The Chen Groups of the Pure Braid Group

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DOI: 10.1090/conm/181/02029
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Abstract
The Chen groups of a group are the lower central series quotients 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 Gröbner basis algorithm. This result shows that, for n ≥ 4, the group P_n is not a direct product of free groups.
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
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
0271-4132/95 $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
k2
=
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
=
n1
Y
i=1
(1 it)=
X
j0
(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 fiber-type hyperplane arrangement A (see [FR1], [OT]), we
have the following.
Theorem 1.3 ([FR1], [FR2]). Let A be a fiber-type 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
j0
(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 fiber-type arrangement A is an
Eilenberg-MacLane space. The factorization of the Poincar´e polynomial of the
group of a fiber-type arrangement may be deduced from the fact that fiber-type
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 fiber-type
arrangement may be realized as an iterated semi-direct 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
n1
o ···o F
2
o F
1
, see [Bi], [Ha].
For a detailed discussion of the homology of iterated semi-direct products of free
groups, see [CS1].
Both the direct product Π
n
= F
n1
×···×F
2
× F
1
, and the semi-direct
product P
n
, may be realized as the fundamental groups of the complements of
(different) fiber-type 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 semi-direct
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
n1
×···×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 3-arrangements 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 fiber-type, 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 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
XL
2
{µ(X)+1},
the rank of the k
th
Chen group of G(A) is given by
θ
k
(G(A)) =
X
XL
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
XL
2
{µ(X) 1}. For the braid arrangement, this degree is 1, while for the
arrangement with group G
n
, the degree is n 2. An arbitrary degree may
be realized by the group of a central 2-arrangement 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 I-adic completion of the Λ-module 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 Λ-module B is the first homology group of
f
M
n
,
the maximal abelian cover of the complement of the braid arrangement. A free
Λ-presentation of B is obtained by comparing the chain complex of
f
M
n
with that
of the universal (abelian) cover of the N-torus, 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
k0
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 I-adic
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 I-adic filtration on Λ, and its associated graded ring,
gr(Λ) =
L
k0
I
k
/I
k+1
. Also, consider the m-adic filtration on Z[[x
i
]], where
m = hx
1
,...,x
N
i, and its associated graded ring, gr(Z[[x
i
]]) =
L
k0
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 Λ-module, called the Alexander invariant of G, see [Ma], [MT],
[Hi], [L].
Let
b
B be the I-adic completion of B. Let
gr(B)=
M
k0
I
k
B/I
k+1
B and gr(
b
B)=
M
k0
m
k
b
B/m
k+1
b
B
be the graded modules associated to the I-adic filtration on B and the m-adic
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 I-adic 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 Hilbert-Serre 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
Λ
b
B 0,
for the Λ-module 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 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 I-adic completion of B.
Simply take
b
Λ
a
b
b
Λ
b
b
B 0, where
b
is obtained from via the Magnus
embedding. Clearly, Im(
b
Ω) =
b
J, the completion of J. Since all the entries of the
matrix for
b
belong to the subring Z[x
i
] Z[[x
i
]], we may restrict
b
to a map
: 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 non-zero 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 N-torus, 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 Λ-module 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
,C
1
N
, and C
2
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
n1,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
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” Λ-free resolution of Z. The
terms of this resolution may be described as follows: C
0
0
,C
0
1
N
is a free
Λ-module with basis {e
i,j
| 1 i<j n}, and for k>1, C
0
k
=
V
k
C
0
1
(
N
k
)
.
The differentials are given by
d
k
(e
i
1
,j
1
∧···∧e
i
k
,j
k
)=
k
X
r=1
(1)
r1
(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 Λ-free 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
ψφ
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
= ψd
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 Λ-module, 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 = Ξ, where Ξ is a Λ-linear
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
)
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 lower-triangular 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
q
r,s,k
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
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 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 Λ-combination 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 I-adic 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
b
Λ
(
n
3
)
b
B 0, where
b
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
to a map : R
(
N
3
)
R
(
n
3
)
. Let J = Im()=
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 R-module 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
n1,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 right-most 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
n2,n1
e
n1,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 S-polynomials 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
S-polynomials 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 S-polynomials 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 S-polynomials are obtained from the ones above by permuting indices.
We are left with computing the S-polynomials 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 R-combination 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
k1
. Note that J
k
R[k 1] = J
k1
,
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
1r<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
1r<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
mα
2,q
,
k q n,0 α m. Hence we have
H(µ
k
1,2
,t)=1+
X
m1
(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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64 DANIEL C. COHEN AND ALEXANDER I. SUCIU
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Daniel C. Cohen, Department of Mathematics, University of California, Davis,
CA, 95616, USA
E-mail address: cohen@math.ucdavis.edu
Alexander I. Suciu, Department of M athematics, Northeastern University,
Boston, M A, 02115, USA
E-mail address: alexsuciu@neu.edu
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