Hyperbolic volumes and zeta values An introduction Matilde N. Laĺın University of Alberta mlalin@math.ulberta.ca http://www.math.ualberta.ca/~mlalin Annual North/South Dialogue in Mathematics University of Calgary, Alberta May 2nd, 2008
The hyperbolic space Hyperbolic Geometry: Lobachevsky, Bolyai, Gauss ( 1830) Beltrami s Half-space model (1868) H n = {(x 1,..., x n 1, x n ) x i R, x n > 0}, ds 2 = dx 2 1 + + dx 2 n x 2 n dv = dx 1... dx n xn, H n = {(x 1,..., x n 1, 0)}.,
The hyperbolic space Hyperbolic Geometry: Lobachevsky, Bolyai, Gauss ( 1830) Beltrami s Half-space model (1868) H n = {(x 1,..., x n 1, x n ) x i R, x n > 0}, ds 2 = dx 2 1 + + dx 2 n x 2 n dv = dx 1... dx n xn, 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. Poincaré (1882): Orientation preserving isometries of H 2 {( a b PSL(2, R) = c d ) } M(2, R) ad bc = 1 / ± I. z = x 1 + x 2 i az + b cz + d.
Geodesics are given by vertical lines and semicircles whose endpoints lie in {x n = 0} and intersect it orthogonally. Poincaré (1882): Orientation preserving isometries of H 2 {( a b PSL(2, R) = 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 PSL(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. Picard (1884): fundamental domain for PSL(2, Z[i]) in H 3 has a finite volume. Humbert (1919) extended this result.
Volumes in H 3 Lobachevsky function: where θ l(θ) = log 2 sin t dt. 0 l(θ) = 1 2 Im ( Li 2 (e 2iθ)), Li 2 (z) = Li 2 (z) = n=1 z z n, z 1. n2 0 log(1 x) dx x. (multivalued) analytic continuation to C \ [1, )
Let be an ideal tetrahedron (vertices in H 3 ). Theorem (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 β γ γ β α α Move a vertex to and use baricentric subdivision to get six simplices with three right dihedral angles.
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. 1 1 z z If ideal vertices are z 1, z 2, z 3, z 4, z 1 1 z 0 1 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 ).
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, }. ( ) 1 D(z) = D(1 z) = D = D( z). z Vol( (z)) = D(z).
Five points in H 3 = P 1 (C), then the sum of the signed volumes of the five possible tetrahedra must be zero: 5 ( 1) i Vol([z 1 : : ẑ i : : z 5 ]) = 0. i=0 Five-term relation D(x) + D(1 xy) + D(y) + D ( ) ( ) 1 y 1 x + D =0. 1 xy 1 xy
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). ζ F (s) = N(A) = O F /A norm. Euler product A ideal 0 P prime 1, Re s > 1, N(A) s 1 1 N(P) s.
Theorem (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) = 2r1 (2π) r2 h F reg F, s 1 ω F DF h F is the class number. ω F is the number of roots of unity in F. reg F is the regulator. lim s 0 s1 r 1 r 2 ζ F (s) = h F reg F ω F.
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 ).
Euler: ζ(2m) = ( 1)m 1 (2π) 2m B m 2(2m)! Klingen, Siegel: F is totally real (r 2 = 0), where r(m) Q. ζ F (2m) = r(m) D F π 2mn, m > 0
Building manifolds Bianchi: F = Q ( d ) d 1 square-free Γ a torsion-free subgroup of PSL (2, O d ), [PSL (2, O d ) : Γ] <. Then H 3 /Γ is an oriented hyperbolic three-manifold. Example: d = 3, O 3 = Z[ω], ω = 1 + 3 2 Riley: [PSL (2, O 3 ) : Γ] = 12 H 3 /Γ diffeomorphic to S 3 \ Fig 8.
Theorem (Essentially Humbert) Vol ( H 3 /PSL(2, O d ) ) = D d Dd 4π 2 ζ Q( d) (2). M hyperbolic 3-manifold { d d 3 mod 4, D d = 4d otherwise. Vol(M) = J D(z j ). j=1 ζ Q( d) (2) = D d Dd 2π 2 J D(z j ). j=1
Example: Vol(S 3 \ Fig 8) = 12 3 3 4π 2 ζ Q( 3) (2) ) ( ) = 3D (e 2iπ 3 = 2D e iπ 3.
Zagier (1986): [F : Q] = r 1 + 2 Γ torsion free subgroup of finite index of the group of units of an order in a quaternion algebra B over F that is ramified at all real places. Vol(H 3 DF /Γ) Q π 2(n 1) ζ F (2). [F : Q] = r 1 + 2r 2, r 2 > 1 Γ discrete subgroup of PSL(2, C) r 2 ( (H 3 Vol ) ) r 2 /Γ Q such that DF π 2(r 1+r 2 ) ζ F (2). ( H 3 ) r 2 /Γ = (z 1 ) (z r2 )
The Bloch group then J Vol(M) = D(z j ), j=1 J z j (1 z j ) = 0 j=1 2 C. 2 C ={x y x x = 0, x 1 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.
Let C(F ) = { [ ] [ ] 1 y 1 x [x] + [1 xy] + [y] + + 1 xy 1 xy Bloch group is x, y F, xy 1}, 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.
Theorem (Zagier, Bloch, Suslin) For a number field [F : Q] = r 1 + 2r 2, B(F ) is finitely generated of rank r 2. ξ 1,... ξ r2 Proof: Q-basis of B(F ) Q. Then ζ F (2) Q DF π 2(r 1+r 2 ) det {D (σ i (ξ j ))} 1 i,j r2. B(F ) is K 3 (F ) Borel s theorem.
Theorem (Zagier, Bloch, Suslin) For a number field [F : Q] = r 1 + 2r 2, B(F ) is finitely generated of rank r 2. ξ 1,... ξ r2 Proof: Q-basis of B(F ) Q. Then ζ F (2) Q DF π 2(r 1+r 2 ) det {D (σ i (ξ j ))} 1 i,j r2. B(F ) is K 3 (F ) Borel s theorem.
Conjecture 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 DF π kn ± det {L k (σ i (ξ j ))} 1 i,j n.
Example F = Q( { [ 5), r 1 = 2, r 2 = 0. ]} [1], 1+ 5 2 basis for B 3 (F ). = L 3 (1) L 3 ( 1+ 5 2 L 3 (1) L 3 ( 1 5 2 ) ) 1 25 ζ(3) 10ζ(3) + 48 5L(3, χ5 ) 1 25 ζ(3) 10ζ(3) 48 5L(3, χ5 ) = 25 5ζ(3)L(3, χ5 ) = 25 5ζF (3). 24 24
Application D Andrea, L. (2007) 1 (2πi) 3 log (1 x)(1 y) z T 3 1 xy dx x dy y dz z = 25 5L(3, χ5 ) π 2