square L-functions at every point on the critical line. Mathematics Subject Classification: 11F67, 11F11. 1. Introduction Let k be a positive even integer and ...
An estimate for symmetric square L-functions on weight aspect By Yuk-Kam Lau Abstract. We establish an mean square estimate on the weight aspect for symmetric square L-functions at every point on the critical line. Mathematics Subject Classification: 11F67, 11F11
1. Introduction Let k be a positive even integer and denote by Bk the set of all normalized Hecke eigenforms for the full modular group. Then Bk is an orthogonal basis of the space of all holomorphic cusp forms of weight k. Suppose that f ∈ Bk P has the Fourier expansion (at the cusp ∞) f (z) = n≥1 λf (n)n(k−1)/2 e(nz) where e(α) = e2πiα . The L-function attached to f is defined as, for 1, L(f, s) = P∞ −s n=1 λf (n)n . Analogously to the Riemann zeta function ζ(s), L(f, s) satisfies a functional equation and has the Euler product L(f, s) =
Y (1 − αp p−s )−1 (1 − βp p−s )−1 p
where αp + βp = λf (p) and αp βp = 1. We define the symmetric square L-function by L(sym2 f, s) =
Y (1 − αp2 p−s )−1 (1 − αp βp p−s )−1 (1 − βp2 p−s )−1 p
= ζ(2s)
∞ X
λf (n2 )n−s .
(1.1)
n=1
The value L(sym2 f, 1) is of particular interest and importance. In fact, it is related to the norm of f (with respect to the Petersson inner product) by L(sym2 f, 1) =
π (4π)k kf k2 2 Γ(k)
(1.2)
where, and in the sequel, Γ(·) denotes the Gamma function. (See [2, p.251].) Using the estimate for kf k in [1], we have k −² ¿ L(sym2 f, 1) ¿ k ² .
1
(1.3)
In [4], Kohnen and Sengupta proved that if one assumes L(sym2 f, 1/2) ≥ 0 for all f ∈ Bk and all k, then for any ² > 0 and for all sufficiently large k, X
L(sym2 f, 1/2) ¿² k 1+² .
(1.4)
f ∈Bk
(The implied constant in ¿∗ depends only on ∗.) In this short note, we shall consider all other points on the critical line and establish unconditionally that (1.4) holds for infinitely many k. We do believe that (1.4) holds unconditionally for all large k but our argument here is insufficient. Our method is an application of a tool in [2] (and in [3] as well). Theorem 1 Let K be a sufficiently large number and ² > 0 be any arbitrarily small constant. Then, for any real number t, X
X
K≤k≤2K k≡0 (2)
f ∈Bk
|L(sym2 f, 1/2 + it)|2 ¿² K 2+² (|t| + 1)3+² .
Theorem 1 yields that (1.4) is true for infinitely many k. More precisely, we have Corollary 2 There exists an even integer K ≤ k ≤ 2K such that X
|L(sym2 f, 1/2 + it)| ¿²,|t| k 1+² .
f ∈Bk
P By Cauchy-Schwarz inequality and f ∈Bk 1 ¿ k, 2 X X |L(sym2 f, 1/2 + it)|2 . |L(sym2 f, 1/2 + it)| ¿ k f ∈Bk
f ∈Bk
Summing both sides over even k with K ≤ k ≤ 2K, we obtain 2 X X |L(sym2 f, 1/2 + it)| ¿ K 3+² . K≤k≤2K k≡0 (2)
f ∈Bk
The corollary follows. In addition, applying the idea in [6], we can obtain the following result. Theorem 3 Let K be a sufficiently large number and ² > 0 be any arbitrarily small constant. We have X K≤k≤2K k≡0 (2)
¯X ζ(2)4 ¯¯ ¯ L(sym2 f, 1) − k ¯ ¿² K 3/2+² . ¯ 4π 2 f ∈Bk
2
Remark. This shows that there exist infinitely many even k such that X
L(sym2 f, 1) =
f ∈Bk
π4 k + O(k 1/2+² ). 864
Analogously to the case for level (in [6]), we expect that it is true for all sufficiently large k.
2. Proof of Theorem 1 Let K ≤ k ≤ 4K be an even integer. Also we take δ > 0 to be any arbitrarily small but fixed constant, and put B = 2 + 1/δ. Define ∆(s) = π −3s/2 Γ(
s+1 s+k−1 s+k )Γ( )Γ( ). 2 2 2
We have ∆(s)L(sym2 f, s) = ∆(1 − s)L(sym2 f, 1 − s), see for example [2, Chapter 13]. As Γ(α)Γ(α + 1/2) = 21−2α π 1/2 Γ(2α), one can obtain by following the argument in [5, Lemma 2.2] that when −B <
