Rigidity of Euler products

arXiv:2103.06464v4 [math.NT] 10 Jul 2021 Rigidity of Euler Products Shin-ya Koyama∗ & Nobushige Kurokawa† July 13, 2021 Abstract We report a simple rigidity theorem for certain Euler products. Key Words: Euler products; Zeta functions AMS Subject Classifications: 11M06, 11M41 Introduction For purely imaginary numbers a, b, c ∈ iR we study the meromorphy of the associated Euler product Y (1 − (pa + pb )p−s + pc−2s )−1 Z abc (s) = p: prime in the family of Euler products Zab = {Z abc (s) | c ∈ iR}. This family Zab contains ζ(s − a)ζ(s − b) = Z ab(a+b) (s), which is a meromorphic function in all s ∈ C (with a functional equation under s ←→ 1 + a + b − s). We prove the converse: Theorem A. If Z abc (s) is meromorphic in all s ∈ C, we have a + b = c and Z abc (s) = ζ(s − a)ζ(s − b). ∗ Department of Biomedical Engineering, Toyo University, 2100 Kujirai, Kawagoe, Saitama, 350-8585, Japan. † Department of Mathematics, Tokyo Institute of Technology, Oh-okayama, Meguro-ku, Tokyo, 152-8551, Japan. 1 This shows rigidity of the Euler product ζ(s − a)ζ(s − b) in the family Zab concerning the meromorphy on the entire C. The next result gives a detailed meromorphy for a + b 6= c. Theorem B. If a + b 6= c, then Z abc (s) has an analytic continuation to Re(s) > 0 as a meromorphic function with the natural boundary Re(s) = 0. More precisely, each point on Re(s) = 0 is a limit point of poles of Z abc (s) in Re(s) > 0. We notice generalizations in §4 in the text. Our theorems follow from results of Kurokawa [4, 5, 6] extending results of Estermann [1]. We remark that our result characterizes ζ(s−a)ζ(s−b) by the meromorphy in all s ∈ C only in contrast to usual “converse theorems” originated by Hamburger [2] and Hecke [3] where the functional equation and the attached automorphic form are important; ζ(s − a)ζ(s − b) corresponds to a Maass wave form studied by Maass [7]. 1 Euler datum We use the triple E = (P, R, α), where P is the set of all prime numbers, R denotes the real numbers, and α is the map α : P → R given by α(p) = log p. Such a triple is a simple example of Euler datum studied in [4, 5, 6]; generalized Euler data are treated there with general “primes” P and general topological groups G instead of R. Let R(R) be the virtual character ring of R defined as ( ) X R(R) = m(a)χa m(a) ∈ Z, m(a) = 0 except for finitely many a , a∈iR where χa is a (continuous) unitary character χa : R → U (1) given by χa (x) = eax for x ∈ R. For a polynomial H(T ) = n X hm T m ∈ 1 + T R(R)[T ] m=0 we denote by L(s, E, H) the Euler product Y L(s, E, H) = Hα(p) (p−s )−1 , p∈P 2 where Hx (T ) = n X hm (x)T m ∈ 1 + T C[T ]. m=0 For example, let a, b, c ∈ iR, then the polynomial H abc (T ) = 1 − (χa + χb )T + χc T 2 in 1 + T R(R)[T ] gives the Euler product Y
−1 L(s, E, H abc ) = 1 − (pa + pb )p−s + pc−2s p∈P since abc = 1 − (χa (log p) + χb (log p))p−s + χc (log p)p−2s Hα(p) = 1 − (pa + pb )p−s + pc−2s . 2 Unitariness and meromorphy Let E = (P, R, α) as in §1 and take a polynomial H(T ) in 1 + T R(R)[T ] of degree n. We say that H(T ) is unitary when there exist functions θj ; R → R satisfying Hx (T ) = (1 − eiθ1 (x) T ) · · · (1 − eiθn (x) T ) for all x. The main theorem proved in [5] gives in this particular situation the following result. Theorem 1. (1) If H(T ) is unitary, then L(s, E, H) is meromorphic in all s ∈ C. (2) If H(T ) is not unitary, then L(s, E, H) is meromorphic in Re(s) > 0 with the natural boundary. Moreover, each point on Re(s) = 0 is a limit point of poles of L(s, E, H) in Re(s) > 0. This theorem was proved in [5] (p.45, §8, Theorem1) since our E = (P, R, α) is nothing but E0 (Q/Q) there. 3 Proof of rigidity After looking Theorem 1 recalled in §2 we see that Theorems A and B in Introduction are both derived from the following result: 3 Theorem 2. Let a, b, c ∈ iR and H abc (T ) = 1 − (χa + χb )T + χc T 2 ∈ 1 + T R(R)[T ]. Then the following conditions are equivalent. (1) a + b = c (2) H(T ) is unitary. Proof. (1)=⇒(2): From a + b = c we get H abc (T ) = 1 − (χa + χb )T + χa χb T 2 = (1 − χa T )(1 − χb T ). This gives Hxabc (T ) = (1 − χa (x)T )(1 − χb (x)T ) = (1 − eax T )(1 − ebx T ) for x ∈ R. Hence H abc (T ) is unitary, by |eax | = |ebx | = 1. (2)=⇒(1): Assume that H abc (T ) is unitary, and set Hxabc (T ) = (1 − eiθ1 (x) T )(1 − eiθ2 (x) T ) with θj : R → R. Then comparing with Hxabc (T ) = 1 − (eax + ebx )T + ecx T 2 we obtain eax + ebx = eiθ1 (x) + eiθ2 (x) , (3.1) ecx = ei(θ1 (x)+θ2 (x)) . (3.2) Note that the complex conjugation of (3.1) gives e−ax + e−bx = e−iθ1 (x) + e−iθ2 (x) . Since e−ax + e−bx = e−(a+b)x (eax + ebx ) and e−iθ1 (x) + e−iθ2 (x) = e−i(θ1 (x)+θ2 (x)) (eiθ1 (x) + eiθ2 (x) ) we obtain the equality e−(a+b)x (eax + ebx ) = e−cx (eax + ebx ) 4 by using (3.1) and (3.2). Hence we get (e(a+b−c)x − 1)(eax + ebx ) = 0 for all x ∈ R. Especially e(a+b−c)x − 1 ax (e + ebx ) = 0 x for all x ∈ R \ {0}. Thus letting x → 0 we obtain the desired equality a + b − c = 0. 4 Generalizations From the proof above it would be easy to see that we have generalizations of Theorems A and B by using results of [4, 5, 6]. Hence we notice simple results only. (1) Dedekind case. Let ζF (s) be the Dedekind zeta function of a finite extension field F of the rational number field Q. Let a, b, c ∈ iR and Y (1 − (N (P )a + N (P )b )N (P )−s + N (P )c−2s )−1 , ZFabc (s) = P ∈Specm(OF ) where P runs over the set Specm(OF ) of maximal ideals of the integer ring OF of F . Then we have exactly the same Theorems A and B chracterizing ζF (s − a)ζF (s − b) among ZFabc (s) by using Theorem 1 of [5, §8] for E0 (F/F ). (2) Selberg case Let ζM (s) be the Selberg (or Ruelle) zeta function Y (1 − N (P )−s )−1 ζM (s) = P ∈Prim(M) of a compact Riemann surface M of genus g ≥ 2, where Prim(M ) denotes the prime geodesics on M with N (P ) = exp(length(P )). Let a, b, c ∈ iR and Y abc (1 − (N (P )a + N (P )b )N (P )−s + N (P )c−2s )−1 . ZM (s) = P ∈Prim(M) 5 Then we have the same Theorems A and B characterizing ζM (s − a)ζM (s − b) abc among ZM (s) by using Theorem 9 of [6, p.232]. (3) More parameters It is possible to generalize the situation with more parameters (or representations). For example, let a, b, c, d ∈ iR and Y (1 − (pa + pb + pc )p−s + (pa+b + pb+c + pc+a)p−2s − pd−3s )−1 . Z abcd (s) = p: prime Then we have the following result by a similar proof: Z abcd (s) is meromorphic in all s ∈ C iff a + b + c = d. This result characterizes ζ(s − a)ζ(s − b)ζ(s − c) among Z abcd (s). Moreover, we have the following Theorem C generalizing Theorems A and B. This characterizes ζ(s − a1 ) · · · ζ(s − an ) for a1 , ..., an ∈ iR with n ≥ 2. Theorem C. For n ≥ 2 and a1 , ..., an , b ∈ iR, let Y
−1 . (1 − pa1 −s ) · · · (1 − pan −s ) + (−1)n (pb − pa1 +···+an )p−ns Z(s) = p: prime Then Z(s) has an analytic continuation to all s ∈ C as a meromorphic function if and only if a1 + · · · + an = b that is Z(s) = ζ(s − a1 ) · · · ζ(s − an ). When a1 + · · · + an 6= b, it holds that Z(s) is meromorphic in Re(s) > 0 with the natural boundary Re(s) = 0. Proof. The method is quite similar to the case of n = 2 treated in the proofs of Theorems A and B. We define H(T ) ∈ 1 + T R(R)[T ] by H(T ) = (1 − χa1 T ) · · · (1 − χan T ) + (−1)n (χb − χa1 +···+an )T n = 1 − (χa1 + · · · + χan )T + · · · + (−1)n χb T n . Then it is sufficient to show the equivalence of (1) a1 + · · · + an = b, and (2) H(T ) is unitary. 6 (1)=⇒(2): If a1 + · · · + an = b, then H(T ) = (1 − χa1 T ) · · · (1 − χan T ), which is unitary. (2)=⇒(1): Suppose that H(T ) is unitary. Then, for x ∈ R Hx (T ) = (1 − eiθ1 (x) T ) · · · (1 − eiθn (x) T ) = 1 − (eiθ1 (x) + · · · + eiθn (x) )T + · · · + (−1)n ei(θ1 (x)+···+θn (x)) T n with θj : R → R. By comparing with Hx (T ) = 1 − (ea1 x + · · · + ean x )T + · · · + (−1)n ebx T n we obtain the following identities for all x ∈ R: (4. 1) ea1 x + · · · + ean x = eiθ1 (x) + · · · + eiθn (x) , (4. n − 1) e(a1 +···+an−1 )x + · · · + e(a2 +···+an )x = ei(θ1 (x)+···+θn−1 (x)) + · · · + ei(θ2 (x)+···+θn (x)) , (4. n) ebx = ei(θ1 (x)+···+θn (x)) , where (4. k) indicates the coefficients of T k in both sides for k = 1, n − 1, n. Now, the complex conjugate of (4. 1) gives (α) e−a1 x + · · · + e−an x = e−iθ1 (x) + · · · + e−iθn (x) . Dividing (4. n − 1) by (4. n) we get e(a1 +···+an−1 −b)x + · · · + e(a2 +···+an −b)x = e−iθ1 (x) + · · · + e−iθn (x) that is (β) e(a1 +···+an −b)x (e−a1 x + · · · + e−an x ) = e−iθ1 (x) + · · · + e−iθn (x) . Then (β) − (α) implies (e(a1 +···+an −b)x − 1)(e−a1 x + · · · + e−an x ) = 0 for all x ∈ R. Hence we obtain e(a1 +···+an −b)x − 1 −a1 x (e + · · · + e−an x ) = 0 x for all x ∈ R \ {0}. Thus, letting x → 0 we have a1 + · · · + an = b. References [1] T. Estermann, On certain functions represented by Dirichlet series, Proc. London Math. Soc. (2) 27 (1928), 435-448. 7 [2] H. Hamburger, Über die Riemannsche Funktionalgleichung der ζ-Funktion, Math. Z. 10 (1921), 240-254. [3] E. Hecke, Über dir Bestimmung Dirichletscher Reihen durch ihre Funktionalgleichung, Math. Ann. 112 (1936), 664–699. [4] N. Kurokawa, On the meromorphy of Euler products, Proc. Japan Acad. 54A (1978), 163-166. [5] , On the meromorphy of Euler products (I), Proc. London Math. Soc (3). 53 (1986), 1-47. [6] , On the meromorphy of Euler products (II), Proc. London Math. Soc (3). 53 (1986), 209-236. [7] H. Maass, Über eine neue Art von nichtanalytischen automorphen Funktionen und die Bestimmung Dirichletscher Reihen durch Funktionalgleichungen, Math. Ann. 121 (1949), 141-183. 8