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SEMI-REGULARITY OF PAIRS OF BOOLEAN
POLYNOMIALS
TIMOTHY J. HODGES AND HARI R. IYER
Abstract. Semi-regular sequences over F2are sequences of homoge-
neous elements of the algebra B(n)=F2[X1, ..., Xn]/(X2
1, ..., X2
n), which
have a given Hilbert series and can be thought of as having as few re-
lations between them as possible. It is believed that most such systems
are semi-regular and this property has important consequences for un-
derstanding the complexity of Gr¨obner basis algorithms such as F4 and
F5 for solving such systems. We investigate the case where the sequence
has length two and give an almost complete description of the number
of semi-regular sequences for each n.
1. Introduction
The concept of F2-semi-regularity (which we will here shorten to semi-
regularity) was introduced in [1, 2] in order to assess the complexity of cer-
tain Gr¨obner basis algorithms, such as the XL algorithm [16] or Faug`ere’s
F4 or F5 algorithms [7, 8], applied to solving systems of non-linear equa-
tions over the Galois field F2. Heuristically, semi-regular systems of equa-
tions are systems for which there are no non-trivial relations between the
equations. Bardet, Faug`ere, Salvy and Yang were able to compute the as-
ymptotic complexity of these algorithms in the case of semi-regular systems,
proving that the complexity was exponential [2]. The motivation behind this
work was to understand the security of certain multi-variate cryptosystems
such as Patarin’s Hidden Field Equation system, since the decryption of
such systems could be performed by solving such systems of equations. Un-
fortunately, it soon became clear that the systems of equations that arose
in multivariate cryptography were not semi-regular. Moreover, despite a
belief that “generic” systems of quadratic equations are semi-regular, lit-
tle progress has been made on proving even the existence of semi-regular
systems of equations. Since systems of polynomial equations over F2arise
Key words and phrases. Semi-regular sequences, finite fields.
2020 Mathematics Subject Classification. 11T55, 12E05, 12E20, 13A02, 13M10, 94A60.
Corresponding author: Hari Iyer, Department of Mathematics, Harvard University,
Cambridge, MA 02138, USA, email: hiyer@college.harvard.edu.
This research did not receive any specific grant from funding agencies in the public,
commercial, or not-for-profit sectors.
We thank the anonymous reviewers for helpful comments and suggestions.
1
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 2
naturally in many diverse settings (such as the solution of the discrete log-
arithm problem [13]), it remains an important goal to understand whether
indeed most such systems are semi-regular.
Set B=F2[X1, . . . , Xn]/(X2
1, . . . , X2
n). Let Vbe an m-dimensional sub-
space of the space B2of homogeneous quadratic elements of B. The space
Vis semi-regular if the Hilbert series of the graded quotient ring B/BV is
given by the polynomial
Tn,m(z) = (1 + z)n
(1 + z2)m
where [P∞
i=0 aizi] denotes the series P∞
i=0 aizitruncated at the first ifor
which ai≤0.
The question we would like to answer in general is: What proportion of
such spaces are semi-regular? The total number of subspaces of dimension
mis well-known - it is the cardinality of the Grassmanian Gr(m, B2). Let
sr(n, m) = |{V∈Gr(m, B2)|Vis semi-regular}|
and let
pn,m =sr(n, m)
|Gr(m, B2)|
That is, pn,m is the proportion of mdimensional spaces that are semi-regular.
It is conjectured that for m=m(n) sufficiently large compared to n(say
m(n)> n/4), this proportion tends to 1 as ntends to infinity. Very little
is known about this conjecture. In particular, it is not even known whether
there are infinitely many nfor which pn,n 6= 0. It was shown in [11] that for
any fixed m, we must have that pn,m = 0 for sufficiently large n. The case
when m= 1 was described in Kruglov’s PhD thesis [12, Lemma 3.12]. We
give a brief review of this case in Section 5.
The purpose of this paper is to describe in detail the case when m= 2
and to give a fairly exact description of which 2-dimensional subspaces are
semi-regular for all possible values of n. In Section 4 we show that no semi-
regular two dimensional subspaces exist for n≥9 (and in more generality
that no semi-regular two dimensional subspaces exist for n≥4(m+ 1)). In
Section 6 we deal with the easy cases when n= 3,4,5 and 7. In the last two
sections we consider the more complicated situations when n= 6 and 8. In
all cases except n= 8 we are able to give the exact value of pn,2; for p8,2
we give a fairly tight bound for the value. Our hope was that analysis of
the special case m= 2 would give insight into the more important general
case. Our results were mixed. It is clear that the rank type (as defined in
Section 3) is an important invariant which may help provide more general
families of semi-regular spaces. On the other hand, as the degree of Tn,2(z)
increased, the problem of determining each of the coefficients of the Hilbert
series became increasingly intricate and frequently required more ad-hoc
methods. This suggests that substantial results on the existence and/or
ubiquity of semi-regular sequences may still be elusive.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 3
Related work: The case when m= 1 was described in Kruglov’s PhD
thesis [12, Lemma 3.12]. We give a brief review of this case in Section 5.
In this case it is relatively easy to give an exact value for pn,1for all n. At
the opposite extreme, recent work by Semaev and Tenti [15] describes the
behavior in the overdetermined case when mis sufficiently large compared
to n. When m > d(n−1)(n−2)/6e, Theorem 1.1 of [15] provides a lower
bound on the proportion of m-dimensional spaces which are semi-regular
and this bound tends to 1 as ntends to infinity. Some general results on
the existence of semi-regular subspaces of Bkfor k≥2 are given in [11]. In
particular it is shown that for all mthere exists an Nmsuch that pn,m = 0
for all n≥Nm. A homological characterization of semi-regularity was given
by Hodges and Molina in [10].
2. Background and Basics
Let F=F2be the field with two elements. Set
B=Bn=F[X1, . . . , Xn]/(X2
1, . . . , X2
n)
(we shall drop the superscript when there is no need to emphasize the num-
ber n); and let xidenote the image of Xiin B. This ring inherits the
structure of a strongly graded ring from the polynomial ring F[X1, . . . , Xn].
That is, if we denote by Bn
kthe span of the monomials xi1. . . xikof de-
gree k, then Bn=Ln
k=0 Bn
kand Bn
kBn
m=Bn
k+m. It is easy to see that
dim Bn
k=n
kand that dim Bn= 2n. The monomials xi=xi1. . . xikform a
basis for Bnso an arbitrary element of Bcan be written as b=Piaixifor
some ai∈F. We define the support of bto be
Supp(b) = {xi|ai6= 0}
For m≤nwe will identify the graded subalgebra of Bngenerated by
x1, . . . , xmwith Bm.
In [2], the concept of a semi-regular sequence of elements of Bwas defined
in the following iterative fashion.
Definition 2.1. Let f1, . . . , fm∈Bbe a sequence of homogeneous polyno-
mials with deg fi=di. Let
Df1,...,fm= min (k|
m
X
i=1
Bk−difi=Bk)
The sequence f1, . . . , fm∈Bis semi-regular if for all i= 1,2, . . . , m and
homogeneous g∈B
gfi∈(f1, . . . , fi−1) and deg(g) + deg(fi)< Df1,...,fm
implies g∈(f1, . . . , fi).
For any series Piaizi∈F[[z]], we denote by Piaizithe truncated series
Pibiziwhere bi=aiif aj>0 for j= 0, . . . , i and bi= 0 otherwise.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 4
Proposition 2.2. [2] Let f1, . . . , fm∈Bbe a sequence of homogeneous
polynomials with deg fi=di. The sequence f1, . . . , fmis semi-regular if and
only if the Hilbert series of the graded ring B/(f1, . . . , fm)is given by
HSB/(f1,...,fm)(z) = (1 + z)n
Qm
i=1(1 + zdi)
This shows that the number Df1,...,fmis the same for any semi-regular
sequence of given multi-degree d= (d1, . . . , dm). We call this number the
degree of regularity of a semi-regular sequence of degree d.
We are interested here in the case where all the fiare quadratic (that is
di= 2 for all i). In this case, Proposition 2.2 implies that if we restrict our
attention to linearly independent sequences, then the semi-regularity of the
sequence depends only on the subspace V⊂B2that they generate and not
on the choice of fi(note that if the sequence is linearly dependent, then it is
never semi-regular so we may disregard this situation). For this reason, we
find it more natural to discuss the semi-regularity of subspaces, rather than
of sequences. Thus a quadratic subspace Vof dimension mis semi-regular
if
HSB/BV (z) = (1 + z)n
(1 + z2)m
Set
Tm,n(z) = (1 + z)n
(1 + z2)m,and Dn,m = 1 + deg Tm,n(z)
So Dn,m is the degree of regularity of an m-dimensional semi-regular space
of homogeneous quadratic elements.
Another way of characterizing semi-regularity is that the only relation
between the fi’s are the trivial ones in degrees less than Dn,m. Consider
the linear maps φj:Bj−2⊗V→Bjgiven by φj(Pibi⊗vi) = Pibivi. Let
Rj(V) = ker φj. Inside Rj(V) there is a subspace of “trivial relations” Tj(V)
spanned by the elements
(1) b(v⊗w−w⊗v) where v, w ∈Vand b∈Bj−4;
(2) b(v⊗v) where v∈Vand b∈Bj−4.
Theorem 2.3. [10, Theorem 3.8] Let Vbe an m-dimensional subspace of
B2and let D=Dn,m. Then Vis semi-regular if and only if
(1) Rj(V) = Tj(V)for all 3≤j < D, and
(2) BD−2V=BD
If {v1, . . . , vm}is a basis for V, then it can be easily shown that
Tk(V) = X
i6=j
Bk−4(vi⊗vj−vj⊗vi) + X
i
Bk−4(vi⊗vi)
We are interested in understanding the proportion of such spaces which
are semi-regular. Note that the set of all m-dimensional subspaces is the
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 5
Grassmannian Gr(m, B2) and that the size of the Grassmanian of m-dimensional
subspaces of the t-dimensional space Wfor t≥mis given by the formula
|Gr(m, W )|=(2t−1)(2t−2) . . . (2t−2m−1)
(2m−1)(2m−2) . . . (2m−2m−1)
Let
sr(n, m) = |{V∈Gr(m, Bn
2)|Vis semi-regular}|
and let
pn,m =sr(n, m)
|Gr(m, Bn
2)|
be the proportion of m-dimensional subspaces which are semi-regular. It is
generally believed that if m(n) is a function of nthat is sufficiently large
relative to n, then
lim
n→∞ pn,m(n)= 1.
For instance, one can conjecture for c > 1/4, that limn→∞ pn,bcnc= 1. By
contrast, we show here that for c < 1/4
lim
n→∞ pn,bcnc= 0
At the other extreme, we can consider spaces such that m= dim Vis large
enough large that deg Tn,m = 2. This is the case if
(n−1)(n−2)
6≤m < n
2
since
(1 + z)n
(1 + z2)m= 1 + nx +n
2−mx2+n
3−mnx3+· · ·
In this case Theorem 2.3 implies that Vis semi-regular if and only if the
map B1⊗V→B3is surjective. In this case, Semaev and Tenti give a lower
bound [15, Theorem 1.1] for pn,m from which one deduces easily that
lim
n→∞ pn,m(n)= 1
if m(n)>d(n−1)(n−2)/6e. However, little is known about the behavior
of pn,m when mis between these bounds; in particular, it is not even known
if pn,n is non-zero for infinitely many n.
The general linear group GL(B1) acts naturally as graded automorphisms
of the algebra B. It therefore acts as permutations of Gr(m, Bn
2). Thus we
can decompose the Grassmannian as a union of GL(B1)-orbits and semi-
regularity is an invariant of these orbits. Under the action of GL(B1) every
element µ∈B2is equivalent to an element of the form x1x2+· · ·+x2i−1x2i,
as shown in Corollary 3.2 below. We call the number 2ithe rank of µand we
often denote it as rk(µ) or rk µ. There is an important connection between
the rank and failure of semi-regularity due to the following result.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 6
Theorem 2.4. [5, Corollary 2.2] If µ∈B2has rank k, then
dim Ann(µ)∩Bd
Bd−2µ=n−k
d−k/22k/2
In particular, Ann(µ)∩Bd)Bd−2µwhen k/2≤d≤n−k/2.
This immediately yields the following condition on the ranks of elements
of a semi-regular space.
Corollary 2.5. If Vis a semi-regular subspace of Bn
2, then Vcontains no
elements of rank kif k/2+2< Dn,m. In particular, in order for there to
exist semi-regular subspaces of dimension m, we must have Dn,m ≤n/2+2.
Proof. Let µ∈Vbe an element of rank kand let V0=hµi. Suppose
d<Dm,n; so Rd(V)/Td(V) = 0. Then by [9, Theorem 2.7]
Rd(V0) = Rd(V0)∩Rd(V) = Rd(V0)∩Td(V) = Td(V0)
so Rd(V0)/Td(V0) = 0. But Rd(V0)/Td(V0) = (Ann(µ)∩Bd−2)/Bd−4µ. So
by Theorem 2.4 Rk/2+2(V0)/Tk/2+2 (V0)6= 0. Hence, k/2+2≥Dn,m .
3. Alternating Matrices and Rank Type
An important invariant of a non-zero homogeneous quadratic element
µ∈B2is its rank: the number msuch that we can write
µ=y1y2+· · · +ym−1ym.
for some y1,...ym∈B1(see Corollary 3.2). From this we can define the
rank type of a two-dimensional subspace to be the set of ranks of its three
non-zero elements. In order to calculate the proportion of subspaces that are
semi-regular we need to count the number of subspaces of each rank type.
Fortunately these numbers can be deduced from some work of Pott, Schmidt
and Zhou [14, Theorem 5] on triples of alternating matrices (A, B, A +B).
In this section we establish the connection between elements of B2and
alternating matrices which enables us to count the subspaces by rank type.
Recall that a matrix A= (aij)∈Mn(F) is alternating if
(1) aii = 0 for all i= 1, . . . , n;
(2) aij =aji for all 1 ≤i, j ≤n
Denote by Altn(F) the space of all alternating matrices. Define Γ : Altn(F)→
B2by: for A= (aij ),
Γ(A) = X
i<j
aijxixj
Define ∆ : B2→Altn(F) by: for µ=Pi<j bij xixj,
∆(µ)=(aij) where aij =(bij if i < j
bji if i > j.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 7
For any P∈GLn(F), denote by σPthe automorphism of Bobtained by
extending the linear isomorphism of B1defined by Pwith respect to the
standard basis x1, . . . , xn.
Theorem 3.1. The maps Γand ∆are mutually inverse linear isomor-
phisms. Moreover, if P∈GLn(F),B∈Mn(F)and A=PTB P , then
Γ(A) = Γ(PTBP ) = σP(Γ(B))
Proof. Routine.
Corollary 3.2. For any µ∈B2there exist linearly independent elements
y1,...y2i∈B1such that
µ=y1y2+· · · +y2i−1y2i.
Moreover the number 2iis equal to the matrix rank of ∆(µ), and is inde-
pendent of the choice of y1,...y2i.
Proof. For any alternating matrix Athere exists a matrix Bwhich consists
of diagonal 2 ×2 blocks of the form 0 1
1 0!and invertible matrix Psuch
that A=PTBP [3, Lemma 10].
Definition 3.3. We call the number 2ithe rank of the element µ, and we
often denote it as rk µ.
Let ν(m) be the number of elements µ∈B2of rank m. Then ν(m) is
the number of alternating matrices in Mn(F) of rank mand this number is
given by the formula [14, Equation 4]
ν(2i) = "t
i#4
i−1
Y
k=0 2
n(n−1)
2t−22k
where
t=jn
2kand "t
i#q
=
i
Y
j=1
qt−j+1 −1
qj−1
Table 1 gives the number of elements of each possible rank rfor n= 3,4,5
and 6
r\n3 4 5 6
2 7 35 155 651
4 0 28 868 18228
6 0 0 0 13888
Table 1. The number of elements of rank rin Bn
2for n=
3,4,5 and 6
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 8
Let V⊂B2be a subspace of dimension 2. In this case V={0, µ, µ0, µ+µ0}
for some µ, µ0∈Bn
2. An important invariant of this space is the triple
Rk(V) = [rk µ, rk µ0,rk µ+µ0]∈N3/Σ3
(that is, the equivalence class of the triple under the action of the symmetric
group S3). We call Rk(V) the rank type of V.
Since ∆(V) is a two-dimensional subspace of the space of alternating
matrices, we may use the formula of Pott, Schmidt and Zhou [14, Theorem
5] which counts the number of triples of alternating matrices (A, B, A +B)
of given matrix rank. Using the ∆ −Γ correspondence this also counts the
number of two-dimensional spaces with given rank type. For instance the
numbers of subspaces of the different rank types when n= 6 are given in
Table 2.
Type Number
[2,2,2] 9,765
[2,2,4] 182,280
[2,4,4] 3,417,750
[2,4,6] 4,666,368
[2,6,6] 2,187,360
[4,4,4] 30,902,536
[4,4,6] 69,995,520
[4,6,6] 54,246,528
[6,6,6] 13,332,480
Total 178,940,587
Table 2. The number of subspaces of B6
2of each rank type
4. An Upper Bound on n
We begin by giving an explicit bound on nabove which there are no m-
dimensional semi-regular subspaces of Bn
2. This improves upon the result
in [11, Theorem 5.1] which established that such a bound always existed. A
version of this result which fully extends [11, Theorem 5.1] is given in the
Appendix.
Lemma 4.1. Given any 06=a∈B, there exists b∈Bsuch that ab =
x1. . . xn. In particular, if ais homogeneous of degree k, we may pick bto
be homogeneous of degree n−k.
Proof. Take a monomial mof smallest length in Suppa. Say after renum-
bering, that m=x1. . . xk. Then m0=xk+1 . . . xnmust annihilate all the
other elements of Supp a. So am0=mm0=x1. . . xn.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 9
Lemma 4.2. If n≥t+j, then Bj∩Ann Bt= 0. Equivalently, Ann Bt∩
Pn−t
i=0 Bi= 0.
Proof. Let 0 6=a∈Bj∩Ann Btwhere j≤n−t, and suppose that a6= 0.
Then by Lemma 4.1 there exists an element b∈Bn−jsuch that ab =
x1. . . xn. But b∈Bn−j=BtBn−j−t, so
ab ∈aBn−j=aBtBn−j−t= 0
contradicting ab 6= 0.
Theorem 4.3. Let Vbe a subspace of B2of dimension mand let D=Dn,m .
If n≥D+ 2m, then BD−2V6=BD; in particular Vis not semi-regular.
Proof. Let B={µ1, . . . , µm}be a basis for V. Choose a subset {µi1, . . . , µis}
which is maximal with respect to
µi1. . . µis6= 0.
Then for any i= 1, . . . , m,µi1. . . µisµi= 0, so µi1. . . µisV= 0. Suppose
that BD−2V=BD. Then
µi1. . . . .µisBD=µi1. . . µisBD−2V=BD−2µi1. . . µisV= 0
This implies that µi1. . . µis∈B2s∩Ann BD. So Lemma 4.2 implies that
n < D + 2s≤D+ 2m. Thus if n≥D+ 2m, then BD−2V6=BDand Vis
not semi-regular.
Unfortunately the behavior of Dn,m is too erratic for this result to give
us an upper bound (for instance, even though Dn,m grows slower than nfor
any fixed m, the difference n−Dn,m is not an increasing function). This
can be rectified somewhat using the following result.
Theorem 4.4. There are no semi-regular m-dimensional subspaces of Bn
2
when n≥4(m+ 1).
Proof. Suppose that n≥4(m+ 1); this implies that n/2 + 2 ≤n−2m. Sup-
pose that there exist semi-regular subspaces of dimension m. By Corollary
2.5, we must have that Dn,m ≤n/2 + 2. So Dn,m ≤n−2mcontradicting
Theorem 4.3.
For small none can always backfill the difference to get more exact an-
swers.
Corollary 4.5. There are no semi-regular subspaces of Bn
2
•of dimension one for n≥7;
•of dimension two for n≥9;
•of dimension three for n≥12;
•of dimension four for n≥14.
Proof. For instance when m= 2, Theorem 4.4 tells us that there are no semi-
regular 2-dimensional subspaces for n≥12. For the cases n= 9,10,11, one
can directly check that Dn,2= 5,6,6 respectively. Thus Dn,2≤n−4
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 10
in these cases and so by Theorem 4.3 there are no 2-dimensional semi-
regular subspaces. The cases m= 3 and 4 follow similarly by observing
that Dn,3= 6,7,7,8 for n= 12,13,14,15; and that Dn,4= 6,7,7,8,8,9 for
n= 14,15,16,17,18,19
This leads to the following interesting conjecture:
Conjecture 4.6. For m6= 2, there exist m-dimensional semi-regular sub-
spaces of Bn
2if and only if n≤Dn,m + 2m.
As we shall see, this conjecture is not true for m= 2. However, this
would seem to be an exceptional case. The validity of this conjecture when
m= 3,4 can be seen in [11, Table 1].
Theorem 4.4 also confirms the need for some condition on cin the Con-
jecture that limn→∞ pn,bcnc= 1.
Corollary 4.7. If c < 1/4, then limn→∞ pn,bcnc= 0.
Proof. If c < 1/4 then there exists an Nsuch that for n>N,cn ≤n/4−1.
So Theorem 4.4 implies that pn,bcnc= 0 for n>N.
5. The Case m= 1
Let us start by briefly reviewing the case when m= 1. If one-dimensional
semi-regular subspaces exist for small n, their Hilbert series and degree of
regularity would be as in Table 1.
n Tn,1(z)Dn,1
3 1 + 3z+ 2z23
4 1 + 4z+ 5z23
5 1 + 5z+ 9z2+ 5z34
6 1 + 6z+ 14z2+ 14z3+z45
7 1 + 7z+ 20z2+ 28z3+ 15z45
Table 3. The polynomials Tn,1(z) and Dn,1= deg Tn,1(z)+1
for n= 2,...,7
Lemma 5.1. Suppose n≥2and let µ∈B2. Then
dim B1µ=(n−2if rk µ= 2
nif rk µ≥4
and
dim B2µ=
n−2
2if rk µ= 2
n
2−5if rk µ= 4
n
2−1if rk µ≥6
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 11
Proof. Note that dim Bdµ= dim Bd−dim Ann(µ)∩Bd. Hence by Theorem
2.4,
dim Bdµ=n
d−n−k
d−k/22k/2−dim Bd−2µ
The result then follows by direct calculation.
Whether or not V={0, µ}is semi-regular depends purely on the rank of
µ.
Theorem 5.2. Let V={0, µ}be a one dimensional subspace of B2.
(1) When n= 3, all one dimensional spaces are semi-regular. So p3,1=
1.
(2) When n= 4,Vis semi-regular if and only if rk µ= 4. So p4,1=
28/63 ≈0.44.
(3) When n= 5,Vis semi-regular if and only if rk µ= 4. So p5,1=
868/1023 ≈0.85
(4) When n= 6,Vis semi-regular if and only if rk µ= 6. So p6,1=
13888/32767 ≈0.42
(5) When n≥7,Vcannot be semi-regular. Thus pn,1= 0 for n≥7.
Proof. In the cases n= 3,4, we have Dn,1= 3, so it suffices to verify the
equality B1V=B3. Since dim B3
3= 1 and dim B4
3= 4, the result follows
immediately from Lemma 5.1. In the case n= 5, we have D5,1= 4, so we
need to verify that the map φ5:B5
1⊗V→B5
3is injective and the map
φ4:B5
2⊗V→B5
4is surjective. Lemma 5.1 implies that these conditions
hold precisely when rk µ= 4. Finally, for n= 6, we need that dim B6
1V= 6,
dim B6
2V= 14 and B6
3V=B3
5. Lemma 5.1 implies that first two conditions
hold only when rk µ= 6. The last condition is easily verified directly when
rk µ= 6.
The figures for the proportions follow from the numbers of elements of
each rank given in Table 1.
6. The Case m= 2 - Preliminaries
We now consider the situation where m= dim V= 2. Then, V=
{0, µ, µ0, µ +µ0}for some µ, µ0∈Bn
2, and recall the definition of the rank
type Rk(V) = [rkµ, rk µ0,rk µ+µ0]. Unfortunately the rank type of a space
Vdoes not determine its equivalence class under the action of GL(B1). For
instance, Examples 7.5 and 7.6 show two subspaces of B6
2of equal rank type
[4,4,4], one of which is not semi-regular while the other is semi-regular;
since automorphisms preserve the linear relations between elements in a
subspace, GL(B1) maps semi-regular spaces to semi-regular spaces, so the
latter examples are in different orbits under the action of GL(B1). However
the rank type does provide an important and useful decomposition of the
Grassmanian Gr(2, B2).
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 12
6.1. The cases n= 3,4,5and 7.From the table below we see that for
n= 3,4,and 5 the degree of regularity of a semi-regular 2-dimensional
subspace of B2would be 3. By Theorem 2.3, when Dn,2= 3, semi-regularity
n Tn,2(z)Dn,2
3 1 + 3z+z23
4 1 + 4z+ 4z23
5 1 + 5z+ 8z23
6 1 + 6z+ 13z2+ 8z34
7 1 + 7z+ 19z2+ 21z34
8 1 + 8z+ 26z2+ 40z3+ 17z45
9 1 + 9z+ 34z2+ 66z3+ 57z45
Table 4. The polynomials Tn,2(z) and Dn,2= deg Tn,2(z)+1
for n= 3,...,9
is equivalent to condition (2), B1V=B3, since condition (1) is null.
Theorem 6.1. If n= 3, then all two dimensional subspaces are semi-
regular.
Proof. In this case dim B3= 1 and B1V6= 0 by Lemma 4.1, so we must
always have B3=B1V.
Theorem 6.2. Let n= 4 and let V⊂B2be a two dimensional subspace.
Then Vis semi-regular if and only if Vcontains an element of rank 4.
Proof. In this case dim B3= 4. If rk µ= 4, then by Lemma 5.1, dim B1µ=
4. So if Vcontains an element of rank 4, we have B1V=B3. On the other
hand, suppose that Vdoes not contain an element of rank 4. Then Vhas
rank type [2,2,2]. Let µ, µ0be a basis for V. Then µ=λ1λ2and µ0=λ0
1λ0
2
for some λ1, λ2, λ0
1, λ0
2∈B1. Let Λ = hλ1, λ2, λ0
1, λ0
2i. If dim Λ = 4, then
the λ’s are linearly independent and µ+µ0would have rank 4; on the other
hand, if dim Λ = 2, then dim Λ2= 1 and V⊂Λ2, a contradiction. Therefore
we must have dim Λ = 3. Hence we can find a one dimensional subspace
V0⊂B1such that B1= Λ ⊕V0. But then
B1V= ΛV+V0V⊂Λ3+V0V
Hence dim B1V≤dim Λ3+ dim V0V≤1 + 2 = 3 and therefore B1V6=B3.
So if Vis semi-regular, it must contain an element of rank 4.
Corollary 6.3. In the case n= 4, the proportion of subspaces of B4
2that
are semi-regular is p4,2= 546/651 ≈0.84.
Proof. The total number of two-dimensional subspaces is |Gr(m, B4
2)|= 651.
From [14, Theorem 5], we have the number of subspaces of type [2,2,2] is
105. So p4,2= (651 −105)/651.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 13
Now consider the case when n= 5. Note that dim B5
3= 10 so B1V=B3
if and only if the map φ3:B5
1⊗V→B5
3is an isomorphism.
Theorem 6.4. The map φ3:B5
1⊗V→B5
3is not surjective for any two
dimensional subspace V⊂B5
2. Hence there are no semi-regular two dimen-
sional subspaces of B5
2.
Proof. We may assume, after appropriate change of variables, that V=
{0, µ, µ0, µ +µ0}where µ∈B4
2and µ0=µ0+λx5for µ0∈B4
2and λ∈B4
1.
Then
B1V=B1µ+B1µ0= (B4
1+Fx5)µ+ (B4
1+Fx5)(µ0+λx5)
⊂(B4
1µ+B4
1µ0)+(Fµ+Fµ0+B4
1λ)x5
Now B5
3=B4
3+B4
2x5, so for this map to be surjective we must have B4
1µ+
B4
1µ0=B4
3and Fµ+Fµ0+B4
1λ=B4
2. However dim B4
1λ≤3, so
dim(Fµ+Fµ0+B4
1λ)≤5<6 = dim B4
2
Thus B1V6=B5
3and Vis not semi-regular.
Next we jump ahead to consider the case when n= 7. Here the degree
of regularity is four. So in order for the space Vto be semi-regular we need
the map φ4:B7
2⊗V→B7
4to be surjective.
Theorem 6.5. The map φ4:B7
2⊗V→B7
4is not surjective for any 2-
dimensional subspace V⊂B7
2. Hence there are no semi-regular two dimen-
sional subspaces of B7
2.
Proof. Pick a basis for V, say {µ, µ0}. After a suitable choice of generators
we can assume that
µ∈B6
2, µ0=µ0+λx7,where µ0∈B6
2, λ ∈B6
1
Then
B7
2V=B7
2µ+B7
2µ0
= (B6
2+B6
1x7)µ+ (B6
2+B6
1x7)(µ0+λx7)
⊂(B6
2µ+B6
2µ0)+(B6
1µ+B6
1µ0+B6
2λ)x7
Suppose that φ4is surjective. Then we must have
B6
1µ+B6
1µ0+B6
2λ=B6
3
If λ= 0, then we would have B6
1µ+B6
1µ0=B6
3which is impossible because
the left hand side has dimension at most 12 and dim B6
3= 20. So λ6= 0.
Consider the map B6→˜
B=B6/(λ)∼
=B5. Denote the images of µand µ0
by ˜µand ˜µ0. Then we would have
˜
B1˜µ+˜
B1˜µ0=˜
B3
But this contradicts Theorem 6.4.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 14
This yields an exact value for pn,2in all cases except n= 6 or 8. In the
next two sections we consider these two remaining cases which are consid-
erably more complicated.
7. The Case m= 2,n= 6
Since D6,2= 4, a two-dimensional space V⊂B6
2is semi-regular if
(1) the map φ3:B6
1⊗V→B6
3is injective; and
(2) the map φ4:B6
2⊗V→B6
4is surjective
Note that dim B6
1= 6, dim B6
2= 15, dim B6
3= 20, and dimB6
4= 15.
Proposition 7.1. If Vcontains an element of rank 2, then Vis not semi-
regular. In particular if Vhas rank type [2,2,2],[2,2,4],[2,4,4] or [2,4,6],
then Vis not semi-regular.
Proof. By Corollary 2.5 a semi-regular two-dimensional subspace of B6
2can
contain no elements of rank less than 2(D6,2−2) = 4.
This leaves the cases where Vhas rank type [4,4,4],[4,4,6],[4,6,6] and
[6,6,6]. In the case where Vcontains an element of rank 6 the surjectivity
condition is easily established.
Lemma 7.2. If Vcontains an element of rank 6, then the map φ4:B6
2⊗
V→B6
4is surjective.
Proof. We may assume that the element of rank 6 is µ=x1x2+x3x4+x5x6.
Then B6
2µcontains all the monomials of B6
4except
x1x2x3x4, x1x2x5x6, x3x4x5x6
In addition it contains
(x1x2+x3x4)x5x6,(x1x2+x5x6)x3x4,(x3x4+x5x6)x1x2
Let µ0be another non-zero element of V. Suppose that we have a monomial
xixj∈Supp(µ0) which is not one of x1x2, x3x4, x5x6. Without loss of gener-
ality suppose it is x1x3. Then x1x2x3x4∈Supp(x2x4µ0). Since B6
2µcontains
all the other monomials involving x2x4,B6
2Vmust contain x1x2x3x4and so
B6
2V=B6
4. Now suppose that Supp(µ0)⊂ {x1x2, x3x4, x5x6}and µ06=µso
µ0is the sum of one or two of these terms. It is easily verified that in this
case again B6
2V=B6
4.
Lemma 7.3. Suppose that n≥6and let Vbe a 2-dimensional subspace of
Bn
2. If Vcontains an element of rank at least 6, then Ann V∩Bn
2= 0. If,
in addition, Vhas no elements of rank 2, then the map φ3:Bn
1⊗V→Bn
3
is injective.
Proof. Suppose that V=hµ, µ0iwhere rk µ≥6 and µ06=µ. Since rk µ≥6,
we know from Lemma 5.1 that Annµ∩B2={0, µ}. Therefore µ0µ6= 0 and
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 15
µ6∈ Ann µ0∩B2. Hence
Ann V∩B2= (Ann µ∩B2)∩(Ann µ0∩B2)
={0, µ} ∩ (Ann µ0∩B2) = {0}
Now assume that rk µ0and rk(µ+µ0) are both at least 4. An element of
Ker φ3is of the form a⊗µ+b⊗µ0where a, b ∈B1and
aµ +bµ0= 0
In this case abµ = 0 and abµ0= 0 so ab ∈Ann V∩B2={0}. Hence
a∈Ann b∩B1={0, b}. If a= 0, then bµ0= 0, so b= 0 since rk µ0≥4.
If a=bthen a(µ+µ0) = 0, so a=b= 0 since rk(µ+µ0)≥4. Thus
Ker φ3= 0.
Theorem 7.4. If Vis a a 2-dimensional subspace of B6
2of rank type
[4,4,6],[4,6,6] or [6,6,6] then Vis semi-regular.
Proof. The injectivity condition follows from Lemma 7.3. The surjectivity
condition follows from Lemma 7.2.
7.1. Spaces of rank type [4,4,4].If Vcontains a rank four element we
can assume this element is of the form µ=x1x2+x3x4. Thus we may
assume that V=hµ, µ0i={0, µ, µ0, µ +µ0}where
µ=x1x2+x3x4
µ0=µ0+λ1x5+λ2x6+x5x6
and µ0∈B4
2,λ1, λ2∈B4
1and ∈ {0,1}.
Example 7.5. If µ0= 0, λ1=x1, λ2=x3and = 0, we get
µ=x1x2+x3x4
µ0=x1x5+x3x6
µ+µ0=x1(x2+x5) + x3(x4+x6)
Note that in this example V⊂B1hx1, x3i. Thus B2V⊂B3hx1, x3iand so
B2Vdoes not contain x2x4x5x6. Thus Vis not semi-regular.
Example 7.6. If µ0=x1x2, λ1=λ2= 0 and = 1, we get
µ=x1x2+x3x4
µ0=x1x2+x5x6
µ+µ0=x3x4+x5x6
One can verify directly that B1V=B1x1x2⊕B1x3x4⊕B1x5x6, which has
dimension 12, so the map φ3is injectve. Thus B2V=B2x1x2+B2x3x4+
B2x5x6=B4since every monomial of length 4 contains one of the subwords
x1x2, x3x4or x5x6. Hence φ4is surjective and Vis semi-regular.
Lemma 7.7. Let Vbe a two-dimensional subspace of rank type [4,4,4]. If
either
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 16
(1) Vis induced (there is a proper subspace W⊂B1such that V⊂W2);
or
(2) there is a two-dimensional subspace Λ⊂B1such that V⊂B1Λ,
then Vis not semi-regular.
Proof. (1) Without loss of generality, we can assume that V⊂B5
2. In this
case,
B6
2V= (B5
2+B5
1x6)V⊂B5
2V+B5
1V x6⊂B5
4+B5
1V x6
Since B6
4=B5
4⊕B5
3x6and B5
1V(B5
3by Theorem 6.4, we cannot have
B6
2V=B6
4.
(2) In this case, as in Example 7.5, B2V⊂B3Λ(B4so Vis not semi-
regular.
Theorem 7.8. Let Vbe a two-dimensional subspace of rank type [4,4,4].
Then Vis semi-regular if and only if it is equivalent to a space of the form
given in Example 7.6
Proof. Suppose that Vis semi-regular. We may assume that Vis generated
by µand µ0of the form
µ=x1x2+x3x4
µ0=µ0+λ1x5+λ2x6+x5x6
where µ0∈B4
2,λ1, λ2∈B4
1and ∈ {0,1}. Let Λ = hλ1, λ2i.
Suppose that = 0. If dimΛ = 1, then µ0=µ0+λx for some λ∈Λ
and x∈ hx5, x6i. So we are in case (1) of Lemma 7.7 with W=B4
1+hxi,
contradicting the semi-regularity of V. Hence we must have dim Λ = 2.
Extend {λ1, λ2}to a basis {λ1, λ2, λ3, λ4}for B4
1. Note that B4
2=λ1B4
1+
λ2B4
1+Fλ3λ4. Therefore, since V6⊂ ΛB1, we must have that either µ0or
µ0+µis of the form λ1a1+λ2a2+λ3λ4for some a1, a2∈B4
1. Assuming
without loss of generality that it is µ0, we have that
µ0=λ1a1+λ2a2+λ3λ4+λ1x5+λ2x6
=λ1(x5+a1) + λ2(x6+a2) + λ3λ4
which is of rank 6 because λ1, λ2, λ3, λ4,(x5+a1),(x6+a2) form a basis for
B6
1. This contradicts the assumption that rk µ0= 4.
Thus we must have = 1. In this case after an appropriate change of
basis, we may assume that λ1=λ2= 0 and µ0=µ0+x5x6. In this case
rk µ0= rk µ0+2, so rk µ0= 2; similarly rk(µ+µ0) = 2. Thus µ=µ0+(µ+µ0)
and up to a linear change of variables we are in the case of Example 7.6.
Theorem 7.9. There are 153,129,088 semi-regular 2-dimensional subspaces
of B6
2. Thus the proportion of such subspaces that are semi-regular is
p6,2=153,129,088
178,940,587 ≈0.86
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 17
Proof. From Proposition 7.1 and Theorem 7.4 it suffices to calculate the
number of spaces of rank type [4,4,4] that are semi-regular. By Theorem
7.8, such spaces are precisely the orbit of the space given in Example 7.6.
The stabilizer of this space in GL6(F) is isomorphic to (GL2(F)×GL2(F)×
GL2(F)) oΣ3which has order 64. Hence the size of the orbit is
20,158,709,760
1,296 = 15,554,560
Adding this number to the total number of subspaces of type [4,4,6],[4,6,6]
or [6,6,6] given in the table, yields the claimed conclusion.
8. The Case n= 8
In this case D8,2= 5, so semi-regularity of a two-dimensional quadratic
subspace Vis equivalent to the following properties
•The map φ3:B1⊗V→B3is injective
•The kernel of φ4:B2⊗V→B4is the trivial kernel T4(V).
•The map φ5:B3⊗V→B5is surjective.
Note that dim B2= 28, dim B3= 56, dim B4= 70, and dim B5= 56.
Throughout this section, unless stated otherwise, Vwill denote a two-
dimensional subspace of B8
2.
Lemma 8.1. Let Vbe a semi-regular two-dimensional subspace of B8
2. Then
Vcontains no non-zero elements of rank less than or equal to 4.
Proof. By Corollary 2.5 a semi-regular two-dimensional subspace of B8
2can
contain no elements of rank less than 2(D8,2−2) = 6.
Thus it remains to investigate semi-regularity when the rank of Vis
[6,6,6],[6,6,8],[6,8,8] or [8,8,8]. Table 5 below gives the numbers of sub-
spaces of the different rank types [14, Theorem 5].
Type Number
[6,6,6] 2,093,462,703,144,960
[6,6,8] 4,719,790,074,101,760
[6,8,8] 3,567,475,986,923,520
[8,8,8] 888,431,072,772,096
Table 5. The number of two dimensional subspaces of B8
2
of Rank Type [6,6,6],[6,6,8],[6,8,8] or [8,8,8].
Note that the injectivity of the map φ3:B1⊗V→B3holds in all such
cases by Lemma 7.3. We can easily eliminate the following special case.
Theorem 8.2. Suppose that there exists a proper subspace W⊂B1such
that V⊂W2. Then the map φ5:B3⊗V→B5is not surjective. Hence V
is not semi-regular.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 18
Proof. Without loss of generality, we may assume W=B7
1and V⊂W2.
Now B3=B7
3⊕B7
2x8, so
B3V=B7
3V+B7
2V x8
By Theorem 6.5, B7
2V(B7
4. Since B5=B7
5⊕B7
4x8we must have B3V(B5
and the map is not surjective.
In this situation (there exists a proper subspace W⊂B1such that V⊂
W2), we shall say that the space Vis induced from W(or just induced if W
is not specified).
Lemma 8.3. Suppose that V=hµ, µ0i. The map φ4:B2⊗V→B4has
trivial kernel if and only if rk µand rk µ0are at least 6and
B2µ∩B2µ0={0, µµ0}
Proof. The trivial kernel of the map φ4:B2⊗V→B4is three dimensional
with basis {µ⊗µ, µ0⊗µ−µ⊗µ0, µ0⊗µ0}.Thus the kernel is trivial if and
only if dim B2V= dim(B2⊗V)−3 = 53.
If rk µ≤4, then by Lemma 5.1 the kernel of the map B2⊗Fµ→B2µhas
dimension at least 5 and so ker φ4cannot be trivial. So suppose that µand
µ0both have rank at least 6. Then dim B2µ= dim B2µ0= 27 by Lemma
5.1. On the other hand B2V=B2µ+B2µ0so the kernel is trivial if and
only if dim B2V= 54 −dim(B2µ∩B2µ0) = 53; that is, dim B2µ∩B2µ0= 1.
Since µµ06= 0 (by Lemma 5.1 again), this is equivalent to B2µ∩B2µ0=
{0, µµ0}.
We now look in detail at the situation where Vcontains an element of
rank 6.
Lemma 8.4. Let µ=y1y2+· · · +ym−1ymbe an element of rank min
Bn
2. Then the space U(µ) = hy1, . . . , ymiis independent of the choice of
y1,...ym.
Proof. Suppose that
µ=y1y2+· · · +ym−1ym=y0
1y0
2+· · · +y0
m−1y0
m
for some y1, . . . , ymand y0
1, . . . , y0
min B1. Since rk µ=m, the y1, . . . , ym
and y0
1, . . . , y0
mmust be linearly independent; hence it suffices to show that
y1, . . . , ym∈ hy0
1, . . . , y0
mi.
Extend y0
1, . . . , y0
mto a basis y0
1, . . . , y0
m, y0
m+1, ...y0
nfor Bn
1. Write
yi=
n
X
j=1
aijy0
j
for some aij ∈F. Suppose that yk6∈ hy0
1, . . . , y0
mi; that is, akl 6= 0 for some
m+ 1 ≤l≤n. After renumbering of the y1, . . . , ymand y0
m+1, ..., y0
nif
necessary, we may assume that a1n= 1. The coefficient of the monomial
y0
jy0
nin y1y2+· · · +ym−1ymis
0 = a1ja2n+a1na2j+· · · +am−1,j amn +am−1,namj
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 19
Hence
a2j=a1ja2n+
m/2
X
k=2
(a2k−1,ja2k,n +a2k−1,na2k,j )
Therefore
y2=
n
X
j=1
a2jy0
j
=
n
X
j=1
a1ja2n+
m/2
X
k=2
(a2k−1,ja2k,n +a2k−1,na2k,j )
y0
j
=
n
X
j=1
a1ja2ny0
j+
m/2
X
k=2
n
X
j=1
a2k−1,ja2k,ny0
j+
n
X
j=1
a2k−1,na2k,j y0
j
=a2ny1+
m/2
X
k=2
a2k,ny2k−1+
m/2
X
k=2
a2k−1,ny2k
contradicting the linear independence of the yi. Hence we must have all
y1, . . . , ym∈ hy0
1, . . . , y0
mi, as required.
Definition 8.5. Let Vbe a non-induced 2-dimensional subspace of B8
2
containing an element µof rank 6. We say that Vis of
(A) Type A with respect to µif V6⊂ U(µ)B1.
(B) Type B with respect to µif V⊂U(µ)B1
Note that the definition of type is dependent on the choice of µ, as the
following example illustrates.
Example 8.6. Let
µ=x1x2+x3x4+x5x6
µ0=x1x2+x3x7+x4x8
and let V=hµ, µ0i. Then µ0∈U(µ)B1, so Vis of Type B with respect to
µbut µ6∈ U(µ0)B1so Vis of Type A with respect to µ0.
The following proposition gives a more explicit description of spaces of
these different types.
Proposition 8.7. Let Vbe a non-induced two dimensional subspace of B2
containing an element µof rank six.
(1) If Vis of Type A with respect to µthen there exists a basis {y1, y2, . . . , y8}
of B1such that V=hµ, µ0iwhere µ=y1y2+y3y4+y5y6and
µ0=µ0+y7y8for some µ0∈B6
2.
(2) If Vis of Type B with respect to µthen there exists a basis {y1, y2, . . . , y8}
of B1such that V=hµ, µ0iwhere µ=y1y2+y3y4+y5y6and
µ0=µ0+λy7+λ0y8for some µ0∈B6
2and some linearly inde-
pendent λ, λ0∈B6
1.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 20
Proof. Since µhas rank six we may choose y1, . . . , y6such that µ=y1y2+
y3y4+y5y6and the yiare linearly independent. Extend {y1, . . . , y6}to a basis
{y1, . . . , y8}for B1. Pick µ0∈V\{0, µ}. Then µ0=µ0+λy7+λ0y8+ηy7y8
where µ0∈B6
2,λ, λ0∈B6
1and η∈F. Clearly Vis of Type A with respect
to µif η= 1 and of type B if η= 0. If η= 1, then
µ0= (µ0+λλ0)+(λ0+y7)(λ+y8)
So replacing y7with λ0+y7and y8with λ0+y8yields the required form. If
η= 0 and dimhλ, λ0i ≤ 1, then Vis induced. So if Vis non-induced and of
Type B we must have dimhλ, λ0i= 2.
Note that if V3µis a Type A space of the form given in (1) above, then
Vhas rank type [6,rk(µ0)+2,rk(µ+µ0) + 2].
We now proceed to count the semi-regular subspaces of each possible rank
type containing a fixed element µof rank 6. We break this into analysis of
the two different possible types, beginning with Type A.
Theorem 8.8. Let Vbe a subspace of rank type [6,6,6],[6,6,8] or [6,8,8].
If Vis of Type A with respect to a rank 6 element µ∈V, then Vis semi-
regular.
Proof. By Proposition 8.7 we can assume that V=hµ, µ0iwhere
µ=x1x2+x3x4+x5x6and µ0=µ0+x7x8
for some µ0∈B6
2; the assumption on the rank type of Vimplies that the
rank of µ0and µ+µ0are both at least 4. We need to prove (i) B2µ∩B2µ0=
{0, µµ0}and (ii) B3V=B5.
(i) B2µ∩B2µ0={0, µµ0}. Suppose that aµ =bµ0∈B2µ∩B2µ0, for some
a, b ∈B2. Let
b=µ1+λ1x7+λ2x8+x7x8, a =µ2+λ3x7+λ4x8+0x7x8
where µ1, µ2∈B6
2,λ1, λ2, λ3, λ4∈B6
1and , 0∈F. Then
0 = aµ +bµ0
= (µ0µ1+µ2µ) + x7(µ0λ1+λ3µ)
+x8(µ0λ2+λ4µ) + x7x8(µ0+µ1+0µ)
So
µ0+µ1=0µ, µ0µ1=µ2µ, λ3µ=λ1µ0, λ4µ=λ2µ0
Then λ1λ3µ=λ2
1µ0= 0. Therefore λ1λ3∈Ann(µ)∩B2={0, µ}. But
µ6=λ1λ3since rk µ= 6, so λ1λ3= 0. Suppose λ1=λ36= 0. Then
λ1(µ+µ0) = 0; but this is impossible since rk(µ+µ0)≥4. If λ16= 0 and
λ3= 0 then we would have λ1µ0= 0 which is again impossible because
rk(µ0)≥4. A similar argument works for the case λ1= 0 and λ36= 0. Thus
we must have λ1=λ3= 0. An analogous argument shows that λ2=λ4= 0
also. Therefore, λ1=λ2=λ3=λ4= 0.
Now consider the first two constraints: µ0+µ1=0µ, µ0µ1=µ2µ.
Consider the two cases:
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 21
0= 1: Then µ0+µ1=µ. So µ2µ=µ0µ1=µ0µ. Hence µ(µ0+µ2)=0
and so µ0+µ2∈Ann(µ)∩B2={0, µ}; that is, µ2∈ {µ0, µ0+µ}. So
a∈ {µ0, µ0+µ}and aµ =µ0µas required.
0= 0: Then µ1=µ0, so µ2µ=µ0µ1= 0. Hence µ2∈ {0, µ}and
aµ = 0.
This proves that B2µ∩B2µ0={0, µµ0}.
(ii) B3V=B5. Recall that B3=B6
3⊕B6
2x7⊕B6
2x8⊕B6
1x7x8so
B3µ=B6
3µ⊕B6
2µx7⊕B6
2µx8⊕B6
1µx7x8
Also
B5=B6
5⊕B6
4x7⊕B6
4x8⊕B6
3x7x8
It is easily seen that any degree 5 monomial is a multiple of µ, so B6
3µ=B6
5.
Since x7µ0=x7µ0,
B3V⊃x7B6
2µ+x7B6
2µ0= (B6
2µ+B6
2µ0)x7=B6
4x7
by Lemma 7.2. Similarly B3V⊃B6
4x8.
Finally, if a∈B6
3then aµ0=aµ0+ax7x8∈B3V. But aµ0∈B6
5⊂B3V,
so ax7x8∈B3Valso. Hence B6
3x7x8⊂B3V. Putting all this together yields
B5=B6
5⊕B6
4x7⊕B6
4x8⊕B6
3x7x8⊂B3V, so B3V=B5as claimed. Hence,
all such Type A spaces are semi-regular.
This enables us to count exactly the number of Type A spaces with respect
to a fixed µ, by rank type.
Theorem 8.9. Let µ=x1x2+x3x4+x5x6. Then
(1) There are 11,796,480 Type A semi-regular subspaces of B8
2contain-
ing µwhich are of type [6,8,8].
(2) There are 31,997,952 Type A semi-regular subspaces of B8
2contain-
ing µwhich are of type [6,8,6].
(3) There are 20,643,840 Type A semi-regular subspaces of B8
2contain-
ing µwhich are of type [6,6,6].
Proof. (1) If Vis of Type A with respect to µ, then there exist λ, λ0∈
hx1, . . . , x6isuch that V=hµ, µ0iand
µ0=µ0+ (λ0+x7)(λ+x8)
for some µ0∈ hx1, . . . , x6i. If Vis of rank type [6,8,8], then hµ, µ0imust be
of rank type [6,6,6]. From Tables 1 and 2 the number of [6,6,6] subspaces
of B6
2is 13,332,480 and the number of elements of B6
2of rank 6 is 13,880.
So the number of [6,6,6] subspaces of B6
2containing µis
3∗13,332,480/13,888 = 2,880
For each such subspace there are 212 choices for λand λ0, yielding a total of
2,880 ∗212 = 11,796,480
[6,6,6] subspaces of Type A containing µ. The numbers in (2) and (3)
are found by a similar calculation using the number of [6,4,6] and [6,4,4]
subspaces (54,246,528 and 69,995,520 respectively).
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 22
This completes our analysis of the Type A case. We now move to the
Type B case, which requires a little more work.
Lemma 8.10. Suppose λ, λ0, κ, κ0∈B1and λand λ0are linearly indepen-
dent. If λκ +λ0κ0= 0, then κ, κ0∈ hλ, λ0i.
Proof. We may assume that λ=x1and λ0=x2. Let κ=Piixiand
κ0=Pi0
ixiwhere i= 0, 0
i∈F. If x1κ+x2κ0= 0, then we must have
i= 0 and 0
i= 0 for i6= 1,2.
Lemma 8.11. Consider elements of the form µ0=µ0+λx7+λ0x8where
µ0∈B6
2and λ, λ0∈B6
1are linearly independent. Then
(1) rk µ0≥6if and only if λλ0µ06= 0;
(2) there are 63 ∗62 ∗29∗28 such elements µ0of rank 8.
Proof. Choose a complementary subspace W⊂B6
1such that B6
1=W⊕
hλ, λ0i. In this case we can write µ0=ν0+κλ +κ0λ0+λλ0where ν0∈W2,
κ, κ0∈Wand ∈F. Then
µ0=ν0+λ(x7+κ+λ0) + λ0(x8+κ0)
and so rk µ0= rk ν+ 4. So rk µ0= 8 if and only if rk ν0= 4. In each case
there are 63 ∗62 choices for λ, λ0, 28choices for κand κ0and 28 choices for
ν0yielding a total of 63 ∗62 ∗29∗28 choices for µ0.
For part (1) observe that rk µ0≥6 if and only if ν06= 0; and this is
equivalent to λλ0µ06= 0.
We can now give a useful characterization of when Type B spaces are
semi-regular.
Theorem 8.12. Suppose that V=hµ, µ0iwhere
µ=x1x2+x3x4+x5x6and µ0=µ0+λx7+λ0x8
for some 06=µ0∈B6
2and some linearly independent λ, λ0∈B6
1. Then Vis
semi-regular if and only if λλ0µ06∈ B6
2µ.
Proof. Suppose that λλ0µ0∈B2µ. We want to show that Vis not semi-
regular. We may assume that µand µ0have rank at least 6 because otherwise
Vis not semi-regular by Theorem 8.1. Clearly λλ0µ0∈B2µ∩B2µ0. We
want to show that λλ0µ06∈ {0, µµ0}. Suppose that λλ0µ0=µµ0. Then
λλ0+µ∈Ann(µ0)∩B6
2={0, µ0}by Lemma 5.1. Hence λλ0∈ {µ, µ +µ0},
contradicting the fact that both µand µ+µ0have rank at least 6. If
λλ0µ0= 0, then λλ0∈Ann(µ0)∩B6
2={0, µ0}, again yielding a contradiction
because the linear independence property of λand λ0implies that λλ06= 0.
So B2µ∩B2µ0){0, µµ0}and Vis not semi-regular by Lemma 8.3.
Now assume that λλ0µ06∈ B2µ. Lemma 8.11 implies that the ranks of µ0
and µ0+µare at least 6. As before, we need to prove (i) B2µ∩B2µ0={0, µµ0}
and (ii) B3V=B5. Set Λ = hλ, λ0i.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 23
(i) B2µ∩B2µ0={0, µµ0}: Suppose aµ =bµ06= 0 where
a=µ2+λ1x7+λ2x8+0x7x8, b =µ1+λ7x7+λ8x8+x7x8,
and µ2, µ1∈B6
2, λ1, λ2, λ7, λ8∈B6
1, , 0∈F. Equating the coefficients of
x7x8on both sides of aµ =bµ0, yields
0µ=µ0+λλ8+λ0λ7.
So
0µ+µ0=λ(x7+λ8) + λ0(x8+λ7).
Since the right hand side has rank at most 4 and the rank of µ, µ0and µ+µ0
are all at least 6, this implies that =0= 0. Hence λλ8+λ0λ7= 0, so by
Lemma 8.10, λ7, λ8∈Λ.
Thus
a=µ2+λ1x7+λ2x8, b =µ1+λ7x7+λ8x8,
Comparing the coefficients of x7, x8and the term that is purely contained
in B6
4yields
µλ1=µ0λ7+µ1λ
µλ2=µ0λ8+µ1λ0
µ0µ1=µµ2
Since λ7∈Λ, λ7λ∈Λ2={0, λλ0}. If λ7λ=λ0λ, then
µλ1λ=µ0λ7λ=µ0λ0λ
contradicting our assumption that µ0λλ0/∈B6
2µ. Therefore λ7λ= 0 and
so λ7∈ {0, λ}. Similarly we obtain λ8∈ {0, λ0}and λ7+λ8∈ {0, λ +λ0}.
Therefore
(λ7, λ8) = (0,0) or (λ, λ0)
Since λ7λ= 0, we also have µλ1λ= 0. Since rk µ= 6, this implies
λ1λ= 0, and so λ1∈ {0, λ}. Similarly we obtain λ2∈ {0, λ0}and λ1+λ2∈
{0, λ +λ0}. Thus
(λ1, λ2) = (0,0) or (λ, λ0)
Suppose λ1=λ2= 0. If λ7=λ8= 0, then λ, λ0∈Ann(µ1), so µ1∈
{0, λλ0}. Since b6= 0, we must have µ16= 0, so λλ0µ0=µ1µ0=µµ2∈B6
2µ,
a contradiction.
Now suppose that (λ7, λ8)=(λ, λ0). Then
(µ0+µ1)λ=µ0λ7+µ1λ= 0 and (µ0+µ1)λ0=µ0λ8+µ1λ0= 0
so λ, λ0∈Ann(µ0+µ1) and µ0+µ1∈ {0, λλ0}. If µ0+µ1= 0, then
b=µ0and bµ0= 0, contradicting our assumption. Thus µ0+µ1=λλ0so
b=µ0+λλ0and λλ0µ0= (b+µ0)µ0=aµ ∈B6
2µ, again a contradiction.
Hence we must have (λ1, λ2)=(λ, λ0). In this case
µλ =µ0λ7+µ1λ
µλ0=µ0λ8+µ1λ0
If (λ7, λ8) = (0,0), then Ann(µ+µ1) contains Λ and therefore µ+µ1∈
{0, λλ0}. If µ+µ1=λλ0, then µ1=µ+λλ0and so µµ2=µ0(µ+λλ0) which
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 24
would imply λλ0µo∈B6
2µ, a contradiction. So µ+µ1= 0, in which case
b=µand bµ0=µµ0as required.
If λ7=λand λ8=λ0, then Ann(µ+µ1+µ0) contains Λ and therefore
µ+µ1+µ0∈ {0, λλ0}. If µ+µ1+µ0=λλ0, then µ1=µ+µ0+λλ0and so
µµ2=µ0(µ+µ0+λλ0) which again implies λλ0µo∈B6
2µ, a contradiction.
So µ+µ1+µ0= 0, or µ=µ0+µ1. Since
b=µ1+λx7+λ0x8=µ+µ0+λx7+λ0x8=µ+µ0
we have that bµ0=µµ0. Thus we have proved that B2µ∩B2µ0={0, µµ0}
In this case {λ, λ0}is linearly independent so we may extend {λ, λ0}to a
basis {λ, λ0, y1, y2, y3, y4}for B6
2. Let Y=hy1, y2, y3, y4i. Then we have that
after a possible change of the xibasis, µ0=µ0+λx7+λ0x8where µ0∈Y.
(ii) B3V=B5: Recall, as in the previous proof, that B3=B6
3⊕x7B6
2⊕
x8B6
2⊕x7x8B6
1; that
B5=B6
5⊕B6
4x7⊕B6
4x8⊕B6
3x7x8
and that B6
5=B6
3µ⊂B3V. The assumption that λλ0µ0=λλ0µ06∈ B6
2µ
implies that B6
2V∩B6
4)B6
2µ. Since dim B6
2µ= 14 = dim B6
4−1, this
implies that B6
2V⊃B6
4. So
B3V⊃(B6
2x7+B6
2x8)V=B6
2V x7+B6
2V x8⊃B6
4x7+B6
4x8.
Thus it remains to show that B3V⊃B6
3x7x8. For b∈B6
2we have that
bx7µ0=bµ0x7+bλ0x7x8
Since bµ0x7∈B4x7⊂B3V, this implies that bλ0x7x8∈B3V. A similar ar-
gument for λyields that B3V⊃(B6
2λ+B6
2λ0)x7x8. Also B3V⊃B6
1x7x8µ0=
B6
1µ0x7x8, and B3V⊃B6
1x7x8µ. Hence
B3V⊃(B6
1µ0+B6
2λ+B6
2λ0+B6
1µ)x7x8
Thus it suffices to show that B6
1µ0+B6
2λ+B6
2λ0+B6
1µ=B6
3. Then we may
write µ=ν+λa+λa0where ν∈Yand a, a0∈B6
1. Then B6
1µ0+B6
2λ+B6
2λ0+
B6
1µ=B6
3is equivalent to Y µ0+Y ν =Y3. Suppose that Y µ0+Y ν 6=Y3.
Then µ0, ν and µ0+νall have rank 2, so we may assume that, after an
appropriate change of basis for Y, that µ0=y1y2and ν=y1y3. In this case
µ=y1y3+λa +λ0a0
and since µhas rank 6, y1, y3, λ, a, λ0, a0must form a basis for B6
1. Let
A=ha, a0i. Then
Λy1µ= Λy1(λa +λ0a0) = y1λλ0A
Since λλ0µ=y1λλ0y3, this yields that
B6
2µ⊃Λy1µ+Fλλ0µ=y1λλ0(A+Fy3) = y1λλ0B6
1= (y1B6
1)λλ0
Hence µ0λλ0=y1y2λλ0∈B6
2µ, contrary to assumption.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 25
Given this condition for Type B sequences to be semi-regular, we now
enumerate such sequences. The following are auxiliary results to help do
this.
Lemma 8.13. Let µ=x1x2+x3x4+x5x6and let λ, λ0be linearly indepen-
dent elements of B6
1. Then there exists a µ1∈B6
2such that λλ0µ16∈ B2µ.
Proof. Recall that dim B6
2µ= 15−1 = 14 by Lemma 5.1 and dimB6
4= 15, so
B6
2µ(B6
4. On the other hand if W=hµ, λλ0i, then by Lemma 7.2 we have
that B6
2W=B6
4. Hence there must exist a µ1∈B6
2with µ1λλ06∈ B6
2µ.
Theorem 8.14. Let µ=x1x2+x3x4+x5x6. There are 63 ∗62 ∗213 =
31,997,952 semi-regular subspaces containing µwhich are of Type B with
respect to µ.
Proof. Recall from Theorem 8.2 that if a space Vis induced then it is
not semi-regular. Let TBbe the set of all non-induced two dimensional
subspaces of B8
2that are Type B with respect to µ. These are spaces of the
form V=hµ, µ0iwhere µ0=µ0+λx7+λ0x8and λ, λ0∈B6
1are linearly
independent. For each pair of linearly independent elements λ, λ0choose
aµ1∈B6
2such that λλ0µ16∈ B2µ. Since dim B6
4/B6
2µ= 1, we have that
λλ0µ0∈B6
2µif and only if λλ0(µ0+µ1)6∈ B6
2µ. Define Φ : TB→ TBby
Φ(hµ, µ0+λx7+λ0x8i) = hµ, µ0+µ1+λx7+λ0x8i
Then Φ2=Iand Φ(V) is semi-regular if and only if Vis not semi-regular.
Hence exactly half of the spaces in TBare semi-regular.
The number of choices for λand λ0is 63∗62; the number of choices for µ0is
215 and these come in pairs, {µ0, µ0+µ}which generate the same subspace.
So the total number of non-induced Type B subspaces is 63 ∗62 ∗214, and
half of these are semi-regular.
At this point we know how many semi-regular spaces there are containing
a given µof rank 6. In order to pass from this local result to a global
result about the number of semi-regular subspaces we need to have a precise
breakdown of these spaces by rank type (because each space can contain 1,2
or 3 elements of rank 6). Finding this breakdown for Type B spaces required
a little more work.
Theorem 8.15. Let λ, λ0, 3, 4, 5, 6, x7, x8be a basis for B8
1and let W=
hλ, λ0, 3, 4, 5, 6i. Let µ0=34+56+λx7+λ0x8and let µ=ν+aλ +
bλ0+ηλλ0∈W2be an element of rank 6 for some η∈F,a, b ∈ h3, 4, 5, 6i
and ν∈ h3, 4, 5, 6i2. Then the two dimensional vector space V=hµ, µ0i
is semi-regular if and only if ν(34+56)6= 0.
Proof. Let µ0=34+56. Suppose that Vis not semi-regular. then by an
earlier result, we know that λλ0µ0=γµ for some γ=e+cλ+dλ0+η0λλ0∈W.
Now
γµ = (ν+aλ +bλ0+ηλλ0)(e+cλ +dλ0+η0λλ0)
=νe + (ae +ν c)λ+ (dν +eb)λ0+ (η0ν+ηe +cb +ad)λλ0
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 26
Comparing coefficients yields
µ0=η0ν+ηe +cb +ad
0 = νe
0 = ae +cν
0 = be +dν
So
νµ0= (η0ν+ηe +cb +ad)ν
=bcν +adν =b(ae) + a(be)=0
Conversely assume that νµ0= 0. Suppose first that η= 1. Then λλ0=
µ+ν+aλ +bλ0and so
λλ0µ0= (µ+ν+aλ +bλ0)µ0=µ0µ+ (aλ +bλ0)µ0
Now µ= (ν+ab) + (λ+b)(λ0+a) and so rk(ν+ab) = 4, since rk µ= 6. Let
W=h3, . . . , 6i. Since rk µ0= 4 also we have W(ν+ab) = W µ0. So there
exist c, d ∈Usuch that aµ0=c(ν+ab) and bµ0=d(ν+ab). But then
[(λ+b)c+ (λ0+a)d]µ= (λ+b)c(ν+ab)+(λ0+a)d(ν+ab)
= (λ+b)aµ0+ (λ0+a)bµ0
= (aλ +bλ0)µ0
So λλ0µ0∈B2µand Vis not semi-regular.
Now suppose η= 0. Then, µ=ν+aλ +bλ0. If a=b,µ=ν+a(λ+λ0)
is expressible in five variables, but rk(µ) = 6, so a6=b. Then we can extend
a, b to a basis a, b, c, d for h3, 4, 5, 6i. Since µis rank 6, it must have a
term not divisible by a or b, so cd ∈Supp(ν)⊂Supp(µ) in the a, b, c, d basis.
Depending on if ac, ad, bc, bd, ab are in Supp(ν), we have µ=ν+aλ +bλ0=
(c+1a+0
1b)(d+2a+0
2b) + ab +aλ +bλ0, for some 1, 0
1, 2, 0
2, ∈F.
Making a coordinate transformation c→c+1a+0
1b, d →d+2a+0
2b, we
get µ=cd +ab +aλ +bλ0, where ha, b, c, di=h3, 4, 5, 6iand ∈ {0,1}
with ν=cd +ab. Note that
(λ0+a)cµ =λλ0ac ∈B2µ
Likewise,
λλ0hac, ad, bc, bdi ∈ B2µ
Also,
λλ0µ=λλ0(ab +cd)∈B2µ
In both cases, whether = 0 or 1, we see therefore that
B2µ⊇λλ0hac, ad, bc, bd, ab +cdi=λλ0Ann(ab+cd) = λλ0Ann(ν)3λλ0µ0
The last inclusion is because νµ0= 0 implies µ0∈Ann(ν). Hence, λλ0µ0∈
B2µ, and Vis not semi-regular.
Lemma 8.16. Let a, b ∈B4
1. Then the following are equivalent
(1) rk(x1x2+ax5+bx6)=6
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 27
(2) x1, x2, a, b, x5, x6are a basis for B6
2
(3) x1, x2, a, b are a basis for B4
2
(4) rk(x1x2+ab) = 4
(5) rk(x1x2+ax5+bx6+x5x6) = 6
Moreover there are 96 possible such choices for the pair a, b.
Proof. The equivalence of the first four conditions is straightforward. For
the last equivalence we note that
x1x2+ax5+bx6+x5x6=x1x2+ab + (a+x6)(b+x5)
Clearly the number of choices of aand bsatisfying (3) is (24−4)(24−8) =
96.
Lemma 8.17. Let µ∈B4
2be an element of rank 4 and let N={ν∈B4
2|
νµ 6= 0}. Then |N|= 32 and Ncontains 12 elements of rank 4 and 20
elements of rank 2. Moreover, if ν∈N, then rk ν= rk(ν+µ).
Proof. Let V=hµ, ν i. Then ν∈Nif and only if V26= 0. The two dimen-
sional subspaces of B4
2of types [4,4,4], [4,4,2] and [4,2,2] are equivalent
up to change of basis to the spaces
[4,4,4] : {0, x1x2+x3x4, x1x2+x1x3+x2x4, x3x4+x1x3+x2x4}
[4,4,2] : {0, x1x2+x3x4, x1x3, x1x3+x1x2+x3x4}
[4,2,2] : {0, x1x2+x3x4, x1x2, x3x4}.
To see this, note first that the latter case [4,2,2] is clear by the rank de-
composition of an element. For [4,4,2], Supp(µ0) and Supp(µ+µ0) cannot
contain only elements of {x1x2, x3x4}, so without loss of generality suppose
µ0is the rank 2 element and that x1x3∈Supp(µ0). Then, since µ0is rank 2,
we can write it as µ0= (x1+a)(x3+b) for a, b ∈ hx2, x4i. In fact, if neces-
sary we can make a change of basis x1→x1+x2or likewise x3→x3+x4
without changing the form of µ, so also assume a∈Fx4and b∈Fx2. If
a=b= 0, we’re done. If a=x4and b=x2, changing x1→x1+x4and
x3→x3+x2again preserves the form of µ=x1x2+x3x4while yielding
µ0=x1x3, so we’re done again. On the other hand, if a=x4and b= 0,
µ+µ0=x1x2+x3x4+ (x1+x4)x3=x1(x2+x3) which is rank 2, contra-
dicting the rank type Rk(V) = [4,4,2], and likewise for a= 0 and b=x3.
The last case is [4,4,4]. In this case, again we can write µ0= (x1+a)(x3+
b) + x2x4for a∈Fx4, b ∈Fx2and each of the 4 choices for (a, b) gives either
the equivalent form of Vabove or a contradiction to the rank type.
Thus V26= 0 if and only if Vis of type [4,4,4] or [4,2,2]. It follows immedi-
ately that rk ν= rk(ν+µ). There are 6 subspaces of type [4,4,4] containing
a given µand 10 of type [4,2,2]. Thus Ncontains 12 elements of rank 4
and 20 elements of rank 2.
Lemma 8.18. Let νbe a rank 4 element of B4
2such that ν6∈ B4
1x1+B4
1x2.
Then there exists a basis x1, x2, y3, y4of B4
1such that ν=x1x2+y3y4.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 28
Proof. Extend x1, x2to an arbitrary basis x1, x2, z3, z4of B4
2. Write ν=
x1x2+z3a+z4b+0z3z4for a, b ∈ hx1, x2i. Since ν /∈B4
1x1+B4
1x2,0= 1.
So, ν= (x1x2+ab) + (z3+b)(z4+a) with x1x2+ab ∈ {0, x1x2}. Then,
since νis rank 4, x1x2+ab =x1x2and ν=x1x2+ (z3+b)(z4+a). Setting
y3=z3+band y4=z4+a,x1, x2, y3, y4is a basis of B4
1(since a, b ∈ hx1, x2i)
such that ν=x1x2+y3y4.
Theorem 8.19. Let Vbe the set of semi-regular two dimensional subspaces
of B6
1B8
1which contain an element of B6
1of rank 6. Then there are
(1) 63 ∗62 ∗29∗28 ∗12 ∗128 spaces in Vof rank type [6,8,8].
(2) 63 ∗62 ∗29∗28 ∗20 ∗192 spaces in Vof rank type [6,6,8];
Proof. Let us fix an element µ0∈B6
1B8
1of rank 8 and count the number of
V∈ V containing µ0. As in the proof of Lemma 8.11, we may assume that
µ0has the form
µ0=µ0+λx7+λ0x8
where B6
1=W⊕ hλ, λ0i, dim W= 4 and µ0∈W2has rank 4. Suppose that
V=hµ, µ0iwhere µ∈B6
2has rank 6. In this case µ=ν+aλ +bλ0+λλ0
for some ν∈W2,a, b ∈Wand ∈F. By Theorem 8.15, Vis semi-regular
if and only if νµ06= 0. By Lemma 8.17, there are 12 choices for νof rank
4 and 20 choices of rank 2. Again by Lemma 8.17, we see that if rk ν= 4,
then rk(ν+µ0) = 4 also, so
µ0+µ= (µ0+ν) + λ(x7+a) + λ0(x8+b) + λλ0
has rank 8. Thus Vis of type [6,8,8]. Similarly, if rk ν= 2, then Vis of
type [6,6,8]. Now we use Lemma 8.16 to count the number of possible µfor
which Vhas each of the two rank types. First consider the case rk ν= 2.
(i) If = 0, then µ=ν+aλ +bλ0and by Lemma 8.16 there are 96 choices
for aand bwhich yield rk µ= 6.
(ii) If = 1, then µ=ν+aλ +bλ0+λλ0. Again by Lemma 8.16 there are
96 choices for aand bwhich yield rkµ= 6..
Thus in the [6,6,8] case, for any given µ0there are 20 choices for νand
192 choice for a, b and . The number of choices for µ0is given by Lemma
8.11 and combining these two results yields (2).
Now consider the case rk ν= 4.
(i) If = 0, then µ=ν+aλ +bλ0. Note that rk µ= 6 implies that
aand bare linearly independent and ν6∈ B4
1a+B4
1b. So by Lemma 8.18
ν=ab +y3y4where a, b, y3, y4is a basis for W. Thus hν, abiis a [4,2,2]
space containing ν. There are ten such subspaces for each ν; each space
yields two choices for ab and each such choice of ab yields 6 choices for a
and b. This yields a total of 120 choices for µ.
(ii) If = 1, then µ=ν+aλ +bλ0+λλ0=ν+ab + (λ+b)(λ0+a).. In this
case rk µ= 6 if and only if rk(ν+ab) = 4. If ab = 0, this is always true and
there are 46 ways to choose aand bsuch that ab = 0. The latter is because
B4
1has 24= 16 elements and ab = 0 implies either a=b(16 choices) or if
a6=b, either a= 0 and b6= 0 (16-1 = 15 choices) or b= 0 and a6= 0 (again
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 29
15 choices), yielding 15 + 15 + 16 = 46 choices. If ab 6= 0, this holds if and
only if hν, abiis of type [4,4,2]. There are 15 such spaces containing a given
element of rank four, so 15 choices for ab; and for each of these, there are 6
different ways of choosing aand b. This yields 136 possibilities for µin this
case.
Thus there is a total of 256 choices for µ. Finally there is a 2-1 corre-
spondence between choices for µand spaces of rank type [6,8,8]
Corollary 8.20. Let µ∈B8
2have rank 6. Then
(1) There are 6,193,152 two-dimensional semi-regular subspaces of B8
2
of type [6,8,8] which are Type B with respect to µ.
(2) There are 15,482,880 two-dimensional semi-regular subspaces of B8
2
of type [6,8,6] which are Type B with respect to µ.
(3) There are 10,321,920 two-dimensional semi-regular subspaces of B8
2
of type [6,6,6] which are Type B with respect to µ.
Proof. Without loss of generality we can assume that µ∈B6
2. In this case
the semi-regular subspaces of Type B with respect to µare exactly the
spaces V∈ V containing µ.
(1) Let
V0={V∈ V | Vhas rank type [6,8,8]}
and let V0
µ={V∈ V0|µ∈V}be the subset of V0consisting of such
spaces containing our fixed element µ. Note that if σ∈GL(B6
1), then
σ(V0
µ) = V0
σ(µ). Since GL(B6
1) acts transitively on the set of all rank 6
elements of B6
2, we have that V=FV0
σ(µ). Thus since B6
2contains 13888
elements of rank 6, |V0|= 13888 ∗ |V 0
µ|. So by Theorem 8.19
|V0
µ|=|V0|/13888 = 63 ∗62 ∗29∗28 ∗12 ∗128/13888 = 6,193,152
A similar argument proves (2). For part (3), notice that the number of semi-
regular subspaces containing µand of Type B with respect to µis 31,997,952
by Theorem 8.14. Since these have either rank type [6,8,8],[6,6,8] or
[6,6,6], the number of the latter type is
31,997,952 −6,193,152 −15,482,880 = 10,321,920
Corollary 8.21. Let µ=x1x2+x3x4+x5x6. Then
(1) There are 17,989,632 two-dimensional semi-regular subspaces of B8
2
of type [6,8,8] containing µ.
(2) There are 47,480,832 two-dimensional semi-regular subspaces of B8
2
of type [6,8,6] containing µ.
(3) There are 30,965,760 two-dimensional semi-regular subspaces of B8
2
of type [6,6,6] containing µ.
Proof. The number of such spaces is just the sum of the number of spaces
which are Type A and Type B with respect to µ. Thus we just add the
numbers in Theorem 8.9, and Corollary 8.20.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 30
Corollary 8.22. There are
(1) 2,697,022,899,486,720 two-dimensional semi-regular subspaces of
B8
2of type [6,8,8]
(2) 3,559,185,957,519,360 two-dimensional semi-regular subspaces of
B8
2of type [6,8,6]
(3) 1,547,472,155,442,200 two-dimensional semi-regular subspaces of
B8
2of type [6,6,6]
Proof. For any element of B8
2of rank 6, there is an automorphism σ∈
GL(B8
1) such that σ(˜µ) = µ. This automorphism then induces a bijection
between the set of semi-regular subspaces of B8
2of type [6,8,6] containing
˜µand the set of semi-regular subspaces of B8
2of type [6,8,6] containing µ.
Since there are 149,920,960 elements of B8
2of rank 6, the total number of
semi-regular subspaces of B8
2of type [6,8,6] is
47,480,832 ∗149,920,960
2= 3,559,185,957,519,360
The other cases are handled similarly.
8.1. Approximation of p8,2.The case when Rk V= [8,8,8] seems to be
even more complex than the Type B case above. Thus we content ourselves
with an approximation of p8,2in this case.
Theorem 8.23. Let p8,2be the proportion of two dimensional subspaces of
B8
2which are semi-regular. Then
0.65 ≤p8,2≤0.72
Proof. We are able to determine the semi-regularity of all but the
888,431,072,772,096 spaces of rank type [8,8,8]. Using Corollary 8.22 we
obtain that the number sr(8,2) of semi-regular 2 dimensional subspaces
satisfies
7,803,681,012,449,280 ≤sr(8,2) ≤8,692,112,085,221,376
Dividing by the total number of 2 dimensional subspaces,
12,009,598,872,103,595 yields the claimed bounds.
9. Hilbert Polynomials
An even more fine-grained understanding can be obtained by looking at
the possible Hilbert polynomials that can arise for B/BV . We list here
(without proof) a complete description of the Hilbert polynomials that can
arise in the cases n= 4,5 and 6. The main determining factor is the rank-
type and whether or not the space is induced.
When n= 4 the situation is simple. When n= 5 we begin to see the
distinction between the induced and non-induced cases. When n= 6, more
subtle distinctions begin to appear. In the types column we have
•i4: Vis induced from a 4 dimensional subspace
•i5: Vis induced from a 5 dimensional subspace
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 31
Type Number HV(z)
[2,2,2] 105 1 + 4z+ 4z2+z3
[2,2,4] 280 1 + 4z+ 4z2
[2,4,4] 210 1 + 4z+ 4z2
[4,4,4] 56 1 + 4z+ 4z2
Total 651
Table 6. Hilbert Polynomials of B/BV by Rank when n= 4
Rank Type Number HV(z)
[2,2,2] 1,085 1 + 5z+ 8z2+ 5z3+z4
[2,2,4] 8,680 1 + 5z+ 8z2+ 4z3
[2,4,4] i 6,510 1 + 5z+ 8z2+ 4z3
[2,4,4] ni 52,080 1 + 5z+ 8z2+ 2z3
[4,4,4] i 1,736 1 + 5z+ 8z2+ 4z3
[4,4,4] ni 104,160 1 + 5z+ 8z2+z3
Total 174,251
Table 7. Hilbert Polynomials of B/BV by Rank and Type
when n= 5
•nin: Vis not induced but not semi-regular
•nis: Vis not induced and is semi-regular
10. Conclusion
We conducted a detailed study of the semi-regularity of two dimensional
quadratic spaces. We found the following values for pn,2, the proportion of
quadratic subspaces that were semi-regular.
Our hope was that this study would shed some light which would en-
able progress towards two of the most glaring open questions concerning
semi-regularity: a) do there exist semi-regular sequences of homogeneous
quadratic elements for all n? and b) is limn→∞ pn,n = 1; i.e., are most se-
quences of nhomogeneous quadratic elements in nvariables semi-regular?
On the positive side, the rank type is an invariant which can be used to
establish certain results easily. It seems possible that the answer to a) can
be found by considering specific spaces of high rank type. On the other
hand the table of Hilbert series in the case n= 6 suggest that getting the
Hilbert series exactly right is a hard thing to control. While most spaces
seem to be close to being semi-regular (in the sense that their Hilbert series
are close to Tn,m(z)), it appears that it will be a highly non-trivial problem
to prove the exact match of dimensions in each degree.
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 32
Rank Type Number HV(z)
[2,2,2] 9,765 1 + 6z+ 13z2+ 13z3+ 6z4+z5
[2,2,4] 182,280 1 + 6z+ 13z2+ 13z3+ 4z4
[2,4,4] i4 136,710 1 + 6z+ 13z2+ 13z3+ 4z4
[2,4,4] i5 3,281,040 1 + 6z+ 13z2+ 10z3+ 2z4
[2,4,6] 4,666,368 1 + 6z+ 13z2+ 10z3
[2,6,6] 2,187,360 1 + 6z+ 13z2+ 10z3
[4,4,4] i4 36,456 1 + 6z+ 13z2+ 13z3+ 4z4
[4,4,4] i5 6,562,080 1 + 6z+ 13z2+ 9z3+z4
[4,4,4] nin 8,749,440 1 + 6z+ 13z2+ 8z3+z4
[4,4,4] nis 15,554,560 1 + 6z+ 13z2+ 8z3
[4,4,6] 69,995,520 1 + 6z+ 13z2+ 8z3
[4,6,6] 54,246,528 1 + 6z+ 13z2+ 8z3
[6,6,6] 13,332,480 1 + 6z+ 13z2+ 8z3
Total 178,940,587
Table 8. Hilbert Polynomials of B/BV by Rank and Type
when n= 6
n3 4 5 6 7 8 ≥9
pn,21.00 0.84 0.00 0.86 0.00 [0.65,0.72] 0.00
Table 9. The proportion pn,2of 2-dimensional subspaces of
B2that are semi-regular
For most applications, it is sufficient to show that the degree of the Hilbert
polynomial is the same as that of a semi-regular system. Proving this should
be significantly easier and would give a more useful result from the point of
view of applications. Thus a weaker but more accessible conjecture would
be that for “most” m-dimensional subspaces BD−2V=BDfor D=Dn,m.
For instance we are able to prove this result in the one case that we were not
able to establish semi-regularity - spaces of rank type [8,8,8] when n= 8.
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Appendix A. The General Upper Bound
Let Vbe an m-dimensional graded subspace of B. Let {µ1, . . . , µm}be a
homogeneous basis for Vand set di=µi. If we assume that d1≤ · · · ≤ dm
then the vector d= (d1, . . . , dm) is independent of the choice of homogeneous
basis. For such a vector d= (d1, . . . , dm) we define
Tn,d(z) = (1 + z)n
Qi(1 + zdi)
and
Dn,d := deg Tn,d(z)
Denote the Hilbert series of the quotient ring B/BV by HSV(z). We say
the space Vis semi-regular if HSV(z) = Tn,d(z).
SEMI-REGULARITY OF PAIRS OF BOOLEAN POLYNOMIALS 34
Theorem A.1. Let Vbe a graded subspace of Bnwith degree vector dand
let d=Pidi. If n≥Dn,d +d, then Vis not semi-regular.
Proof. Let B={µ1, . . . , µm}be a basis for V. Choose an element ξof BV
of maximal degree. Clearly deg ξ≤dand ξµi= 0 for all i. Let D=Dn,d.
If Vis semi-regular, then
BD=X
i
BD−diµi
But then
ξBd=ξX
i
BD−diµi=X
i
BD−diξµi= 0
This implies that ξ∈Bdeg ξ∩Ann BD= 0. So Lemma 4.2 implies that
n<D+ deg ξ≤D+d. Thus if n≥D+d,Vcan not be semi-regular.
Email address, Tim Hodges: timothy.hodges@uc.edu
University of Cincinnati, Cincinnati, OH 45221-0025, USA
Email address, Hari Iyer: hiyer@college.harvard.edu
Harvard College, Cambridge, MA 02138