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On a Heegaard surface homeomorphism obtained by bridge position of a knot

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On a Heegaard surface homeomorphism

obtained by

bridge position of a knot

Shin 0 ya Okazaki (Osaka City University)

岡崎 真也 

(

大阪市立大学

)

December 20 , 2010

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Contents

1 . Background

2 . Main theorem

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1 . Background Fact

Every closed connected orientable 3-manifold M is obtained by the 0-surgery of the 3-sphere S 3 along a link .

We introduce two kinds of genera g bridge ( M ) and

g braid ( M ) .

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Definition (0-surgery of S 3 along L ) L : a link in S 3 .

χ ( L, 0)

We call χ ( L, 0) the 3-manifold obtained by the

0-surgery of S 3 along L.

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bridge( L ): the bridge index of L.

braid( L ): the braid index of L.

Definition (bridge genus and braid genus)

g bridge ( M ) = min { bridge( L ) | χ ( L, 0) = M } . g braid ( M ) = min { braid( L ) | χ ( L, 0) = M } .

We have

g ( M ) g ( M ) .

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Definition (Heegaard splitting)

M : a closed connected orientable 3-manifold .

M = f

H H 0

We call the triple ( H, H 0 ; f ) a Heegaard splitting

of M. The Heegaard genus g H ( M ) of M is the

minimal genus of ∂H.

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Theorem 1 [O . ]

g H ( M ) g bridge ( M ) g braid ( M ) .

The invariants g H , g bridge and g braid are linearly

independent .

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Outline of proof of Theorem 1

S 3 =

B 1 B 2

S 3 = f

H H 0

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Example 1

K : the trivial knot χ ( K, 0) = S 1 × S 2 . Since g H ( M ) = 0 M = S 3 ,

we have g H ( S 1 × S 2 ) 1 .

braid( K ) = 1 g braid ( S 1 × S 2 ) 1 . Therefore , we have

g H ( S 1 × S 2 ) = g bridge ( S 1 × S 2 )

= g braid ( S 1 × S 2 ) = 1 .

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Example 2

L : the Hopf link χ ( L, 0) = S 3 . We have g H ( S 3 ) = 0 .

braid( L ) = 2 g braid ( S 3 ) 2 .

Since g bridge ( M ) = g braid ( M ) = 1

M = S 1 × S 2 ,

we have g bridge ( S 3 ) 2 . Therefore , we have

g H ( S 3 ) = 0 < g bridge ( S 3 ) = g braid ( S 3 ) = 2 .

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2 . Main theorem B 3 : a 3-ball .

h i := D 2 × I ( i = 1 , 2 , . . . , n ) .

H : a genus n handlebody in S 3 s . t . B 3

n i =1 h i . F := ∂H.

H :

h 1

h n h 2

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K : a knot which satisfies condition ( ) . K int H,

h i K = { 0 } × I,

B 3 K is an n -string trivial tangle .

( )

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K 0 : the knot satisfying condition ( ) as following .

H :

h 1

h n h 2

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Objective of this talk

1 . We show that for any K which satisfies con- dition ( ) , K is represented by the product of Suzuki generators from K 0 .

2 . We show that a Heegaard surface homeomor-

phism of χ ( K, 0) in outline of proof of Theorem

1 is represented by the product of Suzuki gen-

erators and two homeomorphisms .

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M ( F ) : the mapping class group of F.

ρ, τ 1 , θ 12 , µ 1 : the homeomorphisms in M ( F ) . ρ

−→ ρ

h 1 h n

h n h 2 h n 1 h 1 τ 1

h 1 τ 1

−→

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θ 12

θ 12

−→

h 1 h 2 h 1 h 2

µ 1

h 1 µ 1

−→

S. Suzuki, On Homeomorphisms of a 3-dimensional handlebody, Canad. J. Math., Vol. 29, (1977), 111–124.

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Theorem 2 [Suzuki]

M ( F ) is generated by ρ, τ 1 , θ 12 , µ 1 .

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ω 1 , ρ 12 , ξ 12 0 : the homeomorphisms in M ( F ) . ω 1

ω 1

−→

h 1 h 1

ρ 12

ρ 12

−→

h 1 h 2 h 1 h 2

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ξ 12 0

ξ 12 0

−→

h 1 h 2 h 1 h 2

ρ, τ 1 , θ 12 , ω 1 , ρ 12 , ξ 12 0 can be extended to homeo-

morphisms of H onto itself . But µ 1 cannot .

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M ( H ) : the mapping class group of H.

ρ, ˜ ω ˜ 1 , ρ ˜ 12 , ξ ˜ 12 0 : the homeomorphisms in M ( H ) . ρ ˜

ρ ˜

−→

h 1 h n

h n h 2 h n 1 h 1 ω ˜ 1

ω ˜ 1

−→

h 1 h 1

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ρ ˜ 12

ρ ˜ 12

−→

h 1 h 2 h 1 h 2

ξ ˜ 12 0

ξ ˜ 12 0

−→

h 1 h 2 h 1 h 2

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Lemma 1

For any n -bridge knot K satisfying ( ) , there ex- ists a power product ˜ φ of ˜ ρ, ω ˜ 1 , ρ ˜ 12 , ξ ˜ 12 0 ∈ M ( H ) s . t . K = ˜ φ ( K 0 ) .

Lemma 2

For any n -braid knot K satisfying ( ) , there ex-

ists a power product ˜ φ of ˜ ρ, ρ ˜ 12 , ω ˜ 1 ξ ˜ 12 0 ω ˜ 1 1 ∈ M ( H )

s . t . K = ˜ φ ( K 0 ) .

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Example 3

K : the trefoil satisfying ( ) .

ω ˜ 1 1 ξ ˜ 0− 12 1

−→ −→

Therefore , K = ξ 0− 12 1 ω 1 1 ( K 0 )

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ψ : an orientation reversing homeomorphism of F onto itself .

−→ ψ

c 1 ψ ( c 1 ) c 2

c n

ψ ( c n ) ψ ( c 2 )

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κ : the homeomorphism in M ( F ) as following .

−→ κ

h 1 h 1

h n h 2 h n h 2

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Theorem 3 [O . ]

K : an n -bridge knot satisfying ( ) . φ ˜ : a power product of ˜ ρ, ω ˜ 1 , ρ ˜ 12 , ξ ˜ 12 0 s . t . K = ˜ φ ( K 0 ) .

φ = ˜ φ | F .

Then χ ( K, 0) has the Heegaard splitting

( H, H 0 ; φκφ 1 ψ ) .

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Theorem 4 [O . ]

K : an n -braid knot satisfying ( ) .

φ ˜ : a power product of ˜ ρ, ρ ˜ 12 , ω ˜ 1 ξ ˜ 12 0 ω ˜ 1 1 s . t . K = ˜ φ ( K 0 ) .

φ = ˜ φ | F .

Then χ ( K, 0) has the Heegaard splitting

( H, H 0 ; φκφ 1 ψ ) .

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Example 4

K : the trefoil satisfying ( ) .

ω ˜ 1 1 ξ ˜ 0− 12 1

−→ −→

Therefore , χ ( K, 0) has the Heegaard splitting

( H, H 0 ; ξ 0− 12 1 ω 1 1 κω 1 ξ 0 12 ψ ) .

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Corollary 1

M : a closed connected orientable 3-manifold . g bridge ( M ) = min { g ( H ) | M = H φκφ

1

ψ H 0 } . Here , φ is a power product of ρ, ω 1 , ρ 12 , ξ 12 0 .

g braid ( M ) = min { g ( H ) | M = H φ

0

κφ

0−1

ψ H 0 } .

Here , φ 0 is a power product of ρ, ρ 12 , ω 1 ξ 12 0 ω 1 1 .

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