{m})Q_{t}\in N_{t}$ and $Y=(pP_{j},I_{m})Q_{u}\in N_{u}$, where $t\neq u$ and $1\leq i,j\leq r$, by \scriptsize\rho\left(\begin{array}[]{c}B_{t}\\ B_{u}\\ \end{array}\right)\geq{\rm rk}\left(\begin{array}[]{c}B_{t}\\ B_{u}\\ \end{array}\right)>m+1 and Lemma 2.4, we have

[TABLE]

Thus, $M^{\prime}:=N_{1}\cup N_{2}\cup\cdots\cup N_{h}$ is an independent set of $G_{p^{s}}$. Consequently

[TABLE]

If $G$ is a vertex-transitive graph, then it is well-known (cf. [9, Lemma 7.2.2]) that $\alpha(G)\leq\frac{|V(G)|}{\omega(G)}$. Since $G_{p^{s}}$ is vertex-transitive, (4.3) and (4.12) imply that $\alpha(G_{p^{s}})\leq p^{(s-1)(m-1)(n-m)}\frac{{n\brack m}_{p}}{{n-m+1\brack 1}_{p}}$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

4.2 Maximum cliques

Theorem 4.6

When $n>2m$, every maximum clique of $G_{p^{s}}(n,m)$ is a star. When $n=2m$, every maximum clique of $G_{p^{s}}(n,m)$ is either a star or a top.

*Proof. *When $s=1$, this theorem is well-known (cf. [6, 13, 22]). From now on we assume that $s\geq 2$ and $n\geq 2m$. Let $E_{ij}=E_{ij}^{m\times(n-m)}$. Recall that $\Gamma(\mathbb{Z}_{p^{s}}^{m\times(n-m)})$ is the bilinear forms graph on $\mathbb{Z}_{p^{s}}^{m\times(n-m)}$.

Let $\mathcal{M}$ be a maximum clique of $G_{p^{s}}(n,m)$. By Theorem 3.1, we may assume with no loss of generality that $A=(0,I_{m})\in\mathcal{M}$. We show that there is a vertex $B\in\mathcal{M}$ such that \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=m+1. Otherwise, for any vertex $B\in\mathcal{M}$ with $B\sim A$, we have \scriptsize m+1=\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)>{\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=m. It is easy to see that $B=(B_{1},I_{m})$ where $B_{1}\in J_{p^{s}}^{m\times(n-m)}$ with $\rho(B_{1})=1$. Thus, there exists a clique $\mathcal{N}$ of $\Gamma(\mathbb{Z}_{p^{s}}^{m\times(n-m)})$ containing [math], such that $\mathcal{N}\subseteq J_{p^{s}}^{m\times(n-m)}$ and $\mathcal{M}=\left\{(X,I_{m}):X\in\mathcal{N}\right\}$. Clearly, for every maximal clique $\mathcal{C}$ containing [math] in $\Gamma(\mathbb{Z}_{p^{s}}^{m\times(n-m)})$, the set $\left\{(X,I_{m}):X\in\mathcal{C}\right\}$ is a clique of $G_{p^{s}}(n,m)$. Since $\mathcal{M}=\left\{(X,I_{m}):X\in\mathcal{N}\right\}$ is a maximum clique, $\mathcal{N}$ must be a maximal clique of $\Gamma(\mathbb{Z}_{p^{s}}^{m\times(n-m)})$, but $\mathcal{N}\subseteq J_{p^{s}}^{m\times(n-m)}$, a contradiction to Lemma 2.10. Therefore, there is vertex $B\in\mathcal{M}$ such that \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=m+1. By Theorem 3.1, we may assume with no loss of generality that

[TABLE]

Let $\pi:\mathbb{Z}_{p^{s}}^{m\times n}\rightarrow\mathbb{Z}_{p}^{m\times n}$ be the natural surjection (2.10). For any vertex $X$ in $\pi(\mathcal{M})$, let $\pi^{-1}(X)$ denote the preimages of $X$ in $\mathcal{M}$, i.e., $\pi^{-1}(X)=\left\{Y\in\mathcal{M}:\pi(Y)=X\right\}$. By Theorem 4.4(ii), when $n>2m$, $\pi(\mathcal{M})$ is a star in $G_{p}(n,m)$. When $n=2m$, $\pi(\mathcal{M})$ is a star or a top in $G_{p}(n,m)$.

Case 1.   $\pi(\mathcal{M})$ is a star in $G_{p}(n,m)$.

We will prove that $\mathcal{M}$ is a star in $G_{p^{s}}(n,m)$. Since $\pi(A)=A$, $\pi(B)=B$ and $A\cap B=(0,I_{m-1})$, (4.7) implies that

[TABLE]

Let $e_{i}$ be the $i$-th row of $I_{n-m+1}$, and let

[TABLE]

By (4.12), we have $\left|\pi^{-1}(A_{i})\right|=p^{(s-1)(n-m)}$. From Lemma 2.11, there is a maximum clique $\mathcal{C}_{i}$ of $\Gamma(\mathbb{Z}_{p^{s-1}}^{m\times(n-m)})$ such that

[TABLE]

$i=1,\ldots,n-m+1$. Here \scriptsize\left(\begin{array}[]{c}\beta_{i1}\\ \beta_{i3}\\ \end{array}\right) or \scriptsize\left(\begin{array}[]{c}\beta_{i2}\\ \beta_{i4}\\ \end{array}\right) may be absent if $i=1$ or $i=n-m+1$.

Let $X_{n-m}\in\pi^{-1}(A_{n-m})$. Then $X_{n-m}\sim A$ and hence \scriptsize\rho\left(\begin{array}[]{c}A\\ X_{n-m}\\ \end{array}\right)=m+1. It follows that $\beta_{n-m,3}=0$. Since $X_{n-m}\sim B$, we have similarly $\beta_{n-m,4}=0$. Therefore, we obtain

[TABLE]

We can write

[TABLE]

By (4.17), we get

[TABLE]

and hence \small\rho\left(\begin{array}[]{cc}0&1\\ A_{21}p&A_{22}p\\ \end{array}\right)=1. Consequently $A_{21}p=0$. Since \small\left(\begin{array}[]{cccc}A_{11}p&A_{12}p&1&0\\ 0&A_{22}p&0&I_{m-1}\\ \end{array}\right)\sim B, it is easy to see that $A_{22}p=0$. Thus

[TABLE]

Let

[TABLE]

By $\left|\pi^{-1}(B_{\alpha})\right|=p^{(s-1)(n-m)}$ and Lemma 2.11, there is a maximum clique $\mathcal{C}_{\alpha}$ of $\Gamma(\mathbb{Z}_{p^{s-1}}^{m\times(n-m)})$, such that

[TABLE]

Let $e_{i}^{\prime}$ be the $i$-th row of $I_{n-m}$. Since

[TABLE]

and (4.1), we get \small\rho\left(\begin{array}[]{c}e_{n-m}^{\prime}\\ B_{2}p\\ \end{array}\right)=1, which implies that $B_{2}p=(0,\beta_{2})$ where $\beta_{2}\in J_{p^{s}}^{(m-1)\times 1}$. By (4.18), we have

[TABLE]

It follows from (4.1) that \small\rho\left(\begin{array}[]{c}\alpha+B_{1}p-\beta\\ B_{2}p\\ \end{array}\right)=1 for all $\beta\in J_{p^{s}}^{1\times(n-m)}$. Taking $\beta=B_{1}p$, we get \small\rho\left(\begin{array}[]{c}\alpha\\ B_{2}p\\ \end{array}\right)=\rho\left(\begin{array}[]{c}\alpha\\ (0,\beta_{2})\\ \end{array}\right)=1. Thus, when $\alpha\neq e_{n-m}^{\prime}$, we have that $B_{2}p=(0,\beta_{2})=0$. When $\alpha=e_{n-m}^{\prime}$,

[TABLE]

By (4.1) we can obtain

[TABLE]

which implies that $B_{2}p=(0,\beta_{2})=0$. Therefore, we always have $B_{2}p=0$. Then we have proved that

[TABLE]

Put

[TABLE]

Then

[TABLE]

Let \scriptsize\left(\begin{array}[]{ccc}\alpha+D_{1}p&D_{2}p&0\\ D_{3}p&D_{4}p&I_{m-1}\\ \end{array}\right)\in\pi^{-1}(D_{\alpha}). Then \scriptsize\left(\begin{array}[]{ccc}\alpha+D_{1}p&D_{2}p&0\\ D_{3}p&D_{4}p&I_{m-1}\\ \end{array}\right)\sim A, and hence \scriptsize\rho\left(\begin{array}[]{c}\alpha+D_{1}p\\ D_{3}p\\ \end{array}\right)=1. Since $\alpha+D_{1}p$ is unimodular, \scriptsize\left(\begin{array}[]{c}\alpha+D_{1}p\\ D_{3}p\\ \end{array}\right) has a minimal factorization \scriptsize\left(\begin{array}[]{c}\alpha+D_{1}p\\ D_{3}p\\ \end{array}\right)=\left(\begin{array}[]{c}1\\ \gamma\\ \end{array}\right)(\alpha+D_{1}p), and hence $D_{3}p=\gamma(\alpha+D_{1}p)$. Therefore, applying elementary row operations of matrix, we can assume

[TABLE]

Put \scriptsize\left(\begin{array}[]{ccc}\alpha+D_{1}p&D_{2}p&0\\ 0&D_{4}^{\prime}p&I_{m-1}\\ \end{array}\right)\in\pi^{-1}(D_{\alpha}). By (4.19), \scriptsize\left(\begin{array}[]{ccc}\alpha+D_{1}p&D_{2}p&0\\ 0&D_{4}^{\prime}p&I_{m-1}\\ \end{array}\right)\sim\left(\begin{array}[]{ccc}\alpha^{\prime}+B_{1}p&1&0\\ 0&0&I_{m-1}\\ \end{array}\right) for all $B_{1}p\in J_{p^{s}}^{n-m}$ and $0\neq\alpha^{\prime}\in\mathbb{Z}_{p}^{n-m}$. Thus, by (4.1) we get \scriptsize\rho\left(\begin{array}[]{cc}\alpha^{\prime}+B_{1}p&1\\ \alpha+D_{1}p&D_{2}p\\ 0&D_{4}^{\prime}p\\ \end{array}\right)=2, for all $0\neq\alpha^{\prime}\in\mathbb{Z}_{p}^{n-m}$. Clearly, there exists $0\neq\alpha^{\prime}\in\mathbb{Z}_{p}^{n-m}$ such that \scriptsize{\rm rk}\left(\begin{array}[]{c}\alpha^{\prime}+B_{1}p\\ \alpha+D_{1}p\\ \end{array}\right)=\rho\left(\begin{array}[]{c}\alpha^{\prime}+B_{1}p\\ \alpha+D_{1}p\\ \end{array}\right)=2, which implies that $D_{4}^{\prime}p=0$. Therefore,

[TABLE]

Let $P=A\cap B=(0,0,I_{m-1})$. By (4.18)-(4.20), we have that $\mathcal{M}=\pi^{-1}(\pi(\mathcal{M}))=[P\rangle_{m}$ is a star in $G_{p^{s}}(n,m)$.

Case 2.   $\pi(\mathcal{M})$ is a top in $G_{p}(n,m)$ with $n=2m$.

We show that $\mathcal{M}$ is a top in $G_{p^{s}}(n,m)$ as follows. Define \mathcal{M}^{\perp}=\left\{X^{\perp}:\mbox{X$is any$m$-subspace in$\mathcal{M}}\right\}. By Lemma 3.7, we have $\pi(\mathcal{M}^{\perp})=(\pi(\mathcal{M}))^{\perp}$. It follows from (4.9) that $(\pi(\mathcal{M}))^{\perp}$ is a star in $G_{p}(n,m)$. By the Case 1, $\mathcal{M}^{\perp}$ is a star in $G_{p^{s}}(n,m)$. Applying (3.8) and (4.9), $\mathcal{M}=(\mathcal{M}^{\perp})^{\perp}$ is a top in $G_{p^{s}}(n,m)$.

Combination Case 1 with Case 2, when $n>2m$, every maximum clique of $G_{p^{s}}(n,m)$ is a star. When $n=2m$, every maximum clique of $G_{p^{s}}(2m,m)$ is either a star or a top. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

5 Automorphisms of Grassmann graph over $\mathbb{Z}_{p^{s}}$

Recall that an endomorphism of a graph $G$ is an adjacency preserving map from $V(G)$ to itself. An automorphism of a graph $G$ is an adjacency preserving bijective map from $V(G)$ to itself. In the algebraic graph theory [9], the characterization of graph automorphism is an important problem. In this section, we determine the automorphisms of $G_{p^{s}}(n,m)$, and our main result is as follows.

Theorem 5.1

Let $n\geq 2m\geq 4$, and let $\varphi$ be an automorphism of $G_{p^{s}}(n,m)$. Then either there exists $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that

[TABLE]

or $n=2m$ and there exists $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that

[TABLE]

Theorem 5.1 also has important significance for the geometry of matrices [12, 25], and it is also the fundamental theorem of the projective geometry of rectangular matrices over $\mathbb{Z}_{p^{s}}$. To prove Theorem 5.1, we need the following knowledge and lemmas.

In $G_{p^{s}}(n,m)$ (where $n>m\geq 1$), two vertices $A,B$ are said to be McCoy adjacent (Mc-adjacent for short), denoted by $A\stackrel{{\scriptstyle mc}}{{\sim}}B$, if \scriptsize\rho\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}A\\ B\\ \end{array}\right)=m+1. Two Mc-adjacent vertices are adjacent but not vice versa. An endomorphism $\varphi$ of $G_{p^{s}}(n,m)$ is called to preserve Mc-adjacency if $A\stackrel{{\scriptstyle mc}}{{\sim}}B$ implies that $\varphi(A)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi(B)$ for any two vertices $A$ and $B$.

Lemma 5.2

Let $n>m\geq 1$, and let $A,B$ be two distinct vertices of $G_{p^{s}}(n,m)$ with $d(A,B)=k$. Then there are vertices $A_{1},\ldots,A_{2k-1}$ of $G_{p^{s}}(n,m)$ such that $A\stackrel{{\scriptstyle mc}}{{\sim}}A_{1}\stackrel{{\scriptstyle mc}}{{\sim}}\cdots\stackrel{{\scriptstyle mc}}{{\sim}}A_{2k-1}\stackrel{{\scriptstyle mc}}{{\sim}}B$.

*Proof. *Let $E_{ij}=E_{ij}^{m\times(n-m)}$. Suppose that $A\sim B$. By Theorem 3.1, there is $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $A=(0,I_{m})U$ and $B=(p^{i}E_{11},I_{m})U$, where $0\leq i\leq{\rm max}\{s-1,1\}$. Put $C=(p^{i}E_{11}+E_{12},I_{m})U$. Then $A\stackrel{{\scriptstyle mc}}{{\sim}}C\stackrel{{\scriptstyle mc}}{{\sim}}B$. Now, let $d(A,B)=k>1$. Then there are vertices $B_{1},\ldots,B_{k-1}$ such that $A\sim B_{1}\sim\cdots\sim B_{k-1}\sim B$. Thus, there are vertices $C_{1},\ldots,C_{k}$ such that $A\stackrel{{\scriptstyle mc}}{{\sim}}C_{1}\stackrel{{\scriptstyle mc}}{{\sim}}B_{1}\cdots\stackrel{{\scriptstyle mc}}{{\sim}}B_{k-1}\stackrel{{\scriptstyle mc}}{{\sim}}C_{k}\stackrel{{\scriptstyle mc}}{{\sim}}B$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Lemma 5.3

Let $n>m\geq 2$. Then $A$ is a vertex of $G_{p^{s}}(n,m)$ if and only if there are two vertices $P_{1},P_{2}\in V(G_{p^{s}}(n,m-1))$ such that $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ and $A=\{A\}=P_{1}\vee P_{2}$.

*Proof. *Let $A$ be a vertex of $G_{p^{s}}(n,m)$. By Theorem 3.1, we may assume with no loss of generality that $A=(0,I_{m})$. Put $P_{1}=(0,I_{m-1}),P_{2}=(0,I_{m-1},0_{m,1})\in V(G_{p^{s}}(n,m-1))$. Then \scriptsize\rho\left(\begin{array}[]{c}P_{1}\\ P_{2}\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}P_{1}\\ P_{2}\\ \end{array}\right)=m, and hence $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$. By Theorem 3.5(ii), we get $A=\{A\}=P_{1}\vee P_{2}$.

Conversely, suppose there are vertices $P_{1},P_{2}\in V(G_{p^{s}}(n,m-1))$ such that $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ and $A=\{A\}=P_{1}\vee P_{2}$. Then $A$ is a fixed $m$-subspace and hence $A$ is a vertex of $G_{p^{s}}(n,m)$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Lemma 5.4

Let $[P_{1}\rangle_{m},[P_{2}\rangle_{m}$ be two distinct stars in $G_{p^{s}}(n,m)$ where $n\geq 2m\geq 4$. Then:

• (i)

$[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\neq\emptyset$* if and only if $P_{1}\sim P_{2}$ (in $G_{p^{s}}(n,m-1)$). Moreover, if $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\neq\emptyset$, then $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}=P_{1}\vee P_{2}$.*

• (ii)

In $G_{p^{s}}(n,m)$, $\left|[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\right|=1$ if and only if $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ (in $G_{p^{s}}(n,m-1)$).

*Proof. *(i).  Assume that $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\neq\emptyset$. Let vertex $C\in[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}$. Then $P_{1},P_{2}\subset C$. It follows from Remark 3.3 that ${\rm dim}(P_{1}\vee P_{2})=m$ and $C\in P_{1}\vee P_{2}$. Thus $P_{1}\sim P_{2}$ (in $G_{p^{s}}(n,m-1)$) and $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\subseteq P_{1}\vee P_{2}$.

Conversely, suppose that $P_{1}\sim P_{2}$ (in $G_{p^{s}}(n,m-1)$). Then (4.1) implies that ${\rm dim}(P_{1}\vee P_{2})=m$. By Remark 3.3, $P_{1}\vee P_{2}\subseteq[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}$ and hence $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\neq\emptyset$. Therefore, $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\neq\emptyset$ if and only if $P_{1}\sim P_{2}$ (in $G_{p^{s}}(n,m-1)$).

Now, we assume $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\neq\emptyset$. Then \scriptsize\rho\left(\begin{array}[]{c}P_{1}\\ P_{2}\\ \end{array}\right)=m={\rm dim}(P_{1}\vee P_{2}) and $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\subseteq P_{1}\vee P_{2}$. By Remark 3.3 again, we have $P_{1}\vee P_{2}\subseteq[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}$. Thus $[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}=P_{1}\vee P_{2}$.

(ii).  Suppose that $\left|[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\right|=1$ in $G_{p^{s}}(n,m)$. By (i), $P_{1}\vee P_{2}$ is a fixed $m$-subspace. By Theorem 3.5(ii), we have \scriptsize\rho\left(\begin{array}[]{c}P_{1}\\ P_{2}\\ \end{array}\right)={\rm rk}\left(\begin{array}[]{c}P_{1}\\ P_{2}\\ \end{array}\right)=m, and hence $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ (in $G_{p^{s}}(n,m-1)$).

Conversely, suppose that $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ (in $G_{p^{s}}(n,m-1)$). Using (i) and Theorem 3.5(ii), we obtain $\left|[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\right|=|P_{1}\vee P_{2}|=1$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Lemma 5.5

In $G_{p^{s}}(n,m)$, if $[P\rangle_{m}\cap\langle Q]_{m}\neq\emptyset$, then $\left|[P\rangle_{m}\cap\langle Q]_{m}\right|=p^{s-1}(p+1)$.

*Proof. *When $s=1$, this lemma is known by [13, Lemma 2.1]. From now on we assume that $s\geq 2$. Suppose $[P\rangle_{m}\cap\langle Q]_{m}\neq\emptyset$. Then $P\subset Q$. By Lemma 2.1, without loss of generality, we may assume that $P=(0,I_{m-1})$. Thus \scriptsize Q=\left(\begin{array}[]{cc}Q_{1}&0\\ 0&I_{m-1}\\ \end{array}\right), where $Q_{1}\in\mathbb{Z}_{p^{s}}^{2\times(n-m+1)}$ has a right inverse. Using appropriate elementary operations of matrix, we may assume with no loss of generality that $Q=(0,I_{m+1})$.

Let vertex $A\in[P\rangle_{m}\cap\langle Q]_{m}$. Then $P\subset A\subset Q$. By (4.7) and (4.8), we have \scriptsize A=\left(\begin{array}[]{ccc}0&\alpha_{1}&0\\ 0&0&I_{m-1}\\ \end{array}\right) where $\alpha_{1}\in\mathbb{Z}_{p^{s}}^{2}$ is unimodular. Since the number of unimodular vectors in $\mathbb{Z}_{p^{s}}^{2}$ is $\frac{\left|\mathbb{Z}_{p^{s}}\right|^{2}-\left|J_{p^{s}}\right|^{2}}{\left|\mathbb{Z}_{p^{s}}^{*}\right|}=p^{s-1}(p+1)$, we obtain $\left|[P\rangle_{m}\cap\langle Q]_{m}\right|=p^{s-1}(p+1)$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Lemma 5.6

Let $n\geq 2m\geq 4$. Suppose $\varphi$ is an automorphism of $G_{p^{s}}(n,m)$ and $\varphi$ maps stars to stars. Let $\varphi([X\rangle_{m})=[X^{\prime}\rangle_{m}$, $X\in V(G_{p^{s}}(n,m-1))$. Define the map $\varphi_{1}:V(G_{p^{s}}(n,m-1))\rightarrow V(G_{p^{s}}(n,m-1))$ by $\varphi_{1}(X)=X^{\prime}$. Then:

• (i)

The map $\varphi_{1}$ is an automorphism of $G_{p^{s}}(n,m-1)$, $\varphi$ is uniquely determined by $\varphi_{1}$, and

[TABLE]

• (ii)

Let $X,Y,W\in V(G_{p^{s}}(n,m-1))$. Then $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ if and only if $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$. Moreover, $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ with $W\subset X\vee Y$, if and only if $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$ with $\varphi_{1}(W)\subset\varphi_{1}(X)\vee\varphi_{1}(Y)$.

*Proof. *(i).  By Lemma 5.4(i), it is easy to see that $\varphi_{1}$ is an injective endomorphism of $G_{p^{s}}(n,m-1)$. Thus Thus $\varphi_{1}$ is an automorphism of $G_{p^{s}}(n,m-1)$.

Put $X,Y\in V(G_{p^{s}}(n,m-1))$ with $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$. From Theorem 3.5(ii) we know that $X\vee Y$ is a fixed $m$-subspace. By Lemma 5.4(i), $X\vee Y=[X\rangle_{m}\cap[Y\rangle_{m}$. Thus $\varphi(X\vee Y)=[\varphi_{1}(X)\rangle_{m}\cap[\varphi_{1}(Y)\rangle_{m}$, and hence $\left|[\varphi_{1}(X)\rangle_{m}\cap[\varphi_{1}(Y)\rangle_{m}\right|=1$. It follows from Lemma 5.4 that $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$ and

[TABLE]

Therefore, (5.3) holds and $\varphi_{1}$ preserves the Mc-adjacency in $G_{p^{s}}(n,m-1)$.

Let $A$ be a vertex of $G_{p^{s}}(n,m)$. By Lemma 5.3, there are two vertices $P_{1},P_{2}\in V(G_{p^{s}}(n,m-1))$ such that $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ and $A=P_{1}\vee P_{2}$. By (5.3), $\varphi$ is uniquely determined by $\varphi_{1}$.

(ii).  Let $\varphi^{-1}$ and $\varphi_{1}^{-1}$ be the inverse maps of $\varphi$ and $\varphi_{1}$, respectively. Then $\varphi^{-1}$ and $\varphi_{1}^{-1}$ are also two graph automorphisms. Since $\varphi([X\rangle_{m})=[\varphi_{1}(X)\rangle_{m}$ for all $X\in V(G_{p^{s}}(n,m-1))$, we have $\varphi^{-1}([Z\rangle_{m})=[\varphi_{1}^{-1}(Z)\rangle_{m}$ for all $Z\in V(G_{p^{s}}(n,m-1))$. By (i), both $\varphi_{1}$ and $\varphi_{1}^{-1}$ preserve the Mc-adjacency in $G_{p^{s}}(n,m-1)$. Therefore, for any $X,Y\in V(G_{p^{s}}(n,m-1))$, it is easy to see that $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ if and only if $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$.

Suppose that $X,Y,W\in V(G_{p^{s}}(n,m-1))$, $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ and $W\subset X\vee Y$. Without losing generality, we assume that $W\neq X$ and $W\neq Y$. Then ${\rm dim}(W\vee Y)={\rm dim}(X\vee Y)=m$. Hence $W\sim Y$. Similarly, $W\sim X$. By (i), we have that $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$, $\varphi_{1}(W)\sim\varphi_{1}(Y)$ and $\varphi_{1}(W)\sim\varphi_{1}(X)$. By Theorem 3.5(ii), $X\vee Y$ and $\varphi_{1}(X)\vee\varphi_{1}(Y)$ are two fixed vertices of $G_{p^{s}}(n,m)$, and $X\vee Y\in W\vee Y$.

Using Lemma 5.4(i), $[X\rangle_{m}\cap[Y\rangle_{m}\in[W\rangle_{m}\cap[Y\rangle_{m}$, and hence

[TABLE]

By Lemma 5.4(i) again, we have $\varphi_{1}(X)\vee\varphi_{1}(Y)\in\varphi_{1}(W)\vee\varphi_{1}(Y)$, it follows from Remark 3.3 that $\varphi_{1}(W)\subset\varphi_{1}(X)\vee\varphi_{1}(Y)$. Thus, if $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ with $W\subset X\vee Y$, then $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$ with $\varphi_{1}(W)\subset\varphi_{1}(X)\vee\varphi_{1}(Y)$. Similarly, considering $\varphi_{1}^{-1}$, we have that $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$ with $\varphi_{1}(W)\subset\varphi_{1}(X)\vee\varphi_{1}(Y)$ implies that $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ with $W\subset X\vee Y$. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Lemma 5.7

Let $\varphi$ be an automorphism of $G_{p^{s}}(2m,m)$ (where $m\geq 2$). If $\varphi$ maps some star to a star, then $\varphi$ maps stars to stars. If $\varphi$ maps some star to a top, then $\varphi$ maps stars to tops.

*Proof. *Case 1.  $\varphi$ maps some star to a star.

Then there is a star $[P_{1}\rangle_{m}$ such that $\varphi([P_{1}\rangle_{m})=[P_{1}^{\prime}\rangle_{m}$. We prove that $\varphi$ maps stars to stars as follows.

Let $[P_{2}\rangle_{m}$ be any star in $G_{p^{s}}(2m,m)$ with $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$ (in $G_{p^{s}}(2m,m-1)$). Note that $\varphi$ carries maximum cliques of $G_{p^{s}}(2m,m)$ to maximum cliques. By Theorem 4.6, $\varphi([P_{2}\rangle_{m})$ is a star or a top. Suppose $\varphi([P_{2}\rangle_{m})$ is a top. Write $\varphi([P_{2}\rangle_{m})=\langle Q]_{m}$. Then

[TABLE]

By Lemma 5.4(ii), we have $\left|\varphi([P_{1}\rangle_{m}\cap[P_{2}\rangle_{m})\right|=\left|[P_{1}\rangle_{m}\cap[P_{2}\rangle_{m}\right|=1$. But by Lemma 5.5 we get $\left|[P_{1}^{\prime}\rangle_{m}\cap\langle Q]_{m}\right|=p^{s-1}(p+1)>1$, a contradiction. Therefore, $\varphi([P_{2}\rangle_{m})$ is a star for any star $[P_{2}\rangle_{m}$ in $G_{p^{s}}(2m,m)$ with $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}P_{2}$.

Now, let $[P\rangle_{m}$ be any star in $G_{p^{s}}(2m,m)$ with $d(P_{1},P)=k>1$. By Lemma 5.2, there are vertices $A_{1},\ldots,A_{2k-1}$ of $G_{p^{s}}(2m,m-1)$ such that $P_{1}\stackrel{{\scriptstyle mc}}{{\sim}}A_{1}\stackrel{{\scriptstyle mc}}{{\sim}}\cdots\stackrel{{\scriptstyle mc}}{{\sim}}A_{2k-1}\stackrel{{\scriptstyle mc}}{{\sim}}P$. Applying the above result, all $\varphi([A_{1}\rangle_{m}),\ldots,\varphi([A_{2k-1}\rangle_{m}),\varphi([P\rangle_{m})$ are stars. Therefore, $\varphi$ maps stars to stars.

Case 2.  $\varphi$ maps some star to a top.

Let $\psi(X)=(\varphi(X))^{\perp}$, $X\in V(G_{p^{s}}(2m,m))$. By (3.6) and (3.8), it is clear that $\psi$ is a surjection from $V(G_{p^{s}}(2m,m))$ to itself. By (3.14), $\psi$ is also an automorphism of $G_{p^{s}}(2m,m)$. Using (4.9), $\psi$ maps some star to a star. It follows from Step 1 that $\psi$ maps stars to stars. Applying (4.9) again, $\varphi$ maps stars to tops. $\hfill\Box\vskip 6.0pt plus 2.0pt minus 2.0pt$

Let $R$ be a commutative local ring and $n\geq 3$. A $1$-subspace of $R^{n}$ is called a line, and a $2$-subspace of $R^{n}$ is called a plane. Let ${\rm P}(R^{n})$ be the set of lines of $R^{n}$, and ${\rm P}(R^{n})$ is called the projective space of $R^{n}$. A bijective map $f:{\rm P}(R^{n})\rightarrow{\rm P}(R^{n})$ is called a projectivity or collineation if $f$ satisfies the following properties: (a)  For lines $L_{1}$ and $L_{2}$, $L_{1}+L_{2}$ (where $L_{1}+L_{2}$ is the direct sum of modules) is a plane if and only if $f(L_{1})+f(L_{2})$ is a plane. (b)  Suppose $L_{1}+L_{2}$ is a plane and $L$ is a line. Then $L\subset L_{1}+L_{2}$ if and only if $f(L)\subset f(L_{1})+f(L_{2})$. Thus, a projectivity is a bijection between lines which preserves planes (cf. [19]).

Remark 5.8

When $R=\mathbb{Z}_{p^{s}}$, for lines $L_{1}$ and $L_{2}$, $L_{1}+L_{2}$ is a plane if and only if $L_{1}\stackrel{{\scriptstyle mc}}{{\sim}}L_{2}$ in $V(G_{p^{s}}(n,1))$. Moreover, if $L_{1}\stackrel{{\scriptstyle mc}}{{\sim}}L_{2}$, then $L_{1}+L_{2}=L_{1}\vee L_{2}$ by Lemma 3.5(ii).

We have the following fundamental theorem of projective geometry [19] over $R$ :

Theorem 5.9

(see [19, Theorem I.13])*  Let $R$ be a commutative local ring and $n\geq 3$. Suppose that $f:{\rm P}(R^{n})\rightarrow{\rm P}(R^{n})$ is a projectivity. Then, there is a semi-linear bijection $\phi:R^{n}\rightarrow R^{n}$ such that $f={\rm P}(\phi)$.*

Proof of Theorem 5.1.  Let $\varphi$ be an automorphism of $G_{p^{s}}(n,m)$ where $n\geq 2m\geq 4$. Since $\varphi$ carries maximum cliques to maximum cliques, Theorem 4.6 and Lemma 5.7 imply that $\varphi$ maps stars to stars, or $\varphi$ maps stars to tops with $n=2m$.

Case 1.   $\varphi$ maps stars to stars.

Let

[TABLE]

Define the map $\varphi_{1}:V(G_{p^{s}}(n,m-1))\rightarrow V(G_{p^{s}}(n,m-1))$ by $\varphi_{1}(X)=X^{\prime}$. By Lemma 5.6, $\varphi_{1}$ is an automorphism of $G_{p^{s}}(n,m-1)$, $\varphi$ is uniquely determined by $\varphi_{1}$, and

[TABLE]

Moreover, for $X,Y,W\in V(G_{p^{s}}(n,m-1))$, $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ if and only if $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$; $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ with $W\subset X\vee Y$ if and only if $\varphi_{1}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{1}(Y)$ with $\varphi_{1}(W)\subset\varphi_{1}(X)\vee\varphi_{1}(Y)$.

Note that every maximum clique of $G_{p^{s}}(n,m-k)$ ($1\leq k\leq m-1$) is a star. The $\varphi_{1}$ maps stars to stars. Put $\varphi_{0}=\varphi$. Similarly, we can define an automorphism $\varphi_{k}:V(G_{p^{s}}(n,m-k))\rightarrow V(G_{p^{s}}(n,m-k))$, such that $\varphi_{k-1}$ is uniquely determined by $\varphi_{k}$ and

[TABLE]

$k=1,\ldots,m-1$. Moreover, for all $X,Y,W\in V(G_{p^{s}}(n,m-k))$, $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ if and only if $\varphi_{k}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{k}(Y)$; $X\stackrel{{\scriptstyle mc}}{{\sim}}Y$ with $W\subset X\vee Y$ if and only if $\varphi_{k}(X)\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{k}(Y)$ with $\varphi_{k}(W)\subset\varphi_{k}(X)\vee\varphi_{k}(Y)$. $k=1,\ldots,m-1$.

Let $R=\mathbb{Z}_{p^{s}}$. Clearly, ${\rm P}(R^{n})=V(G_{p^{s}}(n,1))$ and $\varphi_{m-1}$ is a bijective map from ${\rm P}(R^{n})$ to itself. By Remark 5.8, $\varphi_{m-1}:{\rm P}(R^{n})\rightarrow{\rm P}(R^{n})$ is a projectivity. By Theorem 5.9, there is a semi-linear bijection $\phi:R^{n}\rightarrow R^{n}$ such that $\varphi_{m-1}={\rm P}(\phi)$. Since $1$ generates $\mathbb{Z}_{p^{s}}$, every ring automorphism of $\mathbb{Z}_{p^{s}}$ is the identity mapping. Consequently, every semi-linear bijection $\phi:R^{n}\rightarrow R^{n}$ is a linear bijection. Therefore, there exists $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that

[TABLE]

For any $X\in V(G_{p^{s}}(n,2))$, by Lemma 5.3, there are $X_{1},X_{2}\in V(G_{p^{s}}(n,1))$ such that \scriptsize X=X_{1}\vee X_{2}=\left(\begin{array}[]{c}X_{1}\\ X_{2}\\ \end{array}\right) with $X_{1}\stackrel{{\scriptstyle mc}}{{\sim}}X_{2}$. Since $\varphi_{m-1}(X_{1})\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{m-1}(X_{2})$, from (5.5) and (5.6), we obtain

[TABLE]

Thus

[TABLE]

For any $X\in V(G_{p^{s}}(n,3))$, by Lemma 5.3, there are $X_{1},X_{2}\in V(G_{p^{s}}(n,2))$ such that $\scriptsize X=X_{1}\vee X_{2}$ with $X_{1}\stackrel{{\scriptstyle mc}}{{\sim}}X_{2}$. By Theorems 3.1, 3.5(ii) and (3.2), there is $X_{21}\in V(G_{p^{s}}(n,1))$ such that \scriptsize X_{2}=\left(\begin{array}[]{c}X_{21}\\ X_{22}\\ \end{array}\right) and \scriptsize X=\left(\begin{array}[]{c}X_{1}\\ X_{22}\\ \end{array}\right), where $X_{22}=X_{1}\cap X_{2}$ is a $2$-subspace. Note that $\varphi_{m-2}(X_{1})\stackrel{{\scriptstyle mc}}{{\sim}}\varphi_{m-2}(X_{2})$. Applying (5.5) and (5.7), we have

[TABLE]

Therefore, we get

[TABLE]

Similarly, we can prove that

[TABLE]

Case 2.  $\varphi$ maps stars to tops with $n=2m$.

Let $\psi(X)=(\varphi(X))^{\perp}$, $X\in V(G_{p^{s}}(2m,m))$. By (3.6) and (3.8), $\psi$ is surjection from $V(G_{p^{s}}(2m,m))$ to itself. By (3.14), $\psi$ is also an automorphism of $G_{p^{s}}(2m,m)$. Using (4.9), $\psi$ maps stars to stars. By the Case 1, there exists $U\in GL_{n}(\mathbb{Z}_{p^{s}})$ such that $\psi(X)=XU$, $X\in V(G_{p^{s}}(2m,m))$. Thus

[TABLE]

By above discussion, we complete the proof of Theorem 5.1. $\Box$

Remark 5.10

Suppose that $2\leq m<n<2m$. Then every maximum clique of $G_{p^{s}}(n,m)$ is a top. Recall that $G_{p^{s}}(n,m)$ is isomorphic to $G_{p^{s}}(n,n-m)$. By the bijection $Y\mapsto Y^{\perp}$, it is easy to see that every automorphism $\varphi$ of $G_{p^{s}}(n,m)$ is of the form $\varphi(X)=(XU)^{\perp}$, where $U\in GL_{n}(\mathbb{Z}_{p^{s}})$.