Rigidity of boundary actions Kathryn Mann Brown University The PSL - - PowerPoint PPT Presentation

Rigidity of boundary actions

Kathryn Mann

Brown University

The PSL2(R) character variety

Theorem (Goldman ’88) Hom(π1Σg, PSL2(R)) has 4g − 3 connected components, classified by the Euler number.

• 2g+2

...

• 1

1 ... 2g-2

Hom(π1Σg, G)/G ↔ flat principal G-bundles over Σg

The PSL2(R) character variety

Theorem (Goldman ’88) Hom(π1Σg, PSL2(R))/ PSL has 4g − 3 connected comp’s, classified by the Euler number.

• 2g+2

...

• 1

1 ... 2g-2

Hom(π1Σg, G)/G ↔ flat principal G-bundles over Σg Euler number is a characteristic number of oriented S1 bundles. Topologist’s question: Describe the space of all flat, oriented S1 bundles, i.e. Hom(π1Σg, Homeo(S1))/ Homeo(S1)).

More general character “varieties”

Γ discrete group, G topological group Space of representations up to conjugacy: Hom(Γ, G)/G Problem: quotient space typically not Hausdorff e.g. Hom(Z, SL2(C))/ SL2(C) ↔ trace except ( 1 t

0 1 ) = ( 1 0 0 1 )

Solution: Define “character space” X(Γ, G) := largest Hausdorff quotient of Hom(Γ, G)/G

• for SL(n, C) this *is* characters; for G complex, reductive Lie group, is GIT quotient.
• Ex. (Fricke) X(F2, SL2(C)) = C3
• X(Γ, Homeo(S1))= semi-conjugacy classes of actions of Γ

Rigidity

Definition: ρ : Γ → G is rigid if [ρ] ∈ X(Γ, G) an isolated point. “no nontrivial deformations” Mostow rigidity : Γ = π1(Mn) hyperbolic manifold, n > 2 Γ → SO(n, 1) as cocompact lattice is rigid in X(Γ, SO(n, 1)) Fails for n = 2, M = Σg but ... Theorem (Matsumoto ’87) ρ : π1Σg → Homeo(S1) “boundary action” on

Σg

is rigid in X(π1Σg, Homeo(S1)).

More generally, Definition: ρ : Γ → Homeo(M) is geometric if factors through

Γ ֒ → G ֒ → Homeo(M)

cocompact transitive lattice Lie group

Other examples? Fact: Connected, transitive Lie groups in Homeo(S1) are

• SO(2)
• finite cyclic extensions of PSL2(R)

Z/kZ → G → PSL2(R) Cor.: 1. All groups Γ that act geometrically are (virtually) π1Σg

• 2. Can easily describe all geometric actions of π1Σg on S1.

Theorem (Mann, 2014)

If ρ : π1Σg → Homeo(S1) is geometric, then it is rigid.

Theorem (Mann–Wolff, 2017)

Converse: if ρ ∈ X(π1Σg, Homeo(S1)) is rigid, then is geometric. Consequences

• 1. Know all the isolated points of X(π1Σg, Homeo(S1))

(exponentially many in g)

• 2. Euler number does not distinguish connected components
• 3. New rigidity results for other “boundary” group actions

Guiding principle: analogies with classical character varieties

Proof ingredient: coordinates on X(π1Σg, Homeo(S1))

Proof ingredient: coordinates on X(π1Σg, Homeo(S1)) Analogy: trace coordinates on X(Γ, SL2(C))

Analogy: trace coordinates on X(Γ, SL2(C))

tr : SL2(C) → C

• captures some dynamics

(e.g. tr < 2 ⇔rotation)

• continuous, conjugation invariant

tr(ghg−1) = tr(h)

Reps are determined by trace of finitely many elements. Eg. X(F2, SL2(C)) → C3 ρ → tr(ρ(a)), tr(ρ(b)), tr(ρ(ab)) Level set tr ρ(aba−1b−1) = 4

(image: W. Goldman)

Proof ingredient: coordinates on X(Γ, Homeo(S1))

want: conjugation invariant function on Homeo(S1)

Definition (Poincar´ e)

˜ f ∈ Homeo(R) f ∈ Homeo(S1)

rotation number rot(f ) := lim

n→∞ ˜ f n(0) n

(mod Z)

• rot(f ) ∈ Q ⇔ f has periodic orbit.
• Continuous, conjugation invariant.

Theorem* (Ghys, Matsumoto) ρ ∈ X(Γ, Homeo(S1)) is determined by all rotation numbers rot(ρ(γ))

* technically by cocycle rot ρ(a) + rot ρ(b) − rot ρ(a) ρ(b)

The Euler number

Definition Euler(ρ) =

• P pants
• rot

ρ(a) + rot ρ(b) − rot ρ(a) ρ(b)

• a

(ab)−1 b

The Euler number

Definition Euler(ρ) =

• P pants
• rot

ρ(a) + rot ρ(b) − rot ρ(a) ρ(b)

• a

(ab)−1 b

Comparison

C3 = {tr(ρ(a)), tr(ρ(b)) tr(ρ(ab))} = X(F2, SL2(C))

Comparison

C3 = {tr(ρ(a)), tr(ρ(b)) tr(ρ(ab))} = X(F2, SL2(C))

{rot(ρ(a)), rot(ρ(b)) rot(ρ(ab))} for ρ ∈ X(F2, Homeo(S1))

Calegari & Walker “Ziggurats and Rotation numbers”

Comparison

tr(ρ(aba−1b−1)) = 4

Comparison: Open Questions

Fact: There exist nonconjugate a, b ∈ π1Σg such that tr(ρ(a)) = tr(ρ(b)) for all ρ : π1Σg → SL2(C). Question: Can you find a, b with rot(ρ(a)) = rot(ρ(b)) for all ρ : π1Σg → Homeo(S1)? probably not!

Comparison: Open Questions

Fact: There exist nonconjugate a, b ∈ π1Σg such that tr(ρ(a)) = tr(ρ(b)) for all ρ : π1Σg → SL2(C). Question: Can you find a, b with rot(ρ(a)) = rot(ρ(b)) for all ρ : π1Σg → Homeo(S1)? Question: For which infinite classes of curves C does the “marked rot. spectrum” rot(ρ(γ)), γ ∈ C determine ρ?

C = simple closed curves... bounded self-intersection...?

Problem: Draw the “ziggurat” for ab−2ab Or any general word in a±1, b±1

But some things are known ... Sample Lemma: “Rigidity implies rationality” If ρ ∈ X(π1Σg, Homeo(S1)) is rigid, then rot(ρ(γ)) ∈ Q for all simple closed curves γ in π1Σg.

Proof ingredient 2: bending (twist) deformations

Thurston: ρ : π1Σg → PSL2(R) “bent” into PSL2(C) Bending along simple curves is a way to modify π1Σg → G G any topological group π1Σg = A ∗c B

c

Conjugate ρ : π1Σg → G on A, not on B

Applications

Question (Farb 2006) Consider Aut(π1Σg) acting on ∂∞(π1Σg) = S1. Is this the only faithful action of Aut(π1Σg) on S1? Theorem (M – Wolff, 2018) Any nontrivial ρ : Aut(π1Σg) → Homeo(S1) is semiconjugate to the boundary action. Proof. Look at π1Σg → Aut(π1) → Out(π1) = Modg. Understand Modg–equivariant ρ : π1Σg → Homeo(S1).

each pant contributes same amount to Euler(ρ) etc...

Applications

Question (Farb 2006) Consider Aut(π1Σg) acting on ∂∞(π1Σg) = S1. Is this the only faithful action of Aut(π1Σg) on S1? Theorem (M – Wolff, 2018) Any nontrivial ρ : Aut(π1Σg) → Homeo(S1) is semiconjugate to the boundary action. Problem Let Γ < Aut(π1Σg). Classify Γ-equivariant representations ρ : π1Σg → Homeo(S1) e.g. Γ = φ pseudo-Anosov. Classify π1(M3

φ) → Homeo(S1).

Applications II

Ongoing with J. Bowden: boundary actions in higher dimensions. Mn compact, negative curvature. Then π1(M) → Homeo(Sn) boundary action should be (locally) rigid. Perspective: straightening foliations. Major motivating question: Are there analogs of “trace coords” for other nonlinear rep. spaces?

(reminder) Basic problem

Understand X(π1Σg, Homeo(S1)) = flat S1 bundles over Σg

• 2g+2

... 1 2 ... 2g-2