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arXiv:1504.00356v4 [math.NT] 18 Aug 2015 LACUNARY RECURRENCES FOR EISENSTEIN SERIES MICHAEL H. MERTENS, LARRY ROLEN Abstract. Using results from the theory of modular forms, we reprove and extend a result of Romik about lacunary recurrence relations for Eisenstein series. 1. Introduction It is a classical result from the theory of elliptic functions that the Eisenstein series ∞ X X X 2k dk−1 e2πiτ , (1.1) Gk (τ ) := (mτ + n)−k = 2ζ(k) 1 − B k n=1 2 (m,n)∈Z \{(0,0)} d|n 1 where k ≥ 2 is an even integer , τ is an element of the complex upper half-plane H, and Bk denotes the kth Bernoulli number, satisfy the following recurrence relation X (1.2) (n − 3)(2n − 1)(2n + 1)G2n = 3 (2p − 1)(2q − 1)G2p G2q . p,q≥2 p+q=n From the analytic properties of the zeta function ω(s), which is basically a special case of a Witten zeta function and is essentially a Dirichlet series generating function of the dimensions of irreducible representations of SU(3), Romik [1] derived a new recurrence relation, given by 2n n (4n + 1)! X 2k−1 1 · (1.3) G6n+2 = G2n+2k G4n−2k+2 . 6n 6n + 1 (2n)!2 2n+2k−1 k=1 The most striking difference between the recurrences (1.2) and (1.3) is that in (1.3), only about a third of the previous Eisenstein series are needed, while in (1.2), all Eisenstein series occur. In the end of [1], Romik asked for a direct proof of (1.3) using the theory of modular forms. Here, we provide such a proof, and in particular show that Romik’s example is a natural, and especially symmetric, instance of general relations among products of two The second author thanks the University of Cologne and the DFG for their generous support via the University of Cologne postdoc grant DFG Grant D-72133-G-403-151001011, funded under the Institutional Strategy of the University of Cologne within the German Excellence Initiative. 1Note that for k = 2, we have to fix a certain order of summation to ensure convergence of the defining double series and that the Fourier expansion given is still valid. 1 2 MICHAEL H. MERTENS, LARRY ROLEN Eisenstein series which have been classified in [2] and which are closely related to the theory of period polynomials (see [3, 4]). In particular, we show the following. Theorem 1.1. For all n ∈ N, (1.3) holds. As further examples of such identities, we record two additional lacunary recurrences, i.e. recurrences which use relatively few terms, for the Eisenstein series G6n and G6n+4 , which we give in the following two theorems. Theorem 1.2. For all n ≥ 2, we have 6n + 1 G6n 2n n X 2n + 2k − 1 4n − 2k − 1 2n + 2k − 1 4n − 2k − 1 = +2 G2n+2k G4n−2k . 2n 2n 2n 2n − 2 k=1 Theorem 1.3. For all n ∈ N, we have 6n + 3 6n + 3 +2 G6n+4 2n + 2 2n n+1 X 2n + 2k − 14n − 2k + 3 2n + 2k − 1 4n − 2k + 3 = +2 G2n+2k G4n−2k+4 . 2n 2n 2n 2n + 2 k=1 As an immediate consequence, by considering only the constant terms in the Eisenstein series above, one recovers several of the lacunary recurrences for Bernoulli numbers which were systematically studied by Agoh and Dilcher [5, 6]. As examples of the new recurrences proven, we offer the following special cases of Theorems 1.2 and 1.3. 11G10 = 5G4 G6 , 143G12 = 42G4 G8 + 25G26 , 221G16 = 60G6 G10 + 49G28 , 323G18 = 55G6 G12 + 105G8 G10 , 7429G22 = 1001G8 G14 + 2706G10 G12 , 2185G24 = 182G8 G16 + 546G10 G14 + 363G212 . Acknowledgements. The authors would like to thank Ken Ono for suggesting this project as well as Kathrin Bringmann, Steven J. Miller, Dan Romik, and the anonymous referee for helpful comments. LACUNARY RECURRENCES FOR EISENSTEIN SERIES 3 2. Linear relations among Eisenstein series All linear relations among products of two Eisenstein series Gi Gj and the weight i + j Eisenstein series have been classified in Theorem 1 of [2], which we recall here. Their proof relies on partial fraction decompositions and extends previous work of Zagier [4] and Popa [3]. To explain their results, define for integers r, s ≥ 2 the function Pr,s := Gr Gs + δ2,s ′ δ2,r ′ Gs + G , s r r 1 d where the ′ denotes the renormalized derivative 2πi dτ , δi,j is the usual Kronecker delta symbol, and we set Gr := 0 if r is odd. With this, we can state Theorem 1 in [2] as follows. Theorem 2.1. Let r, s, t ≥ 1 be integers such that k := r + s + t − 1 ≥ 4. Then we have (2.1) X i − 1j − 1 0= (−1)i+r (Pi,j − (−1)j Gk ) t−1 s−1 i+j=k X j − 1h − 1 + (−1)j+s (Ph,j − (−1)h Gk ) r−1 t−1 j+h=k X h − 1 i − 1 + (−1)h+t (Ph,i − (−1)i Gk ). s−1 r−1 h+i=k All linear relations among Gk and P2j,k−2j , j = 1, ..., ⌊ k4 ⌋ are of the form (2.1). In particular, the relations in (1.3) must arise as specializations of the previous theorem. Indeed, by setting r = s = t = 2n + 1 for n ∈ N, hence k = 6n + 2, we obtain the following immediate consequence. Corollary 2.2. The following identity holds for all n ∈ N, 2n+1 X 4n − k + 1 G6n+2 = 2n k=1 n X 2n + 2k − 1 4n − 2k + 1 G2n+2k G4n−2k+2 . 2n 2n 2n + k − 1 2n k=1 Proof. By our choice of parameters, all the sums on (2.1) are equal. Furthermore, the 6n−i+1 occuring in the sum is zero unless we have 2n + 1 ≤ i ≤ 4n + 1. Now product i−1 2n 2n we make the index shift i 7→ k + 2n and bring the terms involving only Gk to the left-hand 4 MICHAEL H. MERTENS, LARRY ROLEN side to obtain 2n+1 X 4n − k + 1 G6n+2 = 2n k=1 2n+1 X 2n + k − 14n − k + 1 (−1)k P2n+k,4n−k+2. 2n 2n 2n + k − 1 2n k=1 Since by definition Pr,s is identically zero if one of r and s is odd, we obtain the desired result. Theorem 1.1 is a consequence of the following identity for binomial coefficients. Lemma 2.3. Let n ≥ 1 and define B(n) := 2n+1 X j=1 2n + j − 1 4n − j + 1 . 2n 2n Then we have that 2n + 2k − 1 4n − 2k + 1 (4n + 1)! 1 1 · = B(n) 6n + 1 (2n)!2 2n 2n for all 1 ≤ k ≤ n. 2n 2k−1 6n 2n+2k−1 Proof. After simplifying the claim by canceling out terms in the binomial expressions, we find that the claim is equivalent to the identity B(n) = 6n+1 2n . We use the following identity due to Hagen and Rothe (see [7]), which is valid for a, b, c, k ∈ N whenever the denominator doesn’t vanish: k X a + bj c − bj a+c a = . (2.2) (a + bj) j k−j k j=0 Setting k = 2n, a = 2n + 1, b = 1, c = 4n, we find X X 2n 2n 2n + 1 + j 6n + 1 4n − j (2n + 1) 2n + j 4n − j = = , (2n + j + 1) j 2n 2n − j 2n 2n j=0 which is equivalent to the claim after a shift in j. j=0 Remark: The identity in Lemma 2.2 of [1], shown basically by the celebrated Wilf-Zeilberger method, is equivalent to the one shown here with more elementary methods. In the above proof, we note that the Eisenstein series of weight 6n + 2 are particularly special as these correspond to a choice of parameters in Theorem 2.1 where r = s = t. To illustrate the power of Theorem 2.1, here we offer similar recurrences for the other residue classes modulo 6. In order to obtain recurrences which are “as lacunary as possible”, one LACUNARY RECURRENCES FOR EISENSTEIN SERIES 5 easily sees from (2.1), that for a given weight k, one has to choose r, s, t all odd and as large as possible, since then most of the binomial coefficients in the sums will vanish. Proof of Theorem 1.2. Choosing (r, s, t) = (2n − 1, 2n + 1, 2n + 1), and hence k = 6n (for n ≥ 2), in (2.1), we obtain 0= + 6n X i − 1 6n − i − 1 i=1 6n X j=1 + 2n j−1 2n − 2 2n (−1)i (Pi,6n−i − (−1)i G6n ) 6n − j − 1 (−1)j (Pj,6n−j − (−1)j G6n ) 2n 6n X h − 1 6n − h − 1 2n h=1 2n − 2 (−1)h (Ph,6n−h − (−1)h G6n ). By reversing the order of summation in the third of these sums, we see that it equals the second one. If we now omit the terms where the binomial coefficients vanish (i.e. those where i ≤ 2n or i ≥ 4n resp. j ≤ 2n − 2 or j ≥ 4n) and shift the summation as in the proof of Corollary 2.2, we obtain the identity 2n+1 2n−1 X 2n + j − 14n − j − 1 X 2n + j − 34n − j + 1 +2 G 6n 2n 2n 2n − 2 2n j=1 = n−1 X k=1 j=1 2n + 2k − 1 4n − 2k − 1 G2n+2k G4n−2k 2n 2n n X 2n + 2k − 1 4n − 2k − 1 +2 G2n+2k G4n−2k . 2n 2n − 2 k=1 The sums on the left hand side can be simplified using (2.2) and by straightforward manipulation of binomial coefficients: For the first sum we can choose a = 2n + 1, b = 1, c = 4n − 2, k = 2n − 2, so that, after an index shift, we have by (2.2) 2n−1 X j=1 2n + j − 1 2n 4n − j − 1 6n − 1 = . 2n 2n − 2 For the second sum we choose a = 2n − 1, b = 1, c = 4n, k = 2n, which gives the identity 2n+1 X 2n + j − 34n − j + 1 6n − 1 = . 2n − 2 2n 2n j=1 6 MICHAEL H. MERTENS, LARRY ROLEN Now we compute directly 6n − 1 6n − 1 (6n − 1)!(2n(2n − 1) + 2 · 4n(4n + 1)) +2 = (2n)!(4n + 1)! 2n − 2 2n (6n − 1)!(36n2 + 6n) 6n + 1 = = . 2n (2n)!(4n + 1)! In order to prove Theorem 1.3, we choose (r, s, t) = (2n + 1, 2n + 1, 2n + 3) (n ≥ 1) in (2.1) and proceed in the same manner as in the proof of Theorem 1.2. Since it would be almost literally the same proof, we omit it here. We just note that in this situation, the factor in front of G6n+4 cannot be simplified to a single binomial coefficient. Competing interests The authors declare that they have no competing interests in the present manuscript. References [1] D. Romik, On the number of n-dimensional representations of SU(3), the Bernoulli numbers, and the Witten zeta function, preprint, http://arxiv.org/abs/1503.03776. [2] M. Hirose, N. Sato, and K. Tasaka, Eisenstein series identities based on partial fraction decomposition, Ramanujan J., DOI 10.1007/s11139-014-9639-7, (2014). [3] A. Popa Rational decomposition of modular forms, Ramanujan J. 26 (2011) no. 3, 419–435. [4] D. Zagier, Values of zeta functions and their applications, in First European Congress of Mathematics, Volume II, Progress in Math. 120 (1994), Birkhäuser-Verlag, Basel, 497–512. [5] T. Agoh and K. Dilcher, Convolution identities and lacunary recurrences for Bernoulli numbers, J. Number Theory 124 (2007), 105–122. [6] T. Agoh and K. Dilcher, Reciprocity relations for Bernoulli numbers, Amer. Math. Monthly 115 (2008), 237–244. [7] W. Chu Elementary proofs for convolution identities of Abel and Hagen-Rothe, Electron. J. Combin. 17 (2010), no. 1, 5 pp. Department of Mathematics and Computer Science, Emory University, 400 Dowman Drive, 30322 Atlanta, GA E-mail address: [email protected] Mathematics Institute University of Cologne, Gyrhofstr. 8b 50931 Cologne E-mail address: [email protected]