Mahler’s measure

of K3 hypersurfaces

Marie José Bertin (Paris 6)

July 12th 2006

I INTRODUCTION

The first interested in was D.H.Lehmer

” On factorization of certain cyclotomic functions” (1933)

with his famous question (still unsolved): does there exist a monic irreducible polynomial P, non cyclotomic, with integer coefficients such that

Ω(P) := _{P (α)=0}^{∏} max(|α|, 1) < Ω(P_{0}) ≃ 1.1762...

where P_{0} is the Lehmer’s polynomial

X^{10} + X^{9} − X^{7} − X^{6} − X^{5} − X^{4} − X^{3} + X + 1?

In fact

Ω(P) = M(P)

the Mahler measure of P (introduced by Mahler in 1962).

The logarithmic Mahler’s measure of a polynomial P is

m(P) := \frac{1}{(2πi)^{n}} ∫_{𝕋^{n}} log | P (x_{1}, ⋯, x_{n}) | \frac{dx_{1}}{x_{1}} ⋯ \frac{dx_{n}}{x_{n}}

and the Mahler’s measure

M(P) = exp(m(P)).

By Jensen’s formula, if P ∈ ℤ[X] is monic, then

M(P) = \begin{matrix}∏ \\ P (α)=0 \end{matrix} max(| α |, 1).

The first partial answer to Lehmer’s question is due to Smyth (1971)

M(P) ≥ M(X^{3} − X − 1) ≃ 1.32..

if P is non reciprocal. The obstruction for Lehmer’s question is therefore the reciprocal polynomials.

Boyd’s limit formula (1981)

m(P (x, x^{N})) ⟶ m(P (x, y))

is a hope to get small measures in one variable from small measures in two variables.

M((x+1)y^{2} +(x^{2} +x+1)y+x(x+1)) = 1.25...

M(y^{2} + (x^{2} + x + 1)y + x^{2}) = 1.28..

are the smallest known measures in two variables.

At the same time Smyth obtained the first explicit Mahler measures:

m(x + y + 1) = L′(χ_{−3}, −1)

m(x + y + z + 1) = \frac{7}{2π^{2}} ζ(3)

Deninger (1996) guessed

m(x + \frac{1}{x} + y + \frac{1}{y} + 1) =^{?} \frac{15}{4π^{2}} L(E, 2) = L′(E, 0)

Since then, an abundant literature in this area, three Conferences on the Mahler measure and developments in many mathematics domains.

QUESTION: Which class of polynomials give such formulae?

II Polynomials in two variables

Nice conjectures (Boyd, Rodriguez-Villegas). See also results by Lalin, Vandervelde, Rogers.

For these conjectures, the main assumption is

P (x, y) “tempered”

Let C the curve defined by P (x, y) = 0.

• If C is of genus 0, we may get πm(P) as a combination of Bloch-Wigner dilogarithm plus eventually a term in ζ_{F} (2).

Here, the Bloch groups of a number field F are involved; hence also K_{2}(F).

• If C is of genus 1, by Jensen’s formula the Mahler measure can be expressed as

m(P) = \frac{1}{(2πi)^{2}} ∫_{|x|=1} ∫_{|y|=1} log |P (x, y)|\frac{dx}{x}\frac{dy}{y}

= \frac{1}{2πi} ∫_{|x|=1} log (max (|y_{1}|, |y_{2}|))\frac{dx}{x}

= \frac{1}{2πi} ∫_{Γ} η_{2}(2)(x, y)

where

η_{2}(2)(x, y) = log |y|di arg x − log |x|di arg y

is a differential form on the variety Γ

Γ = {(x, y) ∈ ℂ^{2}/ (x, y) ∈ C |x| = 1, |y| ≥ 1}.

Now, if Γ generates H_{1}(E, ℤ)^{−} (E the elliptic curve associated to C), then m(P) can be viewed as an elliptic regulator, so, expressed in terms of the elliptic dilogarithm.

More complicated if relaxing the assumption “tempered”, but not desperate!

The polynomial

P := (x^{2} +x−1)y^{2} +(x^{2} +5x+1)y −x^{2} +x+1

is not tempered.

Boyd guessed for m(P) the formula

A := m(P) =^{?} \frac{2}{3} log ϕ + \frac{1}{6}d_{15}

d_{15} = L′(χ_{−15} , −1)

where χ_{−15} is the odd primitive character of conductor 15.

By Bloch’s formula,

\frac{1}{6}L′(χ_{−15} , −1)

= \frac{5}{4π} [D(ζ_{15} ) + D(ζ_{15}^{2} ) + D(ζ_{15}^{4} ) + D(ζ_{15}^{8} )]

Using some parametrization, I got

A = \frac{2}{3} log ϕ − D(−j^{2} \frac{1 − √5}{2} ) − D(j\frac{1 − √5}{2} )

− D(−j\frac{1 + √5}{2} ) − D(j^{2} \frac{1 + √5}{2} )

Numerically

C := −D(−j^{2} \frac{1 − √5}{2} ) − D(j\frac{1 − √5}{2} )

−D(−j\frac{1 + √5}{2} ) − D(j^{2} \frac{1 + √5}{2} )

=^{?}

E := \frac{5}{4}[D(ζ_{15} + D(ζ_{15}^{2} ) + D(ζ_{15}^{4} ) + D(ζ_{15}^{8} )]

So, even if the polynomial is not “tempered”, there is underneath a relation between dilogarithms; hence the objects are living in a Bloch group.

And by Galois descent in the Bloch group, C and E live in B_{2}(ℚ(√−15 )_{ℚ}.

I have not yet guessed to which element in B_{2}(ℚ(√−15 )_{ℚ}, C and E are equal.

III Polynomials defining K3 surfaces

Why K3 surfaces? Because, like elliptic curves, they are Calabi-Yau: they have, up to scalars, a unique holomorphic 2-form (so periods).

My results concern two modular families of K3 surfaces, both with non zero faces (equivalently not “tempered” in 2 variables).

For example one family is

Q_{k} = X + \frac{1}{X} + Y + \frac{1}{Y} + Z + \frac{1}{Z}

+ XY + \frac{1}{XY} + ZY + \frac{1}{ZY} + XY Z + \frac{1}{XY Z} − k.

Computations possible thanks to the following result Theorem 1. (B. 2005) Let k = −(t + \frac{1}{t}) − 2 and

t = \frac{η(3τ)^{4}η(12τ)^{8}η(2τ)^{12}}{η(τ)^{4}η(4τ)^{8}η(6τ)^{12}} .

m(Q_{k}) = \frac{ℑτ}{8π^{3}} \{ ′ ; ∑ ; m,κ ℜ(

(\frac{4}{(mτ + κ)^{3}(mτ̄ + κ)} + \frac{2}{(mτ + κ)^{2}(mτ̄ + κ)^{2}} )

− (\frac{64}{(2mτ + κ)^{3}(2mτ̄ + κ)} + \frac{32}{(2mτ + κ)^{2}(2mτ̄ + κ)^{2}} )

− (\frac{36}{(3mτ + κ)^{3}(3mτ̄ + κ)} + \frac{18}{(3mτ + κ)^{2}(3mτ̄ + κ)^{2}} )

+ (\frac{576}{(6mτ + κ)^{3}(6mτ̄ + κ)} + \frac{288}{(6mτ + κ)^{2}(6mτ̄ + κ)^{2}} ))\}

In 2005, I proved that m(Q_{0}), m(Q_{12} ) and m(P_{2}) are expressed in terms of the L-series of the variety. But the big question was:

In the formula of the previous theorem, where is the part concerning the face?

My understanding of the question is the result of many contributions.

• Boyd guessed numerically

2m(Q_{−36} ) = 4m(Q_{−6}) + m(Q_{0})

• I proved (2004) that the relation is equivalent to a relation is equivalent to a relation between Hecke L-series of orders of a number field.

• To be sure, I showed the relation to Zagier (2005). Zagier proved the relation, interpreting it in terms of modular forms, more precisely Rankin-Cohen brackets.

• I understood (2006) where to find in the formula, the part concerning the face and the part concerning the L-series of the variety.

A new result (2006)

Let P_{10} = xyz(x+y+z)+t^{2}(xy+xz+yz)−10xyzt Then

m(P_{10} ) = 2d_{3} + ∗L − series of X_{10}

if

\frac{9√2}{π^{3}} ( ′ ; ∑(\frac{−1}{(9m^{2} + 2k^{2})^{2}} +\frac{1}{(k^{2} + 18m^{2})^{2}} )) =^{?} 2d_{3}

′ ; ∑ \frac{k^{2} − 18m^{2}}{(k^{2} + 18m^{2})^{3}} + ′ ; ∑ \frac{9m^{2} − 2k^{2}}{(9m^{2} + 2k^{2})^{3}}

=^{?}

\frac{10}{9} ′ ; ∑ \frac{m^{2} − 2k^{2}}{(m^{2} + 2k^{2})^{3}}

d_{3} = \frac{2√3}{π^{3}} ′ ; ∑ \frac{1}{(m^{2} + 3k^{2})^{2}}

These two first relations are proved numerically, the second being related to Rankin-Cohen brackets.