Computing with lattices over group rings of finite groups
Florian Eisele
(joint with L. Margolis)
City, University of London
Groups, Rings and the Yang-Baxter Equation, Spa
June 2017
Lattices
• p: a prime.
• Z(p) = { x ∈ Q | p does not divide y } ⊂ Q
y
• A: a Z(p) -algebra which is free and finitely generated as a Z(p) -module.
• Assume QA := Q ⊗Z A is semisimple.
(p)
Lattices
• p: a prime.
• Z(p) = { x ∈ Q | p does not divide y } ⊂ Q
y
• A: a Z(p) -algebra which is free and finitely generated as a Z(p) -module.
• Assume QA := Q ⊗Z A is semisimple.
(p)
For our purposes: A = Z(p) G for a finite group G , or A = eZ(p) G , where e ∈ Z (QG ) is
an idempotent.
Lattices
• p: a prime.
• Z(p) = { x ∈ Q | p does not divide y } ⊂ Q
y
• A: a Z(p) -algebra which is free and finitely generated as a Z(p) -module.
• Assume QA := Q ⊗Z A is semisimple.
(p)
For our purposes: A = Z(p) G for a finite group G , or A = eZ(p) G , where e ∈ Z (QG ) is
an idempotent.
Definition (Lattice)
Let V be a finite-dimensional QA-module. An A-submodule L 6 V is called a lattice if
• L is finitely generated as a Z(p) -module.
• L spans V as a Q-vector space.
Jordan-Zassenhaus Theorem
For any finite-dimensional QA-module V there are, up to isomorphism, only finitely
many A-lattices L 6 V .
Zassenhaus conjecture via lattices
• G a finite group.
T
• Z|G | = { x ∈ Q | gcd(|G |, y ) = 1} = p||G | Z(p)
y
Zassenhaus conjecture via lattices
• G a finite group.
T
• Z|G | = { x ∈ Q | gcd(|G |, y ) = 1} = p||G | Z(p)
y
Semi-local version of the Zassenhaus conjecture
Let u ∈ Z|G | G be a unit of finite order n. Then there is an invertible a ∈ QG such that
aua−1 = ±g for some g ∈ G .
Zassenhaus conjecture via lattices
• G a finite group.
T
• Z|G | = { x ∈ Q | gcd(|G |, y ) = 1} = p||G | Z(p)
y
Semi-local version of the Zassenhaus conjecture
Let u ∈ Z|G | G be a unit of finite order n. Then there is an invertible a ∈ QG such that
aua−1 = ±g for some g ∈ G .
Lattice version (equivalent)
Let H = G × hx : x n = 1i. If there is a Z|G | H-lattice L such that L|G ∼
= Z|G | G (i.e. it
is free of rank one), then QL ∼
= QM(g ) for some g ∈ G , where
• M(g ) = Z|G | G as a set
• G acts by left-multiplication.
• x acts by multiplication by ±g −1 from the right.
Zassenhaus conjecture via lattices
• G a finite group.
T
• Z|G | = { x ∈ Q | gcd(|G |, y ) = 1} = p||G | Z(p)
y
Semi-local version of the Zassenhaus conjecture
Let u ∈ Z|G | G be a unit of finite order n. Then there is an invertible a ∈ QG such that
aua−1 = ±g for some g ∈ G .
Lattice version (equivalent)
Let H = G × hx : x n = 1i. If there is a Z|G | H-lattice L such that L|G ∼
= Z|G | G (i.e. it
is free of rank one), then QL ∼
= QM(g ) for some g ∈ G , where
• M(g ) = Z|G | G as a set
• G acts by left-multiplication.
• x acts by multiplication by ±g −1 from the right.
Going back and forth
• A given unit u as above turns Z|G | G into an H-module by letting x act by right
multiplication by u −1 .
• Given L as above, fixing an isomorphism ϕ : L|G −→ Z|G | G gives a unit
u = ϕ(x −1 · ϕ−1 (1G ))
Zassenhaus conjecture via lattices (continued)
H = G × hx : x n i
Upshot from the last slide
The (semi-local) Zassenhaus conjecture is equivalent to the non-existence of certain
Z|G | H-lattices L with L|G ∼
= Z|G | G .
Zassenhaus conjecture via lattices (continued)
H = G × hx : x n i
Upshot from the last slide
The (semi-local) Zassenhaus conjecture is equivalent to the non-existence of certain
Z|G | H-lattices L with L|G ∼
= Z|G | G .
For a given (small) group G :
• “Partial augmentations” of a potential unit u
QG -module V in which the
potential Z|G | H-lattice L lies.
• The problem is local:
L exists ⇐⇒ there are Z(p) H-lattices L(p) 6 V with L(p) |G ∼
= Z(p) G for all p | |G |
• If p - n (the order of the unit), then L(p) is projective =⇒ existence easy to check
• By Jordan-Zassenhaus, there are only finitely many isomorphism classes of
lattices in V
finding or disproving the existence of each L(p) is a finite problem
(possibly intractable).
Problem: Given G and V , can we construct L(p) or disprove its existence using a
computer?
Computing with lattices
Given an QA-module V (given by matrices representing the generators of A), one can:
• find an A-lattice L 6 V (given by a basis).
• Compute all maximal sublattices of a given lattice L 6 V
• Compute Hom’s between lattices.
• Compute the radical and top of a lattice.
• Check if two lattices are isomorphic (easy if V is simple, but always possible).
Computing with lattices
Given an QA-module V (given by matrices representing the generators of A), one can:
• find an A-lattice L 6 V (given by a basis).
• Compute all maximal sublattices of a given lattice L 6 V
• Compute Hom’s between lattices.
• Compute the radical and top of a lattice.
• Check if two lattices are isomorphic (easy if V is simple, but always possible).
To check if L|G is free of rank one we can use the following:
Checking for projectivity
An A-lattice L is isomorphic to the projective cover P(S) of the semi-simple module S
if and only if
• rank(L) = rank(P(S))
• L/rad(L) ∼
=S
All of this is implemented in GAP (originally to construct indecomposable projective
modules).
Further reductions
Again: G finite group, H = G × hx : x n i
V a QH-module with V |G ∼
= QG
p a prime dividing n
Wedderburn component by Wedderburn component
Let e ∈ Z (QG ) be an idempotent. Then e · V is a QH-module. We have (trivially)
L 6 V with L|G ∼
= Z(p) G =⇒ eL|G ∼
= eZ(p) G
Further reductions
Again: G finite group, H = G × hx : x n i
V a QH-module with V |G ∼
= QG
p a prime dividing n
Wedderburn component by Wedderburn component
Let e ∈ Z (QG ) be an idempotent. Then e · V is a QH-module. We have (trivially)
L 6 V with L|G ∼
= Z(p) G =⇒ eL|G ∼
= eZ(p) G
This suggests the following strategy:
• For all primitive idempotents e ∈ Z (QG ): Determine all lattices in eV with
restriction eZ(p) G .
• Given orthogonal idempotents e1 , e2 ∈ Z (QG ) and lattices Li 6 ei V with
L i |G ∼
= ei Z(p) G :
• Compute all lattices L in L1 ⊕ L2 with surjective projections onto both summands.
• Pick out those L with L|G ∼
= (e1 + e2 )Z(p) G
Further reductions
Again: G finite group, H = G × hx : x n i
V a QH-module with V |G ∼
= QG
p a prime dividing n
Wedderburn component by Wedderburn component
Let e ∈ Z (QG ) be an idempotent. Then e · V is a QH-module. We have (trivially)
L 6 V with L|G ∼
= Z(p) G =⇒ eL|G ∼
= eZ(p) G
This suggests the following strategy:
• For all primitive idempotents e ∈ Z (QG ): Determine all lattices in eV with
restriction eZ(p) G .
• Given orthogonal idempotents e1 , e2 ∈ Z (QG ) and lattices Li 6 ei V with
L i |G ∼
= ei Z(p) G :
• Compute all lattices L in L1 ⊕ L2 with surjective projections onto both summands.
• Pick out those L with L|G ∼
= (e1 + e2 )Z(p) G
Remarks
• A lattice as above in eV corresponds to a unit in eZ(p) G .
• Given L1 and L2 , a lattice L as desired can only exist if
L1 /radZ(p) G (L1 ) ∼
= L2 /radZ(p) G (L2 )
as Z(p) H-modules
Some partial results
• SG(144, 182): two partial augmentations for units of order n = 6. These units
exist 2-locally but not 3-locally.
Some partial results
• SG(144, 182): two partial augmentations for units of order n = 6. These units
exist 2-locally but not 3-locally.
• SG(160, 234): two partial augmentations for units of order n = 4, one for order
n = 2. At least two of those cannot exist.
Some partial results
• SG(144, 182): two partial augmentations for units of order n = 6. These units
exist 2-locally but not 3-locally.
• SG(160, 234): two partial augmentations for units of order n = 4, one for order
n = 2. At least two of those cannot exist.
• SG(192, 955): partial augmentations for unit of order n = 2. Cannot exist.
Some partial results
• SG(144, 182): two partial augmentations for units of order n = 6. These units
exist 2-locally but not 3-locally.
• SG(160, 234): two partial augmentations for units of order n = 4, one for order
n = 2. At least two of those cannot exist.
• SG(192, 955): partial augmentations for unit of order n = 2. Cannot exist.
• SG(192, 973), SG(192, 974), SG(192, 975), SG(192, 976), SG(192, 1489),
SG(192, 1490): in each case one or two partial augmentations for a unit of order
n = 8.
Analogous situation in each case. Partial result: for SG(192, 1490) we get a unit
u ∈ (1 − e) · Z(2) G
where u is a primitive idempotent in Z (QG ) belonging to a Wedderburn
component of dimension 72 = 2 × 62 .