Lattices Generated by Orbits of Subspaces under Finite Singular Orthogonal Groups II

Abstract

Let 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} be a (2𝜈+𝛿+𝑙)-dimensional vector space over the finite field 𝔽_𝑞. In this paper we assume that 𝔽_𝑞 is a finite field of odd characteristic, and 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞) the singular orthogonal groups of degree 2𝜈+𝛿+𝑙 over 𝔽_𝑞. Let ℳ be any orbit of subspaces under 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞). Denote by ℒ the set of subspaces which are intersections of subspaces in ℳ, where we make the convention that the intersection of an empty set of subspaces of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} is assumed to be 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}. By ordering ℒ by ordinary or reverse inclusion, two lattices are obtained. This paper studies the questions when these lattices ℒ are geometric lattices.

1. Introduction

Let 𝔽_𝑞 be a finite field with 𝑞 elements, where 𝑞 is an odd prime power. We choose a fixed nonsquare element 𝑧 in 𝔽^∗_𝑞∶=𝔽_𝑞⧵{0}. Let 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} be a (2𝜈+𝛿+𝑙)-dimensional row vector space over the finite field 𝔽_𝑞, and let 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞) be one of the singular orthogonal groups of degree 2𝜈+𝛿+𝑙 over 𝔽_𝑞. There is an action of 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞) on 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} defined as follows: 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}×𝑂_{2𝜈+𝛿+𝑙,Δ}𝔽_𝑞⟶𝔽_𝑞^{(2𝜈+𝛿+𝑙)},𝑥₁,𝑥₂,…,𝑥_{2𝜈+𝛿+𝑙},𝑇⟼𝑥₁,𝑥₂,…,𝑥_{2𝜈+𝛿+𝑙}𝑇. (1.1) Let 𝑃 be an 𝑚-dimensional subspace of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}(1≤𝑚≤2𝜈+𝛿+𝑙), and 𝑣₁,𝑣₂,…,𝑣_𝑚 be a basis of 𝑃. Then, the 𝑚×(2𝜈+𝛿+𝑙) matrix: 𝑣₁𝑣₂⋮𝑣_𝑚 (1.2) is called a matrix representation of 𝑃. We usually denote a matrix representation of the 𝑚-dimensional subspace 𝑃 still by 𝑃. The above action induces an action on the set of subspaces of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, that is, a subspace 𝑃 is carried by 𝑇∈𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞) into the subspace 𝑃𝑇. The set of subspaces of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} is partitioned into orbits under 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞). Clearly, {0} and {𝔽_𝑞^{(2𝜈+𝛿+𝑙)}} are two trivial orbits. Let ℳ be any orbit of subspaces under 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞). Denote the set of subspaces which are intersections of subspaces in ℳ by ℒ(ℳ) and call ℒ(ℳ) the set of subspaces generated by ℳ. We agree that the intersection of an empty set of subspaces is 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}. Then, 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}∈ℒ(ℳ). Partially ordering ℒ(ℳ) by ordinary or reverse inclusion, we get two posets and denote them by ℒ_𝑂(ℳ) and ℒ_𝑅(ℳ), respectively. Clearly, for any two elements 𝑃,𝑄∈ℒ_𝑂(ℳ), 𝑃∧𝑄=𝑃∩𝑄,𝑃∨𝑄=∩𝑅∈ℒ_𝑂(ℳ)∶𝑅⊇⟨𝑃,𝑄⟩, (1.3) where ⟨𝑃,𝑄⟩ is a subspace generated by 𝑃 and 𝑄. Therefore, ℒ_𝑂(ℳ) is a finite lattice.

Similarly, for any two elements 𝑃,𝑄∈ℒ_𝑅(ℳ), 𝑃∧𝑄=∩𝑅∈ℒ_𝑅(ℳ)∶𝑅⊇⟨𝑃,𝑄⟩,𝑃∨𝑄=𝑃∩𝑄, (1.4) so ℒ_𝑅(ℳ) is also a finite lattice. Both ℒ_𝑂(ℳ) and ℒ_𝑅(ℳ) are called the lattices generated by ℳ.

The results on the geometricity of lattices generated by subspaces in 𝑑-bounded distance-regular graphs can be found in Guo et al. [1]; on the geometricity and the characteristic polynomial of lattices generated by orbits of flats under finite affine-classical groups can be found in Wang and Feng [2], Wang and Guo [3]; on inclusion relations, the geometricity and the characteristic polynomial of lattices generated by orbits of subspaces under finite nonsingular classical groups and a characterization of subspaces contained in lattices can be found in Huo [4–6], Huo and Wan [7, 8]; on inclusion relations, the geometricity and the characteristic polynomial of lattices generated by orbits of subspaces under finite singular symplectic groups, singular unitary groups, and singular pseudosymplectic groups and a characterization of subspaces contained in lattices can be found in Gao and You [9–12]. In [13], the authors studied the various lattices ℒ_𝑂(ℳ) and ℒ_𝑅(ℳ) generated by different orbits ℳ of subspaces under singular orthogonal group 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞). The study contents include the inclusion relations between different lattices, the characterization of subspaces contained in a given lattice ℒ_𝑅(ℳ) (resp., ℒ_𝑂(ℳ)), and the characteristic polynomial of ℒ_𝑅(ℳ). The purpose of this paper is to study the questions when ℒ_𝑅(ℳ) (resp., ℒ_𝑂(ℳ)) are geometric lattices.

2. Preliminaries

In the following, we recall some definitions and facts on ordered sets and lattices (see [8, 14]).

Let 𝐴 be a partially ordered set, and 𝑎,𝑏∈𝐴. We say that 𝑏  covers 𝑎 and write 𝑎<⋅𝑏, if 𝑎<𝑏 and there exists no 𝑐∈𝐴 such that 𝑎<𝑐<𝑏. An element 𝑚∈𝐴 is called the minimal element if there exists no elements 𝑎∈𝐴 such that 𝑎<𝑚. If 𝐴 has 𝑎 unique minimal element, denote it by 0 and we say that 𝐴 is a poset with 0.

Let 𝐴 be a poset with 0 and 𝑎∈𝐴. If all maximal ascending chains starting from 0 with endpoint 𝑎have the same finite length, this common length is called the rank 𝑟(𝑎) of 𝑎. If rank 𝑟(𝑎) is defined for every 𝑎∈𝐴, 𝐴 is said to have the rank function 𝑟∶𝐴→ℕ, where ℕ is the set consisting of all positive integers and 0.

A poset 𝐴 is said to satisfy the Jordan-Dedekind (JD) condition if any two maximal chains between the same pair of elements of 𝐴 have the same finite length.

Proposition 2.1 ([14, Proposition 2.1]). Let 𝐴 be a poset with 0. If 𝐴 satisfies the JD condition then 𝐴 has the rank function 𝑟∶𝐴→ℕ which satisfies(i)𝑟(0)=0,(ii)𝑎<⋅𝑏⇒𝑟(𝑏)=𝑟(𝑎)+1.Conversely, if 𝐴 admits a function 𝑟∶𝐴→ℕ satisfying (i) and (ii), then 𝐴 satisfies the JD condition with 𝑟 as its rank function.
Let 𝐴 be a poset with 0. An element 𝑎∈𝐴 is called an atom of 𝐴 if 0<⋅𝑎. A lattice 𝐿 with 0 is called an atomic lattice (or a 𝑝𝑜𝑖𝑛𝑡𝑙𝑎𝑡𝑡𝑖𝑐𝑒) if every element 𝑎∈𝐿⧵{0} is a supremum of atoms, that is, 𝑎=sup{𝑏∈𝐿∣0<⋅𝑏≤𝑎}.

Definition 2.2 ([14, page 46]). A lattice 𝐿 is called a semimodular lattice if for all 𝑎,𝑏∈𝐿, 𝑎∧𝑏<⋅𝑎⟹𝑏<⋅𝑎∨𝑏. (2.1)

Proposition 2.3 ([14, Theorem 2.27]). Let 𝐿 be a lattice with 0. Then, 𝐿 is a semimodular lattice if and only if 𝐿 possesses a rank function 𝑟 such that for all 𝑥,𝑦∈𝐿𝑟𝑥∧𝑦+𝑟𝑥∨𝑦≤𝑟(𝑥)+𝑟𝑦. (2.2)

Definition 2.4 ([14, page 52]). A lattice 𝐿 is called a geometric lattice if it is𝐺^′₁an atomic lattice,𝐺^′₂a semimodular lattice,𝐺₃without infinite chains in 𝐿.

According to Definition 2.2, Proposition 2.3, and Definition 2.4, we can obtain the following proposition.

Proposition 2.5. Let 𝐿 be a lattice with 0. Then, 𝐿 is a geometric lattice if and only if𝐺₁for every element 𝑎∈𝐿⧵{0}, 𝑎=sup{𝑏∈𝐿∣0<⋅𝑏≤𝑎},𝐺₂𝐿 possesses a rank function 𝑟 and for all 𝑥,𝑦∈𝐿, (2.2) holds,𝐺₃without infinite chains in 𝐿.

Let 𝑆_{2𝜈+𝛿,Δ}=0𝐼^(𝜈)𝐼^(𝜈)0Δ,𝑆_𝑙=𝑆0^(𝑙), (2.3) where 𝑆=𝑆_{2𝜈+𝛿,Δ}, 𝛿=0,1, or 2, and 1Δ=𝜙,if𝛿=0,1𝑜𝑟𝑧,if𝛿=1,−𝑧,if𝛿=2. (2.4) The set of all (2𝜈+𝛿+𝑙)×(2𝜈+𝛿+𝑙)nonsingular matrices 𝑇 over 𝔽_𝑞 satisfying 𝑇𝑆_𝑙𝑇^𝑡=𝑆_𝑙 (2.5) forms a group which will be called the singular orthogonal group of degree 2𝜈+𝛿+𝑙, rank2𝜈+𝛿, and with definite part Δover 𝔽_𝑞and denoted by 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞).Clearly, 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞)consists of all (2𝜈+𝛿+𝑙)×(2𝜈+𝛿+𝑙)nonsingular matrices of the form: 𝑇𝑇=¹¹𝑇₁₂0𝑇₂₂𝑙,2𝜈+𝛿2𝜈+𝛿𝑙 (2.6) where 𝑇₁₁𝑆𝑇^𝑡₁₁=𝑆, and 𝑇₂₂ is nonsingular.

Two 𝑛×𝑛 matrices 𝐴 and 𝐵 are called to be cogredient if there exists a nonsingular matrix 𝑃 such that 𝑃𝐴𝑃^𝑡=𝐵.

An 𝑚-dimensional subspace 𝑃 is said to be a subspace of type (𝑚,2𝑠+𝛾,𝑠,Γ), if 𝑃𝑆_𝑙𝑃^𝑡 is cogredient to 𝑀(𝑚,2𝑠+𝛾,𝑠,Γ), where the matrix 𝑀(𝑚,2𝑠+𝛾,𝑠,Γ), respectively, is as follows 𝑀(𝑚,2𝑠,𝑠)=0𝐼^(𝑠)𝐼^(𝑠)00^{(𝑚−2𝑠)}𝑀()=,if𝛾=0,𝑚,2𝑠+1,𝑠,10𝐼^(𝑠)𝐼^(𝑠)010^{(𝑚−2𝑠−1)},if𝛾=1 (2.7) or 𝑀(𝑚,2𝑠+1,𝑠,𝑧)=0𝐼^(𝑠)𝐼^(𝑠)0𝑧0^{(𝑚−2𝑠−1)}𝑀()=,if𝛾=1,𝑚,2𝑠+2,𝑠0𝐼^(𝑠)𝐼^(𝑠)010−𝑧_{(𝑚−2𝑠−2)},if𝛾=2. (2.8)

Let 𝑒₁,𝑒₂,…,𝑒_2𝜈+𝛿,𝑒_2𝜈+𝛿+1,…,𝑒_{2𝜈+𝛿+𝑙}be a basis of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, where 𝑒_𝑖=(0,…,0,1,0,…,0), (2.9) 1 is in the 𝑖th position. Denote by 𝐸 the 𝑙-dimensional subspace of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} generated by 𝑒_2𝜈+𝛿+1,𝑒_2𝜈+𝛿+2,…,𝑒_{2𝜈+𝛿+𝑙}. An 𝑚-dimensional subspace 𝑃 is called a subspace of type (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘)if(i)𝑃 is a subspace of type (𝑚,2𝑠+𝛾,𝑠,Γ),(ii)dim(𝑃∩𝐸)=𝑘.

Denote the set of all subspaces of type (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘)in 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} byℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). By [15, Theorem 6.28], we know that  ℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)is nonempty if and only if ,𝑘≤𝑙,2𝑠+𝛾≤𝑚−𝑘≤𝜈+𝑠+min𝛿,𝛾if𝛾≠𝛿or𝛾=𝛿andΓ=Δ,𝜈+𝑠,if𝛾=𝛿=1andΓ≠Δ, (2.10) or ,{},min𝑙,𝑚−2𝑠−𝛾≥𝑘≥max0,𝑚−𝜈−𝑠−min𝛿,𝛾if𝛾≠𝛿or𝛾=𝛿andΓ=Δ,max0,𝑚−𝜈−𝑠if𝛾=𝛿=1andΓ≠Δ. (2.11)

Moreover, if ℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)is nonempty, then it forms an orbit of subspaces under 𝑂_{2𝜈+𝛿+𝑙,Δ}(𝔽_𝑞). Let ℒ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)denote the set of subspaces which are intersections of subspaces in ℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), where we make the convention that the intersection of an empty set of subspaces of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} is assumed to be 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}. Partially ordering ℒ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) by ordinary or reverse inclusion, we get two finite lattices and denote them by ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)and ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), respectively.

The case ℒ_𝑅(𝑚−𝑙,2𝑠+𝛾,𝑠,Γ;2𝜈+𝛿,Δ)has been discussed in [8]. So, we only discuss the case 0≤𝑘<𝑙in this paper.

By [13], we have the following results.

Theorem 2.6. Let 2𝜈+𝛿+𝑙>𝑚≥1,  0≤𝑘<𝑙, assume that (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘) satisfies conditions (2.10) and (2.11). Then, ℒ_𝑅(𝑚,2𝑠+𝑟,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)⊃ℒ_𝑅𝑚₁,2𝑠₁+𝛾₁,𝑠₁,Γ₁,𝑘₁;2𝜈+𝛿+𝑙,Δ (2.12) if and only if 𝑘₁2()𝑚≤𝑘<𝑙,𝑚−𝑘−2₁−𝑘₁≥2𝑠+𝛾−2𝑠₁+𝛾₁+𝛾−𝛾₁≥2𝛾−𝛾₁,𝑖𝑓𝛾₁≠𝛾𝑜𝑟𝛾₁=𝛾𝑎𝑛𝑑Γ₁−=Γ,2𝑠+𝛾2𝑠₁+𝛾₁+2≥4,𝑖𝑓𝛾₁=𝛾=1𝑎𝑛𝑑Γ₁≠Γ. (2.13)

Theorem 2.7. Let 2𝜈+𝛿+𝑙>𝑚≥1,  0≤𝑘<𝑙. Assume that (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘) satisfies condition (2.10), then ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) consists of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} and all the subspaces of type (𝑚₁,2𝑠₁+𝛾₁,𝑠₁,Γ₁,𝑘₁), where (𝑚₁,2𝑠₁+𝛾₁,𝑠₁,Γ₁,𝑘₁)satisfies condition (2.13).

Theorem 2.8. Let 2𝜈+𝛿+𝑙>𝑚≥1,  0≤𝑘<𝑙, and (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘) satisfy ,2𝑠+𝛾≤𝑚−𝑘≤𝜈+𝑠+min𝛿,𝛾𝑖𝑓𝛾≠𝛿𝑜𝑟𝛾=𝛿𝑎𝑛𝑑Γ=Δ,𝜈+𝑠,𝑖𝑓𝛾=𝛿=1𝑎𝑛𝑑Γ≠Δ. (2.14) For any 𝑋∈ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), define 𝑟(𝑋)=dim𝑋,𝑖𝑓𝑋≠𝔽_𝑞^{(2𝜈+𝛿+𝑙)},𝑚+1,𝑖𝑓𝑋=𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, (2.15) then 𝑟∶ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)→ℕ is a rank function of the lattice ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ).

Theorem 2.9. Let 2𝜈+𝛿+𝑙>𝑚≥1,  0≤𝑘<𝑙, and (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘) satisfy (2.14). For any 𝑋∈ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), define 𝑟^′(𝑋)=𝑚+1−dim𝑋,𝑖𝑓𝑋≠𝔽_𝑞^{(2𝜈+𝛿+𝑙)},0,𝑖𝑓𝑋=𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, (2.16) then 𝑟^′∶ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)→ℕ is a rank function of the lattice ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ).

3. The Geometricity of Lattices ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)

Theorem 3.1. Let 2𝜈+𝛿+𝑙>𝑚≥1,0≤𝑘<𝑙, assume that (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘)satisfies conditions (2.10) and (2.11). Then(i)each of ℒ_𝑂(𝑘+1,0,0,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ) and ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)(Γ=1𝑜𝑟𝑧) is a finite geometric lattice, when 𝑘=0, and is a finite atomic lattice, but not a geometric lattice when 0<𝑘<𝑙;(ii)when 2≤𝑚−𝑘≤2𝜈+𝛿−1, ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is a finite atomic lattice, but not a geometric lattice.

Proof. By Theorem 2.8, the rank function of ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is defined by formula (2.15), we will show the condition 𝐺₁ of Proposition 2.5 holds for ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). {0}∈ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) and it is the minimal element, so all 1-dim subspaces in ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) are atoms of ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ).
Let 𝑈∈ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)⧵{{0},𝔽_𝑞^{(2𝜈+𝛿+𝑙)}}, by Theorem 2.7, 𝑈 is a subspace of type (𝑚₁,2𝑠₁+𝛾₁,𝑠₁,Γ₁,𝑘₁) and satisfies condition (2.13). If 𝑚₁=1, then 𝑈 is an atom of ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). Assume 𝑚₁≥2, then 𝑈𝑆_𝑙𝑈^𝑡=𝑆_{2𝑠1+𝛾1,Γ1},0^{(𝑚1−𝑘1−2𝑠1−𝛾1)},0^(𝑘1), (3.1) where Γ₁=𝜙,(1),(𝑧), or [1,−𝑧].
Let 𝑈_𝑖 be an 𝑖th (1≤𝑖≤𝑚₁) row vector of 𝑈, then ⟨𝑈_𝑖⟩ is a subspace of type (1,0,0,𝜙,0),(1,1,0,1,0),(1,1,0,𝑧,0), or (1,0,0,0,1), and ⟨𝑈_𝑖⟩⊂𝑈. By Theorem 2.7, we know ⟨𝑈_𝑖⟩∈ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), so ⟨𝑈_𝑖⟩ is an atom of ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), and 𝑈=∨^𝑚1_𝑖=1⟨𝑈_𝑖⟩, hence, 𝑈 is a union of atoms in ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). Since |ℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)|≥2, there exist 𝑊₁,𝑊₂∈ℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ),𝑊₁≠𝑊₂, such that 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}=𝑊₁∨𝑊₂. 𝑊₁,𝑊₂ are unions of atoms in ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), hence, 𝔽_𝑞^{(2𝜈+𝛿+𝑙)} is a union of atoms in ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), therefore, 𝐺₁ holds.
In the following, we prove (i) and (ii).
The Proof of (i). We only prove the formula (2.2) holds for ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). The other can be obtained in the similar way. We consider two cases:
(a) 𝑘=0. ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) consists of 𝔽_𝑞^{(2𝜈+𝛿+𝑙)},{0} and subspaces of type (1,1,0,Γ,0). Let 𝑈,𝑊∈ℒ_𝑂(1,1,0,Γ,0;2𝜈+𝛿+𝑙,Δ), if 𝑈,𝑊 are 𝔽_𝑞^{(2𝜈+𝛿+𝑙)},{0}, respectively, then 𝑈∨𝑉=𝔽_𝑞^{(2𝜈+𝛿+𝑙)},𝑈∧𝑊={0}, so 𝑟(𝑈∨𝑊)+𝑟(𝑈∧𝑊)=𝑟(𝑈)+𝑟(𝑊). If 𝑈=𝑊 is {0} or 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, the other is a subspace of type (1,1,0,Γ,0), then 𝑈∧𝑊 is {0}or subspace of type (1,1,0,Γ,0), 𝑈∨𝑊 is a subspace of type (1,1,0,Γ,0) or 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, so 𝑟(𝑈∨𝑊)+𝑟(𝑈∧𝑊)=𝑟(𝑈)+𝑟(𝑊). If 𝑈 and 𝑊 are subspaces of type (1,1,0,Γ,0), then 𝑈∧𝑊={0},𝑈∨𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, so 𝑟(𝑈∨𝑊)+𝑟(𝑈∧𝑊)=𝑟(𝑈)+𝑟(𝑊).
Hence, (2.2) holds and ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is a finite geometric lattice when 𝑘=0.
(b) 0<𝑘<𝑙. Let 𝑈=⟨𝑒₁+(Γ/2)𝑒_𝜈+1⟩, 𝑊=⟨𝑒_𝑠+1+(Γ/2)𝑒_𝜈+𝑠+1⟩, where 𝑠≤𝜈−1, then 𝑈,𝑊∈ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). When 𝑞=3(mod4) or 𝑞=1(mod4), then −1 is a nonsquare element or a square element, respectively. Thus, [Γ,Γ] is cogredient to either [1,−𝑧] or 𝑆_2⋅1, and ⟨𝑈,𝑊⟩ is a subspace of type (2,2,0,Γ,0), where Γ=[1,−𝑧], or a subspace of type (2,2,1,𝜙,0). So ⟨𝑈,𝑊⟩∉ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), and we have 𝑈∨𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, 𝑈∧𝑊={0}. By the definition of rank function, 𝑟(𝑈∨𝑊)=𝑘+1+1=𝑘+2,𝑟(𝑈∧𝑊)=0, 𝑟(𝑈)=𝑟(𝑊)=1, we have 𝑟(𝑈∨𝑊)+𝑟(𝑈∧𝑊)=𝑘+2>𝑟(𝑈)+𝑟(𝑊)=2.
Hence, ℒ_𝑂(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is a finite atomic lattice, but not a geometric lattice when 0<𝑘<𝑙.
The Proof of (ii). We will show there exist 𝑈,𝑊∈ℒ_𝑂(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) such that the formula (2.2) does not hold. As to 𝛾=0,1, or 2, we only show the proof of 𝛾=1, others can be obtained in the similar way. We distinguish the following three cases.
(a) 𝛿=0, or 𝛿=1,  Γ≠Δ. Then, the formula (2.10) is changed into 2𝑠+1≤𝑚−𝑘≤𝜈+𝑠. Let 𝜎=𝜈+𝑠−𝑚+𝑘, we distinguish the following two subcases.
(a.1) 𝑚−𝑘−2𝑠−1≥1. From 𝑚−𝑘−2𝑠−1≥1 and 𝑚−𝑘≤𝜈+𝑠, we have 𝑠+2≤𝜈. Let 𝐼𝑈=^(𝑠)0000000000000000𝐼^(𝑠)000000010000Γ/200000000𝐼^(𝜎1)000000000000000000𝐼^(k)0,𝑠11𝜎₁𝜎𝑠11𝜎₁𝑒𝜎𝑘𝑙−𝑘𝑊=_𝑠+2+Γ2𝑒_𝜈+𝑠+2, (3.2) where 𝜎₁=𝑚−𝑘−2𝑠−2, then 𝑈 is a subspace of type (𝑚−1,2𝑠+1,𝑠,Γ,𝑘), 𝑊 is a subspace of type (1,1,0,Γ,0). When 𝑞=3(mod4) or 𝑞=1(mod4), then −1 is a nonsquare element or a square element, respectively, thus [Γ,Γ] is cogredient to either [1,−𝑧] or 𝑆_2⋅1, and ⟨𝑈,𝑊⟩ is a subspace of type (𝑚,2𝑠+2,𝑠,Γ,𝑘) or type (𝑚,2(𝑠+1),𝑠+1,𝜙,𝑘). Consequently, 𝑈,𝑊∈ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), ⟨𝑈,𝑊⟩∉ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). Thus, we have 𝑈∨𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)},  𝑈∧𝑊={0},  𝑟(𝑈∨𝑊)=𝑚+1,𝑟(𝑈∧𝑊)=0,  𝑟(𝑈)=𝑚−1,  𝑟(𝑊)=1.Then, 𝑟(𝑈∨𝑊)+𝑟(𝑈∧𝑊)=𝑚+1>𝑟(𝑈)+𝑟(𝑊)=𝑚−1+1=𝑚. (3.3)
(a.2) 𝑚−𝑘−2𝑠−1=0. From 2≤𝑚−𝑘≤2𝜈+𝛿−1, we have 𝑠+1≤𝜈,  𝑠≥1. Let 𝐼𝑈=^(𝑠−1)000000000000𝐼^(𝑠)000000100Γ/20000000000𝐼^(𝑘)0,𝑒𝑠−111𝜎𝑠1𝜎𝑘𝑙−𝑘𝑊=_𝑠+1−Γ2𝑒_𝜈+𝑠+1, (3.4) then 𝑈 is a subspace of type (𝑚−1,2(𝑠−1)+1,𝑠−1,Γ,𝑘), 𝑊 is a subspace of type (1,1,0,−Γ,0), ⟨𝑈,𝑊⟩ is a subspace of type (𝑚,2𝑠,𝑠,𝜙,𝑘). Consequently, 𝑈,𝑊∈ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), ⟨𝑈,𝑊⟩∉ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). Thus, we have 𝑈∨𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)},  𝑈∧𝑊={0},  𝑟(𝑈∨𝑊)=𝑚+1,  𝑟(𝑈∧𝑊)=0,  𝑟(𝑈)=𝑚−1,  𝑟(𝑊)=1.Then, 𝑟(𝑈∨𝑊)+𝑟(𝑈∧𝑊)=𝑚+1>𝑟(𝑈)+𝑟(𝑊)=𝑚−1+1=𝑚. (3.5) Therefore, there exist 𝑈,𝑊∈ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) such that formula (2.2) does not hold.
(b) 𝛿=1,Γ=Δ. Then, the formula (2.10) is changed into 2𝑠+1≤𝑚−𝑘≤𝜈+𝑠+1. Let 𝜎=𝜈+𝑠−𝑚+𝑘+1, we distinguish the following two subcases.
(b.1) 𝑚−𝑘−2𝑠−1≥1. From 𝑚−𝑘−2𝑠−1≥1, and 2≤𝑚−𝑘≤2𝜈, we have 𝑠+1≤𝜈. Let 𝐼𝑈=^(𝑠)00000000000000𝐼^(𝑠)0000000000000010000𝐼^(𝜎1)00000000000000000𝐼^(𝑘)0,𝑠1𝜎₁𝜎𝑠1𝜎₁𝑒𝜎1𝑘𝑙−𝑘𝑊=_𝑠+1+Δ2𝑒_𝜈+𝑠+1, (3.6) where 𝜎₁=𝑚−𝑘−2𝑠−2, then 𝑈 is a subspace of type (𝑚−1,2𝑠+1,𝑠,Δ,𝑘), 𝑊 is a subspace of type (1,1,0,Δ,0). When 𝑞=3(mod4) or 𝑞=1(mod4), similar to the proof of the case (a.1), ⟨𝑈,𝑊⟩ is a subspace of type (𝑚,2𝑠+2,𝑠,Γ,𝑘) or (𝑚,2(𝑠+1),𝑠+1,𝜙,𝑘). Consequently, 𝑈,𝑊∈ℒ_𝑂(𝑚,2𝑠+1,𝑠,Δ,𝑘;2𝜈+1+𝑙,Δ), ⟨𝑈,𝑊⟩∉ℒ_𝑂(𝑚,2𝑠+1,𝑠,Δ,𝑘;2𝜈+1+𝑙,Δ), and the formula (2.2) does not hold.
(b.2) 𝑚−𝑘−2𝑠−1=0. From 2≤𝑚−𝑘≤2𝜈, we have 𝑠+1≤𝜈. Let 𝐼𝑈=^(𝑠−1)0000000000000𝐼^(𝑠)00000000000010000000000𝐼^(𝑘)0,𝑒𝑠−111𝜎−1𝑠1𝜎−11𝑘𝑙−𝑘𝑊=_𝑠+1+Δ2𝑒_𝜈+𝑠+1, (3.7) then 𝑈 is a subspace of type (𝑚−1,2(𝑠−1)+1,𝑠−1,Δ,𝑘), 𝑊 is a subspace of type (1,1,0,Δ,0), when 𝑞=3(mod4) or 𝑞=1(mod4), ⟨𝑈,𝑊⟩ is subspace of type (𝑚,2(𝑠−1)+2,𝑠−1,Γ,𝑘) or (𝑚,2𝑠,𝑠,𝜙,𝑘). Similar to the proof of the case (a.1), the formula (2.2) does not hold for 𝑈 and 𝑊.
(c) 𝛿=2. Then, the formula (2.10) is changed into 2𝑠+1≤𝑚−𝑘≤𝜈+𝑠+1. Let 𝜎=𝜈+𝑠−𝑚+𝑘+1, we distinguish the following two subcases.
(c.1) 𝑚−𝑘−2𝑠−1≥1. From 𝑚−𝑘−2𝑠−1≥1, and 𝑚−𝑘≤2𝜈+1, we have 𝑠+1≤𝜈. Let 𝐼𝑈=^(𝑠)000000000000000𝐼^(𝑠)000000000000000𝑥𝑦0000𝐼^(𝜎1)0000000000000000000𝐼^(𝑘)0,𝑠1𝜎₁𝜎𝑠1𝜎₁𝑒𝜎11𝑘𝑙−𝑘𝑊=_𝑠+1+Γ2𝑒_𝜈+𝑠+1, (3.8) where 𝜎₁=𝑚−𝑘−2𝑠−2 and 𝑥²−𝑧𝑦²=Γ, then 𝑈 is a subspace of type (𝑚−1,2𝑠+1,𝑠,Γ,𝑘), 𝑊 is a subspace of type (1,1,0,Γ,0). But when 𝑞=3(mod4) or 𝑞=1(mod4), similar to the proof of the case (a.1), ⟨𝑈,𝑊⟩ is a subspace of type (𝑚,2𝑠+2,𝑠,Γ,𝑘) or (𝑚,2(𝑠+1),𝑠+1,𝜙,𝑘). Consequently, 𝑈,𝑊∈ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), ⟨𝑈,𝑊⟩∉ℒ_𝑂(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), and the formula (2.2) does not hold.
(c.2) 𝑚−𝑘−2𝑠−1=0. From 2≤𝑚−𝑘≤2𝜈+1, we have 𝑠≥1 and 𝑚≥3. We choose (𝑎,𝑏) and (𝑐,𝑑) being two linearly independent solutions of the equation 𝑥²−𝑧𝑦²=Γ. Let 𝐼𝑈=^(𝑠−1)00000000000𝐼^(𝑠)0000000000𝑎𝑏000000000𝐼^(𝑘)0,𝑠−11𝜎𝑠𝜎11𝑘𝑙−𝑘𝑊=𝑐𝑒_2𝜈+1+𝑑𝑒_2𝜈+2, (3.9) then 𝑈 is a subspace of type (𝑚−1,2(𝑠−1)+1,𝑠−1,Γ,𝑘), 𝑊 is a subspace of type (1,1,0,Γ,0). Let 1𝐴=𝑎𝑏𝑐𝑑−𝑧𝑎𝑏𝑐𝑑^𝑡, (3.10) because det𝐴=−(𝑎𝑑−𝑏𝑐)²𝑧, hence, 𝐴 is cogredient to [1,−𝑧]. Then, 𝑈𝑊𝑆_𝑙𝑈𝑊^𝑡 (3.11) is cogredient to 𝑆_{2(𝑠−1)+2,Δ},𝑜^{(𝑚−𝑘−2𝑠)},𝑜^(𝑘). (3.12) Therefore, ⟨𝑈,𝑊⟩ is a subspace of type (𝑚,2(𝑠−1)+2,𝑠−1,Γ,𝑘). Similar to the proof of the case (a.2), the formula (2.2) does not hold for 𝑈 and 𝑊.

4. The Geometricity of Lattices ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)

Theorem 4.1. Let 2𝜈+𝛿+𝑙>𝑚≥1,  0≤𝑘<𝑙, assume that (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘) satisfies conditions (2.10) and (2.11). Then,(i)each of ℒ_𝑅(𝑘+1,0,0,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ), ℒ_𝑅(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ)(Γ=1𝑜𝑟𝑧) and ℒ_𝑅(2𝜈+𝛿+𝑘−1,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is a finite geometric lattice when 𝑘=0, and is a finite atomic lattice, but not a geometric lattice when 0<𝑘<𝑙;(ii)when 2≤𝑚−𝑘≤2𝜈+𝛿−2, ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is a finite atomic lattice, but not a geometric lattice.

Proof . By Theorem 2.9, the rank function of ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) is defined by formula (2.16), 𝔽_𝑞^{(2𝜈+𝛿+𝑙)}is the minimal element of ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), all subspaces of type (𝑚,2𝑠+𝛾,𝑠,Γ,𝑘) in ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) are atoms of ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ).
The Proof of (i). By [8], ℒ_𝑅(𝑘+1,0,0,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ), ℒ_R(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), and ℒ_𝑅(2𝜈+𝛿+𝑘−1,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) are finite geometric lattices when 𝑘=0; in the following, we will show that they are finite atomic lattices, but not geometric lattices when 0<𝑘<𝑙.
(a) Let 𝑈=𝑒_𝜈+1,𝑒_2𝜈+𝛿+1,𝑒_2𝜈+𝛿+2,…,𝑒_{2𝜈+𝛿+𝑘},𝑊=𝑒₁,𝑒_2𝜈+𝛿+2,𝑒_2𝜈+𝛿+3,…,𝑒_{2𝜈+𝛿+𝑘+1}. (4.1) Then, both 𝑈 and 𝑊 are subspaces of type (𝑘+1,0,0,𝜙,𝑘), and 𝑈∩𝑊=⟨𝑒_2𝜈+𝛿+2,𝑒_2𝜈+𝛿+3,…,𝑒_{2𝜈+𝛿+𝑘}⟩, ⟨𝑈,𝑊⟩ is a subspace of type (𝑘+3,2,1,𝜙,𝑘+1). Consequently, ⟨𝑈,𝑊⟩∉ℒ_𝑅(𝑘+1,0,0,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ), 𝑟^′(𝑈∧𝑊)=𝑟^′(𝔽_𝑞^{(2𝜈+𝛿+𝑙)})=0,  𝑟^′(𝑈∨𝑊)=𝑟^′(𝑈∩𝑊)=𝑘+2−(𝑘−1)=3,  𝑟^′(𝑈)=𝑟^′(𝑊)=𝑘+2−(𝑘+1)=1. Thus, 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊). (4.2) That is, (2.2) does not hold for 𝑈 and 𝑊. Hence, ℒ_𝑅(𝑘+1,0,0,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ) are not geometric lattices when 0<𝑘<𝑙.
(b) Let 𝑈=𝑒₁+Γ2𝑒_𝜈+1,𝑒_2𝜈+𝛿+1,𝑒_2𝜈+𝛿+2,…,𝑒_{2𝜈+𝛿+𝑘},𝑊=𝑒_𝑠+1+Γ2𝑒_𝜈+𝑠+1,𝑒_2𝜈+𝛿+2,𝑒_2𝜈+𝛿+3,…,𝑒_{2𝜈+𝛿+𝑘+1}. (4.3) Then, both 𝑈 and 𝑊 are subspaces of type (𝑘+1,1,0,Γ,𝑘), and 𝑈∩𝑊=⟨𝑒_2𝜈+𝛿+2,𝑒_2𝜈+𝛿+3,…,𝑒_{2𝜈+𝛿+𝑘}⟩, ⟨𝑈,𝑊⟩ is a subspace of type (𝑘+3,2,0,Γ,𝑘+1) or (𝑘+3,2,1,𝜙,𝑘+1) when 𝑞=3(mod4) or 𝑞=1(mod4). Consequently, ⟨U,𝑊⟩∉ℒ_𝑅(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ),  𝑟^′(𝑈∧𝑊)=𝑟^′(𝔽_𝑞^{(2𝜈+𝛿+𝑙)})=0,  𝑟^′(𝑈∨𝑊)=𝑟^′(𝑈∩𝑊)=𝑘+2−(𝑘−1)=3,  𝑟^′(𝑈)=𝑟^′(𝑊)=𝑘+2−(𝑘+1)=1. Thus, 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊). (4.4) That is, (2.2) does not hold for 𝑈 and 𝑊. Hence, ℒ_𝑅(𝑘+1,1,0,Γ,𝑘;2𝜈+𝛿+𝑙,Δ) are not geometric lattices when 0<𝑘<𝑙.
(c) From the condition (2.10), the following hold.(i)If 𝛾=𝛿=1,Γ≠Δ, then 2𝜈+𝛿−1≤𝜈+𝑠, that is, 𝜈≤𝑠,𝜈=𝑠, hence 2𝜈+1≤2𝜈, and it is a contradiction.(ii)If 𝛾=𝛿,Γ=Δ, then 2𝜈+𝛿−1≤𝜈+s+𝛿, that is, 𝜈−1≤𝑠, hence 𝑠=𝜈, or 𝑠=𝜈−1. When 𝑠=𝜈, from 2𝑠+𝛾≤2𝜈+𝛿−1, we obtain 2𝜈+𝛿≤2𝜈+𝛿−1, and it is a contradiction. When 𝑠=𝜈−1, we have 2𝜈+𝛿−2≤2𝜈+𝛿−1. That is, in this situation, 𝜈−1=𝑠holds.(iii)If 𝛾≠𝛿, then 2𝜈+𝛿−1≤𝜈+𝑠+min{𝛿,𝛾}≤𝜈+𝑠+𝛿, that is, 𝜈−1≤𝑠, hence 𝑠=𝜈, or 𝑠=𝜈−1. When 𝑠=𝜈, we have 2𝜈+𝛾≤2𝜈+𝛿−1, then 𝛾≤𝛿−1. When 𝑠=𝜈−1, we have 2𝜈+𝛾−2≤2𝜈+𝛿−1, then 𝛾−1≤𝛿.
From the discussion above, we know that
(c.1) If 𝑠=𝜈, then 𝛾≤𝛿−1, and we have 𝛿=1,  𝛾=0;  𝛿=2,  𝛾=0, and 𝛿=2,  𝛾=1 three possible cases. For ℒ_𝑅(2𝜈+𝛿+𝑘−1,2𝜈+𝛾,𝜈,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), here we just give the proof of the case 𝛿=2,  𝛾=1, others can be obtained in the similar way. We choose (𝑎,𝑏) and (𝑐,𝑑) being two linearly independent solutions of the equation 𝑥²−𝑧𝑦²=Γ. Let 𝐼𝑈=^(𝜈)000000𝐼^(𝜈)000000𝑎𝑏000000𝐼^(𝑘)0,,𝜈𝜈11𝑘𝑙−𝑘𝑊=00𝑐𝑑0000000001𝜈𝜈11𝑘𝑙−𝑘−11 (4.5) then 𝑈 is a subspace of type (2𝜈+k+1,2𝜈+1,𝜈,Γ,𝑘), 𝑊 is a subspace of type (2,1,0,Γ,1), and ⟨𝑈,𝑊⟩ is a subspace of type (2𝜈+𝑘+3,2𝜈+2,𝜈,Γ,𝑘+1). Consequently, 𝑈,𝑊∈ℒ_𝑅(2𝜈+𝑘+1,2𝜈+1,𝜈,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), ⟨𝑈,𝑊⟩∉ℒ_𝑅(2𝜈+𝑘+1,2𝜈+1,𝜈,Γ,𝑘;2𝜈+𝛿+𝑙,Δ). Thus, we have 𝑈∨𝑊={0},  𝑈∧𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)},  𝑟^′(𝑈∨𝑊)=𝑟^′(𝑈∩𝑊)=2𝜈+𝑘+2,  𝑟^′(𝑈∧𝑊)=0,  𝑟^′(𝑈)=2𝜈+𝑘+2−2𝜈−𝑘−1=1,  𝑟^′(𝑊)=2𝜈+𝑘+2−2=2𝜈+𝑘. Then, 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊). (4.6) That is, (2.2) does not hold for 𝑈 and 𝑊. Hence, ℒ_𝑅(2𝜈+𝑘+1,2𝜈+1,𝜈,1,𝑘;2𝜈+𝛿+𝑙,Δ) are not geometric lattices when 0<𝑘<𝑙.
(c.2) If 𝑠=𝜈−1, then we have 𝛾≠𝛿,𝛾−1≤𝛿; or 𝛾=𝛿, Γ=Δ. As to ℒ_𝑅(2𝜈+𝛿+𝑘−1,2(𝜈−1)+𝛾,𝜈−1,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), we consider 𝛿=0,  𝛿=1, and 𝛿=2 three cases. Here we just give the proof of the case 𝛿=1, and we also discuss the following three subcases:
(c.2.1) 𝛿=1, 𝛾=0. For ℒ_𝑅(2𝜈+𝑘,2(𝜈−1),𝜈−1,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ), let 𝐼𝑈=^(𝜈−1)00000000𝐼^(𝜈)00000000𝐼^(𝑘),,000000001𝜈−11𝜈1𝑘𝑙−𝑘−11𝑊=01000000001000𝜈−11𝜈1𝑘𝑙−𝑘−11 (4.7) then 𝑈 is a subspace of type (2𝜈+𝑘,2(𝜈−1),𝜈−1,𝜙,𝑘+1), 𝑊 is a subspace of type (2,1,0,Δ,0), and ⟨𝑈,𝑊⟩ is a subspace of type (2𝜈+𝑘+2,2𝜈+1,𝜈,Δ,𝑘+1). If 𝜈=1, then 𝑠=0, and as to 𝑊, from the condition (2.10), we obtain 2≤1, that is, it is a contradiction. Consequently, 𝜈≥2, and 𝑈,𝑊∈ℒ_𝑅(2𝜈+𝑘,2(𝜈−1),𝜈−1,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ),⟨𝑈,𝑊⟩∉ℒ_𝑅(2𝜈+𝑘,2(𝜈−1),𝜈−1,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ). Thus, we have 𝑈∨𝑊={0},𝑈∧𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, 𝑟^′(𝑈∨𝑊)=𝑟^′(𝑈∩𝑊)=2𝜈+𝑘+1, 𝑟^′(𝑈∧𝑊)=0, 𝑟^′(𝑈)=2𝜈+𝑘+1−2𝜈−𝑘=1, 𝑟^′(𝑊)=2𝜈+𝑘+1−2=2𝜈+𝑘−1. Then, 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊). (4.8) That is, (2.2) does not hold for 𝑈 and 𝑊. Hence, ℒ_𝑅(2𝜈+𝑘,2(𝜈−1),𝜈−1,𝜙,𝑘;2𝜈+𝛿+𝑙,Δ)are not geometric lattices when 0<𝑘<𝑙.
(c.2.2) 𝛿=1, 𝛾=1, Γ=Δ. For ℒ_𝑅(2𝜈+𝑘,2(𝜈−1)+1,𝜈−1,Δ,𝑘;2𝜈+𝛿+𝑙,Δ), let 𝐼𝑈=^(𝜈−1)00000000𝐼^(𝜈)000000010000000𝐼^(𝑘),,00𝜈−11𝜈1𝑘𝑙−𝑘−11𝑊=01000000001001𝜈−11𝜈1𝑘𝑙−𝑘−11 (4.9) then 𝑈 is a subspace of type (2𝜈+𝑘,2(𝜈−1)+1,𝜈−1,Δ,𝑘), 𝑊 is a subspace of type (2,1,0,Δ,0), and ⟨𝑈,𝑊⟩ is a subspace of type (2𝜈+𝑘+2,2𝜈+1,𝜈,Δ,𝑘+1). Consequently, 𝑈,𝑊∈ℒ_𝑅(2𝜈+𝑘,2(𝜈−1)+1,𝜈−1,Δ,𝑘;2𝜈+𝛿+𝑙,Δ),⟨𝑈,𝑊⟩∉ℒ_𝑅(2𝜈+𝑘,2(𝜈−1)+1,𝜈−1,Δ,𝑘;2𝜈+𝛿+𝑙,Δ). Thus, we have 𝑈∨𝑊={0}, 𝑈∧𝑊=𝔽_𝑞^{(2𝜈+𝛿+𝑙)}, 𝑟^′(𝑈∨𝑊)=𝑟^′(𝑈∩𝑊)=2𝜈+𝑘+1, 𝑟^′(𝑈∧𝑊)=0, 𝑟^′(𝑈)=2𝜈+𝑘+1−2𝜈−𝑘=1, 𝑟^′(𝑊)=2𝜈+𝑘+1−2=2𝜈+𝑘−1. Then, 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊). (4.10) That is, (2.2) does not hold for 𝑈 and 𝑊. Hence, ℒ_𝑅(2𝜈+𝑘,2(𝜈−1)+1,𝜈−1,Δ,𝑘;2𝜈+𝛿+𝑙,Δ)are not geometric lattices when 0<𝑘<𝑙.
(c.2.3) 𝛿=1,𝛾=2. See the proof of the Theorem 7 in [12].
The Proof of (ii). Let 𝑈∈ℳ(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝛿+𝑙,Δ), then 𝑈𝑆_𝑙𝑈^𝑡=Λ₁,0^{𝑚−𝑘−2𝑠−𝛾},0^(𝑘), (4.11) where Λ₁=𝑆_{2𝑠+𝛾,Γ}. Hence, there exists a (2𝜈+𝛿+𝑙−𝑚)×(2𝜈+𝛿+𝑙) matrix 𝑍 such that 𝑈𝑍𝑆_𝑙𝑈𝑍^𝑡=Λ₁,𝑆_{2(𝑚−𝑘−2𝑠−𝛾)},Λ^∗,0^(𝑘),0^{(𝑙−𝑘)}, (4.12) where Λ^∗ takes values in Table 1 as follows.
In Table 1 as follows ∑_𝑖=𝑆_{2(𝜈+𝑠−𝑚+𝑘+𝑖)},𝑖=0,1,or2.
As to 𝛿=0;𝛿=1,Δ=1;𝛿=1,Δ=𝑧, and 𝛿=2 four cases, we only show the proof of the case 𝛿=0, others can be obtained in the similar way. We also distinguish the following three subcases.
(a) If 𝛾=0, then Λ₁=𝑆_2𝑠,Λ^∗=𝑆_{2(𝜈−𝑚+𝑘+𝑠)}. Let 𝑢₁,𝑢₂,…,𝑢_𝑠,𝑣₁,𝑣₂,…,𝑣_𝑠,𝑢_𝑠+1,…,𝑢_{𝑚−𝑘−𝑠},𝑤₁,…,𝑤_𝑘 and 𝑣_𝑠+1,…,𝑣_{𝑚−𝑘−𝑠},𝑢_{𝑚−𝑘−𝑠+1},…,𝑢_𝜈,𝑣_{𝑚−𝑘−𝑠+1},…,𝑣_𝜈,𝑤_𝑘+1,…,𝑤_𝑙 be row vectors of 𝑈 and 𝑍, respectively, 𝑊=⟨𝑣_{𝜈−𝑚+𝑘+𝑠+1},…,𝑣_𝜈−𝑠,𝑢_{𝜈−𝑠+1},…,𝑢_𝜈,𝑣_{𝜈−𝑠+1},…,𝑣_𝜈,𝑤₁,…,𝑤_𝑘⟩, (4.13) then 𝑊∈ℳ(𝑚,2𝑠,𝑠,𝜙,𝑘;2𝜈+𝑙).
From 𝑚−𝑘≤2𝜈−2, we know 𝑠<𝜈. If 𝑚−𝑘=2𝑠, then 𝑚−𝑘−𝑠=𝑠<𝜈, so 𝑢_𝜈,𝑣_𝜈∉𝑈. If 𝑚−𝑘>2𝑠, then 𝑠<𝜈−1, so 𝑣_𝜈−1,𝑣_𝜈∉𝑈. In a word, dim⟨𝑈,𝑊⟩≥𝑚+2,dim(𝑈∩𝑊)≤𝑚−2. That is, 𝑈∧𝑊=𝔽_𝑞^(2𝜈+𝑙), 𝑟^′(𝑈∧𝑊)=0, 𝑟^′(𝑈∨𝑊)≥𝑚+1−(𝑚−2)=3,𝑟^′(𝑈)=𝑟^′(𝑊)=𝑚+1−𝑚=1. Consequently, 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊).
(b) If 𝛾=1, then Λ₁=𝑆_2𝑠+1,Γ,Λ^∗=𝑆_{2(𝜈−𝑚+𝑘+𝑠)+1,−Γ}, and Γ=(1)or(𝑧). Let 𝑢₁,𝑢₂,…,𝑢_𝑠,𝑣₁,𝑣₂,…,𝑣_𝑠,𝜔,𝑢_𝑠+1,…,𝑢_{𝑚−𝑘−𝑠−1},𝑤₁,…,𝑤_𝑘 and 𝑣_𝑠+1,…,𝑣_{𝑚−𝑘−𝑠−1},𝑢_{𝑚−𝑘−𝑠},…,𝑢_𝜈−1,𝑣_{𝑚−𝑘−𝑠},…,𝑣_𝜈−1,𝜔^∗,𝑤_𝑘+1,…,𝑤_𝑙 be row vectors of 𝑈 and 𝑍, respectively 𝑊=𝑣_{𝜈−𝑚+𝑘+𝑠+1},…,𝑣_{𝜈−𝑠−1},𝑢_𝜈−𝑠,…,𝑢_𝜈−2,𝑣_𝜈−𝑠,…,𝑣_𝜈−2,𝜔,𝜔^∗,12Γ𝑢_𝜈−1+𝑣_𝜈−1,𝑤₁,…,𝑤_𝑘, (4.14) because ((1/2)Γ𝑢_𝜈−1+𝑣_𝜈−1)𝑆_2𝜈((1/2)Γ𝑢_𝜈−1+𝑣_𝜈−1)^𝑡=Γ, and 12Γ−12Γ𝜔11𝜔^∗𝑆_2𝜈𝜔𝜔^∗𝑡12Γ−12Γ11^𝑡=𝑆_2⋅1, (4.15) then 𝑊∈ℳ(𝑚,2𝑠+1,𝑠,Γ,𝑘;2𝜈+𝑙). From the conditions 2𝑠+1≤𝑚−𝑘≤2𝜈−2 and 𝑚−𝑘≤𝜈+𝑠, we can obtain 𝑚−𝑘−𝑠−1≤𝜈−1 and 𝑠≤𝜈−1, hence (1/2)Γ𝑢_𝜈−1+𝑣_𝜈−1∉𝑈. Obviously, 𝜔^∗∉𝑈. Similar to the proof of the case (a), 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊).
(c) If 𝛾=2, then Λ₁=𝑆_2𝑠+2,Γ,Λ^∗=𝑆_{2(𝜈−𝑚+𝑘+𝑠)+2,Γ}, and Γ=[1,−𝑧]. Let 𝑢₁,𝑢₂,…,𝑢_𝑠,𝑣₁,𝑣₂,…,𝑣_𝑠,𝜔₁,𝜔₂,𝑢_𝑠+1,…,𝑢_{𝑚−𝑘−𝑠−2},𝑤₁,…,𝑤_𝑘 and 𝑣_𝑠+1,…,𝑣_{𝑚−𝑘−𝑠−2},𝑢_{𝑚−𝑘−𝑠−1},…,𝑢_𝜈−2,𝑣_{𝑚−𝑘−𝑠−1},…,𝑣_𝜈−2,𝜔^∗₁,𝜔^∗₂,𝑤_𝑘+1,…,𝑤_𝑙 be row vectors of 𝑈 and 𝑍, respectively, 𝑊=𝑣_{𝜈−𝑚+𝑘+𝑠+1},…,𝑣_{𝜈−𝑠−2},𝑢_{𝜈−𝑠−1},…,𝑢_𝜈−2,𝑣_{𝜈−𝑠−1},…,𝑣_𝜈−2,𝜔^∗₁,𝜔^∗₂,𝑤₁,…,𝑤_𝑘, (4.16) then 𝑊∈ℳ(𝑚,2𝑠+2,𝑠,Γ,𝑘;2𝜈+𝑙). Obviously, 𝜔^∗₁,𝜔^∗₂∉𝑈. Similar to the proof of the case (a), 𝑟^′(𝑈∧𝑊)+𝑟^′(𝑈∨𝑊)>𝑟^′(𝑈)+𝑟^′(𝑊).
From the discussion above, we know that when 2≤𝑚−𝑘≤2𝜈−2, ℒ_𝑅(𝑚,2𝑠+𝛾,𝑠,Γ,𝑘;2𝜈+𝑙) is a finite atomic lattice, but not a geometric lattice.

Table 1 

Acknowledgments

This work is supported by the National Natural Science Foundation of China under Grant no. 61179026 and the Fundamental Research Funds for the Central Universities under Grant no. ZXH2012K003. The authors would like to thank the referee for his suggestions on simplifying the earlier version of the paper.