Mahler measures and computations with regulators

Transcription

1 Journal of Number Theory ) Mahler measures and computations with regulators Matilde N. Lalín 1 Department of Mathematics, University of British Columbia, Vancouver, BC V6T-1Z, Canada Received 1 December 006; revised 8 February 007 Available online 7 March 007 Communicated by F. Rodriguez-Villegas Abstract In this work we apply the techniques that were developed in [M.N. Lalín, An algebraic integration for Mahler measure, Duke Math. J ), in press] in order to study several examples of multivariable polynomials whose Mahler measure is expressed in terms of special values of the Riemann zeta function or Dirichlet L-series. The examples may be understood in terms of evaluations of regulators. Moreover, we apply the same techniques to the computation of generalized Mahler measures, in the sense of Gon and Oyanagi [Y. Gon, H. Oyanagi, Generalized Mahler measures and multiple sine functions, Internat. J. Math. 15 5) 004) 45 44]. 007 Elsevier Inc. All rights reserved. MSC: 11G55; 19F99 Keywords: Mahler measure; Regulator; Polylogarithms; Riemann zeta function; L-functions; Polynomials 1. Introduction The logarithmic) Mahler measure of a Laurent polynomial P C[x ±1 1,...,x±1 n ] is defined by 1 1 mp ) := log P e πiθ 1,...,e πiθ ) n dθ1 dθ n, 1) 0 0 address: mlalin@math.ubc.ca. 1 This material is partially based upon work supported by the National Science Foundation under agreement No. DMS X/$ see front matter 007 Elsevier Inc. All rights reserved. doi: /j.jnt

2 13 M.N. Lalín / Journal of Number Theory ) = 1 π) n T n log Px 1,...,x n ) dx 1 dx n. ) x 1 x n where T n = x 1 = = x n =1. The Mahler measure of a one-variable polynomial has a simple expression in terms of the roots, due to Jensen s formula. The several-variable case, however, is much more complicated to describe. Many formulas for specific examples have been computed, especially for polynomials in two and three variables. An interesting fact is that many of these formulas express the Mahler measure of a polynomial in terms of special values of the Riemann zeta function or Dirichlet L-series. A typical example is Smyth s formula [Smy81] m1 x y z) = 7 ζ3). 3) π Moreover, L-series of varieties also appear in these kinds of formulas. Inspired by these results, Deninger [Den97] established the relation between Mahler measure and regulators in many cases. More precisely, he wrote for P C[x 1,...,x n ], where mp ) = mp ) 1 πi) n1 Γ η n n)x 1,...,x n ), 4) Γ = Px 1,...,x n ) = 0 x 1 = = x n1 =1, x n 1, and η n n) is Goncharov s regulator, a C form on the generic point of Γ see Definition 1). Boyd [Boy98] computed many more numerical examples that fit into this context. Rodriguez- Villegas [R-V97] applied the ideas of Deninger and developed others to understand and prove many cases in two-variables by making explicit computations with the regulator. This work was, in certain sense, continued by Boyd and Rodriguez-Villegas [BR-V0,BR-V03] for two-variable polynomials. One of the ideas in these works is to identify the cases where η )x 1,x ) is exact and proceed to the computation of the Mahler measure by means of Stokes s Theorem, obtaining special values of the Bloch Wigner dilogarithm. Later Maillot suggested a way to continue these ideas for more variables keeping in mind the cohomological interpretation. In [Lal07] we developed these ideas for some three-variable cases involving evaluations of trilogarithms and shed some light on how these computations could be carried on for more variables. Let us consider the three-variable case. In favorable cases there is a primitive for η 3 3)x 1,x,x 3 ) say ω), one may apply Stokes s Theorem and obtain an integral Γ ω. Maillot proposed a way that desingularizes Γ to certain Γ that has an algebraic description and nontrivial boundary. This process also allows the possibility that ω restricted to Γ ) is exact. If that is the case, we may continue the integration by applying Stokes s Theorem again. In this work we explore the techniques of [Lal07] in application to several concrete examples. Most of the examples that we study here are in three variables, but we also include an example in four variables, and we apply the method to compute some generalized Mahler measures.

3 M.N. Lalín / Journal of Number Theory ) More precisely, we recover formula 3) as well as the following results: m 1 x y 1 1 x y)z ) = 14 3π ζ3) m 1 x)1 y) 1 x)1 y)z ) = 7 π ζ3) m 1 x y 1 x)z ) = 7 log ζ3) π m 1 y)1 x) 1 x)z ) = 8 5π ζ3) m 1 x 1 )1 x) 1 x 1 )1 y)z ) = 4 π 3 Lχ 4, 4) Smyth [Smy0] Lalín [Lal03] Lalín [Lal03] Condon [Con03] Lalín [Lal03,Lal06] Generalized Mahler measures were introduced by Gon and Oyanagi [GO04] who studied their basic properties, computed some examples, and related them to multiple sine functions and special values of Dirichlet L-functions. Given f 1,...,f r C[x ±1 1,...,x±1 n ], the generalized logarithmic) Mahler measure of f 1,...,f r is defined by mf 1,...,f r ) = 1 πi) n T n max log f1,...,log f r dx 1 dx n. 5) x 1 x n Notice that for r = 1 one obtains the classical Mahler measure of f 1. On the other hand, for r =, mf 1,f ) = mf 1 zf ), where z is a variable that is independent of x 1,...,x n. In this work we use regulators to study generalized Mahler measures when f i = Px i ) for a fixed polynomial or rational function) P. We obtain explicit formulas for m 1 x 1 ),...,1 x n ) ), 1 x1 m,..., 1 x ) n, 1 x 1 1 x n and m 1 x 1 x 1 1,...,1 x n xn 1 ). The first case was known to Gon and Oyanagi [GO04]. To our knowledge, the other two cases are new results. Finally, we study the behavior of the generalized Mahler measure in the general case of f i = Px i1,...,x in ) again, P is fixed) when the number of functions f i goes to infinity. The conclusion is that the generalized Mahler measure approaches the sup norm of P in T n.thisis

4 134 M.N. Lalín / Journal of Number Theory ) the content of Proposition 3. This result may be combined with the aforementioned examples in order to compute limits of sums involving zeta values, such as lim m1 m j=1. A construction for regulators 1) j m 1 j ) j)!1 j ) π) j ζj 1) = log. Our goal is to apply the techniques of [Lal07]. In this section we describe our main ingredients. We will be following Goncharov s construction of the regulator on polylogarithmic motivic complexes [Gon0,Gon05]. First recall Zagier s modification of the polylogarithm [Zag91]: n1 L n x) := Re n j=0 j B j j! ) ) j log x Linj x), 6) where B j is the jth Bernoulli number, Li k is the classical polylogarithm and Re k denotes Re or i Im depending on whether n is odd or even. For the record, L n is defined using Re n instead of Re n, where Re n denotes Re or Im depending on whether n is odd or even. Polylogarithms satisfy functional equations such as L n 1 x ) = 1) n1 L n x) for n>1, and L n x) = 1) n1 L n x), L n x) L n x) = L nx ) n1. For n =, one obtains the Bloch Wigner dilogarithm, L x) = Dx) = Im Li x) ) log x arg1 x), 7) which satisfies the well-known five-term relation ) ) 1 y 1 x Dx) D1 xy) Dy) D D = 0. 8) 1 xy 1 xy In particular, Dx) =D1 x). A useful functional equation for n = 3is L 3 x) L 3 1 x) L ) = ζ3). 9) x Finally, for an element z of a field F, we denote by z n the class of z in B n F ) := Z[P 1 F ]/R nf ), where R n F ) is a certain subgroup of Z[P 1 F ] see definition in [Gon95], for instance) that conjecturally consists of all the rational functional equations of the n-polylogarithm in F.

5 M.N. Lalín / Journal of Number Theory ) Given a complex variety X, Goncharov considers certain C forms on the generic point of X, η n n) : n CX) ) Q Ωn1 X η X ), 10) l1 ) η n l) : B nl1 CX) Q CX) ) Q Ωl1 X η X ), l < n. 11) The construction is as follows Definition 1. Let x i be rational functions on X. η n n) : x 1 x n Alt n p 0 p1 log x 1 dlog x j p 1)!n p 1)! j= n j=p di arg x j ), 1) where Alt m Ft 1,...,t m ) := σ S m 1) σ Ft σ1),...,t σm) ). This form has singularities in the zeros and poles of x i. This will not be a problem in the present work, but we refer the reader to [Gon0] for a detailed discussion. Notice that dη n n)x 1,...,x n ) = Re n dx1 x 1 We need more notation for the construction of η n l). For any integers p 1 and k 0, define Then let Definition. We have β k,p := 1) p p 1)! k p 1)! k p 1 j 1 [ p1 ] j=0 L p,q x) := L p x) log q1 x dlog x, p, dx ) n. x n ) kpj B kpj. L 1,q x) := log x dlog 1 x log 1 x dlog x ) log q1 x. η n l) : x nl1 x 1 x l1 L nl1 x) Alt l1 p 0 1 p 1)!l 1 p)! p j=1 dlog x j l1 j=p1 di arg x j )

6 136 M.N. Lalín / Journal of Number Theory ) Notice that 1 k,1 p l1 β k,p L nl1k,k x) log x 1 Alt l1 p 1)!l 1 p)! p dlog x j j= l1 j=p1 di arg x j ). 13) η n 1)x) = L n x), η n n)x 1, 1 x 1,x,...,x n1 ) = dη n n 1)x 1,x...,x n1 ), η n l)x 1,x 1,x,...,x l1 ) = dη n l 1)x 1,x...,x l1 ), l < n. This equality is to be understood in any open set U X where the forms are defined, i.e., excluding the zeros and poles of x i. To be concrete, we describe the two-variable case. We have Moreover, The forms for n = 3are ηx,y) := η )x, y) = log x di argy) log y di argx). 14) η )x, 1 x) = didx). ηx,y,z) := η 3 3)x,y,z) ) 1 = log x dlog y dlog z di arg y di arg z 3 1 log y dlog z dlog x di arg z di arg x 3 η 3 3)x, 1 x,y) = dη 3 )x, y), 1 log z dlog x dlog y di arg x di arg y 3 ωx,y) := η 3 )x, y) = idx)di arg y 1 3 log x dlog 1 x log 1 x dlog x ) log y. ) ), For future reference, observe that ηx,y,z) changes sign under complex conjugation of x,y, and z, and that ωx,y) is invariant under complex conjugation of both x and y. 3. The relation with Mahler measure In this section we summarize the method described in [Lal07]. Let P C[x ± 1,...,x± n ].We may assume, without loss of generality, that P is actually a polynomial. Then we may write

7 M.N. Lalín / Journal of Number Theory ) Px 1,...,x n ) = a d x 1,...,x n1 )x d n a 0x 1,...,x n1 ). Thus, formally, By Jensen s formula, Px 1,...,x n ) = a d x 1,...,x n1 ) mp ) = ma d ) 1 πi) n1 j=1 d xn α j x 1,...,x n1 ) ). j=1 d log α j x 1,...,x n1 ) dx 1 x 1 T n1 dx n1 x n1. From Definition 1, mp ) = ma d ) 1 πi) n1 Γ η n n)x 1,...,x n ), 15) where Γ = Px 1,...,x n ) = 0 x 1 = = x n1 =1, x n 1. Thus we have recovered Deninger s expression 4). We have to compute the integral. Our strategy is as follows. For a given polynomial, we try to prove that η n n)x 1,...,x n ) is exact. In order to achieve that, we try to express x 1 x n as a linear combination of elements of the form y 1 1 y 1 ) y y n1.atthesame time we compute the boundary of Γ. Then we apply Stokes s Theorem in order to obtain a linear combination of integrals of the form η n n 1)y 1,...,y n1 ). Γ We continue by examining the restriction of y 1 y y n1 to each component of Γ and trying to express it as a linear combination of elements of the form z 1 z 1 z z n. This will be different for different components, since η n n 1) is not even closed on Γ itself. If we are successful, then we may obtain a sum of integrals of the form η n n )z 1,...,z n ). Γ Notice that Γ should be empty. However, typically Γ will have singularities. Upon desingularizing Γ, we obtain a set with nontrivial boundary. If we are fortunate enough, we may continue this process until we compute the integral completely. The success of this method depends on the polynomial P. See [Lal07] for a discussion of sufficient conditions and other details.

8 138 M.N. Lalín / Journal of Number Theory ) Examples Now we will apply the machinery that was described in the previous sections in order to understand many examples of Mahler measure formulas in three variables. We will study many of the examples that are known to be related to ζ3) and the trilogarithm. It is hard in practice to apply the technique when the polynomial has degree higher than 1 in the variable x n = x 3 ). From now on we will replace x 1,x,x 3 ) with x,y,z). Hence, all the examples will be written as z = Rx,y) 16) where Rx,y) is a rational function with real coefficients. In computing the boundary Γ, we apply the idea proposed by Maillot [Lal07]. Since we have Γ = z = Rx,y) x = y =1, z 1, Γ = z = Rx,y) x = y = z =1. Notice that R has real coefficients, therefore we can describe Then we may write Γ = z Rx,y) = z 1 R x 1,y 1) = 0 x = y = z =1. where C is the curve defined by γ := Γ = C x = y = z =1, Rx,y)R x 1,y 1) = 1. 17) For the case in four variables we can follow a similar process. In most of the cases the denominator of the rational function R is a product of cyclotomic polynomials. We may multiply Eq. 16) by the denominator in order to obtain a polynomial equation whose Mahler measure is the same as the Mahler measure of z R. We will call this polynomial P. The only exception to this case is P = 1 x y 1 1 x y)z, since the measure of 1 x y is not zero. In what follows, we will compute the logarithmic) Mahler measure of P Smyth s example We are going to start with the simplest example in three variables, which was also due to Smyth [Boy81,Smy0]: π m1 x y z) = 7 ζ3). 18)

9 M.N. Lalín / Journal of Number Theory ) It is easy to see that this problem amounts to computing the Mahler measure of the rational function z 1 x 1 y. We would like to see that ηx,y,z) is exact. The equation with the wedge product yields In other words, x y z = x y 1 x = x y 1 x) x y 1 y) 1 y =x 1 x) y y 1 y) x. ηx,y,z) =ηx,1 x,y) ηy,1 y,x). After performing the first integration, we are left to analyze Equation 17) yields for Γ in this case. When xy = 1 we obtain When x = y we obtain Δ =x y y x. xy 1)x y) = 0 Δ = x x. Δ =x x. One could have problems in the cases when z has a pole or is equal to zero. But those correspond to x = 1ory = 1, and Δ = 0 in these circumstances. We obtain ωx,y) ωy,x) =±ωx,x), which yields mp ) = 1 4π γ ωx,x), where γ = Γ. We now need to check the path of integration γ. Since the equations xy = 1 and x = y intersect in ±1, ±1), there are four paths, which can be parameterized as

10 140 M.N. Lalín / Journal of Number Theory ) Fig. 1. Integration set for 1 x y z. x = e αi, 0 α π, y = x, x 3 ; x = e αi,π α 0, yx = 1, x 3 ; x = e αi, 0 α π, y = x, x 3 ; x = e αi, π α 0, yx = 1, x 3, where the right-hand side column indicates the primitives for the pullback of Δ to each component of γ. Figure 1 shows the integration set for the problem in terms of arg x and arg y. The shaded region corresponds to the original surface Γ, where the first integration is performed, and γ is the boundary of this surface. Finally we obtain mp ) = 1 4π 8 L 3 1) L 3 1) ) = 7 ζ3). 19) π It is important to notice that the orientation for γ is not the one that we could naively expect without taking into account the singularities at ±1, 1). That is to say, if we think of the diagonals in the picture as two circles, the boundary would be zero and we would get the erroneous result mp ) = 0. The fact that different paths in each circle are oriented differently is crucial for the result. The orientation of Γ is not as easy to determine. The simplest way to do this is to choose any orientation for Γ and use the induced orientation in γ. If the chosen orientation for Γ is incorrect, then we obtain mp ) with the wrong sign, but that can be easily corrected since we know mp ) must be positive. 4.. Another example due to Smyth We will now be concerned with another example due to Smyth [Smy0], P = 1 x y 1 1 x y)z.

11 M.N. Lalín / Journal of Number Theory ) But In this case, the equation for the wedge product yields x y z = x y 1 x y 1) x y 1 x y) =x y 1 1 x y 1) x y 1 x y). x y 1 x y) = x y 1 x y) y = x y x y) 1 x y) x y 1 x ) 1 x y) y = x y) 1 x y) x y x ) 1 x ) 1 x y). y y Then we need to analyze Δ = x 1 xy xy 1 x 1 ) y y x y x y x 1 x y). y The technique of Eq. 17) yields for Γ x x 1 ) y y 1) = 0. If y =1, If y = 1, Δ =1 x x) x x 1 x x) x x = 4x x. Δ =1 x x x x) = x x x x). We will use the five-term relation starting with x and x, We obtain by using a =1 a ), x x 1 x) = 0. Δ =x x 1 x) x x x) 1 x) x) =x x) x x) x x x x ).

12 14 M.N. Lalín / Journal of Number Theory ) If x =1, Δ = y y y 1y y) y y y =4y y. then If x = 1, Δ = 1 1 y y 1 ) 1y 1y y y 1 y). y We will use the five-term relation starting with 1 y and 1 y, Now 1 y 1 y 1 y 1 1 y y y y = y = 1 y) y. y) = 1 1 y) 1 y 1 1 y y 1 1 y) y) 1y y) = 1 y y y y ) 1 1 ) y y). y y Then we obtain Δ = 1 y 1y 1 y y y y ) y 1y ) y y) 1 y 1 ) y 1 1 ) y y). y y We may need to take into account the poles or zeros of z. But those are at the points x, y) = ζ 6,ζ 1 6 ), and they do not affect the integration because they are a set of points and the integration is in dimension. We compute the boundary γ see Fig. ) and primitives for pullbacks of Δ. x = e αi, π α 0, y =1, 4x 3 ; x = e αi, 0 α π, y = 1, x 3 x 3 x x 3 ; y = e αi, 0 α π, x =1, 4y 3 ;

13 M.N. Lalín / Journal of Number Theory ) Fig.. Integration set for 1 x y 1 1 x y)z. y = e αi,π α 0, x = 1, 1 y 1 y 3 y 3 y y y 3 y 1 3 y 3; x = e αi,π α 0, y =1, 4x 3 ; x = e αi, 0 α π, y = 1, x 3 x 3 x x 3 ; y = e αi, 0 α π, x =1, 4y 3 ; y = e αi, π α 0, x = 1, 1 y 1 y 3 y 3 y y y 3 y 1 3 y 3. Then we obtain 4π mp ) = 16 L 3 1) L 3 1) ) 4 L 3 1) 3L 3 1) L 3 3) L 3 3) ) = 4L 3 1) 8L 3 1) 8L 3 3) 4L 3 3). It will be necessary to use the identity: which is essentially Lemma 6 in Smyth s [Smy0]. Finally, L 3 3) L 3 3) = 13 ζ3) 0) 6 mp ) = 14 ζ3). 1) 3π It is worth noting that this example involves the evaluation of the element As we noted above, Smyth [Smy0] also encountered this difficulty in his direct computation.

14 144 M.N. Lalín / Journal of Number Theory ) In order to evaluate the element in B 3 Q), we need to use Goncharov s - term relation see [Gon95]). We follow the notation in Zhao [Zha03], and take the relation with a,b,c)= 3, 1, 1), obtaining = Using that 1 3 = and 1 x 3 =x 3, = 0, thus, proving Eq. 0). We owe this argument to the referee of this paper Another three-variable example The following example was first computed in [Lal03]. It is easier to consider the following rational function For the wedge product we have, z 1 x)1 y) 1 x)1 y). x y z = x y 1 x) x y 1 y) x y 1 x) x y 1 y) =x 1 x) y y 1 y) x x) 1 x) y y) 1 y) x. Thus, we need to consider Δ =x y y x x y y x. This time Eq. 17) implies xy 1)x y) = 0. When xy =1, Δ = x x x x). When x =y, Δ =x x x x). The poles or zeros in this case occur with x =±1 and y =±1, but we always obtain Δ = 0 and they do not affect the integration. We now need to check the integration path γ and primitives for pullbacks of Δ.

15 M.N. Lalín / Journal of Number Theory ) Fig. 3. Integration set for 1 x)1 y) 1 x)1 y)z. Therefore, we obtain, x = e αi, 0 α π, y =x 1, x 3 x 3 ; x = e αi,π α 0, y =x, x 3 x 3 ; x = e αi, 0 α π, y =x 1, x 3 x 3 ; x = e αi, π α 0, y =x, x 3 x An example with non-trivial symbol 4π mp ) = 16 L 3 1) L 3 1) ), mp ) = 7 ζ3). ) π Now we will study an example that is of different nature because its symbol in K-theory is not trivial see [Lal07]). In other words, ηx,y,z) is not exact in this case. This example was first computed in [Lal03]. As before, we will consider a simpler form than the one in [Lal03], that is to say the rational function We have Now we use that z 1 x xy. 1 x x y z = x y 1 x xy) x y 1 x). x y 1 x xy) = x) y) 1 x1 y) ) = x1 y) ) y) 1 x1 y) ) 1 y) y) 1 x1 y) ).

16 146 M.N. Lalín / Journal of Number Theory ) Then x y z = x z x 1 x) y) x1 y) ) 1 x1 y) ) y) We need to analyze y) 1 y) 1 x1 y) ). Δ =x y) x1 y) y) y 1 x1 y) ), and then we also need to compute the integral of η,x,z) the nontrivial part in K-theory, i.e., the non-exact part). By Eq. 17) on Γ, For x =1, For y =1, Δ = 0. For xy =1, xy =1, x =1, or y =1. Δ =1 y y) y y) = y y). Δ = 1 1 y y) y y y) y 1 1 ). y But 1 y y = 1 1, y and we may use the five-term relation ignoring torsion) in order to get 1 1 y 1y 1 = 0. 3) y y Then Δ =y y) 1y y) 1 1 y 1y 1 1 ) 1 y y =y y) 1 1 y y y). 1 1 y 1 1 ) y 1 1 ) y ) ) y y

17 M.N. Lalín / Journal of Number Theory ) Fig. 4. Integration set for 1 x xy 1 x)z. There are zeros of z when 1 x xy = 0, and that only can happen in the unit torus if x, y) = 1, 1) and that is just a point. There are poles for x = 1, in this case By the five-term relation 3), Δ = 1 1 y 1 1 y Δ =1 y y) y y). y) y y) y y) y) y y) y y). This integrates to Ω =y 3 y , y 3 y 3 hence to zero when y moves in the unit circle. We now need to check the integration path γ and the primitives for pullbacks of Δ. y = e αi, 0 α π, x =1, y 3 ; y = e αi, π α 0, x =y 1, y 3 1 y y 1 3 y 3; y = e αi, 0 α π, x =1, y 3 ; y = e αi,π α 0, x =y 1, y 3 1 y y 1 3 y 3. We obtain, 4π mp ) = η,x,z) L 3 ) L 3 ) ) L 3 4) L 3 ) L 3 ) ).

18 148 M.N. Lalín / Journal of Number Theory ) But 4 3 = , then 4π mp ) = η,x,z) 16L 3 ). Since =1 3, we obtain 3 = ) = Then 4π mp ) = η,x,z) 14ζ3). We still need to compute η,x,z). B where B =x, y) T π arg x,arg y,arg x arg y π. Since z1 x) = 1 x xy, dz z = z y 1 x dx z1 x) z1 x) dy. But x =1, so dx dargx dargz =Re x dz ) ) dx =Re z x x z1 x) dy. Thus, B η,x,z)= log B dargx dargz =Re log B 1 1 1x xy dx x ) dy. y We need to consider B k=0 1 x xy ) k dx x dy y. Setting x = e iα, y = e iβ, = 0 π π πα π 0 πα π ) 1 eiα )e iβ ) k dβ dα. 4) k=0

19 M.N. Lalín / Journal of Number Theory ) We consider each term in the series separately, 0 π π πα If k = 0, the above integral is 3π. If not, = 0 π = i k = i k = k 1 e iα ) e iβ) k dβ dα. 1 e iα )) k 1) k 1 e ikα ) dα = i ik k k l=0 k l=1 i k ) 0 k e ilα 1 e ikα) dα l π k1 1 1) k ) ) k 1 1) l. l l l=1 0 π 1 e iα ) k 1 e ikα ) dα ) k 1 1) l i 1 ) ) 1)kl l l k l Hence, in order to evaluate the first term in Eq. 4), we need to evaluate k=1 k k k l=1 ) k 1 1) l l l = 1 1) l l=1 l k=l ) k 1 l k k. But k=l ) k λ k l k = λl l! k 1) k l 1)λ kl = λl l1 ) 1 l! λ l1 1 λ k=l = λl l 1)! l! 1 λ) l = λ l l1 λ) l. Using this with λ = 1, k=1 k k k l=1 ) k 1 1) l l l = l=1 The second integral is the same, so as a conclusion, we get B k=0 1 x xy ) k dx x 1 1) l l = 3ζ) = π. dy y =π.

20 150 M.N. Lalín / Journal of Number Theory ) And, B η,x,z)= π log. Finally, the whole formula becomes 4.5. Condon s example mp ) = 7 log ζ3) π. 5) The last and most complex example that we will analyze in three variables was discovered numerically by Boyd and proved by Condon [Con03]. It may be expressed in the following way: 1 y)1 x) z. 1 x The wedge product equation becomes: x y z = x y 1 y) x y 1 x) x y 1 x) = y 1 y) x x) 1 x) y x 1 x) y. Hence we need to consider Equation 17) implies on Γ ) Δ =y x x y x y. ) 1 x y = 1 x 1 y). We now use the five term relation, 1 x x 1 1 x 1 x 1 x x x 1 x x 1 1 x = 0, 1 x = 0. Then Δ =y x 1 x x 1 y y. 1 x 1 x Let us write y = t so that 1 x 1 x =± t 1 t. 6)

21 M.N. Lalín / Journal of Number Theory ) Let us take the -case. Then Thus we may write, x = t t 1 t t 1. Δ = t t t 1 1 t t t 1 t t t 1 t It is clear that a change in the sign of t will simply change the sign of Δ. We will split the integration of Δ into two steps. In order to do that, write so that t. Δ 1 = t t t 1 t t 1, Δ = t t 1 t t 1 t t, Δ = Δ 1 Δ. We will first work with Δ 1.Letϕ = 1 5,soϕ ϕ 1 = 0. By the five-term relation, ϕt ϕ 1)t 1 t 1 ϕt 1 t ϕt 1 ϕ)t 1 t 1 ϕt 1 t Observe that we have 1 ϕ 1)t 1 t = 0, 7) 1 ϕ 1)t 1 t = 0. 8) Δ 1 = t t t 1 t t 1 = t ϕt 1)ϕ 1)t 1) ϕt 1)ϕ 1)t 1) = t 1 ϕt 1 ϕ 1)t t 1 ϕt 1 ϕ 1)t. Now let us apply the five-term relations 7) and 8): =ϕt 1 ϕt 1 ϕ 1)t ϕ 1)t 1 ϕt 1 ϕ 1)t 1 ϕt 1 t 1 ϕt 1 ϕ 1)t 1 ϕ 1)t 1 t ϕt 1 ϕt 1 ϕ 1)t 1 ϕ)t 1 ϕt 1 ϕ 1)t 1 ϕt 1 ϕt 1 ϕ 1)t 1 t 1 ϕ 1)t 1 t 1 ϕt 1 ϕ 1)t 1 ϕt 1 ϕ 1)t.

22 15 M.N. Lalín / Journal of Number Theory ) Then we obtain Δ 1 =1ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) 1 ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) ϕt 1 ϕ 1)t ) ϕt 1 ϕ 1)t ) ϕ 1)t 1 ϕt) 1 ϕ)t 1 ϕt) 1 ϕt 1 t 1 ϕt 1 ϕ 1)t 1 ϕ 1)t 1 t 1 ϕt 1 ϕt 1 ϕ 1)t 1 t 1 ϕ 1)t 1 t Now we will work with Δ. By the five-term relation, 1 ϕt 1 ϕ 1)t 1 ϕt 1 ϕ 1)t. ϕ t 1 1 ϕ 1)t t t 1 1 ϕ 1)t ϕ 1)t 1 t t t 1 = 0, ϕ t ϕ)t t 1 t 1 ϕ 1)t ϕ 1)t 1 t t t 1 = 0, 1 ϕt t 1 ϕ 1) t t 1 ϕt 1 ϕt t 1 t 1 t = 0, 1 ϕt t 1 ϕ 1) t 1 t ϕt 1 ϕt t 1 t 1 t = 0. Applying the above equalities, we obtain Δ = t t 1 t t 1 t t = ϕ t 1 t 1 ϕ)t t 1 ϕ 1)t 1 t ϕ t 1 t ϕ 1)t 1 ϕ 1)t t 1 t ϕt t ϕ t 1 1 ϕt t 1 t ϕt t ϕ t 1 t 1 ϕt 1 t t t. t t We will now use the fact that we will integrate in a set where t =1. Under those circumstances we have the following two identities at the level of the differential form ω): ϕ 1)t t = ϕ 1) t t =ϕt t, 9) 1 ϕ)t t =ϕt t. 30)

23 M.N. Lalín / Journal of Number Theory ) Both identities depend on the fact that t =1 since we are conjugating and using that t = t 1. We will work henceforth on the level of ω i.e., in the coimage of η 3 )). Thus, Δ = ϕ t 1 t ϕ t 1 t 1 ϕ 1)t 1 ϕ 1)t 1 t t 1 t t 1 ϕt 1 ϕt 1 t t 1 t t. We will add Δ 1 and Δ. But first, let us note 1 ϕt 1 ϕt t 1 1 t t1 ϕ 1)t) = ϕ t 1 t t 1 t = t 1 t1 ϕ 1 ϕt ϕ t ϕ 1 t t ϕ = t ϕ 1 ϕt t 1 ϕ t 1 ϕ 1 t Using this and similar identities, we obtain t1 ϕ t ϕ 1 ϕ) ϕ 1). 1 ϕt 1 t 1 ϕ) Δ =1ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) 1 ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) ϕt 1 ϕ 1)t ) ϕt 1 ϕ 1)t ) ϕ 1)t 1 ϕt) 1 ϕ)t 1 ϕt) t ϕ t 1 t ϕ t ϕ 1) ϕ t 1 ϕ t 1 ϕ 1) t ϕ t 1 t ϕ t ϕ 1) ϕ t 1 ϕ t 1 ϕ 1) 1 ϕt 1 ϕ 1)t 1 t ϕ 1) 1 t ϕ 1 ϕt 1 ϕ 1)t 1 t ϕ 1) 1 t ϕ ϕ t 1 t ϕ t 1 t. t ϕ 1) t 1 ϕ 1) t ϕ 1) t 1 ϕ 1) Now observe that 1 ϕt 1 ϕ 1)t 1 ϕt 1 ϕ 1)t 1 t ϕ 1 t ϕ 1 t ϕ 1 t ϕ =ϕt ϕ 1 ϕ)t ϕ ϕt ϕ ϕ 1)t ϕ

24 154 M.N. Lalín / Journal of Number Theory ) by five-term relations 7) and 8). Therefore using 1 ϕ1 = ϕ) Δ =1ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) 1 ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) ϕt 1 ϕ 1)t ) ϕt 1 ϕ 1)t ) ϕ 1)t 1 ϕt) 1 ϕ)t 1 ϕt) t ϕ t 1 t ϕ t ϕ 1) ϕ t 1 ϕ t 1 ϕ 1) t ϕ t ϕ t ϕ 1) t 1 ϕ t 1 ϕ t 1 ϕ 1) t ϕ 1) t 1 ϕ 1) t ϕ 1) t 1 ϕ 1) ϕt ϕ 1 ϕ)t ϕ ϕt ϕ ϕ 1)t ϕ ϕ t 1 t ϕ t 1 t. Next we gather some similar terms together, Observe that ϕ 1)t Δ =1ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) 1 ϕt 1 ϕt) 1 ϕ 1)t 1 ϕ 1)t ) t ϕ t 1 t ϕ t ϕ 1) ϕ t 1 ϕ t 1 ϕ 1) t ϕ t ϕ t ϕ 1) t 1 ϕ t 1 ϕ t 1 ϕ 1) t ϕ 1) t 1 ϕ 1) t ϕ 1) t 1 ϕ 1) ϕt ϕ t) ϕt ϕ t) ϕ 1)t ϕ 1) t ) 1 ϕ)t ϕ 1) t ) ϕ t 1 t ϕ t 1 t. ϕ 1) t ) = ϕ 1)t ϕ 1)t ) ϕ 1)t ϕ t 1). Now conjugate the elements of the second term using ϕ R) = ϕ 1)t ϕ 1)t ) ϕ 1)t 1 ϕ t) = ϕ 1)t ϕ 1)t ) ϕt ϕ t). Hence the last two lines of Δ above equal

25 M.N. Lalín / Journal of Number Theory ) ϕt ϕ t)ϕt ϕ t) ϕ 1)t ϕ 1) t ) 1 ϕ)t ϕ 1) t ) ϕ t 1 t ϕ t 1 t =ϕt ϕ t) ϕ t 1 t ϕt ϕ t) ϕ t 1 t ϕ 1)t ϕ 1)t ) 1 ϕ)t 1 ϕ)t ). We want to simplify the term ϕt ϕ t)ϕ t 1 t. On the one hand, ϕt ϕ t)=ϕt t 1 ϕt 1 ϕt 1). On the other hand, ϕ t 1 t =ϕ t t 1 =ϕ t ϕ t)ϕ t 1 ϕt 1). By the five term relation 1 ϕt 1 ϕ ϕ t 1 ϕ t 1 ϕ 1)t = 0, but 1 ϕ corresponds to zero in the differential since it is a constant real number and DR) = 0. We then get Then Analogously, 1 ϕt 1 ϕt 1) ϕ t 1 ϕt 1) = 1 ϕ t 1 1 ϕt 1) ϕ 1)t 1 ϕt 1) = 1 ϕ t 1 1 ϕ t 1) 1 ϕ t 1 1 ϕ) 1 ϕ 1)t 1 ϕ 1)t ) ϕ 1)t ϕ 1)t ). ϕt ϕ t) ϕ t 1 t =ϕt t ϕ t ϕ t) 1 ϕ t 1 1 ϕ t 1) 1 ϕ t 1 1 ϕ) 1 ϕ 1)t 1 ϕ 1)t ) ϕ 1)t ϕ 1)t ). ϕt ϕ t) ϕ t 1 t =ϕt t ϕ t ϕ t) 1 ϕ t 1 1 ϕ t 1) 1 ϕ t 1 1 ϕ) 1 ϕ 1)t 1 ϕ 1)t ) 1 ϕ)t 1 ϕ)t ).

26 156 M.N. Lalín / Journal of Number Theory ) Thus the last two lines of Δ above reduce to ϕt ϕ t) ϕ t 1 t ϕt ϕ t) ϕ t 1 t ϕ 1)t ϕ 1)t ) 1 ϕ)t 1 ϕ)t ) =ϕt ϕt ϕt ϕ ϕ t ϕ t) 1 ϕ t 1 1 ϕ t 1) 1 ϕ t 1 1 ϕ) 1 ϕ 1)t 1 ϕ 1)t ) 3 ϕ 1)t ϕ 1)t ) ϕt ϕt) ϕt ϕ ϕ t ϕ t) 1 ϕ t 1 1 ϕ t 1) 1 ϕ t 1 1 ϕ) 1 ϕ 1)t 1 ϕ 1)t ) 3 1 ϕ)t 1 ϕ)t ). Next we will see that ϕ t 1 ϕ ϕ t 1 ϕ ϕt ϕ ϕt ϕ corresponds to zero in the level of the differential form. Using that t =1 and d arg ϕ = 0, the differential is 3ω = log 1 ϕ t dlog ϕ t log ϕ t dlog 1 ϕ t log ϕ log 1 ϕ t dlog ϕ t log ϕ t dlog 1 ϕ t log ϕ dlog 1 ϕt log ϕ dlog 1 ϕt = log 1 ϕt dlog ϕ t log ϕ dlog ϕ t log ϕ t dlog 1 ϕt log 1 ϕt dlog ϕ t log ϕ dlog ϕ t log ϕ t dlog 1 ϕt log ϕ dlog t 1 ϕ log ϕ dlog t 1 ϕ = log 1 ϕt dlog ϕt 1 1 log ϕt 1 1 dlog 1 ϕt log 1 ϕt dlog ϕt 1 1 log ϕt 1 1 dlog 1 ϕt = 0. Finally the primitive of Δ is Ω =1ϕt 3 1 ϕ 1)t 3 1 ϕt 3 1 ϕ 1)t 3 t ϕ t ϕ 1) t ϕ t 1 ϕ t 1 ϕ 1) t 1 ϕ which is t ϕ 1) t 1 ϕ 1) ϕt 3 ϕ t 3 1 ϕ t ϕ 1)t 3 3 ϕ 1)t 3 ϕt 3 ϕ t 3 1 ϕ t ϕ 1)t ϕ)t 3, 3

27 M.N. Lalín / Journal of Number Theory ) =1ϕt 3 1 ϕt 3 t ϕ t 1 ϕ 3 t ϕ 1) t 1 ϕ 1) 3 t ϕ t ϕ 1) t 1 ϕ 3 t 1 ϕ 1) 3 ϕt 3 ϕ t 3 1 ϕ t ϕ 1)t 3 3 ϕ 1)t 3 ϕt 3 ϕ t 3 1 ϕ t ϕ 1)t ϕ)t 3. Let us use that x 3 1x =1 3 x 3 and the fact that x 3 = x 3 at the level of the differential. We obtain Ω =41ϕt 3 41 ϕt 3 t ϕ t 1 ϕ 3 t ϕ 1) t 1 ϕ 1) 3 t ϕ t ϕ 1) t 1 ϕ 3 t 1 ϕ 1) 3 ϕt 3 ϕ t 3 1 ϕ t 1 3 ϕt 3 ϕ t 3 1 ϕ t 1 3. Let us note that the poles of z occur with x = 1, which easily implies Δ = 0. Analogously, Δ = 0fory = 1orx =1 which correspond to the zeros of z. We need to describe the integration path. If we let x = e iα, with π α π, and t = eiβ, with π β π. Then condition 6) translates into tan α =±sinβ. The boundaries of the above condition are met when sin β =±1, corresponding to t =±i. Also, by inspection of Eq. 6) we see that t =±1 and x = 1 are also critical points. After carefully analyzing the situation see Fig. 5), and taking into account that Δ = 0 whenever x = 1ort =±1, we conclude that we need to integrate Δ with π β 0 and π β 0. Since Ω is 0 when evaluated for t =±i, and Ω1) =Ω1), we obtain 4π mp ) = 4Ω1) = 4 4L 3 1 ϕ) 4L 3 1 ϕ) L 3 ϕ) L 3 ϕ) L 3 ϕ 1) L 3 ϕ 1) L 3 ϕ) L 3 ϕ) ) = 4 4L 3 1 ϕ) 6L 3 1 ϕ) L 3 ϕ) L 3 ϕ 1) L 3 ϕ) ). Now we use that in order to obtain 1 ϕ 3 = ϕ 1 3 =ϕ 3, ϕ 1 3 =ϕ 3, π mp ) = 6L 3 1 ϕ) 4L 3 ϕ) 4L 3 ϕ) L 3 ϕ).

28 158 M.N. Lalín / Journal of Number Theory ) Fig. 5. Integration set for 1 y)1 x) z1 x). and Then we use ϕ 1 3 ϕ =1 3, ϕ 1 3 ϕ 3 ϕ 3 ϕ 3 =1 3, ϕ 3 1 ϕ =1 3, ϕ 3 ϕ 3 1 ϕ 3 ϕ 3 =1 3. Thus we obtain π mp ) = 8L 3 1) 1L 3 ϕ) 1L 3 ϕ). But 4ϕ 3 4ϕ 3 = ϕ 3 =ϕ 1 3 =1 3 ϕ 3 ϕ 3 implies that Finally we recover Condon s result ϕ 3 ϕ 3 = mp ) = 8 ζ3). 31) 5π Let us note that the computation involves 5, although the final result does not. The same situation applies to Condon s computation in [Con03] in fact, he also needs to use ϕ),

29 M.N. Lalín / Journal of Number Theory ) so it seems that this step is unavoidable as in the case of Smyth s example from Section 4.. Moreover, if we look at the rational number in each of the three-variable formulas we will notice that the denominators are always 1 or except for these two examples of Smyth and Condon. In Smyth s example the denominator is a 3 and in Condon s example the denominator is a 5. Both the appearance of 3 as a denominator together with in the first case, and 5 as a denominator together with elements in BQ 5)) seem to be related phenomena. Gangl suggested that a possible explanation may be that there are torsion elements that we may be overlooking when we tensor by Q as the regulator maps kill torsion). Lastly, this is the first example where we used non-rational functional equations for the polylogarithms, namely D x) =Dx) and L 3 x) = L 3 x). This suggests that for us, the best definition for R n F ) may be all the functional equations of L n instead of all the rational functional equations of L n. Of course, the second definition has the advantage of being more concrete and should be easier to handle An example in four variables We will study an example in four variables: z 1 x)1 x 1) 1 y)1 x 1 ). This example was first computed in [Lal03] in terms of multiple polylogarithms. It was later observed [Lal06] that this particular combination of multiple polylogarithms has a simpler expression in terms of a Dirichlet L-series. Here we relate this Mahler measure directly to the Dirichlet L-series. The wedge product is We integrate and obtain x y x 1 z = x y x 1 1 x)1 x 1) 1 y)1 x 1 ) = x 1 x) y x 1 y 1 y) x x 1 x 1 1 x 1 ) x y x 1 1 x 1 ) x y. Δ =x y x 1 y x x 1 x 1 x y x 1 x y. It is more convenient to eliminate the variable x 1 than z. Notice that x 1 = 1 x) z1 y) 1 x) z1 y). Then the condition whose intersection with T 4 describes Γ becomes 1 x)1 y) x yz ) = 0.

30 160 M.N. Lalín / Journal of Number Theory ) When x =1ory =1, η 4 3)Δ) = 0. Now suppose x =yz. Then we need to integrate Δ = yz y x 1 y yz x 1 x 1 z y x 1 z y. Consider the first two terms, Δ 1 = yz y x 1 y yz x 1 = yz yz x 1 yz z x 1 y y x 1 y z x 1. Use the five-term relation, yz 1 1 z 1 yz y 1 y 1 z y1 z = 0, ) together with the fact that then α := x x 1 = 1 yz z1 y) y z x 1 yz z x 1 = z 1 yz z x 1 1 z = z z x 1 z 1 y) 1 yz = z z x 1 z α z x 1 z x 1 y 1 y1 z ) 1 yz 1 y z x 1 zα z x 1. Conjugate and invert the second term noting that arg α =± π since x 1 =1) Consider the last two terms of Δ: Note that and recall the five term relation: x1 1 x 1 x 1 = 1 x 1 = z z x 1 zα z x 1. Δ =x 1 z y x 1 z y. y = 1 αz zz α), z x 1 1 x1 =α α. 1 x 1 z x 1

31 M.N. Lalín / Journal of Number Theory ) Thus Conjugating the first term, and inverting, Thus Δ =α z 1 αz zz α) α z 1 αz zz α). Δ =α z 1 αz zz α). Δ = yz yz x 1 y y x 1 z z x 1 4zα z 1 α 1 α 4α z 1 αz z α. By inverting and dividing by z: Now notice that α z 1 αz z α =α z 1 αz1 1 αz. zα z 1 α 1 α α z 1 αz1 1 αz =zα z 1 α 1 zα zα z 1 α 1 zα α z 1 α 1 αz α z 1 α 1 αz 1. Conjugate and invert z in the last term zα z 1 α 1 zα zα z 1 α 1 zα α z 1 α 1 αz α z 1 α 1 αz. Now use 1 α α z 1αz 1 αz α z 1 αz 1 α 1 αz 1 z = 0, 1 αz 1 z = 0. 1 αz Then we obtain z z 1 α 1 α 1 zα 1 αz z z 1 α 1 zα 1 α 1 αz z 1 α 1 zα z 1 α 1 zα z1 α) 1 αz z 1 α 1 zα z1 α) 1 αz z 1 α 1 zα.

32 16 M.N. Lalín / Journal of Number Theory ) Then Δ integrates on the component γ 0 =x =yz of Γ )to Ω = yz 3 x 1 y 3 x 1 z 3 x 1 4z 3 1 α 1 α z1 α) 1 zα 4 z 4 z 1 αz 3 1 αz 3 4z 3 1 α 1 α 1 zα 4 z1 α) z 4 z. 1 αz 3 1 αz 3 Now the boundary of γ 0 is x =1ory =1. When x =1 we obtain y =1 and z =±i or α = 0 and x 1 = 1. If y =1, Ω =1 3 x 1 4z 3 1 α 1 zα 4z 3 1 α 1 zα. If we take into account that α moves in the imaginary axis when x 1 moves in the unit circle and that we are evaluating on z =±i, then we just need to evaluate η 4 ) on If α = 0, integrating to Ω =1 3 x 1. Ω =4z 3 z 4z 3 z, Σ 1 =4z 4 4z 4, and the boundary now is z =1. When y =1, then z =±i and x =1orx 1 =1 and α =. In the first case we obtain, as before In the second case we get, integrating to Ω =1 3 x 1. Ω = 8z 3 z 8z 3 z Σ = 8z 4 8z 4. Now we need to take into account the opposite orientations that will cancel the terms with Ω = 1 3 x 1 and add up to Σ = 1z 4 1z 4,

33 M.N. Lalín / Journal of Number Theory ) which must be evaluated between i and i. This yields the final result 5. Generalized Mahler measure mp ) = 4 π 3 Lχ 4, 4). 3) In this section we will apply the algebraic integration to the computation of generalized Mahler measures. We follow the notation from Eq. 5). Let us fix P R[x ±1 ] and take f j = Px j ) so in particular f j R[x j ±1 ] R[x ±1 1,...,x±1 n ]). Suppose, moreover, that P is a monotonic function for 0 arg x π. Write x j = e πiθ j = eθ j ). Then m Px 1 ),...,Px n ) ) = 1 1 max log P eθ 1 ) ),...,log P eθ n ) ) dθ 1 dθ n 1 = n n! 1 log P eθ1 ) ) dθ1 dθ n = n! πi) n 0 θ n θ 1 1 where Γ =0 argx n ) argx 1 ) π. We proceed to compute some examples The case of P = 1 x Γ η n1 n 1) Px 1 ), x 1,...,x n ) Let P = 1 x, then Peθ)) = sinπθ), so it is monotonic on [0,π]. This case was computed by Gon and Oyanagi in [GO04] with the use of multiple sine functions. We start by evaluating η n1 n 1) on 1 x 1 ) x 1 x n. Now this corresponds to an exact form, which we integrate and obtain x 1 x x n. The boundary Γ consists of x 1 =1 and x 1 = x for our purposes the remaining restrictions are 0, e.g., x = x 3 implies x x 3 = 0). Evaluating, 1 x x n x x x 3 x n.

34 164 M.N. Lalín / Journal of Number Theory ) By integrating the term in the right and evaluating in the boundary of Γ x 1 = x ) x =1 and x = x 3, we obtain, 1 x x n 1 3 x 3 x n x 3 3 x 3 x n. Here, as well as in the following pages, and will denote formal versions of and. We will use this notation whenever we have a combination of forms of different dimensions. Eventually, each of these forms will be integrated in a simplex of the right dimension, and the results of each of these integrals will be added to obtain the final result. We continue with the same procedure, until we reach n1 1) n1 1 n1 1) k1 1 k x k x n. k= Now we need to integrate each term. First observe that since the L k is trivial in the real numbers for k even, we just need to consider 1 n1 n1 k=3, kodd 1 k x k x n, where it is understood that the sum is up to n 1ifn is even or n if n is odd. The sum yields Finally, we get n1 k=3, kodd iπ) nk1 n k 1)! Li k1) = n1 k=3, kodd m1 x 1,...,1 x m ) = 1)m1 m)! π m ζm 1) m)! iπ) nk1 1 k1 ) n k 1)! k1 ζk). m 1) j 1 j ) ζj 1), 33) m j)!π) j j=1 m1 m1 x 1,...,1x m1 ) = m 1)! 1) j 1 j ) ζj 1), 34) m j 1)!π) j recovering the result of [GO04]. 5.. The example with P = 1x 1x j=1 Although P = 1x 1x is a rational function, we can still apply the method to this case. Notice that Peθ)) = tanπθ)) so it is monotonic. We start considering 1 x 1 1 x 1 x 1 x n.

35 M.N. Lalín / Journal of Number Theory ) We can find the primitive easily as x 1 x x n x 1 x x n. The boundary consists of x 1 =1 and x 1 = x, and thus we obtain 1 x x n 1 x x n If we continue the integration steps, we reach x x x n x x x n. 1) n1 ) n1 1 n1 1 n1 1) k ) 1 k x k x n 1 k x k x n. k= Taking into account the parity, we just need to consider 1 n1 1 n1 The sum yields n1 k=3, kodd Then we reach the result n1 k=3, kodd iπ) nk1 Lik 1) Li k 1) ) = n k 1)! 1 k x k x n 1 k x k x n. n1 k=3, kodd iπ) nk1 1 k ) ζk). n k 1)!k1 1 x1 m,..., 1 x ) m = 1)m m)!1 m1 ) 1 x 1 1 x m π) m ζm 1) m m)! 1) j 1 j1 ) ζj 1), 35) m j)!π) j 1 x1 m,..., 1 x ) m1 1 x 1 1 x m1 j=1 m1 = m 1)! j=1 1) j 1 j1 ) ζj 1). m j 1)!π) j 36) We remark the specific case 1 x1 m, 1 x ) = 7 1 x 1 1 x π ζ3). It corresponds to the Mahler measure of 1x 1 1x 1 z 1x 1x. Thus we recover formula ).

36 166 M.N. Lalín / Journal of Number Theory ) An example involving the golden ratio We will consider the case of P = 1 x x 1. Observe that P eθ) ) = 1 i sinπθ) = 1 4sin πθ). In this case we need to integrate with 0 arg x i π since that is when P e is monotonic). Then m Px 1 ),...,Px n ) ) ) n = n! η n1 n 1) ) Px 1 ), x 1,...,x n. πi 0 argx n ) argx 1 ) π We start with where ϕ = 1 5. Integrating, we obtain 1 x1 x 1 ) 1 x1 x n = x1 x 1 1 ) x 1 x n = 1 ϕ 1 x 1 ) x1 x n 1 ϕx 1 ) x 1 x n, ) ϕ ϕ 1 x 1 x x1 x n ϕx 1 x x n ϕ x x n. 1 ϕx 1 The boundary is given by x 1 = i and x 1 = x, thus, ϕ 1 i x x n ϕi x x n ϕ 1 x x x n ) ϕ x1 ϕx x x n ϕ x x n. 1 ϕx 1 Integrating once more, we obtain ϕ 1 i x x n ϕi x x n ϕ 1 x 3 x 3 x n ϕx 3 x 3 x n ϕ 1 x ϕ x 3 x n ) ϕ x1 ϕx ϕ x 3 x n ϕ x x n. 1 ϕx 1 Continuing with the same procedure, we reach 1) n1 ϕ 1 n1 ϕ ) n1 n1 1) k1 ϕ 1 i k x ) k x n ϕi k x k x n k=

37 M.N. Lalín / Journal of Number Theory ) ) ϕ x1 ϕ x x n 1 ϕx 1 n 1) k ϕ 1 x k k ϕ x ) k1 x n ϕx k k ϕ x k1 x n. k= Taking into account properties of L k according to the parity of k, =ϕ n1 ϕ n1 n1 k=3,kodd ϕi k x k x n ) ϕ x1 ϕ x x n 1 ϕx 1 n 1) k ϕ 1 x k k ϕ x ) k1 x n ϕx k k ϕ x k1 x n. 37) k= The first sum in 37) yields n1 k=3, kodd The second line in 37) yields 1) n log ϕ 1 i 1 = 1) n1 log ϕ Re n1 iπ) nk1 nk1 n k 1)! L kϕi) = ) dx1 iim dx dx n x 1 ϕ x x n i n1 k=3, kodd 1 dx 1 x 1 ϕ dx dx 1 n x x n i i iπ) nk1 n n k 1)! L k ϕ ). ) ) dx1 iim x 1 ϕ 1 dx dx n x x n ) dx 1 x 1 ϕ 1 dx dx n x x n = 1) n1 log ϕ Re n1 Lin iϕ 1 ) Li n ϕ 1 ) Li n iϕ) Li n ϕ) ). 38) We used the iterated integral notation in the previous equalities. The last sum in 37) yields n k=, keven 1 1) nk1 k B k log k ) ϕ dxk Re dx k1 dx n k! x k ϕ x k1 x n 1) k 1 = 1) n1 n i i ) ) dxk Re x k ϕ 1 dx k1 dx n x k1 x n k=, keven k B k log k ϕ k! 1 Re nk1 i dx k x k ϕ dx k1 dx n x k1 x n

38 168 M.N. Lalín / Journal of Number Theory ) ) dx k x k ϕ 1 dx k1 dx n x k1 x n i = 1) n1 n k=, keven k B k log k ϕ k! Re nk1 Link1 iϕ 1 ) Li nk1 ϕ 1 ) Li nk1 iϕ) Li nk1 ϕ) ). 39) The sum of Eqs. 38) and 39) is = 1) n1 n k=1 k B k log k ϕ k! Re n1 Link1 iϕ) Li nk1 ϕ) ) n 1) n1 k B k log k ϕ 1 Re n1 Link1 iϕ 1 ) Li nk1 ϕ 1 )). k! k=1 Now the generalized Mahler measure is equal to n! πi )n times 1 n L n1 ϕ ) 1 n1 iπ) nk1 n n k 1)! L k ϕ ) k=3, kodd 1) n1 L n1 iϕ) L n1 ϕ) L n1 iϕ 1 ) L n1 ϕ 1 )) 1) n1 Re n1 Lin1 iϕ) Li n1 ϕ) Li n1 iϕ 1 ) Li n1 ϕ 1 )). After simplifying, we obtain: m 1 x 1 x 1 1,...,1 x n xn 1 ) n 1 = n! n k 1)!iπ) k1 L k ϕ ) k=3 ) i n! n Re n1 Lin1 iϕ) Li n1 ϕ) Li n1 iϕ 1 ) Li n1 ϕ 1 )). 40) π A particular case is with n = 1, where we obtain m 1 x 1 x 1 ) 1 =log ϕ = log1 ϕ), thus recovering the expected result from Mahler measure. For n = we obtain m 1 x 1 x 1 1, 1 x x 1 ) = Li3 ϕ ) π Li 3 ϕ )) log ϕ, which recovers the Mahler measure of 1 x 1 x 1 1 z1 x x 1 ).

39 M.N. Lalín / Journal of Number Theory ) A limit formula We end this section with an observation regarding the behavior of the generalized Mahler measure when the number of involved polynomials goes to infinity. Proposition 3. Let P Cx 1,...,x n ) and let f i = Px i,1,...,x i,n ) for i = 1,...,r. Then where stands for the sup norm on T n. lim r mf 1,...,f r ) = log P Proof. For a measurable subset S of T n ) r,leti S denote the integral over S: I S = 1 πi) nr S max log Pxi,1,...,x i,n ) dx 1,1 dx r,n. 1 i r x 1,1 x r,n Then we need to investigate mf 1,...,f r ) = I T n ) r as r goes to infinity. First observe that mf 1,...,f r ) log P. Then we just need to understand the lower bound. Consider the set E ɛ := x 1,...,x n ) T n Px1,...,x n ) P ɛ. Let m ɛ denote the Lebesgue measure of E ɛ normalized so that the torus has measure one). We write T n ) r as a union of sets: T n ) r = E ɛ T n) r1 T n \ E ɛ ) Eɛ T n) r T n \ E ɛ ) r = A T n \ E ɛ ) r. Notice that the set A has measure Therefore, m ɛ 1 m ɛ )m ɛ 1 m ɛ ) m ɛ 1 m ɛ ) r1 m ɛ = 1 1 m ɛ ) r. I A 1 1 m ɛ ) r) log P ɛ ). 41) Outside the set E ɛ we do not have control of the value of log P. This is a problem if the polynomial has zeros in the torus. Let and let m η denote the measure of Z η. Z η = x 1,...,x n ) T n Px1,...,x n ) η,

40 170 M.N. Lalín / Journal of Number Theory ) We write T n \ E ɛ ) r = T n \ E ɛ ) r \ Z r η Z r η. Now On the other hand, I Z r η I T n \E ɛ ) r \Z r η 1 m ɛ ) r m r η) log η. 4) 1 πi) rn Zη r = m r1 η 1 πi) n log Px 1,1,...,x 1,n ) dx 1,1 x 1,1 dx r,n x r,n Z η log Px 1,...,x n ) dx 1 dx n. 43) x 1 x n First choose a small ɛ and large r, soi A becomes close to log P due to inequality 41). Now fix a small η. We have that both 1 m ɛ ) r and m r η are small for r large. If r is large enough, the right side of inequality 4) is close to zero. Then either I T n \E ɛ ) r \Zη r is positive or has a small absolute value. Finally, m r1 η is small for r large and the right side of inequality 43) is close to zero because log P L 1 T n ). Then either I Z r η is positive or has a small absolute value. We reach our conclusion by considering I T n ) r = I A I T n \E ɛ ) r \Zη r I Zη r. We may conclude interesting results. For instance, from Eq. 34) we deduce Acknowledgments lim m1 m j=1 1) j m 1 j ) j)!1 j ) π) j ζj 1) = log. 44) I thank my Ph.D. supervisor Fernando Rodriguez-Villegas for many incredibly helpful and encouraging discussions. I would also like to acknowledge enlightening discussions with David Boyd, Herbert Gangl and Vincent Maillot about Mahler measure, Bloch groups, and cohomology, and with Enrico Bombieri about generalized Mahler measure and more specifically, Proposition 3. I am very grateful to the referee whose dedicated and careful review has much contributed to the clarity of this work including necessary corrections), and for the proof of Eq. 0). Most of this work is part of my Ph.D. dissertation at the Department of Mathematics at the University of Texas at Austin. I am indebted to John Tate and the Harrington fellowship for their generous financial support during my graduate studies. Part of this research was completed while I was a member at the Institute for Advanced Study, a visitor at the Institut des Hautes Études Scientifiques, and a Postdoctoral Fellow at the Pacific Institute for the Mathematical Sciences. I am grateful for their support.

41 M.N. Lalín / Journal of Number Theory ) References [Boy81] D.W. Boyd, Speculations concerning the range of Mahler s measure, Canad. Math. Bull ) [Boy98] D.W. Boyd, Mahler s measure and special values of L-functions, Experiment. Math ) [BR-V0] D.W. Boyd, F. Rodriguez-Villegas, Mahler s measure and the dilogarithm I), Canad. J. Math ) [BR-V03] D.W. Boyd, F. Rodriguez-Villegas, with an appendix by N.M. Dunfield, Mahler s measure and the dilogarithm II), Preprint, July 003. [Con03] J. Condon, Calculation of the Mahler measure of a three variable polynomial, Preprint, October 003. [Den97] C. Deninger, Deligne periods of mixed motives, K-theory and the entropy of certain Z n -actions, J. Amer. Math. Soc. 10 ) 1997) [GO04] Y. Gon, H. Oyanagi, Generalized Mahler measures and multiple sine functions, Internat. J. Math. 15 5) 004) [Gon95] A.B. Goncharov, Geometry of configurations, polylogarithms, and motivic cohomology, Adv. Math. 114 ) 1995) [Gon0] A.B. Goncharov, Explicit regulator maps on polylogarithmic motivic complexes, in: Motives, Polylogarithms and Hodge Theory, Part I, Irvine, CA, 1998, in: Int. Press Lect. Ser., vol. 3, I, Int. Press, Somerville, MA, 00, pp [Gon05] A.B. Goncharov, Regulators, Handbook of K-theory, vols.1,, Springer-Verlag, Berlin, 005, pp [Lal03] M.N. Lalín, Some examples of Mahler measures as multiple polylogarithms, J. Number Theory 103 1) 003) [Lal06] M.N. Lalín, On certain combination of colored multizeta values, J. Ramanujan Math. Soc. 0 1) 006) [Lal07] M.N. Lalín, An algebraic integration for Mahler measure, Duke Math. J ), in press. [R-V97] F. Rodriguez-Villegas, Modular Mahler measures I, in: Topics in Number Theory, University Park, PA, 1997, in: Math. Appl., vol. 467, Kluwer Acad. Publ., Dordrecht, 1999, pp [Smy81] C.J. Smyth, On measures of polynomials in several variables, Bull. Austral. Math. Soc. Ser. A ) Corrigendum with G. Myerson): Bull. Austral. Math. Soc ) [Smy0] C.J. Smyth, An explicit formula for the Mahler measure of a family of 3-variable polynomials, J. Theor. Nombres Bordeaux 14 00) [Zag91] D. Zagier, Polylogarithms, Dedekind Zeta functions, and the algebraic K-theory of fields, in: Arithmetic Algebraic Geometry, Texel, 1989, in: Progr. Math., vol. 89, Birkhäuser Boston, Boston, MA, 1991, pp [Zha03] J. Zhao, Goncharov s relations in Bloch s higher Chow group CH 3 F, 5), J. Number Theory 14 1) 007) 1 5.