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Transcript
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]