An Introduction to the Volume Conjecture and its generalizations, I
Hitoshi Murakami
Tohoku University
Workshop on Volume Conjecture and Related Topics in Knot Theory Indian Institute of Science Education and Research, Pune 19th December, 2018
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 1 / 23 Outline
1 colored Jones polynomial
2 Examples of the colored Jones polynomials
3 Volume conjecture
4 Volume conjecture for the figure-eight knot
5 VC is proved for ...
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 2 / 23 ⟨ ⟩ Kauffman⟨ ⟩ bracket⟨ D⟩ for a knot⟨ diagram⟩ D is defined as = A + A−1 , ⟨ ⟩ ⊔ D = (−A2 − A−2)⟨D⟩, ⟨ ⟩ = 1.
Here ⊔⟨denotes the disjoint union.⟩ Note: ⊔ ⊔ · · · ⊔ = (−A2 − A−2)c−1. | {z } c
colored Jones polynomial Kauffman bracket
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 3 / 23 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1 , ⟨ ⟩ ⊔ D = (−A2 − A−2)⟨D⟩, ⟨ ⟩ = 1.
Here ⊔⟨denotes the disjoint union.⟩ Note: ⊔ ⊔ · · · ⊔ = (−A2 − A−2)c−1. | {z } c
colored Jones polynomial Kauffman bracket
Kauffman bracket ⟨D⟩ for a knot diagram D is defined as
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 3 / 23 ⟨ ⟩ Note: ⊔ ⊔ · · · ⊔ = (−A2 − A−2)c−1. | {z } c
colored Jones polynomial Kauffman bracket
⟨ ⟩ Kauffman⟨ ⟩ bracket⟨ D⟩ for a knot⟨ diagram⟩ D is defined as = A + A−1 , ⟨ ⟩ ⊔ D = (−A2 − A−2)⟨D⟩, ⟨ ⟩ = 1.
Here ⊔ denotes the disjoint union.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 3 / 23 colored Jones polynomial Kauffman bracket
⟨ ⟩ Kauffman⟨ ⟩ bracket⟨ D⟩ for a knot⟨ diagram⟩ D is defined as = A + A−1 , ⟨ ⟩ ⊔ D = (−A2 − A−2)⟨D⟩, ⟨ ⟩ = 1.
Here ⊔⟨denotes the disjoint union.⟩ Note: ⊔ ⊔ · · · ⊔ = (−A2 − A−2)c−1. | {z } c
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 3 / 23 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A2 + + + A−2
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A3 + A + A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ +A + A−1 + A−1 + A−3 .
= A3(−A2 − A−2)2 + A(−A2 − A−2) + A(−A2 − A−2) + A−1 +A(−A2 − A−2) + A−1 + A−1 + A−3(−A2 − A−2) = A7 − A3 − A−5.
colored Jones polynomial Kauffman bracket of the trefoil
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 4 / 23 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A2 + + + A−2
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A3 + A + A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ +A + A−1 + A−1 + A−3 .
= A3(−A2 − A−2)2 + A(−A2 − A−2) + A(−A2 − A−2) + A−1 +A(−A2 − A−2) + A−1 + A−1 + A−3(−A2 − A−2) = A7 − A3 − A−5.
colored Jones polynomial Kauffman bracket of the trefoil ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 4 / 23 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A3 + A + A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ +A + A−1 + A−1 + A−3 .
= A3(−A2 − A−2)2 + A(−A2 − A−2) + A(−A2 − A−2) + A−1 +A(−A2 − A−2) + A−1 + A−1 + A−3(−A2 − A−2) = A7 − A3 − A−5.
colored Jones polynomial Kauffman bracket of the trefoil ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A2 + + + A−2
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 4 / 23 = A3(−A2 − A−2)2 + A(−A2 − A−2) + A(−A2 − A−2) + A−1 +A(−A2 − A−2) + A−1 + A−1 + A−3(−A2 − A−2) = A7 − A3 − A−5.
colored Jones polynomial Kauffman bracket of the trefoil ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A2 + + + A−2
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A3 + A + A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ +A + A−1 + A−1 + A−3 .
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 4 / 23 = A7 − A3 − A−5.
colored Jones polynomial Kauffman bracket of the trefoil ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A2 + + + A−2
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A3 + A + A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ +A + A−1 + A−1 + A−3 .
= A3(−A2 − A−2)2 + A(−A2 − A−2) + A(−A2 − A−2) + A−1 +A(−A2 − A−2) + A−1 + A−1 + A−3(−A2 − A−2)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 4 / 23 colored Jones polynomial Kauffman bracket of the trefoil ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A2 + + + A−2
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ = A3 + A + A + A−1
⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ +A + A−1 + A−1 + A−3 .
= A3(−A2 − A−2)2 + A(−A2 − A−2) + A(−A2 − A−2) + A−1 +A(−A2 − A−2) + A−1 + A−1 + A−3(−A2 − A−2) = A7 − A3 − A−5.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 4 / 23 Definition (writhe) D⃗ : oriented knot diagram. • w(D⃗ ) := ♯ − ♯
Definition K: oriented knot presented by D⃗ . ⃗ D: D without orientation. ⃗ V (K; q) := (−A3)−w(D)⟨D⟩ . q:=A−4
V (K; q) is a knot invariant, called the Jones polynomial.
colored Jones polynomial Jones polynomial
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 5 / 23 Definition K: oriented knot presented by D⃗ . ⃗ D: D without orientation. ⃗ V (K; q) := (−A3)−w(D)⟨D⟩ . q:=A−4
V (K; q) is a knot invariant, called the Jones polynomial.
colored Jones polynomial Jones polynomial
Definition (writhe) D⃗ : oriented knot diagram. • w(D⃗ ) := ♯ − ♯
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 5 / 23 K: oriented knot presented by D⃗ . ⃗ D: D without orientation. ⃗ V (K; q) := (−A3)−w(D)⟨D⟩ . q:=A−4
V (K; q) is a knot invariant, called the Jones polynomial.
colored Jones polynomial Jones polynomial
Definition (writhe) D⃗ : oriented knot diagram. • w(D⃗ ) := ♯ − ♯
Definition
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 5 / 23 colored Jones polynomial Jones polynomial
Definition (writhe) D⃗ : oriented knot diagram. • w(D⃗ ) := ♯ − ♯
Definition K: oriented knot presented by D⃗ . ⃗ D: D without orientation. ⃗ V (K; q) := (−A3)−w(D)⟨D⟩ . q:=A−4
V (K; q) is a knot invariant, called the Jones polynomial.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 5 / 23 w = −3
⇒ J ; q = −A9(A7 − A3 − A−5) = −q−4 + q−3 + q−1. q:=A−4 − − J ; q = q2 − q + 1 − q 1 − q 2.
colored Jones polynomial Example of the Jones polynomials
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 6 / 23
⇒ J ; q = −A9(A7 − A3 − A−5) = −q−4 + q−3 + q−1. q:=A−4 − − J ; q = q2 − q + 1 − q 1 − q 2.
colored Jones polynomial Example of the Jones polynomials
w = −3
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 6 / 23 − − J ; q = q2 − q + 1 − q 1 − q 2.
colored Jones polynomial Example of the Jones polynomials
w = −3
⇒ J ; q = −A9(A7 − A3 − A−5) = −q−4 + q−3 + q−1. q:=A−4
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 6 / 23 colored Jones polynomial Example of the Jones polynomials
w = −3
⇒ J ; q = −A9(A7 − A3 − A−5) = −q−4 + q−3 + q−1. q:=A−4 − − J ; q = q2 − q + 1 − q 1 − q 2.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 6 / 23 2 • 2 − 1 ⇒ := (−A2−A2) =
Definition (Jones–Wenzl idempotent)
k−1 ( ) 1 k k−1 ∆k−2 := 1 − k−2 ∆k−1 1 k−1 ⟨ ⟩ − − k A2(k+1)−A 2(k+1) k with ∆k := ( 1) A2−A−2 = .
k ⇒ = , = (−1)k Ak2+2k k
colored Jones polynomial Jones–Wenzl idempotent
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 7 / 23 2 ⇒ =
Definition (Jones–Wenzl idempotent)
k−1 ( ) 1 k k−1 ∆k−2 := 1 − k−2 ∆k−1 1 k−1 ⟨ ⟩ − − k A2(k+1)−A 2(k+1) k with ∆k := ( 1) A2−A−2 = .
k ⇒ = , = (−1)k Ak2+2k k
colored Jones polynomial Jones–Wenzl idempotent
• 2 − 1 := (−A2−A2)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 7 / 23 Definition (Jones–Wenzl idempotent)
k−1 ( ) 1 k k−1 ∆k−2 := 1 − k−2 ∆k−1 1 k−1 ⟨ ⟩ − − k A2(k+1)−A 2(k+1) k with ∆k := ( 1) A2−A−2 = .
k ⇒ = , = (−1)k Ak2+2k k
colored Jones polynomial Jones–Wenzl idempotent 2 • 2 − 1 ⇒ := (−A2−A2) =
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 7 / 23 k ⇒ = , = (−1)k Ak2+2k k
colored Jones polynomial Jones–Wenzl idempotent 2 • 2 − 1 ⇒ := (−A2−A2) =
Definition (Jones–Wenzl idempotent)
k−1 ( ) 1 k k−1 ∆k−2 := 1 − k−2 ∆k−1 1 k−1 ⟨ ⟩ − − k A2(k+1)−A 2(k+1) k with ∆k := ( 1) A2−A−2 = .
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 7 / 23 = (−1)k Ak2+2k k
colored Jones polynomial Jones–Wenzl idempotent 2 • 2 − 1 ⇒ := (−A2−A2) =
Definition (Jones–Wenzl idempotent)
k−1 ( ) 1 k k−1 ∆k−2 := 1 − k−2 ∆k−1 1 k−1 ⟨ ⟩ − − k A2(k+1)−A 2(k+1) k with ∆k := ( 1) A2−A−2 = .
k ⇒ = ,
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 7 / 23 colored Jones polynomial Jones–Wenzl idempotent 2 • 2 − 1 ⇒ := (−A2−A2) =
Definition (Jones–Wenzl idempotent)
k−1 ( ) 1 k k−1 ∆k−2 := 1 − k−2 ∆k−1 1 k−1 ⟨ ⟩ − − k A2(k+1)−A 2(k+1) k with ∆k := ( 1) A2−A−2 = .
k ⇒ = , = (−1)k Ak2+2k k
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 7 / 23 D : knot diagram.
k ⇒ D : diagram obtained from D by replacing the string with the
Jones–Wenzl idempotent Definition (N-colored Jones polynomial) ⟨ ⟩ ( ) −w(D⃗ ) − N−1 N2−1 N−1 JN (K; q) := ( 1) A D . q:=A4
−1 J2(K; q) = V (K; q ).
colored Jones polynomial colored Jones polynomial
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 8 / 23 k ⇒ D : diagram obtained from D by replacing the string with the
Jones–Wenzl idempotent Definition (N-colored Jones polynomial) ⟨ ⟩ ( ) −w(D⃗ ) − N−1 N2−1 N−1 JN (K; q) := ( 1) A D . q:=A4
−1 J2(K; q) = V (K; q ).
colored Jones polynomial colored Jones polynomial
D : knot diagram.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 8 / 23 Definition (N-colored Jones polynomial) ⟨ ⟩ ( ) −w(D⃗ ) − N−1 N2−1 N−1 JN (K; q) := ( 1) A D . q:=A4
−1 J2(K; q) = V (K; q ).
colored Jones polynomial colored Jones polynomial
D : knot diagram.
k ⇒ D : diagram obtained from D by replacing the string with the
Jones–Wenzl idempotent
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 8 / 23 −1 J2(K; q) = V (K; q ).
colored Jones polynomial colored Jones polynomial
D : knot diagram.
k ⇒ D : diagram obtained from D by replacing the string with the
Jones–Wenzl idempotent Definition (N-colored Jones polynomial) ⟨ ⟩ ( ) −w(D⃗ ) − N−1 N2−1 N−1 JN (K; q) := ( 1) A D . q:=A4
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 8 / 23 colored Jones polynomial colored Jones polynomial
D : knot diagram.
k ⇒ D : diagram obtained from D by replacing the string with the
Jones–Wenzl idempotent Definition (N-colored Jones polynomial) ⟨ ⟩ ( ) −w(D⃗ ) − N−1 N2−1 N−1 JN (K; q) := ( 1) A D . q:=A4
−1 J2(K; q) = V (K; q ). Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 8 / 23 1 3 4 J2( ; q) =q + q − q , 2 5 7 8 9 10 11 J3( ; q) =q + q − q + q − q − q + q , 3 7 10 11 13 14 15 17 19 J4( ; q) =q + q − q + q − q − q + q − q + q + q20 − q21 . . (−1)N−1q3(N2−1)/2 J ( ; q) = N qN/2 − q−N/2 N∑−1 2 × (−1)k q−3(k +k)/2(q(2k+1)/2 − q−(2k+1)/2). k=0 (M. Rosso, V. Jones, H. Morton)
Examples of the colored Jones polynomials colored Jones polynomials of
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 9 / 23 Examples of the colored Jones polynomials colored Jones polynomials of
1 3 4 J2( ; q) =q + q − q , 2 5 7 8 9 10 11 J3( ; q) =q + q − q + q − q − q + q , 3 7 10 11 13 14 15 17 19 J4( ; q) =q + q − q + q − q − q + q − q + q + q20 − q21 . . (−1)N−1q3(N2−1)/2 J ( ; q) = N qN/2 − q−N/2 N∑−1 2 × (−1)k q−3(k +k)/2(q(2k+1)/2 − q−(2k+1)/2). k=0 (M. Rosso, V. Jones, H. Morton)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 9 / 23 2 −1 −2 J2( ) =q − q + 1 − q + q , 6 5 4 3 2 −1 −2 −3 −4 J3( ) =q − q − q + 2q − q − q + 3 − q − q + 2q − q − q−5 + q−6,
12 11 10 8 6 4 2 −2 −4 J4( ) =q − q − q + 2q − 2q + 3q − 3q + 3 − 3q + 3q − 2q−6 + 2q−8 − q−10 − q−11 + q−12 . . N∑−1 ∏j ( )( ) (N−k)/2 −(N−k)/2 (N+k)/2 −(N+k)/2 JN ( ; q) = q − q q − q . j=0 k=1
(K. Habiro, T. Lˆe)
Examples of the colored Jones polynomials colored Jones polynomials of
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 10 / 23 Examples of the colored Jones polynomials colored Jones polynomials of
2 −1 −2 J2( ) =q − q + 1 − q + q , 6 5 4 3 2 −1 −2 −3 −4 J3( ) =q − q − q + 2q − q − q + 3 − q − q + 2q − q − q−5 + q−6,
12 11 10 8 6 4 2 −2 −4 J4( ) =q − q − q + 2q − 2q + 3q − 3q + 3 − 3q + 3q − 2q−6 + 2q−8 − q−10 − q−11 + q−12 . . N∑−1 ∏j ( )( ) (N−k)/2 −(N−k)/2 (N+k)/2 −(N+k)/2 JN ( ; q) = q − q q − q . j=0 k=1
(K. Habiro, T. Lˆe)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 10 / 23 ( √ ) ( √ ) 1 | − | |JN ; exp(2π −1/N) | N log JN ; exp(2π 1/N) 0.6 1000
0.5 800
0.4
600 0.3
400 0.2
200 0.1
20 40 60 80 100 20 40 60 80 100 JN ( ; q) N−1 3(N2−1)/2 N∑−1 (−1) q 2 = (−1)k q−3(k +k)/2(q(2k+1)/2 − q−(2k+1)/2) qN/2 − q−N/2 k=0
Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 11 / 23 ( √ ) 1 | − | N log JN ; exp(2π 1/N) 0.6
0.5
0.4
0.3
0.2
0.1
20 40 60 80 100 JN ( ; q) N−1 3(N2−1)/2 N∑−1 (−1) q 2 = (−1)k q−3(k +k)/2(q(2k+1)/2 − q−(2k+1)/2) qN/2 − q−N/2 k=0
Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
( √ ) |JN ; exp(2π −1/N) |
1000
800
600
400
200
20 40 60 80 100
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 11 / 23 Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
( √ ) ( √ ) 1 | − | |JN ; exp(2π −1/N) | N log JN ; exp(2π 1/N) 0.6 1000
0.5 800
0.4
600 0.3
400 0.2
200 0.1
20 40 60 80 100 20 40 60 80 100 JN ( ; q) N−1 3(N2−1)/2 N∑−1 (−1) q 2 = (−1)k q−3(k +k)/2(q(2k+1)/2 − q−(2k+1)/2) qN/2 − q−N/2 k=0
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 11 / 23 Graph of ( Graph of√ ) ( √ ) 1 | − | |JN ; exp(2π −1/N) |. N log JN ; exp(2π 1/N) .
0.8 6 × 1014
5 × 1014 0.6 4 × 1014
3 × 1014 0.4
2 × 1014 0.2 1 × 1014
20 40 60 80 100 20 40 60 80 100 JN∑( ; q∏) ( )( ) N−1 j (N−k)/2 − −(N−k)/2 (N+k)/2 − −(N+k)/2 = j=0 k=1 q q q q . What is the difference between and ?
Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 12 / 23 ( Graph of √ ) 1 | − | N log JN ; exp(2π 1/N) .
0.8
0.6
0.4
0.2
20 40 60 80 100 JN∑( ; q∏) ( )( ) N−1 j (N−k)/2 − −(N−k)/2 (N+k)/2 − −(N+k)/2 = j=0 k=1 q q q q . What is the difference between and ?
Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
( Graph of√ ) |JN ; exp(2π −1/N) |.
6 × 1014
5 × 1014
4 × 1014
3 × 1014
2 × 1014
1 × 1014
20 40 60 80 100
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 12 / 23 What is the difference between and ?
Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
Graph of ( Graph of√ ) ( √ ) 1 | − | |JN ; exp(2π −1/N) |. N log JN ; exp(2π 1/N) .
0.8 6 × 1014
5 × 1014 0.6 4 × 1014
3 × 1014 0.4
2 × 1014 0.2 1 × 1014
20 40 60 80 100 20 40 60 80 100 JN∑( ; q∏) ( )( ) N−1 j (N−k)/2 − −(N−k)/2 (N+k)/2 − −(N+k)/2 = j=0 k=1 q q q q .
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 12 / 23 Examples of the colored Jones polynomials Colored Jones polynomial at Nth root of unity,
Graph of ( Graph of√ ) ( √ ) 1 | − | |JN ; exp(2π −1/N) |. N log JN ; exp(2π 1/N) .
0.8 6 × 1014
5 × 1014 0.6 4 × 1014
3 × 1014 0.4
2 × 1014 0.2 1 × 1014
20 40 60 80 100 20 40 60 80 100 JN∑( ; q∏) ( )( ) N−1 j (N−k)/2 − −(N−k)/2 (N+k)/2 − −(N+k)/2 = j=0 k=1 q q q q . What is the difference between and ?
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 12 / 23 Definition (Simplicial volume (Gromov norm)) ∑ 3 Vol(S \ K) := Hyperbolic Volume of Hi .
Hi :hyperbolic piece
Definition (Jaco–Shalen–Johannson decomposition) S3 \ K can be uniquely decomposed as (⊔ ) (⊔ ) 3 S \ K = Hi ⊔ Ej
with Hi hyperbolic and Ej Seifert-fibered.
Volume conjecture Volume conjecture Conjecture (Volume Conjecture, R. Kashaev (1997), J. Murakami+H.M. (2001)) K: knot √ log |J (K; exp(2π −1/N))| 2π lim N = Vol(S3 \ K). N→∞ N
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 13 / 23 Definition (Jaco–Shalen–Johannson decomposition) S3 \ K can be uniquely decomposed as (⊔ ) (⊔ ) 3 S \ K = Hi ⊔ Ej
with Hi hyperbolic and Ej Seifert-fibered.
Volume conjecture Volume conjecture Conjecture (Volume Conjecture, R. Kashaev (1997), J. Murakami+H.M. (2001)) K: knot √ log |J (K; exp(2π −1/N))| 2π lim N = Vol(S3 \ K). N→∞ N
Definition (Simplicial volume (Gromov norm)) ∑ 3 Vol(S \ K) := Hyperbolic Volume of Hi .
Hi :hyperbolic piece
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 13 / 23 Volume conjecture Volume conjecture Conjecture (Volume Conjecture, R. Kashaev (1997), J. Murakami+H.M. (2001)) K: knot √ log |J (K; exp(2π −1/N))| 2π lim N = Vol(S3 \ K). N→∞ N
Definition (Simplicial volume (Gromov norm)) ∑ 3 Vol(S \ K) := Hyperbolic Volume of Hi .
Hi :hyperbolic piece
Definition (Jaco–Shalen–Johannson decomposition) S3 \ K can be uniquely decomposed as (⊔ ) (⊔ ) 3 S \ K = Hi ⊔ Ej with Hi hyperbolic and Ej Seifert-fibered.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 13 / 23 = ⊔ .
hyperbolic Seifert fibered Vol = Vol
Volume conjecture Example of JSJ decomposition
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 14 / 23 hyperbolic Seifert fibered Vol = Vol
Volume conjecture Example of JSJ decomposition
= ⊔ .
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 14 / 23 Seifert fibered Vol = Vol
Volume conjecture Example of JSJ decomposition
= ⊔ .
hyperbolic
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 14 / 23 Vol = Vol
Volume conjecture Example of JSJ decomposition
= ⊔ .
hyperbolic Seifert fibered
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 14 / 23 Volume conjecture Example of JSJ decomposition
= ⊔ .
hyperbolic Seifert fibered Vol = Vol
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 14 / 23 Theorem (K. Habiro, T. Lˆe) ( ) N∑−1 ∏j ( )( ) (N−k)/2 −(N−k)/2 (N+k)/2 −(N+k)/2 JN ; q = q − q q − q . j=0 k=1 √ q 7→ exp(2π −1/N)
( √ ) N∑−1 ∏j JN ; exp(2π −1/N) = f (N; k) j=0 k=1
with f (N; k) := 4 sin2(kπ/N).
Volume conjecture for the figure-eight knot Colored Jones polynomial of
Proof of the VC for is given by T. Ekholm in 1999.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 15 / 23 √ q 7→ exp(2π −1/N)
( √ ) N∑−1 ∏j JN ; exp(2π −1/N) = f (N; k) j=0 k=1
with f (N; k) := 4 sin2(kπ/N).
Volume conjecture for the figure-eight knot Colored Jones polynomial of
Proof of the VC for is given by T. Ekholm in 1999. Theorem (K. Habiro, T. Lˆe) ( ) N∑−1 ∏j ( )( ) (N−k)/2 −(N−k)/2 (N+k)/2 −(N+k)/2 JN ; q = q − q q − q . j=0 k=1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 15 / 23 Volume conjecture for the figure-eight knot Colored Jones polynomial of
Proof of the VC for is given by T. Ekholm in 1999. Theorem (K. Habiro, T. Lˆe) ( ) N∑−1 ∏j ( )( ) (N−k)/2 −(N−k)/2 (N+k)/2 −(N+k)/2 JN ; q = q − q q − q . j=0 k=1 √ q 7→ exp(2π −1/N)
( √ ) N∑−1 ∏j JN ; exp(2π −1/N) = f (N; k) j=0 k=1
with f (N; k) := 4 sin2(kπ/N).
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 15 / 23 ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6 ∏ j Put g(N; j) := k=1 f (N; k). j 0 ··· N/6 ··· 5N/6 ··· 1 f (N; k) < 1 1 > 1 1 < 1 g(N; j) 1 ↘ ↗ maximum ↘
Volume conjecture for the figure-eight knot Find the maximum of the summands
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6 ∏ j Put g(N; j) := k=1 f (N; k). j 0 ··· N/6 ··· 5N/6 ··· 1 f (N; k) < 1 1 > 1 1 < 1 g(N; j) 1 ↘ ↗ maximum ↘
Volume conjecture for the figure-eight knot Find the maximum of the summands ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 ∏ j Put g(N; j) := k=1 f (N; k). j 0 ··· N/6 ··· 5N/6 ··· 1 f (N; k) < 1 1 > 1 1 < 1 g(N; j) 1 ↘ ↗ maximum ↘
Volume conjecture for the figure-eight knot Find the maximum of the summands ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 j 0 ··· N/6 ··· 5N/6 ··· 1 f (N; k) < 1 1 > 1 1 < 1 g(N; j) 1 ↘ ↗ maximum ↘
Volume conjecture for the figure-eight knot Find the maximum of the summands ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6 ∏ j Put g(N; j) := k=1 f (N; k).
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 f (N; k) < 1 1 > 1 1 < 1 g(N; j) 1 ↘ ↗ maximum ↘
Volume conjecture for the figure-eight knot Find the maximum of the summands ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6 ∏ j Put g(N; j) := k=1 f (N; k). j 0 ··· N/6 ··· 5N/6 ··· 1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 g(N; j) 1 ↘ ↗ maximum ↘
Volume conjecture for the figure-eight knot Find the maximum of the summands ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6 ∏ j Put g(N; j) := k=1 f (N; k). j 0 ··· N/6 ··· 5N/6 ··· 1 f (N; k) < 1 1 > 1 1 < 1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 Volume conjecture for the figure-eight knot Find the maximum of the summands ( √ ) N∑−1 ∏j 2π −1/N 2 JN ; e = f (N; k) with f (N; k) := 4 sin (kπ/N). j=0 k=1 Graph of f (N; k) f(N;k)
4
1
0 j N/6 5N/6 ∏ j Put g(N; j) := k=1 f (N; k). j 0 ··· N/6 ··· 5N/6 ··· 1 f (N; k) < 1 1 > 1 1 < 1 g(N; j) 1 ↘ ↗ maximum ↘
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 16 / 23 Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6).
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 Volume conjecture for the figure-eight knot Limit of the sum is the limit of the maximum
Maximum of {g(N; j)}0≤j≤N−1 is g(N; 5N/6). ( √ ) ∑ − N−1 JN ; exp(2π 1/N) = j=0 g(N; j) and g(N; j) > 0. ⇓ ( √ ) g(N; 5N/6) ≤ JN ; exp(2π −1/N) ≤ N × g(N; 5N/6) ⇓ log g(N;5N/6) ≤ log JN ≤ log N log g(N;5N/6) N N N + N ⇓ log g(N; 5N/6) log J log N log g(N; 5N/6) lim ≤ lim N ≤ lim + lim N→∞ N N→∞ N N→∞ N N→∞ N ⇓ log J log g(N; 5N/6) lim N = lim N→∞ N N→∞ N
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 17 / 23 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 5N/6 1 ∑ ( ) = 2 lim log 2 sin(kπ/N) N→∞ N ∫ k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 1 ∑ = lim log f (N; k) N→∞ N k=1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 ∫ 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N k=1 k=1
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 2 = − Λ(5π/6) = 0.323066 ..., π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) = log 2 sin x dx π 0
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 = 0.323066 ...,
∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) π 0 π
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 What is Λ(5π/6)?
Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 Volume conjecture for the figure-eight knot Calculation of the limit of the maximum
( √ ) − log JN ; exp(2π 1/N) log g(N; 5N/6) lim = lim N→∞ N N→∞ N 5N/6 5N/6 1 ∑ 1 ∑ ( ) = lim log f (N; k) = 2 lim log 2 sin(kπ/N) N→∞ N N→∞ N ∫ k=1 k=1 2 5π/6 ( ) 2 = log 2 sin x dx = − Λ(5π/6) = 0.323066 ..., π 0 π ∫ − θ | | where Λ(θ) := 0 log 2 sin x dx is the Lobachevsky function. What is Λ(5π/6)?
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 18 / 23 ∫ − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot Lobachevsky function Λ(θ)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine:
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 ( √ ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). 2
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 Volume conjecture for the figure-eight knot
Lobachevsky function Λ(∫θ) − θ | | Some properties of Λ(θ) := 0 log 2 sin x dx. Λ is an odd function and has period π. Λ(2θ) = 2Λ(∑θ) + 2Λ(θ + π/2). n−1 (Λ(nθ) = n k=1 Λ(θ + kπ/n) in general.) The first property is easy. To prove the second, we use the double angle formula of sine: sin(2x) = 2 sin x cos x. ⇒ log |2 sin(2x)| = log |2 sin x| + log |2 sin(x + π/2)|. So we have Λ(5π/6) = −Λ(π/6) Λ(π/3) = 2Λ(π/6) + 2Λ(2π/3) = 2Λ(π/6) − 2Λ(π/3) 3 ⇒ Λ(5π/6) = − Λ(π/3). ( √2 ) ⇒ 2π lim log JN ; exp(2π −1/N) /N = 6Λ(π/3) N→∞
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 19 / 23 What is 6Λ(π/3)? Theorem (W. Thurston)
c e
= C B ∪ D’ B’ b ′ ′ A=A ,B=B , h C’ D C=C ′,D=D′ a A d g A’ f
We can regard both pieces in the right hand side as regular ideal hyperbolic tetrahedra. ⇒ S3 \ possesses a complete hyperbolic structure.
Volume conjecture for the figure-eight knot Decomposition of S 3 \ into two tetrahedra
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 20 / 23 Theorem (W. Thurston)
c e
= C B ∪ D’ B’ b ′ ′ A=A ,B=B , h C’ D C=C ′,D=D′ a A d g A’ f
We can regard both pieces in the right hand side as regular ideal hyperbolic tetrahedra. ⇒ S3 \ possesses a complete hyperbolic structure.
Volume conjecture for the figure-eight knot Decomposition of S 3 \ into two tetrahedra
What is 6Λ(π/3)?
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 20 / 23 ⇒ S3 \ possesses a complete hyperbolic structure.
Volume conjecture for the figure-eight knot Decomposition of S 3 \ into two tetrahedra
What is 6Λ(π/3)? Theorem (W. Thurston)
c e
= C B ∪ D’ B’ b ′ ′ A=A ,B=B , h C’ D C=C ′,D=D′ a A d g A’ f
We can regard both pieces in the right hand side as regular ideal hyperbolic tetrahedra.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 20 / 23 Volume conjecture for the figure-eight knot Decomposition of S 3 \ into two tetrahedra
What is 6Λ(π/3)? Theorem (W. Thurston)
c e
= C B ∪ D’ B’ b ′ ′ A=A ,B=B , h C’ D C=C ′,D=D′ a A d g A’ f
We can regard both pieces in the right hand side as regular ideal hyperbolic tetrahedra. ⇒ S3 \ possesses a complete hyperbolic structure.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 20 / 23 √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t . Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity. We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane. Ideal hyperbolic tetrahedron Top view Ideal hyperbolic tetrahedron is ∆(α, β, γ) defined (up to isometry) by the similarity class of this triangle. β α γ β α γ Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity. We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane. Ideal hyperbolic tetrahedron Top view Ideal hyperbolic tetrahedron is ∆(α, β, γ) defined (up to isometry) by the similarity class of this triangle. β α γ β α γ Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t .
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane. Ideal hyperbolic tetrahedron Top view Ideal hyperbolic tetrahedron is ∆(α, β, γ) defined (up to isometry) by the similarity class of this triangle. β α γ β α γ Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t . Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 Ideal hyperbolic tetrahedron Top view Ideal hyperbolic tetrahedron is ∆(α, β, γ) defined (up to isometry) by the similarity class of this triangle. β α γ β α γ Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t . Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity. We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 Top view Ideal hyperbolic tetrahedron is defined (up to isometry) by the similarity class of this triangle.
β α γ Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t . Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity. We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane. Ideal hyperbolic tetrahedron ∆(α, β, γ)
β α γ
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 Ideal hyperbolic tetrahedron is defined (up to isometry) by the similarity class of this triangle.
Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t . Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity. We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane. Ideal hyperbolic tetrahedron Top view ∆(α, β, γ)
β α γ β α γ
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 Volume conjecture for the figure-eight knot Ideal hyperbolic tetrahedron √ H3 { | } dx2+dy 2+dt2 := (x, y, t) t > 0 : with hyperbolic metric ds := t . Ideal hyperbolic tetrahedron: tetrahedron with geodesic faces with four vertices in the boundary at infinity. We may assume ▶ One vertex is at ∞. ▶ The other three are on xy-plane. Ideal hyperbolic tetrahedron Top view Ideal hyperbolic tetrahedron is ∆(α, β, γ) defined (up to isometry) by the similarity class of this triangle. β α γ β α γ Vol(∆(α, β, γ)) = Λ(α)+Λ(β)+Λ(γ).
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 21 / 23 =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N
Volume conjecture for the figure-eight knot Proof of VC - conclusion
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N =6Λ(π/3)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 ( ) ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has a geometry of surface × circle.
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 Volume conjecture for the figure-eight knot Proof of VC - conclusion
( √ ) log JN ; exp(2π −1/N) 2π lim N→∞ N =6Λ(π/3) =2 Vol(regular ideal hyperbolic tetrahedron) ( ) = Vol S3 \ .
This amazing fact was first observed by R. Kashaev and proved by T. Ekholm. On the other hand the complement of is a Seifert fibered space, that is, it has( a geometry) of surface × circle. ⇒ Vol S3 \ = 0. In fact Kashaev and O. Tirkkonen proved that √ log JN (T (p,q);exp(2π −1/N)) 2π limN→∞ N = 0 for any torus knot T (p, q).
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 22 / 23 figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999))
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)),
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)),
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)),
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23 VC is proved for ... So far the Volume Conjecture is proved for figure-eight knot (hyperbolic) (T. Ekholm (1999)) torus knots (including ) (Seifert fibered) (R. Kashaev + O. Tirkkonen (2000)), iterated torus knots (Seifert fibered + Seifert fibered) (R. van der Veen (2008)), Whitehead doubles of torus knots of (2, 2n + 1) (Seifert fibered + hyperbolic) (H. Zheng (2007)), (2, 2m + 1)-cable of the figure-eight knot (hyperbolic + Seifert fibered) (T. Lˆeand A. Tran (2010)).
52 knot (hyperbolic) (R. Kashaev and Y. Yokota, T. Oh- tsuki) hyperbolic knots with six crossings (T. Ohtsuki and Y. Yokota)
Hitoshi Murakami (Tohoku University ) Volume Conjecture, I IISER, Pune, 19th December, 2018 23 / 23