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トーラス束と準トーラス束上の Sol 構造の具体的構成 と対合の分類

渕上美規

広島大学

2016 12 21

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It is well-known that any torus bundle over the circle with Anosov monodromy and any torus semi-bundle (Sapphire space), which has a double cover homeomorphic to such a torus bundle, admit Sol

structure.

In this talk, I give an explicit construction of Sol structure of these manifolds, and then give a progress report on my project towards determination of their isometry groups and classification of their involutions.

In fact, we describe an algorithm to determine the isometry groups

and all involutions of such torus semi-bundles, through an explicit

example.

(3)

Plan

(1) Definitions of torus bundles and torus semi-bundles (2) Sol geometry

(3) Sol structure on torus bundles

(4) Sol structure on torus semi-bundles

(5) Example

(4)

(1)Definitions of torus bundles and torus semi-bundles

Let A SL(2, Z ).

R 2 A

= //

R 2

T 2 = R 2 / Z 2 A

= // T 2 = R 2 / Z 2 A : T 2 = R 2 / Z 2 T 2 = R 2 / Z 2 ; [x] 7→ [Ax]

Definition (Torus bundle)

M A := T 2 × R /(x, t) (Ax, t + 1) is called the torus bundle over the

circle with monodromy A.

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Definition (Torus semi-bundle)

Let h be a free involution (i.e. h 2 = id M

A

, Fix(h) = ) on the torus bundle which satisfies the following condition.

M A −−−−→ h M A

 

y   y S 1 −−−−→ ¯ h S 1 z −−−−→ z ¯

Then, M(A, h) := M A /h is called a torus semi-bundle.

(6)

ultimate goal ( )

To classify involutions of torus semi-bundles Previous studies

Sakuma [1985] gave a classification of involutions on torus bundles.

Barreto-Goncalves-Vendruscolo [2016] gave a classification of free involutions on torus semi-bundles.

Definition

Let r and r be involutions on a 3-manifold M.

Then r and r are equivalent

⇔ ∃ f : M M homeo. s.t. f r f 1 = r

M f

= //

r

M

r

M f

= //

M If | TrA | ≥ 3, then M A and M (A, h) admit Sol structure.

By the orbifold theorem, ( ) is equivalent to the classification of the order

two elements of Isom(M (A, h)).

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Theorem (Sakuma(1985))

(1)If M A is an orientable torus bundle, then 1 ≤ | Inv(M A ) | ≤ 21.

(2)If M A is a non-orientable torus bundle, then 1 ≤ | Inv(M A ) | ≤ 7.

Here Inv(M ) denotes the set of all equivalence classes of involutions on M, and |S| denotes the cardinality of S.

Example (Sakuma(1985))

For the torus bundle M A with A =

( 89 20 40 9

)

, | Inv(M A ) | = 21.

Theorem (F)

There is an algorithm to determine the set Inv(M (A, h)).

Example For A =

( 89 20 40 9

)

, | Inv(M (A, h)) | = 10.

(8)

(2) Sol geometry

The Lie group Sol is defined to be the semi direct product R 2T R . 1 // R 2 // Sol // R //

oo ι 1

Here,

T : R Aut( R 2 ) z 7→

( e z 0 0 e z

)

If we identify Sol with R 2 × R by

R 2 × R Sol (x, z) 7→ xι(z), then the multiplication is given by

(x, z)(x , z ) = (x + T (z)x , z + z ).

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With respect to a natural Riemann metric on Sol, we have 1 // Sol // Isom(Sol) oo // D(4) // 1

Here D(4) is the dihedral group which consists of 8 maps of R 3 given by

(x, y, z) 7→ (±x, ±y, z) and (x, y, z) 7→ (±y, ±x, −z).

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Definition

A 3-manifold M admits Sol structure

⇔ ∃Γ < Isom(Sol); discrete torsion-free group s.t. M = Sol/Γ

This is equivalent to the existence of the developing map D : M f Sol, where M f is the universal covering of M, and the holonomy homo ρ : π 1 (M ) Isom(Sol).

s.t.

M f D

= //

g

Sol

ρ(g)

M f D

= //

⟲ Sol

g π 1 (M )

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The structure of π 1 (M A )

1 // π 1 (T 2 ) // π 1 (M A ) oo // γ // 1 Z 2

π 1 (M A ) = Z 2 , γ | γnγ 1 = An, n Z 2

The universal cover M f A and the action of π 1 (M A ) M f A = T f 2 × R = R 2 × R

The action of π 1 (M A ) on M f A = R 2 × R is given by n · (x, t) = (x + n, t) (n Z 2 )

γ · (x, t) = (Ax, t + 1).

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(3) Sol structure on torus bundles

Recall that A SL(2, Z ), | TrA | ≥ 3.

Case 1. TrA 3

Choose Q GL + (2, R ), s.t. QAQ −1 =

( e τ 0 0 e τ

)

We define the following developing map D and holonomy homo ρ:

D : M f A Sol (x, t) 7→ (Qx, tτ)

ρ : π 1 (M A ) Sol < Isom(Sol) n 7→ Qn R 2 (n Z 2 )

γ 7→ τ = ι(τ ) 1 // R 2 // Sol // R //

oo ι 1

π 1 (M A ) = Z 2⟨γ⟩

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Then

M f A D

= //

g

Sol

ρ(g)

M f A D

= //

⟲ Sol

∀g π 1 (M A ) = Z 2⟨γ⟩

Put Γ := ρ(π 1 (M A )).

Then we obtain M A = M f A 1 (M A ) = Sol/Γ.

(14)

Case 2. TrA ≤ − 3

Choose Q GL + (2, R ), s.t. QAQ 1 =

( e τ 0 0 e τ

)

We use the developing map D and define the following holonomy homo ρ :

D : M f A Sol (x, t) 7→ (Qx, tτ)

ρ : π 1 (M A ) Isom(Sol)

n 7→ Qn R 2 (n Z 2 ) γ 7→ f τ

where f : Sol Sol; (x, t) 7→ ( x, t).

Then

M f A D

= //

g

Sol ρ (g)

f

M A D

= //

⟲ Sol

g π 1 (M A )

Put Γ := ρ 1 (M A )). Then we obtain M A = M f A 1 (M A ) = Sol/Γ.

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(4)Sol structure on torus semi-bundles

The torus semi-bundle M (A, h) is the union of two copies of twisted I -bundle KI over the Klein bottle glued by a homeomorphism, B , of

∂KI = T 2 .

Here, B : π 1 (∂KI ) π 1 (∂KI) (α, β) 7→ (α, β)B So we put N B := KI

B KI = M (A, h), where B SL(2, Z ).

KI = [0, 1] × S 1 × [ 1, 1]/(0, θ, s) (1, θ, s)

(16)

Observation Let J =

( 1 0 0 1

)

, A = J B 1 J B.

Then the torus semi bundle N B has the torus bundle M A as a double covering.

(Proof)

Note that KI has the double covering S 1 × S 1 × [ 1, 1].

Its covering transformation group is generated by h : S 1 × S 1 × [ 1, 1] S 1 × S 1 × [ 1, 1]

(x, y, t) 7→ (x + 1/2, y, t).

The double covering of N B is the union of two copies of S 1 × S 1 × [ 1, 1]

glued by a homeomorphism of ∂(S 1 × S 1 × [−1, 1]) = T 2 × {−1, 1}.

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T 2 × {−1} ⊂ P

$$

T 2 × [−1, 1]

P

T 2 × {+1}

P +

=

zz

KI

∂KI

∂KI KI

T 2 × {−1} ⊂ P

::

T 2 × [−1, 1] P

OO

T 2 × {+1}

P +

=

dd

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T 2 × {− 1 } ⊂ P

$$

hBh

T 2 × [ 1, 1]

P

T 2 × { +1 } P +

=

zz

h = J

uu

B

KI

∂KI B

∂KI KI

T 2 × {− 1 } ⊂

P

::

T 2 × [ 1, 1] P

OO

T 2 × { +1 } P +

=

dd

h = J

ii

From this picture, the double covering N b B is identified with M A ,

where A = (hBh) 1 B = hB 1 hB = J B 1 J B.

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Remark If B =

( a b c d

)

SL(2, Z ), then A = J B 1 J B =

( 1 + 2bc 2bd 2ac 1 + 2bc

) . Hence, M A = N b B admits Sol structure

⇔ | TrA | = | 2(1 + 2bc) | ≥ 3 bc ̸ = 0, 1.

In the remainder, we assume B = ( a b

c d )

satisfies bc ̸ = 0, 1.

'

&

$

% Recall

M A = M f A 1 (M A ) = Sol/Γ, where Γ := ρ(π 1 (M A )).

M f A D

= //

g π 1 (M A )

Sol

ρ(g)

M f A D

= //

⟲ Sol

The developing homeo D maps (x, t) to (Qx, tτ), where Q GL + (2, R), s.t. QAQ 1 =

( ± e −τ 0 0 ± e τ

)

.

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Let N B be a torus semi-bundle.

For simplicity, we consider the case that B I SL(2, Z /2 Z ).

Let r : M A M A be the covering involution of the double covering M A N B .

By using the fact that

J AJ = J (J B 1 J B)J = B 1 J BJ = A 1 , the involution r lifts to the following involution.

r : R 2 × R = M f A M f A = R 2 × R (x, t) 7→ (J x + 1

2 e 1 , t), where e 1 =

( 1 0 )

.

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In order to put a Sol structure on N B , we have only to show that we can choose the developing map D : M f A Sol so that DrD 1 Isom(Sol).

Note that

DrD 1 : (x, z) 7→ (QJ Q 1 x + 1

2 Qe 1 , z)

M f A D

= //

r

Sol DrD 1

f

M A D

= // Sol Note also that

QJ Q 1

( ± e τ 0 0 ± e τ

)

(QJ Q 1 ) 1 = QJ(Q 1

( ± e τ 0 0 ± e τ

)

Q)J Q 1

= QJ AJ Q 1

= QA −1 Q −1 =

( ± e τ 0 0 ± e τ

) 1

.

(22)

Hence QJ Q 1 = ±

( 0 1/k k 0

)

(k > 0).

Let Q = ( k 0

0 1 )

Q.

Then we have

Q J Q ′− 1 = ( k 0

0 1 )

QJ Q 1

( 1/k 0

0 1

)

= ± ( k 0

0 1

) ( 0 1/k k 0

) ( 1/k 0

0 1

)

= ± ( 0 1

1 0 )

and

Q AQ ′− 1 =

( ± e τ 0 0 ±e τ

)

.

(23)

The map (x, z) 7→

(

± ( 0 1

1 0 )

x, z )

is the element of D(4) < Isom(Sol).

Recall

1 // Sol // Isom(Sol) oo // D(4) // 1

D(4) = { (x, y, z) 7→ ( ± x, ± y, z), (x, y, z) 7→ ( ± y, ± x, z) }

Thus we have DrD 1 Isom(Sol) and M A / DrD 1 = N B , where

M A = Sol/Γ. Hence N B admits Sol structure.

(24)

(5)Example

For A =

( 89 20 40 9

)

, | Inv(M A ) | = 21. (Sakuma [1985])

Let B =

( 5 2 12 5 )

, then M A is the double covering of N B .

The torus semi-bundle N B admits 3 distinct classes of free involutions.

(Barreto-Goncalves-Vendruscolo [2016]) Example | Inv(N B ) | = 10

Step 1 Determine Isom(M A )

Step 2 Choose r Isom(M A ) s.t. M A / r = N B

Step 3 Determine Isom(N B ) and classify the order 2 elements of it

up to conjugacy

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Step 1. (Determine Isom(M A )) Replace A as A = XAX 1 =

( 49 20 120 49 )

. X :=

( 1 0 2 1 )

GL(2, Z ) Let B =

( 5 2 12 5 )

, then J B 1 J B = A.

Note J AJ = A 1 , A = B 2 ( ∵ J B 1 J = B) τ 0 : M f A M f A

(x, t) 7→ (Bx, t + 1 2 )

J : M f A M f A (x, t) 7→ (Jx, t)

E : M f A M f A (x, t) 7→ ( x, t)

1 // Coker(A E) // Isom 0 (M A ) oo // τ 0 | τ 0 2 = 1 // 1

1 // Isom 0 (M A ) // Isom(M A ) oo // J ⟩ ⊕ ⟨− E // 1 J 2 = ( E) 2 = 1

τ 0 0 1 = Bv (v Z 2 )

J vJ = AJ v J τ 0 J = τ 0 1

( E)v( E) = v

( E)τ 0 ( E) = τ 0

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Step 2. (Choose r Isom(M A ) s.t. M A / r = N B ) Coker(A E) = Z 4 Z 24

(¯ 1, ¯ 0) = e 1 , (¯ 0, ¯ 1) = e 2

An isometry (¯ 0, 12)J ¯ is a free involution of M A and it is possible to check M A / (¯ 0, 12)J ¯ = N B .

Step 3. (Determine Isom(N B ) and classify the order 2 elements of it up to conjugacy)

1 // (Z 2 Z 24 ) ⋊ ⟨τ 0 | τ 0 2 = 1⟩ // Isom(N B ) oo // ⟨−E | (−E) 2 = 1⟩ // 1

Isom(N B ) =

f 1 , f 2 , τ 0 , E

f 1 2 = f 2 24 = [f 1 , f 2 ] = τ 0 2 = ( E) 2 = 1, τ 0 f 1 τ 0 = f 1 , τ 0 f 2 τ 0 = f 1 f 2 5 ,

( E)f 1 ( E) = f 1 , ( E)f 2 ( E) = f 2 −1 , ( E)τ 0 ( E) = τ 0

f 1 = (¯ 2, ¯ 0), f 2 = (¯ 0, ¯ 1)

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Thank you for your attention.

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