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

Rigidity of boundary actions Kathryn Mann Brown University The PSL 2 ( R ) character variety Theorem (Goldman 88) Hom( 1 g , PSL 2 ( R )) has 4 g 3 connected components, classified by the Euler number. -2g+2 ... -1 0 1 ...


  1. Rigidity of boundary actions Kathryn Mann Brown University

  2. The PSL 2 ( R ) character variety Theorem (Goldman ’88) Hom( π 1 Σ g , PSL 2 ( R )) has 4 g − 3 connected components, classified by the Euler number. -2g+2 ... -1 0 1 ... 2g-2 Hom( π 1 Σ g , G ) / G ↔ flat principal G -bundles over Σ g

  3. The PSL 2 ( R ) character variety Theorem (Goldman ’88) Hom( π 1 Σ g , PSL 2 ( R )) / PSL has 4 g − 3 connected comp’s, classified by the Euler number. -2g+2 ... -1 0 1 ... 2g-2 Hom( π 1 Σ g , G ) / G ↔ flat principal G -bundles over Σ g Euler number is a characteristic number of oriented S 1 bundles. Topologist’s question: Describe the space of all flat, oriented S 1 bundles, i.e. Hom( π 1 Σ g , Homeo( S 1 )) / Homeo( S 1 )).

  4. 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 , SL 2 ( C )) / SL 2 ( 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 ( F 2 , SL 2 ( C )) = C 3 • X (Γ , Homeo( S 1 ))= semi-conjugacy classes of actions of Γ

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

  6. 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( S 1 ) are • SO(2) • finite cyclic extensions of PSL 2 ( R ) Z / k Z → G → PSL 2 ( R ) Cor.: 1. All groups Γ that act geometrically are (virtually) π 1 Σ g 2. Can easily describe all geometric actions of π 1 Σ g on S 1 .

  7. Theorem (Mann, 2014) If ρ : π 1 Σ g → Homeo( S 1 ) is geometric, then it is rigid. Theorem (Mann–Wolff, 2017) Converse: if ρ ∈ X ( π 1 Σ g , Homeo( S 1 )) is rigid, then is geometric. Consequences 1. Know all the isolated points of X ( π 1 Σ g , Homeo( S 1 )) (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

  8. Proof ingredient: coordinates on X ( π 1 Σ g , Homeo( S 1 ))

  9. Proof ingredient: coordinates on X ( π 1 Σ g , Homeo( S 1 )) Analogy: trace coordinates on X (Γ , SL 2 ( C ))

  10. Analogy: trace coordinates on X (Γ , SL 2 ( C )) tr : SL 2 ( 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. C 3 Eg. X ( F 2 , SL 2 ( C )) → ρ �→ tr( ρ ( a )) , tr( ρ ( b )) , tr( ρ ( ab )) Level set tr ρ ( aba − 1 b − 1 ) = 4 (image: W. Goldman)

  11. Proof ingredient: coordinates on X (Γ , Homeo( S 1 )) want: conjugation invariant function on Homeo( S 1 ) Definition (Poincar´ e) ˜ f ∈ Homeo( R ) f ∈ Homeo( S 1 ) ˜ f n (0) rotation number � rot( f ) := lim (mod Z ) n n →∞ • rot( f ) ∈ Q ⇔ f has periodic orbit. • Continuous, conjugation invariant. Theorem* (Ghys, Matsumoto) ρ ∈ X (Γ , Homeo( S 1 )) is determined by all rotation numbers rot( ρ ( γ )) � � � rot � rot � ρ ( a ) � * technically by cocycle � ρ ( a ) + � ρ ( b ) − � rot ρ ( b )

  12. The Euler number Definition � � � � rot � rot � ρ ( a ) � � ρ ( a ) + � ρ ( b ) − � Euler ( ρ ) = rot ρ ( b ) P pants ( ab ) − 1 a b

  13. The Euler number Definition � � � � rot � rot � ρ ( a ) � � ρ ( a ) + � ρ ( b ) − � Euler ( ρ ) = rot ρ ( b ) P pants ( ab ) − 1 a b

  14. Comparison C 3 = { tr( ρ ( a )) , tr( ρ ( b )) tr( ρ ( ab )) } = X ( F 2 , SL 2 ( C ))

  15. Comparison C 3 = { tr( ρ ( a )) , tr( ρ ( b )) tr( ρ ( ab )) } = X ( F 2 , SL 2 ( C )) { rot( ρ ( a )) , rot( ρ ( b )) rot( ρ ( ab )) } for ρ ∈ X ( F 2 , Homeo( S 1 )) Calegari & Walker “Ziggurats and Rotation numbers”

  16. Comparison tr( ρ ( aba − 1 b − 1 )) = 4

  17. Comparison: Open Questions Fact: There exist nonconjugate a , b ∈ π 1 Σ g such that tr ( ρ ( a )) = tr ( ρ ( b )) for all ρ : π 1 Σ g → SL 2 ( C ). Question: Can you find a , b with rot( ρ ( a )) = rot( ρ ( b )) for all ρ : π 1 Σ g → Homeo( S 1 )? probably not!

  18. Comparison: Open Questions Fact: There exist nonconjugate a , b ∈ π 1 Σ g such that tr ( ρ ( a )) = tr ( ρ ( b )) for all ρ : π 1 Σ g → SL 2 ( C ). Question: Can you find a , b with rot( ρ ( a )) = rot( ρ ( b )) for all ρ : π 1 Σ g → Homeo( S 1 )? 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 − 2 ab Or any general word in a ± 1 , b ± 1

  19. But some things are known ... Sample Lemma: “Rigidity implies rationality” If ρ ∈ X ( π 1 Σ g , Homeo( S 1 )) is rigid, then rot( ρ ( γ )) ∈ Q for all simple closed curves γ in π 1 Σ g .

  20. Proof ingredient 2: bending (twist) deformations Thurston: ρ : π 1 Σ g → PSL 2 ( R ) “bent” into PSL 2 ( C ) Bending along simple curves is a way to modify π 1 Σ g → G G any topological group π 1 Σ g = A ∗ � c � B c Conjugate ρ : π 1 Σ g → G on A , not on B

  21. Applications Question (Farb 2006) Consider Aut ( π 1 Σ g ) acting on ∂ ∞ ( π 1 Σ g ) = S 1 . Is this the only faithful action of Aut ( π 1 Σ g ) on S 1 ? Theorem (M – Wolff, 2018) Any nontrivial ρ : Aut ( π 1 Σ g ) → Homeo( S 1 ) is semiconjugate to the boundary action. Proof. Look at π 1 Σ g → Aut ( π 1 ) → Out ( π 1 ) = Mod g . Understand Mod g –equivariant ρ : π 1 Σ g → Homeo( S 1 ). each pant contributes same amount to Euler ( ρ ) etc...

  22. Applications Question (Farb 2006) Consider Aut ( π 1 Σ g ) acting on ∂ ∞ ( π 1 Σ g ) = S 1 . Is this the only faithful action of Aut ( π 1 Σ g ) on S 1 ? Theorem (M – Wolff, 2018) Any nontrivial ρ : Aut ( π 1 Σ g ) → Homeo( S 1 ) is semiconjugate to the boundary action. Problem Let Γ < Aut ( π 1 Σ g ). Classify Γ-equivariant representations ρ : π 1 Σ g → Homeo( S 1 ) e.g. Γ = � φ � pseudo-Anosov. Classify π 1 ( M 3 φ ) → Homeo( S 1 ).

  23. Applications II Ongoing with J. Bowden: boundary actions in higher dimensions. M n compact, negative curvature. Then π 1 ( M ) → Homeo( S n ) boundary action should be (locally) rigid. Perspective: straightening foliations. Major motivating question: Are there analogs of “trace coords” for other nonlinear rep. spaces?

  24. (reminder) Basic problem Understand X ( π 1 Σ g , Homeo( S 1 )) = flat S 1 bundles over Σ g -2g+2 ... 0 1 2 ... 2g-2

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