Dynamical zeta functions and Kummer congruences

arXiv:math/0309190v1 [math.NT] 11 Sep 2003 DYNAMICAL ZETA FUNCTIONS AND KUMMER CONGRUENCES J. ARIAS DE REYNA Abstract. We establish a connection between the coefficients of Artin-Mazur zeta-functions and Kummer congruences. This allows to settle positively the question of the existence of a map T:X→Xsuch that the number of fixed points of Tnare |E2n|, where E2nare the Euler numbers. Also we solve a problem of Gabcke related to the coefficients of Riemann-Siegel formula. Introduction In this paper we establish a connection between two important topics the Artin- Mazur zeta function and Kummer’s congruences. Some connection between Kummer’s congruences and periodic points are pointed in the paper by Everest, van der Poorten, Puri and Ward [4]. Inspired by the Hasse-Weil zeta function of an algebraic variety over a finite field, Artin and Mazur [2] defined the Artin-Mazur zeta function for an arbitrary map T:X→Xof a topological space X: Z(T;x) := exp ∞ X n=1 Fix Tn nxn!. Where Fix Tnis the number of isolated fixed points of Tn. Manning [11] proved the rationality of the Artin-Mazur zeta function for diffeomorphisms of a smooth compact manifold satisfying Smale axiom A. Following [12], call a sequence a= (an)n≥1of non-negative integers realizable if there is a set Xand a map T:X→Xsuch that anis the number of fixed points of Tn. We must notice that in [12] it is proved that if (an) is realizable, then there exists a compact space Xand a homeomorphism T:X→X, such that an= Fix Tn. Puri and Ward [13] proved that a sequence of non-negative integers (an)n≥1is realizable if and only if Pd|nµ(n/d)adis non negative and divisible by nfor all n≥1. Here µ(n) denotes the well known M¨obius function (see [1]), defined by µ(n) = (−1)kif nis a product of kdifferent prime numbers, and µ(n) = 0 if nis not squarefree. Date: June 26, 2003 . 1991 Mathematics Subject Classification. Primary 11B68, 37C30; Secondary 11B37, 37B10. Key words and phrases. Kummer congruences, Bernoulli numbers, Euler numbers, integer sequences, zeta function. This research was supported by Grant BFM2000-0514. 1 2 J. ARIAS DE REYNA We shall delete the positivity condition, so we shall say that the sequence of integers (an)∞ n=1 is pre-realizable if Pd|nµ(n/d)adis divisible by nfor every natural number. In 1851 Kummer [9] discovered what we call Kummer’s congruences for Bernoulli numbers, (see the book by Nielsen [10]). Carlitz [3] extended these congruences to the generalized Bernoulli numbers of Leopoldt. Some restrictions of Carlitz’s results has been removed by the work of Fresnel [5]. These congruences are important for the definition of the p-adic L-functions. We establish a connection between these concepts that we can formulate as in the following theorem. Theorem 0.1. Let (an)be a sequence that satisfies Kummer congruences for every rational prime, then for every natural number bthe sequence (ab+n)∞ n=1 is prerealizable. This theorem allow us to solve a problem posed by Gabcke [6]. This is connected with the Riemann-Siegel formula. In the investigation of the zeta function of Riemann it is important to compute the values of this function ζ(1/2 + it) at points on the critical line with tvery high. Riemann found a very convenient formula for these computations, yet he does not publish anything about this formula. In 1932 C. L. Siegel was able to recover it from Riemann’s nachlass. Now this formula is known as the Riemann-Siegel formula. To obtain the terms of this formula play a role certain numbers λnthat can be defined by a recurrence relation λ0= 1, (n+ 1)λn+1 = n X k=0 24k+1|E2k+2|λn−k. Here E2ndenotes Euler numbers defined by 1 cosh x= ∞ X n=0 En n!xn. Hence E2n+1 = 0 for n≥0 and E0= 1, E2=−1, E4= 5, E6=−61, E8= 1 385, . . . Gabcke [6] observed that the first six numbers λnare integers and conjectured that this is so for all of them. Gabcke also considers analogous sequences (n) and (µn). Although he does not mention it, the same motivations for his conjecture also supports that these too are integers sequences. We prove all these conjectures. The proof of these assertions was the first motivation of this paper. In [13] Puri and Ward ask if the sequence (|E2n|)n≥1is realizable. As we will see the solution of Gabcke’s problem is connected with this one. We shall prove that in fact it is realizable. Notations: When pis a prime number and man integer we shall put pαkmto indicate that pαis the greatest power of pthat divides m. We indicate this relation also by νp(m) = α. We shall put n⊥mto say that nand mare relatively prime. DYNAMICAL ZETA FUNCTIONS AND KUMMER CONGRUENCES 3 1. Dynamical Zeta Function Theorem 1.1. Let (an)∞ n=1 be a sequence of complex numbers and define the sequence (bn)∞ n=1 by (1.1) nbn=X d|n µ(n/d)ad. Then we have the equality between formal power series ∞ Y n=1 (1 −xn)−bn= exp∞ X n=1 an n. Proof. By the well known M¨obius inversion formula the relation (1.1) is equivalent to (1.2) an=X d|n dbd, therefore we have the following equalities between formal power series log ∞ Y n=1 (1 −xn)−bn=− ∞ X n=1 bnlog(1 −xn) = ∞ X n=1 ∞ X k=1 bn xnk k = ∞ X m=1 xm mX n|m nbn= ∞ X m=1 xm mam. And this is equivalent to the equality we want to prove.  Theorem 1.2. Let (an)∞ n=1 be a sequence of complex numbers and define the sequence (An)∞ n=0 by the recurrence relation A0= 1, (n+ 1)An+1 = n X k=0 An−kak+1, n ≥0. Then we have the equality between formal power series ∞ X n=0 Anxn= exp∞ X n=1 an nxn. Proof. First we have the equality between formal power series ∞ X n=1 nAnxn−1=∞ X n=0 Anxn∞ X n=1 anxn−1 because by the hypothesis the coefficient of xnis equal in both members. Since A0= 1 integrating formally give us log∞ X n=0 Anxn= ∞ X n=1 an nxn. That is equivalent to the relation we wanted to prove.  The following theorem gives various equivalent conditions for a sequence of integers (an)∞ n=1 to be pre-realizable. 4 J. ARIAS DE REYNA Theorem 1.3. Given a sequence (an)n≥1of integers, the following conditions are equivalent: (a) The numbers (bn)n≥1defined by nbn=X d|n µ(n/d)ad are integers for every n∈N. (b) The numbers (An)n≥0defined by A0= 1, (n+ 1)An+1 = n X k=0 An−kak+1, n ≥0 (1.3) are integers for every n≥0. (c) For every prime number pand natural numbers n,αwith p⊥nwe have anpα≡anpα−1(mod pα). Proof. First we prove the equivalence of (a) and (b). (a)=⇒(b). Assume (a). By the definition of the (bn) and Theorem 1.1 we have ∞ Y n=1 (1 −xn)−bn= exp∞ X n=1 an nxn. and by the condition (a) of the theorem the bnare integers. Let (An)∞ n=0 the numbers defined by 1.3, we have to show that they are integers. By Theorem 1.2 these numbers satisfies the relation ∞ X n=0 Anxn= exp∞ X n=1 an nxn. Thus we have ∞ X n=0 Anxn= ∞ Y n=1 (1 −xn)−bn. Expanding this product, since the bnare integers, we get that the Anare also integers. Hence we have that (a) implies (b). (b)=⇒(a). Now, by hypothesis the numbers (An)n≥0are integers. We can determine inductively a unique sequence of integers cnsuch that ∞ X n=0 Anxn= ∞ Y n=1 (1 −xn)−cn. In the first step observe that the coefficients of xin both members must be the same, hence A1=c1. Then observe that (1 −x)c1∞ X n=0 Anxn= 1 + ∞ X n=2 A(2) nxn, where the numbers A(2) nare integers. DYNAMICAL ZETA FUNCTIONS AND KUMMER CONGRUENCES 5 Assume by induction that we have determined integers cj, for j= 1, 2, . . . n−1 such that n−1 Y j=1 (1 −xj)cj∞ X n=0 Anxn= 1 + ∞ X k=n A(n) kxk. Then the A(n) kare integers and we can define cn=A(n) n, that satisfies the induction hypothesis. Now we have ∞ Y j=1 (1 −xj)cj∞ X n=0 Anxn= 1. By the hypothesis and Theorem 1.2 ∞ X n=0 Anxn= exp ∞ X n=1 an nxn!. Therefore ∞ Y n=1 (1 −xn)−cn= ∞ X n=0 Anxn= exp ∞ X n=1 an nxn!. Now take logarithms in both members to obtain ∞ X n=1 cn ∞ X k=1 xkn k= ∞ X n=1 an nxn. Reasoning as in the proof of Theorem 1.1 we get am=X n|m ncn. Therefore by the M¨obius inversion formula cn=bnthe numbers defined on condition (a), and by construction these numbers cnare integers. Thus we have proved (a). (a)=⇒(c). We know that condition (a) is equivalent to the existence of integers bnthat satisfy the equation (1.2). Assume that pis a prime number and nand αnatural numbers such that p⊥n. Then anpα=X d|npα dbd= α X k=0 X d|n dpkbdpk. Analogously anpα−1= α−1 X k=0 X d|n dpkbdpk. Therefore anpα−anpα−1=X d|n dpαbdpα≡0 (mod pα). (c)=⇒(a). Let nbe an integer. We have to show that X d|n µ(n/d)ad is divisible by n. Let pαkn, with α≥1, then n=pαmwith p⊥m. 6 J. ARIAS DE REYNA Since µ(k)6= 0 only when kis squarefree, we get X d|n µ(n/d)ad=X d|m µ(m/d)adpα−X d|m µ(m/d)adpα−1 =X d|m µ(m/d)adpα−adpα−1≡0 (mod pα). The sum is divisible for every primary divisor of n, and therefore divisible by n. 2. Kummer congruences In 1851 Kummer [9] proved the following theorem: Theorem 2.1 (Kummer).Let pbe a prime number. Assume that ∞ X n=0 an xn n!= ∞ X k=0 ck(ebx −eax)k, where a,band the ckare integral (mod p). Then the anare integers (mod p) and for e≥1,n≥1,m≥0, and pe−1(p−1) |wwe have (2.1) n X s=0 (−1)sn sam+sw ≡0 (mod (pm, pne)). The congruences (2.1) are usually called Kummer congruences. We shall say that the sequence (an) satisfies Kummer congruences if we have (2.1) for every prime number p. By Kummer theorem these sequences exist, but we are interested in some particular sequences. Theorem 2.2. The sequence (E2n)∞ n=1 satisfies Kummer congruences. Proof. Since 1 cosh x=2 ex+e−x=2 2 + (ex/2−e−x/2)2, Kummer theorem proves that (En)∞ n=1 satisfies Kummer congruence (2.1) for every odd prime number p. Therefore the sequence (E2n)∞ n=1 satisfies these congruences for every odd prime number p. This reasoning can be found in Kummer [9]. The above procedure does not give the case p= 2, but Fresnel [5] has extended Kummer congruences for Euler number, as we see in the following lines. Let χthe function from Zto {−1,0,1}, such that χ(n) = 0 if nis even, χ(n) = 1 if n≡1 (mod 4) and χ(n) = −1 if n≡3 (mod 4), then the generalized Bernoulli numbers associated to this character, —see [5] for details— are related to Euler numbers as Bn(χ) n=−En−1 2. In [5] p. 319 we found that, when 2ekw, with e≥1 n X s=0 (−1)sn sBm+sw(χ) m+sw ≡0 (mod (2n(e+2),2m−1)). With a change of notation this is equivalent to n X s=0 (−1)sn sE2m+sw 2≡0 (mod (2n(e+2),22m)). DYNAMICAL ZETA FUNCTIONS AND KUMMER CONGRUENCES 7 Obviously this implies that for 2e−1|wwe have n X s=0 (−1)sn sE2(m+sw)≡0 (mod (2ne,2m)).  The above theorem is a model of many more interesting examples. In Carlitz [3], it is proved that if χis a primitive character modf, and fis divisible by at least two distinct rational primes, then Bn(χ)/n is an algebraic integer and n X s=0 (−1)sn sBn+1+sw(χ) n+ 1 + sw ≡0 (mod (pn, pen)), if pe−1(p−1) |w. Thus the sequence (an)∞ n=1, with an=Bn+1(χ)/(n+ 1) satisfies Kummer congruences, if the character χis real. When χis complex the sequence defined by an= Tr(Bn+1(χ)/(n+ 1)) satisfies Kummer congruences. The sequences that satisfies Kummer congruences are pre-realizable, as we will see in the following theorem. Theorem 2.3. Let (an)∞ n=1 a sequence that satisfies Kummer congruences. Then ab+npα≡ab+npα−1(mod pα), for every natural numbers b,n,αand prime number psuch that p⊥n. That is to say that if (an)satisfies Kummer congruences then for every natural number b, the sequence (ab+n)∞ n=1 is pre-realizable. Proof. By (2.1), with n= 1 we have am+pe−1(p−1) ≡am(mod (pm, pe)). Therefore, for every natural number k, and assuming m≥e am+kpe−1(p−1) ≡am(mod pe). Now take m=b+npα−1,k=nand e=α. If b+npα−1≥α, we get ab+npα−1+npα−1(p−1) ≡ab+npα−1(mod pα). Since pα−1≥αfor pprime and α≥1, the condition is satisfied and we get ab+npα≡ab+npα−1(mod pα).  3. Euler numbers as numbers of fixed points We are now in position to solve the problem posed by Puri and Ward in [13], they ask if the sequence (|E2n|)∞ n=1 is realizable. We shall show that this is true. Theorem 3.1. There exists a map T:X→X, such that |E2n|= Fix Tn. 8 J. ARIAS DE REYNA Proof. First we show that |E2n|is a pre-realizable sequence. By Theorem 2.2 the sequence (E2n)∞ n=1 satisfies Kummer congruences. Thus by Theorem 2.3, for p⊥m, E2mpα≡E2mpα−1(mod pα). Therefore, |E2mpα| ≡ |E2mpα−1|(mod pα). By Theorem 1.3 it follows that the numbers bn, defined by nbn=X d|n µ(n/d)|E2d|, are integers. Now we must show that the numbers bnare non negative. To this end we observe that nbn≥ |E2n| − n/2 X d=1 |E2d|. Now we apply the well known formula 1≤π 22d+1 |E2d| (2d)! = 2 ∞ X k=0 (−1)k (2k+ 1)2d+1 ≤2. Thus nbn≥(2n)!2 π2n+1 −2 n/2 X d=1 (2d)!2 π2d+1 ≥(2n)!n2 π2n+1 −2(n)! (2n)! ∞ X d=12 π2d+1o. We can compute the last sum and we get 0.433 ..., therefore nbn≥(2n)!n2 π2n+1 −(n)! (2n)!o. This is positive for n≥2, and we have b1= 1 ≥0.  The first values of these three sequences in this case are the following: an1 5 61 1385 50521 2702765 199360981 . . . bn1 2 20 345 10104 450450 28480140 . . . An1 3 23 371 10515 461869 28969177 2454072147 . . . 4. Solution of Gabcke problem By Theorem 1.3 the assertion of Gabcke, —that the numbers λnare integers—, is equivalent to say that the sequence ak= 24k−3|E2k|is pre-realizable. We shall show that in fact it is realizable. If (an) and (a′ n) are realizable, then the sequence (ana′ n) is also realizable. In fact given T:X→Xand T′:Y→Ysuch that an= Fix Tnand a′ n= Fix T′n, then it is easy to see that T×T′:X×Y→X×Ysatisfies ana′ n= Fix(T×T′). Therefore by Theorem 3.1 what we need is to prove that the sequence 24n−3is realizable. This follows from the following theorem. Theorem 4.1. Let aand b∈Nsuch that b|aand for every prime number p|a p|(a/b). Then the sequence an/b is realizable. DYNAMICAL ZETA FUNCTIONS AND KUMMER CONGRUENCES 9 Proof. By the result of Puri and Ward we must show that the sequence an/b is pre-realizable and that the corresponding bnare non-negative integers. First let p⊥abe a prime number, and let n⊥pand αbe natural numbers. We must show that anpα/b ≡anpα−1/b (mod pα). Since b⊥p, this is equivalent to anpα≡anpα−1(mod pα). Now for α= 1 this is Fermat’s little theorem, and for a general αit follows, by induction, from the fact that for α≥1 if a≡bmod pα, then ap≡bpmod pα+1. Now if p|a, assume that prkaand pskb. By hypothesis we have r≥s+ 1. We have to show that anpα/b ≡anpα−1/b (mod pα), where p⊥nand α≥1. But the two numbers are divisible by prnpα−1−s. All we have to show is that rnpα−1≥s+α. We can assume that n= 1. For α= 1 this is r≥s+ 1 that is true by hypothesis. For other values of α,α≥2 and we have rpα−1=r(pα−1−1) + r≥(α−1) + (s+ 1). Now we define the numbers bnby nbn=X d|n µ(n/d)ad/b. By the previous reasoning we know the bnare integers. If a= 1, it is easy to see that b1= 1 and bn= 0 for n > 1. In other case a≥2 and we have nbn≥1 ban− n/2 X d=1 ad. This is easyly seen to be non-negative.  Corollary 4.2. The sequence (an)∞ n=1, where an= 24n−3is realizable. The three sequences associated to this realizable sequence are an2 32 512 8192 131072 2097152 33554432 536870912 . . . bn2 15 170 2040 26214 349435 4793490 67107840 . . . An1 2 18 204 2550 33660 460020 6440280 91773990 . . . Now we are in position to prove Gabcke’s conjecture. Theorem 4.3. Let λnthe numbers defined by λ0= 1, (n+ 1)λn= n X k=0 24k+1|E2k+2|λn−k,(n≥0), (4.1) nthose defined by 0=−1, (n+ 1)n=− n X k=0 24k+1|E2k+2|n−k,(n≥0), (4.2) and finally let µn= (λn+n)/2. All those numbers are integers.