B ACKGROUND
I NTRODUCTION
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
I
A = L1 ⊃ L2 ⊃ · · · .
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
I
A = L1 ⊃ L2 ⊃ · · · .
I
L2 is spanned by commutators, [a, b], for a, b ∈ A.
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
I
A = L1 ⊃ L2 ⊃ · · · .
I
L2 is spanned by commutators, [a, b], for a, b ∈ A.
I
L3 is spanned by double commtuators, [a, [b, c]].
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
I
A = L1 ⊃ L2 ⊃ · · · .
I
L2 is spanned by commutators, [a, b], for a, b ∈ A.
I
L3 is spanned by double commtuators, [a, [b, c]].
I
L4 is spanned by triple commutators, [a, [b, [c, d]]].
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
I
A = L1 ⊃ L2 ⊃ · · · .
I
L2 is spanned by commutators, [a, b], for a, b ∈ A.
I
L3 is spanned by double commtuators, [a, [b, c]].
I
L4 is spanned by triple commutators, [a, [b, [c, d]]].
I
And so on . . .
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
L OWER CENTRAL SERIES
One way to study interaction between commutative and
non-commutative algebras is via the lower central series:
I
A = L1 ⊃ L2 ⊃ · · · .
I
L2 is spanned by commutators, [a, b], for a, b ∈ A.
I
L3 is spanned by double commtuators, [a, [b, c]].
I
L4 is spanned by triple commutators, [a, [b, [c, d]]].
I
And so on . . .
I
Each successive quotient Bk := Lk /Lk+1 keeps track of more
and more non-commutativity.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
These quotients have surprising descriptions. Consider the free
algebra An over C on x1 , . . . xn .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
These quotients have surprising descriptions. Consider the free
algebra An over C on x1 , . . . xn .
I
B1 (An ) has a basis of cyclic words (e.g. x1 x2 x3 ∼ x3 x1 x2 ).
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
These quotients have surprising descriptions. Consider the free
algebra An over C on x1 , . . . xn .
I
I
B1 (An ) has a basis of cyclic words (e.g. x1 x2 x3 ∼ x3 x1 x2 ).
Feigin-Shoikhet: B2 (An ) ∼
= Ωev (Cn ), the space of
ex
even-degree, exact differential forms on Cn .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
These quotients have surprising descriptions. Consider the free
algebra An over C on x1 , . . . xn .
I
I
B1 (An ) has a basis of cyclic words (e.g. x1 x2 x3 ∼ x3 x1 x2 ).
Feigin-Shoikhet: B2 (An ) ∼
= Ωev (Cn ), the space of
ex
even-degree, exact differential forms on Cn .
I
Etingof and Dobrovolska showed that each Bk (An ), for
k > 2, is also described in differential geometric terms.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
These quotients have surprising descriptions. Consider the free
algebra An over C on x1 , . . . xn .
I
I
B1 (An ) has a basis of cyclic words (e.g. x1 x2 x3 ∼ x3 x1 x2 ).
Feigin-Shoikhet: B2 (An ) ∼
= Ωev (Cn ), the space of
ex
even-degree, exact differential forms on Cn .
I
Etingof and Dobrovolska showed that each Bk (An ), for
k > 2, is also described in differential geometric terms.
I
In particular, for k ≥ 2, each Bk (An ) has polynomial growth.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
These quotients have surprising descriptions. Consider the free
algebra An over C on x1 , . . . xn .
I
I
B1 (An ) has a basis of cyclic words (e.g. x1 x2 x3 ∼ x3 x1 x2 ).
Feigin-Shoikhet: B2 (An ) ∼
= Ωev (Cn ), the space of
ex
even-degree, exact differential forms on Cn .
I
Etingof and Dobrovolska showed that each Bk (An ), for
k > 2, is also described in differential geometric terms.
I
In particular, for k ≥ 2, each Bk (An ) has polynomial growth.
I
The present focus is on extending these methods to
characteristic p, where a geometric approach is less clear.
B ACKGROUND
I NTRODUCTION
Lower central series of free algebras in
characteristic p
Surya Bhupatiraju, William Kuszmaul, Jason Li
MIT PRIMES
May 21, 2011
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
e.g. A2 :
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
e.g. A2 :
1
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
e.g. A2 :
1
xy
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
e.g. A2 :
1
xy
x2 xy yx y2
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
e.g. A2 :
1
xy
x2 xy yx y2
x3 x2 y xyx yx2 y2 x yxy xy2 y3
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
F REE ALGEBRAS
I
Let An be a free algebra on generators x1 , . . . , xn .
I
An is spanned by all words in letters x1 , . . . , xn ,
e.g. A2 :
1
xy
x2 xy yx y2
x3 x2 y xyx yx2 y2 x yxy xy2 y3
I
We can take coefficients in the rational numbers Q, integers
Z, or a finite field Fp .
B ACKGROUND
I NTRODUCTION
C OMMUTATOR
Definition
The commutator, [c, d], for c, d ∈ A, is: [c, d] := cd − dc
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
C OMMUTATOR
Definition
The commutator, [c, d], for c, d ∈ A, is: [c, d] := cd − dc
We can iterate:
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
C OMMUTATOR
Definition
The commutator, [c, d], for c, d ∈ A, is: [c, d] := cd − dc
We can iterate:
I
[b, [c, d]] = [b, cd − dc] = bcd − bdc − cdb + dcb
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
C OMMUTATOR
Definition
The commutator, [c, d], for c, d ∈ A, is: [c, d] := cd − dc
We can iterate:
I
[b, [c, d]] = [b, cd − dc] = bcd − bdc − cdb + dcb
I
[a, [b, [c, d]]] = [a, bcd − bdc − cdb + dcb] =
abcd − abdc − acdb + adcb − bcda + bdca + cdba − dcba
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
L OWER CENTRAL SERIES
Definition
The lower central series filtration of an associative algebra A
A = L1 ⊇ L2 ⊇ L3 ⊇ . . .
is defined recursively by L1 := A, Lk := [A, Lk−1 ], all linear
combinations of expressions [a, b] for a ∈ A, b ∈ Lk−1 .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
L OWER CENTRAL SERIES
Definition
The lower central series filtration of an associative algebra A
A = L1 ⊇ L2 ⊇ L3 ⊇ . . .
is defined recursively by L1 := A, Lk := [A, Lk−1 ], all linear
combinations of expressions [a, b] for a ∈ A, b ∈ Lk−1 .
More explicitly, Lk is all linear combinations of all expressions
[a1 , [a2 , [a3 , [. . . [ak−1 , ak ] . . .]]]] for ai ∈ A.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
L OWER CENTRAL SERIES
Definition
The lower central series filtration of an associative algebra A
A = L1 ⊇ L2 ⊇ L3 ⊇ . . .
is defined recursively by L1 := A, Lk := [A, Lk−1 ], all linear
combinations of expressions [a, b] for a ∈ A, b ∈ Lk−1 .
More explicitly, Lk is all linear combinations of all expressions
[a1 , [a2 , [a3 , [. . . [ak−1 , ak ] . . .]]]] for ai ∈ A.
Definition
The associated graded components Bk to the filtration are
defined as
Bk := Lk /Lk+1 .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
L OWER CENTRAL SERIES
I
Bk are vector spaces when the coefficients are taken in Fp or
Q, and are only abelian groups when coefficients are in Z.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
L OWER CENTRAL SERIES
I
Bk are vector spaces when the coefficients are taken in Fp or
Q, and are only abelian groups when coefficients are in Z.
B1
L1
B2
L2
L3
B3
...
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
H ILBERT S ERIES
Since the spaces we consider are infinite dimensional, we study
them combinatorially via so-called ’Hilbert series’:
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
H ILBERT S ERIES
Since the spaces we consider are infinite dimensional, we study
them combinatorially via so-called ’Hilbert series’:
Definition
A finite m-grading on a vector space V is a direct sum
decomposition:
M
V=
Vk ,
k∈Zm
+
such that each Vk is finite dimensional.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
H ILBERT S ERIES
Since the spaces we consider are infinite dimensional, we study
them combinatorially via so-called ’Hilbert series’:
Definition
A finite m-grading on a vector space V is a direct sum
decomposition:
M
V=
Vk ,
k∈Zm
+
such that each Vk is finite dimensional.
Definition
The multivariable Hilbert series of V is the sum
X
h(V; t1 , . . . , tm ) :=
dim(Vk )tk11 · · · tkmm
k∈Zm
+
B ACKGROUND
I NTRODUCTION
H ILBERT S ERIES
Example
The Hilbert series of A2 is:
X k + l
1
tk1 tl2 =
h(A2 ; t1 , t2 ) :=
k
1 − (t1 + t2 )
m
k∈Z+
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
H ILBERT S ERIES
Example
The Hilbert series of A2 is:
X k + l
1
tk1 tl2 =
h(A2 ; t1 , t2 ) :=
k
1 − (t1 + t2 )
m
k∈Z+
Example
The Hilbert series of C[x, y] is:
h(A2 ; t1 , t2 ) :=
X
k∈Zm
+
tk1 tl2 =
1
(1 − t1 )(1 − t2 )
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T HE P ROBLEM
Problem
Calculate the spaces Bk (An (Fp )), and compare them to Bk (An (Q)).
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T HE P ROBLEM
Problem
Calculate the spaces Bk (An (Fp )), and compare them to Bk (An (Q)).
I
Using the iterative definition of the Lk , and the
programming language MAGMA, we computed the
following Hilbert series:
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T HE P ROBLEM
Problem
Calculate the spaces Bk (An (Fp )), and compare them to Bk (An (Q)).
I
Using the iterative definition of the Lk , and the
programming language MAGMA, we computed the
following Hilbert series:
h(B2 (A2 (Q)); x, y)
xy
= xy + xy2 + x2 y + x3 y + x2 y2 + xy3 + . . . = (1−x)(1−y) .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T HE P ROBLEM
Problem
Calculate the spaces Bk (An (Fp )), and compare them to Bk (An (Q)).
I
Using the iterative definition of the Lk , and the
programming language MAGMA, we computed the
following Hilbert series:
h(B2 (A2 (Q)); x, y)
xy
= xy + xy2 + x2 y + x3 y + x2 y2 + xy3 + . . . = (1−x)(1−y) .
h(B2 (A2 (F2 )); x, y)
xy
= xy + xy2 + x2 y + x3 y + x2 y2 + xy3 + . . . = (1−x)(1−y) .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T HE P ROBLEM
Problem
Calculate the spaces Bk (An (Fp )), and compare them to Bk (An (Q)).
I
Using the iterative definition of the Lk , and the
programming language MAGMA, we computed the
following Hilbert series:
h(B2 (A2 (Q)); x, y)
xy
= xy + xy2 + x2 y + x3 y + x2 y2 + xy3 + . . . = (1−x)(1−y) .
h(B2 (A2 (F2 )); x, y)
xy
= xy + xy2 + x2 y + x3 y + x2 y2 + xy3 + . . . = (1−x)(1−y) .
They appear to coincide.
B ACKGROUND
I NTRODUCTION
D IFFERENCES IN H ILBERT S ERIES
The series h(B2 (A2 (Fp ))), h(B3 (A2 (Fp ))), h(B4 (A2 (Fp ))) are
independent of p in the range we computed.
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
D IFFERENCES IN H ILBERT S ERIES
The series h(B2 (A2 (Fp ))), h(B3 (A2 (Fp ))), h(B4 (A2 (Fp ))) are
independent of p in the range we computed.
I
h(B2 (A3 (Q))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z + 2xyz2 + 2xy2 z2 + 2x2 y2 z2 + · · ·
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
D IFFERENCES IN H ILBERT S ERIES
The series h(B2 (A2 (Fp ))), h(B3 (A2 (Fp ))), h(B4 (A2 (Fp ))) are
independent of p in the range we computed.
I
h(B2 (A3 (Q))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z + 2xyz2 + 2xy2 z2 + 2x2 y2 z2 + · · ·
I
h(B2 (A3 (F2 ))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z + 2xyz2 + 2xy2 z2 + 3x2 y2 z2 + · · ·
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
D IFFERENCES IN H ILBERT S ERIES
The series h(B2 (A2 (Fp ))), h(B3 (A2 (Fp ))), h(B4 (A2 (Fp ))) are
independent of p in the range we computed.
I
h(B2 (A3 (Q))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z + 2xyz2 + 2xy2 z2 + 2x2 y2 z2 + · · ·
I
h(B2 (A3 (F2 ))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z + 2xyz2 + 2xy2 z2 + 3x2 y2 z2 + · · ·
I
Why do these changes occur?
B ACKGROUND
I NTRODUCTION
T ORSION
To explain the varying behavior over different Fp , we work
instead over Z.
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T ORSION
To explain the varying behavior over different Fp , we work
instead over Z.
∼ B2 (An (Z)) ⊗ F.
I We have an isomorphism: B2 (An (F)) =
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T ORSION
To explain the varying behavior over different Fp , we work
instead over Z.
∼ B2 (An (Z)) ⊗ F.
I We have an isomorphism: B2 (An (F)) =
I
B2 (An (Z)) is an abelian group, and may have torsion:
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T ORSION
To explain the varying behavior over different Fp , we work
instead over Z.
∼ B2 (An (Z)) ⊗ F.
I We have an isomorphism: B2 (An (F)) =
I
B2 (An (Z)) is an abelian group, and may have torsion:
Definition
An element g 6= 0 is called an m-torsion element if mg = 0, and
m > 0 is the minimal such.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T ORSION
To explain the varying behavior over different Fp , we work
instead over Z.
∼ B2 (An (Z)) ⊗ F.
I We have an isomorphism: B2 (An (F)) =
I
B2 (An (Z)) is an abelian group, and may have torsion:
Definition
An element g 6= 0 is called an m-torsion element if mg = 0, and
m > 0 is the minimal such.
I
An m-torsion element g ∈ B2 (An (Z)) survives to
B2 (An (Fp )) if, and only if, p divides m.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T ORSION
To explain the varying behavior over different Fp , we work
instead over Z.
∼ B2 (An (Z)) ⊗ F.
I We have an isomorphism: B2 (An (F)) =
I
B2 (An (Z)) is an abelian group, and may have torsion:
Definition
An element g 6= 0 is called an m-torsion element if mg = 0, and
m > 0 is the minimal such.
I
An m-torsion element g ∈ B2 (An (Z)) survives to
B2 (An (Fp )) if, and only if, p divides m.
I
Torsion does not survive to B2 (An (Q)).
B ACKGROUND
I NTRODUCTION
For example,
B2 (A3 (Z))(2,2,2) ∼
= Z ⊕ Z ⊕ Z/2Z.
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
For example,
B2 (A3 (Z))(2,2,2) ∼
= Z ⊕ Z ⊕ Z/2Z.
I
B2 (A3 (Q))(2,2,2) ∼
= (Z ⊕ Z ⊕ Z/2Z) ⊗ Q ∼
= Q ⊕ Q.
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
For example,
B2 (A3 (Z))(2,2,2) ∼
= Z ⊕ Z ⊕ Z/2Z.
I
I
B2 (A3 (Q))(2,2,2) ∼
= (Z ⊕ Z ⊕ Z/2Z) ⊗ Q ∼
= Q ⊕ Q.
h(B2 (A3 (Q))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z2 + 2xy2 z + 2xyz2 + 2x2 y2 z2 + · · ·
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
For example,
B2 (A3 (Z))(2,2,2) ∼
= Z ⊕ Z ⊕ Z/2Z.
I
I
I
B2 (A3 (Q))(2,2,2) ∼
= (Z ⊕ Z ⊕ Z/2Z) ⊗ Q ∼
= Q ⊕ Q.
h(B2 (A3 (Q))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z2 + 2xy2 z + 2xyz2 + 2x2 y2 z2 + · · ·
B2 (A3 (F2 ))(2,2,2) ∼
= (Z ⊕ Z ⊕ Z/2Z) ⊗ F2 ∼
= F2 ⊕ F2 ⊕ F2 .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
For example,
B2 (A3 (Z))(2,2,2) ∼
= Z ⊕ Z ⊕ Z/2Z.
I
I
I
I
B2 (A3 (Q))(2,2,2) ∼
= (Z ⊕ Z ⊕ Z/2Z) ⊗ Q ∼
= Q ⊕ Q.
h(B2 (A3 (Q))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z2 + 2xy2 z + 2xyz2 + 2x2 y2 z2 + · · ·
B2 (A3 (F2 ))(2,2,2) ∼
= (Z ⊕ Z ⊕ Z/2Z) ⊗ F2 ∼
= F2 ⊕ F2 ⊕ F2 .
h(B2 (A3 (F2 ))) = xy + xz + yz + x2 y + x2 z + xy2 + 2xyz +
xz2 + y2 z + yz2 + x2 z2 + y2 z2 + x2 y2 + 2x2 yz + 2x2 y2 z +
2x2 yz2 + 2xy2 z2 + 2xy2 z + 2xyz2 + 3x2 y2 z2 + · · ·
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
I
2-torsion in B2 (A3 (Z))(2,2,2) ,
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
I
2-torsion in B2 (A3 (Z))(2,2,2) ,
I
3-torsion in B2 (A3 (Z))(3,3,3) ,
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
I
2-torsion in B2 (A3 (Z))(2,2,2) ,
I
3-torsion in B2 (A3 (Z))(3,3,3) ,
I
2-torsion in B2 (A3 (Z))(4,2,2) , and its permutations,
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
I
2-torsion in B2 (A3 (Z))(2,2,2) ,
I
3-torsion in B2 (A3 (Z))(3,3,3) ,
I
2-torsion in B2 (A3 (Z))(4,2,2) , and its permutations,
I
2-torsion in B2 (A3 (Z))(4,4,2) , and its permutations,
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
I
2-torsion in B2 (A3 (Z))(2,2,2) ,
I
3-torsion in B2 (A3 (Z))(3,3,3) ,
I
2-torsion in B2 (A3 (Z))(4,2,2) , and its permutations,
I
2-torsion in B2 (A3 (Z))(4,4,2) , and its permutations,
I
2-torsion in B2 (A4 (Z))(2,2,2,2) (3-dimensional).
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
R ESULTS FROM C OMPUTATIONS
Some places where we found torsion are:
I
2-torsion in B2 (A3 (Z))(2,2,2) ,
I
3-torsion in B2 (A3 (Z))(3,3,3) ,
I
2-torsion in B2 (A3 (Z))(4,2,2) , and its permutations,
I
2-torsion in B2 (A3 (Z))(4,4,2) , and its permutations,
I
2-torsion in B2 (A4 (Z))(2,2,2,2) (3-dimensional).
Conjecture
For all a, b, c, the element:
v(a, b, c) = [z, za−1 xb−1 yc−1 [x, y]] ∈ B2 (A3 (Z))(a,b,c) ,
is torsion, of order equal to gcd(a, b, c), and generates the
torsion subgroup of B2 (A3 (Z))(a,b,c) .
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
I
⇒ cv = [z, za−1 [x, xb−1 yc ]].
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
I
⇒ cv = [z, za−1 [x, xb−1 yc ]].
We use the identity in A4 (Z) :
I
[z, w[x, y]] = [[w, y], xz] − [z, [y, wx]] + [x, [w, zy]] + x[z, w]y + [w, z]yx.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
I
⇒ cv = [z, za−1 [x, xb−1 yc ]].
We use the identity in A4 (Z) :
I
[z, w[x, y]] = [[w, y], xz] − [z, [y, wx]] + [x, [w, zy]] + x[z, w]y + [w, z]yx.
I
Set z 7→ z, w 7→ za−1 , x 7→ x, y 7→ xb−1 yc .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
I
⇒ cv = [z, za−1 [x, xb−1 yc ]].
We use the identity in A4 (Z) :
I
[z, w[x, y]] = [[w, y], xz] − [z, [y, wx]] + [x, [w, zy]] + x[z, w]y + [w, z]yx.
I
Set z 7→ z, w 7→ za−1 , x 7→ x, y 7→ xb−1 yc .
I
⇒ cv ∈ L3 (A3 (Z)) ⇒ cv = 0 in B2 (A3 (Z)).
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
I
⇒ cv = [z, za−1 [x, xb−1 yc ]].
We use the identity in A4 (Z) :
I
[z, w[x, y]] = [[w, y], xz] − [z, [y, wx]] + [x, [w, zy]] + x[z, w]y + [w, z]yx.
I
Set z 7→ z, w 7→ za−1 , x 7→ x, y 7→ xb−1 yc .
I
⇒ cv ∈ L3 (A3 (Z)) ⇒ cv = 0 in B2 (A3 (Z)).
I
The claim that av = bv = 0 is proved similarly.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
T OWARDS THE CONJECTURE
It is not hard to prove that gcd(a, b, c) · [z, za−1 xb−1 yc−1 [x, y]] = 0
in B2 (A3 (Z)) :
I
Feigin-Shoikhet: [A, AL3 ] ⊂ L3 .
I
[z, za−1 [x, xb−1 yc ]] = c[z, za−1 xb−1 yc−1 [x, y]] mod L3 .
I
⇒ cv = [z, za−1 [x, xb−1 yc ]].
We use the identity in A4 (Z) :
I
[z, w[x, y]] = [[w, y], xz] − [z, [y, wx]] + [x, [w, zy]] + x[z, w]y + [w, z]yx.
I
Set z 7→ z, w 7→ za−1 , x 7→ x, y 7→ xb−1 yc .
I
⇒ cv ∈ L3 (A3 (Z)) ⇒ cv = 0 in B2 (A3 (Z)).
I
The claim that av = bv = 0 is proved similarly.
However, we still do not know v is nonzero (if gcd(a, b, c) > 1)!
Computation confirms this in small cases.
B ACKGROUND
Future goals:
I NTRODUCTION
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
Future goals:
I
Understand torsion in B2 (An (Z)).
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
Future goals:
I
Understand torsion in B2 (An (Z)).
I
Understand torsion in other Bk (An (Z)).
T HE P ROBLEM
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
Future goals:
I
I
Understand torsion in B2 (An (Z)).
I
Understand torsion in other Bk (An (Z)).
Conjecture
Bk (An (Fp )) has polynomial growth for all k and p (more
precisely the coefficients of the multivariable Hilbert series are
bounded).
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
Future goals:
I
I
Understand torsion in B2 (An (Z)).
I
Understand torsion in other Bk (An (Z)).
Conjecture
Bk (An (Fp )) has polynomial growth for all k and p (more
precisely the coefficients of the multivariable Hilbert series are
bounded).
I
Relate to geometry in characteristic p.
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
Future goals:
I
I
Understand torsion in B2 (An (Z)).
I
Understand torsion in other Bk (An (Z)).
Conjecture
Bk (An (Fp )) has polynomial growth for all k and p (more
precisely the coefficients of the multivariable Hilbert series are
bounded).
I
Relate to geometry in characteristic p.
I
We found no torsion in B2 (A2 (Z)), B3 (A2 (Z)), B4 (A2 (Z)).
We can conjecture there is no torsion in Bk (A2 (Z)). . .
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
Future goals:
I
I
Understand torsion in B2 (An (Z)).
I
Understand torsion in other Bk (An (Z)).
Conjecture
Bk (An (Fp )) has polynomial growth for all k and p (more
precisely the coefficients of the multivariable Hilbert series are
bounded).
I
Relate to geometry in characteristic p.
I
We found no torsion in B2 (A2 (Z)), B3 (A2 (Z)), B4 (A2 (Z)).
We can conjecture there is no torsion in Bk (A2 (Z)). . .
But there exists a 2-torsion element in B5 (A2 (Z))(4,4) !!
B ACKGROUND
I NTRODUCTION
T HE P ROBLEM
A CKNOWLEDGEMENTS
I
Thanks to Martina Balagovic for her thorough edits with
the presentation.
I
Thanks to Pavel Etingof for suggestions and help for many
aspects of the project.
I
Many, many thanks to our mentor David Jordan, for his
countless hours of assistance and help =)
I
Thank you PRIMES!