On a conjecture by Boyd

The zeros of P k (x, y) correspond, generically to a curve of genus 1. Let E k denote the elliptic curve corresponding to the algebraic closure of P k (x, y) = 0.

Recall that the (logarithmic) Mahler measure of a non-zero Laurent polynomial, P (x 1 , . . . , x n ), with complex coefficients is defined as m(P ) = Let us denote m(k) := m(P k ). Boyd computed m(k) for k a positive integer less than or equal to 100 (it is easy to see that the Mahler measure does not depend on the sign of k for this family). He found that m(k)

where r k is a rational number and the question mark stands for an equality that has only been stablished numerically (typically to at least 50 decimal places).

The case with k = 1 (resulting in r k = 1) was considered in detail by Deninger [5], who found an explanation for such a formula by relating it to evaluations of regulators in the context of the Bloch-Beilinson conjectures. Rodriguez-Villegas [8] also considered this family in the context of the Bloch-Beilinson conjectures, including more general cases where k 2 ∈ Q. He was able to prove identities for the cases where the Bloch-Beilinson conjectures are known to be true, such as when E k has complex multiplication.

When the curves E k 1 and E k 2 are isogenous, their L-functions coincide. One can then compare the values in equation ( 1) and conjecture identities of the form r k 2 m(k 1 ) = r k 1 m(k 2 ). For example,

The first identity was proved in [7]. In this note, we prove the second one.

Functional identities for m(k) have been studied by Kurokawa and Ochiai in [6], and by Rogers and the author in [7]. The simplest ones are given as follows:

Theorem 2 We have the following functional equations for m(k):

• [7]: If h = 0, and |h| < 1:

If we set h = 1 √ 2 in both identities, we obtain

Similarly, if we set h = 1 2 , we obtain

Thus, in order to prove ( 2) and ( 3), we need to find one additional equation for each of the above linear systems.

In this section, we sometimes write x k and y k for x and y, so we can distinguish them when we look at different curves.

After the works of Deninger [5] and Rodriguez-Villegas [8], we write

were r k is a period of the regulator in the symbol {x k , y k } ∈ K 2 (E k ). For our purposes, we can reduce to

See [5] and [8] for general details, and [7] for the specific treatment of this particular example.

In our context, it is enough to take into account that

where α is a constant independent of k and D k is the elliptic dilogarithm in E k constructed by Bloch (see [2]).

We will briefly explain the meaning of (x) ⋄ (y). Let E be an elliptic curve with x, y ∈ C(E). Consider the divisors

This is an element in

where the equivalence relation stands for (-T ) ∼ -(T ). Thus, the Mahler measure depends just on D k and (x k ) ⋄ (y k ). For example, if the elliptic curves are isomorphic, D k does not change and the Mahler measure only depends on (x k ) ⋄ (y k ). This idea was discovered by Rodriguez-Villegas [9], and also used by Bertin [1]. We applied this idea again in [7], to isogenous elliptic curves, in order to prove identities like (5).

A Weierstrass model for E k is given by

It is not hard to see that E k (Q(k)) tor ∼ = Z/4Z. To fix notation, we will denote a generator by

Then we have 2P = (0, 0). Eventually, we will perform computations in the curve with parameter k = h + 1 h . In this curve, we will denote

which is a point of order 2. Notice that

In [7] we prove (x) ⋄ (y) = 8(P ).

Consider the isomorphism

which relates two of the curves in equation ( 5). We use this isomorphism to pull the rational functions x, y

On the other hand, it is easy to see that

From the previous section, the problem reduces to finding relations between (P ) and

In order to do that, we will look for elements that are trivial in

In other words, we will find combinations of Steinberg symbols {g, 1 -g} with g ∈ C E 2(h+ 1 h ) , such that the corresponding combination (g) ⋄ (1 -g) yields a linear combination of (P ) and (P + Q). Since {g, 1 -g} is trivial in K-theory, we conclude that (g) ⋄ (1 -g) ∼ 0, yielding a linear combination involving (P ) and (P + Q).

Consider the function

We have

where

In particular, for h = 1 √ 2 , we get

yielding the expected relation.

On the other hand, for h = 1 2 , our function f becomes

In this case, A and B are given by:

.

In particular, we have the relations 2A = 2B = P, B -A = 2P, A + B = -P.

We obtain

We need further relations among the divisors (A), (B). Thus we consider the following function

(X + 4).

We have

The diamond operation yields a new relation:

In order to get more relations, we apply the Galois conjugate,

The last two equations yield (g)⋄(1-g)+(g σ )⋄(1-g σ ) = 6(Q+P )+2(Q+A)+2(Q+B)+2(A)+2(B)-6(P ). Questions that remain open are how to predict identities such as ( 2) and (3) and, more precisely, to list all such i

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