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

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 point lattice) 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,

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 ∈ L

Definition 2.4 ([14, page 52]). A lattice L is called a geometric lattice if it is

•  

  $$ an atomic lattice,

•  

  $$ a semimodular lattice,

•  

  G₃   without infinite chains in L.

Proposition 2.5. Let L be a lattice with 0. Then, L is a geometric lattice if and only if

•  

  G₁   for every element a ∈ 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 in L.

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,

if and only if

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 $$ 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

For any X ∈ ℒ_O(m, 2s + γ, s, Γ, k;   2ν + δ + l, Δ), define 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

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).

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 = 1  or  z) 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 $$, 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

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 $$, hence, U is a union of atoms in ℒ_O(m, 2s + γ, s, Γ, k;   2ν + δ + l, Δ). Since |ℳ(m, 2s + γ, s, Γ, k;   2ν + δ + l, Δ)| ≥ 2, there exist W₁, W₂ ∈ ℳ(m, 2s + γ, s, Γ, k;   2ν + δ + l, Δ), W₁ ≠ W₂, such that $$. W₁, W₂ are unions of atoms in ℒ_O(m, 2s + γ, s, Γ, k;   2ν + δ + l, Δ), hence, $$ 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 $$ and subspaces of type (1, 1, 0, Γ, 0). Let U, W ∈ ℒ_O(1, 1, 0, Γ, 0;   2ν + δ + l, Δ), if U, W are $$, respectively, then $$, so r(U∨W) + r(U∧W) = r(U) + r(W). If U = W is {0} or $$, 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 $$, so r(U∨W) + r(U∧W) = r(U) + r(W). If U and W are subspaces of type (1, 1, 0, Γ, 0), then $$, 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(mod   4) or q = 1(mod   4), 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 = {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

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(mod   4) or q = 1(mod   4), 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 = {0},  r(U∨W) = m + 1, r(U∧W) = 0,  r(U) = m − 1,  r(W) = 1.  Then,

(a.2) m − k − 2s − 1 = 0. From 2 ≤ m − k ≤ 2ν + δ − 1, we have s + 1 ≤ ν,  s ≥ 1. Let

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 = {0},  r(U∨W) = m + 1,  r(U∧W) = 0,  r(U) = m − 1,  r(W) = 1.  Then, 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

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(mod   4) or q = 1(mod   4), 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

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(mod   4) or q = 1(mod   4), 〈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

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(mod   4) or q = 1(mod   4), 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

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 because det   A = −(ad−bc)²z, hence, A is cogredient to [1, −z]. Then, is cogredient to 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.

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 = 1  or  z) 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), $$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

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^′(U∩W) = k + 2 − (k − 1) = 3,  r^′(U) = r^′(W) = k + 2 − (k + 1) = 1. Thus, 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

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(mod   4) or q = 1(mod   4). Consequently, 〈U, W〉 ∉ ℒ_R(k + 1, 1, 0, Γ, k;   2ν + δ + l, Δ),  $$,  r^′(U∨W) = r^′(U∩W) = k + 2 − (k − 1) = 3,  r^′(U) = r^′(W) = k + 2 − (k + 1) = 1. Thus, 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 = s  holds.

• (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

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},  $$,  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, 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

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 $$, 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, 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

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}, $$, 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, 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

where Λ₁ = S_2s+γ,Γ. Hence, there exists a (2ν + δ + l − m) × (2ν + δ + l) matrix Z such that where Λ^* takes values in Table 1 as follows.

In Table 1 as follows $$.

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,

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, $$, 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

because ((1/2)Γu_ν−1 + v_ν−1)S_2ν((1/2)Γu_ν−1 + v_ν−1) ^t = Γ, and 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 $$ be row vectors of U and Z, respectively,

then W ∈ ℳ(m, 2s + 2, s, Γ, k;   2ν + l). Obviously, $$. 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.