On the complex volume of hyperbolic knots Yoshiyuki Yokota Tokyo Metropolitan University

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1 On the complex volume of hyperbolic knots Yoshiyuki Yokota Tokyo Metropolitan University Let M be a closed, oriented, hyperbolic 3-manifold. Then, the Chern-Simons invariant of M is defined by cs(m) = 8π 2 tr(a da + 2 A A A) R/Z, 3 s(m) where A denotes the connection in the orthonormal frame bundle determined by the metric and s(m) is an orthonormal frame field. In this talk, we define the complex volume of M by cv(m) = 2π 2 cs(m) + vol(m) mod 2π 2, which is extended to cusped hyperbolic 3-manifolds modulo π 2.

2 2 Remark. The 3-dimensional hyperbolic space H 3 is the upper half of R 3 endowed with the metric ds 2 = (dx 2 + dy 2 + dt 2 )/t 2. A tetarahedron in H 3 whose 4 vertices are in H 3 = C { } is called ideal. The shape of such a tetrahedron is determined by a complex number called modulus. z /z /( z) 0 z

3 3 Conjecture. Let K be a hyperbolic knot in S 3. If N is large, J K (N; e 2π /N ) e N 2π { 2π 2 cs(s 3 \K)+ vol(s 3 \K)}, where J K (N; q) is the N-colored Jones polynomial of K. This conjecture is still open. However, we can show that J K (N; e 2π /N ) = e N 2π {V (x,...,x n )+O(/N)} dx dx n and the hyperbolicity equations for M = S 3 \ K are given by V x ν = 2π r ν, r ν Z. x ν In this talk, we prove that, if x ν = z ν is the geometric solution, cv(m) = V (z,..., z n ) 2π n r ν log z ν mod π 2. ν=

4 4 Example. Choose a diagram D of a hyperbolic knot K in S 3, and remove an overpass and an underpass of D which are adjacent. Then, we obtain a subgraph G of D with the edge variables x ν s. x5 x3 x2 x4 x

5 5 Put dilogarithm functions on the interior corners of G. x y x y Li 2 (y/x) π 2 /6 π 2 /6 Li 2 (x/y) The potential function V (x, x 2, x 3, x 4, x 5 ) is nothing but the sum of these dilogarithm functions, that is, Li 2 (x /x 4 ) Li 2 (x /x 3 ) + Li 2 (x ) Li 2 (/x 4 ) + Li 2 (x 2 /x 4 ) Li 2 (x 2 ) Li 2 (/x 2 ) + Li 2 (x 5 /x 2 ) Li 2 (x 5 ) Li 2 (/x 5 ) + Li 2 (x 3 /x 5 ) Li 2 (x 3 ) + π 2 /3.

6 6 Then, there is an ideal triangulation S of M = S 3 \ K, such that the hyperbolicity equations for S are given by modulo 2π Z. 0 x V x ln x /x 3 ( x /x 4 )( x ), 0 x 2 V x 2 ln ( x 2)( x 5 /x 2 ) ( x 2 /x 4 )( /x 2 ), 0 x 3 V x 3 ln ( x /x 3 )( x 3 ) x 3 /x 5, 0 x 4 V x 4 ln ( x /x 4 )( x 2 /x 4 ) /x 4, 0 x 5 V x 5 ln ( x 5)( x 3 /x 5 ) ( x 5 /x 2 )( /x 5 )

7 The solutions to the equations above are given by x ± 2.484i +.28 ± 0.392i x ±.29i 0.37 ± 0.68i x 3 = i, i, x ±.29i ± 0.68i.833 x 5.77 ± 0.250i i.379 with r ν 0. Note that these equations satisfy x x 4, x x 3, x, x 4, x 2 x 4, x 2, x 2, x 5 x 2, x 5, x 5, x 3 x 5, x 3 {0,, }. The critical values of V (x, x 2, x 3, x 4, x 5 ) are given by.9099 ± 4.249i,.8538 ±.089i,.20365, and so cv(m) =.9099 ± 4.249i. 7

8 Ê The knot 6 _ li@x_d := PolyLog@2, xd v@x_, y_, z_d := li@xd + li@zd + lia ÅÅÅÅ x E + lia ÅÅÅÅ z E + li@yd - lia ÅÅÅÅ y E - lia ÅÅÅÅ z x E - lia ÅÅÅÅ y z E - ÅÅÅÅÅÅ p2 3 x = 8x, x2, x3<; dv = Table@Exp@x@@iDD! x@@idd v@x, x2, x3dd, 8i, 3<D êê Simplify 9 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ -x + x3 x2 Hx2 - x3l, ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H- + x2l 2 x3, x - x3 z = NSolveA9 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ -x + x3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x x2 - x x3 = x2 Hx2 - x3l ã, ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H- + x2l 2 x3 ã, x - x3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x x2 - x x3 ã =E 88x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<< Table@Im@x@@iDD! x@@idd v@x, x2, x3dd, 8i, 3<D ê. z µ 0-6, µ 0-4, <, µ 0-6, µ 0-4, <, µ 0-7,.264 µ 0-4,.3844 µ 0-5 <, µ 0-7, µ 0-4, µ 0-5 << 8v@x, x2, x3d - 2 p  Log@x3D ê. z@@dd, v@x, x2, x3d ê. z@@3dd< Â, Â<

9 Ê The knot 6 _ 2 li@x_d := PolyLog@2, xd v@x_, y_, z_d := lia ÅÅÅÅ z x E + lia ÅÅÅÅ z E + li@yd - lia ÅÅÅÅ E - li@zd - lia ÅÅÅÅ y x E - li@xd - lia y ÅÅÅÅ z E + ÅÅÅÅÅÅ p2 3 x = 8x, x2, x3<; dv = Table@Exp@x@@iDD! x@@idd v@x, x2, x3dd, 8i, 3<D êê Simplify 9-x + x3, x2 Hx2 - x3l ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H- + x2l 2 x3, x H- + x3l ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅ 2 Hx - x3l Hx2 - x3l = z = NSolveA9-x + x3 ã, x2 Hx2 - x3l ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H- + x2l 2 x3 ã, x H- + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x3l2 ÅÅÅÅÅÅÅÅÅÅÅ Hx - x3l Hx2 - x3l ã =E 88x2 Ø Â, x Ø Â, x3 Ø Â<, 8x2 Ø Â, x Ø Â, x3 Ø Â<, 8x2 Ø , x Ø , x3 Ø 0.579<, 8x2 Ø Â, x Ø Â, x3 Ø Â<, 8x2 Ø Â, x Ø Â, x3 Ø Â<< Table@Im@x@@iDD! x@@idd v@x, x2, x3dd, 8i, 3<D ê. z µ 0-6, µ 0-5, <, µ 0-6,.775 µ 0-5, <, 80, 0., 0.<, µ 0-7, µ 0-6, µ 0-7 <, µ 0-7, µ 0-6, µ 0-7 << 8v@x, x2, x3d - 2 p  Log@x3D ê. z@@dd, v@x, x2, x3d ê. z@@3dd, v@x, x2, x3d ê. z@@4dd< Â, Â, Â<

10 Ê The knot 6 _ 3 v@x_, y_, z_d := lia ÅÅÅÅ y x E + li@xd + li@zd + lia ÅÅÅÅ y E - lia ÅÅÅÅ z E - li@yd - lia ÅÅÅÅ x E - lia ÅÅÅÅ z y E x = 8x, x2, x3<; dv = Table@Exp@x@@iDD! x@@idd v@x, x2, x3dd, 8i, 3<D êê Simplify -x + x2 9 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H- + xl 2, - x H- + x2l ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅ 2 Hx - x2l Hx2 - x3l, x3 H-x2 + x3l ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x2 H- + x3l 2 = -x + x2 z = NSolveA9 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H- + xl ã, - x H- + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x2l2 ÅÅÅÅÅÅÅÅÅÅÅ 2 Hx - x2l Hx2 - x3l ã, x3 H-x2 + x3l ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x2 H- + x3l 2 ã =E 88x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<, 8x3 Ø Â, x Ø Â, x2 Ø Â<< Table@Im@x@@iDD! x@@idd v@x, x2, x3dd, 8i, 3<D ê. z µ 0-6, µ 0-6, µ 0-7 <, µ 0-6, µ 0-6, µ 0-7 <, µ 0-6, µ 0-6, µ 0-6 <, µ 0-6, µ 0-6,.7647 µ 0-6 <, µ 0-5, µ 0-5, µ 0-5 <, µ 0-5,.224 µ 0-5,.5537 µ 0-5 << 8v@x, x2, x3d ê. z@@2dd, v@x, x2, x3d ê. z@@4dd, v@x, x2, x3d ê. z@@5dd< Â, µ Â, Â<

11 8. An ideal triangulation of M We first prepare an ideal octahedron at each crossing of D. + Then, glue the red edges above the crossing, and glue the blue edges below the crossing, where ± denote the poles of S 3.

12 9 This octahedron decomposes into 4 tetrahedra as follows We shall consider the ratios of the edge variables represent the moduli around the vertical edge.

13 0 Gluing them along the edges of D, we obtain S 3 \ (K {± }). δ α γ β a b xν d In the picture above, the moduli of the tetrahedra α, β, γ, δ are represented by a/x ν, x ν /b, c/x ν, x ν /d respectively. c

14 If we collapse the leaf below, the tetrahedra corresponding to the corners are collapsed, and we obtain an ideal triangulation S of M. Note that the tetrahedra in S correspond to the dilogarithm functions in V (x,..., x n ). Suppose that x ν = z ν gives the hyperbolic structure of M.

15 2 The hyperbolicity equations for S can be read as follows. b a x d ( b/x)( d/x) ( a/x)( c/x) = c b a x d ( a/x)( x/c) ( b/x)( x/d) = c b a x d ( x/a)( x/c) ( x/b)( x/d) = Curiously, these equations coincide with the equations ( ) V exp x ν =. x ν Let x ν = z ν be the geometric solution to the equations above. c

16 Remark. The lifts of N(K) form holospheres in H 3. It is easy to draw the triangulations of them induced by S if we know the geometric solution z,..., z n to the hyperbolicity equations. 3

17 4 2. Zickert s formula Let E be an edge of S and Ẽ its lift. Then, the holospheres at Ẽ are interchanged by an element of PSL 2(C) conjugate to ( ) 0 /ξ(e). ξ(e) 0 We call ξ(e) C the edge parameter of E. Note that the ratios of the parameters of the red and blue edges coincides with z ν s. z5 z3 z z2 z4

18 5 Order the vertices of each tetrahedron as follows For a tetrahedron τ in S, define a(τ), b(τ) {0,, 2,..., n} by z a(τ) = ξ(τ 02 )/ξ(τ 03 ), z b(τ) = ξ(τ 2 )/ξ(τ 3 ), where τ ij is the edge of τ between the vertices i and j, and put u(τ) = ln ξ(τ 03 ) ln ξ(τ 3 ) + ln ξ(τ 2 ) ln ξ(τ 02 ), v(τ) = ln ξ(τ 02 ) ln ξ(τ 23 ) + ln ξ(τ 3 ) ln ξ(τ 0 ).

19 6 Then, there exist p τ, q τ Z such that u(τ) = ln z τ + p τ π, v(τ) = ln( z τ ) + q τ π, where we put z τ = z a(τ) /z b(τ). We now define L(τ) by ε(τ)l(τ) = Li 2 (z τ ) + 2 ln z τ ln( z τ ) π π { q τ ln z τ + p τ ln( z τ ) }, where ε(τ) takes or according as τ is right-handed or not. In our case, p τ is always even, and we can write ε(τ)l(τ) Li 2 (z τ ) π u(τ){ v(τ) + 2 ln( z τ ) } mod π 2. Zickert s Theorem. cv(m) τ L(τ) mod π2.

20 7 3. Proof Consider 4 tetrahedra around an edge of G. Let A, B, X, Y be the logarithm of the parameters of red, blue, green, orange edges respectively. δ α γ β a b zν d c

21 8 Let P, Q, R, S be the logarithms of the parameters of α 0 = β 0, α 3 = β 3, γ 02 = δ 02, γ 0 = δ 0 respectively. Then, L(α) + L(β) + L(γ) + L(δ) is equal to Li 2 (a/z ν ) Li 2 (b/z ν ) Li 2 (z ν /c) + Li 2 (z ν /d) { B Q + ξ(α2 ) A }{ A X + Q P + 2 ln( a/z ν ) } { B Q + ξ(β2 ) A }{ A Y + Q P + 2 ln( b/z ν ) } { ξ(γ03 ) B + A R }{ R Y + B S + 2 ln( z ν /c) } + 2{ ξ(δ03 ) B + A R }{ R X + B S + 2 ln( z ν /d) }, and the coefficient of A B, which is congruent to ln z ν, becomes ln( a/z ν ) + ln( b/z ν ) ln( z ν /c) + ln( z ν /d) which should be equal to 2π r ν.

22 We put w + (τ) = 2{ ln ξ(τ03 ) ln ξ(τ 02 ) }{ v(τ) + 2 ln( z τ ) }, w (τ) = 2{ ln ξ(τ2 ) ln ξ(τ 3 ) }{ v(τ) + 2 ln( z τ ) }, so that ε(τ)l(τ) Li 2 (z τ ) π 2 /6 + w + (τ) + w (τ) mod π 2. Then, the above observation implies that w(ν) = ε(τ)w + (τ) + ε(τ)w (τ) a(τ)=ν b(τ)=ν equals 2π r ν ln z ν modulo 4π 2, and τ L(τ) is equal to n V (z,..., z n ) + w(ν) = V (z,..., z n ) 2π n r ν ln z ν modulo π 2. ν= ν= 9