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結び目の数学II (早稲田大学 14号館 B101教室) 2009/12/24 (木), 16:40–17:10.

Gromov hyperbolicity of a variation of the

Gordian complex

In Dae Jong (Osaka City University) joint work with

Kazuhiro Ichihara (Nara University of Education)

(2)

§ 1. Introduction

K := { knots in the 3-sphere } . K , K ′ ∈ K .

K ↔ x K ′ ⇔ K can be deformed into K ′ by once.

Gordian distance d x ( · , · )

³

d x ( K, K ′ ) := min { n | K = K 0 ↔ · · · x ↔ x K n = K ′ } .

µ ´

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Remark 1.1

³

G x = the 1–skelton of the Gordian complex

introduced by [Hirasawa-Uchida ’02].

µ ´

(3)

§ 1. Introduction

K := { knots in the 3-sphere } . K , K ′ ∈ K .

K ↔ x K ′ ⇔ K can be deformed into K ′ by once.

Gordian distance d x ( · , · )

³

d x ( K, K ′ ) := min { n | K = K 0 ↔ · · · x ↔ x K n = K ′ } .

µ ´

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Remark 1.1

³

G x = the 1–skelton of the Gordian complex

introduced by [Hirasawa-Uchida ’02].

µ ´

-14-a

(4)

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Remark 1.1

³

G x = the 1–skelton of the Gordian complex

introduced by [Hirasawa-Uchida ’02].

µ ´

(5)

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Remark 1.1

³

G x = the 1–skelton of the Gordian complex

introduced by [Hirasawa-Uchida ’02].

µ ´

-14-c

(6)

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Remark 1.1

³

G x = the 1–skelton of the Gordian complex

introduced by [Hirasawa-Uchida ’02].

µ ´

(7)

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Remark 1.1

³

G x = the 1–skelton of the Gordian complex

introduced by [Hirasawa-Uchida ’02].

µ ´

-14-e

(8)

Gordian graph G x

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d x ( K, K ′ ) = 1.

µ ´

Proposition 1.2 [Gambaudo-Ghys ’05].

³

The Gordian graph G x is not Gromov hyperbolic.

(We introduce the Gromov hyperbolicity later. )

µ ´

(9)

λ : a local move on knots .

K ↔ λ K ′ ⇔ K can be deformed into K ′ by λ once.

λ -Gordian distance d λ ( · , · )

³

d λ ( K, K ′ ) := min { n | K = K 0 ↔ · · · λ ↔ λ K n = K ′ } .

µ ´

λ -Gordian graph G λ

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d λ ( K, K ′ ) = 1.

µ ´

Problem 1.3.

³

For a given local move λ ,

detect whether G λ is Gromov hyperbolic or not.

µ ´

-13

(10)

λ : a local move on knots .

K ↔ λ K ′ ⇔ K can be deformed into K ′ by λ once.

λ -Gordian distance d λ ( · , · )

³

d λ ( K, K ′ ) := min { n | K = K 0 ↔ · · · λ ↔ λ K n = K ′ } .

µ ´

λ -Gordian graph G λ

³

• { vertices } = K .

• ∃ an edge between K and K ′ ⇔ d λ ( K, K ′ ) = 1.

µ ´

Problem 1.3.

³

For a given local move λ ,

detect whether G λ is Gromov hyperbolic or not.

µ ´

(11)

§ 2. Gromov hyperbolicity

Assumption on graphs

³

Graphs are connected & each edge has length 1.

µ ´

Γ : a graph. N ( γ, ε ) : ε -nbd. of γ ⊂ Γ.

δ -thin

³

T : a triangle with sides s 1 , s 2 , s 3 in Γ.

T is δ -thin ⇔ s i ⊂ N ( s j ∪ s k , δ ) for different i, j, k .

µ ´

-12

(12)

§ 2. Gromov hyperbolicity

Assumption on graphs

³

Graphs are connected & each edge has length 1.

µ ´

Γ : a graph. N ( γ, ε ) : ε -nbd. of γ ⊂ Γ.

δ -thin

³

T : a triangle with sides s 1 , s 2 , s 3 in Γ.

T is δ -thin ⇔ s i ⊂ N ( s j ∪ s k , δ ) for different i, j, k .

µ ´

δ s

1

s

2

s

3

s

3

⊂ N ( s

1

∪ s

2

, δ ).

(13)

Hyperbolicity

³

Γ is δ -hyperbolic (or Gromov hyperbolic) ⇔

any geodesic triangle in Γ is δ -thin for δ > 0.

µ ´

(A geodesic triangle ⇔ each side is a geodesic.) Example 2.1

³

• Any tree is 0-hyperbolic.

• R 2 is not Gromov hyperbolic.

• H 2 is log 3

2 -hyperbolic.

• A graph with finite diameter, r , is r -hyperbolic.

µ ´

Example 2.2 [Masur-Minsky ’99]

³

For a surface S , the curve complex C ( S ) is Gromov hyperbolic. The constant δ depends on S .

µ ´

-11

(14)

Hyperbolicity

³

Γ is δ -hyperbolic (or Gromov hyperbolic) ⇔

any geodesic triangle in Γ is δ -thin for δ > 0.

µ ´

(A geodesic triangle ⇔ each side is a geodesic.) Example 2.1

³

• Any tree is 0-hyperbolic.

• R 2 is not Gromov hyperbolic.

• H 2 is log 3

2 -hyperbolic.

• A graph with finite diameter r is r -hyperbolic.

µ ´

Example 2.2 [Masur-Minsky ’99]

³

For a surface S , the curve complex C ( S ) is Gromov hyperbolic. The constant δ depends on S .

µ ´

(15)

Hyperbolicity

³

Γ is δ -hyperbolic (or Gromov hyperbolic) ⇔

any geodesic triangle in Γ is δ -thin for δ > 0.

µ ´

(A geodesic triangle ⇔ each side is a geodesic.) Example 2.1

³

• Any tree is 0-hyperbolic.

• R 2 is not Gromov hyperbolic.

• H 2 is log 3

2 -hyperbolic.

• A graph with finite diameter r is r -hyperbolic.

µ ´

Example 2.2 [Masur-Minsky ’99]

³

For a surface S , the curve complex C ( S ) is Gromov hyperbolic. The constant δ depends on S .

µ ´

-11-b

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§ 3. ( ι, λ )-Gordian graph

ι : a knot invariant.

K ∼ ι K ′ : ⇔ ι ( K ) = ι ( K ′ ) for K, K ′ ∈ K .

[ K ] ι : the equivalence class of K w.r.t. ∼ ι . K ι := { [ K ] ι | K ∈ K }

( ι, λ )-Gordian graph G λ ι

³

• { vertices } = K ι .

• ∃ an edge between [ K ] ι and [ K ′ ] ι ⇔

∃ J ∈ [ K ] ι , ∃ J ′ ∈ [ K ′ ] ι s.t. d λ ( J, J ′ ) = 1.

µ ´

d λ ι : the metric on G λ ι .

(17)

§ 3. ( ι, λ )-Gordian graph

ι : a knot invariant.

K ∼ ι K ′ : ⇔ ι ( K ) = ι ( K ′ ) for K, K ′ ∈ K .

[ K ] ι : the equivalence class of K w.r.t. ∼ ι . K ι := { [ K ] ι | K ∈ K }

( ι, λ )-Gordian graph G λ ι

³

• { vertices } = K ι .

• ∃ an edge between [ K ] ι and [ K ′ ] ι ⇔

∃ J ∈ [ K ] ι , ∃ J ′ ∈ [ K ′ ] ι s.t. d λ ( J, J ′ ) = 1.

µ ´

d λ ι : the metric on G λ ι .

-10-a

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∇ K : the Conway polynomial of a knot K . ( ∇ K = 1 + a 2 z 2 + · · · + a 2 n z 2 n .)

Proposition 3.1.

³

The ( ∇ , x)-Gordian graph G x ∇ has diameter 2.

µ ´

Proof. Any [ K ] ∇ contains a knot with unknotting number 1 [Kondo], [Sakai] . Thus, diam G x ∇ ≤ 2. On the other hand, d x ∇ ([3 1 ] ∇ , [4 1 ] ∇ ) = 2 [Kawauchi] . ¤

Corollary 3.2.

The graph G x ∇ is Gromov hyperbolic.

Proof. Any graph with finite diameter is

Gromov hyperbolic. ¤

(19)

∇ K : the Conway polynomial of a knot K . ( ∇ K = 1 + a 2 z 2 + · · · + a 2 n z 2 n .)

Proposition 3.1.

³

The ( ∇ , x)-Gordian graph G x ∇ has diameter 2.

µ ´

Proof. Any [ K ] ∇ contains a knot with unknotting number 1 [Kondo], [Sakai] . Thus, diam G x ∇ ≤ 2. On the other hand, d x ∇ ([3 1 ] ∇ , [4 1 ] ∇ ) = 2 [Kawauchi] . ¤

Corollary 3.2.

The graph G x ∇ is Gromov hyperbolic.

Proof. Any graph with finite diameter is

Gromov hyperbolic. ¤

-9-a

(20)

∇ K : the Conway polynomial of a knot K . ( ∇ K = 1 + a 2 z 2 + · · · + a 2 n z 2 n .)

Proposition 3.1.

³

The ( ∇ , x)-Gordian graph G x ∇ has diameter 2.

µ ´

Proof. Any [ K ] ∇ contains a knot with unknotting number 1 [Kondo], [Sakai] . Thus, diam G x ∇ ≤ 2. On the other hand, d x ∇ ([3 1 ] ∇ , [4 1 ] ∇ ) = 2 [Kawauchi] . ¤

Corollary 3.2.

The graph G x ∇ is Gromov hyperbolic.

Proof. Any graph with finite diameter is

Gromov hyperbolic. ¤

(21)

∇ K : the Conway polynomial of a knot K . ( ∇ K = 1 + a 2 z 2 + · · · + a 2 n z 2 n .)

Proposition 3.1.

³

The ( ∇ , x)-Gordian graph G x ∇ has diameter 2.

µ ´

Proof. Any [ K ] ∇ contains a knot with unknotting number 1 [Kondo], [Sakai] . Thus, diam G x ∇ ≤ 2. On the other hand, d x ∇ ([3 1 ] ∇ , [4 1 ] ∇ ) = 2 [Kawauchi] . ¤

Corollary 3.2.

The graph G x ∇ is Gromov hyperbolic.

Proof. Any graph with finite diameter is

Gromov hyperbolic. ¤

-9-c

(22)

Delta-move ∆ [Matveev] [Murakami-Nakanishi]

³

µ ´

We focus on the graph G ∆ ∇ with metric d ∆ ∇ . We call d ∆ ∇ the ( ∇ , ∆)-Gordian distance.

Main Theorem [Ichihara-J.].

The graph G ∆ ∇ is Gromov hyperbolic.

Remark 2.3.

³

The diameter of G ∆ ∇ is infinite.

µ ´

(23)

Delta-move ∆ [Matveev] [Murakami-Nakanishi]

³

µ ´

We focus on the graph G ∆ ∇ with metric d ∆ ∇ . We call d ∆ ∇ the ( ∇ , ∆)-Gordian distance.

Main Theorem [Ichihara-J.].

The graph G ∆ ∇ is Gromov hyperbolic.

Remark 3.3.

³

The diameter of G ∆ ∇ is infinite.

µ ´

-8-a

(24)

A local picture of G ∆ ∇

a

m

( m ≥ 4)

a

2

[ K ]

(25)

§ 4. ( ∇ , ∆)-Gordian distance d ∆ ∇

Lemma 4.1 [Okada ’90].

³

Let K and K ′ be knots with d ∆ ( K, K ′ ) = 1.

Then we have a 2 ( K ) − a 2 ( K ′ ) = ± 1 .

µ ´

Notation: [ K ] = [ K ] ∇ .

Lemma 4.2.

³

For any [ K ] , [ K ′ ], we have the following.

• d ∆ ∇ ([ K ] , [ K ′ ]) ≥ | a 2 ( K ) − a 2 ( K ′ ) | .

• d ∆ ∇ ([ K ] , [ K ′ ]) ≡ | a 2 ( K ) − a 2 ( K ′ ) | mod 2.

µ ´

Recall: For K 1 , K 2 ∈ [ K ], we have ∇ K

1

= ∇ K

2

.

-6

(26)

§ 4. ( ∇ , ∆)-Gordian distance d ∆ ∇

Lemma 4.1 [Okada ’90].

³

Let K and K ′ be knots with d ∆ ( K, K ′ ) = 1.

Then we have a 2 ( K ) − a 2 ( K ′ ) = ± 1 .

µ ´

Notation: [ K ] = [ K ] ∇ .

Lemma 4.2.

³

For any [ K ] , [ K ′ ], we have the following.

• d ∆ ∇ ([ K ] , [ K ′ ]) ≥ | a 2 ( K ) − a 2 ( K ′ ) | .

• d ∆ ∇ ([ K ] , [ K ′ ]) ≡ | a 2 ( K ) − a 2 ( K ′ ) | mod 2.

µ ´

Recall: For K 1 , K 2 ∈ [ K ], we have ∇ K

1

= ∇ K

2

.

(27)

Lemma 4.3.

³

For [ K ] ̸ = [ K ′ ] with a 2 = a 2 ( K ), a ′ 2 = a 2 ( K ′ ), (1) a 2 = a ′ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 2.

(2) | a 2 − a ′ 2 | ≥ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = | a 2 − a ′ 2 | . (3) | a 2 − a ′ 2 | = 1, ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 1 or 3.

µ ´

The proof are achieved by constructing knots which satisfy given conditions. We give the proof later.

… … … … ……

…

…

…

-5

(28)

§ 4. Proof of Main Theorem

V n := { [ K ] ∈ K ∇ | a 2 ( K ) = n } .

S n : the subgraph of G ∆ ∇ induced by V n ∪ V n ± 1 . Lemma 5.1.

³

For [ K ] ∈ K ∇ with a 2 ( K ) = n , N ([ K ] , 3) ⊃ S n .

µ ´

(The proof is immediately obtained by Lemma 4.3.)

am (m ≥ 4)

a2

Vn Vn

−1 Vn+1

S

n

}

[K]

(29)

Main Theorem [Ichihara-J.].

The graph G ∆ ∇ is Gromov hyperbolic.

Proof of Main Theorem There are several cases.

We only show the theorem for a particular case.

Other cases are shown in a similar way.

-3

(30)

T : a geodesic triangle with sides s 1 , s 2 , s 3 . Let

s 1 = x 0 x 1 ∪ x 1 x 2 ∪ · · · ∪ x p − 1 x p , s 2 = y 0 y 1 ∪ y 1 y 2 ∪ · · · ∪ y q − 1 y q , s 3 = z 0 z 1 ∪ z 1 z 2 ∪ · · · ∪ z r − 1 z r ,

where x 0 , . . . , x p , y 0 , . . . , y q , z 0 , . . . , z r ∈ K ∇ with

x 0 = x = z r , y 0 = y = x p , and z 0 = z = y q .

(31)

a

2

a

l≥4

x

1

x

2

x

3

y

3

y

2

y

1

z

3

z

1

z

2

z

4

z

5

z

6

z

7

x

4 =

y

0

z

8 =

x

0

y

4 =

z

0

The figure is an example of T ( p = 4, q = 4, r = 8).

To show: T is 3-thin, namely,

N ( s ( x, y ) ∪ s ( y, z ) , 3) ⊃ s ( z, x ), N ( s ( y, z ) ∪ s ( z, x ) , 3) ⊃ s ( x, y ), and N ( s ( z, x ) ∪ s ( x, y ) , 3) ⊃ s ( y, z ).

-1

(32)

a

2

a

l≥4

x

1

x

2

x

3

y

3

y

2

y

1

z

3

z

1

z

2

z

4

z

5

z

6

z

7

x

4 =

y

0

z

8 =

x

0

y

4 =

z

0

By Lemma 5.1, N ( y j , 3) ⊃ z q − j +1 z q − j , z q − j z q − j − 1 for j = 1 , · · · , q − 1. Thus, we have

N ( s ( y, z ) , 3) ⊃ z q z q − 1 ∪ · · · ∪ z 1 z 0 .

Therefore we have N ( s ( x, y ) ∪ s ( y, z ) , 3) ⊃ s ( z, x ).

(33)

a

2

a

l≥4

x

1

x

2

x

3

y

3

y

2

y

1

z

3

z

1

z

2

z

4

z

5

z

6

z

7

x

4 =

y

0

z

8 =

x

0

y

4 =

z

0

By Lemma 5.1, N ( y j , 3) ⊃ z q − j +1 z q − j , z q − j z q − j − 1 for j = 1 , · · · , q − 1. Thus, we have

N ( s ( y, z ) , 3) ⊃ z q z q − 1 ∪ · · · ∪ z 1 z 0 .

Therefore we have N ( s ( x, y ) ∪ s ( y, z ) , 3) ⊃ s ( z, x ).

-1-b

(34)

a

2

a

l≥4

x

1

x

2

x

3

y

3

y

2

y

1

z

3

z

1

z

2

z

4

z

5

z

6

z

7

x

4 =

y

0

z

8 =

x

0

y

4 =

z

0

Similarly, N ( x j , 3) ⊃ z r − j +1 z r − j , z r − j z r − j − 1 for j = 1 , · · · , p − 1. Thus, we have

N ( s ( x, y ) , 3) ⊃ z r z r − 1 ∪ · · · ∪ z p +1 z p .

Therefore we have N ( s ( x, y ) ∪ s ( y, z ) , 3) ⊃ s ( z, x ).

(35)

a

2

a

l≥4

x

1

x

2

x

3

y

3

y

2

y

1

z

3

z

1

z

2

z

4

z

5

z

6

z

7

x

4 =

y

0

z

8 =

x

0

y

4 =

z

0

Similarly, N ( x j , 3) ⊃ z r − j +1 z r − j , z r − j z r − j − 1 for j = 1 , · · · , p − 1. Thus, we have

N ( s ( x, y ) , 3) ⊃ z r z r − 1 ∪ · · · ∪ z p +1 z p .

Therefore we have N ( s ( x, y ) ∪ s ( y, z ) , 3) ⊃ s ( z, x ).

-1-d

(36)

a

2

a

l≥4

x

1

x

2

x

3

y

3

y

2

y

1

z

3

z

1

z

2

z

4

z

5

z

6

z

7

x

4 =

y

0

z

8 =

x

0

y

4 =

z

0

Similarly, N ( x j , 3) ⊃ z r − j +1 z r − j , z r − j z r − j − 1 for j = 1 , · · · , p − 1. Thus, we have

N ( s ( x, y ) , 3) ⊃ z r z r − 1 ∪ · · · ∪ z p +1 z p .

Therefore we have N ( s ( x, y ) ∪ s ( y, z ) , 3) ⊃ s ( z, x ).

(37)

Remaining two conditions

• N ( s ( y, z ) ∪ s ( z, x ) , 3) ⊃ s ( x, y ) and

• N ( s ( z, x ) ∪ s ( x, y ) , 3) ⊃ s ( y, z )

are shown by the similar argument.

Therefore the geodesic triangle T is 3-thin. ¤

0

(38)

Lemma 4.3.

³

For [ K ] ̸ = [ K ′ ] with a 2 = a 2 ( K ), a ′ 2 = a 2 ( K ′ ), (1) a 2 = a ′ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 2.

(2) | a 2 − a ′ 2 | ≥ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = | a 2 − a ′ 2 | . (3) | a 2 − a ′ 2 | = 1, ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 1 or 3.

µ ´

Proof. Let K m be the twist knot. Then we have

∇ K

m

= 1 + mz 2 , and d ∆ ( K m +1 , K m ) = 1.

(39)

Lemma 4.3.

³

For [ K ] ̸ = [ K ′ ] with a 2 = a 2 ( K ), a ′ 2 = a 2 ( K ′ ), (1) a 2 = a ′ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 2.

(2) | a 2 − a ′ 2 | ≥ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = | a 2 − a ′ 2 | . (3) | a 2 − a ′ 2 | = 1, ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 1 or 3.

µ ´

Proof. Let K m be the twist knot. Then we have

∇ K

m

= 1 + mz 2 , and d ∆ ( K m +1 , K m ) = 1.

…

m-full twists

-3-a

(40)

Let K ± ( α 1 , . . . , α n ) be knots as following.

… … … … ……

…

…

…

α1

}

α2

}

α3

} }

αn−2

}

αn−1

}

αn … … … …

…

…

…

α1

}

α2

}

α3

} }

αn−2

}

αn−1

}

αn

K+(α1… αn) K−(α1… αn)

By [Murakami], [Yamada] , we have

∇ K

±

( α

1

,...,α

n

) = 1 +

∑

n

i =1

( − 1) i − 1 α i z 2 i ,

and d ∆ ( K ± ( α 1 , . . . , α n ) , K α

1

± 1 ) = 1.

(41)

For [ K ] ̸ = [ K ′ ] ∈ K ∇ , let

∇ K = 1 +

∑

n i =1 a 2 i z 2 i , ∇ K

′

= 1 +

∑

m i =1 a ′ 2 i z 2 i , J + = K + ( a 2 , . . . , ( − 1) n − 1 a 2 n ), and

J ± ′ = K ± ( a ′ 2 , . . . , ( − 1) m − 1 a ′ 2 m ).

(1) a 2 = a ′ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≥ 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ 2 by the sequence

of knots J + , K a

2

+1 , J + ′ . ( d ∆ ( J + , J + ′ ) ≤ 2, ∇ J

+

= ∇ K , and ∇ J

′

+

= ∇ K

′

.)

-1

(42)

For [ K ] ̸ = [ K ′ ] ∈ K ∇ , let

∇ K = 1 +

∑

n i =1 a 2 i z 2 i , ∇ K

′

= 1 +

∑

m i =1 a ′ 2 i z 2 i , J + = K + ( a 2 , . . . , ( − 1) n − 1 a 2 n ), and

J ± ′ = K ± ( a ′ 2 , . . . , ( − 1) m − 1 a ′ 2 m ).

(1) a 2 = a ′ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≥ 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ 2 by the sequence

of knots J + , K a

2

+1 , J + ′ . ( d ∆ ( J + , J + ′ ) ≤ 2, ∇ J

+

= ∇ K , and ∇ J

′

+

= ∇ K

′

.)

(43)

For [ K ] ̸ = [ K ′ ] ∈ K ∇ , let

∇ K = 1 +

∑

n i =1 a 2 i z 2 i , ∇ K

′

= 1 +

∑

m i =1 a ′ 2 i z 2 i , J + = K + ( a 2 , . . . , ( − 1) n − 1 a 2 n ), and

J ± ′ = K ± ( a ′ 2 , . . . , ( − 1) m − 1 a ′ 2 m ).

(1) a 2 = a ′ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≥ 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ 2 by the sequence

of knots J + , K a

2

+1 , J + ′ . ( d ∆ ( J + , J + ′ ) ≤ 2, ∇ J

+

= ∇ K , and ∇ J

′

+

= ∇ K

′

.)

-1-b

(44)

(2) | a 2 − a ′ 2 | ≥ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = | a 2 − a ′ 2 | . We may assume that a ′ 2 ≥ a 2 + 2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≥ a ′ 2 − a 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ a ′ 2 − a 2 by the sequence of knots J + , K a

2

+1 , . . . , K a

′

2

− 1 , J − ′ . (3) | a 2 − a ′ 2 | = 1, ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 1 or 3.

We may assume that a ′ 2 = a 2 + 1.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≡ 1 mod 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ 3 by the sequence

of knots J + , K a

2

+1 , K a

2

, J − ′ .

¤ (Lemma 4.3)

(45)

(2) | a 2 − a ′ 2 | ≥ 2 ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = | a 2 − a ′ 2 | . We may assume that a ′ 2 ≥ a 2 + 2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≥ a ′ 2 − a 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ a ′ 2 − a 2 by the sequence of knots J + , K a

2

+1 , . . . , K a

′

2

− 1 , J − ′ . (3) | a 2 − a ′ 2 | = 1, ⇒ d ∆ ∇ ([ K ] , [ K ′ ]) = 1 or 3.

We may assume that a ′ 2 = a 2 + 1.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≡ 1 mod 2 by Lemma 4.2.

• We have d ∆ ∇ ([ K ] , [ K ′ ]) ≤ 3 by the sequence

of knots J + , K a

2

+1 , K a

2

, J − ′ .

¤ (Lemma 4.3)

0-a

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