The Elementary Divisors of the Incidence Matrices of Skew Lines in PG(3,p)

0.13 MB·English

by  Joshua Ducey

THE ELEMENTARY DIVISORS OF THE INCIDENCE
MATRICES OF SKEW LINES IN PG(3,p)
0
1 JOSHUA DUCEY AND PETER SIN
0
2
Abstract. The elementary divisors of the incidence matrices of
n
lines inPG(3,p)arecomputed,wheretwolinesareincidentifand
a
J only if they are skew.

4
1

]
O 1. Introduction
C
Let V be a 4-dimensional vector space over the finite field F of p
. p
h elements, where p is a prime. We declare two 2-dimensional subspaces
t
a R and S to be incident if and only if R ∩ S = {0}. Ordering the
m
2-dimensional subspaces in some arbitrary but fixed manner, we can
[
form the incidence matrix A of this relation. In this paper we compute

1 the elementary divisors of A viewed as an integer matrix. In order to
v
introduce some useful notation, we will view this setup as a special case
1
5 ofa more general situation. For brevity, anr-dimensional subspace will
5
be called an r-subspace in what follows.
2
. More generally, let V be an (n+1)-dimensional vector space over the
1
finite field F , where q = pt is a prime power. Let L denote the set of
0 q r
0 r-subspaces ofV. Thus L denotesthepoints, L denotesthelines, etc.
1 2
1 inP(V). Anr-subspaceR ∈ L andans-subspcace S ∈ L areincident
: r s
v if and only if R ∩ S = {0}. The incidence matrix with rows indexed
i
X by the r-subspaces and columns indexed by the s-subspaces is denoted
r A . We can and do view A as the matrix of a homomorphism of
a r,s r,s
Z
free -modules
η : ZLr → ZLs
r,s
thatsends anr-subspacetothe(formal)sum ofalls-subspaces incident
with it. Then computing the elementary divisors of A is the same
r,s
thing as finding a cyclic decomposition of the cokernel of η .
r,s
These matrices A have been studied by several authors. Their p-
r,s
ranks can be deduced from the work [5], but their elementary divisors
have been computed only in the case that either r or s is equal to
one [4, 2].
Key words and phrases. Incidence matrices, skew lines, elementary divisors.
1
2 JOSHUA DUCEY ANDPETER SIN
We return now and for the remainder of the paper to the case where
V is 4-dimensional over the prime field F , and A = A .
p 2,2
Theorem 1.1. The elementary divisors of the incidence matrix A are
all p-powers, and are as given in the table below.
Elem. Div. Multiplicity
1 p(2p2 +1)/3
p p(3p3 −2p2 +3p−1)/3
p2 p(p+1)(p+2)/3
p3 p(2p2 +1)/3
p4 1
Remark. The fact that the elementary divisors are p-powers follows
from more general work of Andries Brouwer [1]. Indeed, the problem
thatisthesubject ofthispaperwassuggestedbyBrouwertothesecond
author at the recent workshop on invariants of incidence matrices held
at the Banff International Research Station.
Therestofthepaperwillbeorganizedasfollows. Inthenextsection,
the determinant of A will be computed. It will follow at once from this
computation that the cokernel of η is a finite p-group. We will then
2,2
introduce a certain p-filtration of a GL(4,p)-module that allows for a
convenient reformulation of our problem. The proof of the theorem
relies on the submodule structure of this GL(4,p)-module.
2. The Determinant of A
Let x ∈ L . Then we have
2
η (η (x)) = a y
2,2 2,2 x,y
yX∈L2
where
a = |{z ∈ L |z ∩x = {0} and z ∩y = {0}}|
x,y 2
p4, if y = x
= p4 −p3 −p2 +p, if y 6= x and y ∩x = {0}

p4 −p3, if y 6= x and y ∩x 6= {0}
as follows by an elementary counting argument. Keeping in mind that


x is incident with y if and only if x ∩ y = {0}, from the above we
immediately get the relation
A2 = p4I +(p4 −p3 −p2 +p)A+(p4 −p3)(J −A−I),
which reduces to
(2.1) A2 = p3I −(p2 −p)A+(p4 −p3)J,
ELEMENTARY DIVISORS 3
where I andJ denotethe|L |×|L | identity matrixandall-onematrix,
2 2
respectively.
Since all matrices involved in (2.1) are symmetric and commute with
eachother, wecansimultaneously diagonalizetoobtainasystemof|L |
2
equations:
λ2 = p3 −(p2 −p)λ +(p4 −p3)(p2 +1)(p2 +p+1),
1 1
λ2 = p3 −(p2 −p)λ , 2 ≤ i ≤ |L |
i i 2
where λ , λ , . . . are the eigenvalues of A counted with multiplicity.
1 2
Solvingthequadraticequations, weseethatλ iseither p4 orp−p2−p4,
1
and that for each i ≥ 2 we have that λ is either p or −p2. Since the
i
all-one vector is an eigenvector for A with eigenvalue p4, we must have
that λ = p4. Using the fact that the trace of A is zero, one can
1
easily solve for the multiplicities of the remaining eigenvalues of A.
The eigenvalue p has multiplicity p4 + p2 and the eigenvalue −p2 has
multiplicity p3 +p2 +p. This yields:
(2.2) det(A) = p4 ·p(p4+p2) ·(−p2)(p3+p2+p) = (−1)pp(p4+2p3+3p2+2p+4).
3. Some Modules for GL(4,p)
Set G = GL(4,p). If we fix a basis of V then G acts transitively
on the sets L (r = 1,2,3) and the incidence map η : ZL2 → ZL2
r 2,2
is clearly a homomorphism of ZG-permutation modules. Reduction
mod p induces an F G-permutation module homomorphism, denoted
p
η : FL2 → FL2. Define a sequence of ZG-submodules {M } of ZL2
2,2 p p i i≥0
as follows. Put M = ZL2, and for i ≥ 1 put
0
M = {m ∈ ZL2 |η (m) ∈ piZL2}.
i 2,2
Thus we have a descending filtration ZL2 = M ⊇ M ⊇ M ⊇ ··· of
0 1 2
ZL2. For a submodule N of ZL2, denote by N its image in FL2; in other
p
words N = (N +pZL2)/pZL2. We have an induced p-filtration
FL2 = M ⊇ M ⊇ M ⊇ ···
p 0 1 2
of F G-submodules. With a little thought, one sees that for each
p
i ≥ 0 the multiplicity of pi as an elementary divisor of A is precisely
dimFp(Mi/Mi+1). This reformulation of our problem allows us to focus
our attention on the F G-submodule structure of FL2, which we now
p p
do.
For r = 1,2,3 define
Y = a x ∈ FLr a = 0 .
r x p x
( )
xX∈Lr (cid:12) xX∈Lr
(cid:12)
4 JOSHUA DUCEY ANDPETER SIN
We use 1 to denote the element x ∈ ZLr, and the same symbol
x∈Lr
also for its image in FLr (r will always be clear from the context). Since
p P
|L | ≡ 1 (mod p) we have the decomposition
r
(3.1) FLr = F 1⊕Y .
p p r
Thus the essential task is to understand Y . Crucial for our analysis
2
is the F G-submodule structure of Y and Y , which is well-known and
p 1 3
given in the theorem below.
For i = 1,2,3 let S denote the degree i(p − 1) component of the
i
graded algebra S∗(V)/(Vp), the quotient of the symmetric algebra on
V by the ideal generated by p-powers. S is a simple F G-module
i p
with dimension equal to the coefficient of xi(p−1) in the expansion of
(1+x+···+xp−1)4. Explicitly, we have
p(p+1)(p+2)
(3.2) dimFpS1 = dimFpS3 = 6
and
p(2p2 +1)
(3.3) dimFpS2 = 3 .
Theorem 3.1 (Cf. [4], Theorem 2 and the comment following it.).
(1) Y is uniserial with composition series
1
Y = W ⊇ W ⊇ W ⊇ W = {0}
1 1 2 3 4
and simple quotients
W /W ∼= S , for i = 1,2,3.
i i+1 i
(2) Y is uniserial with composition series
3
Y = U ⊇ U ⊇ U ⊇ U = {0}
3 3 2 1 0
and simple quotients
U /U ∼= S , for i = 1,2,3.
i i−1 i
The following two lemmas provide all of the information about the
submodule structure of Y that we will need in order to prove Theo-
2
rem 1.1 in the next section. Before we can state the lemmas we must
first define two important maps. Let φ : ZL1 → ZL2 be the ZG-module
homomorphism defined by
φ(x) = y, for all x ∈ L .
1
yX∈L2
x∩y={0}
ELEMENTARY DIVISORS 5
(So φ is just η .) The other map we call ψ : ZL3 → ZL2, and it is the
1,2
ZG-module homomorphism given by
ψ(x) = y, for all x ∈ L .
3
yX∈L2
x∩y6=y
Denote their reductions mod p by φ and ψ, respectively. Notice how
these maps respect the decomposition (3.1). In particular,
η (1) = φ(1) = ψ(1) = 0.
2,2
Hence
Imη = η (Y ),
2,2 2,2 2
Imφ = φ(Y ),
1
Imψ = ψ(Y ),
3
and these are all submodules of Y . This observation will be used
2
frequently.
Lemma 3.2.
(1) kerη ⊆ M .
2,2 1
(2) F 1⊕(Imφ+Imψ) ⊆ M .
p 2
(3) F 1⊕Imη ⊆ M .
p 2,2 3
(4) F 1 ⊆ M .
p 4
Proof. It follows immediately from the definition of M that, in fact,
1
kerη = M . So (1) holds. Since η (1) = p41, (4) follows.
2,2 1 2,2
We can rewrite equation (2.1) as
(A+(p2 −p)I)A = p3I +(p4 −p3)J.
Then if we define γ : ZL2 → ZL2 by γ(x) = η (x)+(p2 −p)x, for all
2,2
x ∈ L , it follows from the above matrix equation that η (Imγ) ⊆
2 2,2
p3ZL2. Thus Imγ ⊆ M and so Imη = Imη = Imγ ⊆ M . This
3 2,2 2,2 3
establishes (3).
Let x ∈ L . Then we have
1
η (φ(x)) = b y
2,2 x,y
yX∈L2
where
b = |{z ∈ L |z ∩x = {0} and z ∩y = {0}}|
x,y 2
p4, if x∩y 6= {0}
=
p4 −p2, if x∩y = {0}
(
6 JOSHUA DUCEY ANDPETER SIN
as is easily checked. Thus η (φ(x)) ∈ p2ZL2 and so Imφ ⊆ M . By a
2,2 2
dualargument, Imψ ⊆ M establishing (2)andprovingthelemma. (cid:3)
2
Lemma 3.3.
(1) Y has a simple head and socle, and head(Y ) ∼= soc(Y ) ∼= S .
2 2 2 2
(2) Imη = soc(Y ).
2,2 2
(3) (Imφ+Imψ)/soc(Y ) ∼= S ⊕S .
2 1 3
Proof. Since the stabilizer or a 2-subspace has 3 orbits on L , we have
2
(3.4) dimFpEndFpG(FLp2) = 3.
If Y contained the trivial module F then it would follow from the de-
2 p
composition (3.1) that dimFp EndFpG(FLp2) ≥ 4, a contradiction. Thus
the multiplicity of F in soc(Y ) is 0.
p 2
Let L be a simple, non-trivial F G-module. By Frobenius reciprocity
p
we have
HomFpG(L,Y2) ∼= HomFpG(L,FLp2)
∼= HomFpP(L,Fp),
where P is the stabilizer of a 2-subspace. Thus if L is a simple sub-
module of Y , then by dualizing we see that the multiplicity of L in
2
soc(Y2) is dimFp(L∗)P. Therefore P fixes a nonzero vector of L∗, and
it follows from the general theory of modular representations of finite
groups of Lie type [3, 6] that L∗ is determined up to isomorphism by
this property. Since P has a fixed point on S , we thus have S ∼= L∗.
2 2
Furthermore the fixed vector is unique up to scalars, so soc(Y ) is sim-
2
ple. Since S and Y are both self-dual, (1) follows.
2 2
Let π1 and πY2 denote the projections of FLp2 onto Fp1 and Y2, re-
spectively. By (1) we can find an isomorphism α : head(Y ) → soc(Y )
2 2
ofFpG-modules. Then define µ ∈ EndFpG(FLp2) by µ = α◦q◦πY2, where
q denotes taking the quotient onto head(Y2). Clearly {π1,πY2,µ} are
independent and thus give a basis of EndFpG(FLp2) by equation (3.4).
Thus we can write η2,2 = aπ1+bπY2 +cµ for some a,b,c ∈ Fp. Since
Imη ⊆ Y we see that a = 0. Also, the restriction of η to Y has a
2,2 2 2,2 2
nontrivial kernel (it contains, for example, η (x−y), where x,y ∈ L ).
2,2 2
Thus Imη is a proper submodule of Y and so b = 0. So η is a
2,2 2 2,2
nonzero multiple of µ, hence Imη = Imµ = soc(Y ). This proves (2).
2,2 2
Since φ 6= 0 and Imφ ⊆ Y , we have by (1) that soc(Y ) ⊆ Imφ. The
2 2
fact that φ(1) = 0 implies that Imφ is isomorphic to a quotient of Y .
1
∼
Then it follows from Theorem 3.1 and (1) that Imφ = Y /W . Thus
1 3
Imφ has soc(Y ) as its unique proper submodule and
2
∼
Imφ/soc(Y ) = S .
2 1
ELEMENTARY DIVISORS 7
∼
Similarly, Theorem 3.1 and (1) imply that Imψ = Y /U . Thus Imψ
3 1
has soc(Y ) as its unique proper submodule and
2
∼
Imψ/soc(Y ) = S .
2 3
Since Imφ∩Imψ = soc(Y ), (3) follows. (cid:3)
2
4. Proof of Theorem 1.1
For i ≥ 0, let f denote the multiplicity of pi as an elementary divisor
i
of A. Thus fi = dimFp(Mi/Mi+1). For i = 0,1,2,3,4 let ei denote the
claimed multiplicity of pi as listed in the statement of Theorem 1.1.
Then the proof will be completed if we show that e = f for i =
i i
0,1,2,3,4 and f = 0 for i ≥ 5.
i
Since f is precisely the p-rank of A, we have by Lemma 3.3 that
0
f0 = dimFp Imη2,2
= dimFp soc(Y2)
= dimFp S2
p(2p2 +1)
=
3
= e .
0
We proceed by applying Lemma 3.3 andour earlier calculations (3.2)
and (3.3) to compute the dimensions of the modules that occur on the
left hand side of Lemma 3.2. We clearly have
dimFpFp1 = 1 = e4.
Also by Lemma 3.3 we compute
dimFp(Fp1⊕Imη2,2) = 1+dimFpsoc(Y2)
= 1+dimFpS2
p(2p2 +1)
= 1+
3
= e +e
3 4
and
dimFp(Fp1⊕(Imφ+Imψ)) = 1+dimFpsoc(Y2)+dimFp((Imφ+Imψ)/soc(Y2))
= 1+dimFpS2 +dimFp(S1 ⊕S3)
p(2p2 +1) p(p+1)(p+2)
= 1+ +2·
3 6
= e +e +e .
2 3 4
8 JOSHUA DUCEY ANDPETER SIN
A final application of Lemma 3.3 and direct computation gives
dimFp kerη2,2 = dimFpFLp2 −dimFpImη2,2
= dimFpFLp2 −dimFpsoc(Y2)
= dimFpFLp2 −dimFpS2
p(2p2 +1)
= (p2 +p+1)(p2 +1)−
3
= (3p4 +p3 +6p2 +2p+3)/3
= e +e +e +e .
1 2 3 4
ThesecomputationstogetherwiththeinclusionsstatedinLemma3.2
give us the inequalities:
e +e +e +e ≤ f +f +f +f +f +··· ,
1 2 3 4 1 2 3 4 5
e +e +e ≤ f +f +f +f +··· ,
2 3 4 2 3 4 5
e +e ≤ f +f +f +··· ,
3 4 3 4 5
e ≤ f +f +··· .
4 4 5
Summing the left and right hand sides of this system gives us
e +2e +3e +4e ≤ f +2f +3f +4f +5f +···
1 2 3 4 1 2 3 4 5
and therefore
pe1+2e2+3e3+4e4 ≤ pf1+2f2+3f3+4f4+5f5+···.
The right hand side of this inequality is exactly |det(A)|. By direct
calculation and equation (2.2) we find that the left hand side is also
|det(A)|. It follows immediately that we have the equalities
e +e +e +e = f +f +f +f +f +··· ,
1 2 3 4 1 2 3 4 5
e +e +e = f +f +f +f +··· ,
2 3 4 2 3 4 5
e +e = f +f +f +··· ,
3 4 3 4 5
e = f +f +··· ,
4 4 5
and so
e = f ,
1 1
e = f ,
2 2
e = f .
3 3
Finally, since e = 1 it follows easily that f = 1 and f = 0 for
4 4 i
i ≥ 5. (cid:3)
ELEMENTARY DIVISORS 9
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Department ofMathematics, University ofFlorida, Gainesville, FL
32611–8105, USA.
E-mail address: [email protected]
E-mail address: [email protected]