The relational complexity of linear groups acting on subspaces

Saul D. Freedman; Veronica Kelsey; Colva M. Roney-Dougal

doi:10.1515/jgth-2023-0262

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The relational complexity of linear groups acting on subspaces

Saul D. Freedman , Veronica Kelsey and Colva M. Roney-Dougal

Published/Copyright: February 14, 2024

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From the journal Journal of Group Theory Volume 27 Issue 5

Abstract

The relational complexity of a subgroup 𝐺 of Sym ⁢ ( Ω ) is a measure of the way in which the orbits of 𝐺 on Ω k for various 𝑘 determine the original action of 𝐺. Very few precise values of relational complexity are known. This paper determines the exact relational complexity of all groups lying between PSL n ⁢ ( F ) and PGL n ⁢ ( F ) , for an arbitrary field 𝔽, acting on the set of 1-dimensional subspaces of F n . We also bound the relational complexity of all groups lying between PSL n ⁢ ( q ) and P ⁢ Γ ⁢ L n ⁢ ( q ) , and generalise these results to the action on 𝑚-spaces for m ≥ 1 .

1 Introduction

The study of relational complexity began with work of Lachlan in model theory as a way of studying homogeneous relational structures: those in which every isomorphism between induced substructures extends to an automorphism of the whole structure. For the original definition, see, for example, [10]; an equivalent definition in terms of permutation groups was given by Cherlin [1] and, apart from a slight generalisation to group actions, is the one we now present.

Let Ω be an arbitrary set and let 𝐻 be a group acting on Ω. Fix k ∈ Z , and let X := ( x 1 , … , x k ) , Y := ( y 1 , … , y k ) ∈ Ω k . For r ≤ k , we say that 𝑋 and 𝑌 are 𝑟-equivalent under 𝐻, denoted X ∼ H , r Y , if, for every 𝑟-subset of indices { i 1 , … , i r } ⊆ { 1 , … , k } , there exists an h ∈ H such that

( x i 1 h , … , x i r h ) = ( y i 1 , … , y i r ) .

If X ∼ H , k Y , i.e. if Y ∈ X H , then 𝑋 and 𝑌 are equivalent under 𝐻. The relational complexity of 𝐻, denoted RC ⁢ ( H , Ω ) , or RC ⁢ ( H ) when Ω is clear, is the smallest r ≥ 1 such that X ∼ H , r Y implies Y ∈ X H for all X , Y ∈ Ω k and all k ≥ r . Equivalently, RC ⁢ ( H ) is the smallest 𝑟 such that 𝑟-equivalence of tuples implies equivalence of tuples. Note that RC ⁢ ( H ) ≥ 2 if H ≠ 1 and | Ω | > 1 , as 𝑋 or 𝑌 may contain repeated entries.

Calculating the precise relational complexity of a group is often very difficult. A major obstacle is that if K < H ≤ Sym ⁢ ( Ω ) , then there is no uniform relationship between RC ⁢ ( K , Ω ) and RC ⁢ ( H , Ω ) . For example, if n ≥ 4 , then the relational complexities of the regular action of C n and natural actions of A n and S n are 2, n − 1 and 2, respectively. In [1], Cherlin gave three families of finite primitive binary groups (groups with relational complexity two) and conjectured that this list was complete. In a dramatic recent breakthrough, this conjecture was proved by Gill, Liebeck and Spiga in [5]; this monograph also contains an extensive literature review.

In [1, 2], Cherlin determined the exact relational complexity of S n and A n in their actions on 𝑘-subsets of { 1 , … , n } . The relational complexity of the remaining large-base primitive groups is considered in [4]. Looking at finite primitive groups more generally, Gill, Lodà and Spiga proved in [6] that if H ≤ Sym ⁢ ( Ω ) is primitive and not large-base, then RC ⁢ ( H , Ω ) < 9 ⁢ log ⁡ | Ω | + 1 (our logarithms are to the base 2). This bound was tightened by the second and third author in [9] to 5 ⁢ log ⁡ | Ω | + 1 . Both [6] and [9] bounded the relational complexity via base size, and the groups with the largest upper bounds are classical groups acting on subspaces of the natural module, and related product action groups. This motivated us to obtain further information about the relational complexity of these groups; this paper confirms that these bounds are tight, up to constants.

We now fix some notation for use throughout this paper. Let 𝑛 be a positive integer, 𝔽 a (not necessarily finite) field, V = F n , and Ω m = P ⁢ G m ⁢ ( V ) , the set of 𝑚-dimensional subspaces of 𝑉. We shall study the relational complexity of the almost simple groups H ̄ with PSL n ⁢ ( F ) ⁢ ⊴ ⁢ H ̄ ≤ P ⁢ Γ ⁢ L n ⁢ ( F ) , acting on Ω m . We will generally work with the corresponding groups 𝐻 with SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) , as these naturally have the same relational complexity when acting on Ω m .

Several of our results focus on the case F = F q . We begin with the following theorem of Cherlin.

Theorem 1.1

Theorem 1.1 ([1, Example 3])

The relational complexity of GL n ⁢ ( q ) acting on the nonzero vectors of F q n is equal to 𝑛 when q = 2 , and n + 1 when q ≥ 3 . Hence also in the action on 1-spaces, we find that RC ⁢ ( PGL n ⁢ ( 2 ) , Ω 1 ) = n .

More generally, for

PSL n ⁢ ( q ) ⁢ ⊴ ⁢ H ̄ ≤ PGL n ⁢ ( q ) ,

Lodà [11, Corollary 5.2.7] shows that RC ⁢ ( H ̄ , Ω 1 ) < 2 ⁢ log ⁡ | Ω 1 | + 1 . Other results imply an alternative upper bound on RC ⁢ ( H ̄ , Ω 1 ) . We first note that the height of a permutation group 𝐾 on a set Ω, denoted H ⁢ ( K ) or H ⁢ ( K , Ω ) , is the maximum size of a subset Δ of Ω with the property that K ( Γ ) < K ( Δ ) for each Γ ⊊ Δ . It is easy to show (see [6, Lemma 2.1]) that RC ⁢ ( K ) ≤ H ⁢ ( K ) + 1 . By combining this with immediate generalisations of results of Hudson [8, §§5.3–5.4] and Lodà [11, Proposition 5.2.1], we obtain the following (for | F | = 2 , see Theorem 1.1; we also omit a few small exceptional cases for brevity).

Proposition 1.2

Let PSL n ⁢ ( F ) ⁢ ⊴ ⁢ H ̄ ≤ PGL n ⁢ ( F ) and | F | ≥ 3 .

Suppose that n = 2 , with | F | = q ≥ 7 if H ̄ ≠ PGL 2 ⁢ ( F ) . If | F | ≥ 4 , then
H ⁢ ( H ̄ , Ω 1 ) = 3 and RC ⁢ ( H ̄ , Ω 1 ) = n + 2 = 4 ,
whilst RC ⁢ ( PGL 2 ⁢ ( 3 ) , Ω 1 ) = n = 2 .
If n ≥ 3 , then H ⁢ ( H ̄ , Ω 1 ) = 2 ⁢ n − 2 and RC ⁢ ( H ̄ , Ω 1 ) ≤ 2 ⁢ n − 1 .

The following theorem gives the exact relational complexity of groups between PSL n ⁢ ( F ) and PGL n ⁢ ( F ) for n ≥ 3 , acting naturally on 1-spaces.

Theorem A

Let n ≥ 3 , and let 𝔽 be any field. Then the following hold.

We have
RC ⁢ ( PGL n ⁢ ( F ) , Ω 1 ) = { n if ⁢ | F | ≤ 3 , n + 2 if ⁢ | F | ≥ 4 .
If PSL n ⁢ ( F ) ⁢ ⊴ ⁢ H ̄ < PGL n ⁢ ( F ) , then
RC ⁢ ( H ̄ , Ω 1 ) = { 2 ⁢ n − 1 if ⁢ n = 3 , 2 ⁢ n − 2 if ⁢ n ≥ 4 .

For most groups, we see that the relational complexity is very close to the bound in Proposition 1.2 ii. However, the difference between the height and the relational complexity of PGL n ⁢ ( F ) increases with 𝑛 when | F | ≥ 3 . This addresses a recent question of Cherlin and Wiscons (see [5, p. 23]): there exists a family of finite primitive groups that are not large-base, where the difference between height and relational complexity can be arbitrarily large. Theorem A also provides infinitely many examples of almost simple groups H ̄ with RC ⁢ ( Soc ⁢ ( H ̄ ) ) > RC ⁢ ( H ̄ ) .

One way to interpret the gap between the relational complexity of PGL n ⁢ ( F ) and its proper almost simple subgroups with socle PSL n ⁢ ( F ) is to observe that preserving linear dependence and independence is a comparatively “local” phenomenon, requiring information about the images of 𝑛-tuples of subspaces but not (very much) more, whereas restricting determinants requires far richer information. This mimics the difference between the relational complexity of S n and A n in their natural actions, where requiring a map to be a permutation is “local”, but requiring a permutation to be even is a “global” property.

We next bound the relational complexity of the remaining groups with socle PSL n ⁢ ( q ) that act on Ω 1 . For k ∈ Z > 0 , the number of distinct prime divisors of 𝑘 is denoted by ω ⁢ ( k ) , with ω ⁢ ( 1 ) = 0 .

Theorem B

Let H ̄ satisfy PSL n ⁢ ( q ) ≤ H ̄ ≤ P ⁢ Γ ⁢ L n ⁢ ( q ) , and let

e := | H ̄ : H ̄ ∩ PGL n ( q ) | .

Suppose that e > 1 so that q ≥ 4 and H ̄ ≰ PGL n ⁢ ( q ) .

If n = 2 and q ≥ 8 , then
4 + ω ⁢ ( e ) ≥ RC ⁢ ( H ̄ , Ω 1 ) ≥ 4 ,
except that RC ⁢ ( P ⁢ Σ ⁢ L 2 ⁢ ( 9 ) , Ω 1 ) = 3 .
If n ≥ 3 , then
2 ⁢ n − 1 + ω ⁢ ( e ) ≥ RC ⁢ ( H ̄ , Ω 1 ) ≥ { n + 2 always , n + 3 if ⁢ PGL n ⁢ ( q ) < H ̄ , 2 ⁢ n − 2 if ⁢ H ̄ ≤ P ⁢ Σ ⁢ L n ⁢ ( q ) ≠ P ⁢ Γ ⁢ L n ⁢ ( q ) .

In fact, the lower bound of 2 ⁢ n − 2 holds for a larger family of groups; see Proposition 3.7.

Theorem C

Let H ̄ satisfy PSL n ⁢ ( q ) ≤ H ̄ ≤ P ⁢ Γ ⁢ L n ⁢ ( q ) and let

e := | H ̄ : H ̄ ∩ PGL n ( q ) | .

Fix m ∈ { 2 , … , ⌊ n 2 ⌋ } . Then

( m + 1 ) ⁢ n − 2 ⁢ m + 2 + ω ⁢ ( e ) ≥ RC ⁢ ( H ̄ , Ω m ) ≥ m ⁢ n − m 2 + 1 .

GAP [13] calculations using [3] yield RC ⁢ ( P ⁢ Γ ⁢ L 2 ⁢ ( 3 5 ) , Ω 1 ) = 5 = 4 + ω ⁢ ( 5 ) and RC ⁢ ( P ⁢ Γ ⁢ L 4 ⁢ ( 9 ) , Ω 1 ) = 8 = 7 + ω ⁢ ( 2 ) , so the upper bounds of Theorem B cannot be improved in general. On the other hand, RC ⁢ ( P ⁢ Γ ⁢ L 3 ⁢ ( 2 6 ) , Ω 1 ) achieves the lower bound of 6 = 3 + 3 < 7 = 5 + ω ⁢ ( 6 ) . Additionally, RC ⁢ ( PSL 4 ⁢ ( 2 ) , Ω 2 ) achieves the lower bound of 5 from Theorem C, while RC ⁢ ( PSL 4 ⁢ ( 3 ) , Ω 2 ) = 6 and

RC ⁢ ( PGL 4 ⁢ ( 3 ) , Ω 2 ) = RC ⁢ ( PSL 4 ⁢ ( 4 ) , Ω 2 ) = RC ⁢ ( P ⁢ Γ ⁢ L 4 ⁢ ( 4 ) , Ω 2 ) = 8 .

It is straightforward to use our results to bound the relational complexity in terms of the degree. For example, RC ⁢ ( PGL n ⁢ ( q ) , Ω 1 ) < log ⁡ ( | Ω 1 | ) + 3 . Many of our arguments also apply to the case where 𝔽 is an arbitrary field; see Theorem 3.1, Lemmas 3.5 and 3.6, and Propositions 3.7 and 4.1.

This paper is structured as follows. In Section 2, we fix some more notation and prove some elementary lemmas, then prove upper bounds on the relational complexity of the relevant actions on 1-spaces. In Section 3, we shall prove corresponding lower bounds, and then prove Theorems A and B. Finally, in Section 4, we prove Theorem C.

2 Action on 1-spaces: Upper bounds

In this section, we present several preliminary lemmas, and then determine upper bounds for the relational complexity of groups 𝐻, with SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ GL n ⁢ ( F ) , acting on Ω 1 .

We begin with some notation that we will use throughout the remainder of the paper. Let { e 1 , … , e n } be a basis for 𝑉. For a set Γ, a tuple X = ( x i ) i = 1 k ∈ Γ k and a permutation σ ∈ S k , we write X σ to denote the 𝑘-tuple ( x 1 σ − 1 , … , x k σ − 1 ) . For a tuple X ∈ Ω m k , we write ⟨ X ⟩ to denote the subspace of 𝑉 spanned by all entries in 𝑋. For i ∈ { 1 , … , k } , we shall write ( X ∖ x i ) to denote the subtuple of 𝑋 obtained by deleting x i .

In the remainder of this section, let Ω := Ω 1 = P ⁢ G 1 ⁢ ( V ) and let 𝐻 be a group such that

SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ GL n ⁢ ( F ) .

Recall from Theorem 1.1 that RC ⁢ ( GL n ⁢ ( F ) , Ω ) = n when | F | = 2 . Thus we shall assume throughout this section that | F | ≥ 3 and n ≥ 2 .

We write 𝐷 to denote the subgroup of diagonal matrices of GL n ⁢ ( F ) (with respect to the basis { e 1 , … , e n } ), and Δ := { ⟨ e i ⟩ ∣ i ∈ { 1 , … , n } } . Observe that 𝐷 is nontrivial since | F | > 2 , and that D ∩ H is the pointwise stabiliser H ( Δ ) . For a vector v = ∑ i = 1 n α i ⁢ e i ∈ V , the support supp ⁢ ( v ) of 𝑣 is the set

{ i ∈ { 1 , … , n } ∣ α i ≠ 0 } .

Additionally, the support supp ⁢ ( W ) of a subset 𝑊 of 𝑉 is the set ⋃ w ∈ W supp ⁢ ( w ) , and similarly for tuples. In particular, Δ is the set of subspaces of 𝑉 with support of size 1, and supp ⁢ ( W ) = supp ⁢ ( ⟨ W ⟩ ) for all subsets 𝑊 of 𝑉.

2.1 Preliminaries

We begin our study of the action of 𝐻 on Ω with a pair of lemmas that will enable us to consider only tuples of a very restricted form.

Lemma 2.1

Let k ≥ n , and let X , Y ∈ Ω k be such that X ∼ H , n Y . Additionally, let a := dim ⁢ ( ⟨ X ⟩ ) . Then there exist X ′ = ( x 1 ′ , … , x k ′ ) , Y ′ = ( y 1 ′ , … , y k ′ ) ∈ Ω k such that

x i ′ = y i ′ = ⟨ e i ⟩ for i ∈ { 1 , … , a } , and
X ∼ H , r Y if and only if X ′ ∼ H , r Y ′ for each r ∈ { 1 , … , k } .

Proof

Observe that there exists σ ∈ S k such that ⟨ X σ ⟩ = ⟨ x 1 σ − 1 , … , x a σ − 1 ⟩ . Since X ∼ H , n Y and a ≤ n , the definition of 𝑎-equivalence yields X σ ∼ H , a Y σ . Hence there exists an f ∈ H such that x i σ − 1 f = y i σ − 1 for all i ∈ { 1 , … , a } , and so ⟨ Y σ ⟩ = ⟨ y 1 σ − 1 , … , y a σ − 1 ⟩ . Since SL n ⁢ ( F ) is transitive on 𝑛-tuples of linearly independent 1-spaces, there exists h ∈ SL n ⁢ ( F ) ≤ H such that

x i σ − 1 f ⁢ h = y i σ − 1 h = ⟨ e i ⟩ for ⁢ i ∈ { 1 , … , a } .

Define X ′ , Y ′ ∈ Ω k by

x i ′ = x i σ − 1 f ⁢ h and y i ′ = y i σ − 1 h

so that X ′ = X σ ⁢ f ⁢ h and Y ′ = Y σ ⁢ h . Then X ′ ∼ H , r Y ′ if and only if X σ ∼ H , r Y σ , which holds if and only if X ∼ H , r Y . ∎

Lemma 2.2

Let k ≥ r ≥ n , and let X , Y ∈ Ω k be such that X ∼ H , r Y . Additionally, let a := dim ⁢ ( ⟨ X ⟩ ) and assume that a < n . If a = 1 , or if

RC ⁢ ( GL a ⁢ ( F ) , P ⁢ G 1 ⁢ ( F a ) ) ≤ r ,

then Y ∈ X H .

Proof

If a = 1 , then all entries of 𝑋 are equal, so since r ≥ n ≥ 2 , we see that X ∼ H , r Y directly implies Y ∈ X H . We will therefore suppose that a ≥ 2 and RC ⁢ ( GL a ⁢ ( F ) , P ⁢ G 1 ⁢ ( F a ) ) ≤ r . By Lemma 2.1, we may assume without loss of generality that ⟨ X ⟩ = ⟨ Y ⟩ = ⟨ e 1 , … , e a ⟩ . As X ∼ H , r Y and

RC ⁢ ( GL a ⁢ ( F ) , P ⁢ G 1 ⁢ ( F a ) ) ≤ r ,

there exists an element g ∈ GL a ⁢ ( F ) mapping 𝑋 to 𝑌, considered as tuples of subspaces of ⟨ e 1 , … , e a ⟩ . We now let ℎ be the diagonal matrix

diag ⁢ ( det ( g − 1 ) , 1 , … , 1 ) ∈ GL n − a ⁢ ( F )

and observe that g ⊕ h ∈ SL n ⁢ ( F ) maps 𝑋 to 𝑌 and lies in 𝐻, since SL n ⁢ ( F ) lies in 𝐻. Thus Y ∈ X H . ∎

We now begin our study of some particularly nice 𝑘-tuples.

Lemma 2.3

Let k ≥ n + 1 , and let X , Y ∈ Ω k be such that x i = y i = ⟨ e i ⟩ for i ∈ { 1 , … , n } and X ∼ H , n + 1 Y . Then supp ⁢ ( x i ) = supp ⁢ ( y i ) for all i ∈ { 1 , … , k } .

Proof

It is clear that supp ⁢ ( x i ) = { i } = supp ⁢ ( y i ) when i ∈ { 1 , … , n } . Assume therefore that i > n . Since X ∼ H , n + 1 Y , there exists g ∈ H such that

( ⟨ e 1 ⟩ , … , ⟨ e n ⟩ , x i ) g = ( ⟨ e 1 ⟩ , … , ⟨ e n ⟩ , y i ) .

Observe that g ∈ H ( Δ ) = D ∩ H , and so supp ⁢ ( x i ) = supp ⁢ ( y i ) . ∎

Our final introductory lemma collects several elementary observations regarding tuples of subspaces in Δ ̄ := Ω ∖ Δ , the set of 1-dimensional subspaces of support size greater than 1. For r ≥ 1 and A , B ∈ Δ ̄ r , we let M A , B consist of all matrices in M n , n ⁢ ( F ) that fix ⟨ e i ⟩ for 1 ≤ i ≤ n and map a j into b j for 1 ≤ j ≤ r . Notice that all matrices in M A , B are diagonal, and that if g , h ∈ M A , B , then a j g + h = a j g + a j h ≤ b j , so M A , B is a subspace of M n , n . For an n × n matrix g = ( g i ⁢ j ) and a subset 𝐼 of { 1 , … , n } , we write g | I to denote the submatrix of 𝑔 consisting of the rows and columns with indices in 𝐼.

Lemma 2.4

Let r ≥ 1 , and let A = ( a 1 , … , a r ) , B = ( b 1 , … , b r ) ∈ Δ ̄ r .

Let a i and a j be (possibly equal) elements of 𝐴 such that
supp ⁢ ( a i ) ∩ supp ⁢ ( a j ) ≠ ∅ ,
and let g ∈ M A , A . Then g | supp ⁢ ( a i , a j ) is a scalar.
Suppose A ∼ D , 1 B . Then, for 1 ≤ i ≤ r , the space ( M ( a i ) , ( b i ) ) | supp ⁢ ( a i ) is one-dimensional, so the dimension of M ( a i ) , ( b i ) is equal to n + 1 − | supp ⁢ ( a i ) | .
For a subtuple A ′ of 𝐴, let S := { 1 , … , n } ∖ supp ⁢ ( A ′ ) . Then
dim ⁢ ( ( M A ′ , A ′ ) | S ) = | S | .

Proof

Part i is clear. For part ii, by assumption, there is an invertible diagonal matrix mapping a i to b i , so supp ⁢ ( a i ) = supp ⁢ ( b i ) . Let 𝑘 and ℓ be distinct elements of supp ⁢ ( a i ) , which exist as a i ∈ Δ ̄ . If a i g ≤ b i , then e k g = λ ⁢ e k for some λ ∈ F , and the value of 𝜆 completely determines the image μ ⁢ e ℓ of e ℓ under 𝑔. The result follows. Part iii holds since, for all g ∈ M A ′ , A ′ , s ∈ S , λ ∈ F and a ∈ A ′ , the matrix obtained from 𝑔 by adding 𝜆 to its 𝑠-th diagonal entry fixes 𝑎. ∎

2.2 Upper bounds for SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ GL n ⁢ ( F ) on 1-spaces

In this subsection, we will suppose that n ≥ 4 and | F | ≥ 3 , and let 𝐻 be any group such that SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ GL n ⁢ ( F ) . Our main result is Theorem 2.7, which gives upper bounds on RC ⁢ ( H , Ω ) .

Lemma 2.5

If X ∼ H , 2 ⁢ n − 2 Y implies that X ∼ H , 2 ⁢ n − 1 Y for all X , Y ∈ Ω 2 ⁢ n − 1 with x i = y i = ⟨ e i ⟩ for i ∈ { 1 , … , n } , then RC ⁢ ( H , Ω ) ≤ 2 ⁢ n − 2 .

Proof

Let 𝑘 be at least 2 ⁢ n − 1 , and let A , B ∈ Ω k satisfy A ∼ H , 2 ⁢ n − 2 B . Let A ′ be a subtuple of 𝐴 of length 2 ⁢ n − 1 , and B ′ the corresponding ( 2 ⁢ n − 1 ) -subtuple of 𝐵. We shall show that B ′ ∈ A ′ H for all such A ′ and B ′ so that A ∼ H , 2 ⁢ n − 1 B . It will then follow from Proposition 1.2 ii that A ∈ B H , as required.

Let a := dim ⁢ ( ⟨ A ′ ⟩ ) , and suppose first that a < n . We observe from Proposition 1.2 that if a ≥ 2 , then RC ⁢ ( GL a ⁢ ( F ) , P ⁢ G 1 ⁢ ( F a ) ) < 2 ⁢ n − 2 . As A ′ ∼ H , 2 ⁢ n − 2 B ′ , Lemma 2.2 yields B ′ ∈ A ′ H . If instead a = n , then by Lemma 2.1, there exist 𝑋 and 𝑌 in Ω 2 ⁢ n − 1 such that x i = y i = ⟨ e i ⟩ for each i ∈ { 1 , … , n } , and such that, for all r ≥ 1 , the relations A ′ ∼ H , r B ′ and X ∼ H , r Y are equivalent. Now, A ′ ∼ H , 2 ⁢ n − 2 B ′ , so X ∼ H , 2 ⁢ n − 2 Y . If X ∼ H , 2 ⁢ n − 2 Y implies that X ∼ H , 2 ⁢ n − 1 Y , then B ′ ∈ A ′ H , as required. ∎

We shall therefore let 𝑋 and 𝑌 be elements of Ω 2 ⁢ n − 1 with x i = y i = ⟨ e i ⟩ for i ∈ { 1 , … , n } such that X ∼ H , 2 ⁢ n − 2 Y . Additionally, for i ∈ { 1 , … , 2 ⁢ n − 1 } and j ∈ { 1 , … , n } , define

α i ⁢ j , β i ⁢ j ∈ F so that x i = ⟨ ∑ j = 1 n α i ⁢ j ⁢ e j ⟩ and y i = ⟨ ∑ j = 1 n β i ⁢ j ⁢ e j ⟩ .

Lemma 2.6

With the notation above, if at least one of the following holds, then Y ∈ X H .

There exist i , j ∈ { n + 1 , … , 2 ⁢ n − 1 } with i ≠ j and supp ⁢ ( x i ) ⊆ supp ⁢ ( x j ) .
There exists a nonempty R ⊆ { n + 1 , … , 2 ⁢ n − 1 } with | ⋂ i ∈ R supp ⁢ ( x i ) | = 1 .
There exists i ∈ { n + 1 , … , 2 ⁢ n − 1 } such that supp ⁢ ( x i ) ≥ 4 .

Proof

We begin by noting that Lemma 2.3 yields supp ⁢ ( y i ) = supp ⁢ ( x i ) for all i ∈ { 1 , … , 2 ⁢ n − 1 } .

i Since X ∼ H , 2 ⁢ n − 2 Y , there exists an h ∈ H mapping ( X ∖ x i ) to ( Y ∖ y i ) , and such an ℎ is necessarily diagonal, with fixed entries in supp ⁢ ( x j ) (up to scalar multiplication).

Now, let ℓ ∈ { n + 1 , … , 2 ⁢ n − 1 } ∖ { i , j } (this is possible as n ≥ 4 ). There exists an h ′ ∈ H mapping ( X ∖ x ℓ ) to ( Y ∖ y ℓ ) , and as before, each such h ′ is diagonal. Hence every matrix in H ∩ D mapping x j to y j maps x i to y i , and in particular, x i h = y i , and so X h = Y .

ii Let { ℓ } := ⋂ i ∈ R supp ⁢ ( x i ) . Then α i ⁢ ℓ ≠ 0 for all i ∈ R . Since X ∼ H , 2 ⁢ n − 2 Y , there exists h ∈ H such that ( X ∖ x ℓ ) h = ( Y ∖ y ℓ ) . For all k ∈ { 1 , … , n } ∖ { ℓ } , it follows that there exists γ k ∈ F * such that e k h = γ k ⁢ e k . Thus, for each i ∈ R ,

y i = x i h = ⟨ ∑ k ∈ supp ⁢ ( x i ) α i ⁢ k ⁢ e k h ⟩ = ⟨ α i ⁢ ℓ ⁢ e ℓ h + ∑ k ∈ supp ⁢ ( x i ) ∖ { ℓ } α i ⁢ k ⁢ γ k ⁢ e k ⟩ .

Since α i ⁢ ℓ ≠ 0 , we deduce that supp ⁢ ( e ℓ h ) ⊆ supp ⁢ ( y i ) = supp ⁢ ( x i ) . As this holds for all i ∈ R , we obtain supp ⁢ ( e ℓ h ) = { ℓ } . Thus x ℓ h = ⟨ e ℓ ⟩ h = ⟨ e ℓ ⟩ = y ℓ , so X h = Y .

iii Permute the last n − 1 coordinates of 𝑋 and 𝑌 so that supp ⁢ ( x n + 1 ) ≥ 4 . By ii, we may assume that x i ∉ Δ for all i ≥ n + 1 . We define

X n + 1 k := ( x n + 1 , … , x k ) and Y n + 1 k := ( y n + 1 , … , y k )

for each k ∈ { n + 1 , … , 2 ⁢ n − 1 } . As supp ⁢ ( x i ) = supp ⁢ ( y i ) for all 𝑖, we see that X n + 1 k ∼ D , 1 Y n + 1 k , so X n + 1 k and Y n + 1 k satisfy the conditions of Lemma 2.4 ii.

Suppose first that there exists j ∈ { n + 2 , … , 2 ⁢ n − 1 } such that

M X n + 1 j , Y n + 1 j = M X n + 1 j − 1 , Y n + 1 j − 1 .

As X ∼ H , 2 ⁢ n − 2 Y , there exists h ∈ H ∩ D such that ( X ∖ x j ) h = ( Y ∖ y j ) . Hence

h ∈ M X n + 1 j − 1 , Y n + 1 j − 1 , and so h ∈ M X n + 1 j , Y n + 1 j .

Therefore, x j h = y j and X h = Y .

Hence we may assume instead that

M X n + 1 j , Y n + 1 j < M X n + 1 j − 1 , Y n + 1 j − 1 for all ⁢ j ∈ { n + 2 , … , 2 ⁢ n − 1 } .

Then

dim ⁢ ( M X n + 1 j , Y n + 1 j ) ≤ dim ⁢ ( M X n + 1 j − 1 , Y n + 1 j − 1 ) − 1 .

Lemma 2.4 ii yields

dim ⁢ ( M X n + 1 n + 1 , Y n + 1 n + 1 ) ≤ n − 3 ,

and hence

M X n + 1 2 ⁢ n − 2 , Y n + 1 2 ⁢ n − 2 = { 0 } = M X n + 1 2 ⁢ n − 1 , Y n + 1 2 ⁢ n − 2 ,

contradicting our assumption. ∎

We now prove the main result of this subsection.

Theorem 2.7

Suppose that n ≥ 4 and | F | ≥ 3 , and let 𝐻 be any group with SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ GL n ⁢ ( F ) . Then RC ⁢ ( H , Ω ) ≤ 2 ⁢ n − 2 .

Proof

Let X , Y ∈ Ω 2 ⁢ n − 1 be as defined before Lemma 2.6. By Lemma 2.5, it suffices to show that Y ∈ X H , so assume otherwise. We may also assume that all subspaces in 𝑋 are distinct so that

| supp ⁢ ( x i ) | ∈ { 2 , 3 } for each ⁢ i ∈ { n + 1 , … , 2 ⁢ n − 1 }

by Lemma 2.6 iii. For k ∈ { 2 , 3 } , let R k be the set of all i ∈ { n + 1 , … , 2 ⁢ n − 1 } such that | supp ⁢ ( x i ) | = k . Then

(2.1) | R 2 | + | R 3 | = n − 1 .

Observe from Lemma 2.6 i–ii that if i ∈ R 2 , then supp ⁢ ( x i ) ∩ supp ⁢ ( x j ) = ∅ for each j ∈ { n + 1 , … , 2 ⁢ n − 1 } ∖ { i } . Hence 2 ⁢ | R 2 | ≤ n and

(2.2) | U | := | ⋃ j ∈ R 3 supp ⁢ ( x j ) | ≤ | { 1 , … , n } ∖ ( ⋃ ̇ i ∈ R 2 supp ⁢ ( x i ) ) | = n − 2 ⁢ | R 2 | .

Observe next that | R 3 | ≥ 1 , else | R 2 | = n − 1 by (2.1), contradicting 2 ⁢ | R 2 | ≤ n . We shall determine an expression for | U | involving | R 3 | , by considering the maximal subsets 𝑃 of R 3 that correspond to pairwise overlapping supports. To do so, define a relation ∼ on R 3 by i ∼ j if supp ⁢ ( x i ) ∩ supp ⁢ ( x j ) ≠ ∅ , let 𝒫 be the set of equivalence classes of the transitive closure of ∼ , and let P ∈ P . We claim that

| ⋃ c ∈ P supp ⁢ ( x c ) | = 2 + | P | .

By Lemma 2.6 i–ii, | supp ⁢ ( x i ) ∩ supp ⁢ ( x j ) | ∈ { 0 , 2 } for all distinct i , j ∈ R 3 . Thus our claim is clear if | P | ∈ { 1 , 2 } .

If instead | P | ≥ 3 , then there exist distinct c 1 , c 2 , c 3 ∈ P with c 1 ∼ c 2 and c 2 ∼ c 3 . Let I := ⋂ i = 1 3 supp ⁢ ( x c i ) . We observe that | I | ≠ 0 , and so Lemma 2.6 ii shows that 𝐼 has size two and is equal to supp ⁢ ( x c 1 ) ∩ supp ⁢ ( x c 3 ) . Hence c 1 ∼ c 3 and

⋃ i = 1 3 supp ⁢ ( x c i ) = I ∪ ̇ ( ⋃ ̇ i = 1 3 ( supp ⁢ ( x c i ) ∖ I ) ) .

If | P | > 3 , then there exists c 4 ∈ P ∖ { c 1 , c 2 , c 3 } such that, without loss of generality, c 4 ∼ c 1 . As c 1 ∼ c j for each j ∈ { 2 , 3 } , the above argument shows that

⋂ i ∈ { 1 , j , 4 } supp ⁢ ( x c i ) = I and ⋃ i = 1 4 supp ⁢ ( x c i ) = I ∪ ̇ ( ⋃ ̇ i = 1 4 ( supp ⁢ ( x c i ) ∖ I ) ) .

Repeating this argument inductively on | P | shows that

⋃ c ∈ P supp ⁢ ( x c ) = I ∪ ̇ ( ⋃ ̇ c ∈ P ( supp ⁢ ( x c ) ∖ I ) ) ,

which has size 2 + | P | , as claimed.

Finally, let r ≥ 1 be the number of parts of 𝒫. As | R 3 | = ∑ P ∈ P | P | , it follows from our claim that | U | = 2 ⁢ r + | R 3 | ≥ 2 + | R 3 | . Thus (2.2) yields

2 + | R 3 | ≤ n − 2 ⁢ | R 2 | .

Hence 2 ⁢ | R 2 | + | R 3 | ≤ n − 2 < n − 1 , which is equal to | R 2 | + | R 3 | by (2.1), a contradiction. ∎

2.3 Upper bounds for GL n ⁢ ( F ) on 1-spaces

In this subsection, we determine a much smaller upper bound on RC ⁢ ( GL n ⁢ ( F ) , Ω ) via our main result, Theorem 2.12. We shall assume throughout that 𝑛 and | F | are at least 3, and write G := GL n ⁢ ( F ) . Since 𝐷 is the pointwise stabiliser of Δ in 𝐺, we will prove Theorem 2.12 by combining Lemmas 2.1 and 2.2 with information about the action of 𝐷 on 𝑟-tuples 𝐴 and 𝐵 of subspaces in Δ ̄ = Ω ∖ Δ . If these tuples are ( r − 1 ) -equivalent under 𝐷, then by acting on one with a suitable element of Δ ̄ , we may assume that their first r − 1 entries are equal. We shall denote the nonzero entries of elements 𝑔 of 𝐷 by just g 1 , … , g n rather than g 11 , … , g n ⁢ n since 𝑔 is necessarily diagonal.

Lemma 2.8

Let r ≥ 3 , and let A , B ∈ Δ ̄ r be such that

( a 1 , … , a r − 1 ) = ( b 1 , … , b r − 1 ) , A ∼ D , r − 1 B , and B ∉ A D .

Let C = { a 1 , … , a r − 1 } and assume also that supp ⁢ ( C ) = { 1 , … , n } . Then (after reordering the basis for 𝑉 and ( a 1 , … , a r − 1 ) if necessary) the following statements hold.

There exist integers
2 ≤ i 1 < i 2 < ⋯ < i r − 1 = n
such that, for each t ∈ { 1 , … , r − 1 } , supp ⁢ ( a 1 , … , a t ) is equal to { 1 , … , i t } .
Let t ∈ { 1 , … , r − 3 } . Then
supp ⁢ ( a t ) ∩ supp ⁢ ( a u ) = ∅ for all ⁢ u ∈ { t + 2 , … , r − 1 } .
The support of a 2 does not contain 1.
Let t ∈ { 1 , … , r − 1 } . Then i t ∈ supp ⁢ ( a t ) ∩ supp ⁢ ( a t + 1 ) .
Each integer in supp ⁢ ( a r ) lies in the support of a unique subspace in 𝐶.

Proof

We begin by fixing notation related to

a r = ⟨ ∑ ℓ = 1 n α ℓ ⁢ e ℓ ⟩ and b r = ⟨ ∑ ℓ = 1 n β ℓ ⁢ e ℓ ⟩ .

Since A ∼ D , r − 1 B , there exists an element in 𝐷 mapping a r to b r , and so it follows that supp ⁢ ( b r ) = supp ⁢ ( a r ) . On the other hand, B ∉ A D , and so a r ≠ b r . Therefore, by scaling the basis vectors for a r and b r , there exist j , k ∈ { 1 , … , n } such that j < k , α j = β j = 1 , and α k and β k are distinct and nonzero. Reordering { e 1 , … , e n } if necessary, we may assume that j = 1 . Then each element of 𝐷 that maps a r to b r also maps ⟨ e 1 + α k ⁢ e k ⟩ to ⟨ e 1 + β k ⁢ e k ⟩ ; we will use this fact throughout the proof.

i We show first that there is no partition of 𝐶 into proper subsets C ′ and C ′′ such that supp ⁢ ( C ′ ) ∩ supp ⁢ ( C ′′ ) = ∅ , so suppose otherwise, for a contradiction. Then, as | C ′ | < r − 1 and A ∼ D , r − 1 B , there exists an f ∈ D ( C ′ ) such that a r f = b r . Multiplying 𝑓 by a scalar if necessary, we may assume that f 1 = 1 . Then f i = β i / α i for each i ∈ supp ⁢ ( a r ) . Similarly, there exists g ∈ D ( C ′′ ) with the same properties. As supp ⁢ ( C ) ′ ∩ supp ⁢ ( C ) ′′ = ∅ , there exists an h ∈ D such that h | supp ⁢ ( C ) ′ = f | supp ⁢ ( C ) ′ and h | supp ⁢ ( C ) ′′ = g | supp ⁢ ( C ) ′′ . Since supp ⁢ ( C ) = { 1 , … , n } , we observe that h | supp ⁢ ( a r ) = f | supp ⁢ ( a r ) = g | supp ⁢ ( a r ) . Hence a r h = b r . Furthermore, by construction, h ∈ D ( C ′ ) ∩ D ( C ′′ ) = D C . Thus B ∈ A D , a contradiction.

By reordering a 1 , … , a r − 1 if necessary, we may assume that 1 ∈ supp ⁢ ( a 1 ) . Then, by reordering { e 2 , … , e n } if necessary, we may assume that supp ⁢ ( a 1 ) is equal to { 1 , 2 , … , i 1 } for some i 1 ≥ 2 since a 1 ∈ Δ ̄ . Thus the result holds for t = 1 . We will use induction to prove the result in general, and to show that, for all s ∈ { 2 , … , r − 1 } ,

(2.3) there exists ⁢ w ∈ { 1 , … , s − 1 } ⁢ such that ⁢ supp ⁢ ( a s ) ∩ supp ⁢ ( a w ) ≠ ∅ .

Let t ∈ { 2 , … , r − 1 } , let U t − 1 := { a 1 , … , a t − 1 } , and assume inductively that

supp ⁢ ( U t − 1 ) = { 1 , 2 , … , i t − 1 } .

If t ≥ 3 , assume also that (2.3) holds for all s ∈ { 2 , … , t − 1 } . Since 𝐶 cannot be partitioned into two parts whose support has trivial intersection,

supp ⁢ ( a 1 , … , a t − 1 ) ∩ supp ⁢ ( a t , … , a r − 1 ) ≠ ∅ ,

so we may reorder { a t , … , a r − 1 } so that (2.3) holds when s = t .

Suppose for a contradiction that supp ⁢ ( a t ) ⊆ supp ⁢ ( U t − 1 ) . Then (2.3) (applied to each s ∈ { 2 , … , t − 1 } ) and Lemma 2.4 imply that D ( C ) is equal to D ( C ∖ a t ) . Since A ∼ D , r − 1 B , the latter stabiliser contains an element mapping a r to b r . Hence the same is true for D ( C ) , contradicting the fact that B ∉ A D . Therefore, we can reorder { e i t − 1 + 1 , … , e n } so that supp ⁢ ( a t ) contains { i t − 1 + 1 , … , i t } for some i t > i t − 1 , and the result and (2.3) follow by induction. Note in particular that i r − 1 = n since supp ⁢ ( C ) = { 1 , … , n } .

ii Let m ∈ { 1 , … , r − 1 } be such that supp ⁢ ( a m ) contains the integer 𝑘 from the first paragraph of this proof, and let I := { 1 , … , m } . Then, using (2.3) (for each s ∈ I ∖ { 1 } ) and Lemma 2.4 i, we observe that every g ∈ D ( a 1 , … , a m ) satisfies g 1 = g k . Therefore, a r g ≠ b r for all g ∈ D ( a 1 , … , a m ) . As A ∼ D , r − 1 B , we deduce that m = r − 1 . In particular, a r − 1 is the unique subspace in 𝐶 whose support contains 𝑘. Swapping e k and e n if necessary, we may assume that k = n .

Now, for a contradiction, suppose that

supp ⁢ ( a t ) ∩ supp ⁢ ( a u ) ≠ ∅

for some t ∈ { 1 , … , r − 3 } and u ∈ { t + 2 , … , r − 1 } , and assume that 𝑢 is the largest integer with this property. Then (2.3) and the maximality of 𝑢 imply that supp ⁢ ( a s ) ∩ supp ⁢ ( a s − 1 ) ≠ ∅ for all s ∈ { u + 1 , … , r − 1 } . It now follows from Lemma 2.4 i, together with a further application of (2.3) to each s ∈ { 2 , … , t } , that every g ∈ E := D ( a 1 , … , a t , a u , … , a r − 1 ) satisfies g 1 = g n . Therefore, a r g ≠ b r for all g ∈ E . However, | ( a 1 , … , a t , a u , … , a r − 1 ) | < r − 1 , contradicting the fact that A ∼ D , r − 1 B .

iii–iv As in the proof of ii, we may assume that k = n . We observe from ii and (2.3) that supp ⁢ ( a t ) ∩ supp ⁢ ( a t + 1 ) ≠ ∅ for all t ≤ r − 2 . Hence if 1 ∈ supp ⁢ ( a 2 ) , then Lemma 2.4 i shows that every g ∈ D ( a 2 , … , a r − 1 ) satisfies g 1 = g n (since k = n ). This contradicts the fact that A ∼ D , r − 1 B , and so iii holds. Finally, since supp ⁢ ( a t ) ∩ supp ⁢ ( a t + 1 ) ≠ ∅ for each t ≤ r − 2 , we obtain iv by defining i 0 := 1 and reordering the vectors in { e i t − 1 + 1 , … , e i t } if necessary. In particular, for t = r − 1 , the assumption that i r − 1 = n = k ∈ supp ⁢ ( a r ) gives the result.

v Suppose for a contradiction that some ℓ ∈ supp ⁢ ( a r ) lies in the support of more than one subspace in 𝐶. If r = 3 , then ℓ ∈ supp ⁢ ( a 1 ) ∩ supp ⁢ ( a 2 ) and we define t := 2 . If instead r > 3 , then ii implies that ℓ ∈ supp ⁢ ( a t ) for at least one t ∈ { 2 , … , r − 2 } . In either case, we deduce ℓ ≠ 1 since 1 ∉ supp ⁢ ( a 2 ) by iii, and 1 ∉ supp ⁢ ( a u ) for u ∈ { 3 , … , r − 2 } by i–ii. Furthermore, i shows that ℓ ≠ i r − 1 = n .

Suppose first that α ℓ = β ℓ ( ≠ 0 ). Since the supports of a t , a t + 1 , … , a r − 1 consecutively overlap, Lemma 2.4 i shows that g ℓ = g n for each g ∈ D ( a t , … , a r − 1 ) . Since α n ≠ β n , no such 𝑔 maps a r to b r , contradicting the fact that A ∼ D , r − 1 B . Hence α ℓ ≠ β ℓ . However, each g ∈ D a 1 satisfies g 1 = g ℓ if r = 3 , as does each g ∈ D ( a 1 , … , a t ) if r > 3 . Again, no such matrix 𝑔 maps a r to b r , a contradiction. ∎

Recall that 𝐺 denotes GL n ⁢ ( F ) , with n , | F | ≥ 3 . Our next result is a key ingredient in the proof that RC ⁢ ( G , Ω ) is at most n + 2 .

Lemma 2.9

Let r ≥ 2 , and let A , B ∈ Δ ̄ r be such that A ∼ D , r − 1 B and B ∉ A D . Then there exists a subset Γ of Δ of size n + 2 − r such that B ∉ A G ( Γ ) .

Proof

If r = 2 , then set Γ = Δ . Since G ( Γ ) = D and B ∉ A D , we are done. Assume therefore that r ≥ 3 . We will suppose for a contradiction that 𝑛 is the smallest dimension for which the present lemma does not hold, for this value of 𝑟. Since A ∼ D , r − 1 B , we may also assume that ( a 1 , … , a r − 1 ) = ( b 1 , … , b r − 1 ) . Let C = { a 1 , … , a r − 1 } . As B ∉ A D , no element of D ( C ) maps a r to b r . Therefore, B ∉ A G ( Γ ) for a given subset Γ of Δ if and only if no element of G ( Γ ∪ C ) maps a r to b r . We split the remainder of the proof into two cases, depending on whether or not | supp ⁢ ( C ) | = n .

Case | supp ⁢ ( C ) | < n . Let

Δ C := { ⟨ e j ⟩ ∣ j ∈ supp ⁢ ( C ) } ,

let 𝐿 be the subspace ⟨ Δ C ⟩ of 𝑉, and let a ℓ and b ℓ be the projections onto 𝐿 of a r and b r , respectively. Lemma 2.4 iii shows that the diagonal entries corresponding to { 1 , … , n } ∖ supp ⁢ ( C ) of elements of D ( C ) can take any multiset of nonzero values. Since no element of D ( C ) maps a r to b r , it follows that there is no matrix in D ( C ) whose restriction to 𝐿 maps a ℓ to b ℓ . By the minimality of 𝑛, there exists a subset Γ C of Δ C of size | Δ C | + 2 − r such that no element of GL ⁢ ( L ) ( Γ C ∪ C ) maps a ℓ to b ℓ . Setting Γ to be Γ C ∪ ( Δ ∖ Δ C ) so that | Γ | = n + 2 − r , we observe that no element of G ( Γ ∪ C ) maps a r to b r . This is a contradiction, and so the lemma follows in this case.

Case | supp ⁢ ( C ) | = n . In this case, Lemma 2.8 applies, so with the notation of that lemma, let

Γ := Δ ∖ { ⟨ e i 1 ⟩ , … , ⟨ e i r − 2 ⟩ } .

Then | Γ | = n + 2 − r and ⟨ e 1 ⟩ , ⟨ e n ⟩ ∈ Γ since i 1 ≥ 2 and i r − 1 = n .

Let g ∈ G ( Γ ∪ C ) . To complete the proof, we will show that a r g = a r ≠ b r , by showing that g | supp ⁢ ( a r ) is scalar. We will first show that 𝑔 is lower triangular. It is clear that 𝑔 stabilises ⟨ e 1 ⟩ ∈ Γ . Suppose inductively that 𝑔 stabilises ⟨ e 1 , e 2 , … , e s ⟩ for some s ∈ { 1 , … , n − 1 } . If ⟨ e s + 1 ⟩ ∈ Γ , then 𝑔 stabilises

E s + 1 := ⟨ e 1 , e 2 , … , e s ⟩ + ⟨ e s + 1 ⟩ = ⟨ e 1 , e 2 , … , e s + 1 ⟩ .

Otherwise, s + 1 = i t for some t ∈ { 1 , … , r − 2 } , and then Lemma 2.8 i shows that

{ s + 1 } ⊊ supp ⁢ ( a t ) ⊆ { 1 , … , s + 1 } .

In this case, 𝑔 again stabilises ⟨ e 1 , e 2 , … , e s ⟩ + a t = E s + 1 . Hence, by induction, 𝑔 is lower triangular.

Now, let I := { i 1 , … , i r − 1 } , let 𝒰 be the set of integers that each lie in the support of a unique subspace in 𝐶, and let J := I ∪ U . We will show next that g | J is diagonal, by fixing j ∈ J and proving that g k ⁢ j = 0 whenever k > j . First, if ⟨ e k ⟩ ∈ Γ , then it is clear that g k ⁢ j = 0 , and so g n ⁢ j = 0 . Hence we may also assume that k ∈ I ∖ { i r − 1 } .

Suppose inductively that

g i u , j = 0 for some ⁢ u ≥ 2

(the base case here is u = r − 1 so that i u = n ). We will show that if i u − 1 > j , then g i u − 1 , j = 0 . By Lemma 2.8 iv, the indices i u − 1 , i u ∈ supp ⁢ ( a ) u , and furthermore, Lemma 2.8 i–ii shows that supp ⁢ ( a u ) ∩ I = { i u − 1 , i u } . Thus, by the previous paragraph and our inductive assumption,

g k ⁢ j = 0 for all ⁢ k ∈ supp ⁢ ( a u ) ∖ { j , i u − 1 } .

In fact, Lemma 2.8 i–ii shows that each integer in supp ⁢ ( a u ) less than i u − 1 lies in supp ⁢ ( a u − 1 ) . As i u − 1 > j ∈ J , we deduce from the definition of 𝒥 that j ∉ supp ⁢ ( a u ) . Thus g k ⁢ j = 0 for all k ∈ supp ⁢ ( a u ) ∖ { i u − 1 } . As 𝑔 stabilises a u , we deduce that g i u − 1 , j = 0 . Therefore, by induction, g k ⁢ j = 0 for all k ≠ j , and so g | J is diagonal.

Finally, we will show that g | J is scalar. Let j , k ∈ J ∩ supp ⁢ ( a t ) for some t ∈ { 1 , … , r − 1 } . As 𝑔 stabilises a t , and as g | J is diagonal, we deduce that

(2.4) g j ⁢ j = g k ⁢ k .

Now, by Lemma 2.8 iv, i t ∈ supp ⁢ ( a t ) ∩ supp ⁢ ( a t + 1 ) for each t ∈ { 1 , … , r − 2 } , so i t ∈ J ∩ supp ⁢ ( a t ) ∩ supp ⁢ ( a t + 1 ) . Thus, starting from t = 1 and proceeding by induction on 𝑡, it follows from (2.4) that g j ⁢ j = g k ⁢ k for all j , k ∈ J , i.e. g | J is a scalar. Since supp ⁢ ( a r ) ⊆ J by Lemma 2.8 v, we deduce that a r g = a r ≠ b r , as required. ∎

The following lemma is strengthening of Lemma 2.9 in the case | F | = 3 and r = 2 , in which the subset Γ now has size n + 1 − r = n − 1 .

Lemma 2.10

Suppose | F | = 3 , and let A , B ∈ Δ ̄ 2 . Suppose also that A ∼ D , 1 B and B ∉ A D . Then there exists a subset Γ of Δ of size n − 1 such that B ∉ A G ( Γ ) .

Proof

Since A ∼ D , 1 B , without loss of generality, a 1 = b 1 , and there exists an element of 𝐷 mapping a 2 to b 2 . Hence a 2 and b 2 have equal supports. Reordering the basis for 𝑉 if necessary, we may also assume that supp ⁢ ( a 1 ) = { 1 , 2 , … , m } for some m ≥ 2 . Then, by Lemma 2.4, the upper left m × m submatrix of each matrix in D a 1 is a scalar, while the remaining diagonal entries can be chosen independently. As B ∉ A D , no matrix in D a 1 maps a 2 to b 2 . We may therefore assume (by reordering the basis vectors in { e 1 , … , e m } and/or swapping 𝐴 and 𝐵 if necessary) that the projections of a 2 and b 2 onto ⟨ e 1 , e 2 ⟩ are ⟨ e 1 + e 2 ⟩ and ⟨ e 1 − e 2 ⟩ , respectively.

Now, let Γ := Δ ∖ { ⟨ e 2 ⟩ } , let g ∈ G ( Γ ∪ { a 1 } ) , and notice that 𝑔 is diagonal outside of the second row. Write a 1 as

⟨ ∑ i = 1 m α i ⁢ e i ⟩ ,

with α 1 = 1 and α i ≠ 0 for all i ∈ { 2 , … , m } . Since a 1 g = a 1 , we deduce that, without loss of generality, the top left 2 × 2 submatrix of 𝑔 is

( 1 0 g 21 1 + α 2 ⁢ g 21 ) .

Let 𝑣 be the projection of ( e 1 + e 2 ) g onto ⟨ e 1 , e 2 ⟩ . Recall that α 2 ≠ 0 , and note that g 22 ≠ 0 as 𝑔 is invertible. Hence if g 21 = 1 , then α 2 = 1 and v = − e 1 − e 2 ; if g 21 = − 1 , then α 2 = − 1 and v = − e 2 ; and if g 21 = 0 , then v = e 1 + e 2 . Hence, in each case, 𝑣 does not span ⟨ e 1 − e 2 ⟩ = b 2 | ⟨ e 1 , e 2 ⟩ . Therefore, a 2 g ≠ b 2 , and hence B ∉ A G ( Γ ) . ∎

Although the next result holds for all 𝔽, it will only be useful in the case | F | = 3 .

Proposition 2.11

Let X , Y ∈ Ω n + 1 such that X ∼ G , n Y , and suppose ⟨ X ⟩ = V . Then Y ∈ X G .

Proof

As dim ⁢ ( ⟨ X ⟩ ) = n , we may assume by Lemma 2.1 that x i = y i = ⟨ e i ⟩ for i ∈ { 1 , … , n } . Let S := supp ⁢ ( x n + 1 ) and T := supp ⁢ ( y n + 1 ) . We will show that S = T ; it will then follow that there exists an element of D = G ( Δ ) mapping x n + 1 to y n + 1 , and so Y ∈ X G .

If S = { 1 , … , n } = T , then we are done. Otherwise, exchanging 𝑋 and 𝑌 if necessary (note that ⟨ Y ⟩ = V ), we may assume that there exists an element t ∈ { 1 , … , n } ∖ S . Let Γ := Δ ∖ { ⟨ e t ⟩ } . Then, since X ∼ G , n Y , there exists an element of G ( Γ ) mapping x n + 1 to y n + 1 . As G ( Γ ) stabilises each subspace ⟨ e i ⟩ with i ∈ S , it follows that S = T , as required. ∎

We are now able to prove this section’s main theorem.

Theorem 2.12

Suppose that 𝑛 and | F | are at least 3. Then RC ⁢ ( GL n ⁢ ( F ) , Ω ) is at most n + 2 . Moreover, RC ⁢ ( GL n ⁢ ( 3 ) , Ω ) ≤ n .

Proof

Let k ∈ { n , n + 1 , n + 2 } , with k = n + 2 if | F | > 3 . Additionally, let 𝑋 and 𝑌 be tuples in Ω u with u > k and X ∼ G , k Y , where G = GL n ⁢ ( F ) . It suffices to prove that Y ∈ X G . Suppose, for a contradiction, that 𝑛 is the minimal dimension for which the theorem does not hold (for a fixed 𝔽), and that Y ∉ X G . Then, for each m ∈ { 2 , … , n − 1 } , using Proposition 1.2 i in the case m = 2 , we obtain RC ⁢ ( GL m ⁢ ( F ) , P ⁢ G 1 ⁢ ( F m ) ) < k . Since Y ∉ X G , Lemma 2.2 yields ⟨ X ⟩ = V . Hence, by Lemma 2.1, we may assume without loss of generality^[1] that

x i = y i = ⟨ e i ⟩ for ⁢ i ∈ { 1 , … , n } ,

and furthermore that all subspaces in 𝑋 are distinct, so that x i , y i ∈ Δ ̄ for each i ≥ n + 1 .

We will first consider the case k ≥ n + 1 . Since X ∼ G , n + 1 Y , Lemma 2.3 yields supp ⁢ ( x i ) = supp ⁢ ( y i ) for all 𝑖. However, Y ∉ X G . Hence there exist an integer r ≥ 2 and subtuples 𝐴 of 𝑋 and 𝐵 of 𝑌, with A , B ∈ Δ ̄ r , such that ( x 1 , … , x n , a 1 , … , a r ) and ( x 1 , … , x n , b 1 , … , b r ) are ( n + r − 1 ) -equivalent, but not equivalent, under 𝐺. Equivalently, A ∼ D , r − 1 B and B ∉ A D .

If k = n + 2 , then by Lemma 2.9, there exists a set Γ := { ⟨ e i 1 ⟩ , … , ⟨ e i k − r ⟩ } such that B ∉ A G ( Γ ) . However, this means that the subtuples

( x i 1 , … , x i k − r , a 1 , … , a r ) and ( x i 1 , … , x i k − r , b 1 , … , b r )

are not equivalent under 𝐺. This contradicts the assumption that X ∼ G , k Y . Hence, in this case, Y ∈ X G , as required, so RC ⁢ ( G ) ≤ n + 2 . If | F | > 3 , then we are done.

Therefore, assume for the rest of the proof that | F | = 3 and suppose first that k = n + 1 . By the previous paragraph, RC ⁢ ( G ) ≤ n + 2 . Therefore, to prove that RC ⁢ ( G ) ≤ k , it suffices to show that X ∼ G , n + 2 Y whenever X ∼ G , k Y . Thus, by replacing 𝑋 and 𝑌 by suitable subtuples, if necessary, we may assume that u = n + 2 . In this case, r = 2 , and by Lemma 2.10, there exists a subset Γ of Δ of size k − r such that B ∉ A G ( Γ ) . Arguing as in the previous paragraph, this contradicts the assumption that X ∼ G , k Y . Thus RC ⁢ ( G ) ≤ n + 1 .

Finally, suppose that we have k = n . Since RC ⁢ ( G ) ≤ n + 1 , we may assume that u = n + 1 . However, since X ∼ G , n Y and ⟨ X ⟩ = V , Proposition 2.11 shows that Y ∈ X G . Therefore, RC ⁢ ( G ) ≤ n . ∎

3 Action on 1-spaces: Lower bounds

In this section, we again assume that | F | ≥ 3 , and write Ω := Ω 1 = P ⁢ G 1 ⁢ ( V ) . We drop the assumption that n ≥ 3 and permit n = 2 . We shall now prove lower bounds for the relational complexity of each group 𝐻 satisfying

SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) ,

acting on Ω.

For some results in this section, we will assume that F = F q is finite, and when doing so, we fix a primitive element 𝜔, and assume that q = p f for 𝑝 prime. Additionally, we will write

P ⁢ Γ ⁢ L n ⁢ ( q ) / PSL n ⁢ ( q ) = ⟨ δ , ϕ ⟩ , with ⁢ PGL n ⁢ ( q ) / PSL n ⁢ ( q ) = ⟨ δ ⟩ .

Here, the automorphism 𝜙 can be chosen to be induced by the automorphism of GL n ⁢ ( q ) which raises each matrix entry to its 𝑝-th power, and with a slight abuse of notation, we will also write 𝜙 to denote this automorphism of GL n ⁢ ( q ) , and to denote a generator for Aut ⁢ ( F q ) . If 𝔽 is an arbitrary field, then the group Γ ⁢ L n ⁢ ( F ) is still a semi-direct product of GL n ⁢ ( F ) by Aut ⁢ ( F ) (see, for example, [12, Theorem 9.36]), but of course, GL n ⁢ ( F ) / SL n ⁢ ( F ) and Aut ⁢ ( F ) need not be cyclic.

We let Z := Z ⁢ ( GL n ⁢ ( F ) ) and will write I n for the n × n identity matrix, and E i ⁢ j for the n × n matrix with 1 in the ( i , j ) ⁢ -th position and 0 elsewhere. We write A ⊕ B for the block diagonal matrix with blocks 𝐴 and 𝐵.

Our first result is completely general and easy to prove, although we shall later prove much tighter bounds for various special cases.

Theorem 3.1

Let 𝔽 be arbitrary, and let 𝐻 satisfy SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) . Then RC ⁢ ( H , Ω ) ≥ n .

Proof

Define X , Y ∈ Ω n by x i = y i = ⟨ e i ⟩ for i ∈ { 1 , … , n − 1 } , with

x n = ⟨ ∑ i = 1 n e i ⟩ and y n = ⟨ ∑ i = 1 n − 1 e i ⟩ .

Then dim ⁢ ( ⟨ X ⟩ ) = n and dim ⁢ ( ⟨ Y ⟩ ) = n − 1 , so no element of Γ ⁢ L n ⁢ ( F ) maps 𝑋 to 𝑌. Hence Y ∉ X H .

Now, let h ℓ := I n − E ℓ ⁢ n for each ℓ ∈ { 1 , … , n − 1 } , and h n := I n . Then

h ℓ ∈ SL n ⁢ ( F ) ≤ H and ( X ∖ x ℓ ) h ℓ = ( Y ∖ y ℓ ) for each ⁢ ℓ ∈ { 1 , … , n } .

Therefore, X ∼ H , n − 1 Y , and so the result follows. ∎

Our next two results focus on the special cases n = 2 and n = 3 .

Lemma 3.2

Assume that q ≥ 8 , and let 𝐻 satisfy SL 2 ⁢ ( q ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L 2 ⁢ ( q ) . Then RC ⁢ ( H ) ≥ 4 , except that RC ⁢ ( Σ ⁢ L 2 ⁢ ( 9 ) ) = 3 .

Proof

The claim about Σ ⁢ L 2 ⁢ ( 9 ) is an easy computation in GAP using [3], so exclude this group from now on. We divide the proof into two cases. For each, we define X , Y ∈ Ω 4 such that X ∼ H , 3 Y but Y ∉ X H . In both cases, we set ( X ∖ x 4 ) = ( Y ∖ y 4 ) = ( ⟨ e 1 ⟩ , ⟨ e 2 ⟩ , ⟨ e 1 + e 2 ⟩ ) .

Case (a): either 𝑞 is even, or H ≰ ⟨ Z , Σ ⁢ L 2 ⁢ ( q ) ⟩ , where Z = Z ⁢ ( GL n ⁢ ( F ) ) . If 𝑞 is odd, then let α ∈ F p ∗ ∖ { 1 } , and otherwise, let α = ω 3 so that 𝛼 is not in the orbit ω ⟨ ϕ ⟩ . Then let x 4 = ⟨ e 1 + ω ⁢ e 2 ⟩ and y 4 = ⟨ e 1 + α ⁢ e 2 ⟩ .

The stabiliser in 𝐻 of ( X ∖ x 4 ) = ( Y ∖ y 4 ) is contained in ⟨ Z , ϕ ⟩ . As α ∉ ω ⟨ ϕ ⟩ , no element of this stabiliser maps x 4 to y 4 , and so Y ∉ X H . On the other hand, for each j ∈ { 1 , 2 , 3 , 4 } , the matrix g j ∈ GL 2 ⁢ ( q ) given below maps ( X ∖ x j ) to ( Y ∖ y j ) :

g 1 = ( 1 ( α − ω ) ⁢ ( 1 − ω ) − 1 0 1 − ( α − ω ) ⁢ ( 1 − ω ) − 1 ) , g 2 = ( 1 − ( ω ⁢ α − 1 − 1 ) ⁢ ( ω − 1 ) − 1 0 ( ω ⁢ α − 1 − 1 ) ⁢ ( ω − 1 ) − 1 1 ) , g 3 = ( 1 0 0 α ⁢ ω − 1 ) , g 4 = I 2 .

If 𝑞 is even, then some scalar multiple of g j lies in 𝐻 for all 𝑗, so X ∼ H , 3 Y , and we are done. If instead 𝑞 is odd, then our assumption that H ≰ ⟨ Z , Σ ⁢ L 2 ⁢ ( q ) ⟩ implies that 𝐻 contains a scalar multiple of an element diag ⁢ ( ω , 1 ) ⁢ ϕ i for some i ≥ 0 , as diag ⁢ ( ω , 1 ) induces the automorphism 𝛿 of PSL 2 ⁢ ( q ) . Hence, for each 𝑗, there exists ϕ i j ∈ Aut ⁢ ( F q ) such that a scalar multiple of g j ⁢ ϕ i j lies in 𝐻. Since α ∈ F p ∗ , each ϕ i j fixes 𝑌, and thus X ∼ H , 3 Y .

Case (b): 𝑞 is odd and H ≤ ⟨ Z , Σ ⁢ L 2 ⁢ ( q ) ⟩ . Since H ≠ Σ ⁢ L 2 ⁢ ( 9 ) and since Proposition 1.2 i yields the result when H = SL 2 ⁢ ( 9 ) , we may assume that q > 9 . We generalise Hudson’s [8, §5.4] proof that RC ⁢ ( SL 2 ⁢ ( q ) , Ω ) ≥ 4 . First, let

S := F q ∖ { 0 , 1 , − 1 } and T := F q ∖ { 0 , 1 } ,

and for each λ ∈ S , define a map θ λ : T → F q by μ ↦ ( 1 − λ 2 ⁢ μ ) ⁢ ( 1 − μ ) − 1 . We will show that there exist elements λ ∈ S and τ ∈ T satisfying the following conditions:

( τ ) ⁢ θ λ is a square in F q * , and
no automorphism of F q maps 𝜏 to λ 2 ⁢ τ .

It is easy to see that, for each λ ∈ S , the image im ⁢ ( θ λ ) = F q ∖ { 1 , λ 2 } , so the map θ λ is injective, and the preimage of any nonzero square in im ⁢ ( θ λ ) lies in 𝒯 and satisfies condition (i). Hence, for each λ ∈ S , there are precisely ( q − 1 ) / 2 − 2 choices for τ ∈ T satisfying condition (i).

Given λ ∈ S , since λ 2 ≠ 1 , condition (ii) is equivalent to requiring that

λ 2 ⁢ τ ≠ τ p k for all ⁢ k ∈ { 1 , … , f − 1 } ,

i.e. λ 2 ≠ τ p k − 1 for all 𝑘. There are exactly ( q − 3 ) / 2 = ( q − 1 ) / 2 − 1 distinct squares of elements of 𝒮, and precisely ( q − 1 ) / ( p − 1 ) elements in F q * that are ( p − 1 ) -th powers. Hence if p > 3 , then there exists λ ∈ S such that λ 2 is not a ( p − 1 ) -th power in F q . Observe that then λ 2 is not a ( p k − 1 ) -th power for any 𝑘, and so this 𝜆 and any corresponding 𝜏 from the previous paragraph satisfy both conditions.

Suppose instead that p = 3 , and fix λ ∈ S . The number of elements τ ∈ F 3 f * not satisfying (ii), i.e. with λ 2 = τ 3 k − 1 for some k ∈ { 1 , … , f − 1 } , is at most

( 3 − 1 ) + ( 3 2 − 1 ) + ⋯ + ( 3 f − 1 − 1 ) = ( 3 + 3 2 + ⋯ + 3 f − 1 ) − ( f − 1 ) .

On the other hand, we established that the number of elements τ ∈ T satisfying (i) is equal to

( 3 f − 1 ) / 2 − 2 = ( 3 − 1 ) ⁢ ( 1 + 3 + 3 2 + ⋯ + 3 f − 1 ) / 2 − 2 = ( 3 + 3 2 + ⋯ + 3 f − 1 ) − 1 .

Since q > 9 , and hence f > 2 , there again exists τ ∈ T satisfying both conditions.

Finally, fix such a λ ∈ S and τ ∈ T , and complete the definition of X , Y ∈ Ω 4 by setting

x 4 = ⟨ e 1 + τ ⁢ e 2 ⟩ and y 4 = ⟨ e 1 + λ 2 ⁢ τ ⁢ e 2 ⟩ .

The stabiliser in 𝐻 of ( X ∖ x 4 ) = ( Y ∖ y 4 ) is contained in ⟨ Z , ϕ ⟩ . By condition (ii), no such element maps x 4 to y 4 , so Y ∉ X H . However, the proof of [8, Theorem 5.4.6] uses condition (i) to exhibit explicit elements of SL 2 ⁢ ( q ) mapping each 3-tuple of 𝑋 to the corresponding 3-tuple of 𝑌. Therefore, X ∼ H , 3 Y , and the result follows. ∎

Lemma 3.3

Assume that PSL 3 ⁢ ( F ) ≠ PGL 3 ⁢ ( F ) , and let 𝐻 be any group satisfying SL 3 ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L 3 ⁢ ( F ) . If 𝔽 is finite, or if H ≤ GL 3 ⁢ ( F ) , then RC ⁢ ( H ) ≥ 5 .

Proof

If | F | = 4 , then we verify the result in GAP using [3], so assume that | F | ≥ 7 . If 𝔽 is finite, then let λ := ω , whilst if 𝔽 is infinite, then let 𝜆 be any element of F ∗ of multiplicative order at least 3. Define X , Y ∈ Ω 5 by

x i = y i = ⟨ e i ⟩ for ⁢ i ∈ { 1 , 2 , 3 } , x 4 = y 4 = ⟨ e 1 + e 2 + e 3 ⟩ , x 5 = ⟨ e 1 + λ ⁢ e 2 + λ 2 ⁢ e 3 ⟩ , y 5 = ⟨ e 1 + λ − 1 ⁢ e 2 + λ − 2 ⁢ e 3 ⟩

so that x 5 ≠ y 5 .

We first show that Y ∉ X H . The stabiliser in 𝐻 of ( X ∖ x 5 ) = ( Y ∖ y 5 ) lies in H ∩ ⟨ Z , Aut ⁢ ( F ) ⟩ , so if 𝔽 is infinite, then we are done. Assume therefore that F = F q . If x 5 ϕ i = y 5 , then λ p i = λ − 1 = λ p f − 1 . Since i ∈ { 0 , … , f − 1 } and λ = ω , we deduce that ( p , f , i ) ∈ { ( 2 , 2 , 1 ) , ( 3 , 1 , 0 ) } , contradicting q ≥ 7 . Thus Y ∉ X H .

Next, for all 𝔽, we show that X ∼ H , 4 Y . Let

g 1 := ( λ λ + 1 λ + λ − 1 0 − 1 0 0 0 − λ − 1 ) , g 2 := ( − λ 0 0 λ + 1 1 1 + λ − 1 0 0 − λ − 1 ) , g 3 := ( − λ 0 0 0 − 1 0 λ + λ − 1 1 + λ − 1 λ − 1 ) , g 4 := ( λ 2 0 0 0 1 0 0 0 λ − 2 ) , g 5 := I 3 .

Observe that det ( g ℓ ) = 1 for each ℓ ∈ { 1 , … , 5 } , and so g ℓ ∈ SL 3 ⁢ ( F ) ≤ H . It is also easy to check that ( X ∖ x ℓ ) g ℓ = ( Y ∖ y ℓ ) for each ℓ. Thus X ∼ H , 4 Y , and so RC ⁢ ( H ) ≥ 5 . ∎

Our remaining results hold for all sufficiently large 𝑛. The first is specific to GL n ⁢ ( F ) .

Proposition 3.4

If n ≥ 3 and | F | ≥ 4 , then RC ⁢ ( GL n ⁢ ( F ) , Ω ) ≥ n + 2 .

Proof

As | F | ≥ 4 , there exists an element λ ∈ F ∗ such that λ ≠ λ − 1 (so λ ≠ − 1 ). Define X , Y ∈ Ω n + 2 by

x i = y i = ⟨ e i ⟩ for ⁢ i ∈ { 1 , … , n } , x n + 1 = y n + 1 = ⟨ ∑ i = 1 n e i ⟩ , x n + 2 = ⟨ e 1 + λ ⁢ e 2 ⟩ and y n + 2 = ⟨ e 1 + λ − 1 ⁢ e 2 ⟩ .

The stabiliser in GL n ⁢ ( F ) of ( X ∖ x n + 2 ) = ( Y ∖ y n + 2 ) is the group of scalar matrices, so it follows that Y ∉ X GL n ⁢ ( F ) . Additionally, it is easily verified that, for each j ∈ { 1 , … , n + 2 } , the matrix g j ∈ GL n ⁢ ( q ) given below maps ( X ∖ x j ) to ( Y ∖ y j ) :

g 1 = ( λ 1 + λ 0 − 1 ) ⊕ λ ⁢ I n − 2 , g 2 = ( − 1 0 1 + λ − 1 λ − 1 ) ⊕ λ − 1 ⁢ I n − 2 , g n + 1 = diag ⁢ ( λ , λ − 1 , λ , … , λ ) , g j = g n + 1 + ( λ − λ − 1 ) ⁢ E j ⁢ 2 for ⁢ j ∈ { 3 , … , n } , g n + 2 = I n .

Hence X ∼ GL n ⁢ ( F ) , n + 1 Y , and so the result follows. ∎

In the light of Proposition 3.4, the next result in particular bounds the relational complexity of all remaining groups when PSL n ⁢ ( F ) = PGL n ⁢ ( F ) .

Lemma 3.5

Let 𝔽 be arbitrary, assume that n ≥ 3 , and let 𝐻 be any group satisfying GL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) and H ≠ GL n ⁢ ( F ) . Then RC ⁢ ( H ) ≥ n + 3 .

Proof

Since GL n ⁢ ( F ) is a proper subgroup of 𝐻, there exist a nontrivial

ψ ∈ H ∩ Aut ⁢ ( F )

and an element λ ∈ F * with λ ψ ≠ λ . We define X , Y ∈ Ω n + 3 by x i = y i = ⟨ e i ⟩ for i ∈ { 1 , … , n } ,

x n + 1 = y n + 1 = ⟨ ∑ i = 1 n e i ⟩ , x n + 2 = y n + 2 = ⟨ e 1 + e 2 + λ ⁢ e 3 ⟩ , x n + 3 = ⟨ e 1 + λ ⁢ e 2 ⟩ , y n + 3 = ⟨ e 1 + λ ψ ⁢ e 2 ⟩ .

We claim that X ∼ H , n + 2 Y , but Y ∉ X H , from which the result will follow.

The stabiliser in 𝐻 of

( x 1 , … , x n + 1 ) = ( y 1 , … , y n + 1 )

is contained in ⟨ Z , Aut ⁢ ( F ) ⟩ . However, no element of ⟨ Z , Aut ⁢ ( F ) ⟩ maps

( x n + 2 , x n + 3 ) to ( y n + 2 , y n + 3 ) ,

so Y ∉ X H . The reader may verify that, for each j ∈ { 1 , … , n + 3 } , the element h j ∈ ⟨ GL n ⁢ ( F ) , ψ ⟩ ≤ H given below maps ( X ∖ x j ) to ( Y ∖ y j ) , where we define τ := ( λ − 1 ) − 1 (notice that λ ≠ 1 ):

h 1 = ( 1 − τ ⁢ ( λ ψ − λ ) 0 1 + τ ⁢ ( λ ψ − λ ) ) ⊕ I n − 2 ,

h 2 = ( 1 − τ ⁢ ( λ ⁢ ( λ − 1 ) ψ − 1 ) 0 τ ⁢ ( λ ⁢ ( λ − 1 ) ψ − 1 ) 1 ) ⊕ I n − 2 ,

h 3 = ( ( 1 − τ ⁢ ( λ ⁢ ( λ − 1 ) ψ − 1 − 1 ) 0 0 0 1 − τ ⁢ ( λ ⁢ ( λ − 1 ) ψ − 1 − 1 ) 0 τ ⁢ ( λ ⁢ ( λ − 1 ) ψ − 1 − 1 ) τ ⁢ ( λ ⁢ ( λ − 1 ) ψ − 1 − 1 ) 1 ) ⊕ I n − 3 ) ⁢ ψ ,

h j = ( diag ⁢ ( 1 , 1 , λ − 1 ⁢ λ ψ − 1 , 1 , … , 1 ) + ( 1 − λ − 1 ⁢ λ ψ − 1 ) ⁢ E j ⁢ 3 ) ⁢ ψ

for ⁢ j ∈ { 4 , … , n } ,

h n + 1 = diag ⁢ ( 1 , 1 , λ − 1 ⁢ λ ψ − 1 , 1 , … , 1 ) ⁢ ψ , h n + 2 = ψ , h n + 3 = I n .

Hence X ∼ H , n + 2 Y , and the result follows. ∎

Lemma 3.6

Let 𝔽 be arbitrary, assume that n ≥ 4 , and let 𝐻 be any group satisfying SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) and H ≰ GL n ⁢ ( F ) . Then RC ⁢ ( H ) ≥ n + 2 .

Proof

Since H ≰ GL n ⁢ ( F ) , there exist elements h ⁢ ψ ∈ H and λ ∈ F * such that h ∈ GL n ⁢ ( q ) , ψ ∈ Aut ⁢ ( F q ) , and λ ψ ≠ λ . Let X , Y ∈ Ω n + 2 be as in the proof of Lemma 3.5, but supported only on the first n − 1 basis vectors so that ⟨ e n ⟩ lies in neither 𝑋 nor 𝑌, and x n = y n = ⟨ ∑ i = 1 n − 1 e i ⟩ . Just as in that proof, one may check that Y ∉ X H , but X ∼ H , n + 1 Y . ∎

The next result applies, in particular, to all groups 𝐻 such that SL n ⁢ ( F ) ⁢ ⊴ ⁢ H and either H < GL n ⁢ ( F ) or H ≤ Σ ⁢ L n ⁢ ( F ) ≠ Γ ⁢ L n ⁢ ( F ) . We write F × n for the subgroup of F ∗ consisting of 𝑛-th powers, which is the set of possible determinants of scalar matrices in GL n ⁢ ( F ) .

Proposition 3.7

Assume that n ≥ 4 and | F | ≥ 3 , and let 𝐻 be any group satisfying SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) . Assume also that the set

{ det ( g ) ψ ⁢ F × n ∣ g ⁢ ψ ∈ H ⁢ with ⁢ g ∈ GL n ⁢ ( F ) , ψ ∈ Aut ⁢ ( F ) }

is a proper subset of F ∗ / F × n . Then RC ⁢ ( H ) ≥ 2 ⁢ n − 2 .

Proof

By assumption, there exists an α ∈ F ∗ such that α ≠ det ( g ⁢ z ) ψ for all g ⁢ ψ ∈ H and z ∈ Z . Define X , Y ∈ Ω 2 ⁢ n − 2 as follows:

X := ( ⟨ e 2 ⟩ , … , ⟨ e n ⟩ , ⟨ e 1 + e 2 ⟩ , … , ⟨ e 1 + e n ⟩ ) , Y := ( ⟨ e 2 ⟩ , … , ⟨ e n ⟩ , ⟨ α ⁢ e 1 + e 2 ⟩ , … , ⟨ α ⁢ e 1 + e n ⟩ ) .

We show first that Y ∉ X H . Suppose for a contradiction that there exists g ⁢ ψ ∈ H , with g ∈ GL n ⁢ ( F ) and ψ ∈ Aut ⁢ ( F ) , such that X g ⁢ ψ = Y . As g ⁢ ψ fixes ⟨ e 2 ⟩ and ⟨ e 3 ⟩ , and maps ⟨ e 1 + e 2 ⟩ and ⟨ e 1 + e 3 ⟩ to ⟨ α ⁢ e 1 + e 2 ⟩ and ⟨ α ⁢ e 1 + e 3 ⟩ , respectively, we deduce that

e 1 g ⁢ ψ ∈ ⟨ e 1 , e 2 ⟩ ∩ ⟨ e 1 , e 3 ⟩ = ⟨ e 1 ⟩ .

Therefore, we see that ⟨ e i ⟩ g ⁢ ψ = ⟨ e i ⟩ for each i ∈ { 1 , … , n } , and so 𝑔 is diagonal. Let μ := α ψ − 1 . As ⟨ e 1 + e i ⟩ g ⁢ ψ = ⟨ α ⁢ e 1 + e i ⟩ for each i ∈ { 2 , … , n } , we deduce that g = diag ⁢ ( μ , 1 , … , 1 ) ⁢ z for some z ∈ Z . Hence ( det ( g ⁢ z − 1 ) ) ψ = μ ψ = α , a contradiction. Hence Y ∉ X H .

Now, for each i ∈ { 2 , … , n } , let h i := diag ⁢ ( α , 1 , … , 1 , α − 1 , 1 , … , 1 ) , where the α − 1 appears in entry 𝑖. First, for j ∈ { 1 , … , n − 1 } , let k := j + 1 so that x j = y j = ⟨ e k ⟩ . It is easy to verify that h k + ( 1 − α ) ⁢ E k ⁢ 1 has determinant 1 and maps ( X ∖ x j ) to ( Y ∖ y j ) . Finally, for j ∈ { n , … , 2 ⁢ n − 2 } , let k := j + 2 − n so that x j = ⟨ e 1 + e k ⟩ and y j = ⟨ α ⁢ e 1 + e k ⟩ . Then h k has determinant 1 and maps ( X ∖ x j ) to ( Y ∖ y j ) . Thus X ∼ H , 2 ⁢ n − 3 Y , and so RC ⁢ ( H ) ≥ 2 ⁢ n − 2 . ∎

Proof of Theorem A

When | F | = 2 , this result is clear from Theorem 1.1. For the remaining fields 𝔽, the fact that part i gives an upper bound on RC ⁢ ( PGL n ⁢ ( F ) ) is proved in Theorem 2.12, whilst we prove that it gives a lower bound in Theorem 3.1 for | F | = 3 and Proposition 3.4 for | F | ≥ 4 . That part ii gives upper bounds on RC ⁢ ( H ̄ ) is immediate from Theorem 1.2 ii for n = 3 , and from Theorem 2.7 for n ≥ 4 . Lemma 3.3 and Proposition 3.7 show that these are also lower bounds. ∎

Recall that ω ⁢ ( k ) denotes the number of distinct prime divisors of the positive integer 𝑘.

Lemma 3.8

Lemma 3.8 ([7, Lemma 3.1])

Let K ≤ Sym ⁢ ( Γ ) be a finite group with normal subgroup 𝑁 such that K / N is cyclic. Then H ⁢ ( K , Γ ) ≤ H ⁢ ( N , Γ ) + ω ⁢ ( | K / N | ) .

Proof of Theorem B

For the upper bound in i, we combine Proposition 1.2 i with Lemma 3.8 to deduce that H ⁢ ( H ̄ , Ω 1 ) = 3 + ω ⁢ ( e ) , so RC ⁢ ( H ̄ , Ω 1 ) ≤ 4 + ω ⁢ ( e ) . The lower bound (and the case H ̄ = P ⁢ Σ ⁢ L 2 ⁢ ( 9 ) ) is Lemma 3.2.

For the upper bound in part ii, we similarly combine Proposition 1.2 ii with Lemma 3.8. As for the lower bound, first let n = 3 , and notice that, in this case, 2 ⁢ n − 2 = 4 < n + 2 = 5 . If H ̄ properly contains PGL 3 ⁢ ( q ) , then the lower bound of 6 is proved in Lemma 3.5. Otherwise, PSL 3 ⁢ ( q ) ≠ PGL 3 ⁢ ( q ) , and so the lower bound of 5 follows from Lemma 3.3. Now assume that n ≥ 4 . The general lower bound is Lemma 3.6, the bound of n + 3 for groups properly containing PGL n ⁢ ( q ) is Lemma 3.5, and the bound of 2 ⁢ n − 2 is Proposition 3.7. ∎

4 Action on 𝑚-spaces for m ≥ 2

In this section, we consider the action of the group 𝐻 on Ω m = P ⁢ G m ⁢ ( V ) , where SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) , as before, but now 2 ≤ m ≤ n 2 . The main work is to prove a lower bound on RC ⁢ ( H , Ω m ) , as the upper bound follows from existing literature.

Proposition 4.1

Let 𝔽 be arbitrary, let n ≥ 2 ⁢ m ≥ 4 , and let 𝐻 be any group satisfying SL n ⁢ ( F ) ⁢ ⊴ ⁢ H ≤ Γ ⁢ L n ⁢ ( F ) . Then RC ⁢ ( H , Ω m ) ≥ m ⁢ n − m 2 + 1 .

Proof

For each i ∈ { 1 , … , m } and j ∈ { m + 1 , … , n − 1 } , let

B i := { e 1 , e 2 , … , e m } ∖ { e i } , U i ⁢ j := ⟨ B i , e j ⟩ = ⟨ e 1 , … , e i − 1 , e i + 1 , … , e m , e j ⟩ , V i := ⟨ B i , e i + e n ⟩ , and W i := ⟨ B i , e n ⟩

so that U i ⁢ j , V i , W i ∈ Ω m . Define X , Y ∈ Ω m m ⁢ n − m 2 + 1 as follows:

x m ⁢ n − m 2 + 1 := ⟨ e 1 + e 2 , … , e 1 + e m , ∑ i = 1 n e i ⟩ , y m ⁢ n − m 2 + 1 := ⟨ e 1 + e 2 , … , e 1 + e m , − e 1 + ∑ i = m + 1 n e i ⟩ ,

X := ( U 1 ⁢ ( m + 1 ) , U 1 ⁢ ( m + 2 ) , … , U m ⁢ ( n − 1 ) , V 1 , V 2 , … , V m , x m ⁢ n − m 2 + 1 ) , Y := ( U 1 ⁢ ( m + 1 ) , U 1 ⁢ ( m + 2 ) , … , U m ⁢ ( n − 1 ) , W 1 , W 2 , … , W m , y m ⁢ n − m 2 + 1 ) .

We shall first show that Y ∉ X Γ ⁢ L n ⁢ ( F ) , so in particular Y ∉ X H , and then that X ∼ H , m ⁢ n − m 2 Y .

Assume for a contradiction that Y ∈ X Γ ⁢ L n ⁢ ( F ) . Since each subspace in 𝑌 is spanned by vectors of the form

∑ i = 1 n λ i ⁢ e i with ⁢ λ i ∈ { − 1 , 0 , 1 } ,

it follows that there exists g ∈ GL n ⁢ ( F ) with X g = Y . For each i ∈ { 1 , … , m } , choose

k ∈ { 1 , … , m } ∖ { i } .

Then

⟨ e i ⟩ = ⋂ ℓ ∈ { 1 , … , m } ∖ { i } U ℓ ⁢ ( m + 1 ) ∩ V k = ⋂ ℓ ∈ { 1 , … , m } ∖ { i } U ℓ ⁢ ( m + 1 ) ∩ W k ,

so 𝑔 fixes ⟨ e i ⟩ . Similarly, 𝑔 fixes

⟨ e j ⟩ = ⋂ i = 1 m U i ⁢ j for each ⁢ j ∈ { m + 1 , … , n − 1 } .

Therefore, there exist λ 1 , … , λ n ∈ F * and μ 1 , … , μ n − 1 ∈ F such that 𝑔 maps e i to λ i ⁢ e i for all i ∈ { 1 , … , n − 1 } , and maps e n to λ n ⁢ e n + ∑ i = 1 n − 1 μ i ⁢ e i . It now follows that, for each i ∈ { 2 , … , m } , the element 𝑔 maps e 1 + e i ∈ x m ⁢ n − m 2 + 1 to λ 1 ⁢ e 1 + λ i ⁢ e i , which must lie in y m ⁢ n − m 2 + 1 , and hence λ i = λ 1 . Similarly, V i g = W i for each i ∈ { 1 , … , m } , and so W i = ⟨ B i , e n ⟩ contains

( e i + e n ) g = λ 1 ⁢ e i + λ n ⁢ e n + ∑ k = 1 n − 1 μ k ⁢ e k .

Hence μ i = − λ 1 , and μ j = 0 for all j ∈ { m + 1 , … , n − 1 } . It now follows that 𝑔 maps

∑ i = 1 n e i ∈ x m ⁢ n − m 2 + 1 to ∑ i = m + 1 n λ i ⁢ e i ,

which is clearly not in y m ⁢ n − m 2 + 1 , a contradiction. Thus Y ∉ X H .

We now show that X ∼ H , m ⁢ n − m 2 Y , by identifying an element

g ℓ ∈ SL n ⁢ ( F ) ≤ H

that maps ( X ∖ x ℓ ) to ( Y ∖ y ℓ ) for each ℓ ∈ { 1 , … , m ⁢ n − m 2 + 1 } . We divide the proof into three cases, which together account for all values of ℓ. To simplify our expressions, let z := e 1 + e 2 + ⋯ + e m , α 1 := − 1 , and α r := 1 for all r ∈ { 2 , … , m } . In each case, the element g ℓ will be lower unitriangular and so will have determinant 1.

Case (a): ℓ ∈ { 1 , … , m ⁢ ( n − m − 1 ) } . Let r ∈ { 1 , … , m } , s ∈ { m + 1 , … , n − 1 } be such that ℓ = ( n − m − 1 ) ⁢ ( r − 1 ) + ( s − m ) so that x ℓ = y ℓ = U r ⁢ s . Additionally, let g ℓ fix e i for all i ∉ { s , n } , map e s to e s + α r ⁢ e r , and map e n to e n − z . Then g ℓ fixes U i ⁢ j provided ( i , j ) ≠ ( r , s ) , and maps e i + e n ∈ V i to e i + e n − z ∈ W i , and hence V i to W i , for all i ∈ { 1 , … , m } . Finally,

( ∑ i = 1 n e i ) g ℓ = α r ⁢ e r + ∑ i = m + 1 n e i ∈ y m ⁢ n − m 2 + 1 ,

where we have used the fact that

e r + ∑ i = m + 1 n e i = ( e 1 + e r ) + ( − e 1 + ∑ i = m + 1 n e i )

when r > 1 . Hence g ℓ maps x m ⁢ n − m 2 + 1 to y m ⁢ n − m 2 + 1 , as required.

Case (b): ℓ = m ⁢ ( n − m − 1 ) + r , where r ∈ { 1 , … , m } . Here,

x ℓ = V r and y ℓ = W r .

Let g ℓ fix e i for each i ∈ { 1 , … , n − 1 } and map e n to α r ⁢ e r + e n − z . Then g ℓ fixes U i ⁢ j for all 𝑖 and 𝑗, and maps e i + e n ∈ V i to e i + α r ⁢ e r + e n − z ∈ W i , and hence V i to W i , for all i ∈ { 1 , … , m } ∖ { r } . Finally,

( ∑ i = 1 n e i ) g ℓ = α r ⁢ e r + ∑ i = m + 1 n e i ∈ y m ⁢ n − m 2 + 1 ,

as in Case (a).

Case (c): ℓ = m ⁢ n − m 2 + 1 . Let g ℓ fix e i for each i ∈ { 1 , … , n − 1 } , and map e n to e n − z . Then 𝑔 fixes U i ⁢ j for all i , j , and maps e i + e n ∈ V i to e i + e n − z ∈ W i for all 𝑖, as required. ∎

The irredundant base size I ⁢ ( K , Γ ) of a group 𝐾 acting faithfully on a set Γ is the largest size of a tuple ( α 1 , … , α k ) of elements of Γ such that

K > K α 1 > K ( α 1 , α 2 ) > ⋯ > K ( α 1 , … , α k ) = 1 ,

with all inclusions strict. It is clear that I ⁢ ( K , Γ ) is bounded below by the height H ⁢ ( K , Γ ) , which we recall (from Section 1) is bounded below by RC ⁢ ( K , Γ ) − 1 .

Proof of Theorem C

In [9, Theorem 3.1], it is proved that

I ⁢ ( PGL n ⁢ ( F ) , Ω m ) ≤ ( m + 1 ) ⁢ n − 2 ⁢ m + 1 .

Since the irredundant base size of a subgroup is at most the irredundant base size of an overgroup, and the height is at most the irredundant base size, we deduce that H ⁢ ( H ̄ , Ω m ) ≤ ( m + 1 ) ⁢ n − 2 ⁢ m + 1 for all H ̄ ≤ PGL n ⁢ ( F ) . From Lemma 3.8, we then see that, for all H ̄ as in the statement,

H ⁢ ( H ̄ , Ω m ) ≤ ( m + 1 ) ⁢ n − 2 ⁢ m + 1 + ω ⁢ ( e ) ,

and hence the upper bound follows. The lower bound is immediate from Proposition 4.1, so the proof is complete. ∎

Funding source: Engineering and Physical Sciences Research Council

Award Identifier / Grant number: EP/R014604/1

Award Identifier / Grant number: EP/W522422/1

Funding statement: This work was supported by EPSRC grant no. EP/R014604/1, and also partially supported by a grant from the Simons Foundation. The first author was supported by the University of St Andrews (St Leonard’s International Doctoral Fees Scholarship & School of Mathematics and Statistics PhD Funding Scholarship), and by EPSRC grant no. EP/W522422/1. The second author is funded by the Heilbronn Institute.

Acknowledgements

We thank the anonymous referee for their careful reading and helpful comments. The authors would like to thank the Isaac Newton Institute for Mathematical Sciences for support and hospitality during the programme “Groups, representations and applications: New perspectives”, when work on this paper was undertaken.

Communicated by: Andrea Lucchini

References

[1] G. Cherlin, Sporadic homogeneous structures, The Gelfand Mathematical Seminars, 1996–1999, Birkhäuser, Boston (2000), 15–48. 10.1007/978-1-4612-1340-6_2Search in Google Scholar

[2] G. Cherlin, On the relational complexity of a finite permutation group, J. Algebraic Combin. 43 (2016), no. 2, 339–374. 10.1007/s10801-015-0636-8Search in Google Scholar

[3] G. Cherlin and J. Wiscons, RComp (GAP code), 2016, https://webpages.csus.edu/wiscons/research/code/RComp.html. Search in Google Scholar

[4] G. L. Cherlin, G. A. Martin and D. H. Saracino, Arities of permutation groups: Wreath products and 𝑘-sets, J. Combin. Theory Ser. A 74 (1996), no. 2, 249–286. 10.1006/jcta.1996.0050Search in Google Scholar

[5] N. Gill, M. W. Liebeck and P. Spiga, Cherlin’s Conjecture for Finite Primitive Binary Permutation Groups, Lecture Notes in Math. 2302, Springer, Cham, 2022. 10.1007/978-3-030-95956-2Search in Google Scholar

[6] N. Gill, B. Lodà and P. Spiga, On the height and relational complexity of a finite permutation group, Nagoya Math. J. 246 (2022), 372–411. 10.1017/nmj.2021.6Search in Google Scholar

[7] S. Harper, The maximal size of a minimal generating set, Forum Math. Sigma 11 (2023), Paper No. e70. 10.1017/fms.2023.71Search in Google Scholar

[8] S. Hudson, Calculating the height and relational complexity of the primitive actions of PSL 2 ⁢ ( q ) and PGL 2 ⁢ ( q ) , Ph.D. thesis, University of South Wales, 2022. Search in Google Scholar

[9] V. Kelsey and C. M. Roney-Dougal, On relational complexity and base size of finite primitive groups, Pacific J. Math. 318 (2022), no. 1, 89–108. 10.2140/pjm.2022.318.89Search in Google Scholar

[10] A. H. Lachlan, On countable stable structures which are homogeneous for a finite relational language, Israel J. Math. 49 (1984), no. 1–3, 69–153. 10.1007/BF02760647Search in Google Scholar

[11] B. Lodà, The height and relational complexity of finite primitive permutation groups, Ph.D. thesis, University of South Wales, 2020. Search in Google Scholar

[12] J. J. Rotman, An Introduction to the Theory of Groups, 4th ed., Grad. Texts in Math. 148, Springer, New York, 1995. 10.1007/978-1-4612-4176-8Search in Google Scholar

[13] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.12.2, 2022. Search in Google Scholar

Received: 2023-11-22

Revised: 2024-01-04

Published Online: 2024-02-14

Published in Print: 2024-09-01

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