Linear Mahler Measures and Double L-values of Modular Forms

Transcription

1 Linear Mahler Measures and Double L-values of Modular Forms Masha Vlasenko (Trinity College Dublin), Evgeny Shinder (MPIM Bonn) Cologne March 1, 2012

2 The Mahler measure of a Laurent polynomial is defined as m(p) = 1 (2πi) n P(x 1,..., x n ) C[x ±1 1,..., x ±1 n ] x 1 = = x n =1 log P(x 1,..., x n ) dx 1 x 1... dx n x n. This number is obviously a period in the sense of Kontsevich and Zagier. For a monic polynomial in one variable we can compute m(p) by Jensen s formula: 1 2πi x =1 log P(x) dx x = α:p(α)=0 max(0, log α )

3 C. Smyth, 1980: where L(χ 3, s) = m(1 + x 1 + x 2 ) = 3 3 4π L(χ 3, 2) n=1 χ 3 (n) n s = s s 1 5 s +... m(1 + x 1 + x 2 + x 3 ) = 7 2π 2 ζ(3) m(1 + x 1 + x 2 + x 3 + x 4 ) =?

4 Conjecture (F. Rodriguez Villegas): where m(1 + x 1 + x 2 + x 3 + x 4 )? = 6 ( 15 ) 5L(f15, 4) 2πi f 15 = η(3z) 3 η(5z) 3 + η(z) 3 η(15z) 3 is a CM modular form of weight 3 and level 15. This form corresponds to a Galois representation arising from the variety { 1 + x 1 + x 2 + x 3 + x 4 = x x x (1+x 1 +x 2 +x 3 ) ( ) = 1 x 4 = 0 x 1 x 2 x 3 which can be compactified to a K3 surface of Picard rank 20.

5 Mahler s Measures and Differential Equations For a Laurent polynomial P the sequence a m = the free coeffient of P(x 1,..., x n ) m always satisfies a recursion, which can be written as a differential equation for the generating function: a(t) = a m t m L(t, t d dt )a(t) = 0 m=0

6 Observe that m(1 + x x n ) = 1 2 m(p n) where P n = (1 + x x n )(1 + 1 x x n ) and consider the sequence of the free coefficients of the powers of P n : n = 2 a m : 1, 3, 15, 93, n = 3 a m : 1, 4, 28, 256, n = 4 a m : 1, 5, 45, 545,

7 The corresponding differential equations are given by ( d ) L n t, t a(t) = 0 dt L 2 (t, θ) = θ 2 t(10θ θ + 3) + t 2 (9θ θ + 9) L 3 (t, θ) = θ 3 2t(2θ + 1)(5θ 2 + 5θ + 2) + 64t 2 (θ + 1) 3 L 4 (t, θ) = θ 4 t(35θ θ θ θ + 5) + t 2 (θ + 1) 2 (259θ θ + 285) 225t 3 (θ + 1) 2 (θ + 2) 2

8 These differential equations for n = 2, 3 admit modular parametrization: n = 2 : t = η(6z)8 η(z) 4 η(3z) 4 η(2z) 8 = q 4q2 + 10q a = η(2z)6 η(3z) η(z) 3 η(6z) 2 = 1 + 3q + 3q2 +...

9 n = 3 : ( η(2z)η(6z) ) 6 t = a = (η(z)η(3z))4 η(z)η(3z) (η(2z)η(6z)) 2 n = 4 : L 4 (t, t d ) is not a symmetric power, dt hence there is no modular parametrization...

10 The generating function a(t) is related to the Mahler measure by the following trick due to F. Villegas: 1 (2πi) n log( 1 x i =1 t P(x 1,..., x n )) dx 1... dx n x 1 x n t m 1 = log t m (2πi) n P(x 1,..., x n ) m dx 1... dx n m=1 x i =1 x 1 x n t m = log t m a m = ( t d ) 1a(t) dt m=1 hence m(p) = Re ( t d dt ) 1a(t) t=

11 Suppose the differential equation has a modular parametrization, then this reduces to evaluation at the cusp where t = of ( d ) ( 1a(t) t ) [ 1 t = dt Dt D a = D 1 Dt ] t a, where D = q d dq = 1 d 2πi dz, and we reprove C.Smyth s formulas: m(1 + x 1 + x 2 ) = 3 3 4π L(χ 3, 2) m(1 + x 1 + x 2 + x 3 ) = 7 2π 2 ζ(3)

12 Idea: m(p n ) can be also calculated as where b(t) is a solution of m(p n ) = Re ( t d dt ) 1b(t) t= ( d ) L n 1 t, t b(t) = h(t) dt with some simple rational function h(t) in the right-hand side. Recall that L 4 is not modular, but L 3 is!

13 Theorem 1. Consider the following analytic at t = 0 solutions a(t) and b(t) of Then ( d ) L 2 t, t a(t) = 0 a(t) = 1 + 3t +... dt ( d ) 1 L 2 t, t b(t) = b(t) = 1 dt 1 t 9 t t m(p 3 ) = π 3 ( d ) 1[ π 2 t dt 72 a(t) + b(t)] t=.

14 Theorem 2. Consider the solution b(t) = O(t) of where ( d ) L 3 t, t b(t) = h(t) dt h(t) = 3 5Ω 2 10π t(212t t 13) (1 t) π 3 Ω 2 t 1 t, and Ω = 1 30π ( 14 j=1 Γ(j/15)χ K (j) ) 1/4 is the Chowla-Selberg period for the field K = Q( 15). Then m(p 4 ) = ( t d dt ) 1 b(t) t=.

15 Proof: Consider the 1-parametric family of varieties X λ : P n (x 1,..., x n ) = λ and define Ω n (λ) = ω λ X λ { x i =1} where the (n 1)-form ω λ is defined by 1 dx 1 (2πi) n dx n x 1 x n = ω λ dλ. Then Ω n (λ) is a period for this family and satisfies the Picard-Fuchs differential equation ( d ) L n λ, λ Ωn (λ) = 0. dλ

16 We obviously have m(p n ) = (n+1) 2 0 log(λ)ω n (λ)dλ, and by Jensen s formula 1 (2πi) n+1 log 1 + x x n+1 dx 1... dx n+1 x 1 x n+1 or = 1 (2πi) n x i =1 x i =1, 1+x 1 + +x n >1 m(p n+1 ) = (n+1) 2 1 log 1 + x x n dx 1 x 1 log(λ)ω n (λ)dλ.... dx n x n

17 We observe that if Ω(λ) satisfies L ( λ, λ d ) Ω(λ) = 0 dλ then the generating function for the moments along an arbitrary path β b n = λ n Ω(λ)dλ b(t) = b n t n satisfies α n=0 L ( t, t d dt ) b(t) = hβ (t) h α (t) where L(t, θ) = L ( 1/t, θ 1 ) and h α (t) is a simple rational function which depends only on the values of Ω and its derivatives at λ = α and can have a pole only at t = α.

18 In the case of L n we have h α (t) = 0 for α = 0 and α = (n + 1) 2. For the modular parametrization t(z) of L 3 preimages of t = 1 are CM points in the field K = Q( 15), therefore the Chowla-Selberg period for this field appears when we compute the right-hand side h α (t) with α = 1. (The end of the proof.)

19 ... and Double L-values of Modular Forms To compute m(p 3 ) and m(p 4 ) using the above theorems we have to evaluate ( d ) 1a(t) t dt t= and ( d ) 1b(t) t= t dt where a(t), b(t) are solutions of L ( t, t d dt ) ( d ) a(t) = 0 and L t, t b(t) = h(t) dt and L has a modular parametrization, i.e. we are given a modular function t(q) and a modular form a(q) of weight k, where k + 1 is the degree of the differential operator L.

20 Let D = q d dq, and suppose our modular parametrization is such that t = corresponds to q = 1. Then ( d ) ( 1a(t) t ) 1 t dt t= = Dt D q=1 a = D 1 f = L(f, 1) q=1 for the modular form f = Dt a/t of weight k + 2. Here for f = m=0 a mq m the L-function is defined by L(f, s) = m=1 a m m s for sufficiently large Re(s) and by analytic continuation otherwise.

21 Analogously, since the pull-back of L under the modular parametrization is given (up to a contant multiplier) by L ( t, t d dt ) = 1 Dt a Dk+1 1 a we have D k+1 [b/a] = Dt a h(t). In other words, b/a is an Eichler integral of a modular form of weight k + 2. Finally, ( d ) 1b(t) t dt t= = D 1 [f D g] (k+1) = L(g, f, k + 1, 1) q=1 where f = Dt a/t and g = Dt a h(t) are both of weight k + 2.

22 For two forms of weight k g = a n q n f = b m q m n=1 m=0 (g is a cusp form) their L-function is defined by L(g, f, s 1, s 2 ) = n=1 m=1 a n b m n s 1 (n + m) s 2. for Re(s 1 + s 2 ) > 2k, Re(s 2 ) > k and by analytic continuation otherwise. Critical double L-values (0 < s 1, s 2 < k, integers) are periods in the sense of Kontsevich and Zagier.

23 Applying the above strategy in the known case m(p 3 ) = 7ζ(3) we π 2 get the following equality: or 14π 3 π2 ζ(3) L (χ 3, 1) = L(g, f, 2, 1) 2π3 3 3 m(p 3) + π2 8 m(p 2) = L(g, f, 2, 1) where the forms g, f of weight 3 are given by g = q + 4q 2 + q 3 16q = E(z) + 7E(2z) 8E(4z) f = 1 + q 5q 2 + q = E(z) 2E(2z) 8E(4z) with E(z) = χ 3 (d)d 2 q n M 3 (Γ 0 (3), χ 3 ). n 1 d n

24 Analogously, the number m(p 4 ) ( (?= 15 ) 5L(f15 ) 12, 4) 2πi from Villegas conjecture is an expression involving π, the Chowla-Selberg period Ω K and double L-values of meromorphic modular forms with poles at CM points in the same field K = Q( 15).