A Mahler measure of a K3-hypersurface expressed as a Dirichlet L-series
aa r X i v : . [ m a t h . N T ] J a n A MAHLER MEASURE OF A K3-HYPERSURFACE EXPRESSEDAS A DIRICHLET L-SERIES
MARIE JOS´E BERTIN
Abstract.
We present another example of a 3-variable polynomial defining aK3-hypersurface and having a logarithmic Mahler measure expressed in termsof a Dirichlet L-series. Introduction
The logarithmic Mahler measure m ( P ) of a Laurent polynomial P ∈ C [ X ± , ..., X ± n ]is defined by m ( P ) = 1(2 πi ) n Z T n log | P ( x ± , ..., x ± n ) | dx x ... dx n x n where T n is the n-torus { ( x , ..., x n ) ∈ C n / | x | = ... = | x n | = 1 } .For n = 2 and polynomials P defining elliptic curves E , conjectures have beenmade, with proofs in the CM case, by various authors [6], [10], [11]. These conjec-tures give conditions on the polynomial P for getting explicit expressions of m ( P )in terms of the L -series of E . A crucial condition for P is to be “tempered”, thatis the roots of the polynomials of the faces of its Newton polygon are only roots ofunity. This condition is related to the link between m ( P ) and the second group of K -theory, [1], [11].In various papers we obtained results for n = 3 and polynomials P defining K K P ensure theexpression of m ( P ) in terms of the L -series of the K L -series? Our investigations concern two families of polynomials in three variables[2].This result is the second example of a Mahler measure expressed uniquely interms of a Dirichlet L -series.The first example was m ( P ) = m ( X + 1 X + Y + 1 Y + Z + 1 Z ) = d = 3 √ π L ( χ − , , where L ( χ − ,
2) denotes the Dirichlet L -series for the quadratic character χ − at-tached to the imaginary quadratic field Q ( √− L -series of the K Date : October 26, 2018.1991
Mathematics Subject Classification.
11, 14D, 14J.
Key words and phrases.
Modular Mahler measure, Eisenstein-Kronecker’s Series, L -series of K l -adic representations, Livn´e Criterion, Rankin-Cohen brackets. The second example is the following theorem.
Theorem 1.1.
Let Q − the Laurent polynomial Q − = X + 1 X + Y + 1 Y + Z + 1 Z + XY + 1 XY + ZY + 1 ZY + XY Z + 1
XY Z + 3 and define d = 3 √ π L ( χ − , . Then m ( Q − ) = 85 d . In this theorem the evaluation of the modular part needs the use of Livn´e’scriterion [15], since we have to compare two l -adic representations, and also recentresults about Dirichlet L -series [18]. Acknowledgments
The measure m ( Q − ) was guessed numerically some years ago by Boyd [5]. Hisguess and some discussions with Zagier [17] were probably determinant for thediscovery of the proof. So I am pleased to address my grateful thanks to both ofthem. 2. Some facts
The polynomial Q − belong to the family of polynomials Q k whose Mahlermeasure has been studied in a previous paper [2]. Theorem 2.1.
Consider the family of Laurent polynomials Q k = X + 1 X + Y + 1 Y + Z + 1 Z + XY + 1 XY + ZY + 1 ZY + XY Z + 1
XY Z − k. Let k = − ( t + t ) − and define t = η (3 τ ) η (12 τ ) η (2 τ ) η ( τ ) η (4 τ ) η (6 τ ) , where η denotes the Dedekind eta function η ( τ ) = e πiτ Y n ≥ (1 − e πinτ ) . Then
MAHLER MEASURE OF A K3-HYPERSURFACE EXPRESSED AS A DIRICHLET L-SERIES3 m ( Q k ) = ℑ τ π { ′ X m,κ (2(2 ℜ mτ + κ ) ( m ¯ τ + κ ) + 1( mτ + κ ) ( m ¯ τ + κ ) ) − ℜ mτ + κ ) (2 m ¯ τ + κ ) + 1(2 mτ + κ ) (2 m ¯ τ + κ ) ) − ℜ mτ + κ ) (3 m ¯ τ + κ ) + 1(3 mτ + κ ) (3 m ¯ τ + κ ) )+ 288(2 ℜ mτ + κ ) (6 m ¯ τ + κ ) + 1(6 mτ + κ ) (6 m ¯ τ + κ ) )) } Let us recall now the following results.Given a normalised Hecke eigenform f of some level N and weight k = 3, we canassociate a Galois representation [7], [13] ρ f : G al ( ¯ Q / Q ) → G l (2 , Q l ) . To a normalised Hecke newform f can also be associated an L -function L ( f, s )by L ( f, s ) := L ( ρ f , s )(the L -series of the Galois representation ρ f ). Equivalently, if f has a Fourierexpansion f = P n b n q n , then L ( f, s ) is also the Mellin transform of fL ( f, s ) = X n b n n s . Moreover, the series L ( f, s ) has a product expansion L ( f, s ) = X n ≥ b n n s = Y p − b p p − s + χ ( p ) p k − − s where χ ( p ) = 0 if p | N .Concerning the comparison between l -adic representations, Serre’s then Livn´e’sresult can be found for example in [15], [9]. Lemma 2.2.
Let ρ l , ρ ′ l : G Q → Aut V l two rational l -adic representations withTr F p,ρ l = Tr F p,ρ ′ l for a set of primes p of density one (i.e. for all but finitelymany primes). If ρ l and ρ ′ l fit into two strictly compatible systems, the L -functionsassociated to these systems are the same. Then the great idea (Serre [12] , Livn´e [8]) is to replace this set of primes ofdensity one by a finite set.
Definition 1.
A finite set T of primes is said to be an effective test set for a rationalGalois representation ρ l : G Q → Aut V l if the previous lemma holds with the set ofdensity one replaced by T . MARIE JOS´E BERTIN
Definition 2.
Let P denote the set of primes, S a finite subset of P with r elements, S ′ = S ∪ {− } . Define for each t ∈ P , t = 2 and each s ∈ S ′ the function f s ( t ) := 12 (1 + (cid:16) st (cid:17) )and if T ⊂ P , T ∩ S = ∅ , f : T → ( Z / Z ) r +1 such that f ( t ) = ( f s ( t )) s ∈ S ′ . Theorem 2.3. (Livn´e’s criterion) Let ρ and ρ ′ be two -adic G Q -representationswhich are unramified outside a finite set S of primes, satisfyingTr F p,ρ ≡ Tr F p,ρ ′ ≡ mod and det F p,ρ ≡ det F p,ρ ′ ( mod for all p / ∈ S ∪ { } .Any finite set T of rational primes disjoint from S with f ( T ) = ( Z / Z ) r +1 \{ } is an effective test set for ρ with respect to ρ ′ . The K X defined by the polynomial Q − has been studied by Peters,Top and van der Vlugt [9]. In particular they proved the theorem. Theorem 2.4.
There is a system ρ = ( ρ l ) of 2-dimensional l -adic representationsof G Q = Gal ( ¯ Q / Q ) ρ l : G Q → Aut H trc ( ˜ X, Q l ) . The system ρ = ( ρ l ) has an L -function L ( s, ρ ) = Y p =3 , − A p p − s + (cid:0) p (cid:1) p p − s . This L -function is the L -function of the modular form f + = gθ ∈ S (15 , (cid:0) . (cid:1) ) where θ = X m,n ∈ Z q m + mn +4 n g = η ( z ) η (3 z ) η (5 z ) η (15 z ) and η is the Dedekind eta function. The Mellin transform P b n n s of f + satisfies b p = A p for p = 3 , , where A p can be computed as follows. • If p ≡ or mod. , find an integral solution of the equation x + xy +4 y = p . Then A p = 2 x − y + 2 xy . • If p ≡ or mod. , find an integral solution of the equation x + xy +2 y = p . Then A p = x + 8 xy + y . MAHLER MEASURE OF A K3-HYPERSURFACE EXPRESSED AS A DIRICHLET L-SERIES5 Proof of theorem 1
The proof follows from three propositions.
Proposition 1. m ( Q − ) = 3 √ π ′ X m ′ ,κ (cid:18) k − m ′ ( m ′ + 15 κ ) + − k + 3 m ′ (3 m ′ + 5 κ ) (cid:19) + (cid:18)
12 2 m ′ + 2 m ′ κ − κ ( m ′ + m ′ κ + 4 κ ) + 12 m ′ + 8 m ′ κ + κ (2 m ′ + m ′ κ + 2 κ ) (cid:19) + 6 √ π ′ X m ′ ,κ (cid:18) m ′ + 15 κ ) − m ′ + 5 κ ) (cid:19) + (cid:18) m ′ + m ′ κ + 2 κ ) − m ′ + m ′ κ + 4 κ ) (cid:19) Proof.
Define D jτ = ( mjτ + κ )( mj ¯ τ + κ ) . So m ( Q k ) = ℑ τ π ′ X m,κ [2 ( m ( τ + ¯ τ ) + 2 κ ) D τ + − D τ −
32 (2 m ( τ + ¯ τ ) + 2 κ ) D τ + 32 D τ −
18 (3 m ( τ + ¯ τ ) + 2 κ ) D τ + 18 D τ + 288 (6 m ( τ + ¯ τ ) + 2 κ ) D τ − D τ ]If k = −
3, then τ = − √− and D τ = 124 ( m − mκ + 24 κ ) = 124 ( m ′ + 15 κ ) with m ′ = m − κD τ = 16 ( m − mκ + 6 κ ) = 16 ( m ′ + m ′ κ + 4 κ ) with m ′ = m − κD τ = 18 (3 m − mκ + 8 κ ) = 18 (3 m ′ + 5 κ ) with m ′ = m − κD τ = 12 (3 m − mκ + 2 κ ) = 12 (2 m + mκ + 2 κ ′ ) with κ ′ = κ − m. Thus m ( Q − ) = √ × π ′ X m ′ ,κ ( A + A + A + A ) . Now A can be written A = (24) (cid:18) − m ′ + 15 κ − m ′ κ ( m ′ + 15 κ ) + 2( m ′ + 15 κ ) (cid:19) MARIE JOS´E BERTIN and ′ X m ′ ,κ A = (24) ′ X m ′ ,κ (cid:18) k − m ′ ( m ′ + 15 κ ) + 2( m ′ + 15 κ ) (cid:19) . Then, we get A = (24) (cid:18) m ′ + 16 m ′ κ + 4 κ ( m ′ + m ′ κ + 4 κ ) − m ′ + m ′ κ + 4 κ ) (cid:19) Now with the change of variable κ = κ ′ − m ′ we put the denominators of A symmetric with respect to m ′ and κ ′ . So A = (24) (cid:18) − m ′ + 8 m ′ κ ′ + 4 κ ′ (4 m ′ − m ′ κ ′ + 4 κ ′ ) − m ′ − m ′ κ ′ + 4 κ ) (cid:19) that is A = (24) (cid:18) − m ′ + 16 m ′ κ ′ − κ ′ (4 m ′ − m ′ κ ′ + 4 κ ′ ) − m ′ − m ′ κ ′ + 4 κ ) (cid:19) and coming back to variables m ′ and κ , A = (24) (cid:18)
12 2 m ′ + 2 m ′ κ − κ ( m ′ + m ′ κ + 4 κ ) − m ′ + m ′ κ + 4 κ ) (cid:19) . The same way we obtain, A = (24) (cid:18) m ′ + 30 m ′ κ − κ (3 m ′ + 5 κ ) − m ′ + 5 κ ) (cid:19) or A = (24) (cid:18) m ′ − κ (3 m ′ + 5 κ ) − m ′ + 5 κ ) (cid:19) . Finally using the same tricks as for A , we obtain A = (24) (cid:18) m + 8 mκ ′ + κ ′ (2 m + mκ ′ + 2 κ ′ ) + 2(2 m + mκ ′ + 2 κ ′ ) (cid:19) . (cid:3) From proposition 1. we notice that the Mahler measure is expressed as a sum ofa modular part3 √ π ′ X m ′ ,κ (cid:18) k − m ′ ( m ′ + 15 κ ) + − k + 3 m ′ (3 m ′ + 5 κ ) (cid:19) + (cid:18)
12 2 m ′ + 2 m ′ κ − κ ( m ′ + m ′ κ + 4 κ ) + 12 m ′ + 8 m ′ κ + κ (2 m ′ + m ′ κ + 2 κ ) (cid:19) and a part related to a Dirichlet L -series+ 6 √ π ′ X m ′ ,κ (cid:18) m ′ + 15 κ ) − m ′ + 5 κ ) (cid:19) + (cid:18) m ′ + m ′ κ + 2 κ ) − m ′ + m ′ κ + 4 κ ) (cid:19) . To prove that the modular part is 0, we observe first that
MAHLER MEASURE OF A K3-HYPERSURFACE EXPRESSED AS A DIRICHLET L-SERIES7 L ( f , s ) = 12 ′ X r,s r − k (3 r + 5 k ) s and L ( f , s ) = 12 ′ X r,s r − k ( r + 15 k ) s are the Mellin transform of the two weight 3 modular forms f = 12 X r,s ∈ Z (5 r − k ) q r +5 k f = 12 X r,s ∈ Z ( r − k ) q r +15 k . Then using theorem 2 . ′ X (cid:18)
14 2 m ′ + 2 m ′ κ − κ ( m ′ + m ′ κ + 4 κ ) s + 14 m + 8 mκ ′ + κ ′ (2 m + mκ ′ + 2 κ ′ ) s (cid:19) = L ( f + , s )is the L -series attached to the modular K X . Proposition 2. ′ X m,k (cid:18) − k + m ( m + 15 k ) + 5 k − m (3 m + 5 k ) (cid:19) = ′ X m,k (cid:18)
12 2 m + 2 mk − k ( m + mk + 4 k ) + 12 m + 8 mk + k (2 m + mk + 2 k ) (cid:19) . Proof.
Let a a rational integer and denote θ a = P n ∈ Z q an the weight 1 / (4). Denote f := [ θ , θ ] f := [ θ , θ ]the Rankin-Cohen brackets wich are modular forms of weight 3 for Γ.Recall that, if f and g are modular forms of respective weight k and l for acongruence subgroup, then its Rankin-Cohen bracket is the modular form of weight k + l + 2 defined by [ g, h ] := kgh ′ − lg ′ h. Thus we get the two weight 3 modular forms f = 12 X r,s ∈ Z (5 r − k ) q r +5 k f = 12 X r,s ∈ Z ( r − k ) q r +15 k . So to compare L ( f , s ) + L ( f , s ) = P A ( n ) n s and L ( f + , s ) = P A ( n ) n s we applyLivn´e’s criterion.First we determine an effective test set T for the respective representations T = { , , , , , , , , , , , , , , } . Then we compute the corresponding A ( p ) and A ( p ).p 7 11 13 17 19 23 29 31 41 43 53 61 71 73 83 A ( p ) 0 0 0 -14 -22 34 0 2 0 0 -86 -118 0 0 154 A ( p ) 0 0 0 -14 -22 34 0 2 0 0 -86 -118 0 0 154This achieves the proof of the proposition. (cid:3) MARIE JOS´E BERTIN
Proposition 3. √ π ′ X m,k m + 15 k ) − m + 5 k ) + 1(2 m + mk + 2 k ) − m + mk + 4 k ) = 85 d Proof.
We denote L f ( s ) := L ( χ f , s )the Dirichlet’s L -series for the character χ f attached to the quadratic field Q ( √ f ).The proof follows from a lemma. Lemma 3.1. (1) ′ X m,k (cid:18) m + mk + k ) s + 1 m + mk + 4 k ) s (cid:19) = 2 ζ ( s ) L − ( s )(2) ′ X m,k (cid:18) m + 5 k ) s + 1( m + 15 k ) s (cid:19) = 2(1 + 12 s − − s − ) ζ ( s ) L − ( s )(3) ′ X m,k (cid:18) m + mk + 4 k ) s − m + mk + 2 k ) s (cid:19) = 2 L − ( s ) L ( s )(4) ′ X m,k (cid:18) m + 15 k ) s − m + 5 k ) s (cid:19) = 2(1 + 12 s − + 12 s − ) L − ( s ) L ( s ) Proof.
The assertion (1) follows from the result [16] ′ X (cid:18) m + mk + k ) s + 1 m + mk + 4 k ) s (cid:19) = ζ Q ( √− ( s )and the formula ζ Q ( √− ( s ) = ζ ( s ) L − ( s ) . The assertion (2) follows from results of K. Williams [14] and Zucker [18]. TakingWilliams’s notations we set φ ( q ) := + ∞ X −∞ q n and get φ ( q ) φ ( q ) + φ ( q ) φ ( q ) = 2 + X n ≥ a n ( − q n − q n , MAHLER MEASURE OF A K3-HYPERSURFACE EXPRESSED AS A DIRICHLET L-SERIES9 where a n ( −
60) = n ≡ , , , , , , (mod. 60)2 if n ≡ , , , , , , , , , , , , , , , − n ≡ , , , , , , , , , , , , , , , . As explained in [18], often we may get Q ( a, b, c ; s ) = ′ X am + bmn + cn ) s in terms of L ± h when expressing them as Mellin transforms of products of variousJacobi functions θ ( q ) for different arguments.More precisely, Q (1 , , λ ; s ) = 1Γ( s ) Z ∞ t s − ′ X e − ( m t + λn t ) dt = 1Γ( s ) Z ∞ ( θ ( q ) θ ( q λ ) − dt where e − t = q and θ ( q ) = 1 + 2 q + 2 q + 2 q + . . . ;thus writing θ ( q ) θ ( q λ ) − P n ≥ a n q n − q n , very often theintegral is given in terms of L -series.So we get Q (1 , , s ) + Q (3 , , s ) = 1Γ( s ) Z ∞ t s − ( θ ( q ) θ ( q ) + θ ( q ) θ ( q ) − dt = 1Γ( s ) Z + ∞ t s − ( X n ≥ a n ( − e − tn − e − tn ) dt. Since Γ( s ) = Z + ∞ e − y y s − dy making the change variable nt = y , it follows1Γ( s ) Z + ∞ t s − e − tn − e − tn dt = Z + ∞ (cid:16) yn (cid:17) s − e − y − e − y dyn = 1Γ( s ) 1 n s Z + ∞ y s − e y − dy = 1 n s ζ ( s ) . Thus Q (1 , , s ) + Q (3 , , s ) = ζ ( s ) X n ≥ a n ( −
60) 1 n s . But L − ( s ) = 11 s − s − s − s + 117 s + 119 s + 123 s + 131 s − s − s − s + 147 s + 149 s + 153 s − s + . . . (mod. 60)and L − ( s ) = 11 s + 12 s + 14 s − s + 18 s − s − s − s + 116 s + 117 s + 119 s − s + . . . (mod. 15) . So,12 X n ≥ a n ( −
60) 1 n s = L − ( s ) + 12 s ( − s + 14 s + 17 s + 18 s + 111 s + 113 s − s + 116 s − s − s − s − s − s − s + 129 s + . . . ) (mod. 30) . Let us define L − ( s ) := X n ≥ χ − ( n ) n s = L + ( s ) + L − ( s )where L + ( s ) = X n ≥ , n pair χ − ( n ) n s L − ( s ) = X n ≥ , n impair χ − ( n ) n s . Obviously, L + ( s ) = 12 s L − ( s ) , L − ( s ) = L − ( s ) , L − ( s ) = L − ( s ) + 12 s L − ( s ) . Thus, 12 X n ≥ a n ( − n s = L − ( s ) + 12 s ( L + ( s ) − L − ( s ))= (1 + 12 s − − s − ) L − ( s ) . From this last equality we deduce the formula (2).From [19] we get Q (1 , , s ) = ζ ( s ) L − ( s ) + L − ( s ) L ( s )so from formula (1) we obtain the formula (3).Equality (4) derives from a formula by Zucker and Robertson [19] giving Q (1 , , s ) = (1 − s − + 12 s − ) ζ ( s ) L − ( s )+ (1 + 12 s − + 12 s − ) L − ( s ) L ( s ) . So, thanks to formula (2)
MAHLER MEASURE OF A K3-HYPERSURFACE EXPRESSED AS A DIRICHLET L-SERIES11 Q (1 , , s ) − Q (3 , , s ) = 2 Q (1 , , s ) − ( Q (1 , , s ) + Q (3 , , s ))= 2(1 + 12 s − + 12 s − ) L − ( s ) L ( s ) (cid:3) By substracting (3) to (4) for s = 2 and using [18] L = 4 π √ , we get the proposition. (cid:3) The proof of theorem 1.1 is just a combination of the three propositions.
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