# Lattices Generated by Two Orbits of Subspaces under Finite Singular Symplectic Groups

Authors: Xuemei Liu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

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.


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## Body

## 1. Introduction

In the following we recall some definitions and facts on ordered sets and lattices (see [1]).LetP denote a finite set. A partial order on P is a binary relation ≤ on P such that(1)
a≤a for any a∈P.(2)
a≤b and b≤a implies a=b.(3)
a≤b and b≤c implies a≤c.By apartial ordered set (orposet 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,≤).LetP 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)∑a≤c≤bμ(a,c)={1,ifa=b,0,otherwise.For any two elementsa,b∈P, we say acoversb, 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 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(1)
r(0)=0,(2)
r(a)=r(b)+1, whenever b<·a.LetP be a finite poset with 0 and 1. The polynomial
(2)χ(P,x)=∑a∈Pμ(0,a)xr(1)-r(a),
is called the characteristic polynomialof P, where r is the rank function of P.A posetP 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
(3)r(a∧b)+r(a∨b)≤r(a)+r(b),∀a,b∈P.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(4)Kν,l=(0I(ν)-I(ν)00(l)).LetFq 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 degree2ν+l over Fq, denoted by Sp2ν+l,2ν(Fq), consists of all (2ν+l)×(2ν+l) nonsingular matrices T over Fq satisfying TKlTt=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 PKlPt 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].For1≤m1≤m2≤2ν-1, 0≤k1≤k2≤l, let L1(m1,s1,k1;2ν+l,ν), (resp., L2(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)⊆L1(m1,s1,k1;2ν+l,ν), (resp., M(m1,s1,k1;2ν+l,ν)⊆L2(m2,s2,k2;2ν+l,ν)). Suppose L(m1,s1,k1;m2,s2,k2;2ν+l,ν) denotes the intersection of L1(m1,s1,k1;2ν+l,ν) and L2(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 LO(m1,s1,k1;m2,s2,k2;2ν+l,ν) or LR(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 inL(m1,s1,k1;m2,s2,k2;2ν+l,ν)

Lemma 1.
Let2ν+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(5)PKνPT=(ks00)2sm+1-(k+1)-2sk+1.
Write P=(P11P120P22)2νlm+1-(k+1)k+1, where P22=(P22′z1z2)k-111. Then  P1=(P11P120P22′0z1) and P2=(P11P120P22′0z2)m+1-(k+1)k-11 are subspace of type (m,s,k), such that P=P1+P2.Lemma 2.
Let2ν+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 thatPKνPT=(P0x1y1)Kν(P0x1y1)T=(ks0k10)2sm+1-2s-k-22k, write P=(P11P120P22′)2νlm+1-kk, where P11=(P0x1y1)m+1-k-211,x1 and y1 are the (m+1-(k+1))-th and (m-k+1)-th row vectors of P11, respectively.
P
12
=
(
P
0
′
x
1
′
y
1
′
)
m
+
1
-
k
-
2
1
1
,
x
1
′ and y1′ are the (m+1-(k+1))-th and (m-k+1)-th row vectors of P12, respectively.
Then(6)P1=(P0P0′x1x1′0P22)m+1-k-21k,P2=(P0P0′y1y1′0P22)m+1-k-21k
are subspace of type (m,s,k), such that P=P1+P2.Theorem 3.
Let2ν+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
(7)(i)s=s1=ν,0≤k1≤k≤l,(8)(ii)s<ν,k1=k=l,m-m1≥s-s1≥0,(9)(iii)s<ν,0≤k1≤k≤l,(m-k)-(m1-k1)≥s-s1≥0.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, Ifs1=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), 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.
Ifs1<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, Assume Q=(Q11Q120Q22)2νlm-1l where rank Q11=m-l, rank Q22=l and Q11KQt=M(m-l,s). Since P∩E⊂Q∩E, there exists a l×l matrix P22 with rank P22=l, such that P22Q22 is a matrix representation of subspace P∩E. Thus we can assume P=(P11P120P22)2νlm1-1l, where P11 is a matrix with rank  m1-l. Because P is a subspaces of type (m1,s1,l), we can assume P11M(m-l,s)P11t=M(m1-l,s). Then Q11 can be considered 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.
Let2ν+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.
For1≤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 Theorems3 and 4, it is directed.


## 3. The Geometricity of LatticesLO(m1,s1,k1;m2,s2,k2;2ν + l,ν) and LR(m1,s1,k1;m2,s2,k2;2ν + l,ν)

Lemma 6 (see [3]).
If0<k<l, then(i)
LR(2ν+k,ν,k;2ν+l,ν)≃LR(k,l), LO(2ν+k,ν,k;2ν+l,ν)≃LO(k,l),(ii)
LR(k,0,k;2ν+l,ν)≃LR(k,l), LO(k,0,k;2ν+l,ν)≃LO(k,l).Lemma 7 (see [3]).
If0≤s<ν and 2s≤m-l≤ν+s, then(i)
LR(m,s,l;2ν+l,ν)≃LR(m-l,s;2ν), LO(m,s,l;2ν+l,ν)≃LO(m-l,s;2ν),(ii)
LR(m,s,0;2ν+l,ν)≃LR(m,s;2ν), LO(m,s,0;2ν+l,ν)≃LO(m,s;2ν).Theorem 8.
Let2ν+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)
LO(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)
LO(k,0,k;2ν+k,ν,k;2ν+l,ν) is a finite geometric lattice if and only if k=1, l-1,(iii)
LO(m1,s1,k1;m2,s2,k2;2ν+l,ν) is not a geometric lattice when 2≤m1-k1≤m2-k2≤2ν-2.Proof.
For anyX∈LO(m1,s1,k1;m2,s2,k2;2ν+l,ν), define
(10)rO(X)={0,ifX=0,m2-m1+2,ifX=Fq(2ν+l)dimX-m1+1,otherwise.,
Then rO is the rank function of LO(m1,s1,k1;m2,s2,k2;2ν+l,ν).
(i) For latticeLO(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν).
Ifk=0, by Lemma 7, LO(1,0,0;2ν-l,ν-1,0;2ν+l,ν)≃LO(1,0,2ν-1,ν-1;2ν), by [12], LO(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν) is a finite geometric lattice.
Ifk=l, by Lemma 7, LO(l+1,0,l;2ν-l+l,ν-1,l;2ν+l,ν)≃LO(1,0,2ν-1,ν-1;2ν), by [12], LO(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν) is a finite geometric lattice.
If0 < 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〉∈LO(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
(11)rO(U∧W)+rO(U∨W)=3>rO(U)+rO(W)=2.
That is LO(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν) is not a geometric lattice when 0<k<l.
(ii) For latticeLO(k,0,k;2ν+k,ν,k;2ν+l,ν), by Lemma 6LO(k,0,k;2ν+k,ν,k;2ν+l,ν)≃LO(k,l).
Ifk=1, l-1, LO(1,l) and LO(l-1,l) is a geometric lattice.
If2≤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〉∈LO(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
(12)rO(U∧W)+rO(U∨W)=3>rO(U)+rO(W)=2.
That is LO(k,0,k;2ν+k,ν,k;2ν+l,ν) is not a geometric lattice when 2≤k≤l-2.
(iii) For latticeLO(m1,s1,k1;m2,s2,k2;2ν+l,ν), when 2≤m1-k1≤m2-k2≤2ν-2.
Case (a).k1≤k2≤l.
(a1)s2>0, Let(13)U=(I(s2-1)000s2-10000000000I(s2)0000I(m2-k2-2s2)00000000000I(k2)1m2-k2-2s2ν+s2-m2+k2s2m2-k2-2s2ν+s2-m2+k2k20000)l-k2,W=(00s2-1001000000000000I(k2)1m2-k2-2s21ν+s2-m2+k2s2m2-k2-2s2ν+s2-m2+k2k200)l-k2.
ThenU 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
(14)rO(U∨W)+rO(U∧W)>rO(U)+rO(W).
Hence, LO(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.
Whens2=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(15)U=(I(s2)000xys200000000000I(s2)0000I(m2-k2-2s2)00000000000000I(k2)m2-k2-2s2-11ν+s2-m2+k2s2m2-k2-2s2-11ν+s2-m2+k2k20000)l-k2,W=(00xys200100000000I(k2)1ν-s2-1s21ν-s2-1k200)l-k2.
ThenU 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
(16)rO(U∨W)+rO(U∧W)>rO(U)+rO(W).
Hence,LO(m1,s1,k1;m2,s2,k2;2ν+l,ν) is not a geometric lattice when (a2).
Case (b).k1=k2=l.
In this caseLO(m1,s1,l;m2,s2,l;2ν+l,ν)≃LO(m1-l,s1;m2-l,s2;2ν), LO(m1,s1,l;m2,s2,l;2ν+l,ν) is not a geometric lattice.
HenceLO(m1,s1,k1;m2,s2,k2;2ν+l,ν) is not a geometric lattice when 2≤m1-k1≤m2-k2≤2ν-2.Theorem 9.
Let2ν+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)
LR(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)
LR(k,0,k;2ν+k,ν,k;2ν+l,ν) is a finite geometric lattice if and only if k=1, l-1,(iii)
LR(m1,s1,k1;m2,s2,k2;2ν+l,ν) is not a geometric lattice when 2≤m1-k1≤m2-k2≤2ν-2.Proof.
(i) ForX∈LR(m1,s1,k1;m2,s2,k2;2ν+l,ν), define
(17)rR(X)={0,ifX=Fq((2ν+l)),m2-m1+2,ifX=0,m2+1-dimX,otherwise.
For latticeLR(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν).
Ifk=0,  by Lemma 7, LR(1,0,0;2ν-l,ν-1,0;2ν+l,ν)≃LR(1,0,2ν-1,ν-1;2ν), by [12],  LR(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν) is a finite geometric lattice.
Ifk=l, by Lemma 7, LR(l+1,0,l;2ν-l+l,ν-1,l;2ν+l,ν)≃LR(1,0,2ν-1,ν-1;2ν), by [12],  LR(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν) is a finite geometric lattice.
If0 < 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
(18)r(U∧W)+r(U∨W)=3>r(U)+r(W)=2.
Hence,LR(k+1,0,k;2ν-1+k,ν-1,k;2ν+l,ν) is not a geometric lattice when 0<k<l.
(ii) ifk=0, l-1, by Lemma 7   LR(k,0,k;2ν+k,ν,k;2ν+l,ν)≃LR(k,l) when k=1, l-1, LR(k,0,k;2ν+k,ν,k;2ν+l,ν) is a geometric lattice.
If2≤k≤l-2, by [7], LR(k,l) is not a geometric lattice.
(iii) Case (a). ifm1-k1<ν+s1, then ν+s1-m1+k1-1≥0.
Let(19)U=(I(s1)0000s100000000000I(s1)0000100000000I(m1-k1-2s1)0000000000000I(k1)1m1-k1-2s1ν+s1-m1+k1-1s11m1-k1-2s1ν+s1-m1+k1k100000)l-k1,W=(000001I(s1)00000000000I(s1)000000100000I(m1-k1-2s1)0000000000000I(k1)s1m1-k1-2s1ν+s1-m1+k1-11s1m1-k1-2s1ν+s1-m1+k1-1k100000)l-k1.
ThenU,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〉∈LR(m1,s1,k1;m2,s2,k2;2ν+l,ν), U∩W∉LR(m1,s1,k1;m2,s2,k2;2ν+l,ν).
SorR(U∨W)=m2-m1+2, rR(U∧W)=m2-m1-1, rR(U)=m2+1-(m1+1)=m2-m1=rR(W). We have
(20)rR(U∧W)+rR(U∨W)=2m2-2m1+1>rR(U)+rR(W)=2m2-2m1.
Hence LR(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(21)U=(I(s1)000000s100000000000000I(s1)000001000000000010000000000I(m1-k1-2s1)00000000000001000000000000I(k1)11m1-k1-2s1ν+s1-m1+k1s111m1-k1-2s1ν+s1-m1+k1k10000000)l-k1,W=(I(s1)000000s100000000000000I(s1)000000000010000010000000000I(m1-k1-2s1)00000000000001000000000000I(k1)11m1-k1-2s1ν+s1-m1+k1s111m1-k1-2s1ν+s1-m1+k1k10000000)l-k1.
ThenU, 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).
SorR(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.
HenceLR(m1,s1,k1;m2,s2,k2;2ν+l,ν) is not a geometric lattice when (b).
HenceLR(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 LatticeLR(m1,s1,k1;m2,s2,k2;2ν+l,ν)

In this section we compute the characteristic polynomial of the latticeLR(m1,s1,k1;m2,s2,k2;2ν+l,ν).Theorem 10.
Let2ν+l>0, Assume that (m,s,k) satisfies 0≤k≤l, 2s≤m-k≤ν+s and 0<m<2ν+l. Then
(22)χ(LR(m1,s1,k1;m2,s2,k2;2ν+l,ν),t)=tm2-m1+2-∑k=k1k‍∑s=s1s2∑m=m2-(k2-k)+(s2-s)+1m1-(k-k1)+(s-s1)‍N(m,s,k;2ν+l,ν)gm(t)+∑k=0k1‍∑s=0s1-1‍∑m=2s+kν+s+k‍N(m,s,k;2ν+l,ν)gm(t)+∑k=k2l‍∑s=s2+1ν‍∑m=2s+kν+s+k‍N(m,s,k;2ν+l,ν)gm(t),
where gm(t)=∏i=0m-1‍(t-qi).Proof.
Define(23)rR(X)={0,ifX=Fq(2ν+l),m2-m1+2,ifX={0},m2+1-dim(X),otherwise.
ThenrR is the rank function on LR(m1,s1,k1;m2,s2,k2;2ν+l,ν). Let V=Fq(2ν+l), LO=LR(2ν+l,ν), L=LR(m1,s1,k1;m2,s2,k2;2ν+l,ν) we have
(24)χ(LO,t)=∏i=02ν+l-1(t-qi).
For any P∈L, define
(25)LP={Q∈L∣Q⊂P}={Q∈L∣Q≥P},LOP={Q∈LO∣Q⊂P}={Q∈LO∣Q≥P}.
Clearly, LV=L, LP=LOP, when P≠{0}, P≠V(26)χ(LV,t)=χ(L,t)=∑P∈L‍μ(0,P)tr(1)-r(P)=∑P∈L‍μ(0,P)tr(0)-r(P).
By inversion to Möbius we have
(27)χ(L,t)=χ(LV,t)=tm2-m1+2-∑P∈L∖{V}‍χ(LP,t),
by Theorem 5 we have
(28)χ(LR(m1,s1,k1;m2,s2,k2;2ν+l,ν),t)=tm2-m1+2-∑k=k1k2‍∑s=s1s2‍∑m=m2-(k2-k)+(s2-s)+1m1-(k-k1)+(s-s1)‍N(m,s,k;2ν+l,ν)gm(t)+∑k=0k1‍∑s=0s1-1‍∑m=2s+kν+s+k‍N(m,s,k;2ν+l,ν)gm(t)+∑k=k2l‍∑s=s2+1ν‍∑m=2s+kν+s+k‍N(m,s,k;2ν+l,ν)gm(t).



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*Source: 732808-2013-02-21.xml*