Polylogarithms and Hyperbolic volumes Matilde N. Laĺın University of British Columbia and PIMS, Max-Planck-Institut für Mathematik, University of Alberta mlalin@math.ubc.ca http://www.math.ubc.ca/~mlalin July 2nd, 2007 1
The hyperbolic space Beltrami s Half-space model: H n = {(x 1,..., x n 1, x n ) x i R, x n > 0}, ds 2 = dx2 1 +... + dx2 n x 2, n dv = dx 1... dx n x n, H n = {(x 1,..., x n 1, 0)}. Geodesics are given by vertical lines and semicircles whose endpoints lie in {x n = 0} and intersect it orthogonally. 2
Orientation preserving isometries of H 2 P SL(2, R) = {( a b c d ) M(2, R) ad bc = 1 } /±I. z = x 1 + x 2 i az + b cz + d. Orientation preserving isometries of H 3 is P SL(2, C). H 3 = {z = x 1 + x 2 i + x 3 j x 3 > 0}, subspace of quaternions (i 2 = j 2 = k 2 = 1, ij = ji = k). z (az+b)(cz+d) 1 = (az+b)( z c+ d) cz+d 2. Poincaré: study of discrete groups of hyperbolic isometries. 3
Volumes in H 3 Picard: fundamental domain for P SL(2, Z[i]) in H 3 has a finite volume. Humbert extended this result. Lobachevsky function: l(θ) = θ 0 log 2 sin t dt. l(θ) = 1 2 Im ( Li 2 ( e 2iθ )), where Li 2 (z) = n=1 z n n2, z 1. z Li 2 (z) = log(1 x)dx 0 x. (multivalued) analytic continuation to C\[1, ) 4
Let be an ideal tetrahedron (vertices in H 3 ). Theorem 1 (Milnor, after Lobachevsky) The volume of an ideal tetrahedron with dihedral angles α, β, and γ is given by Vol( ) = l(α) + l(β) + l(γ). α β γ γ β γ β α α Move a vertex to and use baricentric subdivision to get six simplices with three right dihedral angles. 5
Triangle with angles α, β, γ, defined up to similarity. Let (z) be the tetrahedron determined up to transformations by any of z, 1 1 z, 1 1 z. z 1 1 z z 1 1 z 0 1 If ideal vertices are z 1, z 2, z 3, z 4, z = [z 1 : z 2 : z 3 : z 4 ] = (z 3 z 2 )(z 4 z 1 ) (z 3 z 1 )(z 4 z 2 ). 6
Bloch-Wigner dilogarithm D(z) = Im(Li 2 (z) + log z log(1 z)). Continuous in P 1 (C), real-analytic in P 1 (C) \ {0, 1, }. D(z) = D(1 z) = D ( ) 1 z = D( z). Five-term relation: D(x)+D(1 xy)+d(y)+d ( 1 y 1 xy ) +D ( 1 x 1 xy ) =0. ( ( ) z ( 1 z 1 ) ( (1 z) 1 )) D(z) = 1 2 D z + D 1 z 1 + D (1 z) 1. Vol( (z)) = D(z). 7
Five points in H 3 = P 1 (C), then the sum of the signed volumes of the five possible tetrahedra must be zero: 5 i=0 ( 1) i Vol([z 1 :... : ẑ i :... : z 5 ]) = 0. D(x)+D(1 xy)+d(y)+d ( 1 y 1 xy ) +D ( 1 x 1 xy ) =0. 8
Dedekind ζ-function F number field [F : Q] = n = r 1 + 2r 2 τ 1,..., τ r1 real embeddings σ 1,..., σ r2 a set of complex embeddings (one for each pair of conjugate embeddings). Dedekind ζ-function ζ F (s) = A ideal 0 N(A) = O F /A norm. 1 N(A) s, Re s > 1, Euler product P prime 1 1 N(P) s. 9
Theorem 2 (Dirichlet s class number formula) ζ F (s) extends meromorphically to C with only one simple pole at s = 1 with where lim (s 1)ζ F (s) = 2r 1(2π) r 2h F reg F. s 1 ω F D F h F is the class number. ω F is the number of roots of unity in F. D F is the discriminant. reg F is the regulator. 10
Regulator {u 1,..., u r1 +r 2 1} basis for O F modulo torsion L(u i ) := (log τ 1 u i,..., log τ r1 u i, 2 log σ 1 u i,..., 2 log σ r2 1u i ) reg F is the determinant of the matrix. = (up to a sign) the volume of fundamental domain for L(O F ). Functional equation ξ F (s) = ξ F (1 s), ξ F (s) = ( ) s DF 2 ( ) s r1 Γ Γ(s) r 2 4 r 2π n ζ F (s). 2 lim s 0 s1 r 1 r 2 ζ F (s) = h F reg F. ω F 11
Euler: ζ(2m) = ( 1)m 1 (2π) 2m B m 2(2m)! Klingen, Siegel: F is totally real (r 2 = 0), ζ F (2m) = r(m) D F π 2mn for m > 0. where r(m) Q. 12
Building manifolds Bianchi, Humbert : F = Q ( d ) d 1 square-free Γ a torsion-free subgroup of P SL (2, O d ), [P SL (2, O d ) : Γ] <. Then H 3 /Γ is an oriented hyperbolic threemanifold. Example: d = 3, O 3 = Z[ω], ω = 1+ 3 2. Riley: [P SL (2, O 3 ) : Γ] = 12. H 3 /Γ diffeomorphic to S 3 \ Fig 8. 13
Theorem 3 (Essentially Humbert) Vol ( H 3 /P SL(2, O d ) ) = D d Dd 4π 2 ζ Q( d) (2). D d = { d d 3 mod 4, 4d otherwise. M hyperbolic 3-manifold Vol(M) = J j=1 D(z j ). ζ Q ( d) = D d Dd 2π 2 J j=1 D(z j ). 14
Example: Vol(S 3 \ Fig 8) = 12 3 3 4π 2 ζ Q( 3) (2) = 3D ( e 2iπ 3 ) = 2D ( e iπ 3 ). Zagier: [F : Q] = r 1 + 2, Γ group of units of an order in a quaternion algebra B over F that is ramified at all real places such that M = H 3 /Γ has volume D F Q π 2(n 1)ζ F (2). 15
[F : Q] = r 1 + 2r 2, r 2 > 1, Zagier: discrete subgroup Γ of P SL(2, C) r 2 such that M = ( H 3) r 2 /Γ has volume Q D F π 2(r 1+r 2 ) ζ F (2). M = (z 1 )... (z r2 ) 16
The Bloch group Vol(M) = J j=1 D(z j ), then J j=1 z j (1 z j ) = 0 2 C. 2 C ={x y x x = 0, x1 x 2 y = x 1 y + x 2 y} Vol(M) = D(ξ M ), where ξ M A( Q), and A(F ) = { ni [z i ] Z[F ] ni z i (1 z i ) = 0 }. 17
Let { [ 1 y ] [ 1 x ] C(F ) = [x] + [1 xy] + [y] + 1 xy + 1 xy x, y F, xy 1}, Bloch group is B(F ) = A(F )/C(F ). D : B(C) R well-defined function, Vol(M) = D(ξ M ) for some ξ M B( Q), independently of the triangulation. Then ζ F (2) = D F π 2(n 1) D(ξ M ) for r 2 = 1. 18
The K-theory connection R ring. K 0 (R), K 1 (R) = R, etc Borel: K n (F ) Q free abelian and dim Q (K n (F ) Q) = 0 n 2 even, r 1 + r 2 n 1 mod 4, r 2 n 3 mod 4. Let n + = r 1 + r 2 and n = r 2. 19
The regulator map reg m : K 2m 1 (C) R, is such that K 2m 1 (F ) K 2m 1 (R) r 1 K 2m 1 (C) r 2 R n. K 2m 1 (F )/torsion goes to a cocompact lattice of R n. Covolume is a rational multiple of D F ζ F (m)/π kn ±. m = 2, Bloch and Suslin: B(F ) is essentially K 3 (F ), and D is the regulator. K 3 (F ) reg 2 R r 2 φ F B(F ) (D σ 1,...,D σ r2 ) 20
Theorem 4 For a number field [F : Q] = r 1 + 2r 2, B(F ) is finitely generated of rank r 2. ξ 1,... ξ r2 Q-basis of B(F ) Q. Then ζ F (2) Q D F π 2(r 1+r 2 ) det { D ( σ i ( ξj ))} 1 i,j r 2. 21
Zagier s conjecture Generalize to values ζ F (m). k-polylogarithm Li k (z) = n=1 z n nk, z 1. It has an analytic continuation to C \ [1, ). L k (z) = Re k k 1 j=0 2 j B j j! log j z Li k j (z), Re k = Re or Im depending on whether k is odd or even, and B j is the jth Bernoulli number. It is continuous in P 1 (C), real analytic in P 1 (C)\ {0, 1, }. L k (z) = ( 1) k 1 L k ( 1 z ). 22
Generalized Bloch groups B k (F ) = A k (F )/C k (F ). C k (F ) set of functional equations of the kth polylogarithm and A k (F ) is the set of allowed elements. A k (F ):= { ξ Z[F ] ι φ (ξ) C k 1 (F ) φ Hom(F, Z) } where ι φ ( n i [x i ]) = n i φ(x i )[x i ]. C k (F ) := {ξ A k (F ) L k (σ(ξ)) = 0 σ Σ F }. 23
Conjecture 5 Let F be a number field. Let n + = r 1 + r 2, n = r 2, and = ( 1) k 1. Then B k (F ) is finitely generated of rank n. ξ 1,... ξ n Q-basis of B k (F ) Q. Then ζ F (k) Q D F π kn ± det { ( ( ))} L k σi ξj 1 i,j n. 24
Example F = Q( 5), r 1 = 2, r 2 = 0. { [ [1], 1+ ]} 5 2 A 3 (F ), basis for B 3 (F ). L 3 ( L 3 (1) L 3 (1) 1+ 5 2 ) L 3 ( 1 5 2 ) = ζ(3) 1 10ζ(3) + 25 48 5L(3, χ5 ) ζ(3) 1 10ζ(3) 25 48 5L(3, χ5 ) = 25 5ζ(3)L(3, χ5 ) = 25 5ζF (3). 24 24 25
More K-theory Expectation: K 2m 1 (F ) and B m (F ) should be isomorphic. The regulator map should be given by L m. de Jeu, Beilinson Deligne: there is a map K 2m 1 (F ) B m (F ). Goncharov: map is surjective for m = 3. 26
Goncharov G F (k) : G k (F ) G k 1 F Gk 2 2 F...... G 2 k 2 F k F G k (F ) = Z[F ]/C k (F ) [x] x 1... x l = [x] x x 1... x l. H 1 (G F (k)) B k (F ). Goncharov conjectures H i (G F (k) Q) gr γ k K 2k i(f ) Q. 27
Volumes in higher dimensions Orthoscheme in H n : simplex bounded by hyperplanes H 0,..., H n such that H i H j i j > 1 Theorem 6 (Schläfli s formula) R H n is an orthoscheme, dvol n (R) = 1 n 1 n j=1 Vol n 2 (F j )dα j. where F j = R H j 1 H j, α j is the angle attached at F j, and Vol 0 (F j ) = 1. Vol 2m-simplex Vol dimension 2m 1 and lower. dimension 5: sum of trilogarithmic expressions (Böhn, Müller, Kellerhals, and Goncharov). 28
Goncharov: M (2m 1)-dimensional hyperbolic manifold of finite volume. Theorem 7 There is a γ M K 2m 1 ( Q) Q such that Vol(M) = reg m (γ M ). Conjecture 8 There is a ξ M such that B m ( Q) Q Vol(M) = L m (ξ M ). Goncharov: true for dimension 5. 29