Functional equations for Mahler measures of genus-one curves (joint with Mat Rogers, University of British Columbia) Matilde N.

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1 Functional equations for Mahler measures of genus-one curves joint with Mat Rogers, University of British Columbia Matilde N. Laĺın Max-Planck-Institut für Mathematik, CMI and mostly UBC, PIMS August 9th, 006 1

2 Mahler measure of one-variable polynomials Pierce 1918: P Z[x] monic, P x = i x α i n = i α n i 1 P x = x n = n 1

3 Lehmer 1933: lim n α n+1 1 α n 1 = { α if α > 1 1 if α < 1 For P x = a i x α i MP = a i max{1, α i } mp = log MP = log a + i log + α i 3

4 Kronecker s Lemma: P Z[x], P 0, mp = 0 P x = x n φ i x Lehmer s Question 1933: Does there exist C > 0 such that P x Z[x] Is mp = 0 or mp > C?? mx 10 + x 9 x 7 x 6 x 5 x 4 x 3 + x + 1 the best possible? = = 1, 794, 37, 140, 357 4

5 Mahler measure of multivariable polynomials P C[x ±1 1,..., x±1 n ], the logarithmic Mahler measure is : mp = = log P 0 0 eπiθ 1,..., e πiθn dθ 1... dθ n 1 πi n log P x 1,..., x n dx 1... dx n T n x 1 x n Jensen s formula: 1 0 log eπiθ α dθ = log + α recovers one-variable case. 5

6 mp Q = mp + mq mp 0 if P has integral coefficients. α Q, P α is its minimal polynomial over Q, then mp α = [Qα : Q] hα, where h is the logarithmic Weil height. Boyd Lawton: P C[x 1,..., x n ] lim k... lim kn mp x, x k,..., x k n = mp x 1,..., x n 6

7 Examples Smyth 1981 m1 + x + y = 3 3 4π Lχ 3, = L χ 3, 1 Smyth1981 m1 + x + y + z = 7 π ζ3 D Andrea & L 003 m z1 xy 1 x1 y = 4 5ζ Q 5 3 ζ3π Boyd & L 005 mx +1+x+1y+x 1z = Lχ 4, + 1 π 8π ζ3 7

8 Examples in several variables L 003 m x1 1 x 1 x3 1 + x x 1 + x 3 z = 4 π 3Lχ 4, 4 + Lχ 4, π m x1 1 + x x4 1 + x 4 z = 6 14 π4ζ5 + 3π ζ3 m 1 + x + 1 x1 1 + x yz = 4 π 3Lχ 4, 4 m 1 + x + 1 x1 1 x 1 + x x 1 + yz = 93 π 4ζ5 8

9 The measures of a family of genus-one curves Boyd, Deninger, Rodriguez-Villegas m x + 1x + y + 1y + k? = L E k, 0 k N 0, 4 s k x + 1x + y + 1y + 4 m = L χ 4, 1 E k determined by x + 1 x + y + 1 y = 0. L-functions Beilinson s conjectures k = 4 CM case m x + 1 x + y + 1 y + 4 = L E 4, 0 k = 3 modular curve X 0 4 m x + 1 x + y + 1 y + 3 = 5 L E 3, 0 9

10 mk := m x + 1x + y + 1y + k Theorem 1 Rodriguez-Villegas mk = Re = Re 16y µ π m,n πiµ + where je k = j 1 4µ χ 4 m m + n4µ m + n4 µ n=1 d n χ 4 dd qn n q = e πiµ 16 = q k = exp π F 1 1, 1 ; 1, F 1, 1 ; 1, 16 k and y µ is the imaginary part of µ. k 10

11 Theorem For h R, m4h 4 + m h = m h + 1. h For h R, h < 1, m h m h Corollary 3 ih + 1 ih m8 = 4m = 8 5 m3 = m 4 h. 11

12 The elliptic regulator F field. Matsumoto Theorem: K F = {a, b}, a, b F / bilinear, {a, 1 a} F with discrete valuation v and maximal ideal M Tame symbol x, y v 1 vxvyxvy mod M yvx F = QE, E/Q elliptic curve, S E Q determines a valuation, v : K QE QS. We have 0 K E Q K QE Q S E Q QS Q 1

13 x, y QE, assume trivial tame symbols ηx, y := log x d arg y log y d arg x 1-form on EC \ S for any loop γ EC \ S consider γ, ηx, y = 1 ηx, y π γ Regulator map Beilinson, Bloch: for γ H 1 E, Z. r : K E Q H 1 E, R {x, y} {γ γ, ηx, y} H 1 E, R dual of H 1 E, Z. Follows from ηx, 1 x = ddx, Dx = ImLi x + arg1 x log x is the Bloch-Wigner dilogarithm We may think of γ H 1 E, Z. 13

14 Bloch regulator function given by Kronecker- Eisenstein series R τ e πiα = y τ π m,n Z e πibn am mτ + n m τ + n if α = a + bτ and y τ is the imaginary part of τ. Let Jz = log z log 1 z, EC = C/Z + τz = C /q Z where z mod Λ = Z + τz is identified with e iπz. J τ z := n=0 Jzq n n=1 + 1 log z 3 log q B 3 log q Jz 1 q n B 3 x = x 3 3 x + 1 x third Bernoulli polynomial. 14

15 Elliptic dilogarithm D τ z := n Z Dzq n Regulator function given by R τ = D τ ij τ Z[EC] = Z[EC]/ [ P ] [P ]. R τ is an odd function, Z[EC] C. x = m i a i, y = n j b j. CE CE Z[EC] x y = m i n j a i b j. 15

16 Theorem 4 Deninger x, y are non-constant functions in QE with {x, y} K E, Ω ηx, y = Im R τ x y γ y τ Ω 0 where Ω 0 is the real period and Ω = γ ω. Idea: i[ω] [ ω] generates H 1 E/R, R, therefore, EC ηx, y ω = αi ηx, y = αi[ω] [ ω] γ Beilinson proves ηx, y = αi ImΩ EC ω ω = αω 0 y τ EC ηx, y ω = Ω 0R τ x y 16

17 The relation with Mahler measures Deninger 1 mk Z r{x, y}γ π yp k x, y = y y 1 xy y x, mk = 1 πi T 1log+ y 1 x +log + y x dx x. By Jensen s formula respect to y. mk = 1 πi log y dx T1 x = 1 π ηx, y, T1 T 1 H 1 E, Z. 17

18 Modularity for the regulator α β Let γ δ that b a SL Z and let τ = ατ+β γτ+δ, such = δ γ β α b a Then: R τ e πia +b τ = 1 γ τ + δ R τ e πia+bτ. 18

19 Functional equations for the regulator From Let p prime, Jz = p x p =z 1+χ 4 pp J 4τ e πiτ = Jx p 1 j=0 +χ 4 pj 4pτ e πipτ pj 4τ+j p e πiτ+j p In particular, p =, J 4τ e πiτ = J τ e πiτ + J τ+1 e πiτ+1 Also: J τ+1 e πiτ = J τ e πiτ J τ e πiτ 19

20 Idea of Proof Weierstrass form: x + 1 x + y + 1 y + k = 0 x = kx Y XX 1 y = kx + Y XX 1. Y = X X + k 4 X + 1. P = 1, k, torsion point of order 4. x = P P 3P + O, y = P P + 3P + O. x y = 4P 4 P = 8P. 0

21 P 1 4 mod Z + τz k R τ = iy τ k R, k > 4, τ = 1 + iy τ k R, k < 4 Understand cycle [ x = 1] H 1 E, Z Ω = τω 0 k R 1

22 mk = 4 π Im τ y τ R τ i, k R Take SL Z. mk = 4 τ πy τ J 1 τ e πi 4τ If we let µ = 1 4τ, then mk = 1 J 4µ e πiµ 1 = Im R 4µ e πiµ πy µ πy µ = Re 16y µ π m,n χ 4 m m + n4µ m + n4 µ

23 First: J 4µ e πiµ = J µ e πiµ + J µ+1 e πiµ+1 1 J 4µ e πiµ = 1 J µ e πiµ + 1 J µ e πiµ y 4µ y µ y µ set τ = µ 1, for h < 1 so µ ir Dτ i = D τ i + 1 J y µ+1 µ+1 Second: 1 y µ J µ+1 e πiµ e πiµ+1 = 1yµ J µ e πiµ 1yµ J µ e πiµ Set τ = 1 µ and use D τ 1 1 i = D τ i J y µ+1 µ+1 e πiµ+1 3

24 Putting things together, D τ i = Dτ i + D τ 1 i this is the second equality. It turns out that mk = Re πiµ πi where eµ = 1 4 µ n=1 d n i ez 1dz χ 4 dd q n is an Eisenstein series. Hence the equations can be also deduced from identities of Hecke operators. 4

25 Parameter k. k 16 q = q k = exp π F 1 1, 1 ; 1, F 1, 1 ; 1, 16 k Second degree modular equation, h < 1, h R, h ih q q h 1 + h h 1 + h = q = q h 4. ih 1 h. Then the equation with J becomes h ih m q 1 + h +m q 1 h =m q h 4. m h + 1 h + m ih + 1 ih = m 4 h. 5

26 The identity with h = 1 m + m8 = m 3 m 3 + m i = m8 f = Y X in RE 3. Y X = P + P + Q 3O, 1 Y X = P + Q + 3P + Q 3O, Q = 1 h, 0 has order. 6

27 f 1 f = 6P 10P +Q 6P 10P +Q. φ : E 3 E i X, Y X, iy But r i {x, y} = r 3 {x φ, y φ} x φ y φ = 8P + Q 6r 3 {x, y} = 10r i {x, y} and 3m3 = 5mi. Consequently, m8 = 8 5 m3 m = 5 m3 7

28 Other families Hesse family ha 3 = m x 3 + y xy a studied by Rodriguez-Villegas hu 3 = j=0 h ξj 3 u 1 + ξ j u small 3 u More complicated equations for examples studied by Stienstra: m x + 1y + 1x + y xy t and Zagier and Stienstra: m x + y + 1x + 1y + 1 xy t 8