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

Authors: You Gao; XinZhi Fu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

Let𝔽q(2ν+δ+l) be a (2ν+δ+l)-dimensional vector space over the finite field 𝔽q. In this paper we assume that 𝔽q is a finite field of odd characteristic, and O2ν+δ+l,Δ(𝔽q) the singular orthogonal groups of degree 2ν+δ+l over 𝔽q. Let ℳ be any orbit of subspaces under O2ν+δ+l,Δ(𝔽q). 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 𝔽q(2ν+δ+l) is assumed to be 𝔽q(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.


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## Body

## 1. Introduction

Let𝔽q be a finite field with q elements, where q is an odd prime power. We choose a fixed nonsquare element z in 𝔽q*:=𝔽q∖{0}. Let 𝔽q(2ν+δ+l) be a (2ν+δ+l)-dimensional row vector space over the finite field 𝔽q, and let O2ν+δ+l,Δ(𝔽q) be one of the singular orthogonal groups of degree 2ν+δ+l over 𝔽q. There is an action of O2ν+δ+l,Δ(𝔽q) on 𝔽q(2ν+δ+l) defined as follows:
(1.1)Fq(2ν+δ+l)×O2ν+δ+l,Δ(Fq)⟶Fq(2ν+δ+l),((x1,x2,…,x2ν+δ+l),T)⟼(x1,x2,…,x2ν+δ+l)T.
Let P be an m-dimensional subspace of 𝔽q(2ν+δ+l)(1≤m≤2ν+δ+l), and v1,v2,…,vm be a basis of P. Then, the m×(2ν+δ+l) matrix:
(1.2)(v1v2⋮vm)
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 𝔽q(2ν+δ+l), that is, a subspace P is carried by T∈O2ν+δ+l,Δ(𝔽q) into the subspace PT. The set of subspaces of 𝔽q(2ν+δ+l) is partitioned into orbits under O2ν+δ+l,Δ(𝔽q). Clearly, {0} and {𝔽q(2ν+δ+l)} are two trivial orbits. Let ℳ be any orbit of subspaces under O2ν+δ+l,Δ(𝔽q). 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 𝔽q(2ν+δ+l). Then, 𝔽q(2ν+δ+l)∈ℒ(ℳ). Partially ordering ℒ(ℳ) by ordinary or reverse inclusion, we get two posets and denote them by ℒO(ℳ) and ℒR(ℳ), respectively. Clearly, for any two elements P,Q∈ℒO(ℳ),
(1.3)P∧Q=P∩Q,P∨Q=∩{R∈LO(M):R⊇〈P,Q〉},
where 〈P,Q〉 is a subspace generated by P and Q. Therefore, ℒO(ℳ) is a finite lattice.Similarly, for any two elementsP,Q∈ℒR(ℳ),
(1.4)P∧Q=∩{R∈LR(M):R⊇〈P,Q〉},P∨Q=P∩Q,
so ℒR(ℳ) is also a finite lattice. Both ℒO(ℳ) and ℒR(ℳ) are called the lattices generated by ℳ.The results on the geometricity of lattices generated by subspaces ind-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 ℒO(ℳ) and ℒR(ℳ) generated by different orbits ℳ of subspaces under singular orthogonal group O2ν+δ+l,Δ(𝔽q). The study contents include the inclusion relations between different lattices, the characterization of subspaces contained in a given lattice ℒR(ℳ) (resp., ℒO(ℳ)), and the characteristic polynomial of ℒR(ℳ). The purpose of this paper is to study the questions when ℒR(ℳ) (resp., ℒO(ℳ)) are geometric lattices.


## 2. Preliminaries

In the following, we recall some definitions and facts on ordered sets and lattices (see [8, 14]).LetA be a partially ordered set, and a,b∈A. We say that b  coversa 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.LetA 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 rankr(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 posetA 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]).
LetA 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.
LetA 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 latticeL is called a semimodular lattice if for all a,b∈L,
(2.1)a∧b<⋅a⟹b<⋅a∨b.Proposition 2.3 ([14, Theorem 2.27]).
LetL 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∈L(2.2)r(x∧y)+r(x∨y)≤r(x)+r(y).Definition 2.4 ([14, page 52]).
A latticeL is called a geometric lattice if it isG1′
an atomic lattice,G2′
a semimodular lattice,G3
without infinite chains inL.According to Definition2.2, Proposition 2.3, and Definition 2.4, we can obtain the following proposition.Proposition 2.5.
LetL be a lattice with 0. Then, L is a geometric lattice if and only ifG1
for every elementa∈L∖{0}, a=sup{b∈L∣0<·b≤a},G2
L possesses a rank function r and for all x,y∈L, (2.2) holds,G3
without infinite chains inL.Let(2.3)S2ν+δ,Δ=(0I(ν)I(ν)0Δ),Sl=(S0(l)),
where S=S2ν+δ,Δ, δ=0,1, or 2, and
(2.4)Δ={ϕ,ifδ=0,1orz,ifδ=1,(1-z),ifδ=2.
The set of all (2ν+δ+l)×(2ν+δ+l)nonsingular matrices T over 𝔽q satisfying
(2.5)TSlTt=Sl
forms a group which will be called the singular orthogonal group of degree 2ν+δ+l, rank2ν+δ, and with definite part Δover 𝔽qand denoted by O2ν+δ+l,Δ(𝔽q).Clearly, O2ν+δ+l,Δ(𝔽q)consists of all (2ν+δ+l)×(2ν+δ+l)nonsingular matrices of the form:
(2.6)T=(T11T120T22)2ν+δl,2ν+δl
where T11ST11t=S, and T22 is nonsingular.Twon×n matrices A and B are called to be cogredient if there exists a nonsingular matrix P such that PAPt=B.Anm-dimensional subspace P is said to be a subspace of type(m,2s+γ,s,Γ), if PSlPt is cogredient to M(m,2s+γ,s,Γ), where the matrix M(m,2s+γ,s,Γ), respectively, is as follows
(2.7)M(m,2s,s)=(0I(s)I(s)00(m-2s)),ifγ=0,M(m,2s+1,s,1)=(0I(s)I(s)010(m-2s-1)),ifγ=1
or
(2.8)M(m,2s+1,s,z)=(0I(s)I(s)0z0(m-2s-1)),ifγ=1,M(m,2s+2,s)=(0I(s)I(s)01-z0(m-2s-2)),ifγ=2.Lete1,e2,…,e2ν+δ,e2ν+δ+1,…,e2ν+δ+lbe a basis of 𝔽q(2ν+δ+l), where
(2.9)ei=(0,…,0,1,0,…,0),
1 is in the ith position. Denote by E the l-dimensional subspace of 𝔽q(2ν+δ+l) generated by e2ν+δ+1,e2ν+δ+2,…,e2ν+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 𝔽q(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
(2.10)k≤l,2s+γ≤m-k≤{ν+s+min⁡{δ,γ},ifγ≠δorγ=δandΓ=Δ,ν+s,ifγ=δ=1andΓ≠Δ,
or
(2.11)min⁡{l,m-2s-γ}≥k≥{max⁡{0,m-ν-s-min⁡{δ,γ}},ifγ≠δorγ=δandΓ=Δ,max⁡{0,m-ν-s},ifγ=δ=1andΓ≠Δ.Moreover, ifℳ(m,2s+γ,s,Γ,k;2ν+δ+l,Δ)is nonempty, then it forms an orbit of subspaces under O2ν+δ+l,Δ(𝔽q). 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 𝔽q(2ν+δ+l) is assumed to be 𝔽q(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 ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ)and ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), respectively.The caseℒR(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.
Let2ν+δ+l>m≥1,  0≤k<l, assume that (m,2s+γ,s,Γ,k) satisfies conditions (2.10) and (2.11). Then,
(2.12)LR(m,2s+r,s,Γ,k;2ν+δ+l,Δ)⊃LR(m1,2s1+γ1,s1,Γ1,k1;2ν+δ+l,Δ)
if and only if
(2.13)k1≤k<l,2(m-k)-2(m1-k1)≥{(2s+γ)-(2s1+γ1)+|γ-γ1|≥2|γ-γ1|,ifγ1≠γorγ1=γandΓ1=Γ,(2s+γ)-(2s1+γ1)+2≥4,ifγ1=γ=1andΓ1≠Γ.Theorem 2.7.
Let2ν+δ+l>m≥1,  0≤k<l. Assume that (m,2s+γ,s,Γ,k) satisfies condition (2.10), then ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) consists of 𝔽q(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.
Let2ν+δ+l>m≥1,  0≤k<l, and (m,2s+γ,s,Γ,k) satisfy
(2.14)2s+γ≤m-k≤{ν+s+min⁡{δ,γ},ifγ≠δorγ=δandΓ=Δ,ν+s,ifγ=δ=1andΓ≠Δ.
For any X∈ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), define
(2.15)r(X)={dim⁡X,ifX≠Fq(2ν+δ+l),m+1,ifX=Fq(2ν+δ+l),
then r:ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ)→ℕ is a rank function of the lattice ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ).Theorem 2.9.
Let2ν+δ+l>m≥1,  0≤k<l, and (m,2s+γ,s,Γ,k) satisfy (2.14). For any X∈ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), define
(2.16)r′(X)={m+1-dim⁡X,ifX≠Fq(2ν+δ+l),0,ifX=Fq(2ν+δ+l),
then r′:ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ)→ℕ is a rank function of the lattice ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ).


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

Theorem 3.1.
Let2ν+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 ofℒO(k+1,0,0,ϕ,k;2ν+δ+l,Δ) and ℒO(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)
when2≤m-k≤2ν+δ-1, ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) is a finite atomic lattice, but not a geometric lattice.Proof.
By Theorem2.8, the rank function of ℒO(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 ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ). {0}∈ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) and it is the minimal element, so all 1-dim subspaces in ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) are atoms of ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ).
LetU∈ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ)∖{{0},𝔽q(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 ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ). Assume m1≥2, then
(3.1)USlUt=[S2s1+γ1,Γ1,0(m1-k1-2s1-γ1),0(k1)],
where Γ1=ϕ,(1),(z), or [1,-z].
LetUi be an ith (1≤i≤m1) row vector of U, then 〈Ui〉 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 〈Ui〉⊂U. By Theorem 2.7, we know 〈Ui〉∈ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), so 〈Ui〉 is an atom of ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), and U=∨i=1m1〈Ui〉, hence, U is a union of atoms in ℒO(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 𝔽q(2ν+δ+l)=W1∨W2. W1,W2 are unions of atoms in ℒO(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), hence, 𝔽q(2ν+δ+l) is a union of atoms in ℒO(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 ℒO(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. ℒO(k+1,1,0,Γ,k;2ν+δ+l,Δ) consists of 𝔽q(2ν+δ+l),{0} and subspaces of type (1,1,0,Γ,0). Let U,W∈ℒO(1,1,0,Γ,0;2ν+δ+l,Δ), if U,W are 𝔽q(2ν+δ+l),{0}, respectively, then U∨V=𝔽q(2ν+δ+l),U∧W={0}, so r(U∨W)+r(U∧W)=r(U)+r(W). If U=W is {0} or 𝔽q(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 𝔽q(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=𝔽q(2ν+δ+l), so r(U∨W)+r(U∧W)=r(U)+r(W).
Hence, (2.2) holds and ℒO(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)eν+1〉, W=〈es+1+(Γ/2)eν+s+1〉, where s≤ν-1, then U,W∈ℒO(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 S2·1, 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〉∉ℒO(k+1,1,0,Γ,k;2ν+δ+l,Δ), and we have U∨W=𝔽q(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,ℒO(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∈ℒO(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
(3.2)U=(I(s)0000000000000000I(s)000000010000Γ/200000000I(σ1)000000000000000000I(k)0),s11σ1σs11σ1σkl-kW=〈es+2+(Γ2)eν+s+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 S2·1, 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∈ℒO(m,2s+1,s,Γ,k;2ν+δ+l,Δ), 〈U,W〉∉ℒO(m,2s+1,s,Γ,k;2ν+δ+l,Δ). Thus, we have U∨W=𝔽q(2ν+δ+l),  U∧W={0},  r(U∨W)=m+1,r(U∧W)=0,  r(U)=m-1,  r(W)=1.Then,
(3.3)r(U∨W)+r(U∧W)=m+1>r(U)+r(W)=m-1+1=m.
(a.2)m-k-2s-1=0. From 2≤m-k≤2ν+δ-1, we have s+1≤ν,  s≥1. Let
(3.4)U=(I(s-1)000000000000I(s)000000100Γ/20000000000I(k)0),s-111σs1σkl-kW=〈es+1-(Γ2)eν+s+1〉,
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∈ℒO(m,2s+1,s,Γ,k;2ν+δ+l,Δ), 〈U,W〉∉ℒO(m,2s+1,s,Γ,k;2ν+δ+l,Δ). Thus, we have U∨W=𝔽q(2ν+δ+l),  U∧W={0},  r(U∨W)=m+1,  r(U∧W)=0,  r(U)=m-1,  r(W)=1.Then,
(3.5)r(U∨W)+r(U∧W)=m+1>r(U)+r(W)=m-1+1=m.
Therefore, there exist U,W∈ℒO(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
(3.6)U=(I(s)00000000000000I(s)0000000000000010000I(σ1)00000000000000000I(k)0),s1σ1σs1σ1σ1kl-kW=〈es+1+(Δ2)eν+s+1〉,
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∈ℒO(m,2s+1,s,Δ,k;2ν+1+l,Δ), 〈U,W〉∉ℒO(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
(3.7)U=(I(s-1)0000000000000I(s)00000000000010000000000I(k)0),s-111σ-1s1σ-11kl-kW=〈es+1+(Δ2)eν+s+1〉,
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
(3.8)U=(I(s)000000000000000I(s)000000000000000xy0000I(σ1)0000000000000000000I(k)0),s1σ1σs1σ1σ11kl-kW=〈es+1+(Γ2)eν+s+1〉,
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∈ℒO(m,2s+1,s,Γ,k;2ν+δ+l,Δ), 〈U,W〉∉ℒO(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
(3.9)U=(I(s-1)00000000000I(s)0000000000ab000000000I(k)0),s-11σsσ11kl-kW=〈ce2ν+1+de2ν+2〉,
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
(3.10)A=(abcd)(1-z)(abcd)t,
because detA=-(ad-bc)2z, hence, A is cogredient to [1,-z]. Then,
(3.11)(UW)Sl(UW)t
is cogredient to
(3.12)[S2(s-1)+2,Δ,o(m-k-2s),o(k)].
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 LatticesℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ)

Theorem 4.1.
Let2ν+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 ofℒR(k+1,0,0,ϕ,k;2ν+δ+l,Δ), ℒR(k+1,1,0,Γ,k;2ν+δ+l,Δ)(Γ=1orz) and ℒR(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)
when2≤m-k≤2ν+δ-2, ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) is a finite atomic lattice, but not a geometric lattice.Proof .
By Theorem2.9, the rank function of ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) is defined by formula (2.16), 𝔽q(2ν+δ+l)is the minimal element of ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ), all subspaces of type (m,2s+γ,s,Γ,k) in ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ) are atoms of ℒR(m,2s+γ,s,Γ,k;2ν+δ+l,Δ).
The Proof of (i). By [8], ℒR(k+1,0,0,ϕ,k;2ν+δ+l,Δ), ℒR(k+1,1,0,Γ,k;2ν+δ+l,Δ), and ℒR(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(4.1)U=〈eν+1,e2ν+δ+1,e2ν+δ+2,…,e2ν+δ+k〉,W=〈e1,e2ν+δ+2,e2ν+δ+3,…,e2ν+δ+k+1〉.
Then, both U and W are subspaces of type (k+1,0,0,ϕ,k), and U∩W=〈e2ν+δ+2,e2ν+δ+3,…,e2ν+δ+k〉, 〈U,W〉 is a subspace of type (k+3,2,1,ϕ,k+1). Consequently, 〈U,W〉∉ℒR(k+1,0,0,ϕ,k;2ν+δ+l,Δ), r′(U∧W)=r′(𝔽q(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,
(4.2)r′(U∧W)+r′(U∨W)>r′(U)+r′(W).
That is, (2.2) does not hold for U and W. Hence, ℒR(k+1,0,0,ϕ,k;2ν+δ+l,Δ) are not geometric lattices when 0<k<l.
(b) Let(4.3)U=〈e1+(Γ2)eν+1,e2ν+δ+1,e2ν+δ+2,…,e2ν+δ+k〉,W=〈es+1+(Γ2)eν+s+1,e2ν+δ+2,e2ν+δ+3,…,e2ν+δ+k+1〉.
Then, both U and W are subspaces of type (k+1,1,0,Γ,k), and U∩W=〈e2ν+δ+2,e2ν+δ+3,…,e2ν+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〉∉ℒR(k+1,1,0,Γ,k;2ν+δ+l,Δ),  r′(U∧W)=r′(𝔽q(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,
(4.4)r′(U∧W)+r′(U∨W)>r′(U)+r′(W).
That is, (2.2) does not hold for U and W. Hence, ℒR(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)
Ifγ=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)
Ifγ=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)
Ifγ≠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) Ifs=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 ℒR(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
(4.5)U=(I(ν)000000I(ν)000000ab000000I(k)0),νν11kl-kW=(00cd0000000001),νν11kl-k-11
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∈ℒR(2ν+k+1,2ν+1,ν,Γ,k;2ν+δ+l,Δ), 〈U,W〉∉ℒR(2ν+k+1,2ν+1,ν,Γ,k;2ν+δ+l,Δ). Thus, we have U∨W={0},  U∧W=𝔽q(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,
(4.6)r′(U∧W)+r′(U∨W)>r′(U)+r′(W).
That is, (2.2) does not hold for U and W. Hence, ℒR(2ν+k+1,2ν+1,ν,1,k;2ν+δ+l,Δ) are not geometric lattices when 0<k<l.
(c.2) Ifs=ν-1, then we have γ≠δ,γ-1≤δ; or γ=δ, Γ=Δ. As to ℒR(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 ℒR(2ν+k,2(ν-1),ν-1,ϕ,k;2ν+δ+l,Δ), let
(4.7)U=(I(ν-1)00000000I(ν)00000000I(k)000000001),ν-11ν1kl-k-11W=(01000000001000),ν-11ν1kl-k-11
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∈ℒR(2ν+k,2(ν-1),ν-1,ϕ,k;2ν+δ+l,Δ),〈U,W〉∉ℒR(2ν+k,2(ν-1),ν-1,ϕ,k;2ν+δ+l,Δ). Thus, we have U∨W={0},U∧W=𝔽q(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,
(4.8)r′(U∧W)+r′(U∨W)>r′(U)+r′(W).
That is, (2.2) does not hold for U and W. Hence, ℒR(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 ℒR(2ν+k,2(ν-1)+1,ν-1,Δ,k;2ν+δ+l,Δ), let
(4.9)U=(I(ν-1)00000000I(ν)000000010000000I(k)00),ν-11ν1kl-k-11W=(01000000001001),ν-11ν1kl-k-11
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∈ℒR(2ν+k,2(ν-1)+1,ν-1,Δ,k;2ν+δ+l,Δ),〈U,W〉∉ℒR(2ν+k,2(ν-1)+1,ν-1,Δ,k;2ν+δ+l,Δ). Thus, we have U∨W={0}, U∧W=𝔽q(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,
(4.10)r′(U∧W)+r′(U∨W)>r′(U)+r′(W).
That is, (2.2) does not hold for U and W. Hence, ℒR(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
(4.11)USlUt=[Λ1,0m-k-2s-γ,0(k)],
where Λ1=S2s+γ,Γ. Hence, there exists a (2ν+δ+l-m)×(2ν+δ+l) matrix Z such that
(4.12)(UZ)Sl(UZ)t=[Λ1,S2(m-k-2s-γ),Λ*,0(k),0(l-k)],
where Λ* takes values in Table 1 as follows.
In Table1 as follows ∑i=S2(ν+s-m+k+i),i=0,1,or2.
As toδ=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) Ifγ=0, then Λ1=S2s,Λ*=S2(ν-m+k+s). Let u1,u2,…,us,v1,v2,…,vs,us+1,…,um-k-s,w1,…,wk and vs+1,…,vm-k-s,um-k-s+1,…,uν,vm-k-s+1,…,vν,wk+1,…,wl be row vectors of U and Z, respectively,
(4.13)W=〈vν-m+k+s+1,…,vν-s,uν-s+1,…,uν,vν-s+1,…,vν,w1,…,wk〉,
then W∈ℳ(m,2s,s,ϕ,k;2ν+l).
Fromm-k≤2ν-2, we know s<ν. If m-k=2s, then m-k-s=s<ν, so uν,vν∉U. If m-k>2s, then s<ν-1, so vν-1,vν∉U. In a word, dim〈U,W〉≥m+2,dim(U∩W)≤m-2. That is, U∧W=𝔽q(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) Ifγ=1, then Λ1=S2s+1,Γ,Λ*=S2(ν-m+k+s)+1,-Γ, and Γ=(1)or(z). Let u1,u2,…,us,v1,v2,…,vs,ω,us+1,…,um-k-s-1,w1,…,wk and vs+1,…,vm-k-s-1,um-k-s,…,uν-1,vm-k-s,…,vν-1,ω*,wk+1,…,wl be row vectors of U and Z, respectively
(4.14)W=〈(12)vν-m+k+s+1,…,vν-s-1,uν-s,…,uν-2,vν-s,…,vν-2,ω,ω*,(12)Γuν-1+vν-1,w1,…,wk〉,
because ((1/2)Γuν-1+vν-1)S2ν((1/2)Γuν-1+vν-1)t=Γ, and
(4.15)((12)Γ(-12)Γ11)(ωω*)S2ν(ωω*)t((12)Γ(-12)Γ11)t=S2⋅1,
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)Γuν-1+vν-1∉U. Obviously, ω*∉U. Similar to the proof of the case (a), r′(U∧W)+r′(U∨W)>r′(U)+r′(W).
(c) Ifγ=2, then Λ1=S2s+2,Γ,Λ*=S2(ν-m+k+s)+2,Γ, and Γ=[1,-z]. Let u1,u2,…,us,v1,v2,…,vs,ω1,ω2,us+1,…,um-k-s-2,w1,…,wk and vs+1,…,vm-k-s-2,um-k-s-1,…,uν-2,vm-k-s-1,…,vν-2,ω1*,ω2*,wk+1,…,wl be row vectors of U and Z, respectively,
(4.16)W=〈vν-m+k+s+1,…,vν-s-2,uν-s-1,…,uν-2,vν-s-1,…,vν-2,ω1*,ω2*,w1,…,wk〉,
then W∈ℳ(m,2s+2,s,Γ,k;2ν+l). Obviously, ω1*,ω2*∉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 when2≤m-k≤2ν-2, ℒR(m,2s+γ,s,Γ,k;2ν+l) is a finite atomic lattice, but not a geometric lattice.Table 1
δ
=
0
δ
=
1
,
Δ
=
1
δ
=
1
,
Δ
=
z
δ
=
2
γ
=
0
Σ
0
[
Σ
0
,
1
]
[
Σ
0
,
z
]
[
Σ
0
,
1
,
-
z
]
γ
=
1
,
Γ
=
1
[
Σ
0
,
-
1
]
Σ
1
[
Σ
0
,
-
1
,
z
]
[
Σ
1
,
-
z
]
γ
=
1
,
Γ
=
z
[
Σ
0
,
-
z
]
[
Σ
0
,
1
,
-
z
]
Σ
1
[
Σ
1
,
-
1
]
γ
=
2
[
Σ
0
,
1
,
-
z
]
[
Σ
1
,
z
]
[
Σ
1
,
1
]
Σ
2



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*Source: 387132-2012-08-14.xml*