Automorphisms of the subspace sum graphs on a vector space
Fenglei Tian, Dein Wong

TL;DR
This paper fully characterizes the automorphisms of the subspace sum graph on a finite-dimensional vector space, extending previous structural studies by providing a complete description of its symmetries.
Contribution
It offers the first complete characterization of automorphisms of the subspace sum graph, a problem left open in prior research.
Findings
Automorphisms are fully characterized.
The structure of the automorphism group is described.
Results extend understanding of the graph's symmetries.
Abstract
The subspace sum graph on a finite dimensional vector space was introduced by Das [Subspace Sum Graph of a Vector Space, arXiv:1702.08245], recently. The vertex set of consists of all the nontrivial proper subspaces of and two distinct vertices and are adjacent if and only if . In that paper, some structural indices (e.g., diameter, girth, connectivity, domination number, clique number and chromatic number) were studied, but the characterization of automorphisms of was left as one of further research topics. Motivated by this, we in this paper characterize the automorphisms of completely.
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Taxonomy
TopicsAdvanced Graph Theory Research · Graph Labeling and Dimension Problems · Cooperative Communication and Network Coding
Automorphisms of the subspace sum graphs on a vector space
Fenglei Tian, Dein Wong
School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China. Corresponding author. E-mail address: [email protected]. Supported by “the Fundamental Research Funds for the Central Universities (No. 2017BSCXB53)”.
Abstract: The subspace sum graph on a finite dimensional vector space was introduced by Das [Subspace Sum Graph of a Vector Space, arXiv:1702.08245], recently. The vertex set of consists of all the nontrivial proper subspaces of and two distinct vertices and are adjacent if and only if . In that paper, some structural indices (e.g., diameter, girth, connectivity, domination number, clique number and chromatic number) were studied, but the characterization of automorphisms of was left as one of further research topics. Motivated by this, we in this paper characterize the automorphisms of completely.
Keywords: Automorphisms; Subspace sum graphs; Vector space
AMS classification: 05C25; 05C69; 20H20
1 Introduction
Up to now, there are some abstract graphs defined on certain algebraic objects, which are proposed from the motivation that the properties of those algebraic structures can be revealed by studying the graphs associated with them. For example, the zero-divisor graphs defined on finite rings [4], commutative rings [6] or noncommutative rings [3], regular graphs or total graphs on a commutative ring [2, 5], cozero-divisor graph of a commutative ring [1] and so on. For various graphs defined on finite dimensional vector space, readers can refer to [9, 10, 11, 12] for details.
Let be a finite dimensional vector space of dimension greater than one over a field . Recently, the subspace sum graph on was introduced by Das [8], the vertex set of which is composed of all the nontrivial proper subspaces of and two vertices are adjacent, written as , if and only if . In [8], the author investigated the diameter, girth, connectivity, domination number, clique number, chromatic number of and the properties of with the base field being finite.
Since automorphisms of graphs can reveal the relationship among their vertices, then they can help analyze the structure of graphs and let us go further to achieve the motivation stated at the beginning. Moreover, automorphisms of graphs are also of importance in algebraic graph theory. Thus, characterizing the automorphisms of graphs also have attracted many attentions. Let be a ring and the nonzero zero divisor set of . The zero-divisor graph of is defined as a graph with vertex set and for vertices , there is a directed edge from to if . Anderson and Livingston [6] proved that the automorphism group of () is a direct product of some symmetric groups. Wong et al. [15] characterized the automorphisms of the zero-divisor graph, whose vertex set consists of all rank one upper triangular matrices over a finite field. Applying the main results of [15], Wang [13, 14] obtained the automorphisms of the zero-divisor graph on upper triangular matrix ring or full matrix ring with a finite base field, respectively. Inspired by these results, we can ask a natural problem: how about the automorphisms of a subspace sum graph? In addition, this is also an open problem for further research on subspace sum graphs in [8].
Hence, in present paper we focus on this problem and address it completely (see Section 3). Prior to presenting the proof of it, some lemmas are demonstrated in Section 2.
2 Preliminaries and Lemmas
Let be a finite field of order and with a prime integer. Throughout, we let be a finite dimensional vector space with the base field . The dimension of a subspace of is denoted by . Denote by the vertex set of . From the definition of graph , it is clear that if , then the vertex set is empty. If , one can easily see that is a complete graph (also see Theorem 3.1 of [8]). Hence, in the following, we always set . Let be a vector with the -th component 1 and the others 0. Then constructs a basis of and any vector can be uniquely expressed as with .
First, we present two standard automorphisms of as follows.
Invertible linear transformation
Let be an invertible matrix of order over . Define a mapping from to itself as
[TABLE]
Clearly, is an invertible linear transformation of , which naturally generates an automorphism of , also written as , such that for .
Field automorphism
Let be an automorphism of the base field and a mapping from to itself such that
[TABLE]
Then one can easily check that the mapping on generated by , also denoted by , sending to is an automorphism of .
Next, we give some lemmas which will be used later.
Lemma 2.1**.**
(Theorem 6.1, [8])* Let be a -dimensional subspace of over the base field . Then the degree of in , denoted by , is , where*
[TABLE]
From Lemma 2.1, it is clear that only depends on the dimension of , and if are of distinct dimensions , then . Thus Lemma 2.2 follows from Lemma 2.1 immediately.
Lemma 2.2**.**
Let be an automorphism of . Then sends each -dimensional subspace of to a subspace of equal dimension, where .
Let be a subset of and the subspace spanned by be . The -dimensional subspace spanned by a nonzero vector is denoted by .
Lemma 2.3**.**
Let be an automorphism of and a -dimensional subspace (). Suppose is a basis of and for . Then we have the following conclusions.
- (i)
* for *; 2. (ii)
* are linearly independent*; 3. (iii)
.
Proof. First, we show the proof of (i). Assume on the contrary that there exists which is not a subspace of . From Lemma 2.2, we see , then let be a basis of . From the assumption, we obtain that are linearly independent. Extend to a basis of . Let be a -dimensional subspace, then and . Thus is not adjacent to , i.e., . Furthermore, , and thus . Since is an automorphism, so is . Then applying to and , we get and , which contradicts with . As a result, the assumption does not hold, and the conclusion follows.
Second, We will apply induction on to complete the proof of (ii). First, if and are not linearly independent, then with . Thus , that is, , a contradiction.
Now suppose the conclusion holds for any and we will show the case when . Let () are linearly independent vectors and for . Extend to a basis of . Let and , then
[TABLE]
From the induction hypothesis, we know that are linearly independent. If are not linearly independent, then can be linearly expressed by , that is, with and . Then
[TABLE]
In addition, by the result of (i), for . Hence
[TABLE]
Consequently, we derive that
[TABLE]
which implies that , a contradiction. So, we say are linearly independent.
At last, we prove the result of (iii). Combining (i) and (ii), we see that and , which complete the proof.
The following lemma is a more general result of Lemma 2.3 (i).
Lemma 2.4**.**
Let be an automorphism of and () a -dimensional subspace. Suppose is a nonzero vector, we obtain
- (i)
if , then ; 2. (ii)
if is not a subspace of , then is not a subspace of .
Proof. Here we mainly give the proof of (i) and the other can be proved similarly. Since , then can be linearly expressed by . Let and for . From Lemma 2.3 (iii), . If is not a subspace of , i.e., is not a subspace of , then are linearly independent. Let be a -dimensional subspace. As is also an automorphism of , applying to together with Lemma 2.3, we obtain that
[TABLE]
which contradicts with Lemma 2.2.
By and , we respectively denote the identity matrix of order and the matrix unit of order with the -entry 1 and others 0. We use to denote the permutation matrix obtained from by permuting the -th row and the -th row (particularly, ). For an 1-dimensional subspace , since with , the expression can be not unique, but if we let the first nonzero component of be 1, then the expression become unique. For convenience, in the following, we always suppose that each 1-dimensional subspace is of this form.
Lemma 2.5**.**
Let be an automorphism of . Then there exists an invertible matrix such that the automorphism fixes each ().
Proof. Let and with and . Then we take
[TABLE]
It is obvious that . Set . Further, suppose . Since and are linearly independent, so are and from Lemma 2.3 (ii). Then we can write and . If (), then we take
[TABLE]
and one can easily check that fixes . Set . Note that also fixes .
Let and . It follows from Lemma 2.3 that . Since , by Lemma 2.4 (ii), we have is not a subspace of (i.e., ). Thus we can write and . Similarly, we can take an invertible matrix such that fixes and , . Proceeding in this method, we can find a matrix sequence such that fixes every (). Let , then is what we want.
3 Automorphisms of the subspace sum graph
In this section, we characterize the automorphisms of the subspace sum graph . Let be a finite dimensional vector space over a finite base field with .
Theorem 3.1**.**
A mapping on the vertex set of is an automorphism of if and only if can be uniquely decomposed as , where and are two standard automorphisms defined as above.
Proof. The proof for the sufficiency part is clear. In what follows, we present the proof for the necessity part. From Lemma 2.5, we can find an invertible matrix such that fixes every (). Let . The remaining proof is divided into the following claims.
Claim 1. For a nonzero vector with , assume , where with . Then if and only if ().
Let for . Since fixes every for , then from Lemma 2.3 (iii), for . If , then . Further, from Lemma 2.4 (i), , which implies that . By applying to the above, we obtain whenever .
It follows from Claim 1 that with and . Thus we define a function over such that and if and only if . Now we investigate the properties of .
Claim 2. * for .*
From Claim 1, suppose . For , let
[TABLE]
Then from Lemmas 2.3 and 2.5,
[TABLE]
Since , then after applying , we derive from Lemma 2.4(i) that
[TABLE]
which indicates that . Notice that , then the conclusion follows.
Claim 3. Suppose and , then
- (i)
; 2. (ii)
; 3. (iii)
.
Applying on , together with Lemmas 2.4 and 2.5 and Claim 2, we obtain
[TABLE]
which points out that . Taking respectively , and in (i), one can easily deduce the results of (ii) and (iii).
Claim 4. Let , then for and ,
- (i)
; 2. (ii)
* and *; 3. (iii)
* and *; 4. (iv)
; 5. (v)
* is an automorphism of *.
From Claim 3 (iii), we see for that
[TABLE]
Then
[TABLE]
which proves (i).
It follows from the definition of that . Then , and thus , which establishes (ii).
Since , after applying together with Lemma 2.4 and Claim 2, it follows that . Thus we obtain
[TABLE]
which, together with Claim 3 (i), imply that . Further, we obtain
[TABLE]
and if . The proof of (iii) is finished.
Note that , then if in (iv), the result is clear. Now suppose . Applying on , together with Lemma 2.4 and Claim 2, we deduce
[TABLE]
which indicates that
[TABLE]
As a consequence, holds.
Combining (i)-(iv), (v) follows immediately.
Claim 5. There exists a diagonal matrix such that fixes each 1-dimensional subspace.
From Claims 3 and 4, . Then the result of Claim 2 can be rewritten as
[TABLE]
Let and , then
[TABLE]
and further it is clear that fixes each 1-dimensional subspace. Set .
Now we are in a position to complete the proof.
Let be any -dimensional nontrivial proper subspace of and a basis of with . From Claim 5, fixes every 1-dimensional subspace, then fixes (). Using Lemma 2.3, we obtain fixes , i.e., fixes every vertex of . As a result, is the identity mapping. Thus
[TABLE]
with . Next, we show the decomposition of is unique. Suppose there are two distinct decompositions, , which implies that . Since fixes each , so does . Then we see is a diagonal matrix. Similarly, also fixes every for , from which it follows that is a scalar matrix. So, is a nonzero scalar multiple of , and thus . As a consequence, .
The proof of Theorem 3.1 is completed.
Note that the automorphism group of with is a cyclic group of order . Then the following result follows from Theorem 3.1 immediately.
Corollary 3.2**.**
The automorphism group of is isomorphic to , where is the quotient group of all the invertible matrices over to the normal subgroup of all the nonzero scalar matrices over .
Acknowledgement
The authors thank the anonymous referees for a careful and thorough reading of the paper and for valuable comments, and also thank the support of the Fundamental Research Funds for the Central Universities (No. 2017BSCXB53).
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