on the twisted alexander polynomial for metabelian sl2(c)-...

On the twisted Alexander polynomial formetabelian SL2(C)–representations

with the adjoint action

Yoshikazu Yamaguchi

JSPS Research fellow (PD)Tokyo Institute of Technology

RIMS Seminar“Twisted topological invariants and topology of

low-dimensional manifolds”

Yoshikazu Yamaguchi The twisted Alexander poly. for metabelian reps.

Basic notion and notation

The twisted Alexander polynomial

The twistedAlexanderpolynomial

=A refinement of ∆K (t)(the Alexander polynomial)with ρ : π1 → GL(V )

Notation

EK := S3 \ N(K ) a knot exterior,

Ad ◦ ρ : π1(EK )ρ−→ SL2(C)

Ad−−→ Aut(sl2(C))

γ 7→ ρ(γ) 7→ Adρ(γ) : v 7→ ρ(γ)vρ(γ)−1

sl2(C) = C(

0 10 0

)

⊕ C(

1 00 −1

)

⊕ C(

0 01 0

)

The adjoint action Ad gives a connection with the charactervariety Hom(π1(EK ),SL2(C))//SL2(C).


Metabelian representations

Definition of metabelian reps.

ρ : π1(EK ) → SL2(C) is metabelian

⇐⇒ ρ([π1(EK ), π1(EK )]) ⊂ SL2(C) an abelian subgroup.

Remark

ρ : π1(EK ) → SL2(C) is abelian⇐⇒ ρ(π1(EK )) ⊂ SL2(C) an abelian subgroup,

⇐⇒ ρ([π1(EK ), π1(EK )]) = {1},

⇐⇒π1(EK ) SL2(C)

H1(EK ; Z) π1(EK )/[π1(EK ), π1(EK )] ≃ 〈µ〉

ρ

ρ is determined by only ρ(µ).

We focus on non–abelian representations.


Definitions of reducible and irreducible reps.

ρ : π1(EK ) → SL2(C)

ρ : reducible

Def⇐⇒ ∃L ⊂ C2 s.t. ρ(g)(L) ⊂ L (∀g ∈ π1(EK ))

by taking conjugation,

ρ : π1(EK ) →{(

a ∗0 a−1

)∣

∣

∣

∣

a ∈ C \ {0}}

⊂ SL2(C)

ρ : irreducible

Def⇐⇒ ρ : not reducible.


Reducible and irreducible reps. in metabelian ones

ρ : π1(EK ) → SL2(C) metabelianρ : reducible

ρ : [π1(EK ), π1(EK )] →{

±(

1 ω0 1

)∣

∣

∣

∣

ω ∈ C

}

⊂ SL2(C)

since Im ρ ⊂{(

a ∗0 a−1

)∣

∣

∣

∣

a ∈ C \ {0}}

and

A,B ∈ SL2(C) : upper triangular ⇒ ABA−1B−1 =

(

1 ∗0 1

)

ρ : irreducible

ρ : [π1(EK ), π1(EK )] →{(

a 00 a−1

)∣

∣

∣

∣

a ∈ C \ {0}}

⊂ SL2(C)

ρ(µ) =

(

0 1−1 0

)

by F. Nagasato.


Background

ρ : π1(EK ) → SL2(C) metabelian

ρ : reducible∆K (t) appears in the twisted Alexander polynomial.

−→ Hyperbolic torsion at “bifurcation points”in Hom(π1(EK ,SL2(C)))//SL2(C).

ρ : irreducible

“Does ∆K (t) appear in the twisted Alexander?”


Main Theorem

Theorem

Suppose that

ρ : π1(EK ) → SL2(C) s.t.{

an irred. metabelian and;“longitude–regular ”.

Then(the twisted Alexander poly.)

∆α⊗Ad◦ρEK

(t) ·= (t − 1)∆K (−t)P(t),

where ∆K (t) is the Alexander polynomial of K .


Hierarchy of metabelian representations

Remark

ρ : π1(EK ) → SL2(C) reducible =⇒ ρ : metabelian

(∵) Im ρ ⊂{( a ∗

0 a−1

) ∣

∣ a 6= 0}

⇒ ρ([π1(EK ), π1(EK )]) ⊂{

±(

1 ∗0 1

)}

We have the following hierarchy of metabelian representations:

abelian ⊂ reducible ⊂ metabelian

roots of ∆K (t) ∆K (−1)

difference difference


The details of reducible metabelian representations

Existence of reducible representations (G. Burde, G. de Rham)

∃ρ : π1(EK ) → SL2(C) : reducible and non–abelianif and only if∆K (λ2) = 0 where λ is an eigenvalue of ρ(µ).

{the characters of non–abelian reducible representations}= Cabel ∩ Cnon–abel in Hom(π1(EK ),SL2(C))//SL2(C),

where

Cabel is the component of abelian characters and

Cnon–abel is the components of non–abelian ones.


The twisted Alexander polynomial of reducibleSL2(C)-representations

Theorem

ρ : π1(EK ) → SL2(C) reducible and non–abelian,

corresponding to λ ∈ C s.t.{

λ is an eigenvalue of ρ(µ) and∆K (λ2) = 0

,

then

∆α⊗Ad◦ρEK

(t) =∆K (λ2t)(λ2t − 1)

· ∆K (t)(t − 1)

· ∆K (λ−2t)(λ−2t − 1)

.


The details of irreducible metabelian representations

Existence of irreducible metabelian reps. (Nagasato)

♯{the characters of irereducible metabelian reps.}=

|∆K (−1)| − 12

.

ι ∈ H1(EK ; Z2) induces

an involution ι̂ on Hom(π1(EK ),SL2(C))//SL2(C).

Remark (Nagasato & Y.)

{the characters of irereducible metabelian reps.}= {the fixed point set of ι̂} in Hom(π1(EK ),SL2(C))//SL2(C).

Remark

A higher rank analogy was given by H. Boden and S. Friedl.


Main Theorem (again)

Theorem

Suppose that

ρ : π1(EK ) → SL2(C) s.t.{

an irred. metabelian and;“longitude–regular ”.

Then(the twisted Alexander poly.)

∆α⊗Ad◦ρEK

(t) ·= (t − 1)∆K (−t)P(t),

where ∆K (t) is the Alexander polynomial of K .


The details of ∆α⊗Ad◦ρEK

(t)

Homomorphisms

ρ : π1(EK ) → SL2(C) is metabelian

⇐⇒ ρ([π1(EK ), π1(EK )])(⊂ SL2(C)) is abelian,

Suppose that ρ([π1(EK ), π1(EK )]) 6= {1}.α : π1(EK ) → π1(EK )/[π1(EK ), π1(EK )] ≃ H1(EK ) = 〈t〉s.t. α(µ) = t

the twisted Alexander poly.

∆α⊗Ad◦ρEK

(t) =

det

(

α⊗ Ad ◦ ρ(

∂ri∂gj

)

i , j 6=1

)

det (α⊗ Ad ◦ ρ (g1 − 1))

from a presentation π1(EK ) = 〈g1,g2, . . . ,gk | r1, . . . , rk−1〉.


Main tools

We need a “good” presentation of π1(EK )for metabelian reps.X-S. Lin introduced a suitable presentation of π1(EK )by using a free Seifert surface of K .a Seifert surface S is free⇔ S3 = S × [−1,1] ∪ S3 \ S × [−1,1]

: a Heegaard splitting.

Figure: a free Seifert surface of the trefoil knot


Lin’s presentation

By using a free Seifert surface S with genus 2g,

π1(EK ) = 〈µ, x1, . . . , x2g |µa+i µ

−1 = a−i (i = 1, . . . 2g)〉

where

xi is a closed loop corresponding to 1–handle inS3 \ S × [−1,1],

a±i is a word in x1, . . . , x2g , corresponding to closed loops

in the spine of S.

Figure: a free Seifert surface of the trefoil knot


Remark & Examples of Lin’s presentation

RemarkThe generators x1, . . . , x2g are null–homologous,

i.e., xi ∈ [π1(EK ), π1(EK )].

K = trefoil knot

π1(EK ) =

⟨

µ, x1, x2

∣

∣

∣

∣

µx1µ−1 = x1x−1

2 ,

µx−12 x1µ

−1 = x−12

⟩

K = figure eight knot

π1(EK ) =

⟨

µ, x1, x2

∣

∣

∣

∣

µx1µ−1 = x1x−1

2 ,

µx2x1µ−1 = x2

⟩


Explicit form of metabelian reps.

X-S. Lin, F. Nagasato

The correspondence

µ 7→(

0 1−1 0

)

, xi 7→(

ξi 00 ξ−1

i

)

gives a metabelian rep.

They gives all representatives of conj. classes ofmetabelian reps.

♯ (conj. classes of irred. metabelian reps) =|∆K (−1)| − 1

2.


Sketch of proof

Ad ◦ ρ : π1(EK ) → Aut(sl2(C))

= Aut(C(

0 10 0

)

⊕ C(

1 00 −1

)

⊕ C(

0 01 0

)

)

µ 7→

−1−1

−1

xi 7→

ξ2

1ξ−2

The subspace C(

1 00 −1

)

is invariant space.

Decomposition of Ad ◦ ρ

Ad ◦ ρ = ρ2 ⊕ ρ1

where ρ1 is 1–dim. rep. and ρ2 is 2–dim. rep.


Observation about ∆α⊗Ad◦ρEK

(t)

∆α⊗Ad◦ρEK

(t) = ∆α⊗ρ2EK

(t) · ∆α⊗ρ1EK

(t)

= Q(t) · ∆K (−t)(−t − 1)

Wada Milnor

= (t − 1)(t + 1)P(t) · ∆K (−t)−t − 1

longitude–regular& inv. of conj.

for R–torsion

= −(t − 1) · P(t) · ∆K (−t).

Remark

The P(t) satisfies that P(t) = P(−t).


Details in the proof

The representations ρ1 and ρ2 are given by

ρ1 : µ 7→ −1, xi 7→ 1,

ρ2 : µ 7→(

0 −1−1 0

)

, xi 7→(

ξ2i 00 ξ−2

i

)

.

Wada’s criterion

ψ : π1(EK ) → GLn(R), R : UFDIf ∃γ ∈ [π1(EK ), π1(EK )] s. t. 1 is not an eigenvalue of ψ(γ),⇒ ∆α⊗ψ

EK(t) is a Laurent polynomial.

ρ2 : xi 7→(

ξ2i 00 ξ−2

i

)

(xi ∈ [π1(EK ), π1(EK )])

Wada’s criterion−−−−−−−−−−−−−→ ∆α⊗ρ2EK

(t) = Q(t) (Laurent polynomial)


Decomposition of Q(t)

ρ : longitude-regular ⇒ ∆α⊗Ad◦ρEK

(1) = 0,

∆α⊗Ad◦ρEK

(1) = Q(1) · ∆K (−1)

−2= 0 ⇒ Q(1) = 0.

Set C =

(√−1 00 −

√−1

)

,

Cρ2C−1 : µ 7→(

0 11 0

)

= −ρ(µ), xi 7→(

ξ2i 11 ξ−2

i

)

= ρ(xi)

inv. of conj.−−−−−−−−−→ Q(t) = Q(−t).

ThereforeQ(t) = (t − 1)(t + 1)P(t)


Observation about ∆α⊗Ad◦ρEK

(t) (again)

∆α⊗Ad◦ρEK

(t) = ∆α⊗ρ2EK

(t) · ∆α⊗ρ1EK

(t)

= Q(t) · ∆K (−t)(−t − 1)

Wada Milnor

= (t − 1)(t + 1)P(t) · ∆K (−t)−t − 1

longitude–regular& inv. of conj.

for R–torsion

= −(t − 1) · P(t) · ∆K (−t).

Remark

The P(t) satisfies that P(t) = P(−t).


Examples

trefoil knot ∆K (t) = t2 − t + 1 ( |∆K (−1)|−12 = 1).

∃1 irred. metabelian rep. (up to conjugate)Lin presentation

π1(EK ) =

⟨

µ, x1, x2

∣

∣

∣

∣

µx1µ−1 = x1x−1

2 ,

µx−12 x1µ

−1 = x−12

⟩

ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ3 00 ζ−1

3

)

, ρ(x2) =

(

ζ23 00 ζ−2

3

)

where ζ3 = e2π

√

−13

⇒ ∆α⊗Ad◦ρEK

(t) = (t − 1)∆K (−t).


figure eight knot

Figure: a free Seifert surface of the figure eight knot


figure eight knot ∆K (t) = t2 − 3t + 1 ( |∆K (−1)|−12 = 2).

∃2 irred. metabelian reps. (up to conjugate)Lin presentation

π1(EK ) =

⟨

µ, x1, x2

∣

∣

∣

∣

µx1µ−1 = x1x−1

2 ,µx2x1µ

−1 = x2

⟩

(1) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ5 00 ζ−1

5

)

, ρ(x2) =

(

ζ25 00 ζ−2

5

)

(2) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ25 00 ζ−2

5

)

, ρ(x2) =

(

ζ45 00 ζ−4

5

)

where ζ5 = e2π

√

−15


(t) = (t − 1)∆K (−t).(the both reps. have the same result)


52 knot

Figure: a free Seifert surface of 52 knot


52 knot ∆K (t) = 2t2 − 3t + 2 ( |∆K (−1)|−12 = 3).

∃3 irred. metabelian rep. (up to conjugate)Lin presentation

π1(EK ) =

⟨

µ, x1, x2

∣

∣

∣

∣

µx1µ−1 = x1x−1

2 ,

µx−22 x1µ

−1 = x−22

⟩

(1) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ7 00 ζ−1

7

)

, ρ(x2) =

(

ζ27 00 ζ−2

7

)

(2) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ27 00 ζ−2

7

)

, ρ(x2) =

(

ζ47 00 ζ−4

7

)

(3) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ37 00 ζ−3

7

)

, ρ(x2) =

(

ζ67 00 ζ−6

7

)

where ζ7 = e2π

√

−17


(1) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ7 00 ζ−1

7

)

, ρ(x2) =

(

ζ27 00 ζ−2

7

)


(t) = (t−1)(2+e6π√−1/7+e−6π

√−1/7)∆K (−t),

(2) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ27 00 ζ−2

7

)

, ρ(x2) =

(

ζ47 00 ζ−4

7

)


(t) = (t−1)(2+e2π√−1/7+e−2π

√−1/7)∆K (−t),

(3) ρ(µ) =

(

0 1−1 0

)

, ρ(x1) =

(

ζ37 00 ζ−3

7

)

, ρ(x2) =

(

ζ67 00 ζ−6

7

)


(t) = (t−1)(2+e4π√−1/7+e−4π

√−1/7)∆K (−t).


on the twisted alexander polynomial for metabelian sl2(c)-...

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