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)
§ 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 ´
§ 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
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 ´
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
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 ´
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
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 ´
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
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 ´
§ 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
§ 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
1s
2s
3s
3⊂ N ( s
1∪ s
2, δ ).
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
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 ´
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
§ 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 .
§ 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
∇ 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. ¤
∇ 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
∇ 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. ¤
∇ 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
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 ´
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
A local picture of G ∆ ∇
a
m( m ≥ 4)
a
2[ K ]
§ 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
§ 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.
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
§ 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]
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
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 .
a
2a
l≥4x
1x
2x
3y
3y
2y
1z
3z
1z
2z
4z
5z
6z
7x
4 =y
0z
8 =x
0y
4 =z
0The 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
a
2a
l≥4x
1x
2x
3y
3y
2y
1z
3z
1z
2z
4z
5z
6z
7x
4 =y
0z
8 =x
0y
4 =z
0By 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 ).
a
2a
l≥4x
1x
2x
3y
3y
2y
1z
3z
1z
2z
4z
5z
6z
7x
4 =y
0z
8 =x
0y
4 =z
0By 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
a
2a
l≥4x
1x
2x
3y
3y
2y
1z
3z
1z
2z
4z
5z
6z
7x
4 =y
0z
8 =x
0y
4 =z
0Similarly, 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 ).
a
2a
l≥4x
1x
2x
3y
3y
2y
1z
3z
1z
2z
4z
5z
6z
7x
4 =y
0z
8 =x
0y
4 =z
0Similarly, 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
a
2a
l≥4x
1x
2x
3y
3y
2y
1z
3z
1z
2z
4z
5z
6z
7x
4 =y
0z
8 =x
0y
4 =z
0Similarly, 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 ).
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
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.
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
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.
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
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
′.)
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
(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)
(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