Abstract commensurators of lattices in Lie groups
Daniel Studenmund
University of Chicago
Spring Topology and Dynamics Conference, 24 March 2013
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Definition and Examples
Partial automorphisms
Let Γ be an infinite group.
Definition
A partial automorphism of Γ is an isomorphism φ : H → K between finite
index subgroups H, K ≤ Γ.
Example: Let Γ = Z. Then 2n 7→ 3n defines an isomorphism
φ1 : 2Z → 3Z.
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Definition and Examples
Equivalence of partial automorphisms
Define φ1 ∼ φ2 if there is some finite index subgroup Λ ≤ Γ so that
φ1 Λ = φ2 Λ .
Example: Let φ1 : 2Z → 3Z map 2n 7→ 3n and
φ2 : 4Z → 6Z map
4n 7→ 6n. Then φ1 ∼ φ2 because φ1 4Z = φ2 4Z
Remark: Can’t compose
φ1 : 2Z → 3Z
2n 7→ 3n
and
φ−1
2 : 6Z → 4Z
6n 7→ 4n
But φ1 ∼ φ2 , so can say [φ1 ] ◦ [φ−1
2 ] = [Id].
Key Point: Can compose any two partial automorphisms up to
equivalence.
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Definition and Examples
Abstract commensurator
Definition
The abstract commensurator of Γ, denoted Comm(Γ), is the group of
equivalence classes of partial automorphisms of Γ.
.
Write Γ1 = Γ2 if Γ1 and Γ2 contain isomorphic finite index subgroups.
.
Remark: If Γ1 = Γ2 , then Comm(Γ1 ) = Comm(Γ2 ).
Example
Comm(Z) = Q∗
Proof: [φ] ∈ Comm(Z) satisfying φ(m) = n corresponds to
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
n
m
∈ Q∗ .
24 March 2013
Definition and Examples
Examples
Exercise: Comm(Zn ) = GLn (Q)
(Ivanov) Mapping class groups:
Comm(Mod± (Σg )) = Mod± (Σg ) for g ≥ 2
(Farb–Handel) Outer automorphisms of free groups:
Comm(Out(Fn )) = Out(Fn ) for n ≥ 4
(Leininger–Margalit) Braid groups:
Comm(Bn ) = Mod± (Σ0,n+1 ) n Q× n Q∞
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
for n ≥ 4
24 March 2013
Question and Outline
Motivating question
Question
Suppose Γ is a lattice in a Lie group G . What is Comm(Γ)?
Example: G = Rn and Γ = Zn .
Aut(Zn ) = GLn (Z)
Comm(Zn ) = GLn (Q)
Aut(Rn ) = GLn (R)
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Question and Outline
Outline of approach
Question
Suppose Γ is a lattice in a Lie group G . What is Comm(Γ)?
Levi Decomposition: If G an (algebraically) simply-connected Lie group,
G = G sol o G ss ,
where G sol is the maximal connected solvable normal subgroup,
and G ss is semisimple. Decompose question:
1
Semisimple case
2
Solvable case
3
Combine semisimple and solvable cases
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Semisimple case
Step 1: Lattices in semisimple groups
Example (Margulis, Borel)
Let G = PGLn (R) and Γ = PGLn (Z) for n ≥ 3. Then
.
Comm(PGLn (Z)) = PGLn (Q).
Theorem (Mostow, Prasad, Margulis, Borel)
Let G be a ‘nice’ semisimple Lie group without center, without compact
factors, not PSL2 (R), and Γ ≤ G an irreducible lattice. Then
.
Γ = G (Z) ⇐⇒ Comm(Γ) = G (Q)
.
Γ non-arithmetic ⇐⇒ Comm(Γ) = Γ.
Main tool: Strong rigidity: Commensurations of Γ extend to
automorphisms of G .
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Solvable case
Step 2: Lattices in solvable groups
Theorem (S.)
Suppose Γ is a lattice in a simply-connected solvable Lie group G . Then
there is some linear algebraic group A such that Comm(Γ) = A(Q) and
.
Aut(Γ) = A(Z).
Remark: (Malcev) If G nilpotent then A(R) = Aut(G ).
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Solvable case
Good example: lattice in SOL
Let Γ :=
Z2
oM Z, where M =
2 1
.
1 1
Fact: Every [φ] ∈ Comm(Γ) satisfies φ(H ∩ Z2 ) = K ∩ Z2 .
Restriction induces a map
Comm(Γ) → Comm(Z2 ) = GL2 (Q).
Proposition
There is (virtually) an exact sequence
1 → Q2 → Comm(Γ) → CGL2 (Q) (M) → 1.
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Solvable case
Illustrating difficulty
First attempt at proof: As in semisimple case, try to use rigidity of Γ in
G to extend elements of Comm(Γ) to automorphisms of G .
Key Difficulty: Rigidity fails!
^ and p : SO(2)
^ → SO(2). Let
Example: Let G = R2 o SO(2)
2
−1
Γ = Z o p (0). Then
Γ = Z3
Automorphism swapping a Z factor with p −1 (0) does not extend to
an automorphism of G .
Comm(Γ) = GL3 (Q) while Aut(G ) 6= GL3 (R)
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013
Solvable case
Overcoming the difficulty
Goal: Find linear algebraic group A such that Comm(Γ) = A(Q).
Steps:
1
(Virtually) embed Γ in an algebraic group H with
Comm(Γ) → Aut(H) (the algebraic hull, using ideas of Mostow,
Raghunathan)
2
Find an algebraic subgroup A ≤ Aut(H) such that Comm(Γ) = A(Q)
(techniques of Baues–Grunewald)
Thank you!
Daniel Studenmund (UChicago)
Commensurations of lattices in Lie groups
24 March 2013