Basics of PSL(2 , C ) a b | a, b, c, d ∈ C , ad − bc = 1 • SL(2 , C ) = c d • PSL(2 , C ) = SL(2 , C ) / {± I } • PGL(2 , C ) ∼ 1 = PSL(2 , C ) by A �→ √ det A A • PSL(2 , C ) acts on C P 1 = C ∪{∞} by linear fractional trans- formation: · z = az + b a b c d cz + d 7-a
Basics of PSL(2 , C ) a b | a, b, c, d ∈ C , ad − bc = 1 • SL(2 , C ) = c d • PSL(2 , C ) = SL(2 , C ) / {± I } • PGL(2 , C ) ∼ 1 = PSL(2 , C ) by A �→ √ det A A • PSL(2 , C ) acts on C P 1 = C ∪{∞} by linear fractional trans- formation: · z = az + b a b c d cz + d 7-b
Basics of PSL(2 , C ) a b | a, b, c, d ∈ C , ad − bc = 1 • SL(2 , C ) = c d • PSL(2 , C ) = SL(2 , C ) / {± I } • PGL(2 , C ) ∼ 1 = PSL(2 , C ) by A �→ √ det A A • PSL(2 , C ) acts on C P 1 = C ∪{∞} by linear fractional trans- formation: · z = az + b a b c d cz + d 7-c
Lem A ( x 1 , x 2 , x 3 ) : distinct 3 points of C P 1 ( x ′ 1 , x ′ 2 , x ′ 3 ) : other distinct 3 points of C P 1 There exists a unique A ∈ PSL(2 , C ) s.t. A · x i = x ′ i . In fact, A is explicitly given by ( ) 1 a 11 a 12 A = √ a 21 a 22 ( x 1 − x 2 )( x 2 − x 3 )( x 3 − x 1 )( x ′ 1 − x ′ 2 )( x ′ 2 − x ′ 3 )( x ′ 3 − x ′ 1 ) where a 11 = x 1 x ′ 1 ( x ′ 2 − x ′ 3 ) + x 2 x ′ 2 ( x ′ 3 − x ′ 1 ) + x 3 x ′ 3 ( x ′ 1 − x ′ 2 ) , a 12 = x 1 x 2 x ′ 3 ( x ′ 1 − x ′ 2 ) + x 2 x 3 x ′ 1 ( x ′ 2 − x ′ 3 ) + x 3 x 1 x ′ 2 ( x ′ 3 − x ′ 1 ) , a 21 = x 1 ( x ′ 2 − x ′ 3 ) + x 2 ( x ′ 3 − x ′ 1 ) + x 3 ( x ′ 1 − x ′ 2 ) , a 22 = x 1 x ′ 1 ( x 2 − x 3 ) + x 2 x ′ 2 ( x 3 − x 1 ) + x 3 x ′ 3 ( x 1 − x 2 ) . 8
Lem A ( x 1 , x 2 , x 3 ) : distinct 3 points of C P 1 ( x ′ 1 , x ′ 2 , x ′ 3 ) : other distinct 3 points of C P 1 There exists a unique A ∈ PSL(2 , C ) s.t. A · x i = x ′ i . In fact, A is explicitly given by ( ) 1 a 11 a 12 A = √ a 21 a 22 ( x 1 − x 2 )( x 2 − x 3 )( x 3 − x 1 )( x ′ 1 − x ′ 2 )( x ′ 2 − x ′ 3 )( x ′ 3 − x ′ 1 ) where a 11 = x 1 x ′ 1 ( x ′ 2 − x ′ 3 ) + x 2 x ′ 2 ( x ′ 3 − x ′ 1 ) + x 3 x ′ 3 ( x ′ 1 − x ′ 2 ) , a 12 = x 1 x 2 x ′ 3 ( x ′ 1 − x ′ 2 ) + x 2 x 3 x ′ 1 ( x ′ 2 − x ′ 3 ) + x 3 x 1 x ′ 2 ( x ′ 3 − x ′ 1 ) , a 21 = x 1 ( x ′ 2 − x ′ 3 ) + x 2 ( x ′ 3 − x ′ 1 ) + x 3 ( x ′ 1 − x ′ 2 ) , a 22 = x 1 x ′ 1 ( x 2 − x 3 ) + x 2 x ′ 2 ( x 3 − x 1 ) + x 3 x ′ 3 ( x 1 − x 2 ) . 8-a
Proof of Lem A Consider a linear fractional transformation which sends ( x 1 , x 2 , x 3 ) to ( ∞ , 0 , 1). This is given by z �→ z − x 2 · x 3 − x 1 z − x 1 x 3 − x 2 uniquely. Thus the matrix is given by ( x 3 − x 1 ) − ( x 3 − x 1 ) x 2 ( x 3 − x 2 ) − ( x 3 − x 2 ) x 1 For general case, consider the composition A − 1 2 A 1 : ∼ ∼ = = ( x ′ 1 , x ′ 2 , x ′ ( x 1 , x 2 , x 3 ) ( ∞ , 0 , 1) 3 ) − − → ← − − � A 1 A 2 9
Fact A ∈ SL(2 , C ) e ∈ C \ { 0 , ± 1 } : one of the eigenvalues of A x, y ∈ C P 1 : fixed points of A . (Assume x � = y .) Suppose x corresponds to the eigenvector of e . (Equivalently, assume x is the attractive fixed point if | e | > 1.) Then A is uniquely determined by e , x , y . − 1 0 e x y x y A = 0 e − 1 1 1 1 1 ex − e − 1 y − ( e − e − 1 ) xy 1 = e − e − 1 − ey + e − 1 x x − y =: M ( e ; x, y ) 10
E.g. 0 e • M ( e ; ∞ , 0) = . 0 e − 1 As a linear fractional transformation t �→ e 2 t . e − 1 0 • M ( e ; 0 , ∞ ) = . 0 e As a linear fractional transformation t �→ e − 2 t . ex − e − 1 y − ( e − e − 1 ) xy 1 M ( e ; x, y ) = e − e − 1 − ey + e − 1 x x − y 11
Lem B x , y : distinct points on C P 1 . z 1 , z 2 : points on C P 1 different from x and y . ( z 1 and z 2 may coincide.) Then there exists a unique t ∈ C ∗ up to sign such that M ( t ; x, y ) sends z 1 to z 2 . Proof Since M ( t ; x, y ) · z 1 = ( tx − t − 1 y ) z 1 − ( t − t − 1 ) xy = z 2 , ( t − t − 1 ) z 1 − ty + t − 1 x we have t 2 = ( x − z 1 )( y − z 2 ) ( x − z 2 )( y − z 1 ) = [ y : x : z 1 : z 2 ] . Therefore t is well-defined up to sign. � 12
Lem B x , y : distinct points on C P 1 . z 1 , z 2 : points on C P 1 different from x and y . ( z 1 and z 2 may coincide.) Then there exists a unique t ∈ C ∗ up to sign such that M ( t ; x, y ) sends z 1 to z 2 . Proof Since M ( t ; x, y ) · z 1 = ( tx − t − 1 y ) z 1 − ( t − t − 1 ) xy = z 2 , ( t − t − 1 ) z 1 − ty + t − 1 x we have t 2 = ( x − z 1 )( y − z 2 ) ( x − z 2 )( y − z 1 ) = [ y : x : z 1 : z 2 ] . Therefore t is well-defined up to sign. � 12-a
SL(2 , C ) - and PSL(2 , C ) - character variety • S : a compact orientable surface • ρ : π 1 ( S ) → SL(2 , C ) (or PSL(2 , C )) is reducible if ρ ( π 1 ( S )) fixes a point on C P 1 . Otherwise ρ is called irreducible. • SL(2 , C ) acts on SL(2 , C )-representations by conjugation. • { ρ : π 1 ( S ) → SL(2 , C ) | irred. reps } / ∼ conj can be regarded as a subset of the SL(2 , C )-character variety X SL ( S ). • { ρ : π 1 ( S ) → PSL(2 , C ) | irred. reps } / ∼ conj can be re- garded as a subset of the PSL(2 , C )-character variety X PSL ( S ). 13
SL(2 , C ) - and PSL(2 , C ) - character variety • S : a compact orientable surface • ρ : π 1 ( S ) → SL(2 , C ) (or PSL(2 , C )) is reducible if ρ ( π 1 ( S )) fixes a point on C P 1 . Otherwise ρ is called irreducible. • SL(2 , C ) acts on SL(2 , C )-representations by conjugation. • { ρ : π 1 ( S ) → SL(2 , C ) | irred. reps } / ∼ conj can be regarded as a subset of the SL(2 , C )-character variety X SL ( S ). • { ρ : π 1 ( S ) → PSL(2 , C ) | irred. reps } / ∼ conj can be re- garded as a subset of the PSL(2 , C )-character variety X PSL ( S ). 13-a
SL(2 , C ) - and PSL(2 , C ) - character variety • S : a compact orientable surface • ρ : π 1 ( S ) → SL(2 , C ) (or PSL(2 , C )) is reducible if ρ ( π 1 ( S )) fixes a point on C P 1 . Otherwise ρ is called irreducible. • SL(2 , C ) acts on SL(2 , C )-representations by conjugation. • { ρ : π 1 ( S ) → SL(2 , C ) | irred. reps } / ∼ conj can be regarded as a subset of the SL(2 , C )-character variety X SL ( S ). • { ρ : π 1 ( S ) → PSL(2 , C ) | irred. reps } / ∼ conj can be re- garded as a subset of the PSL(2 , C )-character variety X PSL ( S ). 13-b
SL(2 , C ) - and PSL(2 , C ) - character variety • S : a compact orientable surface • ρ : π 1 ( S ) → SL(2 , C ) (or PSL(2 , C )) is reducible if ρ ( π 1 ( S )) fixes a point on C P 1 . Otherwise ρ is called irreducible. • SL(2 , C ) acts on SL(2 , C )-representations by conjugation. • { ρ : π 1 ( S ) → SL(2 , C ) | irred. reps } / ∼ conj can be regarded as a subset of the SL(2 , C )-character variety X SL ( S ). • { ρ : π 1 ( S ) → PSL(2 , C ) | irred. reps } / ∼ conj can be re- garded as a subset of the PSL(2 , C )-character variety X PSL ( S ). 13-c
SL(2 , C ) - and PSL(2 , C ) - character variety • S : a compact orientable surface • ρ : π 1 ( S ) → SL(2 , C ) (or PSL(2 , C )) is reducible if ρ ( π 1 ( S )) fixes a point on C P 1 . Otherwise ρ is called irreducible. • SL(2 , C ) acts on SL(2 , C )-representations by conjugation. • { ρ : π 1 ( S ) → SL(2 , C ) | irred. reps } / ∼ conj can be regarded as a subset of the SL(2 , C )-character variety X SL ( S ). • { ρ : π 1 ( S ) → PSL(2 , C ) | irred. reps } / ∼ conj can be re- garded as a subset of the PSL(2 , C )-character variety X PSL ( S ). 13-d
Representations of π 1 ( P ) Let P be a pair of pants (3 holed sphere). Take generators γ i of π 1 ( P ) as: γ 1 ∗ γ 3 γ 2 They satisfy γ 1 γ 2 γ 3 = 1. We will parametrize reps ρ : π 1 ( P ) → SL(2 , C ) satisfying (i) ρ ( γ i ) has two fixed points for each i , (ii) irreducible. 14
Representations of π 1 ( P ) Let P be a pair of pants (3 holed sphere). Take generators γ i of π 1 ( P ) as: γ 1 ∗ γ 3 γ 2 They satisfy γ 1 γ 2 γ 3 = 1. We will parametrize reps ρ : π 1 ( P ) → SL(2 , C ) satisfying (i) ρ ( γ i ) has two fixed points for each i , (ii) irreducible. 14-a
Representations of π 1 ( P ) Let P be a pair of pants (3 holed sphere). Take generators γ i of π 1 ( P ) as: γ 1 ∗ γ 3 γ 2 They satisfy γ 1 γ 2 γ 3 = 1. We will parametrize reps ρ : π 1 ( P ) → SL(2 , C ) satisfying (i) ρ ( γ i ) has two fixed points for each i , (ii) irreducible. 14-b
Representations of π 1 ( P ) e i : one of the eigenvalues of ρ ( γ i ) x i , y i : fixed points of ρ ( γ i ). Assume x i is attractive if | e i | > 1. Prop 1 ρ is uniquely determined by e i and x i . In fact, y i = e 2 i e i +2 x i +2 ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) for i = 1 , 2 , 3 (mod 3), and thus ( ) 1 a 11 a 12 ρ ( γ i ) = , a 21 a 22 e i e i +2 ( x i +1 − x i )( x i +2 − x i ) a 11 = e 2 i e i +2 x i ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) , a 12 = x i ( e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i +1 ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 )) , a 21 = e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) , a 22 = e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 ) . Conversely, if neither e 1 = e 2 e 3 , e 2 = e 3 e 1 , e 3 = e 1 e 2 nor e 1 e 2 e 3 = 1, then the above ρ is an irreducible rep. 15
Representations of π 1 ( P ) e i : one of the eigenvalues of ρ ( γ i ) x i , y i : fixed points of ρ ( γ i ). Assume x i is attractive if | e i | > 1. Prop 1 ρ is uniquely determined by e i and x i . In fact, y i = e 2 i e i +2 x i +2 ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) for i = 1 , 2 , 3 (mod 3), and thus ( ) 1 a 11 a 12 ρ ( γ i ) = , a 21 a 22 e i e i +2 ( x i +1 − x i )( x i +2 − x i ) a 11 = e 2 i e i +2 x i ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) , a 12 = x i ( e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i +1 ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 )) , a 21 = e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) , a 22 = e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 ) . Conversely, if neither e 1 = e 2 e 3 , e 2 = e 3 e 1 , e 3 = e 1 e 2 nor e 1 e 2 e 3 = 1, then the above ρ is an irreducible rep. 15-a
Representations of π 1 ( P ) e i : one of the eigenvalues of ρ ( γ i ) x i , y i : fixed points of ρ ( γ i ). Assume x i is attractive if | e i | > 1. Prop 1 ρ is uniquely determined by e i and x i . In fact, y i = e 2 i e i +2 x i +2 ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) for i = 1 , 2 , 3 (mod 3), and thus ( ) 1 a 11 a 12 ρ ( γ i ) = , a 21 a 22 e i e i +2 ( x i +1 − x i )( x i +2 − x i ) a 11 = e 2 i e i +2 x i ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) , a 12 = x i ( e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i +1 ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 )) , a 21 = e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) , a 22 = e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 ) . Conversely, if neither e 1 = e 2 e 3 , e 2 = e 3 e 1 , e 3 = e 1 e 2 nor e 1 e 2 e 3 = 1, then the above ρ is an irreducible rep. 15-b
Proof of Prop 1 Assume that ( x 1 , x 2 , x 3 ) = (0 , ∞ , 1). Then ρ ( γ i ) are uniquely ex − e − 1 y − ( e − e − 1 ) xy 1 determined by Fact M ( e ; x, y ) = e − e − 1 − ey + e − 1 x x − y e − 1 0 e 2 ( e − 1 − e 2 ) y 2 1 , 2 ρ ( γ 1 ) = ρ ( γ 2 ) = , e − 1 1 − e 1 e − 1 0 e 1 2 y 1 e − 1 3 y 3 − e 3 ( e 3 − e − 1 1 3 ) y 3 . ρ ( γ 3 ) = e − 1 e 3 y 3 − e − 1 y 3 − 1 − e 3 3 3 From the identity ρ ( γ 1 ) ρ ( γ 2 ) = ρ ( γ 3 ) − 1 , we have e 1 − e − 1 y 2 = e 2 − e 1 e − 1 y 3 = e 2 − e 1 e − 1 1 3 3 y 1 = , , . e − 1 2 e 3 − e − 1 e 2 − e − 1 e 2 − e 1 e 3 1 2 For general case, use Lem A . 16
Proof of Prop 1 Conversely, we can show that the above ρ is a homomorphism for any e i � = 0 and distinct triple x 1 , x 2 , x 3 . To make sure that ρ is irreducible, we have to check that y i ’s are different from x i ’s and distinct each other. We can show that e 1 = e 2 e 3 iff x 1 = y 2 = y 3 , e 2 = e 3 e 1 iff y 1 = x 2 = y 3 , e 3 = e 1 e 2 iff y 1 = y 2 = x 3 , e 1 e 2 e 3 = 1 iff y 1 = y 2 = y 3 . � 17
Proof of Prop 1 Conversely, we can show that the above ρ is a homomorphism for any e i � = 0 and distinct triple x 1 , x 2 , x 3 . To make sure that ρ is irreducible, we have to check that y i ’s are different from x i ’s and distinct each other. We can show that e 1 = e 2 e 3 iff x 1 = y 2 = y 3 , e 2 = e 3 e 1 iff y 1 = x 2 = y 3 , e 3 = e 1 e 2 iff y 1 = y 2 = x 3 , e 1 e 2 e 3 = 1 iff y 1 = y 2 = y 3 . � 17-a
Proof of Prop 1 Conversely, we can show that the above ρ is a homomorphism for any e i � = 0 and distinct triple x 1 , x 2 , x 3 . To make sure that ρ is irreducible, we have to check that y i ’s are different from x i ’s and distinct each other. We can show that e 1 = e 2 e 3 iff x 1 = y 2 = y 3 , e 2 = e 3 e 1 iff y 1 = x 2 = y 3 , e 3 = e 1 e 2 iff y 1 = y 2 = x 3 , e 1 e 2 e 3 = 1 iff y 1 = y 2 = y 3 . � 17-b
Remark • ρ is completely determined (not up to conjugacy) by e i and x i ( i = 1 , 2 , 3). • The conjugacy class of ρ is determined by e i ( i = 1 , 2 , 3), since it is determined by tr( γ i ) = e i + e − 1 . i • ρ also gives a PSL(2 , C )-rep. Any other lift to SL(2 , C )-rep is obtained by the action of H 1 ( P ; Z 2 ) ∼ = Hom( π 1 ( P ) , Z 2 ) ( e 1 , e 2 , e 3 ) �→ ( ε 1 e 1 , ε 2 e 2 .ε 3 e 3 ) where ε i = ± 1 s.t. ε 1 ε 2 ε 3 = 1. 18
Remark • ρ is completely determined (not up to conjugacy) by e i and x i ( i = 1 , 2 , 3). • The conjugacy class of ρ is determined by e i ( i = 1 , 2 , 3), since it is determined by tr( γ i ) = e i + e − 1 . i • ρ also gives a PSL(2 , C )-rep. Any other lift to SL(2 , C )-rep is obtained by the action of H 1 ( P ; Z 2 ) ∼ = Hom( π 1 ( P ) , Z 2 ) ( e 1 , e 2 , e 3 ) �→ ( ε 1 e 1 , ε 2 e 2 .ε 3 e 3 ) where ε i = ± 1 s.t. ε 1 ε 2 ε 3 = 1. 18-a
Remark • ρ is completely determined (not up to conjugacy) by e i and x i ( i = 1 , 2 , 3). • The conjugacy class of ρ is determined by e i ( i = 1 , 2 , 3), since it is determined by tr( γ i ) = e i + e − 1 . i • ρ also gives a PSL(2 , C )-rep. Any other lift to SL(2 , C )-rep is obtained by the action of H 1 ( P ; Z 2 ) ∼ = Hom( π 1 ( P ) , Z 2 ) ( e 1 , e 2 , e 3 ) �→ ( ε 1 e 1 , ε 2 e 2 .ε 3 e 3 ) where ε i = ± 1 s.t. ε 1 ε 2 ε 3 = 1. 18-b
Observation Let C 3 ⊃ E = { ( e 1 , e 2 , e 3 ) | e i � = 0 , ± 1 , e s 1 1 e s 2 2 e s 3 3 � = 1 for any s i = ± 1 } . Then we have E/ ( Z 2 ) 3 injective − − − − − − → X SL ( P ) where ( Z 2 ) 3 acts on E as e i �→ e i − 1 . We also have ( E/ ( Z 2 ) 3 ) / ( Z 2 ) 2 injective − − − − − − → X PSL ( P ) where ( Z 2 ) 2 acts on E as ( e 1 , e 2 , e 3 ) �→ ( ε 1 e 1 , ε 2 e 2 , ε 3 e 3 ) for ε i = ± 1 s.t. ε 1 ε 2 ε 3 = 1. 19
� � � � � � � Observation Let X = { ( x 1 , x 2 , x 3 ) | x i � = x j ( i � = j ) } ⊂ C 3 , then inj Hom( π 1 ( P ) , SL(2 , C )) E × X pr E inj E/ ( Z 2 ) 3 X SL ( P ) inj ( E/ ( Z 2 ) 3 ) / ( Z 2 ) 2 � X PSL ( P ) The images of the horizontal maps (irreducible representations s.t. ρ ( γ i )’s have two fixed points) are open and dense. 20
Twist parameter S = P ∪ P ′ : a four holed sphere P ′ ρ : π 1 ( S ) → SL(2 , C ) : a rep whose restrictions γ 5 e 5 e 4 to π 1 ( P ) and π 1 ( P ′ ) satisfies (i), (ii). γ 4 Let e i be one of the eigenvalues of ρ ( γ i ) for ( e 1 , t 1 ) i = 1 , . . . , 5 and x i (resp. y i ) be the fixed point γ 1 γ 2 corresponding to e i (resp. e − 1 ). i e 3 By Lem B , there exists a unique t 1 satisfying e 2 γ 3 P M ( √− t 1 ; x 1 , y 1 ) · x 2 = x 5 . We call t 1 the twist parameter. To define t 1 , we assign an (oriented) graph dual to the pants decomposition. 21
Twist parameter S = P ∪ P ′ : a four holed sphere P ′ ρ : π 1 ( S ) → SL(2 , C ) : a rep whose restrictions γ 5 e 5 e 4 to π 1 ( P ) and π 1 ( P ′ ) satisfies (i), (ii). γ 4 Let e i be one of the eigenvalues of ρ ( γ i ) for ( e 1 , t 1 ) i = 1 , . . . , 5 and x i (resp. y i ) be the fixed point γ 1 γ 2 corresponding to e i (resp. e − 1 ). i e 3 By Lem B , there exists a unique t 1 satisfying e 2 γ 3 P M ( √− t 1 ; x 1 , y 1 ) · x 2 = x 5 . We call t 1 the twist parameter. To define t 1 , we assign an (oriented) graph dual to the pants decomposition. 21-a
Twist parameter S = P ∪ P ′ : a four holed sphere P ′ ρ : π 1 ( S ) → SL(2 , C ) : a rep whose restrictions γ 5 e 5 e 4 to π 1 ( P ) and π 1 ( P ′ ) satisfies (i), (ii). γ 4 Let e i be one of the eigenvalues of ρ ( γ i ) for ( e 1 , t 1 ) i = 1 , . . . , 5 and x i (resp. y i ) be the fixed point γ 1 γ 2 corresponding to e i (resp. e − 1 ). i e 3 By Lem B , there exists a unique t 1 satisfying e 2 γ 3 P M ( √− t 1 ; x 1 , y 1 ) · x 2 = x 5 . We call t 1 the twist parameter. To define t 1 , we assign an (oriented) graph dual to the pants decomposition. 21-b
Twist parameter S = P ∪ P ′ : a four holed sphere P ′ ρ : π 1 ( S ) → SL(2 , C ) : a rep whose restrictions γ 5 e 5 e 4 to π 1 ( P ) and π 1 ( P ′ ) satisfies (i), (ii). γ 4 Let e i be one of the eigenvalues of ρ ( γ i ) for ( e 1 , t 1 ) i = 1 , . . . , 5 and x i (resp. y i ) be the fixed point γ 1 γ 2 corresponding to e i (resp. e − 1 ). i e 3 By Lem B , there exists a unique t 1 satisfying e 2 γ 3 P M ( √− t 1 ; x 1 , y 1 ) · x 2 = x 5 . We call t 1 the twist parameter. To define t 1 , we assign an (oriented) graph dual to the pants decomposition. 21-c
Remark Once we fix choices of the eigenvalues e 1 , . . . , e 5 , the twist parameter t 1 is a conjugacy invariant. It is also well-defined for PSL(2 , C )-representations. 22
x 5 e 5 e 4 Prop 2 x 4 e i : one of the eigenvalues of ρ ( γ i ) ( e 1 , t 1 ) x 1 x i : the fixed point corresponding to e i x 2 t 1 : twist parameter e 3 e 2 Then we have x 3 x 4 = { e 1 ( − ( e 2 − e 1 e 3 )( e 5 − e 1 e 4 ) t 1 + e 3 ( e 1 e 5 − e 4 )) x 1 ( x 2 − x 3 ) + e 12 e 2 ( e 1 e 5 − e 4 ) x 2 ( x 3 − x 1 ) + e 2 ( e 1 e 5 − e 4 ) x 3 ( x 1 − x 2 ) } / { e 1 ( − ( e 2 − e 1 e 3 )( e 5 − e 1 e 4 ) t 1 + e 3 ( e 1 e 5 − e 4 ))( x 2 − x 3 ) + e 12 e 2 ( e 1 e 5 − e 4 )( x 3 − x 1 ) + e 2 ( e 1 e 5 − e 4 )( x 1 − x 2 ) } x 5 = ( − ( e 2 − e 1 e 3 ) t 1 + e 1 e 3 ) x 1 ( x 2 − x 3 ) + e 12 e 2 x 2 ( x 3 − x 1 ) + e 2 x 3 ( x 1 − x 2 ) ( − ( e 2 − e 1 e 3 ) t 1 + e 1 e 3 )( x 2 − x 3 ) + e 12 e 2 ( x 3 − x 1 ) + e 2 ( x 1 − x 2 ) This means that x 4 and x 5 are uniquely determined by x 1 , x 2 , x 3 ∈ C P 1 , e 1 , . . . , e 5 and t 1 . 23
Remark • Conversely x 2 and x 3 are determined by x 1 , x 4 , x 5 ∈ C P 1 , e 1 , . . . , e 5 and t 1 : x 2 = ( − ( e 4 − e 1 e 5 ) t 1 − 1 + e 1 e 5 ) x 1 ( x 5 − x 4 ) + e 12 e 4 x 5 ( x 4 − x 1 ) + e 4 x 4 ( x 1 − x 5 ) ( − ( e 4 − e 1 e 5 ) t 1 − 1 + e 1 e 5 )( x 5 − x 4 ) + e 12 e 4 ( x 4 − x 1 ) + e 4 ( x 1 − x 5 ) x 3 = { e 1 ( − ( e 4 − e 1 e 5 )( e 2 − 1 − e 1 e 3 − 1 ) t 1 − 1 + e 5 ( e 1 e 2 − 1 − e 3 − 1 )) x 1 ( x 5 − x 4 ) 1 e 4 ( e 1 e 2 − 1 − e 3 − 1 ) x 5 ( x 4 − x 1 ) + e 4 ( e 1 e 2 − 1 − e 3 − 1 ) x 4 ( x 1 − x 5 ) } / + e 2 { e 1 ( − ( e 4 − e 1 e 5 )( e 3 − e 1 e 2 ) t 1 − 1 + e 5 ( e 1 e 3 − e 2 ))( x 5 − x 4 ) + e 2 1 e 4 ( e 1 e 3 − e 2 )( x 4 − x 1 ) + e 4 ( e 1 e 3 − e 2 )( x 1 − x 5 ) } • From Prop 1 and Prop 2 , ρ is uniquely determined by x 1 , x 2 , x 3 ∈ C P 1 , e 1 , . . . , e 5 and t 1 . The conjugacy class of ρ is uniquely determined by e 1 , . . . , e 5 and t 1 . 24
Remark • Moreover a conjugacy class of π 1 ( b -holed sphere) → SL(2 , C ) is completely determined by these eigenvalue parameters e i and twist parameters t i by Prop 1 and Prop 2 . x 5 x 2 x 7 x 4 x 1 x 3 x 6 (We assign ( e i , t i ) for each ‘interior’ edge and e i for each ‘boundary’ edge of the dual graph.) 25
Remark • Moreover a conjugacy class of π 1 ( b -holed sphere) → SL(2 , C ) is completely determined by these eigenvalue parameters e i and twist parameters t i by Prop 1 and Prop 2 . x 5 x 2 x 7 x 4 x 1 x 3 x 6 (We assign ( e i , t i ) for each ‘interior’ edge and e i for each ‘boundary’ edge of the dual graph.) 25-a
Remark • Moreover a conjugacy class of π 1 ( b -holed sphere) → SL(2 , C ) is completely determined by these eigenvalue parameters e i and twist parameters t i by Prop 1 and Prop 2 . x 5 x 2 x 7 x 4 x 1 x 3 x 6 (We assign ( e i , t i ) for each ‘interior’ edge and e i for each ‘boundary’ edge of the dual graph.) 25-b
Parametrization S : a surface of genus g . Given C ⊂ S : a pants decomposition (set of maximal disjoint scc.) G : an (oriented) graph dual to C We will parametrize the reps into PSL(2 , C ) satisfying, (i) ρ ( c ) has two fixed points for each scc c ⊂ C , (ii) restriction to each pair of pants is irreducible. 26
Parameters Assign an eigenvalue parameter e i and a twist parameter t i to each edge of G . Let P be the set of triples of scc’s of C which are the boundary of a component of S \ C . We let E ( S, C ) = { ( e 1 , . . . , e 3 g − 3 ) | e i � = 0 , ± 1 , e s 1 i e s 2 j e s 3 k � = 1 for ( i, j, k ) ∈ P} . We will reconstruct a representation from ( e 1 , . . . , e 3 g − 3 ) ∈ E ( S, C ) and t i ∈ C \ { 0 } . 27
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). ( e 2 , t 2 ) ( e 3 , t 3 ) ( e 1 , t 1 ) 28
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). ( e 2 , t 2 ) Take a maximal tree T in the dual graph G . ( e 3 , t 3 ) ( e 1 , t 1 ) 29
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). Cut S along the scc’s correspond- ing to the edges G \ T , we obtain a 2 g -holed sphere S 0 . 30
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). α 3 α 4 Take α i , α i + g ∈ π 1 ( S 0 ) for each edge of G \ T . They satisfy α i 1 . . . α i 2 g = 1 . (Eg. α 3 α 1 α 2 α 4 = 1 on the left Figure) α 1 α 2 31
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). Take β 1 , . . . , β g ∈ π 1 ( S ) for each edge of G \ T . β 2 β 1 32
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). α 3 α 4 π 1 ( S ) has the following presenta- tion: β 2 β 1 α 1 α 2 α g + i − 1 = β i − 1 α i β i � � α 1 , . . . , α 2 g, β 1 . . . β g | α i 1 . . . α i 2 g = 1 , g ∏ ± 1 , β i k ± 1 ] = 1 � = � α 1 , . . . , α g , β 1 . . . , β g | [ α i k k 33
Matrix generators For a dual graph, we give a presentation of π 1 ( S ). α 3 α 4 π 1 ( S ) has the following presenta- tion: β 2 β 1 α 1 α 2 (Eg. on the above Figure, � α 1 , . . . , α 4 , β 1 , β 2 | α 3 α 1 α 2 α 4 = 1 , α 3 − 1 = β 1 − 1 α 1 β 1 , α 4 − 1 = β 2 − 1 α 2 β 2 � = � α 1 , α 2 , β 1 , β 2 | [ β 1 − 1 , α − 1 1 ][ α 2 , β 2 − 1 ] = 1 � ) 34
We give matrices corresponding to these generators. Step 1 Compute sufficiently many number of fixed points for � G (universal cover of G ) by using Prop 2 . x 8 x 7 x 9 x 6 e 1 e 2 ( e 2 , t 2 ) x 5 x 4 ( e 3 , t 3 ) ( e 1 , t 1 ) ( e 3 , t 3 ) x 3 x 1 x 2 35
Step 1 Compute sufficiently many number of fixed points for � G by using Prop 2 . x 8 x 7 x 9 x 6 e 1 e 2 ( e 2 , t 2 ) x 5 x 4 ( e 3 , t 3 ) ( e 1 , t 1 ) ( e 3 , t 3 ) x 3 x 1 x 2 36
Step 1 Compute sufficiently many number of fixed points for � G by using Prop 2 . x 8 x 7 x 9 x 6 e 1 e 2 ( e 2 , t 2 ) x 5 x 4 ( e 3 , t 3 ) ( e 1 , t 1 ) ( e 3 , t 3 ) x 3 x 1 x 2 36-a
Step 1 Compute sufficiently many number of fixed points for � G by using Prop 2 . x 8 x 7 x 9 x 6 e 1 e 2 ( e 2 , t 2 ) x 5 x 4 ( e 3 , t 3 ) ( e 1 , t 1 ) ( e 3 , t 3 ) x 3 x 1 x 2 36-b
Step 1 Compute sufficiently many number of fixed points for � G by using Prop 2 . x 8 x 7 x 9 x 6 e 1 e 2 ( e 2 , t 2 ) x 5 x 4 ( e 3 , t 3 ) ( e 1 , t 1 ) ( e 3 , t 3 ) x 3 x 1 x 2 36-c
Step 2 Using Prop 1 , compute the matrices for α i ’s. Recall Prop 1 : ( ) 1 a 11 a 12 ρ ( γ i ) = , a 21 a 22 e i e i +2 ( x i +1 − x i )( x i +2 − x i ) a 11 = e 2 i e i +2 x i ( x i − x i +1 ) + e i +2 x i +1 ( x i +2 − x i ) + e i e i +1 x i ( x i +1 − x i +2 ) , a 12 = x i ( e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i +1 ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 )) , a 21 = e 2 i e i +2 ( x i − x i +1 ) + e i +2 ( x i +2 − x i ) + e i e i +1 ( x i +1 − x i +2 ) , a 22 = e 2 i e i +2 x i +2 ( x i +1 − x i ) + e i +2 x i ( x i − x i +2 ) + e i e i +1 x i ( x i +2 − x i +1 ) , (SL(2 , C ) matrix from eigs e 1 , e 2 , e 3 and fixed pts x 1 , x 2 , x 3 .) e 1 x 1 x 2 e 3 x 3 e 2 37
Step 3 Using Lem A , compute the matrices for β i ’s. Recall Lem A : There exists a unique matrix in PSL(2 , C ) s.t. ( x 1 , x 2 , x 3 ) �→ ( x ′ 1 , x ′ 2 , x ′ 3 ) x 1 x ′ x ′ x 3 1 2 x 2 x ′ 3 This kind of matrix conjugating ρ ( α i ) to ρ ( α g + i ) − 1 . 38
� � � Thus we can reconstruct a PSL(2 , C )-representation from the eigenvalue and twist parameters. In other words, we have obtained a map E ( S, C ) × ( C \ { 0 } ) 3 g − 3 → X PSL ( S ) . If we take a covering space Y corresponding to the signs of ρ ( β i ), we obtain the following diagram X SL ( S ) Y H 1 ( G ; Z 2 ) H 1 ( S ; Z 2 ) E ( S, C ) × ( C \ { 0 } ) 3 g − 3 � X PSL ( S ) (Here the covering group H 1 ( G ; Z 2 ) ∼ = ( Z 2 ) g .) 39
� � � Thus we can reconstruct a PSL(2 , C )-representation from the eigenvalue and twist parameters. In other words, we have obtained a map E ( S, C ) × ( C \ { 0 } ) 3 g − 3 → X PSL ( S ) . If we take a covering space Y corresponding to the signs of ρ ( β i ), we obtain the following diagram X SL ( S ) Y H 1 ( G ; Z 2 ) H 1 ( S ; Z 2 ) E ( S, C ) × ( C \ { 0 } ) 3 g − 3 � X PSL ( S ) (Here the covering group H 1 ( G ; Z 2 ) ∼ = ( Z 2 ) g .) 39-a
P ′ γ 5 e 5 e 4 Examples: γ 4 4 holed sphere ( e 1 , t 1 ) γ 1 γ 2 We apply Prop 2 for ( x 1 , x 2 , x 3 ) = ( ∞ , 1 , 0), then we have e 3 e 2 γ 3 P x 4 = e 1 ( e 2 − e 1 e 3 )( e 5 − e 1 e 4 ) t 1 − e 1 ( e 3 − e 1 e 2 )( e 1 e 5 − e 4 ) , ( e 12 − 1) e 2 ( e 1 e 5 − e 4 ) x 5 = ( e 2 − e 1 e 3 ) t 1 − e 1 ( e 3 − e 1 e 2 ) . ( e 12 − 1) e 2 40
By Prop 1 , we have, e 3 − e 1 e 1 e 3 + e 2 + 1 e 3 − 1 e 1 e 2 − e 1 e 2 e 2 ρ ( γ 1 ) = ρ ( γ 2 ) = , e 2 − e 1 e 1 1 0 e 1 e 3 e 3 1 0 a 11 a 12 e 3 , ρ ( γ 3 ) = , ρ ( γ 4 ) = e 3 − e 2 1 a 21 a 22 e 1 e 3 − e 1 ( e 52 + 1) a 11 = e 12 ( e 4 + e 4 − 1 ) + ( e 1 e 4 e 5 − 1)( e 1 e 2 − e 3 )( e 1 e 5 − e 4 ) , ( e 12 − 1) ( e 12 − 1) e 5 ( e 12 − 1)( e 1 e 3 − e 2 ) e 4 e 5 t 1 − e 1 a 12 = ( e 12 − 1) 2 e 2 ( e 1 e 3 − e 2 ) e 4 e 5 t 1 × (( e 1 e 3 e 5 + e 1 e 2 e 4 )( t 1 + 1) − e 2 e 5 ( e 12 + t 1 ) − e 3 e 4 (1 + e 12 t 1 )) × (( e 1 e 3 e 4 e 5 + e 1 e 2 )( t 1 + 1) − e 2 e 4 e 5 ( e 12 + t 1 ) − e 3 (1 + e 12 t 1 )) , a 21 = e 2 ( e 1 e 5 − e 4 )( e 1 e 4 e 5 − 1) , e 1 ( e 1 e 3 − e 2 ) e 4 e 5 t 1 + e 1 ( e 52 + 1) a 22 = − ( e 4 + e 4 − 1 ) − ( e 1 e 4 e 5 − 1)( e 1 e 2 − e 3 )( e 1 e 5 − e 4 ) . ( e 12 − 1) ( e 12 − 1) e 5 ( e 12 − 1)( e 1 e 3 − e 2 ) e 4 e 5 t 1
P ′ γ 5 e 5 e 4 For example, we have γ 4 tr( ρ ( γ 3 γ 4 )) = a 1 ( e 1 , t 1 ) , γ 1 a 2 γ 2 e 3 e 2 γ 3 P a 1 = − ( e 2 e 3 − e 1 )( e 1 e 3 − e 2 ) ( e 4 e 5 − e 1 )( e 1 e 4 − e 5 ) t 1 e 1 e 2 e 3 e 1 e 4 e 5 − (1 − e 1 e 2 e 3 )( e 1 e 2 − e 3 ) (1 − e 1 e 4 e 5 )( e 1 e 5 − e 4 ) 1 e 1 e 2 e 3 e 1 e 4 e 5 t 1 + χ 1 ( χ 3 χ 5 + χ 2 χ 4 ) − 2( χ 2 χ 5 + χ 3 χ 4 ) , a 2 =( e 1 − e 1 − 1 ) 2 . where χ i = e i + e i − 1 . 41
δ 1 α 2 One holed torus ( e 1 , t 1 ) Define a pants decomposition, a dual e 2 graph and the parameters e 1 , e 2 , t 1 as β 1 in the right Figure. Then we have α 1 e 1 e − 1 − e − 1 1 e − 1 , 1 2 ρ ( α 1 ) = e − 1 0 1 ( e 2 − e 2 1 1 ) t 1 + ( e 2 − 1) ( t 1 + 1)(1 − e 2 ) ρ ( β 1 ) = √− e 2 t 1 ( e 2 − e 2 ( e 2 e 2 ( e 2 1 − 1) 1 − 1) 1 − 1) 42
( e 1 , t 1 ) ( e 3 , t 3 ) ( e 2 , t 2 ) Closed surface of genus 2 ( ) e − 1 0 1 ρ ( α 1 ) = , − e 1 + e − 1 2 e 3 e 1 ( ) e 1 e − 1 e 2 − e 1 e − 1 3 3 β 2 ρ ( α 2 ) = , β 1 − e − 1 + e 1 e − 1 e 2 + e − 1 − e 1 e − 1 2 3 2 3 α 1 α 2 ( ) ( ) 1 1 a 11 a 12 b 11 b 12 ρ ( β 1 ) = √ t 1 t 3 , ρ ( β 2 ) = √ t 2 t 3 , ( e 22 − 1) e 3 a 21 a 22 b 21 b 22 a 12 = − ( e 2 e 3 − e 1 )( t 3 + 1) a 21 = e 1 ( t 1 + 1)( e 1 e 2 − e 3 ) a 11 = 1 , , e 1 ( e 32 − 1) ( e 12 − 1) e 2 a 22 = ( e 1 e 2 e 3 − 1)( e 1 e 3 − e 2 ) t 1 t 3 − ( e 1 e 2 − e 3 )( e 2 e 3 − e 1 )( t 1 + t 3 + 1) , ( e 12 − 1) e 2 ( e 32 − 1) b 11 = ( e 1 e 2 − e 3 ) t 2 − e 2 ( e 2 e 3 − e 1 ) , b 12 = − ( e 2 e 3 − e 1 )( e 3 ( e 1 e 2 e 3 − 1) t 2 t 3 + ( e 3 − e 1 e 2 ) t 2 + e 2 e 3 ( e 2 − e 1 e 3 ) t 3 − e 2 ( e 1 − e 2 e 3 )) / ( e 1 ( e 32 − 1)) , b 21 = ( e 1 e 2 − e 3 )( t 2 + 1) , b 22 = − ( e 3 ( e 1 e 2 e 3 − 1)( e 2 e 3 − e 1 ) t 2 t 3 − e 3 ( e 1 e 2 − e 3 )( e 1 e 3 − e 2 ) t 3 + ( e 1 e 2 − e 3 )( e 2 e 3 − e 1 )(1 + t 2 )) / ( e 1 ( e 32 − 1)) . 43
Action of ( Z / 2 Z ) 3 g − 3 For each pair of pants, ( Z 2 ) 3 acts on the eigenvalues as e 1 �→ e 1 − 1 , e 2 �→ e 2 − 1 , e 3 �→ e 3 − 1 . In general the action affects on the twist parameters: e 5 e 4 ( e 1 , t 1 ) �→ ( e 1 − 1 , t 1 − 1 ) , ( e 2 , t 1 ) �→ ( e 2 − 1 , e 2 e 3 − e 1 · e 1 e 3 − e 2 t 1 ) , 1 − e 1 e 2 e 3 e 1 e 2 − e 3 e 3 �→ e − 1 3 , ( e 1 , t 1 ) e 4 �→ e 4 − 1 , ( e 5 , t 1 ) �→ ( e 5 − 1 , e 4 e 5 − e 1 · e 1 e 4 − e 5 t 1 ) . e 3 e 2 1 − e 1 e 4 e 5 e 1 e 5 − e 4 44
Globally ( Z 2 ) 3 g − 3 acts on the parameter space E ( S, C ) × T where T = ( C \ { 0 } ) 3 g − 3 corresponds to the twist parameters. The map E ( S, C ) × T → X PSL ( S ) induces ( E ( S, C ) × T ) / ( Z 2 ) 3 g − 3 → X PSL ( S ) . This is not injective since we can change the signs of the eigen- value parameters as ( e i , e j , e k ) �→ ( ε i e i , ε j e j , ε k e k ) for ε i ε j ε k = 1 if ( e i , e j , e k ) belongs to a pair of pants. Globally the group is isomorphic to H 1 ( G ; Z 2 ) ∼ = ( Z 2 ) g . Theorem (( E ( S, C ) × T ) / ( Z 2 ) 3 g − 3 ) / ( Z 2 ) g injective − − − − − − → X PSL ( S ) . 45
Globally ( Z 2 ) 3 g − 3 acts on the parameter space E ( S, C ) × T where T = ( C \ { 0 } ) 3 g − 3 corresponds to the twist parameters. The map E ( S, C ) × T → X PSL ( S ) induces ( E ( S, C ) × T ) / ( Z 2 ) 3 g − 3 → X PSL ( S ) . This is not injective since we can change the signs of the eigen- value parameters as ( e i , e j , e k ) �→ ( ε i e i , ε j e j , ε k e k ) for ε i ε j ε k = 1 if ( e i , e j , e k ) belongs to a pair of pants. Globally the group is isomorphic to H 1 ( G ; Z 2 ) ∼ = ( Z 2 ) g . Theorem (( E ( S, C ) × T ) / ( Z 2 ) 3 g − 3 ) / ( Z 2 ) g injective − − − − − − → X PSL ( S ) . 45-a
Globally ( Z 2 ) 3 g − 3 acts on the parameter space E ( S, C ) × T where T = ( C \ { 0 } ) 3 g − 3 corresponds to the twist parameters. The map E ( S, C ) × T → X PSL ( S ) induces ( E ( S, C ) × T ) / ( Z 2 ) 3 g − 3 → X PSL ( S ) . This is not injective since we can change the signs of the eigen- value parameters as ( e i , e j , e k ) �→ ( ε i e i , ε j e j , ε k e k ) for ε i ε j ε k = 1 if ( e i , e j , e k ) belongs to a pair of pants. Globally the group is isomorphic to H 1 ( G ; Z 2 ) ∼ = ( Z 2 ) g . Theorem (( E ( S, C ) × T ) / ( Z 2 ) 3 g − 3 ) / ( Z 2 ) g injective − − − − − − → X PSL ( S ) . 45-b
� � � � This map gives a parametrization of the representations sat- isfying (i) and (ii). In particular, it contains all quasi-Fuchsian representations. As a summary, we have the following diagram: ( Z 2 ) 3 g − 3 X SL ( S ) Y H 1 ( G ; Z 2 ) H 1 ( S ; Z 2 ) E ( S, C ) × T H 1 ( G ; Z 2 ) ( Z 2 ) 3 g − 3 ( E ( S, C ) × T/H 1 ( G ; Z 2 )) � X PSL ( S ) The horizontal maps induce injections after taking quotient. 46
Coordinate change Our coordinates depend on the choice of a pants decompo- sition with a dual oriented graph. We will give a formula for coordinate change. Prop Any two pants decomposition with dual graphs are re- lated by the following five types moves. Transformation for- mulas for such moves are given as follows. 47
(I) Reverse orientation Reverse the orientation of an edge of the dual graph. e 5 e 4 e 5 e 4 ( e 1 − 1 , e 1 e 2 − e 3 e 1 e 5 − e 4 e 1 e 4 − e 5 t 1 − 1 ) ( e 1 , t 1 ) e 1 e 3 − e 2 e 3 e 3 e 2 e 2 48
(II) Dehn twist Change the dual graph by a (left or right) Dehn twist along a pants curve. (III) Vertex move For a vertex of the dual graph, change the edges adjacent to the vertex by their right half-twists as: These moves are (locally) expressed by compositions of the following formula: e 5 e 4 e 5 e 4 ( e 1 , − e 1 ( e 1 e 3 − e 2 ) ( e 1 , t 1 ) e 1 e 2 − e 3 t 1 ) e 3 e 3 e 2 e 2 49
(IV) Graph automorphism Just change the variables by per- mutations. (V) Elementary move On a subsurface homeomorphic to a one-holed torus or a four-holed sphere, we define the moves by: (a clockwise rotation of angle π/ 2) 50
The transformation formula for type (V) move is compli- cated... e 12 e 11 e 12 e 11 ( e 4 , t ′ 4 ) ( e 4 , t 4 ) ( e ′ 1 , t ′ 1 ) e 10 e 10 e 13 e 13 ( e 5 , t ′ ( e 1 , t 1 ) ( e 5 , t 5 ) 5 ) ( e 3 , t ′ 3 ) ( e 3 , t 3 ) e 9 e 9 e 6 e 6 ( e 2 , t 2 ) ( e 2 , t ′ 2 ) e 7 e 8 e 7 e 8 e 2 e ′ ( e 1 e 2 − e 3 )( e 1 e 3 − e 2 )( t 1 + 1) 1 − e 5 t ′ 2 = t 2 , e 2 e 5 − e ′ ( e 2 e 3 − e 1 )( e 1 e 3 − e 2 ) t 1 + (1 − e 1 e 2 e 3 )( e 1 e 2 − e 3 ) 1 3 = ( e 2 e 3 − e 1 )(( e 1 e 3 − e 2 )( e 1 e 4 − e 5 ) t 1 + ( e 1 e 2 − e 3 )( e 1 e 5 − e 4 )) t ′ ( e 1 e 3 − e 2 )(( e 2 e 3 − e 1 )( e 1 e 4 − e 5 ) t 1 + (1 − e 1 e 2 e 3 )( e 1 e 5 − e 4 )) t 3 , e 3 e ′ 4 = ( e 1 e 3 − e 2 )( e 1 e 4 − e 5 ) t 1 + ( e 1 e 2 − e 3 )( e 1 e 5 − e 4 ) 1 − e 4 t ′ t 4 , 1 − e 3 e 4 e ′ ( e 1 e 3 − e 2 )( e 4 e 5 − e 1 ) t 1 + ( e 1 e 2 − e 3 )(1 − e 1 e 4 e 5 ) 1 ( e 1 e 5 − e 4 )( e 1 e 4 e 5 − 1)( t 1 + 1) t ′ 5 = ( e 1 − e 4 e 5 )( e 1 e 4 − e 5 ) t 1 + ( e 1 e 4 e 5 − 1)( e 1 e 5 − e 4 ) t 5 , 51
where e ′ 1 is one of the solution of x 2 − tr( ρ ( γ 3 γ 4 )) x + 1 = 0 where ( 1 − ( e 2 e 3 − e 1 )( e 1 e 3 − e 2 ) ( e 4 e 5 − e 1 )( e 1 e 4 − e 5 ) tr( ρ ( γ 3 γ 4 )) = t 1 ( e 1 − e 1 − 1 ) 2 e 1 e 2 e 3 e 1 e 4 e 5 − (1 − e 1 e 2 e 3 )( e 1 e 2 − e 3 ) (1 − e 1 e 4 e 5 )( e 1 e 5 − e 4 ) 1 e 1 e 2 e 3 e 1 e 4 e 5 t 1 + ( e 1 + e 1 − 1 ) ( ( e 3 + e 3 − 1 )( e 5 + e 5 − 1 ) + ( e 2 + e 2 − 1 )( e 4 + e 4 − 1 ) ) ( e 2 + e 2 − 1 )( e 5 + e 5 − 1 ) + ( e 3 + e 3 − 1 )( e 4 + e 4 − 1 ) )) − 2 ( . And e ′ e ′ 1 1 e 5 e 2 1 e 3 e 4 t ′ 1 = − 1 ( e 5 e 2 − e ′ 1 )( e ′ ( e 3 e 4 − e ′ 1 )( e ′ e ′ 1 + e ′ 1 e 2 − e 5 ) 1 e 3 − e 4 ) 1 ( − 1 ) ( − 1 tr( ρ ( γ 2 γ 4 ) ) ( e ′ 1 − e ′ e ′ 1 tr( ρ ( γ 3 γ 5 ) − e ′ 1 1 − 1 ) ( 5 ) ) − ( e ′ 1 + e ′ ( e 2 + e − 1 2 )( e 3 + e − 1 3 ) + ( e 4 + e − 1 4 )( e 5 + e − 1 1 4 ) )) + 2 ( ( e 3 + e − 1 3 )( e 5 + e − 1 5 ) + ( e 2 + e − 1 2 )( e 4 + e − 1 I omit the one-holed torus case. 52
Geometric meaning of the twist parameters It is easy to see that { ( e i , t i ) ∈ R 2(3 g − 3) | e i < − 1 , t i > 0 } corresponds to the Fuchsian representations. Restrict to this subset, we can interpret our twist parameter as: x 5 e 5 e 4 x 4 x 1 ( e 1 , t 1 ) x 2 log | t 1 | e 2 e 3 x 3 53
On the other hand, usual F-N twist parameters are defined as: log | t FN | 1 Prop Let t FN be the exponential of the usual F-N twist pa- 1 rameter. Then we have � � � ( e 1 e 3 − e 2 )( e 2 e 3 − e 1 )( e 1 e 4 − e 5 )( e 4 e 5 − e 1 ) � t FN = ( e 1 e 2 − e 3 )( e 1 e 2 e 3 − 1)( e 1 e 5 − e 4 )( e 1 e 4 e 5 − 1) t 1 . 1 54
Remark is invariant under the action of ( Z / 2 Z ) 3 g − 3 up to sign. It t FN i might be a better parametrization although I do not know any appropriate choice of signs. The traces of the four holed sphere seem to be much simpler. ( √ 1 ( χ 2 1 + χ 2 2 + χ 2 tr( ρ ( γ 3 γ 4 )) = − 3 − χ 1 χ 2 χ 3 − 4) × ( e i − e i − 1 ) 2 √ ( χ 2 1 + χ 2 4 + χ 2 5 − χ 1 χ 4 χ 5 − 4)( t FN + ( t FN ) − 1 ) 1 1 ) + χ 1 ( χ 3 χ 5 + χ 2 χ 4 ) − 2( χ 2 χ 5 + χ 3 χ 4 ) where χ i = e i + e i − 1 . 55
Remark is invariant under the action of ( Z / 2 Z ) 3 g − 3 up to sign. It t FN i might be a better parametrization although I do not know any appropriate choice of signs. The traces of the four holed sphere seem to be much simpler. ( 1 − ( e 2 e 3 − e 1 )( e 1 e 3 − e 2 ) ( e 4 e 5 − e 1 )( e 1 e 4 − e 5 ) tr( ρ ( γ 3 γ 4 )) = t 1 ( e 1 − e 1 − 1 ) 2 e 1 e 2 e 3 e 1 e 4 e 5 − (1 − e 1 e 2 e 3 )( e 1 e 2 − e 3 ) (1 − e 1 e 4 e 5 )( e 1 e 5 − e 4 ) 1 e 1 e 2 e 3 e 1 e 4 e 5 t 1 ) + χ 1 ( χ 3 χ 5 + χ 2 χ 4 ) − 2( χ 2 χ 5 + χ 3 χ 4 ) where χ i = e i + e i − 1 . 55-a
Developing map S : a bordered surface (mainly interested in a pair of pants) T : ideal triangulation of S S of S has an ideal triangulation � � The universal cover T lifted from T . T → C P 1 is called a developing map. An equivariant map f : ∂ ∞ � v 1 � γ 2 v 3 c 1 c 1 γ 1 � γ 1 v 2 ∆ 1 p v i ∈ ∂ ∞ � γ 1 T, � γ 2 v 2 ∆ 0 ∗ γ 3 f ( v i ) ∈ C P 1 . � � c 3 c 2 c 2 c 3 γ 3 γ 2 v 3 γ 3 v 1 By Lem A , a developing map determines a representation π 1 ( S ) → PSL(2 , C ). 56
T → C P 1 , we as- For a developing map f : ∂ ∞ � v 1 sign to each edge of T a complex number defined by the cross ratio [ f ( v 0 ) , f ( v 1 ) , f ( v 2 ) , f ( v 3 )] where v 2 v 3 ( v 0 , v 1 , v 2 ) and ( v 0 , v 1 , v 3 ) are ideal triangles of � T v 0 as in the right figure. Conversely, a developing map is completely determined by these complex parameters: Lem C x 0 , x 1 , x 2 : distinct points of C P 1 , z ∈ C \ { 0 } Then there exists a unique x 3 ∈ C P 1 s.t. [ x 0 , x 1 , x 2 , x 3 ] = z . (Set x 3 = x 0 ( x 2 − x 1 ) − zx 1 ( x 2 − x 0 ) .) ( x 2 − x 1 ) − z ( x 2 − x 0 )) 57
T to C P 1 by Lem C inductively: We can develop � x 0 C P 1 x 5 x 3 x 1 x 2 x 4 58
T to C P 1 by Lem C inductively: We can develop � x 0 C P 1 x 5 x 3 x 1 x 2 x 4 58-a
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