On the volume conjecture for quantum 6j symbols

Transcription

1 On the volume conjecture for quantum 6j symbols Jun Murakami Waseda University July 27, 2016 Workshop on Teichmüller and Grothendieck-Teichmüller theories Chern Institute of Mathematics, Nankai University

2 Contents Quantum Invariants Volume Conjecture Quantum 6j symbol 2/22

3 Overview Quantum Invariants Volume Conjecture Quantum 6j symbol 3/22

4 Jones Polynomial Jones polynomial (1984) ` U q (sl 2 ) t`1 V K`(t) ` t V K+ (t) = t 1=2 ` t`1=2 V K0 K + : K` : K 0 : jk 1 j : Cf. Alexander polynomial r K+ (z)`r K`(z) = z r K0 (z) HOMFLY-PT polynomial (1987) ` U q (sl n ) a`1 P K`(t) ` a P K+ (t) = t 1=2 ` t`1=2 P K0 Kauffman polynomial (1987) ` U q (so n ); U q (sp 2n ) a`1 F jk`j(t) ` a F jk+j(t) = t 1=2 ` t`1=2 F jk0j ` F jk1j 4/22

5 Overview Quantum Invariants Volume Conjecture Quantum 6j symbol 5/22

6 Kashaev s conjecture Let L be a knot and K N (L) be the quantum invariant introduced by Kashaev. Then 4 1 : 5 2 : 6 1 : 2 ı lim N!1 N log jk N(L)j = Vol(S 3 n L): N`1 X K N (4 1 ) = j(q) k j 2, ky (q) k = (1 ` q j ) k=0 j=1 `! N! K N (5 2 ) = X k»l K N (6 1 ) = X k+l»m q`k(l+1)=2 (q) 2 l (q`1 ) k `! N!1 q (m`k`l)(m`k+1)=2 j(q) mj 2 (q) k (q`1 ) l `! N!1 K N (L) is equal to the colored Jones inv. V N L (q) for N-dim. rep. of U q (sl 2 ) at q = exp 2ı p`1=n. (H. Murakami-J.M.) 6/22

7 Generalizations of Kashaev s Conjecture s N = exp(ı p`1=n), q N = s 2 N. Volume Conjecture. (H. Murakami-J.M.) L : a knot 2 ı lim N!1 N log V N L (q N) = VolGr (S 3 n L); where Vol Gr is Gromov s simplicial volume. Complesified Volume Conjecture (H.Murakami-J.M.-M.Okamoto-T.Takata-Y.Yokota) L : hyperbolic knot 2 ı q lim N!1 N log V N L (q N) = Vol(S 3 n L) + `1 CS(S 3 n L): 7/22

8 Proof for the figure-eight knot (Ekholm) K : figure-eight knot, s N = exp(ı p`1=n), q N = s 2 N. N`1 X jy V N K (s N) = (s N`k N j=0 k=1 N`1 X ` s`n+k N )(s N+k N ` s`n`k N ) = jy j=0 k=1 4 sin 2 ık N : Let a j = Q j ık k=1 4 sin2 N and a j max be the max. term of a j. Since a jmax» V N K (s N)» N a jmax, 2ı log a lim jmax N!1 N 2ı log V» lim N K (sn) N!1 N 2ı log(n a» lim jmax ) N!1 N 2ı log(a = lim jmax ) : N!1 N a j is decreasing for small j s, increasing for middle j s and then decreasing for large j s. It takes the maximal at j 5N 6. Since a 0 = 1, a N`1 = N 2, a j is maximum at j 5N 6. Therefore 2ı log(a jmax ) lim N N!1 4 Z 5ı 6 0 4ı P 5N=6 k=1 log(2 sin ık=n) = lim N!1 N! log(2 sinx) dx = `4 5 ı 6 = = 2: ::: 8/22

9 Volume potential function Dilogarithm function Analytic continuation of Z x log(1`u) Li 2 (x) = ` du = x+ x 2 3 u 2 +x + (0 < x < 1): It is a multi-valued function and its branches are q li 2 (z) = Li 2 (z) + 2 k ı `1 log z + 4 l ı 2 (k; l 2 Z) Volume potential function U(x 1 ; x 2 ; : : : ) The function obtained by replacing (q) k in U q (sl 2 )-invariant by Li 2 (x) (x = q k N ). Saddle points of the volume potential function i = 0 (i = 1; 2; : : : ). These equations correspond to the glueing equation of the tetrahedral decomposition (Yokota). 9/22

10 Hyperbolic volume from the saddle point K : hyperbolic knot, U : Volume potential function of V N K (q N) (colored Jones) Hyperbolic volume! For some x (0) 1, x 2, : : : satisfying exp = i U(x (0) 1 ; x (0) 2 ; : : : ) ` log x i i x1=x (0) 1 ; i is the hyperbolic volume of the complement of K (Yokota, J. Cho). Optimistic calculation (H.Murakami, arxiv:math/ ) M : 3-manifold fi N (M) : Witten-Reshetikhin-Turaev invariant Check that fi N (M) `! hyperoblic volume of M for M obtained by surgery along the figure-eight knot. Optimistic conjecture U q (sl 2 ) inariant `! hyperbolic volume 10/22

11 q N `! q 2 N Attention! The U q (sl 2 )-invariant grows exponentially only for Kashaev s invariant and its deformation. For other U q (sl 2 )-invariants, they does not grow exponentially. Renovation by Q. Chen-T. Yang arxiv: Replace q N = exp(2ı p`1=n) by q 2 N. Various U q (sl 2 )-invariants grow exponentially and the growth rates are given by the hyperbolic volume of the corresponding geometric objects. Including: WRT invariant, Turaev-Viro inv. for 3-mfds, Kirillov-Reshetikhin invariant for knotted graphs 11/22

12 Overview Quantum Invariants Volume Conjecture Quantum 6j symbol 12/22

13 Quantum 6j symbol The quantum 6j symbol is introduced to express V m,! X ȷ ff V l V k,! A i j l m,! V m k n i V n,! A V i V j V k q n V i V j V k for representations of U q (sl 2 ) by Kirillov-Reshetikhin. fkg = fq 1=2 ` q`1=2 g; fkg! = fkg fk ` 1g : : : f1g; fc + 1g W 1 (e) = f1g ; W f c1+c2+c3 + 1g! 2 2(f ) = f1g f `c1+c2+c3 g!f c1`c2+c3 g!f c1+c2`c3 g! ; ȷ ff «c1 c 2 c 5 c1 c ȷ ff 2 c 5 RW c1 c 2 c 5 c 4 c 3 c 6 q c 4 c 3 c 6 = p = p c 4 c 3 c ; 6 q W1(c5) W 1(c 6) W2(1;2;5) W 2(1;3;6) W 2(2;4;6) W 2(3;4;5) «min(ax 1;:::;A 3) c1 c 2 c 5 (`1) k fk + 1g! = c 4 c 3 c 6 f1g fa 1 ` kg! : : : fa 3 ` kg! fk ` B 1g! : : : fk ` B ; 4g! k=max(b 1; ;B 4) where 2 A 1 = c 1 + c 2 + c 3 + c 4 ; 2 B 1 = c 1 + c 2 + c 5 ; 2 A 2 = c 1 + c 4 + c 5 + c 6; 2 B 2 = c 1 + c 3 + c 6; 2 A 3 = c 2 + c 3 + c 5 + c 6 ; 2 B 3 = c 2 + c 4 + c 6 ; 2 B 4 = c 3 + c 4 + c 5: Here we assume c 1,, c 6 are admissible, i.e. A i, B j, A i ` B j 2 Z 0. 13/22

14 Turaev-Viro invariant State sum on a tetrahedral decomposition Let s N = exp( ıp`1 `k sk N`sN ), [k] =, [k]! = [k][k ` 1] [1], N s N`s `1 N I N = f0; 1; : : : ; N ` 2g. Turaev-Viro invariant Let M be a closed 3-manifold and let be a tetrahedral decomposition of M. Let T, F, E be the set of tetrahedrons, faces, edges of. Then TV N (M) = X : E! I N admissible Y W 1 (e) Y W 2 (f )`1 Y W 3 (t): e2e f 2F t2t TV! potential fun.! saddle pt.! hyp. volume Quantum 6j symbol quantum 6j! potential symbol function! saddle hyperbolic volume! point of tetrahedron (J. M.-M. Yano) 14/22

15 Chen-Yang s invariant and q `! q 2 Let s N = exp(ı p`1=n) and q N = s 2 N. Chen-Yang s invariant Extend TV-invariant for 3-manifolds with boundary by using ideal tetrahedrons and truncated tetrahedrons. Conjecture (Q.Chen-T.Yang, arxiv: ) Let M be a 3-manifold N be a positive odd integer, CY N (M) be Chen-Yang s invariant, fi N (M) be the WRT invariant, and TV N (M) be the Turaev-Viro invariant of M. Then we have 2 ı 1. lim CY N!1 N log N (M)j sn!s 2 = Vol(M) + N «4 ı 2. lim fi N!1 N log N (M)j sn!s 2 = Vol(M) + N «q `1 CS(M) q `1 CS(M) 2 ı 3. lim TV N!1 N log N (M)j sn!s 2 = Vol(M); N since 2. and the fact that TV N (M) = jfi N (M)j 2. «15/22

16 Volume conjecture for the quantum 6j symbol Conjecture T : hyperbolic tetrahedron with dihedral angles 1, : : :, 6, N : positive odd integer, a (N) 2 ı i : admissible sequences with lim N!1 N a(n) i = ı ` i, (1» i» 6) Then 8 2 ı lim N!1 N log < a (N) 1 a (N) 2 a (N) 9 RW = 5 : a (N) = Vol(T ): ; 4 a (N) 3 a (N) 6 Theorem The above conjecture is true if all the vertices are truncated, i.e. the sum of the three dihedral angles sharing a vertex is less than ı. q 2 N 16/22

17 Proof Use the same idea for the proof for figure-eight knot. If the terms of the sum are all positive, then only the maximum term contributes to the limit. Recall ȷ ff RW c1 c 2 c 5 c 4 c 3 c 6 q = «c1 c 2 c 5 c 4 c 3 c 6 p ; W2(c1;c 2;c 5) W 2(c 1;c 3;c 6) W 2(c 2;c 4;c 6) W 2(c 3;c 4;c 5) f c1+c2+c3 + 1g! 2 W 2(f ) = f1g f `c1+c2+c3 g!f c1`c2+c3 g!f c1+c2`c3 g! ; «min(ax 1;:::;A 3) c1 c 2 c 5 (`1) k fk + 1g! = c 4 c 3 c 6 f1g fa 1 ` kg! : : : fa 3 ` kg! fk ` B 1 g! : : : fk ` B 4 g! ; k=max(b 1; ;B 4) where 2 A 1 = c 1 + c 2 + c 3 + c 4; 2 B 1 = c 1 + c 2 + c 5; 2 A 2 = c 1 + c 4 + c 5 + c 6 ; 2 B 2 = c 1 + c 3 + c 6 ; 2 A 3 = c 2 + c 3 + c 5 + c 6 ; 2 B 3 = c 2 + c 4 + c 6 ; 2 B 4 = c 3 + c 4 + c 5: Now apply the above idea to a(n) 1 a (N) 2 a (N) 1 5 a (N) 4 a (N) 3 a (N) 6 A. 17/22

18 Proof (sign) where k = Recall that a (N) 1 a (N) 2 a (N)! 5 a (N) 4 a (N) 3 a (N) 6 = min(a 1;:::;A 3) X k=max(b 1; ;B 4) (`1) k fk + 1g! f1g fa 1 ` kg! : : : fa 3 ` kg! fk ` B 1g! : : : fk ` B 4g! ; 2 A 1 = a (N) 1 + a (N) 2 + a (N) 3 + a (N) 4 ; 2 B 1 = a (N) 1 + a (N) 2 + a (N) 5 ; 2 A 2 = a (N) 1 + a (N) 4 + a (N) 5 + a (N) 6 ; 2 B 2 = a (N) 1 + a (N) 3 + a (N) 6 ; 2 A 3 = a (N) 2 + a (N) 3 + a (N) 5 + a (N) 6 ; 2 B 3 = a (N) 2 + a (N) 4 + a (N) 6 ; s N = exp ı p`1 ; fjg = s 2 j `s`2 j N N N = 2 p`1 sin 2 j ı N ; k 2 B 4 = a (N) 3 + a (N) 4 + a (N) 5 : ( Imfjg > 0 if 0 < j < N 2, Imfjg < 0 if N 2 < j < N: 2 ı Since lim N!1 N a(n) = ı ` i i and » ı,, we have B i > N=2, A i < N and the terms of the denominator of k are all positive. Hence the range of k is contained in N=2 < k < N and sign fk+1g! p`1 k+1 = (`1)(N+1)=2 : 18/22

19 Proof ( kmax ) a (N) 1 a (N) 2 a (N)! 5 a (N) 4 a (N) 3 a (N) 6 = min(a 1;:::;A 3) X k=max(b 1; ;B 4) where (`1) k fk + 1g! k = f1g fa 1 ` kg! : : : fa 3 ` kg! fk ` B 1 g! : : : fk ` B 4 g! ; fjg = s 2 j N ` s`2 j N = 2 p`1 ( sin 2 j ı N ; Im fjg > 0 if 0 < j < N 2, Im fjg < 0 if N 2 < j < N: The terms of the denominator of k are all positive. Hence the range of k is contained in N=2 < k < N and sign fk+1g! p`1 k+1 = (`1) (N+1)=2 ; sign k = (`1) (N+1)=2 : Therefore 2 ı lim N!1 N X j kmax j» k» N j k max j ; k 0 a(n) 1 a (N) 2 a (N) 1 5 a (N) 4 a (N) 3 a (N) A = lim 2 ı N!1 N log j k max j : 6 k 19/22

20 Proof (k max ) a (N) 1 a (N) 2 a (N)! 5 a (N) 4 a (N) 3 a (N) 6 = min(a 1;:::;A 3) X k=max(b 1; ;B 4) where (`1) k fk + 1g! k = f1g fa 1 ` kg! : : : fa 3 ` kg! fk ` B 1 g! : : : fk ` B 4 g! ; fjg = s 2 j N ` s`2 j N = 2 p`1 ( sin 2 j ı N ; Im fjg > 0 if 0 < j < N 2, Im fjg < 0 if N 2 < j < N: The ratio k+1 k = ` fk + 2] fa 1 ` kg : : : fa 3 ` kg fk + 1 ` B 1 g : : : fk + 1 ` B 4 g is positive and bigger than 1 if k is small and less than 1 if k is big, and is monotonically decreasing between max(b 1 ; B 2 ; B 3 ; B 4 ) and min(a 1 ; A 2 ; A 3 ). Hence k has a maximum value at k max which satisfies kmax+1 k max 1 and so `fk max + 2g fa 1 ` k max g : : : fa 3 ` k max g fk max + 1 ` B 1 g : : : fk max + 1 ` B 4 g 1: 20/22 k

21 Proof (hyperbolic volume) Now consider about lim N!1 where ȷ ff RW c1 c 2 c 5 c 4 c 3 c 6 q 2 N 2 ı N = ( log a (N) 1 a (N) 2 a (N) ) RW 5 a (N) 4 a (N) 3 a (N) 6 q 2 «N c1 c 2 c 5 c 4 c 3 c 6 p ; W2(c1;c 2;c 5) W 2(c 1;c 3;c 6) W 2(c 2;c 4;c 6) W 2(c 3;c 4;c 5) f c1+c2+c3 + 1g! 2 W 2 (f ) = f1g f `c1+c2+c3 g!f c1`c2+c3 g!f c1+c2`c3 g! : By using sectional mensuration, we have kx 2 ı 2 ı lim log jfkg!j = lim log 2 sin 2 ı j N!1 N N!1 N N = j=1 Z x j2 sin tj dt = (x) = Im Li 2(e 2 p`1 x ); 0 where x = 2 ı k N. So, replace fkg! by Li 2, then we get the volume formula for a truncated tetrahedron given by A. Ushijima, A volume formula for generalised hyperbolic tetrahedra. Non-Euclidean geometries, , Math. Appl. (N. Y.), 581, Springer, New York, /22

22 A generalizations Compact tetrahedron There are some oscillating terms in k, but they seems to be small and negligible. 22/22

23 A generalizations Compact tetrahedron There are some oscillating terms in k, but they seems to be small and negligible ık N ; + ı n o RW 2ı k : : : n o RW : 101 log 2ı k : : : n o RW k : : : s=exp 101 2ıi : 301 log 2ı k : : : k : : : s=exp 301 2ıi : 1001 log k : : : s=exp ıi : Vol(T jı` j ) 22/22