Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 732808, 7 pages
http://dx.doi.org/10.1155/2013/732808
Research Article
Lattices Generated by Two Orbits of Subspaces under Finite
Singular Symplectic Groups
Xuemei Liu
College of Science, Civil Aviation University of China, Tianjin 300300, China
Correspondence should be addressed to Xuemei Liu; xm-liu771216@163.com
Received 29 September 2012; Revised 18 November 2012; Accepted 21 November 2012
Academic Editor: P. G. L. Leach
Copyright © 2013 Xuemei Liu. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In the paper titled “Lattices generated by two orbits of subspaces under finite classical group” by Wang and Guo. The subspaces in the
lattices are characterized and the geometricity is classified. In this paper, the result above is generalized to singular symplectic space.
This paper characterizes the subspaces in these lattices, classifies their geometricity, and computes their characteristic polynomials.
1. Introduction
In the following we recall some definitions and facts on
ordered sets and lattices (see [1]).
Let 𝑃 denote a finite set. A partial order on 𝑃 is a binary
relation ≤ on 𝑃 such that
(1) 𝑟(0) = 0,
(2) 𝑟(𝑎) = 𝑟(𝑏) + 1, whenever 𝑏 < ⋅𝑎.
Let 𝑃 be a finite poset with 0 and 1. The polynomial
(1) 𝑎 ≤ 𝑎 for any 𝑎 ∈ 𝑃.
(2) 𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑎 implies 𝑎 = 𝑏.
𝜒 (𝑃, 𝑥) = ∑ 𝜇 (0, 𝑎) 𝑥𝑟(1)−𝑟(𝑎) ,
(3) 𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑐 implies 𝑎 ≤ 𝑐.
𝑎∈𝑃
By a partial ordered set (or poset for short), we mean a
pair (𝑃, ≤), where 𝑃 is a finite set and ≤ is a partial order on
𝑃. As usual, we write 𝑎 < 𝑏 whenever 𝑎 ≤ 𝑏 and 𝑎 ≠ 𝑏. By
abusing notation, we will suppress reference to ≤, and just
write 𝑃 instead of (𝑃, ≤).
Let 𝑃 be a poset and let 𝑅 be a commutative ring with the
identical element. A binary function 𝜇(𝑎, 𝑏) on 𝑃 with values
in 𝑅 is said to be the Möbius function of 𝑃 if
1,
∑ 𝜇 (𝑎, 𝑐) = {
0,
𝑎≤𝑐≤𝑏
with 0, (resp., 1). Let 𝑃 be a finite poset with 0. By a rank
function on 𝑃, we mean a function 𝑟 from 𝑃 to the set of all
the nonnegative integers such that
if 𝑎 = 𝑏,
otherwise.
(1)
For any two elements 𝑎, 𝑏 ∈ 𝑃, we say 𝑎 covers 𝑏, denoted
by 𝑏 < ⋅𝑎, if 𝑏 < 𝑎 and there exists no 𝑐 ∈ 𝑃 such that 𝑏 < 𝑐 < 𝑎.
An element 𝑚 of 𝑃 is said to be minimal, (resp., maximal)
whenever there is no element 𝑎 ∈ 𝑃 such that 𝑎 < 𝑚, (resp.,
𝑎 > 𝑚). If 𝑃 has a unique minimal, (resp., maximal) element,
then we denote it by 0, (resp., 1) and say that 𝑃 is a poset
(2)
is called the characteristic polynomial of 𝑃, where 𝑟 is the rank
function of 𝑃.
A poset 𝑃 is said to be a lattice if both 𝑎 ∨ 𝑏 := sup{𝑎, 𝑏}
and 𝑎∧𝑏 := inf{𝑎, 𝑏} exist for any two elements 𝑎, 𝑏 ∈ 𝑃. Let 𝑃
be a finite lattice with 0. By an atom in 𝑃, we mean an element
in 𝑃 covering 0. We say 𝑃 is atomic lattice if any element in
𝑃 \ {0} is a union of atoms. A finite atomic lattice 𝑃 is said to
be a geometric lattice if 𝑃 admits a rank function 𝑟 satisfying
𝑟 (𝑎 ∧ 𝑏) + 𝑟 (𝑎 ∨ 𝑏) ≤ 𝑟 (𝑎) + 𝑟 (𝑏) ,
∀𝑎, 𝑏 ∈ 𝑃.
(3)
In this section we will introduce the concepts of subspaces
of type (𝑚, 𝑠, 𝑘) in singular symplectic spaces. Notation and
terminologies will be adopted from Wan’s book [2].
We always assume that
𝐾𝜈,𝑙
0 𝐼(𝜈)
= (−𝐼(𝜈) 0
0(𝑙)
).
(4)
2
Journal of Applied Mathematics
Let 𝐹𝑞 be a finite field with 𝑞 elements, where 𝑞 is a prime
power, and let 𝐸 denote the subspace of 𝐹𝑞(2𝜈+𝑙) generated
by 𝑒2𝜈+1 , 𝑒2𝜈+2 , . . . , 𝑒2𝜈+𝑙 , where 𝑒𝑖 is the row vector in 𝐹𝑞(2𝜈+𝑙)
whose 𝑖th coordinate is 1 and all other coordinates are 0.
The singular symplectic group of degree 2𝜈 + 𝑙 over 𝐹𝑞 ,
denoted by 𝑆𝑝2𝜈+𝑙,2𝜈 (𝐹𝑞 ), consists of all (2𝜈 + 𝑙) × (2𝜈 + 𝑙)
nonsingular matrices 𝑇 over 𝐹𝑞 satisfying 𝑇𝐾𝑙 𝑇𝑡 = 𝐾𝑙 . The
row vector space 𝐹𝑞(2𝜈+𝑙) together with the right multiplication
action of 𝑆𝑝2𝜈+𝑙,2𝜈 (𝐹𝑞 ) is called the (2𝜈 + 𝑙)-dimensional singular symplectic space over 𝐹𝑞 . An 𝑚-dimensional subspace
𝑃 in the (2𝜈+𝑙)-dimensional singular symplectic space is said
to be of type (𝑚, 𝑠, 𝑘), if 𝑃𝐾𝑙 𝑃𝑡 is of rank 2𝑠 and dim(𝑃 ∩
𝐸) = 𝑘. In particular, subspaces of type (𝑚, 0, 0) are called
𝑚-dimensional totally isotropic subspaces. Clearly, singular
symplectic group 𝑆𝑝2𝜈+𝑙,2𝜈 (𝐹𝑞 ) is transitive on the set of all
subspaces of the same type in 𝐹𝑞(2𝜈+𝑙) , see [2, Theorems 3.22].
The results on the lattices generated by one orbit of
subspaces under finite classical groups may be found in Gao
and You [3], Huo et al. [4–6], Huo and Wan [7], Orlik and
Solomon [8], Wang and Feng [9], Wang and Guo [10], and
Wang and Li [11].
For 1 ≤ 𝑚1 ≤ 𝑚2 ≤ 2𝜈 − 1, 0 ≤ 𝑘1 ≤ 𝑘2 ≤ 𝑙, let
𝐿 1 (𝑚1 , 𝑠1 , 𝑘1 ; 2𝜈 + 𝑙, 𝜈), (resp., 𝐿 2 (𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈)) denote
the set of all subspaces which are sums (resp. intersections) of
subspaces in 𝑀(𝑚1 , 𝑠1 , 𝑘1 ; 2𝜈 + 𝑙, 𝜈), (resp., 𝑀(𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 +
𝑙, 𝜈)) such that 𝑀(𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) ⊆ 𝐿 1 (𝑚1 , 𝑠1 , 𝑘1 ; 2𝜈 +
𝑙, 𝜈), (resp., 𝑀(𝑚1 , 𝑠1 , 𝑘1 ; 2𝜈 + 𝑙, 𝜈) ⊆ 𝐿 2 (𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈)).
Suppose 𝐿(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+𝑙, 𝜈) denotes the intersection of 𝐿 1 (𝑚1 , 𝑠1 , 𝑘1 ; 2𝜈+𝑙, 𝜈) and 𝐿 2 (𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+𝑙, 𝜈) containing 0 and 𝐹𝑞(2𝜈+𝑙) . By ordering 𝐿(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 +
𝑙, 𝜈) by ordinary or reverse inclusion, two families of atomic
lattices are obtained, denoted by 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+
𝑙, 𝜈) or 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈), respectively. Wang
and Guo [12] discussed the geometricity of the two lattices
when 𝑙 = 0. In this paper, we generalized their result to
general case, characterizes the subspaces in these lattices
in Section 2, classify their geometricity in Section 3, and
computes their characteristic polynomials in Section 4.
2. Characterization of Subspaces Contained in
𝐿(𝑚1 ,𝑠1 ,𝑘1 ;𝑚2 ,𝑠2 ,𝑘2 ;2𝜈+𝑙,𝜈)
Lemma 1. Let 2𝜈 + 𝑙 > 0, 0 ≤ 𝑘 ≤ 𝑙, 2𝑠 ≤ 𝑚 − 𝑘 ≤ 𝜈 + 𝑠 and
𝑚 − 𝑘 ≥ 1. For any subspace 𝑃 of type (𝑚 + 1, 𝑠, 𝑘 + 1), there are
two subspaces 𝑃1 and 𝑃2 of type (𝑚, 𝑠, 𝑘) such that 𝑃 = 𝑃1 + 𝑃2 .
Proof. Assume that
𝑃𝐾𝜈 𝑃𝑇 = (
Write 𝑃 = ( 𝑃011
𝑃1 = (
𝑃11 𝑃12
󸀠
0 𝑃22
0 𝑧1
𝑃12
𝑃22
2𝜈 𝑙
𝑘𝑠
0
2𝑠
) 𝑚 + 1 − (𝑘 + 1) − 2𝑠 .
0
𝑘+1
) 𝑚+1−(𝑘+1)
, where 𝑃22 = (
𝑘+1
) and 𝑃2 = (
𝑃11 𝑃12
󸀠
0 𝑃22
0 𝑧2
)
𝑚+1−(𝑘+1)
𝑘−1
1
type (𝑚, 𝑠, 𝑘), such that 𝑃 = 𝑃1 + 𝑃2 .
󸀠
𝑃22
𝑧1
𝑧2
)
𝑘−1
1 .
1
(5)
Then
are subspace of
Lemma 2. Let 2𝜈 + 𝑙 > 0, 0 ≤ 𝑘 ≤ 𝑙, 2𝑠 ≤ 𝑚 − 𝑘 ≤ 𝜈 + 𝑠 and
𝑚 ≠ 2𝜈 + 𝑙. For any subspace 𝑃 of type (𝑚 + 1, 𝑠 + 1, 𝑘), there are
two subspaces 𝑃1 and 𝑃2 of type (𝑚, 𝑠, 𝑘) such that 𝑃 = 𝑃1 + 𝑃2 .
Proof. Assume that 𝑃𝐾𝜈 𝑃𝑇
𝑘𝑠
2𝑠
𝑚+1−2𝑠−𝑘−2 ,
2
𝑘
0
𝑃0 𝑚+1−𝑘−2
𝑃11 =( 𝑥1 ) 1 , 𝑥1 and
𝑦1
1
(
0
𝑘1
)
write
=
𝑃 =(
𝑃0
𝑃0
𝑦1
𝑦1
𝑇
( 𝑥1 ) 𝐾𝜈 ( 𝑥1 )
𝑃11 𝑃12 𝑚+1−𝑘
󸀠 )
,
𝑘
0 𝑃22
2𝜈
=
where
𝑙
𝑦1 are the (𝑚 + 1 − (𝑘 + 1))-th and
(𝑚 − 𝑘 + 1)-th row vectors of 𝑃11 , respectively.
𝑃0󸀠
𝑃12 =( 𝑥1󸀠 )
𝑦1󸀠
𝑚+1−𝑘−2 󸀠
,𝑥
1
1
1
and 𝑦1󸀠 are the (𝑚 + 1 − (𝑘 + 1))-th
and (𝑚 − 𝑘 + 1)-th row vectors of 𝑃12 , respectively.
Then
𝑃0 𝑃0󸀠
𝑚+1−𝑘−2
1
𝑃1 = (𝑥1 𝑥1󸀠 )
𝑘,
0 𝑃22
𝑃0 𝑃0󸀠
𝑚+1−𝑘−2
1
𝑃2 = (𝑦1 𝑦1󸀠 )
𝑘
0 𝑃22
(6)
are subspace of type (𝑚, 𝑠, 𝑘), such that 𝑃 = 𝑃1 + 𝑃2 .
Theorem 3. Let 2𝜈 + 𝑙 > 0, assume that (𝑚, 𝑠, 𝑘), (𝑚1 , 𝑠1 , 𝑘1 )
satisfy 0 ≤ 𝑘 ≤ 𝑙, 2𝑠 ≤ 𝑚 − 𝑘 ≤ 𝜈 + 𝑠, 0 ≤ 𝑘1 ≤ 𝑙, 2𝑠 ≤ 𝑚1 − 𝑘1 ≤
𝜈 + 𝑠. For any subspace 𝑃 of type (𝑚, 𝑠, 𝑘), there are subspace
𝑃1 , 𝑃2 , . . . , 𝑃𝑖 of type (𝑚1 , 𝑠1 , 𝑘1 ) such that 𝑃 = 𝑃1 + 𝑃2 + ⋅ ⋅ ⋅ + 𝑃𝑖
if and only if
(i) 𝑠 = 𝑠1 = 𝜈,
(ii) 𝑠 < 𝜈,
0 ≤ 𝑘1 ≤ 𝑘 ≤ 𝑙,
𝑘1 = 𝑘 = 𝑙,
𝑚 − 𝑚1 ≥ 𝑠 − 𝑠1 ≥ 0,
(iii) 𝑠 < 𝜈,
0 ≤ 𝑘1 ≤ 𝑘 ≤ 𝑙,
(𝑚 − 𝑘) − (𝑚1 − 𝑘1 ) ≥ 𝑠 − 𝑠1 ≥ 0.
(7)
(8)
(9)
Proof. Suppose that (𝑚, 𝑠, 𝑘) and (𝑚1 , 𝑠1 , 𝑘1 ) satisfy condition
(7). Let 𝑘 − 𝑘1 = ℎ(ℎ ≥ 0), since 𝑠 = 𝑠1 = 𝜈, 𝑚 = 2𝜈 +
𝑘, 𝑚1 = 2𝜈 + 𝑘1 , 𝑚 − 𝑚1 = 𝑘 − 𝑘1 = ℎ, By Lemma 1 the
desired result follows. Suppose (𝑚, 𝑠, 𝑘) and (𝑚1 , 𝑠1 , 𝑘1 ) satisfy
condition (8). Let 𝑠 − 𝑠1 = 𝑡, 𝑚 − 𝑚1 = 𝑡 + 𝑡󸀠 (𝑡, 𝑡󸀠 ≥ 0).
By Lemma 1, any subspace of type (𝑚, 𝑠, 𝑙) is the sum some
subspaces of type (𝑚 − 𝑡, 𝑠, 𝑙). By Lemma 2, any subspace of
type (𝑚 − 𝑡, 𝑠, 𝑙) is the sum some subspaces of type (𝑚1 , 𝑠1 , 𝑙).
Hence the desired result follows.
Suppose (𝑚, 𝑠, 𝑘) and (𝑚1 , 𝑠1 , 𝑘1 ) satisfy condition (9).
Let 𝑠 − 𝑠1 = 𝑡, 𝑘 − 𝑘1 = ℎ, 𝑚 − 𝑚1 = 𝑡 + ℎ +
𝑡󸀠 (𝑡, 𝑡󸀠 , ℎ ≥ 0), By Lemma 1, we have any subspace of type
(𝑚, 𝑠, 𝑘) is the sum some subspace of type (𝑚 − 𝑡󸀠 , 𝑠, 𝑘). By
Lemma 2, we have any subspace of type (𝑚 − 𝑡󸀠 , 𝑠, 𝑘) is the
sum some subspace of type (𝑚 − 𝑡󸀠 − 𝑡, 𝑠1 , 𝑘). By Lemma 1,
we have any subspace of type (𝑚 − 𝑡󸀠 − 𝑡, 𝑠1 , 𝑘) is the sum
some subspace of type (𝑚1 , 𝑠1 , 𝑘1 ). Hence the desired result
follows.
Conversely, If 𝑠1 = 𝜈, then 𝑠 = 𝜈, 𝑘1 ≤ 𝑙. Let 𝑄 ∈
𝑀(𝑚, 𝜈, 𝑘; 2𝜈 + 𝑙, 𝜈), there exists 𝑃 ∈ 𝑀(𝑚1 , 𝜈, 𝑘1 ; 2𝜈 + 𝑙, 𝜈),
Journal of Applied Mathematics
such that 𝑃 ⊂ 𝑄, hence 𝑃 ∩ 𝐸 ⊂ 𝑄 ∩ 𝐸, 𝑘1 = dim(𝑃 ∩ 𝐸) ≤
dim(𝑄 ∩ 𝐸) = 𝑘. Therefore, 𝑠 = 𝑠1 = 𝜈, 0 ≤ 𝑘1 ≤ 𝑘 ≤ 𝑙,
condition (7) hold.
If 𝑠1 < 𝜈, let 𝑃 ∈ 𝑀(𝑚1 , 𝜈, 𝑘1 ; 2𝜈 + 𝑙, 𝜈), 𝑄 ∈ 𝑀(𝑚, 𝜈, 𝑘 :
2𝜈 + 𝑙, 𝜈), such that 𝑃 ⊂ 𝑄, then 𝑃 ∩ 𝐸 ⊂ 𝑄 ∩ 𝐸, and 𝑘1 =
dim(𝑃 ∩ 𝐸) ≤ dim(𝑄 ∩ 𝐸) = 𝑘. If 𝑘1 = 𝑙, then 𝑘 = 𝑙 and 𝑚1 ≥ 𝑙,
𝑄12 𝑚−1
) 𝑙 where rank 𝑄11 = 𝑚 − 𝑙, rank
Assume 𝑄 = ( 𝑄011 𝑄
22
2𝜈 𝑙
𝑄22 = 𝑙 and 𝑄11 𝐾𝑄𝑡 = 𝑀(𝑚 − 𝑙, 𝑠). Since 𝑃 ∩ 𝐸 ⊂ 𝑄 ∩ 𝐸, there
exists a 𝑙 × 𝑙 matrix 𝑃22 with rank 𝑃22 = 𝑙, such that 𝑃22 𝑄22 is a
matrix representation of subspace 𝑃 ∩ 𝐸. Thus we can assume
𝑃 = ( 𝑃011 𝑃𝑃1222 ) 𝑚1𝑙−1 , where 𝑃11 is a matrix with rank 𝑚1 − 𝑙.
2𝜈 𝑙
Because 𝑃 is a subspaces of type (𝑚1 , 𝑠1 , 𝑙), we can assume
𝑡
= 𝑀(𝑚1 − 𝑙, 𝑠). Then 𝑄11 can be considered
𝑃11 𝑀(𝑚 − 𝑙, 𝑠)𝑃11
as a subspace of type (𝑚1 − 𝑙, 𝑠1 ) in singular symplectic space
𝐹𝑞(2𝑠+(𝑚−2𝑠−𝑙)) . Hence 𝑚1 − 𝑙 − 𝑠 − 𝑠1 ≤ 𝑚 − 2𝑠 − 𝑙, condition (8)
hold. Similarly we also can prove condition (9) hold.
Theorem 4. Let 2𝜈 + 𝑙 > 0, assume that (𝑚, 𝑠, 𝑘) and
(𝑚2 , 𝑠2 , 𝑘2 ) satisfy 0 ≤ 𝑘 ≤ 𝑙, 2𝑠 ≤ 𝑚 − 𝑘 ≤ 𝜈 + 𝑠, 0 ≤ 𝑘2 ≤ 𝑙,
2𝑠2 ≤ 𝑚2 − 𝑘2 ≤ 𝜈 + 𝑠. For any subspace 𝑃 of type (𝑚, 𝑠, 𝑘),
there are subspace 𝑃1 , 𝑃2 , . . . , 𝑃𝑗 of type (𝑚2 , 𝑠2 , 𝑘2 ) such that
𝑃 = 𝑃1 ∩ 𝑃2 ∩ ⋅ ⋅ ⋅ ∩ 𝑃𝑗 if and only if
(i) 𝑠 = 𝑠2 = 𝜈, 0 ≤ 𝑘2 ≤ 𝑘 ≤ 𝑙,
(ii) 𝑠 < 𝜈, 𝑘2 = 𝑘 = 𝑙 and 𝑚2 − 𝑚 ≥ 𝑠2 − 𝑠 ≥ 0,
(iii) 𝑠 < 𝜈, 𝑘 ≤ 𝑘2 ≤ 𝑙 and (𝑚2 − 𝑘2 ) − (𝑚 − 𝑘) ≥ 𝑠2 − 𝑠 ≥ 0.
Proof. By [3], it is directed.
Theorem 5. For 1 ≤ 𝑚1 ≤ 𝑚2 ≤ 2𝜈 + 𝑙, 𝐿(𝑚1 , 𝑠1 , 𝑘1 ;
𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) consist of {0}, 𝐹𝑞(2𝜈+𝑙) and all subspaces of
type (𝑚, 𝑠, 𝑘) in 𝐹𝑞(2𝜈+𝑙) such that
(i) 𝑠 = 𝑠1 = 𝑠2 = 𝜈, 0 ≤ 𝑘1 ≤ 𝑘2 ≤ 𝑘 ≤ 𝑙,
(ii) 𝑠 < 𝜈, 𝑘1 = 𝑘 = 𝑘2 = 𝑙 and 𝑚 − 𝑚1 ≥ 𝑠 − 𝑠1 ≥ 0,
𝑚2 − 𝑚 ≥ 𝑠2 − 𝑠 ≥ 0,
(iii) 𝑠 < 𝜈, 𝑘1 ≤ 𝑘 ≤ 𝑘2 ≤ 𝑙 and (𝑚2 − 𝑘2 ) − (𝑚 − 𝑘) ≥
𝑠2 − 𝑠 ≥ 0, (𝑚 − 𝑘) − (𝑚1 − 𝑘1 ) ≥ 𝑠 − 𝑠1 ≥ 0.
Proof. By Theorems 3 and 4, it is directed.
3. The Geometricity of Lattices
𝐿 𝑂(𝑚1 ,𝑠1 ,𝑘1 ;𝑚2 ,𝑠2 ,𝑘2 ;2𝜈 + 𝑙,𝜈) and
𝐿 𝑅(𝑚1 ,𝑠1 ,𝑘1 ;𝑚2 ,𝑠2 ,𝑘2 ;2𝜈 + 𝑙,𝜈)
Lemma 6 (see [3]). If 0 < 𝑘 < 𝑙, then
(i) 𝐿 𝑅 (2𝜈+𝑘, 𝜈, 𝑘; 2𝜈+𝑙, 𝜈) ≃ 𝐿 𝑅 (𝑘, 𝑙), 𝐿 𝑂(2𝜈+𝑘, 𝜈, 𝑘; 2𝜈+
𝑙, 𝜈) ≃ 𝐿 𝑂(𝑘, 𝑙),
3
(ii) 𝐿 𝑅 (𝑚, 𝑠, 0; 2𝜈 + 𝑙, 𝜈) ≃ 𝐿 𝑅 (𝑚, 𝑠; 2𝜈), 𝐿 𝑂(𝑚, 𝑠, 0; 2𝜈 +
𝑙, 𝜈) ≃ 𝐿 𝑂(𝑚, 𝑠; 2𝜈).
Theorem 8. Let 2𝜈 + 𝑙 > 0. Assume that (𝑚1 , 𝑠1 , 𝑘1 ),
(𝑚2 , 𝑠2 , 𝑘2 ) satisfies 0 ≤ 𝑘1 ≤ 𝑙, 2𝑠1 ≤ 𝑚1 − 𝑘1 ≤ 𝜈 + 𝑠1 ,
0 ≤ 𝑘2 ≤ 𝑙, 2𝑠2 ≤ 𝑚2 − 𝑘2 ≤ 𝜈 + 𝑠2 and 1 ≤ 𝑚1 ≤ 𝑚2 < 2𝜈 + 𝑙.
Then
(i) 𝐿 𝑂(𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite
geometric lattice if and only if 𝑘 = 0, 𝑙,
(ii) 𝐿 𝑂(𝑘, 0, 𝑘; 2𝜈 + 𝑘, 𝜈, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite geometric
lattice if and only if 𝑘 = 1, 𝑙 − 1,
(iii) 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) is not a geometric
lattice when 2 ≤ 𝑚1 − 𝑘1 ≤ 𝑚2 − 𝑘2 ≤ 2𝜈 − 2.
Proof. For any 𝑋 ∈ 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈), define
0,
if 𝑋 = 0,
{
{
if 𝑋 = 𝐹𝑞(2𝜈+𝑙) ,
𝑟𝑂 (𝑋) = {𝑚2 − 𝑚1 + 2,
{
{dim 𝑋 − 𝑚1 + 1, otherwise.
(10)
Then 𝑟𝑂 is the rank function of 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 +
𝑙, 𝜈).
(i) For lattice 𝐿 𝑂(𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈).
If 𝑘 = 0, by Lemma 7, 𝐿 𝑂(1, 0, 0; 2𝜈 − 𝑙, 𝜈 − 1, 0; 2𝜈 + 𝑙, 𝜈) ≃
𝐿 𝑂(1, 0, 2𝜈 − 1, 𝜈 − 1; 2𝜈), by [12], 𝐿 𝑂(𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 −
1, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite geometric lattice.
If 𝑘 = 𝑙, by Lemma 7, 𝐿 𝑂(𝑙 + 1, 0, 𝑙; 2𝜈 − 𝑙 + 𝑙, 𝜈 − 1, 𝑙; 2𝜈 +
𝑙, 𝜈) ≃ 𝐿 𝑂(1, 0, 2𝜈 − 1, 𝜈 − 1; 2𝜈), by [12], 𝐿 𝑂(𝑘 + 1, 0, 𝑘; 2𝜈 −
1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite geometric lattice.
If 0 < 𝑘 < 𝑙. Let 𝑈 = ⟨𝑒1 , 𝑒2𝜈+1 , . . . , 𝑒2𝜈+𝑘 ⟩ and 𝑊 =
⟨𝑒𝜈+1 , 𝑒2𝜈+2 , . . . , 𝑒2𝜈+𝑘+1 ⟩, then 𝑈 and 𝑊 both are of type (𝑘 +
1, 0, 𝑘), ⟨𝑈, 𝑊⟩ is of type (𝑘 + 3, 1, 𝑘 + 1), ⟨𝑈, 𝑊⟩ ∈ 𝐿 𝑂(𝑘 +
1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈), 𝑈 ∩ 𝑊 is of type
(𝑘 − 1, 0, 𝑘 − 1), hence 𝑟𝑂(𝑈 ∧ 𝑊) = 0, 𝑟𝑂(𝑈 ∨ 𝑊) = 3,
𝑟𝑂(𝑈) = 𝑟𝑂(𝑊) = 1. We have
𝑟𝑂 (𝑈 ∧ 𝑊) + 𝑟𝑂 (𝑈 ∨ 𝑊) = 3 > 𝑟𝑂 (𝑈) + 𝑟𝑂 (𝑊) = 2. (11)
That is 𝐿 𝑂(𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈) is not a
geometric lattice when 0 < 𝑘 < 𝑙.
(ii) For lattice 𝐿 𝑂(𝑘, 0, 𝑘; 2𝜈 + 𝑘, 𝜈, 𝑘; 2𝜈 + 𝑙, 𝜈), by
Lemma 6 𝐿 𝑂(𝑘, 0, 𝑘; 2𝜈 + 𝑘, 𝜈, 𝑘; 2𝜈 + 𝑙, 𝜈) ≃ 𝐿 𝑂(𝑘, 𝑙).
If 𝑘 = 1, 𝑙−1, 𝐿 𝑂(1, 𝑙) and 𝐿 𝑂(𝑙−1, 𝑙) is a geometric lattice.
If 2 ≤ 𝑘 ≤ 𝑙 − 2, let 𝑣1 , 𝑣2 , . . . , 𝑣𝑙 be a basis of 𝐹𝑞(𝑙) .
Since 2 ≤ 𝑘 ≤ 𝑙 − 2, we can take 𝑈 = ⟨𝑣1 , 𝑣2 , . . . , 𝑣𝑘 ⟩,
𝑊 = ⟨𝑣3 , 𝑣4 , . . . , 𝑣𝑘+2 ⟩ ∈ 𝐿 𝑂(𝑘, 𝑙). Hence 𝑈 ∧ 𝑊 = {0},
𝑈 ∨ 𝑊 = ⟨𝑈, 𝑊⟩, 𝑟𝑂(𝑈 ∧ 𝑊) = 0, 𝑟𝑂(𝑈 ∨ 𝑊) = 3,
𝑟𝑂(𝑈) = 𝑟𝑂(𝑊) = 1. We have
(ii) 𝐿 𝑅 (𝑘, 0, 𝑘; 2𝜈 + 𝑙, 𝜈) ≃ 𝐿 𝑅 (𝑘, 𝑙), 𝐿 𝑂(𝑘, 0, 𝑘; 2𝜈 + 𝑙, 𝜈) ≃
𝐿 𝑂(𝑘, 𝑙).
𝑟𝑂 (𝑈 ∧ 𝑊) + 𝑟𝑂 (𝑈 ∨ 𝑊) = 3 > 𝑟𝑂 (𝑈) + 𝑟𝑂 (𝑊) = 2. (12)
Lemma 7 (see [3]). If 0 ≤ 𝑠 < 𝜈 and 2𝑠 ≤ 𝑚 − 𝑙 ≤ 𝜈 + 𝑠, then
That is 𝐿 𝑂(𝑘, 0, 𝑘; 2𝜈+𝑘, 𝜈, 𝑘; 2𝜈+𝑙, 𝜈) is not a geometric lattice
when 2 ≤ 𝑘 ≤ 𝑙 − 2.
(iii) For lattice 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+𝑙, 𝜈), when 2 ≤
𝑚1 − 𝑘1 ≤ 𝑚2 − 𝑘2 ≤ 2𝜈 − 2.
(i) 𝐿 𝑅 (𝑚, 𝑠, 𝑙; 2𝜈 + 𝑙, 𝜈) ≃ 𝐿 𝑅 (𝑚 − 𝑙, 𝑠; 2𝜈), 𝐿 𝑂(𝑚, 𝑠, 𝑙; 2𝜈 +
𝑙, 𝜈) ≃ 𝐿 𝑂(𝑚 − 𝑙, 𝑠; 2𝜈),
4
Journal of Applied Mathematics
Case (a). 𝑘1 ≤ 𝑘2 ≤ 𝑙.
(a1 ) 𝑠2 > 0, Let
𝐼(𝑠2 −1)
0
(
𝑈=
0
0
𝑠2 −1
𝑊=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
𝐼(𝑠2 )
0
)
0
,
0
0
0
0
0
0 𝐼(𝑚2 −𝑘2 −2𝑠2 )
0
0
0
0
0
0
𝐼(𝑘2 ) 0
1 𝑚2 −𝑘2 −2𝑠2 𝜈+𝑠2 −𝑚2 +𝑘2 𝑠2 𝑚2 −𝑘2 −2𝑠2 𝜈+𝑠2 −𝑚2 +𝑘2 𝑘2 𝑙−𝑘2
(13)
0
1
0
0
0
0
0
0
0 0
)
0 0
0
0
0
0
0
0
𝐼(𝑘2 ) 0 .
𝑠2 −1 1 𝑚2 −𝑘2 −2𝑠2 1 𝜈+𝑠2 −𝑚2 +𝑘2 𝑠2 𝑚2 −𝑘2 −2𝑠2 𝜈+𝑠2 −𝑚2 +𝑘2 𝑘2 𝑙−𝑘2
(
Then 𝑈 is a of type (𝑚2 − 1, 𝑠2 − 1, 𝑘2 ), 𝑊 is a of type
(𝑘2 + 1, 0, 𝑘2 ), ⟨𝑈, 𝑊⟩ is a of type (𝑚2 , 𝑠2 − 1, 𝑘2 ), 𝑟𝑂(𝑈 ∨ 𝑊) =
𝑟𝑂(𝐹𝑞(2𝜈+𝑙) ) = 𝑚2 − 𝑚1 + 2, 𝑟𝑂(𝑈 ∧ 𝑊) = 𝑘2 − 𝑚1 + 1,
𝑟𝑂(𝑈) = 𝑚2 − 1 − 𝑚1 + 1 = 𝑚2 − 𝑚1 , 𝑟𝑂(𝑊) = 𝑘2 + 1 − 𝑚1 + 1 =
𝑘2 − 𝑚1 + 2, 𝑟𝑂(𝑈 ∨ 𝑊) + 𝑟𝑂(𝑈 ∧ 𝑊) = 𝑚2 − 2𝑚1 + 𝑘2 + 3,
𝑟𝑂(𝑈) + 𝑟𝑂(𝑊) = 𝑚2 − 2𝑚1 + 𝑘2 + 2. We have
𝐼(𝑠2)
0
(
𝑈=
0
0
𝑠2
𝑟𝑂 (𝑈 ∨ 𝑊) + 𝑟𝑂 (𝑈 ∧ 𝑊) > 𝑟𝑂 (𝑈) + 𝑟𝑂 (𝑊) .
Hence, 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) is not a geometric
lattice when (a1 ).
(a2 ) 𝑠2 > 0, 𝑚2 − 𝑘2 = 𝜈 + 𝑠2 or 𝑠2 = 0.
When 𝑠2 = 0, we have 𝑚2 −𝑘2 −2𝑠2 ≥ 1, when 𝑠2 > 0, 𝑚2 −𝑘2 =
𝜈 + 𝑠2 , we have 𝜈 − 2 ≥ 𝑠2 , 𝑚2 − 𝑘2 − 2𝑠2 − 1 = 𝜈 − 𝑠2 − 1 ≥ 1.
Let
0
0
0
0
0
0
0
0
0
𝐼(𝑠2)
0
0
0
0
𝐼(𝑚2−𝑘2−2𝑠2)
0
0
0
0
0
𝑚2 −𝑘2 −2𝑠2 −1 1 𝜈+𝑠2 −𝑚2 +𝑘2 𝑠2 𝑚2 −𝑘2 −2𝑠2 −1
𝑊=
(
(14)
0
0
0 0
0
0
0 0
)
,
0
0
0 0
(𝑘2)
0
0
0
𝐼
1 𝜈+𝑠2 −𝑚2 +𝑘2 𝑘2 𝑙−𝑘2
(15)
0
1 0
0
0
0
0 0
)
(𝑘2 )
0 .
0 0
0
0 0
0
𝐼
𝑠2 1 𝜈 − 𝑠2 − 1 𝑠2 1 𝜈 − 𝑠2 − 1 𝑘2 𝑙 − 𝑘2
Then 𝑈 is a of type (𝑚2 − 1, 𝑠2 , 𝑘2 ), 𝑊 is a of type (𝑘2 +
1, 0, 𝑘2 ), ⟨𝑈, 𝑊⟩ is a of type (𝑚2 , 𝑠2 + 1, 𝑘2 ), clearly, we have
𝑟𝑂 (𝑈 ∨ 𝑊) + 𝑟𝑂 (𝑈 ∧ 𝑊) > 𝑟𝑂 (𝑈) + 𝑟𝑂 (𝑊) .
(16)
Hence, 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) is not a geometric
lattice when (a2 ).
Case (b). 𝑘1 = 𝑘2 = 𝑙.
In this case 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑙; 𝑚2 , 𝑠2 , 𝑙; 2𝜈 + 𝑙, 𝜈) ≃ 𝐿 𝑂(𝑚1 −
𝑙, 𝑠1 ; 𝑚2 − 𝑙, 𝑠2 ; 2𝜈), 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑙; 𝑚2 , 𝑠2 , 𝑙; 2𝜈 + 𝑙, 𝜈) is not a
geometric lattice.
Hence 𝐿 𝑂(𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+𝑙, 𝜈) is not a geometric
lattice when 2 ≤ 𝑚1 − 𝑘1 ≤ 𝑚2 − 𝑘2 ≤ 2𝜈 − 2.
Theorem 9. Let 2𝜈 + 𝑙 > 0. Assume that (𝑚1 , 𝑠1 , 𝑘1 ),
(𝑚2 , 𝑠2 , 𝑘2 ) satisfies 0 ≤ 𝑘1 ≤ 𝑙, 2𝑠1 ≤ 𝑚1 − 𝑘1 ≤ 𝜈 + 𝑠1 ,
0 ≤ 𝑘2 ≤ 𝑙, 2𝑠2 ≤ 𝑚2 − 𝑘2 ≤ 𝜈 + 𝑠2 and 1 ≤ 𝑚1 ≤ 𝑚2 < 2𝜈 + 𝑙.
Then
(i) 𝐿 𝑅 (𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite
geometric lattice if and only if 𝑘 = 0, 𝑙,
(ii) 𝐿 𝑅 (𝑘, 0, 𝑘; 2𝜈 + 𝑘, 𝜈, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite geometric
lattice if and only if 𝑘 = 1, 𝑙 − 1,
(iii) 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) is not a geometric
lattice when 2 ≤ 𝑚1 − 𝑘1 ≤ 𝑚2 − 𝑘2 ≤ 2𝜈 − 2.
Proof. (i) For 𝑋 ∈ 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈), define
𝑟𝑅(𝑋)
0,
{
{
= {𝑚2 − 𝑚1 + 2,
{
{𝑚2 + 1 − dim 𝑋,
if 𝑋 = 𝐹𝑞((2𝜈+𝑙)) ,
if 𝑋 = 0,
otherwise.
(17)
For lattice 𝐿 𝑅 (𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈).
If 𝑘 = 0, by Lemma 7, 𝐿 𝑅 (1, 0, 0; 2𝜈 − 𝑙, 𝜈 − 1, 0; 2𝜈 + 𝑙, 𝜈) ≃
𝐿 𝑅 (1, 0, 2𝜈 − 1, 𝜈 − 1; 2𝜈), by [12], 𝐿 𝑅 (𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 −
1, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite geometric lattice.
If 𝑘 = 𝑙, by Lemma 7, 𝐿 𝑅 (𝑙 + 1, 0, 𝑙; 2𝜈 − 𝑙 + 𝑙, 𝜈 − 1, 𝑙; 2𝜈 +
𝑙, 𝜈) ≃ 𝐿 𝑅 (1, 0, 2𝜈 − 1, 𝜈 − 1; 2𝜈), by [12], 𝐿 𝑅 (𝑘 + 1, 0, 𝑘; 2𝜈 −
1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈) is a finite geometric lattice.
If 0 < 𝑘 < 𝑙, let 𝑈 = ⟨𝑒1 , . . . , 𝑒𝜈−1 , 𝑒𝜈+1 , . . . ,
𝑒2𝜈 , 𝑒2𝜈+1 , . . . , 𝑒2𝜈+𝑘 ⟩, 𝑊 = ⟨𝑒2 , . . . , 𝑒𝜈 , 𝑒𝜈+2 , . . . , 𝑒2𝜈+1 , 𝑒2𝜈+2 ,
. . . , 𝑒2𝜈+𝑘 ⟩. Then 𝑈, 𝑊 both are of type (2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘),
𝑈∧𝑊 = ⟨𝑈, 𝑊⟩ is of type (2𝜈+1+𝑘, 𝜈, 𝑘+1), 𝑈∨𝑊 = 𝑈∩𝑊
is of type (2𝜈 − 3 + 𝑘, 𝜈, 𝑘). We have
𝑟 (𝑈 ∧ 𝑊) + 𝑟 (𝑈 ∨ 𝑊) = 3 > 𝑟 (𝑈) + 𝑟 (𝑊) = 2.
(18)
Journal of Applied Mathematics
5
Hence, 𝐿 𝑅 (𝑘 + 1, 0, 𝑘; 2𝜈 − 1 + 𝑘, 𝜈 − 1, 𝑘; 2𝜈 + 𝑙, 𝜈) is not
a geometric lattice when 0 < 𝑘 < 𝑙.
𝐼(𝑠1)
0
( 0
𝑈=
0
0
𝑠1
0
0
0
0
0
0
0
𝐼(𝑠1 )
1
0
0
0
0
0
0 𝐼(𝑚1 −𝑘1 −2𝑠1 )
0
0
0
0
1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 −1 𝑠1
0
0
𝐼(𝑠1)
0
0
0
0
0
0
0
0 𝐼(𝑚1 −𝑘1 −2𝑠1 )
0
0
0
𝑠1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 −1
0
0
(0
𝑊=
0
0
1
(ii) if 𝑘 = 0, 𝑙 − 1, by Lemma 7 𝐿 𝑅 (𝑘, 0, 𝑘; 2𝜈 + 𝑘, 𝜈, 𝑘; 2𝜈 +
𝑙, 𝜈) ≃ 𝐿 𝑅 (𝑘, 𝑙) when 𝑘 = 1, 𝑙−1, 𝐿 𝑅 (𝑘, 0, 𝑘; 2𝜈+𝑘, 𝜈, 𝑘; 2𝜈+𝑙, 𝜈)
is a geometric lattice.
If 2 ≤ 𝑘 ≤ 𝑙 − 2, by [7], 𝐿 𝑅 (𝑘, 𝑙) is not a geometric lattice.
(iii) Case (a). if 𝑚1 −𝑘1 < 𝜈+𝑠1 , then 𝜈+𝑠1 −𝑚1 +𝑘1 −1 ≥ 0.
Let
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
𝐼(𝑘1)
1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 𝑘1
0 0
0
0
0
0
0
0
0 𝐼(𝑠1 )
1 0
0
0
0
0 0
0
0
0
0 0
0
0
𝐼(𝑘1)
1 𝑠1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 −1 𝑘1
Then 𝑈, 𝑊 are of type (𝑚1 + 1, 𝑠1 , 𝑘1 ), 𝑈 ∩ 𝑊 is of
type (𝑚1 , 𝑠1 − 1, 𝑘1 ), ⟨𝑈, 𝑊⟩ is of type (𝑚1 + 1, 𝑠1 , 𝑘1 ),
⟨𝑈, 𝑊⟩ ∈ 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈), 𝑈 ∩ 𝑊 ∉
𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈).
So 𝑟𝑅 (𝑈 ∨ 𝑊) = 𝑚2 − 𝑚1 + 2, 𝑟𝑅 (𝑈 ∧ 𝑊) = 𝑚2 − 𝑚1 − 1,
𝑟𝑅 (𝑈) = 𝑚2 + 1 − (𝑚1 + 1) = 𝑚2 − 𝑚1 = 𝑟𝑅 (𝑊). We have
𝑟𝑅 (𝑈 ∧ 𝑊) + 𝑟𝑅 (𝑈 ∨ 𝑊)
= 2𝑚2 − 2𝑚1 + 1
0
0
0)
,
0
0
𝑙−𝑘1
0
0
0)
.
0
0
𝑙−𝑘1
> 𝑟𝑅 (𝑈) + 𝑟𝑅 (𝑊)
= 2𝑚2 − 2𝑚1 .
(20)
Hence 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) is not a geometric
lattice when (a).
Case (b). 𝑚1 − 𝑘1 = 𝜈 + 𝑠1 , from 𝑚1 − 𝑘1 = 𝜈 + 𝑠1 , and
2 ≤ 𝑚1 − 𝑘1 ≤ 2𝜈 − 2, we have 𝜈 − 2 ≥ 𝑠1 , 𝑚1 − 𝑘1 − 2𝑠1 − 2 =
𝜈 − 𝑠1 − 2 ≥ 0.
Let
𝐼(𝑠1)
0
( 0
( 0
𝑈= (
0
0
( 0
𝑠1
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
𝐼(𝑠1)
0
0
0
0
1
0
0
0
0
0
0 𝐼(𝑚1−𝑘1−2𝑠1)
0
0
0
0
0
0
0
0
1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 𝑠1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
𝐼(𝑘1)
1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 𝑘1
0
0
0)
0)
),
0
0
0)
𝑙−𝑘1
𝐼(𝑠1)
0
( 0
( 0
𝑊= (
0
0
( 0
𝑠1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
𝐼(𝑠1)
0
0
0
0
1
0
0
0
0
0
0 𝐼(𝑚1−𝑘1−2𝑠1)
0
0
0
0
0
0
0
0
1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 𝑠1
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
𝐼(𝑘1)
1 𝑚1 −𝑘1 −2𝑠1 𝜈+𝑠1 −𝑚1 +𝑘1 𝑘1
0
0
0)
0)
).
0
0
0)
𝑙−𝑘1
Then 𝑈, 𝑊 are of type (𝑚1 + 1, 𝑠1 + 1, 𝑘1 ), 𝑈 ∩ 𝑊 is of
type (𝑚1 , 𝑠1 + 1, 𝑘1 ), and ⟨𝑈, 𝑊⟩ is of type (𝑚1 + 2, 𝑠1 + 2,
𝑘1 ).
(19)
(21)
So 𝑟𝑅 (𝑈 ∨ 𝑊) = 𝑚2 − 𝑚1 , 𝑟𝑅 (𝑈 ∧ 𝑊) = 𝑚2 + 1 − (𝑚1 + 2) =
𝑚2 − 𝑚1 − 1, 𝑟𝑅 (𝑈 ∧ 𝑊) + 𝑟𝑅 (𝑈 ∨ 𝑊) = 2𝑚2 − 2𝑚1 + 1 >
𝑟𝑅 (𝑈) + 𝑟𝑅 (𝑊) = 2𝑚2 − 2𝑚1 .
6
Journal of Applied Mathematics
Hence 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+𝑙, 𝜈) is not a geometric
lattice when (b).
Hence 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈+𝑙, 𝜈) is not a geometric
lattice when 2 ≤ 𝑚1 − 𝑘1 ≤ 𝑚2 − 𝑘2 ≤ 2𝜈 − 2.
4. Characteristic Polynomial of Lattice
𝐿 𝑅(𝑚1 ,𝑠1 ,𝑘1 ;𝑚2 ,𝑠2 ,𝑘2 ;2𝜈+𝑙,𝜈)
= 𝑡𝑚2 −𝑚1 +2
𝑘2
Theorem 10. Let 2𝜈 + 𝑙 > 0, Assume that (𝑚, 𝑠, 𝑘) satisfies
0 ≤ 𝑘 ≤ 𝑙, 2𝑠 ≤ 𝑚 − 𝑘 ≤ 𝜈 + 𝑠 and 0 < 𝑚 < 2𝜈 + 𝑙. Then
𝜒 (𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) , 𝑡)
𝑠2
𝑚1 −(𝑘−𝑘1 )+(𝑠−𝑠1 )
∑
𝑘=𝑘1 𝑠=𝑠1 𝑚=𝑚2 −(𝑘2 −𝑘)+(𝑠2 −𝑠)+1
𝑁 (𝑚, 𝑠, 𝑘; 2𝜈 + 𝑙, 𝜈) 𝑔𝑚 (𝑡)
𝑘1 𝑠1 −1 𝜈+𝑠+𝑘
+∑ ∑
∑ 𝑁 (𝑚, 𝑠, 𝑘; 2𝜈 + 𝑙, 𝜈) 𝑔𝑚 (𝑡)
𝑘=0 𝑠=0 𝑚=2𝑠+𝑘
𝑙
𝜈
+ ∑ ∑
𝜈+𝑠+𝑘
∑ 𝑁 (𝑚, 𝑠, 𝑘; 2𝜈 + 𝑙, 𝜈) 𝑔𝑚 (𝑡) .
𝑘=𝑘2 𝑠=𝑠2 +1 𝑚=2𝑠+𝑘
𝑚2 −𝑚1 +2
=𝑡
(28)
𝑠2
− ∑ ∑
𝑚1 −(𝑘−𝑘1 )+(𝑠−𝑠1 )
𝑁 (𝑚, 𝑠, 𝑘; 2𝜈 + 𝑙, 𝜈) 𝑔𝑚 (𝑡)
∑
𝑘=𝑘1 𝑠=𝑠1 𝑚=𝑚2 −(𝑘2 −𝑘)+(𝑠2 −𝑠)+1
Acknowledgments
𝑘1 𝑠1 −1 𝜈+𝑠+𝑘
+∑ ∑
∑ 𝑁 (𝑚, 𝑠, 𝑘; 2𝜈 + 𝑙, 𝜈) 𝑔𝑚 (𝑡)
This work is supported by the National Natural Science Foundation of China under Grant no. 61179026 and Supported by
the Fundamental Research Funds for the Central Universities
(ZXH2012 K003).
𝑘=0 𝑠=0 𝑚=2𝑠+𝑘
𝑙
𝜒 (𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) , 𝑡)
− ∑ ∑
In this section we compute the characteristic polynomial of
the lattice 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈).
𝑘
by Theorem 5 we have
𝜈
+ ∑ ∑
𝜈+𝑠+𝑘
∑ 𝑁 (𝑚, 𝑠, 𝑘; 2𝜈 + 𝑙, 𝜈) 𝑔𝑚 (𝑡) ,
𝑘=𝑘2 𝑠=𝑠2 +1 𝑚=2𝑠+𝑘
(22)
𝑖
where 𝑔𝑚 (𝑡) = ∏𝑚−1
𝑖=0 (𝑡 − 𝑞 ).
Proof. Define
0,
if 𝑋 = 𝐹𝑞(2𝜈+𝑙) ,
{
{
{
{
𝑟𝑅 (𝑋) = {𝑚2 − 𝑚1 + 2,
if 𝑋 = {0} ,
{
{
{
{𝑚2 + 1 − dim (𝑋) , otherwise.
(23)
Then 𝑟𝑅 is the rank function on 𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 ,
𝑘2 ; 2𝜈 + 𝑙, 𝜈). Let 𝑉 = 𝐹𝑞(2𝜈+𝑙) , 𝐿 𝑂 = 𝐿 𝑅 (2𝜈 + 𝑙, 𝜈), 𝐿 =
𝐿 𝑅 (𝑚1 , 𝑠1 , 𝑘1 ; 𝑚2 , 𝑠2 , 𝑘2 ; 2𝜈 + 𝑙, 𝜈) we have
2𝜈+𝑙−1
𝜒 (𝐿 𝑂, 𝑡) = ∏ (𝑡 − 𝑞𝑖 ) .
(24)
𝑖=0
For any 𝑃 ∈ 𝐿, define
𝐿𝑃 = {𝑄 ∈ 𝐿 | 𝑄 ⊂ 𝑃} = {𝑄 ∈ 𝐿 | 𝑄 ≥ 𝑃} ,
𝐿𝑃𝑂 = {𝑄 ∈ 𝐿 𝑂 | 𝑄 ⊂ 𝑃} = {𝑄 ∈ 𝐿 𝑂 | 𝑄 ≥ 𝑃} .
(25)
Clearly, 𝐿𝑉 = 𝐿, 𝐿𝑃 = 𝐿𝑃𝑂, when 𝑃 ≠ {0}, 𝑃 ≠ 𝑉
𝜒 (𝐿𝑉, 𝑡) = 𝜒 (𝐿, 𝑡) = ∑ 𝜇 (0, 𝑃) 𝑡𝑟(1)−𝑟(𝑃)
𝑃∈𝐿
= ∑ 𝜇 (0, 𝑃) 𝑡𝑟(0)−𝑟(𝑃) .
(26)
𝑃∈𝐿
By inversion to Möbius we have
𝜒 (𝐿, 𝑡) = 𝜒 (𝐿𝑉, 𝑡) = 𝑡𝑚2 −𝑚1 +2 − ∑ 𝜒 (𝐿𝑃 , 𝑡) ,
𝑃∈𝐿\{𝑉}
(27)
References
[1] M. Aigner, Combinatorial Theory, vol. 234, Springer, Berlin,
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[2] Z. X. Wan, Geometry of Classical Groups Over Finite Fields,
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