SELBERG ZETA-FUNCTION ASSOCIATED TO COMPACT RIEMANN SURFACE IS PRIME

. Let Z ( s ) be the Selberg zeta-function associated to a compact Riemann surface. We consider decompositions Z ( s ) = f ( h ( s )), where f and h are meromorphic functions, and show that such decompositions can only be trivial.


Introduction
We continue the investigation of decompositions of the Selberg zeta-function which was started in Garunkštis and Steuding [6]. First we reproduce required definitions. Let s = σ +it be a complex variable and X a compact Riemann surface of genus g ≥ 2 with constant negative curvature −1. The surface X can be written as a quotient Γ\H, where Γ ⊂ PSL(2, R) is a strictly hyperbolic Fuchsian group and H is the upper half-plane of C. Then the Selberg zeta-function associated with X = Γ\H is defined by (see Hejhal [ (1.1) Here {P 0 } is the conjugacy class of a primitive hyperbolic element P 0 of Γ and N (P 0 ) = α 2 if the eigenvalues of P 0 are α and α −1 with |α| > 1.  Liao and Yang [10] showed that the Riemann zeta-function is prime. In [6] the following theorem is proved. Here we complete Theorem A.  The proof of Lemma 1.3 is based on the distribution of zeros of Z(s) − a, a ∈ C, (such zeros are called a-points of Z(s)) and of zeros of Z (s) in the left half-plane of C. These zeros are described below.
For the trivial zeros of Z (s), Theorem 1 from [5] together with the equality Z(s) = Z(s) give the following proposition.
For more about the zeros of the derivative of the Selberg zeta-function see [4,11,12].
For the a-points of Z(s) we will prove the following two statements. Proposition 1.5. Let b > 0 and 1/6 < r < 1/2. Then there exists a negative On the other hand, Proposition 1.4 implies that, for sufficiently large negative n, a neighborhood of n + 1/2 contains a double zero of Z(s) − Z(n + 1/2).
Using Proposition 1.5 and the particular kind of polynomials P (z) = z k + C we can easily demonstrate the main idea of the proof of Lemma 1.3. Indeed, let where C = 0 and h(s) is an entire function. Then all zeros of Z(s) − C are at least of order k. By Proposition 1.5 we see that k = 1 and Lemma 1.3 is true for this particular kind of polynomials. To consider the general case we will need the following consequence of Proposition 1.5. In the last corollary, 1/3 can be replaced by any number r, 1/6 < r < 1/2. Various properties of a-points of Selberg zeta-functions were considered in [2,3].
The next section contains the proofs of Proposition 1.5, Corollary 1.6, and Lemma 1.3.

Proofs
Proof of Proposition 1.5. We have (see [ It is known ([5, Lemma 6]) that, for t ≥ 0 and s not an integer, where the integration is along the straight line segment joining the origin to s − 1/2 if s is not on the real line; if s is on the real line, and not an integer, we define the integral by the requirement of continuity as s is approached from the upper half-plane; furthermore, the branch of the logarithm is chosen such that −π/2 ≤ log 1 + e 2πi(s−1/2) ≤ π/2.
Proof. Let deg P = k ≥ 2. Conversely to the statement of the lemma, suppose that the roots of P (z) − c are simple for all c = 0. Then (P (z) − c) = P (z) has no common roots with P (z) − c for any c = 0. Therefore, for any root z j , j ∈ {1, . . . , k − 1}, of P (z), we have P (z j ) = 0. This is possible only if P (z) = a(z − z 1 ) k and z j = z 1 , for all j ∈ {2, . . . , k − 1}. The contradiction obtained proves the lemma.
Proof of Lemma 1.3. Note that P cannot be a constant polynomial. To obtain a contradiction, assume that Z(s) = P (h(s)) and the polynomial P , deg P = k, has at least two different roots. Then Lemma 2.1 implies the existence of a 1 such that P (a 1 ) = 0 and P (a 1 ) = 0. Therefore we can write where k 1 ≥ 2 and k 1 + · · · + k m = k. In view of Proposition 1.5 there are infinitely many zeros of Z(s) − P (a 1 ) each of which lies at a distance smaller than 1/3 from some negative integer. Thus there are an infinite subset S of these zeros and a j defined by (2.6) such that h(ρ) − a j = 0 for ρ ∈ S. If k j ≥ 2 then the zeros ρ are multiple zeros of Z(s) − P (a 1 ) and this contradicts Proposition 1.5. Hence k j = 1, P (a j ) = 0, and by (2.6) we see that j ≥ 2. Therefore there is a continuous function a : [0, 1] → C, such that a(0) = a j , a(1) = a 1 , and P (a(x)) = 0 for x ∈ [0, 1). (2.7) By Corollary 1.6 there is a continuous function ψ : [0, 1] → C such that ψ(0) ∈ S, Z(ψ(x)) = P (a(x)), (2.8) and, for x ∈ [0, 1] and some large negative integer n, |ψ(x) − n| < 1/3.