Contemporary Mathematics
Volume 181, 1995
The Chen Groups of the Pure Braid Group
DANIEL C. COHEN AND ALEXANDER I. SUCIU
Abstract.
The Chen groups of a group are the lower central series quo
tients of its maximal metabelian quotient. We show that the Chen groups
of the pure braid group P
n
are free abelian, and we compute their ranks.
The computation of these Chen groups reduces to the computation of the
Hilbert series of a certain graded module over a polynomial ring, and the
latter is carried out by means of a Groebner basis algorithm. This result
shows that, for n ≥ 4, the group P
n
is not a direct product of free groups.
1. Introduction
The Chen groups of a group G are the lower central series quotients of G
modulo its second commutator subgroup G
00
. These groups were introduced by
Chen in [Ch], so as to provide a computable approximation to the lower central
series quotients of a link group. Although apparently weaker invariants than
the lower central series quotients of G itself, the lower central series quotients
of G/G
00
sometimes provide more subtle information about the structure of the
group G. In this paper, we illustrate this point by computing the Chen groups
of the pure braid group P
n
, for every n ≥ 2. Our results show that, unlike the
lower central series quotients of P
n
, the Chen groups are not determined by the
exponents of the braid arrangement.
Theorem 1.1. The Chen groups of the pure braid group P
n
are free abelian.
The rank, θ
k
, of the k
th
Chen group of P
n
is given by
θ
1
=
µ
n
2
¶
,θ
2
=
µ
n
3
¶
, and θ
k
=(k − 1) ·
µ
n +1
4
¶
for k ≥ 3.
1991 Mathematics Subject Classiﬁcation. Primary 20F36, 20F14, 32S25; Secondary 13D40,
13P10, 57M05.
Key words and phrases. Chen group, pure braid group, Groebner basis, Hilbert series.
The second author was partially supported by N.S.F. grant DMS–9103556
c
°1995 American Mathematical Society
02714132/95 $1.00 + $.25 per page
45
46 DANIEL C. COHEN AND ALEXANDER I. SUCIU
Remark. The ranks of the Chen groups of P
n
(for k ≥ 2) are given by the
generating series
∞
X
k=2
θ
k
t
k−2
=
µ
n +1
4
¶
·
1
(1 − t)
2
−
µ
n
4
¶
.
Throughout the paper, we use the convention
¡
p
q
¢
=0ifp<q.
For any group G, let Γ
k
(G) denote its k
th
lower central series subgroup, de
ﬁned inductively by Γ
1
(G)=G and Γ
k+1
(G)=[Γ
k
(G),G] for k ≥ 1. The
projection of G onto its maximal metabelian quotient G/G
00
induces an epimor
phism
Γ
k
(G)
Γ
k+1
(G)
³
Γ
k
(G/G
00
)
Γ
k+1
(G/G
00
)
from the k
th
lower central series quotient of G to the k
th
Chen group of G.For
1 ≤ k ≤ 3, it is easy to see that this epimorphism is in fact an isomorphism.
In particular, the ﬁrst three Chen groups of P
n
and the ﬁrst three lower central
series quotients of P
n
are identical. The lower central series quotients of the
pure braid group were computed by Kohno, using Sullivan’s minimal models
technique.
Theorem 1.2 ([K]). The lower central series quotients of the pure braid
group P
n
are free abelian. Their ranks, φ
k
, are given by the equality
∞
Y
k=1
(1 − t
k
)
φ
k
=
n−1
Y
i=1
(1 − it)=
X
j≥0
(−1)
j
b
j
(P
n
)t
j
in Z[[t]],
where b
j
(P
n
) denotes the j
th
betti number of P
n
.
The relation between the ranks of the lower central series quotients and the
betti numbers described above holds in greater generality (see [KO] for one
generalization). For instance, if G = G(A) is the fundamental group of the
complement of a ﬁbertype hyperplane arrangement A (see [FR1], [OT]), we
have the following.
Theorem 1.3 ([FR1], [FR2]). Let A be a ﬁbertype hyperplane arrangement
with exponents {d
1
,d
2
,...,d
`
}, let G denote the fundamental group of the com
plement of A, and let φ
k
denote the rank of the k
th
lower central series quotient
of G. Then the lower central series quotients of G are free abelian and, in Z[[t]],
we have
∞
Y
k=1
(1 − t
k
)
φ
k
=
`
Y
i=1
(1 − d
i
t)=
X
j≥0
(−1)
j
b
j
(G)t
j
.
Implicit in the above result are the computation of the betti numbers of the
group G, and the factorization of the Poincar´e polynomial of G. In the case
where G = P
n
, these computations were carried out by Arnol’d [A], who also
determined the structure of the cohomology ring of P
n
. For a generalization to
the conﬁguration spaces of n points in R
m
, see Cohen [Co]. The cohomology
THE CHEN GROUPS OF THE PURE BRAID GROUP 47
of the complement of an arbitrary arrangement was found by Brieskorn [Br];
and Orlik and Solomon [OS] subsequently provided a combinatorial description
of the cohomology ring. These results suﬃce for the computation of the coho
mology of G(A), since the complement of any ﬁbertype arrangement A is an
EilenbergMacLane space. The factorization of the Poincar´e polynomial of the
group of a ﬁbertype arrangement may be deduced from the fact that ﬁbertype
arrangements are supersolvable, see [T].
Alternatively, one can make explicit use of the structure of the group G to
compute its betti numbers and Poincar´e polynomial. The group of any ﬁbertype
arrangement may be realized as an iterated semidirect product of free groups
(see e.g. [OT]). In particular, the group of the braid arrangement P
n
= {H
i,j
=
ker(z
i
− z
j
)} in C
n
may be realized as P
n
= F
n−1
o ···o F
2
o F
1
, see [Bi], [Ha].
For a detailed discussion of the homology of iterated semidirect products of free
groups, see [CS1].
Both the direct product Π
n
= F
n−1
×···×F
2
× F
1
, and the semidirect
product P
n
, may be realized as the fundamental groups of the complements of
(diﬀerent) ﬁbertype arrangements with exponents {1, 2,...,n− 1}. The above
results show that the betti numbers, and the ranks of the lower central series
quotients of P
n
are equal to those of Π
n
. Thus neither homology nor the lower
central series can distinguish between the direct product Π
n
and the semidirect
product P
n
.(Forn ≤ 3 this is not surprising, as P
2
∼
=
F
1
and P
3
∼
=
F
2
× F
1
.)
However, these groups can be distinguished by means of their respective Chen
groups. First, it should be noted that the Chen groups of a group G are invariants
of isomorphism type for G, since the derived series and the lower central series
subgroups of a group are characteristic. The Chen groups of a (single) free group
are known ([Ch], [Mu]), and the Chen groups of a direct product of free groups
can therefore be easily computed, see [CS3].
Theorem 1.4. Let G = F
d
1
×···×F
d
`
be a direct product of free groups.
Then the Chen groups of G are free abelian. The rank, θ
k
, of the k
th
Chen group
of G is given by
θ
1
=
`
X
i=1
d
i
, and by θ
k
=(k − 1) ·
`
X
i=1
µ
k + d
i
− 2
k
¶
for k ≥ 2.
In particular, the ranks of the Chen groups of Π
n
= F
n−1
×···×F
1
are
θ
1
=
µ
n
2
¶
, and θ
k
=(k − 1) ·
µ
k + n − 2
k +1
¶
for k ≥ 2.
Together, Theorem 1.1 and Theorem 1.4 yield
Corollary 1.5. For n ≥ 4, the groups P
n
/P
00
n
and Π
n
/Π
00
n
are not isomor
phic.
48 DANIEL C. COHEN AND ALEXANDER I. SUCIU
Corollary 1.6. For n ≥ 4, the groups P
n
and Π
n
are not isomorphic.
Corollary 1.6 may also be obtained by analyzing the cohomology rings of the
groups P
n
and Π
n
. This is not an obvious task, for these graded rings are typi
cally given in terms of generators and relations; nevertheless, Falk [Fa] managed
to deﬁne a new invariant that does distinguish these rings. The corollary may
also be deduced from the Tits conjecture for P
5
, proved in [DLS].
The techniques we employ (see below and [CS3]) may be generalized so as to
yield an algorithm for computing the Chen groups of the group of an arbitrary
hyperplane arrangement. We have carried out this algorithm for a number of ar
rangements. The most striking examples we have encountered are the following.
Example ([CS3]). Consider the 3arrangements A
1
and A
2
with deﬁning
polynomials
Q(A
1
)=xyz(x − y)(x − z)(x − 2z)(x − 3z)(y − z)(x − y − z)
and
Q(A
2
)=xyz(x − y)(x − z)(x − 2z)(x − 3z)(y − z)(x − y − 2z).
Both these arrangements are ﬁbertype, with exponents {1, 4, 4}. Thus, G
1
=
G(A
1
) and G
2
= G(A
2
) have isomorphic homology groups and lower central
series quotients. These arrangements were introduced in [Fa], where it is shown
that an invariant even ﬁner than the ranks of the lower central series quotients
cannot distinguish between (the cohomology algebras of) these arrangements.
Since the lower central series quotients of G
1
and G
2
are isomorphic, so are
the ﬁrst three Chen groups. Explicitly, we have θ
1
(G
1
)=θ
1
(G
2
)=9,θ
2
(G
1
)=
θ
2
(G
2
) = 12, and θ
3
(G
1
)=θ
3
(G
2
) = 40 by Theorem 1.3. However, for k ≥ 4,
we have
θ
k
(G
1
)=
1
2
(k − 1)(k
2
+3k + 24) and θ
k
(G
2
)=
1
2
(k − 1)(k
2
+3k + 22).
Therefore G
1
À G
2
, and the arrangements are topologically distinct.
The ranks of the Chen groups of all the arrangements we have considered
are given by a pleasant combinatorial formula which we now describe. For an
arbitrary central arrangement A with group G = G(A), let L = L(A) denote
the intersection lattice of A, let µ : L → Z be the M¨obius functions of L, and
let L
2
be the set of rank 2 elements in L (see [OT] for standards deﬁnitions and
notational conventions concerning hyperplane arrangements). Deﬁne β(A)tobe
the number of subarrangements B⊆Athat are lattice isomorphic to P
4
, the
braid arrangement with group P
4
.
Conjecture. Let A be a central arrangement. For k ≥ max
X∈L
2
{µ(X)+1},
the rank of the k
th
Chen group of G(A) is given by
θ
k
(G(A)) =
X
X∈L
2
(k − 1)
µ
k + µ(X) − 2
k
¶
+ β(A)(k − 1).
THE CHEN GROUPS OF THE PURE BRAID GROUP 49
We leave it to the reader to verify that this conjecture holds for all arrangements
considered in this paper.
Implicit in the above conjecture is the assertion that, for k suﬃciently large,
the rank of the k
th
Chen group of G(A) is given by a polynomial in k of degree
max
X∈L
2
{µ(X) − 1}. For the braid arrangement, this degree is 1, while for the
arrangement with group G =Π
n
, the degree is n − 2. An arbitrary degree may
be realized by the group of a central 2arrangement of the appropriate number
of hyperplanes. For a more general context in which this assertion ﬁts (the
determination of the “lower central dimension” of a certain class of metabelian
groups), see Baumslag [Ba].
The structure of this paper is as follows:
In section 2, we show how to reduce the computation of the Chen groups of
an arrangement to a problem in commutative algebra. This follows the approach
taken by Massey in [Ma] for the computation of the Chen groups of a classical
link: For any group G with G/G
0
= Z
N
, the nilpotent completion of G/G
00
corresponds to the Iadic completion of the Λmodule B = G
0
/G
00
, where I is
the augmentation ideal of the group ring Λ = ZZ
N
. Thus, the computation of
the Chen groups reduces to the computation of gr(
b
B), the associated graded
module over the polynomial ring gr(
b
Λ)
∼
=
Z[x
1
,...,x
N
].
In section 3, we ﬁnd a presentation for B = P
0
n
/P
00
n
. This is accomplished
topologically as follows. The Λmodule B is the ﬁrst homology group of
f
M
n
,
the maximal abelian cover of the complement of the braid arrangement. A free
Λpresentation of B is obtained by comparing the chain complex of
f
M
n
with that
of the universal (abelian) cover of the Ntorus, where N =
¡
n
2
¢
is the number of
generators of P
n
(the cardinality of the braid arrangement.)
In section 4, a presentation for gr(
b
B) is obtained. This is done by ﬁnding a
Groebner basis for the module generated by the rows of the presentation matrix
for B. The computation is much easier than one might expect: The theoretical
upper bound for the degrees of the polynomial entries in this Groebner basis is
doubly exponential in N, whereas the actual polynomials we ﬁnd are at most
cubics.
Finally, in section 5, the ranks of the Chen groups of P
n
are computed from the
coeﬃcients of the Hilbert series,
P
k≥0
rank
³
gr(
b
B)
(k)
´
t
k
, of the graded module
gr(
b
B).
2. Outline of the Proof
Our approach to the computation of the Chen groups of P
n
follows that of
Massey, who studied an analogous problem for link groups in [Ma] (see also
[MT].) We start by outlining Massey’s setup in a context which covers both link
groups and hyperplane arrangements groups.
First, consider the free abelian group Z
N
, and ﬁx generators A
1
,...,A
N
.We
then can identify Λ = ZZ
N
, the group ring of Z
N
, with Z[A
i
,A
−1
i
], the ring of
50 DANIEL C. COHEN AND ALEXANDER I. SUCIU
Laurent polynomials in the variables A
i
. This ring can be viewed as a subring of
Z[[x
i
]], the ring of formal power series in the variables x
1
,...,x
N
. The “Magnus
embedding” (see [Ma], and compare [MKS]), Z[A
±1
i
] → Z[[x
i
]], is given by
A
i
7→ 1 − x
i
and A
−1
i
7→
P
∞
k=0
x
k
i
. Notice that the image of the polynomial
subring Z[A
i
] under this map is contained in Z[x
i
], the polynomial ring in the
variables x
1
,...,x
N
.
Next, let : ZZ
N
→ Z be the augmentation map, and I =ker be the
augmentation ideal of Λ. Consider the completion of Λ relative to the Iadic
topology,
b
Λ = lim
←−
Λ/I
k
. Then, the Magnus embedding extends to a ring isomor
phism
b
Λ
∼
−→ Z[[x
i
]].
Finally, consider the Iadic ﬁltration on Λ, and its associated graded ring,
gr(Λ) =
L
k≥0
I
k
/I
k+1
. Also, consider the madic ﬁltration on Z[[x
i
]], where
m = hx
1
,...,x
N
i, and its associated graded ring, gr(Z[[x
i
]]) =
L
k≥0
m
k
/m
k+1
∼
=
Z[x
i
]. Then, the Magnus embedding induces a graded ring isomorphism gr(Λ)
∼
−→
gr(Z[[x
i
]]).
Now let G be a group with abelianization G/G
0
∼
=
Z
N
. Then each homology
group of G
0
supports the structure of a module over the ring Λ = ZZ
N
. The
main object of our study is the ﬁrst homology group of G
0
,
B = B(G)=G
0
/G
00
,
viewed as a Λmodule, called the Alexander invariant of G, see [Ma], [MT],
[Hi], [L].
Let
b
B be the Iadic completion of B. Let
gr(B)=
M
k≥0
I
k
B/I
k+1
B and gr(
b
B)=
M
k≥0
m
k
b
B/m
k+1
b
B
be the graded modules associated to the Iadic ﬁltration on B and the madic
ﬁltration on
b
B, respectively. Then, the canonical map B →
b
B induces an iso
morphism gr(B)
∼
−→ gr(
b
B) of graded modules over the graded polynomial ring
Z[x
i
]. Massey [Ma] observed that
I
k
B
I
k+1
B
=
Γ
k+2
(G/G
00
)
Γ
k+3
(G/G
00
)
.
Combining these facts, we can restate Massey’s result as follows:
Theorem 2.1 ([Ma]). The generating series for the ranks of the Chen groups
of G,
∞
X
k=0
θ
k+2
t
k
,
equals the Hilbert series of the graded module associated to the Iadic completion
of B(G),
H(gr(
b
B),t)=
∞
X
k=0
rank(m
k
b
B/m
k+1
b
B)t
k
.
THE CHEN GROUPS OF THE PURE BRAID GROUP 51
An immediate consequence of this theorem is that, for k suﬃciently large, the
rank θ
k
of the k
th
Chen group of G is given by a polynomial in k. Indeed, this
is just the HilbertSerre polynomial of gr(
b
B), see [ZS].
Thus, the determination of the Chen groups of G has been reduced to the
determination of the graded module gr(
b
B). In order to achieve this, we ﬁrst
need a ﬁnite presentation,
Λ
a
Ω
−→ Λ
b
→ B → 0,
for the Λmodule B. In the case where G is the group of a complexiﬁed real hy
perplane arrangement, we have an algorithm for doing this based on the “Randell
presentation” ([R]) of G, see [CS2]. In the case where G = P
n
, we ﬁnd a simpli
ﬁed presentation for B in section 3. A useful feature of the matrix of Ω is that
all its entries are actual polynomials in the variables A
i
. This can also be done
in general, by replacing the generators e
s
of the free module Λ
b
by suitable mul
tiples λ
s
e
s
, if necessary. Denote by J = Im(Ω) the submodule of Λ
b
generated
by the rows of the matrix of Ω.
It is now an easy task to ﬁnd a presentation for the Iadic completion of B.
Simply take
b
Λ
a
b
Ω
−→
b
Λ
b
→
b
B → 0, where
b
Ω is obtained from Ω via the Magnus
embedding. Clearly, Im(
b
Ω) =
b
J, the completion of J. Since all the entries of the
matrix for
b
Ω belong to the subring Z[x
i
] ⊂ Z[[x
i
]], we may restrict
b
Ω to a map
Ω : Z[x
i
]
a
→ Z[x
i
]
b
, whose image,
b
J ∩ Z[x
i
]
b
, we will denote by J.
Next, we must ﬁnd a presentation for the associated graded module gr(
b
B)=
gr
³
Z[[x
i
]]
b
/
b
J
´
. This module is known to be isomorphic to Z[x
i
]
b
/lt(
b
J), where
lt(
b
J) is the submodule of Z[x
i
]
b
consisting of lowest degree homogeneous forms
of elements in
b
J (the “leading terms” of elements in
b
J), see e.g. [ZS]. From the
deﬁnitions, we have lt(
b
J)=lt(J). Thus, gr(
b
B)=Z[x
i
]
b
/lt(J), and we are left
with ﬁnding a ﬁnite generating set for the module lt(J).
Such a set is provided by Mora’s algorithm ([Mo]) for ﬁnding the tangent cone
of an aﬃne variety at the origin, see [CLO], [BW] for detailed explanations.
A script that implements this algorithm using the symbolic algebra package
Macaulay was written by Michael Stillman. We gratefully acknowledge the use
of this script, and thank Tony Iarrobino for guiding us to it. Essentially, we must
ﬁnd a (minimal) Groebner basis G = {g
1
,...,g
c
} for the module J, with respect
to a suitable monomial ordering. Then, lt(J) has Groebner basis lt(G)=
{lt(g
1
),...,lt(g
c
)}, out of which we can extract a minimal Groebner basis H =
{h
1
,...,h
d
}. Putting all these facts together, we obtain the following.
Theorem 2.2. The Z[x
i
]module gr(
b
B) has a ﬁnite presentation given by
Z[x
i
]
d
Θ
−→ Z[x
i
]
b
→ gr(
b
B) → 0,
where the matrix of Θ has rows h
1
,...,h
d
deﬁned above.
In the case where G = P
n
, we use Buchberger’s algorithm to ﬁnd a Groebner
basis G for J in section 4. The determination of the Groebner basis H for lt(J)
52 DANIEL C. COHEN AND ALEXANDER I. SUCIU
and the computation of the Hilbert series for gr(
b
B) is carried out in section 5.
It follows from Theorems 2.1 and 5.6 that the ranks of the Chen groups of P
n
are as stated in Theorem 1.1.
To ﬁnish the proof of Theorem 1.1 we only have to show that the Chen groups
of P
n
are free abelian. But this follows from the discussion above and a careful
look at the various changes of bases that we perform. Indeed, we always work
over Z, and never divide by nonzero integers diﬀerent from ±1.
The Chen groups of a link group may have torsion, as the examples in [Ma]
illustrate. We do not know whether the Chen groups of an arrangement group
are always torsion free.
3. The Presentation Matrix
We brieﬂy sketch our algorithm for ﬁnding a presentation of the module B =
B(G) in the case where G = P
n
is the group of pure braids on n strings. For a
detailed discussion of this algorithm in the case where G is the fundamental group
of the complement of an arbitrary complexiﬁed real hyperplane arrangement, see
[CS2].
Recall that the pure braid group P
n
may be realized as the fundamental group
of the complement M
n
of the hyperplane arrangement P
n
= {H
i,j
= ker(z
i
−z
j
)}
in C
n
. It is easy to show that the complement M
n
may be realized as a linear
slice of the complex Ntorus, T =(C
∗
)
N
, where N =
¡
n
2
¢
is the cardinality
of P
n
(the number of generators of P
n
). Let
f
M
n
and
e
T denote the universal
abelian covers of M
n
and T respectively. Let Λ = ZZ
N
denote the group ring of
Z
N
= H
1
(M
n
)=H
1
(T ). We wish to identify the Λmodule B = H
1
(
f
M
n
).
For that, let C = C
•
(
f
M
n
) denote the augmented chain complex of
f
M
n
. This
complex is of the form
···→C
2
∂
2
−→ C
1
∂
1
−→ C
0
²
−→ Z → 0,
where C
0
=Λ,C
1
=Λ
N
, and C
2
=Λ
ρ
(here ρ = b
2
(M
n
) is the number
of relations in P
n
). We specify the ﬁrst several diﬀerentials of this complex
using the “Burau presentation” of P
n
(which may be obtained from the standard
presentation of P
n
found in [Bi], [MKS], [Ha]). The pure braid group P
n
has
generators A
i,j
,1≤ i<j≤ n, and relations
A
i,k
A
j,k
A
i,j
= A
i,j
A
i,k
A
j,k
= A
j,k
A
i,j
A
i,k
i<j<k,
A
k,l
A
i,j
= A
i,j
A
k,l
,A
i,l
A
j,k
= A
j,k
A
i,l
,A
j,l
A
A
j,k
i,k
= A
A
j,k
i,k
A
j,l
i<j<k<l,
where u
v
= v
−1
uv. The ﬁrst two diﬀerentials of this complex are
∂
1
=(1− A
1,2
1 − A
1,3
1 − A
2,3
... 1 − A
n−1,n
)
T
(where (•)
T
denotes the transpose), and ∂
2
=
µ
∂R
∂A
i,j
¶
, the abelianization of
the Jacobian matrix of Fox derivatives of the relations {R} of P
n
, where R runs
through the relations speciﬁed above.
THE CHEN GROUPS OF THE PURE BRAID GROUP 53
At this point, it is an easy task to ﬁnd a presentation for the Alexander
module A = A(P
n
), which is deﬁned as H
1
(
f
M
n
,p
−1
(∗)), where p :
f
M
n
→ M
n
is
the covering map, and ∗ is the basepoint of M
n
, see [Ma], [Hi]. Indeed, A is
the cokernel of ∂
2
:Λ
ρ
→ Λ
N
. This module, the Alexander invariant, and the
augmentation ideal comprise the Crowell exact sequence 0 → B → A → I → 0.
We now ﬁnd a presentation for B by comparing the chain complexes C
•
(
f
M
n
)
and C
•
(
e
T ).
The chain complex C
0
= C
•
(
e
T ) is the “standard” Λfree resolution of Z. The
terms of this resolution may be described as follows: C
0
0
=Λ,C
0
1
=Λ
N
is a free
Λmodule with basis {e
i,j
 1 ≤ i<j≤ n}, and for k>1, C
0
k
=
V
k
C
0
1
=Λ
(
N
k
)
.
The diﬀerentials are given by
d
k
(e
i
1
,j
1
∧···∧e
i
k
,j
k
)=
k
X
r=1
(−1)
r−1
(1 − A
i
r
,j
r
)(e
i
1
,j
1
∧···∧ˆe
i
r
,j
r
∧···∧e
i
k
,j
k
).
The natural inclusion M
n
→ T induces a chain map φ : C → C
0
covering the
identity map Z → Z. The map φ is not unique, but its chain homotopy class is
unique. Clearly we may identify C
0
and C
0
0
, so set φ
0
=id:Λ→ Λ. Furthermore,
since H
1
(M
n
)=H
1
(T ), we identify C
1
and C
0
1
, and set φ
1
=id:Λ
N
→ Λ
N
.
Letting R denote a generic relation in P
n
, we deﬁne φ
2
: C
2
→ C
0
2
by
φ
2
(R)=
A
j,k
e
i,j
e
i,k
+ e
j,k
e
i,k
if R is A
i,k
A
j,k
A
i,j
= A
j,k
A
i,j
A
i,k
,
A
i,k
e
i,j
e
j,k
+ e
i,j
e
i,k
if R is A
i,k
A
j,k
A
i,j
= A
i,j
A
i,k
A
j,k
,
e
i,j
e
k,l
if R is A
k,l
A
i,j
= A
i,j
A
k,l
,
e
j,k
e
i,l
if R is A
i,l
A
j,k
= A
j,k
A
i,l
,
A
−1
j,k
(A
j,l
− 1)e
j,k
e
i,k
+ e
i,k
e
j,l
if R is A
j,l
A
A
j,k
i,k
= A
A
j,k
i,k
A
j,l
.
Proposition 3.1. We have d
2
◦ φ
2
= ∂
2
.
The proof of this result is an exercise in Fox calculus which we omit.
Now deﬁne a map ψ : C
0
2
→ C
0
2
by
ψ(e
i,k
e
j,l
)=e
i,k
e
j,l
+(1− A
j,l
)(A
−1
j,k
e
j,k
e
i,k
− e
i,j
e
i,k
+ A
i,k
e
i,j
e
j,k
),
ψ(e
j,k
e
i,k
)=e
j,k
e
i,k
− A
j,k
e
i,j
e
i,k
+ A
i,k
A
j,k
e
i,j
e
j,k
,
ψ(e
i,j
e
i,k
)=e
i,j
e
i,k
− A
i,k
e
i,j
e
j,k
,
for i<j<k<l, and ψ(e
p,q
e
r,s
)=e
p,q
e
r,s
otherwise.
Proposition 3.2. The map ψ is an isomorphism, and we have
ψ ◦ φ
2
(R)=
e
j,k
e
i,k
if R denotes A
i,k
A
j,k
A
i,j
= A
j,k
A
i,j
A
i,k
,
e
i,j
e
i,k
if R denotes A
i,k
A
j,k
A
i,j
= A
i,j
A
i,k
A
j,k
,
e
i,j
e
k,l
if R denotes A
k,l
A
i,j
= A
i,j
A
k,l
,
e
j,k
e
i,l
if R denotes A
i,l
A
j,k
= A
j,k
A
i,l
,
e
i,k
e
j,l
if R denotes A
j,l
A
A
j,k
i,k
= A
A
j,k
i,k
A
j,l
.
54 DANIEL C. COHEN AND ALEXANDER I. SUCIU
Proof. With respect to the ordering
e
1,2
e
2,3
<e
1,2
e
1,3
<e
2,3
e
1,3
< ···
···<e
1,2
e
1,n
<e
2,3
e
1,n
<e
1,3
e
1,n
< ···<e
2,n
e
1,n
of the basis elements of C
0
2
, the matrix of ψ is lower triangular, with ones on the
diagonal, hence is an isomorphism. The second assertion of the proposition is a
routine computation. ¤
Corollary 3.3. The map φ
2
: C
2
→ C
0
2
is injective.
Restricting our attention to the truncated complex C
2
→ C
1
→ C
0
(which
we continue to denote by C), we observe that the chain map φ : C → C
0
is
injective. Letting Q denote the quotient, we have an exact sequence of complexes
C → C
0
→ Q, with C
0
acyclic. Note that since C
i
= C
0
i
for i = 0 and i =1,
we have Q
0
= 0 and Q
1
= 0. Passing to homology, we obtain H
i
(C)=H
i+1
(Q)
since H
i
(C
0
) = 0. In particular, we have B = H
1
(C)=H
2
(Q). We obtain
our presentation of B by modifying the standard Λfree resolution of Z (i.e. the
complex C
0
) as indicated in the commutative diagram
0 −−−−→ C
2
∂
2
−−−−→ C
1
d
1
−−−−→ C
0
²
−−−−→ Z
y
y
ψ◦φ
2
y
=
y
=
y
=
···
d
5
−−−−→ C
0
4
d
4
−−−−→ C
0
3
˜
d
3
−−−−→ C
0
2
˜
d
2
−−−−→ C
0
1
d
1
−−−−→ C
0
0
²
−−−−→ Z
y
=
y
=
y
π
y
···
d
5
−−−−→ Q
4
d
4
−−−−→ Q
3
∆
−−−−→ Q
2
−−−−→ 0
where denotes the augmentation map,
˜
d
2
= d
2
◦ψ
−1
,
˜
d
3
= ψ◦d
3
, π : C
0
2
→ Q
2
=
C
0
2
/ Im(ψ ◦ φ
2
) denotes the natural projection, and ∆ = π ◦
˜
d
3
. Since Im(ψ ◦ φ
2
)
is a direct summand of C
0
2
, the quotient Q
2
= span{e
i,j
e
j,k
 1 ≤ i<j<k≤ n}
is a free Λmodule, and ∆ : Q
3
→ Q
2
provides a presentation for B = H
2
(Q).
Carrying out the various steps indicated above, we obtain
Theorem 3.4. The Alexander invariant B = B(P
n
) of the pure braid group
P
n
has a presentation Λ
a
∆
−→ Λ
b
→ B → 0, with b =
¡
N
2
¢
− b
2
(P
n
)=
¡
n
3
¢
generators, and a =
¡
N
3
¢
relations. This presentation is given by ∆=π ◦ ψ ◦ d
3
,
where
d
3
(e
r,s
e
i,j
e
k,l
)=(1− A
k,l
)e
r,s
e
i,j
− (1 − A
i,j
)e
r,s
e
k,l
+(1− A
r,s
)e
i,j
e
k,l
and
π ◦ ψ(e
r,s
e
i,j
)=
(A
r,s
− 1)A
i,j
e
i,s
e
s,j
if r<i<s<j,
−A
r,j
e
r,s
e
s,j
if r = i<s<j,
A
i,s
A
r,s
e
i,r
e
r,s
if i<r<s= j,
e
r,s
e
s,j
if r<s= i<j,
0 if r<s<i<j or i<r<s<j.
THE CHEN GROUPS OF THE PURE BRAID GROUP 55
A long sequence of elementary row operations (which we suppress) has the
eﬀect of replacing the map ∆ by the map Ω = ∆ ◦ Ξ, where Ξ is a Λlinear
automorphism of Λ
a
. This yields the following simpliﬁed presentation for B.
Theorem 3.5. The Alexander invariant B of the pure braid group P
n
has a
presentation Λ
(
N
3
)
Ω
−→ Λ
(
n
3
)
→ B → 0, where N =
¡
n
2
¢
and
Ω(e
r,s
e
i,j
e
k,l
)=
(1 − A
r,s
A
r,j
A
s,j
)e
r,s
e
s,j
if r = k<s= i<j= l,
(1 − A
j,l
A
r,l
A
s,l
)e
r,s
e
s,j
if r = k<s= i<j<l,
(1 − A
k,j
)e
k,r
e
r,s
+(1− A
k,j
)e
r,s
e
s,j
if k<r= i<s<j= l,
(1 − A
r,l
)e
i,r
e
r,s
+(1− A
i,s
)e
r,s
e
s,l
if i<r<s= j = k<l,
(1 − A
j,l
)e
r,s
e
s,j
+(1− A
r,s
)e
s,j
e
j,l
if r<s= i<j= k<l,
A
s,l
(A
r,l
− 1)e
r,s
e
s,j
+(1− A
s,j
)e
r,j
e
j,l
if r<s= i = k<j<l,
A
k,j
(A
k,j
− 1)e
r,k
e
k,s
+(1− A
k,j
)e
r,s
e
s,j
if r<k<s= i<j= l,
A
s,l
(A
j,l
− 1)e
r,s
e
s,j
+(1− A
r,s
)e
r,j
e
j,l
if r = i = k<s<j<l,
A
r,l
A
s,l
(1 − A
i,l
)e
i,r
e
r,s
+(1− A
r,s
)e
i,r
e
r,l
if i<r= k<s= j<l,
A
s,l
A
j,l
(1 − A
s,l
)e
r,s
e
s,j
+(1− A
r,j
)e
r,s
e
s,l
if r = i<s= k<j<l,
A
s,j
A
i,j
(1 − A
i,j
)e
r,s
e
s,i
+(1− A
i,j
)e
r,s
e
s,j
if r<s= k<i<j= l,
(1 − A
k,l
)e
r,s
e
s,j
if r<s= i<j<k<l,
(1 − A
k,l
)e
r,s
e
s,j
if r<s= i<k<j<l,
(1 − A
k,l
)e
r,s
e
s,j
if r = i<k<s<j<l,
(1 − A
k,l
)e
r,s
e
s,j
if k<r<s= i<j<l,
(1 − A
r,s
)e
i,j
e
j,l
if r<s<i<j= k<l,
(1 − A
r,s
)e
i,j
e
j,l
if r<i= k<s<j<l,
(1 − A
r,s
)e
i,j
e
j,l
if i<r<s<j= k<l,
(1 − A
i,j
)e
r,s
e
s,l
if r<s= k<i<j<l,
(1 − A
i,j
)e
r,s
e
s,l
if r = k<i<s<j<l,
(1 − A
r,s
)e
k,i
e
i,j
if r<k<i<s<j= l,
0 otherwise.
When written in a suitably ordered basis, the matrix of Ω takes a particularly
nice “block lowertriangular form.” For example, the presentation matrix for
B(P
4
) is:
Ω=
1 − A
1,2
A
1,3
A
2,3
000
1 − A
1,4
A
2,4
A
3,4
000
01− A
2,3
A
2,4
A
3,4
00
1 − A
3,4
1 − A
1,2
00
1 − A
2,4
1 − A
1,3
00
1 − A
1,4
1 − A
1,4
00
001− A
1,3
A
1,4
A
3,4
0
A
2,4
(A
3,4
− 1) 0 1 − A
1,2
0
A
2,4
(A
2,4
− 1) 0 1 − A
2,4
0
A
2,4
(A
1,4
− 1) 0 1 − A
2,3
0
0001− A
1,2
A
1,4
A
2,4
A
2,4
A
3,4
(1 − A
3,4
)0 0 1− A
3,4
A
2,4
A
3,4
(1 − A
2,4
)0 0 1− A
1,3
A
2,4
A
3,4
(1 − A
1,4
)0 0 1− A
2,3
.
Let J = Im(Ω) denote the submodule of relations of Ω. In other words, J is
the submodule of Λ
(
n
3
)
generated by the rows of the presentation matrix of Ω.
56 DANIEL C. COHEN AND ALEXANDER I. SUCIU
Corollary 3.6. The module J is generated by
G
1
=
(
γ
r,s,k
=(1− A
r,s
A
r,k
A
s,k
)e
r,s
e
s,k
,
γ
q
r,s,k
=(1− A
r,q
A
s,q
A
k,q
)e
r,s
e
s,k
for k<q≤ n
)
,
G
2
=
(1 − A
k,q
)e
r,s
e
s,k
+(1− A
r,s
)e
s,k
e
k,q
,
(1 − A
s,q
)e
r,s
e
s,k
+(1− A
r,k
)e
s,k
e
k,q
,
(1 − A
r,q
)e
r,s
e
s,k
+(1− A
r,q
)e
s,k
e
k,q
,
−A
s,q
(1 − A
k,q
)e
r,s
e
s,k
+(1− A
r,s
)e
r,k
e
k,q
,
−A
s,q
(1 − A
s,q
)e
r,s
e
s,k
+(1− A
s,q
)e
r,k
e
k,q
,
−A
s,q
(1 − A
r,q
)e
r,s
e
s,k
+(1− A
s,k
)e
r,k
e
k,q
,
A
s,q
A
k,q
(1 − A
k,q
)e
r,s
e
s,k
+(1− A
k,q
)e
r,s
e
s,q
,
A
s,q
A
k,q
(1 − A
s,q
)e
r,s
e
s,k
+(1− A
r,k
)e
r,s
e
s,q
,
A
s,q
A
k,q
(1 − A
r,q
)e
r,s
e
s,k
+(1− A
s,k
)e
r,s
e
s,q
for k<q≤ n
,
G
3
=
n
γ
p,q
r,s,k
=(1− A
p,q
)e
r,s
e
s,k
for p, q /∈{r, s} and k<p<q≤ n
o
,
for 1 ≤ r<s<k≤ n.
The elements γ
r,s,k
,γ
q
r,s,k
,γ
p,q
r,s,k
belong to the submodule J
k
r,s
= J ∩Λ·e
r,s
e
s,k
,
but they do not generate this module. Consider the following sets:
G
4
=
½
γ
i,j,q
r,s,k
=(1− A
i,q
)(−A
t,u
+ A
j,q
)e
r,s
e
s,k
for i ∈{r, s, k}, {j, t, u} = {r, s, k} and k<q≤ n
¾
,
G
5
=
½
γ
i,j,p,q
r,s,k
=(1− A
i,p
)(1 − A
j,q
)e
r,s
e
s,k
for i, j ∈{r, s, k} and k<p<q≤ n
¾
,
where 1 ≤ r<s<k≤ n. It is readily seen that the elements γ
i,j,k
r,s,k
,γ
i,j,p,q
r,s,k
also
belong to the submodule J
k
r,s
. The reason for introducing these sets will become
apparent in the next section, where we shall make use of the following fact.
Proposition 3.7. Let d be a 2×2 minor of Ω that involves the column e
r,s
e
s,k
and one of the columns e
s,k
e
k,q
, e
r,k
e
k,q
,ore
r,s
e
s,q
. Then d · e
r,s
e
s,k
belongs to
the submodule J
k
r,s
, and can be expressed as a Λcombination of elements from
G
k
r,s
=(G
1
∪ G
3
∪ G
4
∪ G
5
) ∩ Λ · e
r,s
e
s,k
.
Proof. This is a straightforward computation. For example,
(1 − A
k,q
)(1 − A
s,k
A
s,q
A
k,q
)e
r,s
e
s,k
=(1− A
k,q
) · γ
q
r,s,k
+ A
s,q
A
k,q
· γ
k,r,s
r,s,k
,
and
((1 − A
r,k
)(1 − A
k,q
) − (1 − A
r,s
)(1 − A
s,q
))e
r,s
e
s,k
= γ
k,s,q
r,s,k
− γ
s,k,q
r,s,k
.
The other 2 × 2 minors are handled similarly. ¤
THE CHEN GROUPS OF THE PURE BRAID GROUP 57
4. The Groebner Basis
Let
b
B be the Iadic completion of B, viewed as a module over
b
Λ
∼
=
Z[[x
i,j
]],
1 ≤ i<j≤ n. This module has a presentation
b
Λ
(
N
3
)
b
Ω
−→
b
Λ
(
n
3
)
→
b
B → 0, where
b
Ω
is obtained from Ω via the Magnus embedding. Let
b
J = Im(
b
Ω) be the submodule
of
b
Λ
(
n
3
)
generated by the rows of the matrix of
b
Ω. As noted before, the entries of
this matrix belong to the polynomial subring R = Z[x
i,j
], and so we may restrict
b
Ω to a map Ω : R
(
N
3
)
→ R
(
n
3
)
. Let J = Im(Ω)=
b
J ∩ R
(
n
3
)
. We can view the
elements of J as linear forms in the variables e
r,s
e
s,k
,1≤ r<s<k≤ n, with
coeﬃcients in R. The purpose of this section is to identify a (minimal) Groebner
basis G = {g
1
,...,g
c
} for the module J, with respect to a suitable monomial
ordering on R[e
r,s
e
s,k
]. This means that hin(J)i, the module generated by the
initial terms of elements in J, has generating set {in(g
1
),...,in(g
c
)}, see e.g.
[CLO], [BW].
It follows from Corollary 3.6 that the Rmodule J is generated by G
1
∪G
2
∪G
3
,
where G
i
is the image of G
i
in Z[x
i,j
] under the assignment A
i,j
7→ 1 − x
i,j
.
It will turn out that the required Groebner basis is the larger generating set,
G = G
1
∪G
3
∪G
4
∪G
5
∪G
2
, where
G
1
=
½
g
r,s,k
=(1− (1 − x
r,s
)(1 − x
r,k
)(1 − x
s,k
))e
r,s
e
s,k
,
g
q
r,s,k
=(1− (1 − x
r,q
)(1 − x
s,q
)(1 − x
k,q
))e
r,s
e
s,k
for k<q≤ n
¾
,
G
3
= { g
p,q
r,s,k
= x
p,q
e
r,s
e
s,k
for p, q /∈{r, s} and k<p<q≤ n } ,
G
4
=
½
g
i,j,q
r,s,k
= x
i,q
(x
t,u
− x
j,q
)e
r,s
e
s,k
for i ∈{r, s, k}, {j, t, u} = {r, s, k} and k<q≤ n
¾
,
G
5
= { g
i,j,p,q
r,s,k
= x
i,p
x
j,q
e
r,s
e
s,k
for i, j ∈{r, s, k} and k<p<q≤ n } ,
G
2
=
x
k,q
e
r,s
e
s,k
+ x
r,s
e
s,k
e
k,q
,
x
s,q
e
r,s
e
s,k
+ x
r,k
e
s,k
e
k,q
,
x
r,q
e
r,s
e
s,k
+ x
r,q
e
s,k
e
k,q
,
−x
k,q
(1 − x
s,q
)e
r,s
e
s,k
+ x
r,s
e
r,k
e
k,q
,
−x
s,q
(1 − x
s,q
)e
r,s
e
s,k
+ x
s,q
e
r,k
e
k,q
,
−x
r,q
(1 − x
s,q
)e
r,s
e
s,k
+ x
s,k
e
r,k
e
k,q
,
x
k,q
(1 − x
s,q
)(1 − x
k,q
)e
r,s
e
s,k
+ x
k,q
e
r,s
e
s,q
,
x
s,q
(1 − x
s,q
)(1 − x
k,q
)e
r,s
e
s,k
+ x
r,k
e
r,s
e
s,q
,
x
r,q
(1 − x
s,q
)(1 − x
k,q
)e
r,s
e
s,k
+ x
s,k
e
r,s
e
s,q
for k<q≤ n
,
for 1 ≤ r<s<k≤ n.
We ﬁrst have to deﬁne a monomial ordering on our polynomial ring. Start by
58 DANIEL C. COHEN AND ALEXANDER I. SUCIU
ordering the variables in R = Z[x
i,j
] as follows:
x
1,2
>x
2,3
>x
1,3
>x
3,4
>x
2,4
>x
1,4
> ······>x
n−1,n
> ···>x
1,n
,
and extend this ordering to the graded reverse lexicographic (“grevlex”) order on
the set of monomials in R. Recall brieﬂy how this is done (see [CLO] for details).
Let x
α
denote an arbitrary monomial in the variables x
i,j
, and let deg(x
α
)be
its multidegree. We say that x
α
>x
β
if deg(α) > deg(β), or deg(α) = deg(β)
and, in α − β, the rightmost entry is negative. For a polynomial f ∈ R, we will
denote by in(f) its highest term c
α
x
α
, and by deg(f) its multidegree α.
Finally, extend the grevlex ordering on R to one on R[e
r,s
e
s,k
] by requiring
that x
i,j
<e
r,s
e
s,k
and
e
1,2
e
2,3
<e
2,3
e
3,4
<e
1,3
e
3,4
<e
1,2
e
2,4
< ···<e
n−2,n−1
e
n−1,n
< ···<e
1,2
e
2,n
.
Theorem 4.1. The elements of G constitute a minimal Groebner basis for
the module J with respect to the above monomial ordering.
Proof. To prove that G = {g
1
,...,g
c
} is a Groebner basis for J, it is enough
to check that all the Spolynomials S(g
i
,g
j
) either vanish or reduce to zero
modulo G, see [CLO]. Recall that the syzygy polynomial of f and g is
S(f,g)=
LCM(in(f), in(g))
in(f)
· f −
LCM(in(f), in(g))
in(g)
· g,
and that f reduces to zero modulo G if there is f can be written as f = a
1
g
1
+
a
2
g
2
+ ···+ a
c
g
c
, with deg(a
l
g
l
) ≤ deg f . There is an even stronger criterion
(see [BW]): The set G = {g
1
,...,g
c
} is a Groebner basis if, for all i, j, either
S(g
i
,g
j
) = 0, or S(g
i
,g
j
)=a
1
g
1
+ ···+a
c
g
c
and in(a
l
g
l
) < LCM(in(g
i
), in(g
j
)).
If g
i
and g
j
are monomials, then clearly S(g
i
,g
j
) = 0. Thus, all the S
polynomials from G
3
∪G
5
vanish.
If the initial terms of g
i
and g
j
have no variables in common, then S(g
i
,g
j
)
reduces to 0 mod G (“Buchberger’s First Criterion”). Thus, all the following
Spolynomials vanish modulo G: those from G
1
, those involving elements from
G
3
, those of the form S(g
r,s,k
,g
i,j,p,q
r,s,k
), and those of the form S(g
i,j,q
r,s,k
,g
l,l,q
r,s,k
) with
l 6= i, l 6= j.
Next, we ﬁnd standard representations for the Spolynomials from G
1
∪G
3
∪
G
4
∪G
5
not covered by the arguments above.
S(g
r,s,k
,g
r,k,q
r,s,k
)=−x
s,k
· g
q
r,s,k
+(x
r,s
− x
k,q
) · g
r,r,q
r,s,k
+(1− x
k,q
) · g
s,r,q
r,s,k
+(−1+x
r,s
+ x
s,k
− x
s,k
x
k,q
) · g
r,s,q
r,s,k
+ g
k,r,q
r,s,k
+(−1+x
r,q
+ x
s,q
) · g
r,k,q
r,s,k
,
S(g
q
r,s,k
,g
r,k,q
r,s,k
)=x
k,q
· g
q
r,s,k
+(1− x
k,q
) · g
s,k,q
r,s,k
+ g
k,k,q
r,s,k
+(1− x
s,q
− x
k,q
) · g
r,k,q
r,s,k
,
S(g
q
r,s,k
,g
k,r,p,q
r,s,k
)=(1− x
s,q
− x
k,q
) · g
k,r,p,q
r,s,k
+(1− x
k,q
) · g
k,s,p,q
r,s,k
+ g
k,k,p,q
r,s,k
,
THE CHEN GROUPS OF THE PURE BRAID GROUP 59
S(g
r,k,q
r,s,k
,g
s,k,q
r,s,k
)=x
r,q
· (g
s,k,q
r,s,k
− g
k,s,q
r,s,k
),
S(g
r,k,q
r,s,k
,g
k,r,p,q
r,s,k
)=−x
r,q
· g
k,k,p,q
r,s,k
.
It is readily veriﬁed that all these polynomials reduce to 0 modulo G. The
remaining Spolynomials are obtained from the ones above by permuting indices.
We are left with computing the Spolynomials involving at least one element
from G
2
. For simplicity, let us write a typical element in G
2
as ge
1
+ xe
2
, where
e
1
= e
r,s
e
s,k
, e
2
= e
t,u
e
u,q
, with {t, u}⊂{r, s, k}, and x = x
i,j
is a (linear)
monomial with indices {i, j}⊂{r, s, k, q}, but {i, j}6⊂{t, u, q}. Since e
1
<e
2
,
we have
S(ge
1
+ xe
2
,g
0
e
1
+ x
0
e
2
)=(x
0
g − xg
0
)e
1
,
and this clearly reduces to 0 mod G by Proposition 3.7.
Now let fe
2
be an arbitrary element from G∩R · e
t,u
e
u,q
. Notice that the
polynomial f does not involve the variable x. Hence:
S(ge
1
+ xe
2
,fe
2
)=in(f) · (ge
1
+ xe
2
) − x · fe
2
= fge
1
− (f − in(f)) · (ge
1
+ xe
2
),
where fge
1
can be written as an Rcombination of elements in G by Proposi
tion 3.7. We claim that S(ge
1
+xe
2
,fe
2
) satisﬁes the criterion mentioned above.
There are two possibilities to consider: If in(ge
1
+ xe
2
)=ge
1
, then we are
done, since ge
1
and fe
2
involve separate variables, as can be easily checked. If
in(ge
1
+ xe
2
)=xe
2
, then LCM(in(ge
1
+ xe
2
), in(fe
2
)) = in(f)xe
2
, in(fge
1
)=
in(f)in(g)e
1
< in(f)xe
2
, and in(f − in(f)) · (ge
1
+ xe
2
)=in(f − in(f ))xe
2
<
in(f)xe
2
. This proves the claim.
The Groebner basis G is minimal in the sense that each of its elements has
leading coeﬃcient 1, and, for all g
i
∈G, in(g
i
) /∈hin(G−{g
i
})i. ¤
5. The Hilbert Series
We now pass from the completion
b
Λ
∼
=
Z[[x
i,j
]] to the associated graded ring
gr(
b
Λ)
∼
=
Z[x
i,j
], and compute the Hilbert series of the graded module gr(
b
B)
associated to
b
B. We ﬁrst ﬁnd a minimal Groebner basis H for J := lt(J), the
module over R = Z[x
i,j
] of leading forms of J.
Let H = H
1
∪H
3
∪H
4
∪H
5
∪H
2
, where
H
1
=
½
h
r,s,k
= lt(g
r,s,k
)=(x
r,s
+ x
r,k
+ x
s,k
)e
r,s
e
s,k
,
h
q
r,s,k
= lt(g
q
r,s,k
)=(x
r,q
+ x
s,q
+ x
k,q
)e
r,s
e
s,k
for k<q≤ n
¾
,
H
3
= { h
p,q
r,s,k
= lt(g
p,q
r,s,k
)=x
p,q
e
r,s
e
s,k
for k<p<q≤ n } ,
H
4
=
½
h
i,i,q
r,s,k
= lt(g
i,j,q
r,s,k
)=x
i,q
(x
t,k
− x
j,q
)e
r,s
e
s,k
for i, j ∈{r, s}, {j, t} = {r, s}, and k<q≤ n
¾
,
60 DANIEL C. COHEN AND ALEXANDER I. SUCIU
H
5
=
½
h
i,j,p,q
r,s,k
= lt(g
i,j,p,q
r,s,k
)=x
i,p
x
j,q
e
r,s
e
s,k
for i, j ∈{r, s} and k<p<q≤ n
¾
,
H
2
=
x
s,k
e
p,r
e
r,s
+ x
p,r
e
r,s
e
s,k
, −x
s,k
e
r,p
e
p,s
+ x
r,p
e
r,s
e
s,k
,
x
r,k
e
p,r
e
r,s
+ x
p,s
e
r,s
e
s,k
, −x
p,k
e
r,p
e
p,s
+ x
p,s
e
r,s
e
s,k
,
x
p,k
e
p,r
e
r,s
+ x
p,k
e
r,s
e
s,k
, −x
r,k
e
r,p
e
p,s
+ x
p,k
e
r,s
e
s,k
,
x
p,k
e
r,s
e
s,p
+ x
p,k
e
r,s
e
s,k
,x
s,k
e
r,s
e
s,p
+ x
r,p
e
r,s
e
s,k
,
x
r,k
e
r,s
e
s,p
+ x
s,p
e
r,s
e
s,k
for 1 ≤ p<k, p/∈{r, s}
,
for 1 ≤ r<s<k≤ n. Note that H
1
= lt(G
1
), H
3
= lt(G
3
), H
4
⊂ lt(G
4
), and
H
5
⊂ lt(G
5
). Also observe that H
2
= lt(G
2
) consists of the leading terms of G
2
,
but with slightly diﬀerent indexing.
Proposition 5.1. The elements of H constitute a minimal Groebner basis
for the module J = lt(J).
Proof. By Theorem 4.1 and the remarks in section 2, the elements of the
set lt(G) form a Groebner basis for J . However, this basis is not minimal.
Compute:
lt(g
r,k,q
r,s,k
)=x
r,q
· h
r,s,k
− x
r,q
· h
q
r,s,k
− h
r,s,q
r,s,k
− h
r,r,q
r,s,k
,
lt(g
s,k,q
r,s,k
)=x
s,q
· h
r,s,k
− x
s,q
· h
q
r,s,k
− h
s,r,q
r,s,k
− h
s,s,q
r,s,k
,
lt(g
k,s,q
r,s,k
)=(x
r,k
− x
s,q
) · h
q
r,s,k
− h
r,s,q
r,s,k
− h
s,s,q
r,s,k
,
lt(g
k,r,q
r,s,k
)=(x
s,k
− x
r,q
) · h
q
r,s,k
− h
s,r,q
r,s,k
− h
r,r,q
r,s,k
,
lt(g
k,k,q
r,s,k
)=x
k,q
· h
r,s,k
+ h
r,r,q
r,s,k
+ h
s,s,q
r,s,k
+ h
r,s,q
r,s,k
+ h
s,r,q
r,s,k
+(x
r,q
+ x
s,q
− x
r,k
− x
s,k
− x
k,q
) · h
q
r,s,k
,
lt(g
k,s,p,q
r,s,k
)=x
s,q
· h
p
r,s,k
− h
r,s,p,q
r,s,k
− h
s,s,p,q
r,s,k
,
lt(g
k,r,p,q
r,s,k
)=x
r,q
· h
p
r,s,k
− h
r,r,p,q
r,s,k
− h
s,r,p,q
r,s,k
,
lt(g
s,k,p,q
r,s,k
)=x
s,p
· h
q
r,s,k
− h
s,r,p,q
r,s,k
− h
s,s,p,q
r,s,k
,
lt(g
r,k,p,q
r,s,k
)=x
r,p
· h
q
r,s,k
− h
r,r,p,q
r,s,k
− h
r,s,p,q
r,s,k
,
lt(g
k,k,p,q
r,s,k
)=x
k,p
· h
q
r,s,k
+ h
r,s,p,q
r,s,k
+ h
s,r,p,q
r,s,k
+ h
r,r,p,q
r,s,k
+ h
s,s,p,q
r,s,k
− (x
r,q
+ x
s,q
) · h
p
r,s,k
.
Since the elements of H have pairwise distinct initial terms, and each has leading
coeﬃcient 1, they form a minimal Groebner basis. ¤
In order to facilitate the Hilbert series computation, we now deﬁne a ﬁltration
of the module J = lt(J). For each k,3≤ k ≤ n, let J
k
= J∩R{e
r,s
e
s,j

1 ≤ r<s<j≤ k}, and let Q
k
= J
k
/J
k−1
. Note that J
k
∩ R[k − 1] = J
k−1
,
where R[]=R{e
r,s
e
s,j
 1 ≤ r<s<j≤ }. A routine diagram chase
yields an inclusion Q
k
→ R[k]/R[k − 1], so we view Q
k
as a submodule of
R{e
r,s
e
s,k
 1 ≤ r<s<k}.
THE CHEN GROUPS OF THE PURE BRAID GROUP 61
For each ﬁxed {r, s} with r<s<k, let Q
k
r,s
= Q
k
∩ R · e
r,s
e
s,k
. We view the
module Q
k
r,s
as an ideal in R in the obvious manner. We then have
Proposition 5.2.
(i) Q
k
=
M
1≤r<s<k
Q
k
r,s
.
(ii) The Hilbert series, H(gr(
b
B),t), of the graded module gr(
b
B) is given by
H(gr(
b
B),t)=
n
X
k=3
H(Q
k
,t).
Furthermore, the Hilbert series of the module Q
k
is given by
H(Q
k
,t)=
X
1≤r<s<k
H(Q
k
r,s
,t),
the sum of the Hilbert series of the ideals Q
k
r,s
.
Thus the problem is reduced to computing the Hilbert series of the ideals
Q
k
r,s
. First we ﬁnd minimal Groebner bases for these ideals. Suppressing the
index e
r,s
e
s,k
, let
H
k
r,s
=
h
r,s,k
= x
r,s
+ x
r,k
+ x
s,k
,
h
q
r,s,k
= x
r,q
+ x
s,q
+ x
k,q
for k<q≤ n,
h
i,j,q
r,s,k
= x
i,q
(x
t,k
− x
j,q
) for i ∈{r, s}, {j, t} = {r, s}, k<q≤ n,
h
i,j,p,q
r,s,k
= x
i,p
x
j,q
for i, j ∈{r, s} and k<p<q≤ n,
h
p,q
r,s,k
= x
p,q
for p/∈{r, s, k}, p<q, and k ≤ q ≤ n
.
Proposition 5.3. The elements of H
k
r,s
constitute a minimal Groebner basis
for the ideal Q
k
r,s
.
Proof. This follows from Proposition 5.1 and the deﬁnition of the ideal
Q
k
r,s
. ¤
Proposition 5.4. For 1 ≤ r<s<kand 1 ≤ u<v<k, we have
H(Q
k
r,s
,t)=H(Q
k
u,v
,t).
Proof. The automorphism of the polynomial ring R = Z[x
i,j
], 1 ≤ i<j≤
n, deﬁned by interchanging the pairs of indeterminates {x
r,s
,x
u,v
}, {x
r,l
,x
u,l
},
and {x
s,l
,x
v,l
} for each l, k ≤ l ≤ n, maps Q
k
r,s
isomorphically onto Q
k
u,v
. ¤
Proposition 5.5. For 1 ≤ r<s<k, the Hilbert series of the ideal Q
k
r,s
is
given by
H(Q
k
r,s
,t)=
1+2(n − k)t − (n − k)t
2
(1 − t)
2
.
Proof. By the above result, it suﬃces to consider Q
k
1,2
. This ideal is gener
ated by the elements of the set H
k
1,2
.
62 DANIEL C. COHEN AND ALEXANDER I. SUCIU
Deﬁne a linear coordinate change by
y
i,j
=
x
1,2
+ x
1,k
+ x
2,k
if (i, j)=(1, 2),
x
1,q
+ x
2,q
+ x
k,q
if (i, j)=(k,q), k<q≤ n,
x
i,j
otherwise.
In Z[y
i,j
], 1 ≤ i<j≤ n, the indeterminates y
p,q
, for (p, q) 6=(1,l), (2,l) and
k ≤ l ≤ n, are generators of the ideal Q
k
1,2
. Thus, we may restrict our attention
to the ring Z[y
1,k
,y
2,k
,...,y
1,n
,y
2,n
], and consider the ideal µ
k
1,2
generated by
y
1,p
y
1,q
,y
1,p
y
2,q
,y
2,p
y
1,q
,y
2,p
y
2,q
for k<p<q≤ n,
(y
1,k
− y
2,q
)y
1,q
, (y
1,k
− y
2,q
)y
2,q
for k<q≤ n,
(y
2,k
− y
1,q
)y
1,q
, (y
2,k
− y
1,q
)y
2,q
for k<q≤ n.
Deﬁne another linear coordinate change by
z
1,k
= y
1,k
−
n
X
p=k+1
y
2,p
,z
2,k
= y
2,k
−
n
X
p=k+1
y
1,p
, and z
i,j
= y
i,j
otherwise.
Then in Z[z
1,k
,z
2,k
,...,z
1,n
,z
2,n
], the ideal µ
k
1,2
is generated by
z
1,p
z
1,q
,z
1,p
z
2,q
,z
2,p
z
1,q
,z
2,p
z
2,q
for k<p<q≤ n,
z
1,q
(z
1,k
+
n
X
p=k+1
p6=q
z
2,p
),z
1,q
(z
2,k
+
n
X
p=k+1
p6=q
z
1,p
) for k<q≤ n,
z
2,q
(z
1,k
+
n
X
p=k+1
p6=q
z
2,p
),z
2,q
(z
2,k
+
n
X
p=k+1
p6=q
z
1,p
) for k<q≤ n.
Since z
1,q
z
1,p
, z
1,q
z
2,p
, z
2,q
z
1,p
, z
2,q
z
2,p
are generators of µ
k
1,2
for k<p<q≤ n,
we observe that the ideal µ
k
1,2
is in fact generated by
z
1,p
z
1,q
,z
1,p
z
2,q
,z
2,p
z
1,q
,z
2,p
z
2,q
for k ≤ p<q≤ n.
We now ﬁnd the Hilbert series of µ
k
1,2
. By the above remarks, we have
H(µ
k
1,2
,t)=H(Q
k
1,2
,t). As µ
k
1,2
is a monomial ideal, the m
th
term of the Hilbert
series of µ
k
1,2
is given by number of monomials in Z[z
1,k
,z
2,k
,...,z
1,n
,z
2,n
] of to
tal degree m not in µ
k
1,2
. Evidently, these monomials are of the form z
α
1,q
z
m−α
2,q
,
k ≤ q ≤ n,0≤ α ≤ m. Hence we have
H(µ
k
1,2
,t)=1+
X
m≥1
(m + 1)(n − k +1)t
m
=
1+2(n − k)t − (n − k)t
2
(1 − t)
2
. ¤
Combining the above results, we obtain
THE CHEN GROUPS OF THE PURE BRAID GROUP 63
Theorem 5.6. The Hilbert series of the graded module gr(
b
B) is given by
H(gr(
b
B),t)=
n
X
k=3
µ
k − 1
2
¶
·
1+2(n − k)t − (n − k)t
2
(1 − t)
2
=
µ
n +1
4
¶
·
1
(1 − t)
2
−
µ
n
4
¶
. ¤
Note added in proof. We have recently found a counterexample to the conjecture stated in
the Introduction, see [CS3].
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Daniel C. Cohen, Department of Mathematics, University of California, Davis,
CA, 95616, USA
Email address: cohen@math.ucdavis.edu
Alexander I. Suciu, Department of M athematics, Northeastern University,
Boston, M A, 02115, USA
Email address: alexsuciu@neu.edu