On Successive Quotients of Lower Central Series ... - MIT Mathematics
On Successive Quotients of Lower Central Series Ideals for Finitely Generated Algebras Kathleen Zhou Under the direction of Teng Fei and Dr. Pavel Etingof (MIT)
Abstract Ni (A)
This paper examines the behavior of the successive quotients ideals
Mi (A)
Li (A)
by
of a nitely generated associative algebra
A
over
Z.
We de ne the lower central series
L1 (A) = A, Li+1 (A) = [A, Li (A)], Mi (A) = A · Li (A) · A,
We decompose the
Ni
of the lower central series
and
Ni (A) = Mi (A)/Mi+1 (A).
into its free and torsion components using the structure theorem of nitely
generated abelian groups, and we examine patterns in the ranks and torsion of various homogeneous relations, including and
xm + y m .
x2
in multiple variables,
q -polynomial
a complete description of
N2
for
for algebras with
relation
yx − qxy ,
In order to do this, we create data tables with the ranks and torsion of various
previously uncalculated, based on calculations done in the program
ranks of
Ni
Ni
for the
q -polynomial
algebra,
Magma.
Zhx, yi (yx − qxy)
Ni ,
This paper includes and a proof for the
Ahx, yi/(xm +y m ), which provides insight into how changing the coe cient or degree
of a relation aects rank and torsion, as well as general patterns for which primes appear in torsion.
1
1
Introduction Li (A),
For the past several years, algebraists have been studying the lower central series are successive subspaces of an associative algebra
A
formed from the commutators of
A.
which
Thus, the
lower central series and its related objects can be used to measure the noncommutativity of algebras. We consider the successive quotients of the two-sided ideals quotients
Ni .
We study the structure and properties of
Mi
Ni
generated by
Li ,
and we call these
to better understand the structure of
associative algebras. Lower central series quotients of free associative algebras were rst studied by Feigin and Shoikhet [4].
They looked at the successive quotients
was an isomorphism between the space
A/M3 (A)
isomorphism is essential in proving the pattern of
Bi = Li /Li+1 ,
and concluded that there
and the space of even dierential forms.
N2 (A2 /(xm + y m ))
found in this paper.
This The
study of quotients continued with Etingof, Kim, and Ma [3], who completely described the quotient
A/Mi (A)
for
i = 4.
The study of at
B2
Bi
was continued in the work of Balagovi¢ and Balasubramanian [1], who looked
in the quotient of a free algebra.
B2 (A2 /(xd + y d )),
In particular, they provided a complete description of
which is similar to the results found in this paper for
While the structure of
Bi
studied. Kerchev [5] studied
N2 (A2 /(xm + y m )).
has been studied in multiple papers, the quotients
Ni
for free algebras and computed
Ni
even for free algebras, and
have been less
Ni (An ) for several values of i and n.
However, there is still much work to be done in studying the structure of has never been calculated for
Ni
Ni
Ni .
In particular, torsion
have not been studied for algebras
with relations. In this paper, we study the behavior of
Ni
for an associative algebra
Ni
Zhx, yi
with various re-
lations.
We also compute the ranks and torsion of various
using a computer program called
Magma.
The process of collecting data is explained in Section 3. Several patterns suggested by the
data are contained in Section 4. In Section 5 we provide a complete description of with the relation structure of
N2
yx − qxy = 0,
with the relation
also known as
xm + y m
q -polynomial
Ni
for algebras
algebras. The results concerning the
is proven in Section 6.
We begin with preliminary background to better understand the algebraic objects of study.
2
2
Preliminary Background
2.1 Associative Algebras and Their Lower Central Series De nition 2.1. operation by
1,
Let
A
(a, b) 7→ a · b,
then
A
is a
be a vector space over a eld which is also written as
ab.
If
k
with a bilinear associative multiplication
A
also has a multiplicative identity, denoted
unital associative algebra.
A free algebra is a unital associative algebra that is generated by a set of generators with no relation. In this paper, we are interested in associative algebras that are not necessarily free, more precisely, algebras with homogeneous relations. An algebra of
A by the ideal generated by the relation P .
This bracket operation satis es
[a, a] = 0
A/hP i
will denote the quotient algebra
We de ne a bracket operation on
A by [a, b] = a·b−b·a.
and the Jacobi identity:
[a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0. A vector space with a bilinear bracket operation are satis ed is called a
[a, b] such that the Jacobi identity and [a, a] = 0
Lie algebra, so any associative algebra is also a Lie algebra with [a, b] =
a · b − b · a.
De nition 2.2. Let A be a Lie algebra. and
Li+1 (A) = [A, Li (A)],
span([c, d]) such that abbreviate
Li (A)
as
De ne a series of Lie ideals inductively such that
where the bracket of two subspaces
c ∈ C, d ∈ D.
C
This series of Lie ideals is the
and
[C, D] =
lower central series of A.
Z
by using
Z-modules
We
(i.e. Abelian groups) instead of vector
spaces. Similarly, we can use this de nition over any commutative ring Denote the two-sided ideals generated by each
It is easy to see that
De nition 2.4.
is de ned as
Li .
We may make this de nition over
De nition 2.3.
D
L1 (A) = A
Mi = A · L i
De ne the
Ni
Li
by
R.
Mi ,
i.e.
Mi = A · Li · A.
. Using this de nition, we can de ne the quotients
to be the successive quotients
Ni .
Mi /Mi+1 .
Now we introduce the idea of grading, which is crucial to representing data eectively.
De nition 2.5.
Let
A
be a module over a commutative ring
a direct sum decomposition into submodules then
A
is a
graded algebra.
M
Ai .
If
A
k.
The module
A
is
is an algebra such that
graded if A has Ai · Aj ⊂ Ai+j ,
i≥0
3
Example 1.
the grading is by the degree of the polynomial. grading from part of
Ni
k[x1 , . . . , xn ],
The simplest example of a graded algebra is a polynomial ring
A.
We observe that
Our study is simpli ed if we look at
at degree
d
will be denoted as
Ni [d],
Ni
Ni
where
is graded, as it inherits its
by its degree, which is denoted by
which is a nitely generated
d.
The
k -module.
2.2 Torsion and Class cation of Finitely Generated Abelian Groups De nition 2.6. positive integer
a
is an
An element
n.
n-torsion
a
of an Abelian group
G
torsion element if n · a = 0 for some
is a
Conventionally, 0 is also considered a torsion element. In this case, we say that
G
element. All of the torsion elements in
form a subgroup of
G.
To clarify the concept of torsion, we oer a simple example.
Example 2.
Consider the group
G = Z6 = Z2 ⊕ Z3 .
This group has
2-torsion, 3-torsion,
and
6-
torsion. 0, 2, and 4 are 3-torsion elements, 0, 3 are 2-torsion elements, and all elements are 6-torsion elements. All 2 and 3-torsion elements are also 6-torsion elements.
The idea of torsion becomes especially important due to the Structure Theorem of Finitely Generated Abelian groups, which states that groups can be separated into their free and torsion components.
Theorem 2.1 (Structure Theorem of Finitely Generated Abelian Groups). Every nitely generated Abelian group G is isomorphic to a nite direct sum of in nite cyclic groups and cyclic groups of order pn , for various primes p. This decomposition is unique up to order of summands. The theorem can be restated as
G∼ = F ⊕ T, where
F
is the free component, which is isomorphic to
Zr
for some
component, consisting of a nite sum of cyclic groups of order
r
is called the
pn
r ∈ Z,
and
T
for various primes
is the torsion
p.
In this case,
rank of the free component, known simply as rank.
The goal of the project is to determine the structure in the ranks and torsion of
Ni
for algebras over
Z by studying patterns
Ni .
2.3 Sample Calculations for Zhx, yi We provide a set of sample calculations to illustrate how consider the free associative algebra
Zhx, yi.
L i , Mi ,
and
Ni
are constructed.
We
Because it is not known whether torsion exists in
4
Ni
for free algebras generated by 2 variables, we focus on calculating the ranks of
Li
calculate the bases of the rst few By de nition,
L 1 = A,
L1
so
is spanned by
L2
[x, xy] + [x, yx] = [y, x2 ].
from the basis.
Similarly,
remain in the basis of
The results of the basis of
Li [d] L1 L2 L3
Thus,
and
[y, x2 ]
Li
L3 [3]
Calculating
L1
L2 [2]
However, both
[x, x2 ]
and
L2
is
is 1, the
is
[x, y],
Thus, the potential basis vectors are
[y, y 2 ].
[y, y 2 ]
as
[x, xy],
are 0. In
is not linearly independent and can be removed
[y, xy] + [y, yx] = [x, y 2 ],
L2 [3].
The next row
is 2. The only term in the basis of
[L1 [1], L1 [2]].
[y, xy], [x, yx], [x, y 2 ], [y, x2 ], [y, yx], [x, x2 ], addition,
x, y .
As the minimum degree of a non-trivial part of
minimum degree of a non-trivial part of
[y, x] = −[x, y]. L2 [3]
First, we
in low degrees.
spanned by all the monomials in
[A, L1 ].
formed from the set of all
Ni .
[x, y 2 ]
and
can be eliminated.
is more straightforward, as
Only 4 terms
L3 [3] = [L1 [1], L2 [2]].
can be found in Table 1, where the top row indicates the degree
1
2
3
x, y 0
x2 , xy , yx, y 2 [x, y]
0
0
x3 , x2 y , xyx, yx2 , y 3 , y 2 x, yxy , xy 2 [x, xy], [y, xy], [x, yx], [y, yx] [x, [x, y]], [y, [x, y]]
Li The Mi can be constructed from Li , as Mi = A · Li · A = A · Li .
d.
Table 1: Bases for
as
Li [i]
Li [i] = Mi [i],
must be multiplied by scalars on both sides for the minimum non-trivial degree
M1 = A ,
and
M2 [2]
is also easy to compute.
M2 [3] = L1 [0] · L2 [3] + L1 [1] · L2 [2]. [y, xy], [x, yx], [y, yx], x[x, y], contains
and
[x, xy], [y, xy], [x, yx], Mi [d] M1 M2 M3
M2 [3]
Then,
M3 [3] = L3 [3],
2
3
0
x2 , xy , yx, y 2 [x, y]
0
0
x3 , x2 y , xyx, yx2 , y 3 , y 2 x, yxy , xy 2 [x, xy], [y, xy], [x, yx], [y, yx] [x, [x, y]], [y, [x, y]]
=
are
[x, xy],
Mi
which are the cardinalities of the basis of each
rank(Mi [d])
M2 [3]
so Table 2 is complete.
x, y
rank(Ni [d])
Thus,
Eliminating linearly dependent terms, the basis of
[y, yx].
Ni [d],
i.
is slightly more complicated.
1
Now we calculate the ranks of
Ni = Mi /Mi+1 ,
M2 [3]
Therefore, the possible terms in the basis of
y[x, y].
and
Calculating
Table 2: Bases for
Ni [d]
By this de nition,
− rank(Mi+1 [d]).
Ni [d].
As
Thus, computing the ranks of each
becomes a simple subtraction problem. The ranks are shown in Table 3.
Ni [d] N1 N2
1
2
3
2
3
4
0
1
2
Table 3: Free Ranks for
Ni 5
3
Data Collection
We rst compile data tables of the ranks and torsion of
Ni
for various relations. By changing the
number of variables, coe cients, or degree of the relations, we can nd patterns and form conjectures about the behavior of the ranks and torsion of
Ni .
Data is collected by running computations in
Magma
[2]. This code was run for many relations
over the integers, and the outputs were then organized into table form by grading. The left column displays the
Ni ,
while the top row is organized by grading (degree). Each term in the data table
includes the rank, which is displayed outside of the parentheses, and the torsion, which is displayed within the parentheses. Here, we provide an example of how the data was processed and organized into tables. example below shows how to format the output in
Example 3.
Magma
The
to a data table.
The output code
$N_ 4 $ & 8 16(Abelian Group isomorphic to Z/2 + Z/2 + Z/4 + Z/4) is expressed as
16(22 · 42 )
for
N4 [8]
in a table.
The expression has a rank of 16, which is the last numbered output before the parentheses. The torsion is slightly more di cult to express. The output data in the parentheses represents the direct sum of many cyclic groups. While prime power cyclic groups do not need to be further decomposed, other groups can be decomposed into coprime components. For example, into
Z3 × Z4 × Z5
by the result in group theory that states
Z60
Zmn = Zm × Zn
for
can be decomposed
m, n
coprime. The
data is decomposed into prime powers for the tables. These components can then be combined through exponent rules (eg. Z/2 + Z/2 given in the output is expressed in the table as is
22 ).
Thus, the nal form of the term in the data table for
N4 [8]
16(22 · 42 ). We look for patterns within these data tables, then try to prove them.
4
Observations of Patterns in Data
After compiling tables of algebras with various relations, we nd several patterns in the ranks and torsion of the
Ni . 6
4.1 Patterns in Zhx1 , . . . , xn i/(x21 ) Interesting patterns arise for the algebra with the relation
x21 = 0 with a number of variables, shown
in Table 4. With two variables, it seems that the torsion and ranks of that is to say, the torsion and ranks do not chnage as
Ni [i + 2]
the stabilization of the ranks from
Ni [d] N2 N3 N4 N5 N6 N7 N8 N9
d
Ni
stabilize rather quickly
increases. Reading across the rows shows
and a stabilizing torsion.
2
3
4
5
6
7
8
9
1
1(2)
1(2)
1(2)
1(2)
1(2)
1(2)
1(2)
0
2
3(2)
3(2 )
3(2 )
3(2 )
3(2 )
3(2 )
4 3(2 )
5 3(2 )
5 3(2 )
5 3(2 )
7(2 )
7(2
2
0
0
2
2 3(2 )
0
0
0
4
2
3
2
7
2
9
10
0
0
0
0
5
5 9(2 )
0
0
0
0
0
9
· 3) 12 9(2 · 3 · 5) 7 18(2 )
0
0
0
0
0
0
12
· 3) · 3 · 5) 19 · 32 · 5) 19(2 12 25(2 )
0
0
0
0
0
0
0
20
2 Table 4: x1
= 0,
· 3)
2
7(2
7(2
16 9(2
two variables
This yields a conjecture about the stabilization of the ranks and torsion of
Ni .
Conjecture 4.1.1. For Zhx,yi/(x2 ), Ni [j] ∼ = Ni [j + 1] for j ≥ 2i − 1. It would be interesting to know how soon the ranks stabilize for algebras with more generators, and which primes will ultimately appear in torsion.
4.2 Patterns in Zhx, yi/(yx − qxy) We notice that the ranks are non-zero only for the diagonal
Ni [i], and these ranks are equal to i − 1,
as seen in Tables 9 through 11, found in Appendix A.1. Additionally, xing a
Ni [d]
for
i < d
are the same:
(Zq−1 )d−1 .
Along the diagonal
pattern with more primes. We are able to completely describe the diagonal
i = d,
i = d,
Ni
4.3 Patterns in
the torsion in all
there is a more interesting
in this case, and we show that on
all primes will eventually appear, except those that divide
description and proof of the result on the torsion in
d,
Zhx, yi/(yx − qxy)
q.
A more detailed
are provided in Section 5.
Zhx,yi/(xm +y m )
By examining the algebras Zhx,yi/(xm +y m ), we discover several patterns occurring across tables in for
i = 2, 3, 4.
The complete set of data can be found in Tables 12 to 14 in Appendix A.2. For
the ranks follow a palindromic pattern similar to the one found in 5, where the left column indicates values of
B2
Ni
N2 ,
[1], which is shown in Table
m.
7
Proposition 4.1.
N2 [d]
2
2
1
3 4 5
3
4
5
6
1
2
1
1
2
3
2
1
1
2
3
4
3
7
8
2
1
N2 [d] with the relation xm + y m = 0 In the algebra Zhx, yi/(xm + y m ), rank(N2 [k]) = k − 1 for k < m, Table 5: Ranks for
rank(N2 [k])
=
2m − k − 1 for m ≤ k ≤ 2m − 2, and 0 for all other values of k . This proposition will be proven in Section 6. The patterns developing in the ranks of
N3
and
arithmetic sequences. The values for the ranks of showing values of
are displayed in Table 6, with the left column
N3 [d]
3
4
5
6
7
8
9
10
2
2
1
0
0
0
0
0
0
3
2
5
4
1
0
0
0
0
4
2
5
8
7
4
1
0
0
5
2
5
8
11
10
7
4
1
Conjecture 4.3.1. In the algebra
While
N3
are almost palindromic, and we see pseudo-
m.
Table 6: Ranks for
rank(N3 [k])
N4
N3 [d]
with the relation
Zhx, yi/(xm + y m ),
xm + y m = 0 = 3k − 7 for k ≤ m + 1,
rank(N3 [k])
= 6m − 3d + 1 for m + 1 < k < 2m + 1, and 0 for all other values of k .
N2
and
N3
seem to have easily generalizable patterns,
N4
is slightly more complicated.
The bolded numbers in Table 7 are the ones that remain consistent as
N4 [d]
4
2
2
3 4 5
5
6
7
8
9
d
increases.
10
3 7 4 3 8 13 10 4 3 8 14 19 16 10 4
Table 7: Ranks for
N4 [d]
with the relation
xm + y m = 0
Conjecture 4.3.2. The ranks of N4 will become stable outside of the diagonal d = m + 2, where d is the degree of the grading. We expect
rank(N4 (A2 /(x
m
+ y m ))[2m − k]) to stabilize for large m
and xed k ≥ 0. This rank vanishes for k < 0.
8
5
Complete Description of Ni (Zhx, yi/(yx − qxy))
We consider the speci c algebra
Zhx, yi
with the relation
yx − qxy ,
algebra. A clear pattern emerges in the ranks and torsion of the 10, which are located in Appendix A.1. The torsion in along the diagonal
i=d
Ni [d]
for
Ni ,
also known as a
q -polynomial
as shown in Tables 9 through
i < d−1
is
(Zq−1 )d−1 .
The torsion
has a more interesting pattern. To understand this pattern, we use a ner
grading on the degrees, de ning
x
to have degree
h1, 0i
and
y
to have degree
h0, 1i
where for degree
hu, vi, d = u + v .
Theorem 5.1. Let
A = Zhx, yi/(yx − qxy), where q ∈ Z, q 6= ±1. The
i = j . Otherwise, the rank is 0. The torsion in Ni [d], also written as M (Zq−1 )d−1 . Along the diagonal i = d − 1, Tor(Ni [d]) = Zq(u,v) −1 .
rank(Ni [j])
= i − 1 for
, for i < d − 1 is
Tor(Ni [d])
u+v=d
Note.
In Theorem 5.1 and its proof, the greatest common divisor of
Now we give some preliminary information for the proof. spaces
Lk
and
Mk .
We rst note that because
result of any bracket operation as a sum of (see Table 8 for element of
Lk
Lk [hu, vi]
De nition 5.1.
j 6= i
as
and
hu, vi
Ni [j] = i − 1
L1 L2 L3 L4
if
h0, 1i x
h0, 2i y2
0 0 0
is denoted as
(u, v).
First, we consider the bases of the
with some coe cients. We rst construct a table
keeping in mind that
yx = qxy ).
For
k > 1,
the basis
k xu y v . Su,v k xu y v Lk [hu, vi] ⊂ span Su,v
we include only the coe cients of the bases,
hv, ui
v
and
under the relation, we can express the
is the largest possible integer such that
k > 1,
and
k = u + v,
will be denoted by
k Su,v
In the table for
u < v,
on the diagonal
xu y v
yx = qxy
u
are symmetric. There is no torsion if
k , Su,v
u, v = 0.
.
and the terms in which
The rank of
Ni [j] = 0
if
j = i. h0, 3i y3
0
h1, 1i xy (q − 1)
0
0
0
h1, 2i xy 2 (q − 1) (q − 1)2
0
0
0
0
0
Table 8: Construction of We consider the bases along the diagonal
h0, 4i y4 0 0 0
h1, 3i xy 3 (q − 1) (q − 1)2 (q − 1)3
h2, 2i x2 y 2 (q 2 − 1) (q − 1) · (q 2 − 1) (q − 1)2 · (q 2 − 1)
Lk for the relation yx − qxy = 0 k = u + v and nd that there is
a pattern (Lemma
5.1).
9
Lemma 5.1. For the algebra Zhx, yi
k for k = u + v , (yx − qxy), with q ∈ Z and q 6= ±1, and Su,v
is k Su,v = (q − 1)u+v−2 · (q (u,v) − 1).
(5.1)
To prove this lemma, we include a few known facts in number theory:
Fact 1. (λm − 1, λn − 1) = λ(m,n) − 1 Fact 2. (a, b) = 1 =⇒ (a, bc) = (a, c) Fact 3. Proof.
i j , h h
= 1 ⇐⇒ (i, j) = h
We prove Lemma 5.1 by induction on
u, v .
The base case is satis ed, as
2 S1,1 = (q − 1)1+1−2 (q (1,1) − 1) = q − 1. Because
k satis es Lk [hu, vi] = [x, Lk [hu − 1, vi]] + [y, Lk−1 [hu, v − 1i]], Su,v k−1 k−1 k (q u − 1) . (q v − 1), Su,v−1 Su,v = Su−1,v
the recursive equation (5.2)
Assuming that
k−1 = (q − 1)u+v−3 (q (u−1,v) − 1), Su−1,v k−1 = (q − 1)u+v−3 (q (u,v−1) − 1), Su,v−1 we want to show that
k Su,v = (q − 1)u+v−3 (q (u−1),v − 1)(q v − 1), (q − 1)u+v−3 (q (u,v−1) − 1)(q u − 1) = (q − 1)u+v−2 (q (u,v) − 1). We can pull out the expression
(q − 1)u+v−3 ,
(5.3)
as it is common to both of the components of the
greatest common divisor. Thus, equation (5.3) is reduced to
k Su,v = (q − 1)u+v−3 (q (u−1,v) − 1)(q v − 1), (q (u,v−1) − 1)(q u − 1) .
(5.4)
We set the following:
q (u−1,v) − 1 = α, q v − 1 = β, q (u,v−1) − 1 = γ, q u − 1 = δ.
10
Using Facts 3 and 1, we have
(α, γδ) = q − 1 ⇐⇒
α γδ , q−1 q−1
=1
(5.5)
and
γδ α , q−1 q−1
= 1 =⇒
γδ ,β q−1
=
γδ αβ , q−1 q−1
(5.6)
We also note that
β,
γ γδ = 1 =⇒ β, = (β, δ). q−1 q−1 α γ By Fact 1, (α, γ) = q − 1, as ((u − 1, v), (u, v − 1)) = 1. Thus, = 1. , q−1 q−1 γδ α γδ (5.5), ,β = , β is true. q−1 q−1 q−1
By equation
= 1. By Fact 1, (β, δ) = (q (u,v) −1). Thus, by equa(β, γ) = q−1 by Fact 1, γ γδ αβ γδ (u,v) (β, δ) = β, δ = (q − 1). By equation (5.6), , = β, = q−1 q−1 q−1 q−1
Because tion (5.7),
γ β, q−1
(5.7)
q (u,v) − 1.
It follows by Fact 3 that
(αβ, γδ) = (q − 1)(q (u,v) − 1).
We can now prove Theorem 5.1.
Proof.
Mk [hu, vi]
We denote the basis of
the terms for
Lk
as
k xu y v , Tu,v
and note that for
on the diagonal are of the lowest possible degree. Thus,
constants on either side to form Now we consider
k Tu,v
Lk
as
must be multiplied by
Mk = A · Lk · A.
u + v > k.
for
k = Tk , k = u + v , Su,v u,v
We note that
Mu+v−1 [hu, vi] = Lu+v−1 [hu, vi] + x · Lu+v−1 [hu − 1, vi] + y · Lu+v−1 [hu, v − 1i]. Set
T1 xu y v
1i],
where
Thus,
to be the basis of
u+v−1 T1 = Su−1,v
(T1 , T2 )xu y v
1)u+v−2 xu y v
spans
k = (q − 1)k−1 Tu,v
if
spans
x · Lu+v−1 [hu − 1, vi] and T2 xu y v
and
u+v−1 T2 = Su,v−1 .
Mu+v−1 [hu, vi].
Mu+v−1 [hu, vi]
It is true that
It is known that
by Lemma 5.1 and
to be the basis of of
T1 xu y v , T2 xu y v
y · Lu+v−1 [hu, v −
span
Mu+v−1 [hu, vi].
(T1 , T2 ) = (q − 1)u+v−2 .
u+v−1 = (q − 1)u+v−2 . Tu,v
Then,
(q −
This suggests that
u + v > k.
Using this information, we can calculate the torsion
Ni .
Using the bases of
Mk ,
we divide to get
11
Tor (Nu+v−1 [hu, vi])
u+v xu y v Z · Tu,v
=
u,v−1 u v Z · Tu,v x y
= Zq(u,v) −1 .
Tor (Nk−1 [k])
Summing over all
M
=
u, v
yields
Zq(u,v) −1 .
u+v=k The ranks are easy to verify given the bases. The space
k k S1,k−1 xy k−1 , . . . , Sk−1,1 xk−1 y and above the diagonal
with
i = d,
Mk [k] is a free Abelian group with basis
q 6= ±1, while Mk+1 [k] = 0. Mk [k]
the ranks of
and
Nk [k] is free of rank k − 1.
Below
are the same, so the rank of
Nk [k]
So
Mk [k + 1]
is 0.
Corollary 5.1. All primes except those that divide q appear in the torsion of Ni [i + 1] for some i. Proof.
Given that the
that divide
q
Ni
has
q (u,v) − 1 -torsion, by Fermat's little theorem, all primes except those
will appear in the torsion of
Ni .
With Theorem 5.1 and Corollary 5.1, we now have a clearer idea of how the coe cients of a relation aect the ranks and torsion of
6
Ni .
Ranks of N2 (Zhx, yi/(xm + y m ))
We wish to nd the basis of
N2 (A2 /(xm +y m )) in order to prove Proposition 4.1, where A2 = Qhx, yi.
To do so, we use the short exact sequence First, we nd the generators of using the isomorphism
A/M3 .
0 → N2 → A/M3 → A/M2 → 0,
A = A2/(xm +ym ).
We then prove the linear independence of these generators by
A2 /M3 ∼ = Ωeven (Q2 )∗
the result for the ranks of
where
found in Feigin and Shoikhet's paper [4], thus proving
N2 (Zhx, yi/(xm + y m )).
6.1 Generators of A/M3 We consider
A = Qhx, yi/(xm + y m )
m − 1, 0 ≤ j
and
xi y j u
relations are satis ed in Because
xm + y m = 0
in
for
with
u = [x, y].
0 ≤ i, j ≤ m − 1
span
We want to show that
A/M3 .
the relation also holds in
for
0 ≤ i ≤
To do this, we show that the following
A/M3 : u2 = 0, [u, x] = [u, y] = 0, xm + y m = 0, A,
xi y j
and
xm−1 u = y m−1 u = 0.
A/M3 .
Lemma 6.1. In A/M3 , u2 = 0. Proof. u2 = [x, y] · [x, y] = [x, y] · (xy − yx).
12
Because
[x, y] · xy = [[x, y], x]y + x[x, y]y , u2 = [[x, y], x]y + x[x, y]y − [x, y]yx.
Because
x[x, y]y − [x, y]yx = −[[x, y]y, x], u2 = [[x, y], x]y − [[x, y]y, x].
Because
[[x, y], x]y ∈ M3
and
−[[x, y]y, x] ∈ M3 ,
the relation
u2 = 0
is satis ed in
A/M3 .
Lemma 6.2. In A/M3 , the relations [u, x] = [u, y] = 0 hold. Proof. [u, x] = [[x, y], x].
Because
[[x, y], x] ∈ M3 , [u, x] = 0
in
A/M3 .
Similarly,
[u, y] = 0.
Lemma 6.3. In A/M3 , xm−1 u = ym−1 u = 0. Proof.
We know that
We will show that each other by
0 = [xm + y m , x]
[x, y m ] = my m−1 u
[u, y] = 0,
because
in
A2 /M3
xm + y m = 0.
Additionally,
through induction. Because
[xm + y m , x] = [x, y m ]. u
and
y
commute with
the base case is satis ed:
0 = [x, y 2 ] = xy 2 − y 2 x = xy 2 − yxy + yxy − y 2 x = uy + yu = 2yu, since
[y, u] = 0 in A2 /M3 .
[x, y k ]
Now we assume that for some integer
k , [x, y k ] = ky k−1 u.
We can expand
as follows:
k
k
k
k−1 X
k
0 = [x, y ] = xy − y x = xy + k+1 ] Now we consider [x, y
0 = [x, y
k+1
=
xy k+1 k
xy +
]=
−
−y i xy k−i + y i xy k−i − y k x.
i=1 k+1 y . We expand and factor:
k−1 X
i
−y xy
k−i
i
+ y xy
k−i
! − y x y + y k xy − y k+1 x. k
i=1 Thus,
[x, y k+1 ] = [x, y k ]y + y k u = ky k u + y k u = (k + 1)y k u = 0 since
m
[y, u] = 0
in
A2 /M3 .
We have proved through induction that
is some positive integer,
xm−1 u = 0
y m−1 u = 0
is satis ed in
since we are working over
Q.
Because
Similarly,
is also satis ed.
From Lemma 6.1, 6.2, and 6.3, we know that
A/M3 .
A/M3 ,
[x, y m ] = my m−1 u = 0.
Since
u
commutes with
the end of all expressions in
x
and
A/M3 .
y,
xi y j
for
0 ≤ i < m and xi y j u for i, j < m − 1 span
we can assume without loss of generality that
Thus, the degree of
u
can either be 0 or 1, as for
u
appears at
a ≥ 2, ua = 0.
13
Calculating the number of generators for each degree, we nd that the dimension (from the data) for that degree equals the number of generators, predicted by Proposition 4.1. We now show the linear independence of these generators.
6.2 The Basis of A/M3 We consider the short exact sequence
0 → M2 /M3 → A/M3 → A/M2 → 0, and let
f
be the surjection
Q[x, y]/(xm + y m ).
A/M3 → A/M2 .
A/M2
is the abelianization of
Now we wish to prove that the generators
xi y j
and
xi y j u
A,
so
A/M2 =
are linearly inde-
pendent.
Lemma 6.4. The images of xi yj in A/M2 are linearly independent for 0 ≤ i < m. Proof.
Cij xi y j = 0
P
If
However, this is impossible, as
xi y j
Cij
for
constants that are not all zero,
i < m.
are linearly independent in
Thus, if
P
Cij xi y j = 0,
all
xm + y m Cij
divides
P
Cij xi y j
.
must be zero, proving that
A/M2 .
Lemma 6.5. The spaces spanned by
xi y j for 0 ≤ i < m and xi y j u for i, j < m − 1 have 0
intersection. Proof.
Let
v
be a common vector in the spaces spanned by the two sets of generators. Because
a linear combination of some by
xi y j ,
xi y j
xi y j u ∈ ker f , f (v) = 0.
so it must be a linear combination of
is linearly independent in
A/M2 ,
xi y j .
so in order for
We know that
v
v
is
is also in the space spanned
However, we have proved by Lemma 6.4 that
f (v) = 0,
we must have
v = 0.
Lemma 6.6. The generators xi yj u for i, j < m − 1 are linearly independent in A/M3 . To prove this lemma, we work with
dierential form of
x
and
y.
α
is of the form
We assign
f0 + f3 dx ∧ dy .
dx
and
dx ∧ dy .
f
dierential forms
f0 + f1 dx + f2 dy + f3 (dx ∧ dy),
where
fi
over
Q2 .
A
are polynomials
to be of degree one, so even dierential forms are of the form
Ωeven (Q2 ).
dx ∧ dx = dy ∧ dy = 0 If
the space of all
We write the space of dierential forms of degree
dierential forms is denoted by such that
dy
Ω(Q2 ),
and
We de ne a distributive
dx ∧ dy = −dy ∧ dx.
is a polynomial, we write
f ∧α
as
fα
k
as
Ωk ,
so the space of even
wedge product, ∧, on Ω(Q2 )
Functions commute with
for any form
α.
dx, dy ,
and
The wedge product makes
14
Ω(Q2 )
a noncommutative ring. Now we de ne a linear map
df = (∂f /∂x)dx + (∂f /∂y)dy ,
df
and
d : Ωi → Ωi+1 .
If
f ∈ Q[x, y],
then
is of degree 1. We have the following properties:
d(f dx) = df ∧ dx = −(∂f /∂y) dx ∧ dy, d(dα) = 0, d(α ∧ β) = dα ∧ β + (−1)deg(α) (α ∧ dβ).
We also de ne an associative asterisk operation, The even dierential forms with this
∗
∗,
such that
α ∗ β = α ∧ β + (−1)deg(α) (dα ∧ dβ). Ωeven (Q2 )∗ .
operation form a subring, denoted by
Now we
can prove Lemma 6.6.
Proof.
We want to prove the linear independence of
A2 /(xm + y m ).
We know that
A/M3 =
xi y j u
A2 /M3 (A2 )/hP i.
for
If
i, j < m − 1
xi y j u
Because
Ωeven (Q2 )∗
φ : A/M3 → Ωeven (Q2 )∗
the ideal generated by
φ(xm + y m ),
which is spanned by
dierential forms. It is easy to calculate that
Ωeven (Q2 )∗ that
either a power of
xm−1
or
y m−1 ,
prove linear independence. Let
dα = dfo , dβ = dg0 ,
where
f
φ(x) = x
A=
A/M3 ,
then
and
We consider
α ∗ (xm + y m ) ∗ β ,
g
and
where
α
φ(y) = y .
φ(xi y j u)
and
β
so the images of
φ(xi y j u),
and
are even
xi y j u
in
Thus, we only need to show
are polynomials, and each term in
ensuring no overlap with
where
α = f0 + f1 dx ∧ dy , β = g0 + g1 dx ∧ dy ,
dγ = m(xm−1 dx + y m−1 dy).
and
A/M3 .
2xi y j , i, j < m − 1.
where
A2 /M3 (A2 ).
φ(xi y j u) = 2xi y j dx ∧ dy ,
are forms of degree 2, with coe cient
α ∗ (xm + y m ) ∗ β = f + g dx ∧ dy ,
in
[4], where
is not a quotient, it is easier to study than
A/M3 ,
is independent in
span(xi y j u) ∩ hP i = 0, where hP i is the ideal generated by xm + y m We consider the isomorphism
in
i, j < m − 1,
and
g
has
which will
γ = (xm + y m ).
Thus,
Then,
α ∗ γ = α ∧ γ + dα ∧ dγ = γf0 + df0 ∧ dγ + γf1 dx ∧ dy. From this, we know that
d(α ∗ γ) = d(f0 γ) = γdf0 + f0 dγ ,
(α ∗ γ) ∗ β = (α ∗ γ) ∧ β + d(α ∗ γ) ∧ dβ .
by the chain rule. Now we wish to nd
We calculate:
(α ∗ γ) ∧ β = γf0 g0 + g0 df0 ∧ dγ + γg0 f1 dx ∧ dy. Now we calculate
d(α ∗ γ) ∧ dβ : d(α ∗ γ) ∧ dβ = γdf0 ∧ dg0 + f0 dγ ∧ dg0 .
15
We calculate:
α ∗ γ ∗ β = γf0 g0 + γ(g0 f1 dx ∧ dy + df0 ∧ dg0 ) + dγ ∧ (f0 dg0 − g0 df0 ). g dx ∧ dy = γ(g0 f1 dx ∧ dy + df0 ∧ dg0 ) + dγ ∧ (f0 dg0 − g0 df0 )
It is easy to see that each term of has a power of
xm−1
or
y m−1 ,
and the ideal generated by in
as
γ = xm + y m
φ(xm + y m )
dγ = m(xm−1 dx + y m−1 dy).
and
have no intersection in
Ωeven (Q2 )∗ ,
Thus,
xi y j u
so
φ(xi y j u)
is independent
A/M3 . By Lemmas 6.4, 6.5, and 6.6, we have proved that
are the basis of
A/M3 ,
and
xi y j u,
xi y j
for
0≤i<m
and using the short exact sequence,
We now count the number of generators by degree in
N2 (A)
and
xi y j u
xi y j u
for
i, j < m − 1 N2 (A).
is the basis of
to verify Proposition 4.1.
6.3 Counting xi yj u in N2 (A) xi y j u
We wish to count the number of generators degree 1, and
u
for
i, j < m − 1
is of degree 2. The maximum degree of
xi y j u
is
in
N2 ,
2m − 2,
where
as
x
and
y
are of
i, j < m − 1.
Let
k
be the degree in which we are counting generators. We count by two cases and use the short exact sequence to nd
dim(N2 [k]).
Case 1. k ≥ 2m − 1 Because the maximum degree of
xi y j u
is
2m − 2,
there will be no
xi y j u
for
k ≥ 2m − 1.
Case 2. k ≤ 2m − 2 For the maximum degree that
i + j = k − 2.
2m − 2, i = j = m − 2.
So we wish to solve for
k − 2 − i ≤ m − 2 and i ≥ k − m. k−m≤i≤m−2
7
has
For
2m − k − 1
i
and
j
for
Because
is of degree 2, we also know
0 ≤ i, j ≤ m − 2.
k < m, 0 ≤ i ≤ k − 2 has k − 1 solutions.
Because For
j ≤ m − 2,
m ≤ k ≤ 2m − 2,
solutions, which con rms the results of Proposition 4.1.
Conclusion
We have carefully examined the ranks and torsion of program in
Magma,
Ni
for several classes of relations.
x2
Using a
we generated data which we examined for patterns in the structure of
particular, we looked at homogeneous relations in two variables (x and
u
m +y m ),
Ni .
In
q -polynomials (yx−qxy ),
in multiple variables. We discovered patterns in all of these relations and provided a full
proof of the ranks of
N2 (A2 /(xm + y m ))
and a complete description of the behavior of
Ni
for the
16
q -polynomial
algebra.
These results illustrate how the degree and coe cients in a homogeneous
relation can aect the ranks and torsion of
Ni
of the algebra with that relation, particularly with
regard to which primes appear in torsion. There remains much to study about the structure of pattern in the ranks of the
N3
and
N4
Ni .
For example, the pseudo-arithmetic
(Conjectures 4.3.1 and 4.3.2) with relation
be proved, perhaps with previous results [3]. The pattern found in one found in
B2 ,
which suggests a natural isomorphism
B2 → N2 ,
that this mapping is an isomorphism by using the description of In addition, the behavior of torsion can be studied for
N2
is yet to
for that relation matches the
at least over
A/M3
x2 = 0
xm + y m
Q.
We plan to show
by generators and relations.
of multiple variables, and in
particular, Conjecture 4.1.1 could be proved. More work could be done in discovering which primes appear in the torsion, and why they appear. The study of
Ni
and its structure has applications in the study of commutativity in groups
and rings, and it can be used to study maps between groups. cohomology. Studying associative algebra.
Ni
It also has connections in cyclic
can help us build a better understanding of the structure of a general
Outside of its mathematical implications, the lower central series has wider
applications, as noncommutative algebras often appear in quantum theory, so studying lower central series ideas can help build our fundamental understanding of the universe, as well as bring about the technological advancements quantum theory promises.
17
8
Acknowledgments
I would like to acknowledge my mentor Mr. Teng Fei and Professor Pavel Etingof (Massachusetts Institute of Technology), who suggested the problem. Their guidance has been invaluable in the completion of this project.
I would also like to thank the MIT math department for being so
welcoming, particular Tanya Khovanova, who provided me with excellent direction on math research. I would like to thank Dr. John Rickert for his help in writing papers, as well as all the feedback he gave me. In addition, I would like to thank Sitan Chen for his time in helping me revise my paper. I would like to acknowledge the Center for Excellence in Education and the Research Science Institute (RSI) for providing me the opportunity to conduct research. Finally, I would like to extend my gratitude to Dr. Samuel Gilbert, Dr. Elisabeth Vrahoupoulou, Mr. Ken Panos from Aerojet, and Mr. Zachary Lemnios, Dr. Laura Adol e, Dr John Fischer, and Dr. Robin Sta n from the United States Department of Defense for sponsoring my stay at RSI and making my research possible.
18
A
Data Tables
The tables contain the free and torsion components of
A.1 Ni [d] N2 N3 N4 N5 N6 N7 N8 N9 N10
2
3
4
5
6
7
8
9
10
1
0(32 )
0(33 )
0(34 )
0(35 )
0(36 )
0(37 )
0(38 )
0(39 )
0 0 0 0 0 0
3
2
0(3 )
0
0
0(3
0
0
0
0(3 )
8
0(3 )
0(3 )
7
8
0(3 )
6
0
0(3 )
7
0(3 )
0
8
0(3 )
6 6
5
0(3 )
7
0(3 )
· 9)
8
0(3 )
6
0(3 ) 4
7
0(3 )
5
0
0
6
0(3 )
4
0
0
0(3 ) 0(3 )
0
0
5
4
3
0
4
0(3
0
7
0(3 ) 8
· 5)
0(3 ) 7
7
0(3
· 9)
11 0(310 )
9
0(310 )
9
0(310 )
9
0(310 )
9
0(310 )
9
0(310 )
0(3 ) 0(3 ) 0(3 ) 0(3 ) 0(3 ) 0(3 ) 9
0
0
0
0
0
0
8
0(3
0
0
0
0
0
0
0
0
9
2
3
4 2
5 3
6
0(310 )
yx + 2xy 7
5
0(310 )
· 11)
8 6
9
0(4 )
0(4 )
0(4 )
0(4 )
0(4 )
0(4 )
0(4 )
0(49 )
0
2
0(42
0(44 )
0(45 )
0(46 )
0(47 )
0(48 )
0(49 )
0
0
3
0(44 )
0(45 )
0(46 )
0(47 )
0(48 )
0(49 )
0
0
0
4
0(43
0(46 )
0(47 )
0(48 )
0(49 )
0
0
0
0
5
0(46 )
0(47 )
0(48 )
0(49 )
0
0
0
0
0
6
0(44
0(48 )
0(49 )
0
0
0
0
0
0
7
0(48
0(49 )
0
0
0
0
0
0
0
8
0(45
0
0
0
0
0
0
0
0
9
0
0
0
0
0
0
0
0
0
· 7 · 8)
7
10
1
· 8)
4
0(310 )
9
0
Table 10:
Ni [d] N2 N3 N4 N5 N6 N7 N8
with the relation used in the caption.
Zhx,yi/(yx−qxy)
Table 9:
Ni [d] N2 N3 N4 N5 N6 N7 N8 N9 N10 N11
Ni [d],
8
· 5 · 82 · 16)
· 72 )
· 84 · 61)
yx + 3xy
2
3
4
5
6
7
8
9
10
11
1
0(52 )
0(53 )
0(54 )
0(55 )
0(56 )
0(57 )
0(58 )
0(59 )
0(510 )
0
2
0(3 · 53 )
0(54 )
0(55 )
0(56 )
0(57 )
0(58 )
0(59 )
0(510 )
0 0 0 0 0
0 0 0 0 0
3 0 0 0 0
4
0(5 ) 4 0 0 0
5
6
0(5 ) 2
0(3
0(5 ) 5
· 5 · 13)
6
0(5 ) 6
5
0(5 )
0
6
0
0 Table 11:
7
8
0(5 )
0(5 )
7
0(5 )
7
3
0(3 7
0(5 ) 7
· 5 · 17)
0(5 ) 0(5
· 13
0(510 )
9
0(510 )
9
0(510 )
0(5 )
8 8
0(510 )
9
0(5 )
8
0(5 )
0(510 )
9
0(5 )
8
0(5 )
9
0(5 ) 2
)
0(5 )
yx + 4xy
19
A.2 Zhx, yi/(xm + ym ) Ni [d] N2 N3 N4 N5 N6 N7 N8 N9
1
2
3
0
1
0(2 )
0(2 )
0
0
2
0
0
4
2
0
5
6
2
0(2 )
1(2 )
2
2
7
2
0(2 )
0(2 )
4
4 0(2 )
8
2
0(2 )
2
0 (2 )
0(2 )
4
5 0(2 )
2
0(2 )
4
0(2 )
6 0(2 )
0(2 )
6 0(2 )
0(2 )
4 6
0
0
0
0
4
3 2(2 )
7
0
0
0
0
0
3
0(2 )
0(2 )
0
0
0
0
0
0
6
3(2 )
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
6
7 4
2 Table 12: x
Ni [d] N2 N3 N4 N5 N6 N7 N8
1
2
3
4
0
1
2
1(3 )
0(3 )
5
0
0
2
5
2
6
7
3
0(3 )
3
4(3 )
4
1(3 )
8
3
8
0(3 )
9
9
0(3 )
37 )
0(3 )
0
0
0
3
7(2)
4 4(2
0
0
0
0
6
16(2 )
11(2
0
0
0
0
0
9
22(2 )
0
0
0
0
0
0
18
0
0
0
0
0
0
0
· 2
4 0(2
9
3
0(3 )
·
314 )
0(2
6
· 314 )
· 330 ) 17 · 322 ) 11(2 7 45(2 )
+
· 7
315 )
3
0(3 )
9
0(3 )
15 0(3 )
27
· 335 ) 17 · 343 ) 0(2 21 · 339 ) 30(2 14 3(2 )
2(2
5
3 Table 13: x
Ni [d] N2 N3 N4 N5 N6 N7 N8
+ y2
0(2
4
y3
1
2
3
4
5
6
7
8
9
0
1
2
3
2(42 )
1(43 )
0(44 )
0(44 )
0(44
0
0
2
5
8
7(2 · 43 )
4(22
1(24
0(24
7
1
0
0
0
3
8
13(2 · 5)
0
0
0
0
6
18
· 46 ) 10(2 · 32 · 44 · 52 ) 30(22 · 52 )
0
0
0
0
0
9
30
· 47 ) 4(2 · 34 · 412 · 52 ) 26(213 · 34 · 48 · 55 ) 49(24 · 43 · 55 )
0
0
0
0
0
0
18
63
0
0
0
0
0
0
0
8
4 Table 14: x
30
+
· 48 ) 0(2 0 · 34 · 414 ) 12(226 · 32 · 418 · 56 ) 38(226 · 310 · 420 · 512 ) 106(212 · 43 · 510 ) 110(22 )
y4
20
References [1] Martina Balagovic and Anirudha Balasubramanian. associative algebra.
On the lower central series of a graded
Journal of Algebra, 328:287300, 2011.
[2] Wieb Bosma, John Cannon, and Catherine Playoust. The magma algebra system. i. the user language.
J. Symbolic Output., 24(3-4):235265, 1997.
[3] Pavel Etingof, John Kim, and Xiaoguang Ma. On universal lie nilpotent associative algebras.
Journal of Algebra, 321:697703, 2009.
[4] Boris Feigin and Boris Shoikhet.
On
[A, a]/[a, [a, a]]
commutators of free associative algebras.
and on a
wn -action
on the consecutive
Math. Res. Lett., 14(5):781795, 2007.
[5] George Kerchev. On the ltration of a free algebra by its associative lower central series.
print, arXiv(1101.5741v1), 2011.
pre-
Abstract Ni (A)
This paper examines the behavior of the successive quotients ideals
Mi (A)
Li (A)
by
of a nitely generated associative algebra
A
over
Z.
We de ne the lower central series
L1 (A) = A, Li+1 (A) = [A, Li (A)], Mi (A) = A · Li (A) · A,
We decompose the
Ni
of the lower central series
and
Ni (A) = Mi (A)/Mi+1 (A).
into its free and torsion components using the structure theorem of nitely
generated abelian groups, and we examine patterns in the ranks and torsion of various homogeneous relations, including and
xm + y m .
x2
in multiple variables,
q -polynomial
a complete description of
N2
for
for algebras with
relation
yx − qxy ,
In order to do this, we create data tables with the ranks and torsion of various
previously uncalculated, based on calculations done in the program
ranks of
Ni
Ni
for the
q -polynomial
algebra,
Magma.
Zhx, yi (yx − qxy)
Ni ,
This paper includes and a proof for the
Ahx, yi/(xm +y m ), which provides insight into how changing the coe cient or degree
of a relation aects rank and torsion, as well as general patterns for which primes appear in torsion.
1
1
Introduction Li (A),
For the past several years, algebraists have been studying the lower central series are successive subspaces of an associative algebra
A
formed from the commutators of
A.
which
Thus, the
lower central series and its related objects can be used to measure the noncommutativity of algebras. We consider the successive quotients of the two-sided ideals quotients
Ni .
We study the structure and properties of
Mi
Ni
generated by
Li ,
and we call these
to better understand the structure of
associative algebras. Lower central series quotients of free associative algebras were rst studied by Feigin and Shoikhet [4].
They looked at the successive quotients
was an isomorphism between the space
A/M3 (A)
isomorphism is essential in proving the pattern of
Bi = Li /Li+1 ,
and concluded that there
and the space of even dierential forms.
N2 (A2 /(xm + y m ))
found in this paper.
This The
study of quotients continued with Etingof, Kim, and Ma [3], who completely described the quotient
A/Mi (A)
for
i = 4.
The study of at
B2
Bi
was continued in the work of Balagovi¢ and Balasubramanian [1], who looked
in the quotient of a free algebra.
B2 (A2 /(xd + y d )),
In particular, they provided a complete description of
which is similar to the results found in this paper for
While the structure of
Bi
studied. Kerchev [5] studied
N2 (A2 /(xm + y m )).
has been studied in multiple papers, the quotients
Ni
for free algebras and computed
Ni
even for free algebras, and
have been less
Ni (An ) for several values of i and n.
However, there is still much work to be done in studying the structure of has never been calculated for
Ni
Ni
Ni .
In particular, torsion
have not been studied for algebras
with relations. In this paper, we study the behavior of
Ni
for an associative algebra
Ni
Zhx, yi
with various re-
lations.
We also compute the ranks and torsion of various
using a computer program called
Magma.
The process of collecting data is explained in Section 3. Several patterns suggested by the
data are contained in Section 4. In Section 5 we provide a complete description of with the relation structure of
N2
yx − qxy = 0,
with the relation
also known as
xm + y m
q -polynomial
Ni
for algebras
algebras. The results concerning the
is proven in Section 6.
We begin with preliminary background to better understand the algebraic objects of study.
2
2
Preliminary Background
2.1 Associative Algebras and Their Lower Central Series De nition 2.1. operation by
1,
Let
A
(a, b) 7→ a · b,
then
A
is a
be a vector space over a eld which is also written as
ab.
If
k
with a bilinear associative multiplication
A
also has a multiplicative identity, denoted
unital associative algebra.
A free algebra is a unital associative algebra that is generated by a set of generators with no relation. In this paper, we are interested in associative algebras that are not necessarily free, more precisely, algebras with homogeneous relations. An algebra of
A by the ideal generated by the relation P .
This bracket operation satis es
[a, a] = 0
A/hP i
will denote the quotient algebra
We de ne a bracket operation on
A by [a, b] = a·b−b·a.
and the Jacobi identity:
[a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0. A vector space with a bilinear bracket operation are satis ed is called a
[a, b] such that the Jacobi identity and [a, a] = 0
Lie algebra, so any associative algebra is also a Lie algebra with [a, b] =
a · b − b · a.
De nition 2.2. Let A be a Lie algebra. and
Li+1 (A) = [A, Li (A)],
span([c, d]) such that abbreviate
Li (A)
as
De ne a series of Lie ideals inductively such that
where the bracket of two subspaces
c ∈ C, d ∈ D.
C
This series of Lie ideals is the
and
[C, D] =
lower central series of A.
Z
by using
Z-modules
We
(i.e. Abelian groups) instead of vector
spaces. Similarly, we can use this de nition over any commutative ring Denote the two-sided ideals generated by each
It is easy to see that
De nition 2.4.
is de ned as
Li .
We may make this de nition over
De nition 2.3.
D
L1 (A) = A
Mi = A · L i
De ne the
Ni
Li
by
R.
Mi ,
i.e.
Mi = A · Li · A.
. Using this de nition, we can de ne the quotients
to be the successive quotients
Ni .
Mi /Mi+1 .
Now we introduce the idea of grading, which is crucial to representing data eectively.
De nition 2.5.
Let
A
be a module over a commutative ring
a direct sum decomposition into submodules then
A
is a
graded algebra.
M
Ai .
If
A
k.
The module
A
is
is an algebra such that
graded if A has Ai · Aj ⊂ Ai+j ,
i≥0
3
Example 1.
the grading is by the degree of the polynomial. grading from part of
Ni
k[x1 , . . . , xn ],
The simplest example of a graded algebra is a polynomial ring
A.
We observe that
Our study is simpli ed if we look at
at degree
d
will be denoted as
Ni [d],
Ni
Ni
where
is graded, as it inherits its
by its degree, which is denoted by
which is a nitely generated
d.
The
k -module.
2.2 Torsion and Class cation of Finitely Generated Abelian Groups De nition 2.6. positive integer
a
is an
An element
n.
n-torsion
a
of an Abelian group
G
torsion element if n · a = 0 for some
is a
Conventionally, 0 is also considered a torsion element. In this case, we say that
G
element. All of the torsion elements in
form a subgroup of
G.
To clarify the concept of torsion, we oer a simple example.
Example 2.
Consider the group
G = Z6 = Z2 ⊕ Z3 .
This group has
2-torsion, 3-torsion,
and
6-
torsion. 0, 2, and 4 are 3-torsion elements, 0, 3 are 2-torsion elements, and all elements are 6-torsion elements. All 2 and 3-torsion elements are also 6-torsion elements.
The idea of torsion becomes especially important due to the Structure Theorem of Finitely Generated Abelian groups, which states that groups can be separated into their free and torsion components.
Theorem 2.1 (Structure Theorem of Finitely Generated Abelian Groups). Every nitely generated Abelian group G is isomorphic to a nite direct sum of in nite cyclic groups and cyclic groups of order pn , for various primes p. This decomposition is unique up to order of summands. The theorem can be restated as
G∼ = F ⊕ T, where
F
is the free component, which is isomorphic to
Zr
for some
component, consisting of a nite sum of cyclic groups of order
r
is called the
pn
r ∈ Z,
and
T
for various primes
is the torsion
p.
In this case,
rank of the free component, known simply as rank.
The goal of the project is to determine the structure in the ranks and torsion of
Ni
for algebras over
Z by studying patterns
Ni .
2.3 Sample Calculations for Zhx, yi We provide a set of sample calculations to illustrate how consider the free associative algebra
Zhx, yi.
L i , Mi ,
and
Ni
are constructed.
We
Because it is not known whether torsion exists in
4
Ni
for free algebras generated by 2 variables, we focus on calculating the ranks of
Li
calculate the bases of the rst few By de nition,
L 1 = A,
L1
so
is spanned by
L2
[x, xy] + [x, yx] = [y, x2 ].
from the basis.
Similarly,
remain in the basis of
The results of the basis of
Li [d] L1 L2 L3
Thus,
and
[y, x2 ]
Li
L3 [3]
Calculating
L1
L2 [2]
However, both
[x, x2 ]
and
L2
is
is 1, the
is
[x, y],
Thus, the potential basis vectors are
[y, y 2 ].
[y, y 2 ]
as
[x, xy],
are 0. In
is not linearly independent and can be removed
[y, xy] + [y, yx] = [x, y 2 ],
L2 [3].
The next row
is 2. The only term in the basis of
[L1 [1], L1 [2]].
[y, xy], [x, yx], [x, y 2 ], [y, x2 ], [y, yx], [x, x2 ], addition,
x, y .
As the minimum degree of a non-trivial part of
minimum degree of a non-trivial part of
[y, x] = −[x, y]. L2 [3]
First, we
in low degrees.
spanned by all the monomials in
[A, L1 ].
formed from the set of all
Ni .
[x, y 2 ]
and
can be eliminated.
is more straightforward, as
Only 4 terms
L3 [3] = [L1 [1], L2 [2]].
can be found in Table 1, where the top row indicates the degree
1
2
3
x, y 0
x2 , xy , yx, y 2 [x, y]
0
0
x3 , x2 y , xyx, yx2 , y 3 , y 2 x, yxy , xy 2 [x, xy], [y, xy], [x, yx], [y, yx] [x, [x, y]], [y, [x, y]]
Li The Mi can be constructed from Li , as Mi = A · Li · A = A · Li .
d.
Table 1: Bases for
as
Li [i]
Li [i] = Mi [i],
must be multiplied by scalars on both sides for the minimum non-trivial degree
M1 = A ,
and
M2 [2]
is also easy to compute.
M2 [3] = L1 [0] · L2 [3] + L1 [1] · L2 [2]. [y, xy], [x, yx], [y, yx], x[x, y], contains
and
[x, xy], [y, xy], [x, yx], Mi [d] M1 M2 M3
M2 [3]
Then,
M3 [3] = L3 [3],
2
3
0
x2 , xy , yx, y 2 [x, y]
0
0
x3 , x2 y , xyx, yx2 , y 3 , y 2 x, yxy , xy 2 [x, xy], [y, xy], [x, yx], [y, yx] [x, [x, y]], [y, [x, y]]
=
are
[x, xy],
Mi
which are the cardinalities of the basis of each
rank(Mi [d])
M2 [3]
so Table 2 is complete.
x, y
rank(Ni [d])
Thus,
Eliminating linearly dependent terms, the basis of
[y, yx].
Ni [d],
i.
is slightly more complicated.
1
Now we calculate the ranks of
Ni = Mi /Mi+1 ,
M2 [3]
Therefore, the possible terms in the basis of
y[x, y].
and
Calculating
Table 2: Bases for
Ni [d]
By this de nition,
− rank(Mi+1 [d]).
Ni [d].
As
Thus, computing the ranks of each
becomes a simple subtraction problem. The ranks are shown in Table 3.
Ni [d] N1 N2
1
2
3
2
3
4
0
1
2
Table 3: Free Ranks for
Ni 5
3
Data Collection
We rst compile data tables of the ranks and torsion of
Ni
for various relations. By changing the
number of variables, coe cients, or degree of the relations, we can nd patterns and form conjectures about the behavior of the ranks and torsion of
Ni .
Data is collected by running computations in
Magma
[2]. This code was run for many relations
over the integers, and the outputs were then organized into table form by grading. The left column displays the
Ni ,
while the top row is organized by grading (degree). Each term in the data table
includes the rank, which is displayed outside of the parentheses, and the torsion, which is displayed within the parentheses. Here, we provide an example of how the data was processed and organized into tables. example below shows how to format the output in
Example 3.
Magma
The
to a data table.
The output code
$N_ 4 $ & 8 16(Abelian Group isomorphic to Z/2 + Z/2 + Z/4 + Z/4) is expressed as
16(22 · 42 )
for
N4 [8]
in a table.
The expression has a rank of 16, which is the last numbered output before the parentheses. The torsion is slightly more di cult to express. The output data in the parentheses represents the direct sum of many cyclic groups. While prime power cyclic groups do not need to be further decomposed, other groups can be decomposed into coprime components. For example, into
Z3 × Z4 × Z5
by the result in group theory that states
Z60
Zmn = Zm × Zn
for
can be decomposed
m, n
coprime. The
data is decomposed into prime powers for the tables. These components can then be combined through exponent rules (eg. Z/2 + Z/2 given in the output is expressed in the table as is
22 ).
Thus, the nal form of the term in the data table for
N4 [8]
16(22 · 42 ). We look for patterns within these data tables, then try to prove them.
4
Observations of Patterns in Data
After compiling tables of algebras with various relations, we nd several patterns in the ranks and torsion of the
Ni . 6
4.1 Patterns in Zhx1 , . . . , xn i/(x21 ) Interesting patterns arise for the algebra with the relation
x21 = 0 with a number of variables, shown
in Table 4. With two variables, it seems that the torsion and ranks of that is to say, the torsion and ranks do not chnage as
Ni [i + 2]
the stabilization of the ranks from
Ni [d] N2 N3 N4 N5 N6 N7 N8 N9
d
Ni
stabilize rather quickly
increases. Reading across the rows shows
and a stabilizing torsion.
2
3
4
5
6
7
8
9
1
1(2)
1(2)
1(2)
1(2)
1(2)
1(2)
1(2)
0
2
3(2)
3(2 )
3(2 )
3(2 )
3(2 )
3(2 )
4 3(2 )
5 3(2 )
5 3(2 )
5 3(2 )
7(2 )
7(2
2
0
0
2
2 3(2 )
0
0
0
4
2
3
2
7
2
9
10
0
0
0
0
5
5 9(2 )
0
0
0
0
0
9
· 3) 12 9(2 · 3 · 5) 7 18(2 )
0
0
0
0
0
0
12
· 3) · 3 · 5) 19 · 32 · 5) 19(2 12 25(2 )
0
0
0
0
0
0
0
20
2 Table 4: x1
= 0,
· 3)
2
7(2
7(2
16 9(2
two variables
This yields a conjecture about the stabilization of the ranks and torsion of
Ni .
Conjecture 4.1.1. For Zhx,yi/(x2 ), Ni [j] ∼ = Ni [j + 1] for j ≥ 2i − 1. It would be interesting to know how soon the ranks stabilize for algebras with more generators, and which primes will ultimately appear in torsion.
4.2 Patterns in Zhx, yi/(yx − qxy) We notice that the ranks are non-zero only for the diagonal
Ni [i], and these ranks are equal to i − 1,
as seen in Tables 9 through 11, found in Appendix A.1. Additionally, xing a
Ni [d]
for
i < d
are the same:
(Zq−1 )d−1 .
Along the diagonal
pattern with more primes. We are able to completely describe the diagonal
i = d,
i = d,
Ni
4.3 Patterns in
the torsion in all
there is a more interesting
in this case, and we show that on
all primes will eventually appear, except those that divide
description and proof of the result on the torsion in
d,
Zhx, yi/(yx − qxy)
q.
A more detailed
are provided in Section 5.
Zhx,yi/(xm +y m )
By examining the algebras Zhx,yi/(xm +y m ), we discover several patterns occurring across tables in for
i = 2, 3, 4.
The complete set of data can be found in Tables 12 to 14 in Appendix A.2. For
the ranks follow a palindromic pattern similar to the one found in 5, where the left column indicates values of
B2
Ni
N2 ,
[1], which is shown in Table
m.
7
Proposition 4.1.
N2 [d]
2
2
1
3 4 5
3
4
5
6
1
2
1
1
2
3
2
1
1
2
3
4
3
7
8
2
1
N2 [d] with the relation xm + y m = 0 In the algebra Zhx, yi/(xm + y m ), rank(N2 [k]) = k − 1 for k < m, Table 5: Ranks for
rank(N2 [k])
=
2m − k − 1 for m ≤ k ≤ 2m − 2, and 0 for all other values of k . This proposition will be proven in Section 6. The patterns developing in the ranks of
N3
and
arithmetic sequences. The values for the ranks of showing values of
are displayed in Table 6, with the left column
N3 [d]
3
4
5
6
7
8
9
10
2
2
1
0
0
0
0
0
0
3
2
5
4
1
0
0
0
0
4
2
5
8
7
4
1
0
0
5
2
5
8
11
10
7
4
1
Conjecture 4.3.1. In the algebra
While
N3
are almost palindromic, and we see pseudo-
m.
Table 6: Ranks for
rank(N3 [k])
N4
N3 [d]
with the relation
Zhx, yi/(xm + y m ),
xm + y m = 0 = 3k − 7 for k ≤ m + 1,
rank(N3 [k])
= 6m − 3d + 1 for m + 1 < k < 2m + 1, and 0 for all other values of k .
N2
and
N3
seem to have easily generalizable patterns,
N4
is slightly more complicated.
The bolded numbers in Table 7 are the ones that remain consistent as
N4 [d]
4
2
2
3 4 5
5
6
7
8
9
d
increases.
10
3 7 4 3 8 13 10 4 3 8 14 19 16 10 4
Table 7: Ranks for
N4 [d]
with the relation
xm + y m = 0
Conjecture 4.3.2. The ranks of N4 will become stable outside of the diagonal d = m + 2, where d is the degree of the grading. We expect
rank(N4 (A2 /(x
m
+ y m ))[2m − k]) to stabilize for large m
and xed k ≥ 0. This rank vanishes for k < 0.
8
5
Complete Description of Ni (Zhx, yi/(yx − qxy))
We consider the speci c algebra
Zhx, yi
with the relation
yx − qxy ,
algebra. A clear pattern emerges in the ranks and torsion of the 10, which are located in Appendix A.1. The torsion in along the diagonal
i=d
Ni [d]
for
Ni ,
also known as a
q -polynomial
as shown in Tables 9 through
i < d−1
is
(Zq−1 )d−1 .
The torsion
has a more interesting pattern. To understand this pattern, we use a ner
grading on the degrees, de ning
x
to have degree
h1, 0i
and
y
to have degree
h0, 1i
where for degree
hu, vi, d = u + v .
Theorem 5.1. Let
A = Zhx, yi/(yx − qxy), where q ∈ Z, q 6= ±1. The
i = j . Otherwise, the rank is 0. The torsion in Ni [d], also written as M (Zq−1 )d−1 . Along the diagonal i = d − 1, Tor(Ni [d]) = Zq(u,v) −1 .
rank(Ni [j])
= i − 1 for
, for i < d − 1 is
Tor(Ni [d])
u+v=d
Note.
In Theorem 5.1 and its proof, the greatest common divisor of
Now we give some preliminary information for the proof. spaces
Lk
and
Mk .
We rst note that because
result of any bracket operation as a sum of (see Table 8 for element of
Lk
Lk [hu, vi]
De nition 5.1.
j 6= i
as
and
hu, vi
Ni [j] = i − 1
L1 L2 L3 L4
if
h0, 1i x
h0, 2i y2
0 0 0
is denoted as
(u, v).
First, we consider the bases of the
with some coe cients. We rst construct a table
keeping in mind that
yx = qxy ).
For
k > 1,
the basis
k xu y v . Su,v k xu y v Lk [hu, vi] ⊂ span Su,v
we include only the coe cients of the bases,
hv, ui
v
and
under the relation, we can express the
is the largest possible integer such that
k > 1,
and
k = u + v,
will be denoted by
k Su,v
In the table for
u < v,
on the diagonal
xu y v
yx = qxy
u
are symmetric. There is no torsion if
k , Su,v
u, v = 0.
.
and the terms in which
The rank of
Ni [j] = 0
if
j = i. h0, 3i y3
0
h1, 1i xy (q − 1)
0
0
0
h1, 2i xy 2 (q − 1) (q − 1)2
0
0
0
0
0
Table 8: Construction of We consider the bases along the diagonal
h0, 4i y4 0 0 0
h1, 3i xy 3 (q − 1) (q − 1)2 (q − 1)3
h2, 2i x2 y 2 (q 2 − 1) (q − 1) · (q 2 − 1) (q − 1)2 · (q 2 − 1)
Lk for the relation yx − qxy = 0 k = u + v and nd that there is
a pattern (Lemma
5.1).
9
Lemma 5.1. For the algebra Zhx, yi
k for k = u + v , (yx − qxy), with q ∈ Z and q 6= ±1, and Su,v
is k Su,v = (q − 1)u+v−2 · (q (u,v) − 1).
(5.1)
To prove this lemma, we include a few known facts in number theory:
Fact 1. (λm − 1, λn − 1) = λ(m,n) − 1 Fact 2. (a, b) = 1 =⇒ (a, bc) = (a, c) Fact 3. Proof.
i j , h h
= 1 ⇐⇒ (i, j) = h
We prove Lemma 5.1 by induction on
u, v .
The base case is satis ed, as
2 S1,1 = (q − 1)1+1−2 (q (1,1) − 1) = q − 1. Because
k satis es Lk [hu, vi] = [x, Lk [hu − 1, vi]] + [y, Lk−1 [hu, v − 1i]], Su,v k−1 k−1 k (q u − 1) . (q v − 1), Su,v−1 Su,v = Su−1,v
the recursive equation (5.2)
Assuming that
k−1 = (q − 1)u+v−3 (q (u−1,v) − 1), Su−1,v k−1 = (q − 1)u+v−3 (q (u,v−1) − 1), Su,v−1 we want to show that
k Su,v = (q − 1)u+v−3 (q (u−1),v − 1)(q v − 1), (q − 1)u+v−3 (q (u,v−1) − 1)(q u − 1) = (q − 1)u+v−2 (q (u,v) − 1). We can pull out the expression
(q − 1)u+v−3 ,
(5.3)
as it is common to both of the components of the
greatest common divisor. Thus, equation (5.3) is reduced to
k Su,v = (q − 1)u+v−3 (q (u−1,v) − 1)(q v − 1), (q (u,v−1) − 1)(q u − 1) .
(5.4)
We set the following:
q (u−1,v) − 1 = α, q v − 1 = β, q (u,v−1) − 1 = γ, q u − 1 = δ.
10
Using Facts 3 and 1, we have
(α, γδ) = q − 1 ⇐⇒
α γδ , q−1 q−1
=1
(5.5)
and
γδ α , q−1 q−1
= 1 =⇒
γδ ,β q−1
=
γδ αβ , q−1 q−1
(5.6)
We also note that
β,
γ γδ = 1 =⇒ β, = (β, δ). q−1 q−1 α γ By Fact 1, (α, γ) = q − 1, as ((u − 1, v), (u, v − 1)) = 1. Thus, = 1. , q−1 q−1 γδ α γδ (5.5), ,β = , β is true. q−1 q−1 q−1
By equation
= 1. By Fact 1, (β, δ) = (q (u,v) −1). Thus, by equa(β, γ) = q−1 by Fact 1, γ γδ αβ γδ (u,v) (β, δ) = β, δ = (q − 1). By equation (5.6), , = β, = q−1 q−1 q−1 q−1
Because tion (5.7),
γ β, q−1
(5.7)
q (u,v) − 1.
It follows by Fact 3 that
(αβ, γδ) = (q − 1)(q (u,v) − 1).
We can now prove Theorem 5.1.
Proof.
Mk [hu, vi]
We denote the basis of
the terms for
Lk
as
k xu y v , Tu,v
and note that for
on the diagonal are of the lowest possible degree. Thus,
constants on either side to form Now we consider
k Tu,v
Lk
as
must be multiplied by
Mk = A · Lk · A.
u + v > k.
for
k = Tk , k = u + v , Su,v u,v
We note that
Mu+v−1 [hu, vi] = Lu+v−1 [hu, vi] + x · Lu+v−1 [hu − 1, vi] + y · Lu+v−1 [hu, v − 1i]. Set
T1 xu y v
1i],
where
Thus,
to be the basis of
u+v−1 T1 = Su−1,v
(T1 , T2 )xu y v
1)u+v−2 xu y v
spans
k = (q − 1)k−1 Tu,v
if
spans
x · Lu+v−1 [hu − 1, vi] and T2 xu y v
and
u+v−1 T2 = Su,v−1 .
Mu+v−1 [hu, vi].
Mu+v−1 [hu, vi]
It is true that
It is known that
by Lemma 5.1 and
to be the basis of of
T1 xu y v , T2 xu y v
y · Lu+v−1 [hu, v −
span
Mu+v−1 [hu, vi].
(T1 , T2 ) = (q − 1)u+v−2 .
u+v−1 = (q − 1)u+v−2 . Tu,v
Then,
(q −
This suggests that
u + v > k.
Using this information, we can calculate the torsion
Ni .
Using the bases of
Mk ,
we divide to get
11
Tor (Nu+v−1 [hu, vi])
u+v xu y v Z · Tu,v
=
u,v−1 u v Z · Tu,v x y
= Zq(u,v) −1 .
Tor (Nk−1 [k])
Summing over all
M
=
u, v
yields
Zq(u,v) −1 .
u+v=k The ranks are easy to verify given the bases. The space
k k S1,k−1 xy k−1 , . . . , Sk−1,1 xk−1 y and above the diagonal
with
i = d,
Mk [k] is a free Abelian group with basis
q 6= ±1, while Mk+1 [k] = 0. Mk [k]
the ranks of
and
Nk [k] is free of rank k − 1.
Below
are the same, so the rank of
Nk [k]
So
Mk [k + 1]
is 0.
Corollary 5.1. All primes except those that divide q appear in the torsion of Ni [i + 1] for some i. Proof.
Given that the
that divide
q
Ni
has
q (u,v) − 1 -torsion, by Fermat's little theorem, all primes except those
will appear in the torsion of
Ni .
With Theorem 5.1 and Corollary 5.1, we now have a clearer idea of how the coe cients of a relation aect the ranks and torsion of
6
Ni .
Ranks of N2 (Zhx, yi/(xm + y m ))
We wish to nd the basis of
N2 (A2 /(xm +y m )) in order to prove Proposition 4.1, where A2 = Qhx, yi.
To do so, we use the short exact sequence First, we nd the generators of using the isomorphism
A/M3 .
0 → N2 → A/M3 → A/M2 → 0,
A = A2/(xm +ym ).
We then prove the linear independence of these generators by
A2 /M3 ∼ = Ωeven (Q2 )∗
the result for the ranks of
where
found in Feigin and Shoikhet's paper [4], thus proving
N2 (Zhx, yi/(xm + y m )).
6.1 Generators of A/M3 We consider
A = Qhx, yi/(xm + y m )
m − 1, 0 ≤ j
and
xi y j u
relations are satis ed in Because
xm + y m = 0
in
for
with
u = [x, y].
0 ≤ i, j ≤ m − 1
span
We want to show that
A/M3 .
the relation also holds in
for
0 ≤ i ≤
To do this, we show that the following
A/M3 : u2 = 0, [u, x] = [u, y] = 0, xm + y m = 0, A,
xi y j
and
xm−1 u = y m−1 u = 0.
A/M3 .
Lemma 6.1. In A/M3 , u2 = 0. Proof. u2 = [x, y] · [x, y] = [x, y] · (xy − yx).
12
Because
[x, y] · xy = [[x, y], x]y + x[x, y]y , u2 = [[x, y], x]y + x[x, y]y − [x, y]yx.
Because
x[x, y]y − [x, y]yx = −[[x, y]y, x], u2 = [[x, y], x]y − [[x, y]y, x].
Because
[[x, y], x]y ∈ M3
and
−[[x, y]y, x] ∈ M3 ,
the relation
u2 = 0
is satis ed in
A/M3 .
Lemma 6.2. In A/M3 , the relations [u, x] = [u, y] = 0 hold. Proof. [u, x] = [[x, y], x].
Because
[[x, y], x] ∈ M3 , [u, x] = 0
in
A/M3 .
Similarly,
[u, y] = 0.
Lemma 6.3. In A/M3 , xm−1 u = ym−1 u = 0. Proof.
We know that
We will show that each other by
0 = [xm + y m , x]
[x, y m ] = my m−1 u
[u, y] = 0,
because
in
A2 /M3
xm + y m = 0.
Additionally,
through induction. Because
[xm + y m , x] = [x, y m ]. u
and
y
commute with
the base case is satis ed:
0 = [x, y 2 ] = xy 2 − y 2 x = xy 2 − yxy + yxy − y 2 x = uy + yu = 2yu, since
[y, u] = 0 in A2 /M3 .
[x, y k ]
Now we assume that for some integer
k , [x, y k ] = ky k−1 u.
We can expand
as follows:
k
k
k
k−1 X
k
0 = [x, y ] = xy − y x = xy + k+1 ] Now we consider [x, y
0 = [x, y
k+1
=
xy k+1 k
xy +
]=
−
−y i xy k−i + y i xy k−i − y k x.
i=1 k+1 y . We expand and factor:
k−1 X
i
−y xy
k−i
i
+ y xy
k−i
! − y x y + y k xy − y k+1 x. k
i=1 Thus,
[x, y k+1 ] = [x, y k ]y + y k u = ky k u + y k u = (k + 1)y k u = 0 since
m
[y, u] = 0
in
A2 /M3 .
We have proved through induction that
is some positive integer,
xm−1 u = 0
y m−1 u = 0
is satis ed in
since we are working over
Q.
Because
Similarly,
is also satis ed.
From Lemma 6.1, 6.2, and 6.3, we know that
A/M3 .
A/M3 ,
[x, y m ] = my m−1 u = 0.
Since
u
commutes with
the end of all expressions in
x
and
A/M3 .
y,
xi y j
for
0 ≤ i < m and xi y j u for i, j < m − 1 span
we can assume without loss of generality that
Thus, the degree of
u
can either be 0 or 1, as for
u
appears at
a ≥ 2, ua = 0.
13
Calculating the number of generators for each degree, we nd that the dimension (from the data) for that degree equals the number of generators, predicted by Proposition 4.1. We now show the linear independence of these generators.
6.2 The Basis of A/M3 We consider the short exact sequence
0 → M2 /M3 → A/M3 → A/M2 → 0, and let
f
be the surjection
Q[x, y]/(xm + y m ).
A/M3 → A/M2 .
A/M2
is the abelianization of
Now we wish to prove that the generators
xi y j
and
xi y j u
A,
so
A/M2 =
are linearly inde-
pendent.
Lemma 6.4. The images of xi yj in A/M2 are linearly independent for 0 ≤ i < m. Proof.
Cij xi y j = 0
P
If
However, this is impossible, as
xi y j
Cij
for
constants that are not all zero,
i < m.
are linearly independent in
Thus, if
P
Cij xi y j = 0,
all
xm + y m Cij
divides
P
Cij xi y j
.
must be zero, proving that
A/M2 .
Lemma 6.5. The spaces spanned by
xi y j for 0 ≤ i < m and xi y j u for i, j < m − 1 have 0
intersection. Proof.
Let
v
be a common vector in the spaces spanned by the two sets of generators. Because
a linear combination of some by
xi y j ,
xi y j
xi y j u ∈ ker f , f (v) = 0.
so it must be a linear combination of
is linearly independent in
A/M2 ,
xi y j .
so in order for
We know that
v
v
is
is also in the space spanned
However, we have proved by Lemma 6.4 that
f (v) = 0,
we must have
v = 0.
Lemma 6.6. The generators xi yj u for i, j < m − 1 are linearly independent in A/M3 . To prove this lemma, we work with
dierential form of
x
and
y.
α
is of the form
We assign
f0 + f3 dx ∧ dy .
dx
and
dx ∧ dy .
f
dierential forms
f0 + f1 dx + f2 dy + f3 (dx ∧ dy),
where
fi
over
Q2 .
A
are polynomials
to be of degree one, so even dierential forms are of the form
Ωeven (Q2 ).
dx ∧ dx = dy ∧ dy = 0 If
the space of all
We write the space of dierential forms of degree
dierential forms is denoted by such that
dy
Ω(Q2 ),
and
We de ne a distributive
dx ∧ dy = −dy ∧ dx.
is a polynomial, we write
f ∧α
as
fα
k
as
Ωk ,
so the space of even
wedge product, ∧, on Ω(Q2 )
Functions commute with
for any form
α.
dx, dy ,
and
The wedge product makes
14
Ω(Q2 )
a noncommutative ring. Now we de ne a linear map
df = (∂f /∂x)dx + (∂f /∂y)dy ,
df
and
d : Ωi → Ωi+1 .
If
f ∈ Q[x, y],
then
is of degree 1. We have the following properties:
d(f dx) = df ∧ dx = −(∂f /∂y) dx ∧ dy, d(dα) = 0, d(α ∧ β) = dα ∧ β + (−1)deg(α) (α ∧ dβ).
We also de ne an associative asterisk operation, The even dierential forms with this
∗
∗,
such that
α ∗ β = α ∧ β + (−1)deg(α) (dα ∧ dβ). Ωeven (Q2 )∗ .
operation form a subring, denoted by
Now we
can prove Lemma 6.6.
Proof.
We want to prove the linear independence of
A2 /(xm + y m ).
We know that
A/M3 =
xi y j u
A2 /M3 (A2 )/hP i.
for
If
i, j < m − 1
xi y j u
Because
Ωeven (Q2 )∗
φ : A/M3 → Ωeven (Q2 )∗
the ideal generated by
φ(xm + y m ),
which is spanned by
dierential forms. It is easy to calculate that
Ωeven (Q2 )∗ that
either a power of
xm−1
or
y m−1 ,
prove linear independence. Let
dα = dfo , dβ = dg0 ,
where
f
φ(x) = x
A=
A/M3 ,
then
and
We consider
α ∗ (xm + y m ) ∗ β ,
g
and
where
α
φ(y) = y .
φ(xi y j u)
and
β
so the images of
φ(xi y j u),
and
are even
xi y j u
in
Thus, we only need to show
are polynomials, and each term in
ensuring no overlap with
where
α = f0 + f1 dx ∧ dy , β = g0 + g1 dx ∧ dy ,
dγ = m(xm−1 dx + y m−1 dy).
and
A/M3 .
2xi y j , i, j < m − 1.
where
A2 /M3 (A2 ).
φ(xi y j u) = 2xi y j dx ∧ dy ,
are forms of degree 2, with coe cient
α ∗ (xm + y m ) ∗ β = f + g dx ∧ dy ,
in
[4], where
is not a quotient, it is easier to study than
A/M3 ,
is independent in
span(xi y j u) ∩ hP i = 0, where hP i is the ideal generated by xm + y m We consider the isomorphism
in
i, j < m − 1,
and
g
has
which will
γ = (xm + y m ).
Thus,
Then,
α ∗ γ = α ∧ γ + dα ∧ dγ = γf0 + df0 ∧ dγ + γf1 dx ∧ dy. From this, we know that
d(α ∗ γ) = d(f0 γ) = γdf0 + f0 dγ ,
(α ∗ γ) ∗ β = (α ∗ γ) ∧ β + d(α ∗ γ) ∧ dβ .
by the chain rule. Now we wish to nd
We calculate:
(α ∗ γ) ∧ β = γf0 g0 + g0 df0 ∧ dγ + γg0 f1 dx ∧ dy. Now we calculate
d(α ∗ γ) ∧ dβ : d(α ∗ γ) ∧ dβ = γdf0 ∧ dg0 + f0 dγ ∧ dg0 .
15
We calculate:
α ∗ γ ∗ β = γf0 g0 + γ(g0 f1 dx ∧ dy + df0 ∧ dg0 ) + dγ ∧ (f0 dg0 − g0 df0 ). g dx ∧ dy = γ(g0 f1 dx ∧ dy + df0 ∧ dg0 ) + dγ ∧ (f0 dg0 − g0 df0 )
It is easy to see that each term of has a power of
xm−1
or
y m−1 ,
and the ideal generated by in
as
γ = xm + y m
φ(xm + y m )
dγ = m(xm−1 dx + y m−1 dy).
and
have no intersection in
Ωeven (Q2 )∗ ,
Thus,
xi y j u
so
φ(xi y j u)
is independent
A/M3 . By Lemmas 6.4, 6.5, and 6.6, we have proved that
are the basis of
A/M3 ,
and
xi y j u,
xi y j
for
0≤i<m
and using the short exact sequence,
We now count the number of generators by degree in
N2 (A)
and
xi y j u
xi y j u
for
i, j < m − 1 N2 (A).
is the basis of
to verify Proposition 4.1.
6.3 Counting xi yj u in N2 (A) xi y j u
We wish to count the number of generators degree 1, and
u
for
i, j < m − 1
is of degree 2. The maximum degree of
xi y j u
is
in
N2 ,
2m − 2,
where
as
x
and
y
are of
i, j < m − 1.
Let
k
be the degree in which we are counting generators. We count by two cases and use the short exact sequence to nd
dim(N2 [k]).
Case 1. k ≥ 2m − 1 Because the maximum degree of
xi y j u
is
2m − 2,
there will be no
xi y j u
for
k ≥ 2m − 1.
Case 2. k ≤ 2m − 2 For the maximum degree that
i + j = k − 2.
2m − 2, i = j = m − 2.
So we wish to solve for
k − 2 − i ≤ m − 2 and i ≥ k − m. k−m≤i≤m−2
7
has
For
2m − k − 1
i
and
j
for
Because
is of degree 2, we also know
0 ≤ i, j ≤ m − 2.
k < m, 0 ≤ i ≤ k − 2 has k − 1 solutions.
Because For
j ≤ m − 2,
m ≤ k ≤ 2m − 2,
solutions, which con rms the results of Proposition 4.1.
Conclusion
We have carefully examined the ranks and torsion of program in
Magma,
Ni
for several classes of relations.
x2
Using a
we generated data which we examined for patterns in the structure of
particular, we looked at homogeneous relations in two variables (x and
u
m +y m ),
Ni .
In
q -polynomials (yx−qxy ),
in multiple variables. We discovered patterns in all of these relations and provided a full
proof of the ranks of
N2 (A2 /(xm + y m ))
and a complete description of the behavior of
Ni
for the
16
q -polynomial
algebra.
These results illustrate how the degree and coe cients in a homogeneous
relation can aect the ranks and torsion of
Ni
of the algebra with that relation, particularly with
regard to which primes appear in torsion. There remains much to study about the structure of pattern in the ranks of the
N3
and
N4
Ni .
For example, the pseudo-arithmetic
(Conjectures 4.3.1 and 4.3.2) with relation
be proved, perhaps with previous results [3]. The pattern found in one found in
B2 ,
which suggests a natural isomorphism
B2 → N2 ,
that this mapping is an isomorphism by using the description of In addition, the behavior of torsion can be studied for
N2
is yet to
for that relation matches the
at least over
A/M3
x2 = 0
xm + y m
Q.
We plan to show
by generators and relations.
of multiple variables, and in
particular, Conjecture 4.1.1 could be proved. More work could be done in discovering which primes appear in the torsion, and why they appear. The study of
Ni
and its structure has applications in the study of commutativity in groups
and rings, and it can be used to study maps between groups. cohomology. Studying associative algebra.
Ni
It also has connections in cyclic
can help us build a better understanding of the structure of a general
Outside of its mathematical implications, the lower central series has wider
applications, as noncommutative algebras often appear in quantum theory, so studying lower central series ideas can help build our fundamental understanding of the universe, as well as bring about the technological advancements quantum theory promises.
17
8
Acknowledgments
I would like to acknowledge my mentor Mr. Teng Fei and Professor Pavel Etingof (Massachusetts Institute of Technology), who suggested the problem. Their guidance has been invaluable in the completion of this project.
I would also like to thank the MIT math department for being so
welcoming, particular Tanya Khovanova, who provided me with excellent direction on math research. I would like to thank Dr. John Rickert for his help in writing papers, as well as all the feedback he gave me. In addition, I would like to thank Sitan Chen for his time in helping me revise my paper. I would like to acknowledge the Center for Excellence in Education and the Research Science Institute (RSI) for providing me the opportunity to conduct research. Finally, I would like to extend my gratitude to Dr. Samuel Gilbert, Dr. Elisabeth Vrahoupoulou, Mr. Ken Panos from Aerojet, and Mr. Zachary Lemnios, Dr. Laura Adol e, Dr John Fischer, and Dr. Robin Sta n from the United States Department of Defense for sponsoring my stay at RSI and making my research possible.
18
A
Data Tables
The tables contain the free and torsion components of
A.1 Ni [d] N2 N3 N4 N5 N6 N7 N8 N9 N10
2
3
4
5
6
7
8
9
10
1
0(32 )
0(33 )
0(34 )
0(35 )
0(36 )
0(37 )
0(38 )
0(39 )
0 0 0 0 0 0
3
2
0(3 )
0
0
0(3
0
0
0
0(3 )
8
0(3 )
0(3 )
7
8
0(3 )
6
0
0(3 )
7
0(3 )
0
8
0(3 )
6 6
5
0(3 )
7
0(3 )
· 9)
8
0(3 )
6
0(3 ) 4
7
0(3 )
5
0
0
6
0(3 )
4
0
0
0(3 ) 0(3 )
0
0
5
4
3
0
4
0(3
0
7
0(3 ) 8
· 5)
0(3 ) 7
7
0(3
· 9)
11 0(310 )
9
0(310 )
9
0(310 )
9
0(310 )
9
0(310 )
9
0(310 )
0(3 ) 0(3 ) 0(3 ) 0(3 ) 0(3 ) 0(3 ) 9
0
0
0
0
0
0
8
0(3
0
0
0
0
0
0
0
0
9
2
3
4 2
5 3
6
0(310 )
yx + 2xy 7
5
0(310 )
· 11)
8 6
9
0(4 )
0(4 )
0(4 )
0(4 )
0(4 )
0(4 )
0(4 )
0(49 )
0
2
0(42
0(44 )
0(45 )
0(46 )
0(47 )
0(48 )
0(49 )
0
0
3
0(44 )
0(45 )
0(46 )
0(47 )
0(48 )
0(49 )
0
0
0
4
0(43
0(46 )
0(47 )
0(48 )
0(49 )
0
0
0
0
5
0(46 )
0(47 )
0(48 )
0(49 )
0
0
0
0
0
6
0(44
0(48 )
0(49 )
0
0
0
0
0
0
7
0(48
0(49 )
0
0
0
0
0
0
0
8
0(45
0
0
0
0
0
0
0
0
9
0
0
0
0
0
0
0
0
0
· 7 · 8)
7
10
1
· 8)
4
0(310 )
9
0
Table 10:
Ni [d] N2 N3 N4 N5 N6 N7 N8
with the relation used in the caption.
Zhx,yi/(yx−qxy)
Table 9:
Ni [d] N2 N3 N4 N5 N6 N7 N8 N9 N10 N11
Ni [d],
8
· 5 · 82 · 16)
· 72 )
· 84 · 61)
yx + 3xy
2
3
4
5
6
7
8
9
10
11
1
0(52 )
0(53 )
0(54 )
0(55 )
0(56 )
0(57 )
0(58 )
0(59 )
0(510 )
0
2
0(3 · 53 )
0(54 )
0(55 )
0(56 )
0(57 )
0(58 )
0(59 )
0(510 )
0 0 0 0 0
0 0 0 0 0
3 0 0 0 0
4
0(5 ) 4 0 0 0
5
6
0(5 ) 2
0(3
0(5 ) 5
· 5 · 13)
6
0(5 ) 6
5
0(5 )
0
6
0
0 Table 11:
7
8
0(5 )
0(5 )
7
0(5 )
7
3
0(3 7
0(5 ) 7
· 5 · 17)
0(5 ) 0(5
· 13
0(510 )
9
0(510 )
9
0(510 )
0(5 )
8 8
0(510 )
9
0(5 )
8
0(5 )
0(510 )
9
0(5 )
8
0(5 )
9
0(5 ) 2
)
0(5 )
yx + 4xy
19
A.2 Zhx, yi/(xm + ym ) Ni [d] N2 N3 N4 N5 N6 N7 N8 N9
1
2
3
0
1
0(2 )
0(2 )
0
0
2
0
0
4
2
0
5
6
2
0(2 )
1(2 )
2
2
7
2
0(2 )
0(2 )
4
4 0(2 )
8
2
0(2 )
2
0 (2 )
0(2 )
4
5 0(2 )
2
0(2 )
4
0(2 )
6 0(2 )
0(2 )
6 0(2 )
0(2 )
4 6
0
0
0
0
4
3 2(2 )
7
0
0
0
0
0
3
0(2 )
0(2 )
0
0
0
0
0
0
6
3(2 )
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
6
7 4
2 Table 12: x
Ni [d] N2 N3 N4 N5 N6 N7 N8
1
2
3
4
0
1
2
1(3 )
0(3 )
5
0
0
2
5
2
6
7
3
0(3 )
3
4(3 )
4
1(3 )
8
3
8
0(3 )
9
9
0(3 )
37 )
0(3 )
0
0
0
3
7(2)
4 4(2
0
0
0
0
6
16(2 )
11(2
0
0
0
0
0
9
22(2 )
0
0
0
0
0
0
18
0
0
0
0
0
0
0
· 2
4 0(2
9
3
0(3 )
·
314 )
0(2
6
· 314 )
· 330 ) 17 · 322 ) 11(2 7 45(2 )
+
· 7
315 )
3
0(3 )
9
0(3 )
15 0(3 )
27
· 335 ) 17 · 343 ) 0(2 21 · 339 ) 30(2 14 3(2 )
2(2
5
3 Table 13: x
Ni [d] N2 N3 N4 N5 N6 N7 N8
+ y2
0(2
4
y3
1
2
3
4
5
6
7
8
9
0
1
2
3
2(42 )
1(43 )
0(44 )
0(44 )
0(44
0
0
2
5
8
7(2 · 43 )
4(22
1(24
0(24
7
1
0
0
0
3
8
13(2 · 5)
0
0
0
0
6
18
· 46 ) 10(2 · 32 · 44 · 52 ) 30(22 · 52 )
0
0
0
0
0
9
30
· 47 ) 4(2 · 34 · 412 · 52 ) 26(213 · 34 · 48 · 55 ) 49(24 · 43 · 55 )
0
0
0
0
0
0
18
63
0
0
0
0
0
0
0
8
4 Table 14: x
30
+
· 48 ) 0(2 0 · 34 · 414 ) 12(226 · 32 · 418 · 56 ) 38(226 · 310 · 420 · 512 ) 106(212 · 43 · 510 ) 110(22 )
y4
20
References [1] Martina Balagovic and Anirudha Balasubramanian. associative algebra.
On the lower central series of a graded
Journal of Algebra, 328:287300, 2011.
[2] Wieb Bosma, John Cannon, and Catherine Playoust. The magma algebra system. i. the user language.
J. Symbolic Output., 24(3-4):235265, 1997.
[3] Pavel Etingof, John Kim, and Xiaoguang Ma. On universal lie nilpotent associative algebras.
Journal of Algebra, 321:697703, 2009.
[4] Boris Feigin and Boris Shoikhet.
On
[A, a]/[a, [a, a]]
commutators of free associative algebras.
and on a
wn -action
on the consecutive
Math. Res. Lett., 14(5):781795, 2007.
[5] George Kerchev. On the ltration of a free algebra by its associative lower central series.
print, arXiv(1101.5741v1), 2011.
pre-
