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On the Kuznetsov formula
Journal of Functional Analysis 268 (2015) 869–886
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Journal of Functional Analysis www.elsevier.com/locate/jfa
On the Kuznetsov formula Fernando Chamizo ∗,1 , Dulcinea Raboso Departamento de Matemáticas, Universidad Autónoma de Madrid and ICMAT, 28049 Madrid, Spain
a r t i c l e
i n f o
Article history: Received 16 July 2014 Accepted 10 November 2014 Available online 18 November 2014 Communicated by Alain Connes MSC: 11F72 11F30 30F35 58C40 Keywords: Kuznetsov formula Pretrace formula Fourier coefficients of Maass forms
a b s t r a c t The Kuznetsov formula provides a deep connection between the spectral theory in hyperbolic Riemann surfaces and some exponential sums of arithmetic nature that has been extremely fruitful in modern number theory. Unfortunately the application of the Kuznetsov formula is by no means easy in practice because it involves oscillatory integral transforms with kernels given by special functions in non-standard ranges. In this paper we introduce a new formulation of the Kuznetsov formula that rules out these complications reducing the integral transforms to something almost as simple as a composition of two Fourier transforms. This formulation admits a surprisingly short and clean proof that does not require any knowledge about special functions, solving in this way another of the disadvantages of the classical approach. Moreover the reversed formula becomes more natural and in the negative case it reduces to a direct application of Fourier inversion. We also show that our approach is more convenient in applications and gives some freedom to play with explicit test functions. © 2014 Elsevier Inc. All rights reserved.
* Corresponding author. 1
E-mail addresses: [email protected] (F. Chamizo), [email protected] (D. Raboso). Partially supported by the grant MTM2011-22851 of the Ministerio de Ciencia e Innovación (Spain).
http://dx.doi.org/10.1016/j.jfa.2014.11.002 0022-1236/© 2014 Elsevier Inc. All rights reserved.
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1. Introduction To fix the notation and to best appreciate the significance of our result, we start with a short review of some basic points in the spectral theory of automorphic forms. We follow mainly the notation of [7]. An automorphic kernel K(z, w) on Poincaré’s upper half plane H is defined by K(z, w) =
|z − w|2 , k u(γz, w) with u(z, w) = 4zw
(1)
γ∈Γ
where k : [0, ∞) → C and Γ is a Fuchsian group of the first kind. Here u(z, w) is related to the hyperbolic distance ρ(z, w) induced in H by the hyperbolic metric ds2 = y −2 (dx2 + dy 2 ) through the simple formula cosh ρ(z, w) = 1 + 2u(z, w).
(2)
A key result in the spectral theory of automorphic forms is the so-called pretrace formula that establishes the spectral expansion of the automorphic kernels. For the full modular group Γ = PSL2 (Z) it reads
K(z, w) =
j
1 h(tj )uj (z)uj (w) + 4π
∞ h(t)E(z, 1/2 + it)E(w, 1/2 + it) dt,
(3)
−∞
2 where {uj }∞ j=0 is an orthonormal system in L (PSL2 (Z)\H) of Maass forms (u0 is the constant eigenfunction) spanning the point spectrum of the Laplace–Beltrami operator, with respective eigenvalues λj = 14 + t2j ≥ 0; E(z, s) is the Eisenstein series, the analytic extension in s of
E(z, s) =
(γz)s =
γ∈Γ∞ \Γ
1 2
∞ c,d=−∞ gcd(c,d)=1
ys , |cz + d|2s
and the function h is the Selberg transform of k, more commonly known as the Selberg/Harish-Chandra transform, a kind of hyperbolic analogue of the Fourier transform, which is defined by h(t) =
1 k u(z, i) (z) 2 +it dμ.
(4)
H
Here dμ is the invariant Haar measure corresponding to the hyperbolic metric, i.e., dμ = y −2 dxdy, with x = (z) and y = (z).
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With a simple change of variables [7, §1.8] one can also perform the following three steps ∞ q(v) = v
r , g(r) = 2q sinh 2
k(u) du √ , u−v
2
∞ eirt g(r) dr,
h(t) =
(5)
−∞
that can be inverted with 1 g(r) = 2π
∞ e
irt
h(t) dt,
−∞
∞
√ 1 q(v) = g 2 sinh−1 v , 2
k(u) = u
−q (v) dv √ . π v−u
(6)
For u = 0 this simplifies to [11] 1 k(0) = 4π
∞ th(t) tanh(πt) dt.
(7)
−∞
Once we know that k can be recovered from h, the convergence issues in (1) and (3) are avoided under the following regularity condition on h: RC. For some η ∈ R+ , the function h is holomorphic in the strip |(t)| ≤ 1/2 + η, and satisfies h(t) = O(|t|−2−η ) in it. Two outstanding results in the spectral theory of automorphic forms are the Selberg trace formula and the Kuznetsov formula. The first one establishes a neat relation between eigenvalues and lengths of closed geodesics and has some direct arithmetic interpretations (e.g. [15]). But the Kuznetsov formula [12] has provided the most fruitful interplay between spectral theory and number theory. Symbolically, it reads Snm = Anm
for n, m ∈ Z − {0}
where Snm is the spectral expression j≥1
1 h(tj )νj (n)νj (m) + 4π
∞ h(t)ηt (n)ηt (m) dt,
(8)
−∞
where {νj (n)} are the scaled Fourier coefficients of the Maass cusp forms and {ηt(n)} of the Eisenstein series, with h satisfying RC. On the other hand Anm is the arithmetic expression δmn π
∞ t tanh(πt)h(t) dt + −∞
∞ 1 c=1
c
S(m, n; c)H
4π
|mn| , c
(9)
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where S(m, n; c) are the Kloosterman sums, δmn = 1 if m = n and is zero otherwise, and H is an integral transform given by ∞
th(t) J2it (x) dt if mn > 0 cosh(πt)
(10)
th(t)K2it (x) sinh(πt) dt if mn < 0,
(11)
H(x) = 2i −∞
and 4 H(x) = π
∞ −∞
with Jν and Kν the Bessel functions which have the following integral representation ∞ Kν (x) =
e−x cosh v cosh(νv) dv
(12)
0
and Jν (x) =
1 π
π cos(νθ − x sin θ) dθ −
sin(νπ) π
0
∞ eνθ−x sinh θ dθ.
(13)
0
The sequences {νj (n)}n=0 and {ηt (n)}n=0 are the scaled Fourier coefficients (the scaling is as in [7, §8.1]), in the following sense νj (n) = ρj (n)
π cosh(πtj )
and ηt (n) = τt (n)
π , cosh(πtj )
(14)
where ρj (n) and ηt (n) are the actual Fourier coefficients: 1
1 uj (z)e(−nx) dx = cntj (y)ρj (n),
0
1 E z, + it e(−nx) dx = cnt (y)τt (n), 2
(15)
0
where z = x +iy and cnt (y) = y 1/2 Kit (2π|n|y). In other words, the full Fourier expansion of a cusp form uj is uj (z) =
ρj (n)cntj (y)e(nx),
n=0
and of the Eisenstein series is 1 E z, + it = y 1/2+it + η(t)y 1/2−it + τt (n)cnt (y)e(nx). 2 n=0
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For the sake of simplicity we have only considered the case of the full modular group Γ = PSL2 (Z), as Kuznetsov did himself. In the general case, one simply has to add the contribution of the Eisenstein series corresponding to the different cusps and change the definition of the Kloosterman sums accordingly (see [7] for the general case and [5] for the case Γ0 (q) fully worked out). To our knowledge, there are three models of the proof of the Kuznetsov formula. Firstly the one given in the original paper [12] (see also the independent work [1]) that uses Poincaré series (see Chapter 16 of [8] for a short version). An interesting variation due to Motohashi [13,14] employs some Barnes like integrals (products of Γ functions) as a shortcut. Secondly one can use the automorphic Green’s function, the kernel of the resolvent operator, as in [7]. Finally, there is a proof [4] in the framework of representation theory, revealing the Kuznetsov formula as the “relative trace formula” developed by Jacquet [9]. During the revision of this paper the anonymous referee kindly pointed out the existence of an unpublished proof by Zagier using the pretrace formula (as we do). We have not been able to get the original proof but it seems that it is reproduced in the first section of [10] and a kind of generalization to GL(n) due to X. Li appears in Chapter 11 of [6] (we thank again the anonymous referee for this reference). All of these proofs depend heavily on more or less involved properties of special functions. In principle this is natural because the very statement of the Kuznetsov formula contains Bessel transforms with imaginary orders. The drawback of the presence of these special functions is that in applications one has to deal with cumbersome oscillatory integrals. In this paper we state the Kuznetsov formula in a form that admits a short derivation from the pretrace formula (as in the proof by Zagier) and involves almost no use of special functions. In fact we only use as definitions ∞ Kit (x) =
e 0
−x cosh v
cos(tv) dv
1 and J0 (x) = 2π
2π cos(x cos θ) dθ,
(16)
0
and no prior knowledge about special functions. We only appeal to basic arguments from real analysis. The statement and the proof are completely contained in Section 2. Beyond the simplified statement and proof, we want to emphasize the practical and theoretical advantages of our formulation. It reduces complicated oscillatory integral transforms to something that is close to the composition of two Fourier transforms (we illustrate this point in Section 5), not requiring to study the subtleties of the behavior of the Bessel functions of imaginary orders (for instance the so-called transition range). With our formulation the cases mn > 0 and mn < 0 are covered by the same transform while in the previous works one has to distinguish two Kuznetsov formulas. As we show in Section 3, the reversed formula becomes more natural and simpler. The classical approach, appeals to the Kontorovich–Lebedev inversion formula in the case mn < 0 and in our approach it reduces to Fourier inversion without further considerations. In Section 4 we prove the
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equivalence of our formulation with the classical one. We put special care in using only the integral representations (12) and (13) and arguments involving simple Fourier and complex analysis (residue theorem). In Section 5 we illustrate some of the aforementioned advantages. Our statement replaces the original Bessel transforms by a Hankel transform of a Selberg transform. We show that the latter can be estimated by a Fourier transform. The former is in fact the Fourier transform of a radial function and even disregarding its oscillation, one obtains a highly nontrivial conclusion. As an illustration we deduce without effort a sharpened form of a known result about smoothed averages of Fourier coefficients. We also discuss briefly the choice of explicit test functions. 2. A new form of the Kuznetsov formula As we have mentioned, in our formulation the original Bessel transforms are replaced by a single Hankel transform of a Selberg transform (the Hankel transform has appeared in previous works as an analytic tool to deal with a part of a certain term in the proof of the reversed formula when mn > 0). Theorem 2.1. Under the regularity condition RC, for Snm as in (8) with n and m non-zero integers, we have Snm = 4δmn k(0) +
∞ 1 c=1
c
S(m, n; c)G
4π
|mn| , c
where ∞ G(x) = 4πx
√ k(r)J0 (x r + ) dr
0
with = 0 if mn < 0 and = 1 if mn > 0. We need the following known lemma whose proof reduces to a calculation [7, §5.2]. We include the proof for the sake of completeness. Lemma 2.2. The Fourier coefficient anm of the automorphic kernel (1) as a function of (z) and −(w), i.e., 1 1 anm = 0
K x + (z)i, x + (w)i e −nx + mx dxdx ,
0
is given by ∞ anm = δmn −∞
∞ 1 S(m, n; c)bnm , e(−nx)k u x + (z)i, (w)i dx + c c=1
(17)
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where ∞ ∞ bnm = c
e −nx + mx k u
−∞ −∞
−c−2 , x + (w)i dxdx . x + (z)i
Proof. The Fourier expansion will be derived from the decomposition of Γ into double cosets with respect to the stability group of the single cusp ∞, ∞ c−1
Γ =T∪
T ωd/c T,
c=1 d=0 gcd(c,d)=1
∗ ∗ where T denotes the group of integral translation and ωd/c = c d ∈ Γ is a choice of the representative of the double coset. The terms associated with translations can be handled in a simple way. Write r = (z − w), then by the Poisson summation formula, ∞ ∞ k u(γz, w) = k u(z + n, w) = k u r + (z)i + n, (w)i n=−∞
γ∈T
∞
=
n=−∞
∞ e(nr)
n=−∞
e(−nx)k u x + (z)i, (w)i dx
−∞
and we obtain the first part of the coefficient formula. In all other cases, again by the Poisson summation formula,
k u(γz, w)
γ∈Γ −T
=
∞ c−1
∞
k u ωd/c (z + N ), w − M
N,M =−∞ c=1 d=0 gcd(c,d)=1
=
∞ c−1 ∞ ∞ e n(z) − m(w) e −nx + mx Kcd x, x dxdx ,
∞
n,m=−∞
c=1 d=0 −∞ −∞ gcd(c,d)=1
where Kcd x, x = k u
−c−2 , x − d/c + i(w) x + d/c + i(z)
and d is the inverse of d modulo c. After a translation, we obtain the result. 2
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876
Lemma 2.3. We have ∞ 2 Kit (2πy) dy =
π . 8 cosh(πt)
0
Proof. By (16), the integral equals 1 8π
∞ ∞ −∞ −∞
cos(t(u + v)) dudv. cosh u + cosh v
Now, the change u = π(r + s) and v = π(r − s) separates variables giving known single integrals (for instance by the residue theorem, see Section 4) which yields the result. 2 Proof of Theorem 2.1. Let anm be as in (17) with (z) = y/|n| and = y/|m|, by (w) ∞ the pretrace formula (3), (15), (14) and Lemma 2.3, we have that 8 |mn| 0 anm y −1 dy equals Snm . Then, we must prove ∞ ∞ 4π |mn| dy 1 = 4δmn k(0) + S(m, n; c)G 8 |mn| anm y c c c=1 0
for n, m = 0. The term with the coefficient δmn comes from its counterpart in the formula for anm in Lemma 2.2 since with the change of variables x = 2|n|−1 yv ∞ ∞ 8|n| 0 −∞
∞ ∞ y dxdy y = 16 e(−nx)k u x + i , i e(−2yv)k v 2 dvdy |n| |n| y 0 −∞
which equals 4k(0) by Fourier inversion. We now consider the contribution of bnm in Lemma 2.2. Write λ = |mn|/c, with a simple change of variables, bnm contributes to ∞ 8 |mn| 0 y −1 anm dy as 8 λ
∞ ∞ ∞ 0 −∞ −∞
m n −λ2 x k u , x + iy y −1 e − x + dxdx dy |n| |m| x + iy
π
2 ∞ ∞ = 16λ
0 −∞ −π 2
k s2 +
mn −1 v+v sin θ cos 2πλ 2s cos θ − |mn|
(1 − v 2 )2 dsdvdθ, 4v 2
where the last expression is derived by taking x = λv sin θ,
y = λv cos θ
and x =
2syv − x . v2
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If mn > 0, by the change v = w + π
2 ∞ ∞
877
√ w2 + 1 we can write the above integral as
cos 4πλ s cos θ − w2 + 1 sin θ k s2 + w2 dsdwdθ
−∞ −∞ −π 2
1 = 2
2π ∞ ∞
cos 4πλ s2 + w2 + 1 cos θ k s2 + w2 dsdwdθ.
0 −∞ −∞
A similar analysis applies to the case mn < 0, which leads to 1 2
2π ∞ ∞
cos 4πλ s2 + w2 cos θ k s2 + w2 dsdwdθ.
0 −∞ −∞
Now, it is sufficient to use polar coordinates and the integral representation of J0 in (16) to obtain G(x). 2 3. The reversed Kuznetsov formula It is an elementary fact that the Fourier transform of f ( x2 + y 2 ) is given by ∞ 2 2 2π 0 rf (r)J0 (2πr ξ + η ) dr. Renaming the variables and using that the Fourier transform is involutive for even functions, one gets at once the inversion formula for the Hankel transform ∞ F (x) =
∞ rf (r)J0 (rx) dr
implies
0
f (r) =
xF (x)J0 (rx) dx.
(18)
0
Note that G in Theorem 2.1 can be written as a transform of this kind ∞ G(x) = 8πx
rk r2 − J0 (rx) dr.
The asymmetry of the reversed formula for mn > 0 comes from the fact that the interval [0, 1] is missing for = 1 and we have to compensate for this part. This appears in Kuznetsov’s original approach [12, (6.19)] but is hidden under involved integral transforms. With our formulation it becomes more transparent and natural. In this asymmetric case mn > 0 of the reversed formula, for the purpose of applying Petersson’s formulas, it is convenient to consider Bessel functions of integral orders Jn(t) that, according to (13), are defined as the Fourier coefficients of eit sin(2πθ) ; i.e., 1 eit sin(2πθ) e(−nθ) dθ.
Jn (t) = 0
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To ensure convergence in the reversed formula, one needs some regularity conditions on the test function f , and that f (0) = 0. Here we use strong regularity conditions in our proof for simplicity, although these are certainly stronger than strictly necessary; specifically that f (k) (0) = 0 for all k even and hence f can be extended to a regular odd function on R. Kuznetsov imposes f (0) = 0 but this seems to be unnecessary [7]. In general, regularity matters are less important for applications. Theorem 3.1. Let f ∈ C0∞ (R) be odd and as in Theorem 2.1. Then ∞ 1 c=1
c
S(m, n; c)f
4π
∞ |mn| δmn = Snm − f (t)J0 (t) dt c 2π 0
+
∞ 1 c=1
c
S(m, n; c)V
4π
|mn| , c
where Snm is as in (8), with h the Selberg transform of the function k(r) = ∞ √ 1 8π 0 f (x)J0 (x r + ) dx, and V , which only appears when mn > 0, admits the representations ∞ 1 V (x) = x
rf (t)J0 (rt)J0 (rx) drdt 0
0
and V (x) = 2
∞ j=1
∞ (2j − 1)J2j−1 (x)
t−1 f (t)J2j−1 (t) dt.
0
Remark 3.2. By a formula due to H. Petersson [7], for m, n > 0 √ ∞ (2j − 2)! 1 4π mn j √ S(m, n; c)J2j−1 , alj (m)alj (n) = δmn + 2π(−1) c c (4π mn)2j−1 c=1 l
where alj (n) is the n-th Fourier coefficient of the l-th element of an orthonormal basis (with respect to the Petersson inner product) of the classical cusp forms of weight 2j. Using this formula, one can completely rule out Kloosterman sums in the term containing V . If we think of classical modular forms as harmonic functions (corresponding to the zero eigenvalue of Laplacian), one obtains a truly reversed formula containing Kloosterman sums in one side and spectral information in the other side. In practice the term involving V gives a minor contribution. ∞ √ 1 Proof. Define k(r) = 8π f (x)J0 (x r + ) dx. Recall that k(r) a constant) 0 is (up to the radial component of the Fourier transform of the function f ( x2 + y 2 )/ x2 + y 2 ∈
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√ C0∞ evaluated at r + . Then we have that k is a rapidly decreasing function, and consequently g in (5) is also rapidly decreasing, in fact with exponential decay, and by partial integration we conclude that RC holds true for the corresponding Selberg ∞ transform h. Note that 8πk(0) = 0 f (x)J0 (x) dx for = 1. In any case, by (18) f can be recovered as ∞ f (x) = 4πx
√ k(r)J0 (x r + ) dr,
−
and by Theorem 2.1 it only remains to prove that, with = 1, 0 V (x) = 4πx
√ k(r)J0 (x r + 1) dr
−1
for both expressions of V . The definition of k gives, after a change of variables, 0 4πx
∞ 1
√
k(r)J0 (x r + 1) dr = x
−1
rf (y)J0 (ry)J0 (rx) drdy, 0
0
which is the first expression. The second one would follow from the following Bessel function identity 1 xy
rJ0 (rx)J0 (ry) dr = 2
∞
(2j − 1)J2j−1 (x)J2j−1 (y).
(19)
j=1
0
Indeed, the definition of Jn (t) implies eiy sin θ = Jn (y)e(nθ), and using this Fourier expansion as a generating function, one easily deduces Jn−1 (y) − Jn+1 (y) = 2Jn (y)
and Jn−1 (y) + Jn+1 (y) =
2n Jn (y), y
which leads to the relation Jn−1 (z)Jn−1 (w) − Jn+1 (z)Jn+1 (w) =
2n 2n Jn (w)Jn (z) + Jn (z)Jn (w) w z
and gives a telescoping series when summing over (positive) odd numbers n. Namely zwJ0 (z)J0 (w) = 2
n zJn (w)Jn (z) − wJn (z)Jn (w) .
n odd
Substituting z = ry, w = rx, and integrating against r−1 in the interval [0, 1], we obtain (19). 2
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4. Equivalence with the classic Kuznetsov formula This section is devoted to the proof of the equivalence between the classical Kuznetsov formula and Theorem 2.1. To see this it suffices to prove that H equals G, by the very definition of H we are forced to use some special functions. Theorem 4.1. We have G(x) = H(x), for all x > 0. ∞ Proof. Define the sine transform SF (ξ) = 0 F (x) sin(ξx) dx. By Fourier inversion it is enough to prove SG = SH . It will be convenient to consider the formula d SG f (ξ) f (ξ) = − dξ
∞
dx . G(x) cos xf (ξ) x
(20)
0
If mn < 0, using ∞ 0
⎧ 1 , ⎪ ⎨ β 2 −α2 J0 (xβ) cos(αx) dx = ∞, ⎪ ⎩ 0,
0 < α2 < β 2 , β 2 = α2 , 0 < β 2 < α2 ,
and (5), we can manipulate (20) with f (ξ) = sinh ξ, changing the order of integration to get
SG (sinh ξ) cosh ξ = −4π
d dξ
∞ sinh2 ξ
k(r) r − sinh ξ 2
dr = −4πg (2ξ).
(21)
If mn > 0, we consider two cases. First, if ξ > 1 then f (ξ) = cosh ξ in (20) also leads to −4πg (2ξ), as in (21), and a similar procedure for ξ < 1 and f (ξ) = sin ξ yields d SG (sin ξ) cos ξ = −4π dξ
∞ 0
k(r) dr. r + cos2 ξ
(22)
Once we have simplified SG , we are going to check that coincides with SH . For mn < 0, using the definition (11), 4 SH (sinh ξ) = π
∞
th(t) sinh(πt)
−∞
∞
K2it (x) sin(x sinh ξ) dx dt. 0
By (12), the term inside the parenthesis is equal to
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∞ 2 sinh ξ
π sin(2tξ) cos(2tv) sinh(πt) dv = , cosh(2v) + cosh(2ξ) 2 cosh ξ
881
(23)
0
where the last equality follows after the change w = e2v and using the residue theorem over a keyhole contour. The case mn < 0 is then settled because ∞ SH (sinh ξ) cosh ξ = 2
th(t) sin(2tξ) dt = −4πg (2ξ).
−∞
If mn > 0, the definition (10) gives ∞ SH (η) = −∞
i th(t) cosh(πt)
∞
J2it (x) − J−2it (x) sin(xη) dx dt
0
and by (13), we may write the term inside the parentheses as π ∞
d 2 π cosh(πt) dη
sinh(2tθ) sin(x sin θ) cos(xη) 0
+
4η π
∞ 0
dx dθ x
0
cos(2tθ) sinh(πt) dθ. sinh2 θ + η 2
(24)
If η > 1, the first integral vanishes and writing η = cosh ξ, the second integral is the same as the one appearing in (23). Hence SH (cosh ξ) sinh ξ = −4πg (2ξ). If 0 < η < 1, taking η = sin ξ, (24) equals d 2 cosh(πt) cos ξ dξ
π−ξ
∞
ξ
0
4 sinh(πt) sinh(2tθ) dθ + π
cos(2tθ) dθ. sinh2 θ + sin2 ξ
Now, the first integral is immediate and the second one can be computed following the same steps as in (23). Combining these results we get ∞ SH (sin ξ) cos ξ = − −∞
By (5), the previous integral is
d sinh(2tξ) dt = − th(t) cosh(πt) dξ
∞ h(t) −∞
cosh(2tξ) dt. cosh(πt)
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∞
∞ g(r)
−∞
−∞
cos(rt) cosh(2tξ) dtdr = 8 cosh(πt)
∞
−∞
q(sinh2 r) cos ξ cosh r dr cosh(2r) + cos(2ξ)
∞ ∞ =4 0 w
k(r) cos ξ drdw . (w + cos2 ξ) w(r − w)
Exchanging the order of integration, the inner integral can be computed using residue theory by integration over a dumbbell contour around [0, r], which leads to (22), and this finishes the proof. 2 5. Some estimates and examples Our new statement for the Kuznetsov formula not only shows advantages in its proof, it also eases estimations in practice and the search of explicit examples. The following result shows that the Selberg transform can be estimated as a Fourier transform evaluated in the hyperbolic distance. Lemma 5.1. Let h be a function satisfying the regularity conditions RC and assume ∞ j t h(t)eirt dt Bj (r) for j = 0, 1, where B0 (r) and B1 (r)/ sinh r are non-increasing −∞ functions for r > 0, then k2 (u) B0 (ρ)B1 (ρ)/ sinh ρ, where ρ and u are linked by (2). Proof. By (6), we have q(sinh2 v2 ) B0 (v) and q (sinh2 v2 ) B1 (v)/ sinh v. For ξ > u, write ∞ k(u) = u
−q (v) dv √ = π v−u
∞ ξ + = I0 (ξ) + I1 (ξ). ξ
u
By the second mean value theorem 1 I0 (ξ) √ ξ−u
ξ ξ
B0 (ρ) q (v) dv √ ξ−u
and B1 (ρ) I1 (ξ) sinh ρ
ξ √ u
Choosing ξ = u +
B0 B1
B1 (ρ) dv ξ − u. sinh ρ v−u
sinh ρ the result follows. 2
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For instance, for the Gaussian choice h(t) = e−t /T we can take B0 (r) = T e−T r /4 2 2 and B1 (r) = rT 3 e−T r /4 , getting 2 2 u cosh r − 1 T 2 e−T r /4 , (25) k 2 sinh u 2
2
2 2
which is in fact sharp and plays a role in the large sieve inequalities of [2]. In general, using the Weil bound [8, Corollary 11.12], the relation (7) and J0 (t) (1 + |t|)−1/2 , we deduce at once Proposition 5.2. For Snm as in (8) and with the notation of Lemma 5.1, we have δmn Snm − π
∞ t tanh(πt)h(t) dt |mn|
1/2
∞ τ (c) c=1
−∞
with
∞ I(λ) =
c3/2
1/2
(m, n, c)
mn I c2
B0 (v)B1 (v) sinh v dv, 1 + (|λ| cosh v + λ)1/2
0
where τ (c) is the number of divisors of c and (m, n, c) represents the greatest common divisor. To illustrate its applicability we give an estimate for smoothed sums of Fourier coefficients. Corollary 5.3. Given δ > 0, we have ∞ 2 −t2 /T 2 1 νj (n) e j ηt (n)2 e−t2 /T 2 dt = π −1 T 2 + O |n|1/2+δ T , + 4π j −∞
for any n = 0 and T > 1. This improves (16.56) in [8]. We think that the argument employed there does not match the exponent in (16.55), that can be lowered, but our application of Proposition 5.2 gives anyway a better result. Proof of Corollary 5.3. Choose h(t) = e−t /T and n = m in Proposition 5.2. As in the 2 2 2 2 derivation of (25), we can take B0 (r) = T e−T r /4 and B1 (r) = rT 3 e−T r /4 . Then we have 2 ∞ 1/2 −T 2 v2 +v v e n 2 I 2 T dv T 1/2 min 1, c1/2 n−1/2 c 1 + ev/2 (n/c)1/2 2
0
that gives the expected error term.
2
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Finally, note that tanh(πt) = ±1 + O(e−π|t| ) where the sign of ±1 coincides with that of t and consequently the main term is π −1 T 2 + O(1). 2 Another advantage provided by the new expression is to get some explicit instances of Kuznetsov formula. That is, given k and its Selberg transform, finding the corresponding G function. There are some pairs of k and h having closed formulas, this is employed in [3] to produce some unusual approximate formulas. We use some of these pairs in our statement with special emphasis in giving simple and self-contained proofs. Forexample, taking k(r) = e−μr with μ > 0, the corresponding Selberg transform is μ/2 4e π/μKit (μ/2) and using the Taylor expansion of the cosine in (16) we have ∞
x2n G(x) = 2x (−1) (2n)! n=0
∞
n
e
−μr n
2π cos2n θ dθ.
r dr
0
0
Both integrals are simple, and the resulting series is the Taylor expansion of 2 4πxμ−1 e−x /4μ . If we consider functions of the type √ αβ
k(r) = e− 4 (α + β)2 + 4αβr
(α+β)2 +4αβr
,
where α, β > 0, we have h(t) = Kit (α)Kit (β) and after a change of variables we can write G as πx(α + β) √ 2 αβ
∞ e
−β(α+β)y
J0
x(α + β) 2 √ y − 1 dy. 2 αβ
1
It follows from (16) that 2π 1 J0 γ y 2 − 1 = eiγy cos θ−γ sin θ dθ 2π 0
so, after integrating in y, it is sufficient to apply the residue theorem to obtain G(x) =
πx 4αβ + x2
α+β √ αβ
e− 2
4αβ+x2
.
In other examples, like functions of the type k(r) = (1 + r)−μ with μ > 1, it may be helpful to use the Hankel transform. In this case, the Selberg transform is h(t) =
4π Γ (μ − 1/2 + it)Γ (μ − 1/2 − it). Γ 2 (μ)
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On the other hand, observe that the generalized binomial theorem gives
1 + r2
−μ
=
∞ Γ (μ + n) (−1)n r2n . n!Γ (μ) n=0
Furthermore, by (12) we have ∞
∞ x
μ+2n
Kμ−1 (x) dx = Γ (μ + 2n + 1)
0
cosh((μ − 1)v) dv (cosh v)μ+2n+1
0 μ+2n−1
=2
n!Γ (μ + n).
To see this, just write the hyperbolic cosine as a sum of exponential and apply the change v = ± 12 log 1−y y in the resulting integrals. Substituting this expression into the above, gives the expansion of J0 (or the Taylor series of the cosine as was done in the first example) and finally we obtain
1+r
2 −μ
∞ =
xμ Kμ−1 (x)J0 (xr) dx. 2μ−1 Γ (μ)
0
Now, the inversion formula for the Hankel transform gives G(x) = In fact, if μ − the above as
1 2
πxμ Kμ−1 (x). 2μ−4 Γ (μ)
∈ Z+ we can take the series expansion of the Bessel function to write
μ−3/2 πe−x (μ − 1/2)! (μ − 3/2 + k)! (2x)μ−1/2−k . 8(2μ − 1)! k!(μ − 3/2 − k)! k=0
Other instances of “exact” Hankel transforms can be found in mathematical tables but rarely do they correspond to an explicit function h. Acknowledgments This paper is part of the Ph.D. Thesis of the second author to be defended at Universidad Autónoma de Madrid. We acknowledge the research environment provided by the Mathematics Department and the ICMAT. We are indebted to N. Pitt and A. Ubis for some comments and suggestions. We are also very grateful to the anonymous referee for pointing out the unpublished proof by Zagier and its extension to GL(n).
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References [1] R.W. Bruggeman, Fourier coefficients of cusp forms, Invent. Math. 45 (1) (1978) 1–18. [2] F. Chamizo, The large sieve in Riemann surfaces, Acta Arith. 77 (4) (1996) 303–313. [3] F. Chamizo, D. Raboso, S. Ruiz-Cabello, Exotic approximate identities and Maass forms, Acta Arith. 159 (1) (2013) 27–46. [4] J.W. Cogdell, I. Piatetski-Shapiro, The Arithmetic and Spectral Analysis of Poincaré Series, Perspect. Math., vol. 13, Academic Press Inc., Boston, MA, 1990. [5] J.-M. Deshouillers, H. Iwaniec, Kloosterman sums and Fourier coefficients of cusp forms, Invent. Math. 70 (2) (1982/1983) 219–288, http://dx.doi.org/10.1007/BF01390728. [6] D. Goldfeld, Automorphic Forms and L-Functions for the Group GL(n, R), Cambridge Stud. Adv. Math., vol. 99, Cambridge University Press, Cambridge, 2006, with an appendix by Kevin A. Broughan. [7] H. Iwaniec, Spectral Methods of Automorphic Forms, 2nd edition, Grad. Stud. Math., vol. 53, American Mathematical Society, Providence, RI, 2002. [8] H. Iwaniec, E. Kowalski, Analytic Number Theory, Amer. Math. Soc. Colloq. Publ., vol. 53, American Mathematical Society, Providence, RI, 2004. [9] H. Jacquet, Automorphic spectrum of symmetric spaces, in: Representation Theory and Automorphic Forms, Edinburgh, 1996, in: Proc. Sympos. Pure Math., vol. 61, Amer. Math. Soc., Providence, RI, 1997, pp. 443–455. [10] D. Joyner, On the Kuznetsov–Bruggeman formula for a Hilbert modular surface having one cusp, Math. Z. 203 (1) (1990) 59–104, http://dx.doi.org/10.1007/BF02570723. [11] T. Kubota, Elementary Theory of Eisenstein Series, Kodansha Ltd., Tokyo, 1973. [12] N.V. Kuznecov, The Petersson conjecture for cusp forms of weight zero and the Linnik conjecture. Sums of Kloosterman sums, Mat. Sb. (N.S.) 111(153) (3) (1980) 334–383, 479. [13] Y. Motohashi, On Kuznetsov’s trace formulae, in: Analytic Number Theory, vol. 2, Allerton Park, IL, 1995, in: Progr. Math., vol. 139, Birkhäuser Boston, Boston, MA, 1996, pp. 641–667. [14] Y. Motohashi, Spectral Theory of the Riemann Zeta-Function, Cambridge Tracts in Math., vol. 127, Cambridge University Press, Cambridge, 1997. [15] P. Sarnak, Class numbers of indefinite binary quadratic forms, J. Number Theory 15 (2) (1982) 229–247, http://dx.doi.org/10.1016/0022-314X(82)90028-2.
Contents lists available at ScienceDirect
Journal of Functional Analysis www.elsevier.com/locate/jfa
On the Kuznetsov formula Fernando Chamizo ∗,1 , Dulcinea Raboso Departamento de Matemáticas, Universidad Autónoma de Madrid and ICMAT, 28049 Madrid, Spain
a r t i c l e
i n f o
Article history: Received 16 July 2014 Accepted 10 November 2014 Available online 18 November 2014 Communicated by Alain Connes MSC: 11F72 11F30 30F35 58C40 Keywords: Kuznetsov formula Pretrace formula Fourier coefficients of Maass forms
a b s t r a c t The Kuznetsov formula provides a deep connection between the spectral theory in hyperbolic Riemann surfaces and some exponential sums of arithmetic nature that has been extremely fruitful in modern number theory. Unfortunately the application of the Kuznetsov formula is by no means easy in practice because it involves oscillatory integral transforms with kernels given by special functions in non-standard ranges. In this paper we introduce a new formulation of the Kuznetsov formula that rules out these complications reducing the integral transforms to something almost as simple as a composition of two Fourier transforms. This formulation admits a surprisingly short and clean proof that does not require any knowledge about special functions, solving in this way another of the disadvantages of the classical approach. Moreover the reversed formula becomes more natural and in the negative case it reduces to a direct application of Fourier inversion. We also show that our approach is more convenient in applications and gives some freedom to play with explicit test functions. © 2014 Elsevier Inc. All rights reserved.
* Corresponding author. 1
E-mail addresses: [email protected] (F. Chamizo), [email protected] (D. Raboso). Partially supported by the grant MTM2011-22851 of the Ministerio de Ciencia e Innovación (Spain).
http://dx.doi.org/10.1016/j.jfa.2014.11.002 0022-1236/© 2014 Elsevier Inc. All rights reserved.
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1. Introduction To fix the notation and to best appreciate the significance of our result, we start with a short review of some basic points in the spectral theory of automorphic forms. We follow mainly the notation of [7]. An automorphic kernel K(z, w) on Poincaré’s upper half plane H is defined by K(z, w) =
|z − w|2 , k u(γz, w) with u(z, w) = 4zw
(1)
γ∈Γ
where k : [0, ∞) → C and Γ is a Fuchsian group of the first kind. Here u(z, w) is related to the hyperbolic distance ρ(z, w) induced in H by the hyperbolic metric ds2 = y −2 (dx2 + dy 2 ) through the simple formula cosh ρ(z, w) = 1 + 2u(z, w).
(2)
A key result in the spectral theory of automorphic forms is the so-called pretrace formula that establishes the spectral expansion of the automorphic kernels. For the full modular group Γ = PSL2 (Z) it reads
K(z, w) =
j
1 h(tj )uj (z)uj (w) + 4π
∞ h(t)E(z, 1/2 + it)E(w, 1/2 + it) dt,
(3)
−∞
2 where {uj }∞ j=0 is an orthonormal system in L (PSL2 (Z)\H) of Maass forms (u0 is the constant eigenfunction) spanning the point spectrum of the Laplace–Beltrami operator, with respective eigenvalues λj = 14 + t2j ≥ 0; E(z, s) is the Eisenstein series, the analytic extension in s of
E(z, s) =
(γz)s =
γ∈Γ∞ \Γ
1 2
∞ c,d=−∞ gcd(c,d)=1
ys , |cz + d|2s
and the function h is the Selberg transform of k, more commonly known as the Selberg/Harish-Chandra transform, a kind of hyperbolic analogue of the Fourier transform, which is defined by h(t) =
1 k u(z, i) (z) 2 +it dμ.
(4)
H
Here dμ is the invariant Haar measure corresponding to the hyperbolic metric, i.e., dμ = y −2 dxdy, with x = (z) and y = (z).
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With a simple change of variables [7, §1.8] one can also perform the following three steps ∞ q(v) = v
r , g(r) = 2q sinh 2
k(u) du √ , u−v
2
∞ eirt g(r) dr,
h(t) =
(5)
−∞
that can be inverted with 1 g(r) = 2π
∞ e
irt
h(t) dt,
−∞
∞
√ 1 q(v) = g 2 sinh−1 v , 2
k(u) = u
−q (v) dv √ . π v−u
(6)
For u = 0 this simplifies to [11] 1 k(0) = 4π
∞ th(t) tanh(πt) dt.
(7)
−∞
Once we know that k can be recovered from h, the convergence issues in (1) and (3) are avoided under the following regularity condition on h: RC. For some η ∈ R+ , the function h is holomorphic in the strip |(t)| ≤ 1/2 + η, and satisfies h(t) = O(|t|−2−η ) in it. Two outstanding results in the spectral theory of automorphic forms are the Selberg trace formula and the Kuznetsov formula. The first one establishes a neat relation between eigenvalues and lengths of closed geodesics and has some direct arithmetic interpretations (e.g. [15]). But the Kuznetsov formula [12] has provided the most fruitful interplay between spectral theory and number theory. Symbolically, it reads Snm = Anm
for n, m ∈ Z − {0}
where Snm is the spectral expression j≥1
1 h(tj )νj (n)νj (m) + 4π
∞ h(t)ηt (n)ηt (m) dt,
(8)
−∞
where {νj (n)} are the scaled Fourier coefficients of the Maass cusp forms and {ηt(n)} of the Eisenstein series, with h satisfying RC. On the other hand Anm is the arithmetic expression δmn π
∞ t tanh(πt)h(t) dt + −∞
∞ 1 c=1
c
S(m, n; c)H
4π
|mn| , c
(9)
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where S(m, n; c) are the Kloosterman sums, δmn = 1 if m = n and is zero otherwise, and H is an integral transform given by ∞
th(t) J2it (x) dt if mn > 0 cosh(πt)
(10)
th(t)K2it (x) sinh(πt) dt if mn < 0,
(11)
H(x) = 2i −∞
and 4 H(x) = π
∞ −∞
with Jν and Kν the Bessel functions which have the following integral representation ∞ Kν (x) =
e−x cosh v cosh(νv) dv
(12)
0
and Jν (x) =
1 π
π cos(νθ − x sin θ) dθ −
sin(νπ) π
0
∞ eνθ−x sinh θ dθ.
(13)
0
The sequences {νj (n)}n=0 and {ηt (n)}n=0 are the scaled Fourier coefficients (the scaling is as in [7, §8.1]), in the following sense νj (n) = ρj (n)
π cosh(πtj )
and ηt (n) = τt (n)
π , cosh(πtj )
(14)
where ρj (n) and ηt (n) are the actual Fourier coefficients: 1
1 uj (z)e(−nx) dx = cntj (y)ρj (n),
0
1 E z, + it e(−nx) dx = cnt (y)τt (n), 2
(15)
0
where z = x +iy and cnt (y) = y 1/2 Kit (2π|n|y). In other words, the full Fourier expansion of a cusp form uj is uj (z) =
ρj (n)cntj (y)e(nx),
n=0
and of the Eisenstein series is 1 E z, + it = y 1/2+it + η(t)y 1/2−it + τt (n)cnt (y)e(nx). 2 n=0
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For the sake of simplicity we have only considered the case of the full modular group Γ = PSL2 (Z), as Kuznetsov did himself. In the general case, one simply has to add the contribution of the Eisenstein series corresponding to the different cusps and change the definition of the Kloosterman sums accordingly (see [7] for the general case and [5] for the case Γ0 (q) fully worked out). To our knowledge, there are three models of the proof of the Kuznetsov formula. Firstly the one given in the original paper [12] (see also the independent work [1]) that uses Poincaré series (see Chapter 16 of [8] for a short version). An interesting variation due to Motohashi [13,14] employs some Barnes like integrals (products of Γ functions) as a shortcut. Secondly one can use the automorphic Green’s function, the kernel of the resolvent operator, as in [7]. Finally, there is a proof [4] in the framework of representation theory, revealing the Kuznetsov formula as the “relative trace formula” developed by Jacquet [9]. During the revision of this paper the anonymous referee kindly pointed out the existence of an unpublished proof by Zagier using the pretrace formula (as we do). We have not been able to get the original proof but it seems that it is reproduced in the first section of [10] and a kind of generalization to GL(n) due to X. Li appears in Chapter 11 of [6] (we thank again the anonymous referee for this reference). All of these proofs depend heavily on more or less involved properties of special functions. In principle this is natural because the very statement of the Kuznetsov formula contains Bessel transforms with imaginary orders. The drawback of the presence of these special functions is that in applications one has to deal with cumbersome oscillatory integrals. In this paper we state the Kuznetsov formula in a form that admits a short derivation from the pretrace formula (as in the proof by Zagier) and involves almost no use of special functions. In fact we only use as definitions ∞ Kit (x) =
e 0
−x cosh v
cos(tv) dv
1 and J0 (x) = 2π
2π cos(x cos θ) dθ,
(16)
0
and no prior knowledge about special functions. We only appeal to basic arguments from real analysis. The statement and the proof are completely contained in Section 2. Beyond the simplified statement and proof, we want to emphasize the practical and theoretical advantages of our formulation. It reduces complicated oscillatory integral transforms to something that is close to the composition of two Fourier transforms (we illustrate this point in Section 5), not requiring to study the subtleties of the behavior of the Bessel functions of imaginary orders (for instance the so-called transition range). With our formulation the cases mn > 0 and mn < 0 are covered by the same transform while in the previous works one has to distinguish two Kuznetsov formulas. As we show in Section 3, the reversed formula becomes more natural and simpler. The classical approach, appeals to the Kontorovich–Lebedev inversion formula in the case mn < 0 and in our approach it reduces to Fourier inversion without further considerations. In Section 4 we prove the
F. Chamizo, D. Raboso / Journal of Functional Analysis 268 (2015) 869–886
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equivalence of our formulation with the classical one. We put special care in using only the integral representations (12) and (13) and arguments involving simple Fourier and complex analysis (residue theorem). In Section 5 we illustrate some of the aforementioned advantages. Our statement replaces the original Bessel transforms by a Hankel transform of a Selberg transform. We show that the latter can be estimated by a Fourier transform. The former is in fact the Fourier transform of a radial function and even disregarding its oscillation, one obtains a highly nontrivial conclusion. As an illustration we deduce without effort a sharpened form of a known result about smoothed averages of Fourier coefficients. We also discuss briefly the choice of explicit test functions. 2. A new form of the Kuznetsov formula As we have mentioned, in our formulation the original Bessel transforms are replaced by a single Hankel transform of a Selberg transform (the Hankel transform has appeared in previous works as an analytic tool to deal with a part of a certain term in the proof of the reversed formula when mn > 0). Theorem 2.1. Under the regularity condition RC, for Snm as in (8) with n and m non-zero integers, we have Snm = 4δmn k(0) +
∞ 1 c=1
c
S(m, n; c)G
4π
|mn| , c
where ∞ G(x) = 4πx
√ k(r)J0 (x r + ) dr
0
with = 0 if mn < 0 and = 1 if mn > 0. We need the following known lemma whose proof reduces to a calculation [7, §5.2]. We include the proof for the sake of completeness. Lemma 2.2. The Fourier coefficient anm of the automorphic kernel (1) as a function of (z) and −(w), i.e., 1 1 anm = 0
K x + (z)i, x + (w)i e −nx + mx dxdx ,
0
is given by ∞ anm = δmn −∞
∞ 1 S(m, n; c)bnm , e(−nx)k u x + (z)i, (w)i dx + c c=1
(17)
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875
where ∞ ∞ bnm = c
e −nx + mx k u
−∞ −∞
−c−2 , x + (w)i dxdx . x + (z)i
Proof. The Fourier expansion will be derived from the decomposition of Γ into double cosets with respect to the stability group of the single cusp ∞, ∞ c−1
Γ =T∪
T ωd/c T,
c=1 d=0 gcd(c,d)=1
∗ ∗ where T denotes the group of integral translation and ωd/c = c d ∈ Γ is a choice of the representative of the double coset. The terms associated with translations can be handled in a simple way. Write r = (z − w), then by the Poisson summation formula, ∞ ∞ k u(γz, w) = k u(z + n, w) = k u r + (z)i + n, (w)i n=−∞
γ∈T
∞
=
n=−∞
∞ e(nr)
n=−∞
e(−nx)k u x + (z)i, (w)i dx
−∞
and we obtain the first part of the coefficient formula. In all other cases, again by the Poisson summation formula,
k u(γz, w)
γ∈Γ −T
=
∞ c−1
∞
k u ωd/c (z + N ), w − M
N,M =−∞ c=1 d=0 gcd(c,d)=1
=
∞ c−1 ∞ ∞ e n(z) − m(w) e −nx + mx Kcd x, x dxdx ,
∞
n,m=−∞
c=1 d=0 −∞ −∞ gcd(c,d)=1
where Kcd x, x = k u
−c−2 , x − d/c + i(w) x + d/c + i(z)
and d is the inverse of d modulo c. After a translation, we obtain the result. 2
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876
Lemma 2.3. We have ∞ 2 Kit (2πy) dy =
π . 8 cosh(πt)
0
Proof. By (16), the integral equals 1 8π
∞ ∞ −∞ −∞
cos(t(u + v)) dudv. cosh u + cosh v
Now, the change u = π(r + s) and v = π(r − s) separates variables giving known single integrals (for instance by the residue theorem, see Section 4) which yields the result. 2 Proof of Theorem 2.1. Let anm be as in (17) with (z) = y/|n| and = y/|m|, by (w) ∞ the pretrace formula (3), (15), (14) and Lemma 2.3, we have that 8 |mn| 0 anm y −1 dy equals Snm . Then, we must prove ∞ ∞ 4π |mn| dy 1 = 4δmn k(0) + S(m, n; c)G 8 |mn| anm y c c c=1 0
for n, m = 0. The term with the coefficient δmn comes from its counterpart in the formula for anm in Lemma 2.2 since with the change of variables x = 2|n|−1 yv ∞ ∞ 8|n| 0 −∞
∞ ∞ y dxdy y = 16 e(−nx)k u x + i , i e(−2yv)k v 2 dvdy |n| |n| y 0 −∞
which equals 4k(0) by Fourier inversion. We now consider the contribution of bnm in Lemma 2.2. Write λ = |mn|/c, with a simple change of variables, bnm contributes to ∞ 8 |mn| 0 y −1 anm dy as 8 λ
∞ ∞ ∞ 0 −∞ −∞
m n −λ2 x k u , x + iy y −1 e − x + dxdx dy |n| |m| x + iy
π
2 ∞ ∞ = 16λ
0 −∞ −π 2
k s2 +
mn −1 v+v sin θ cos 2πλ 2s cos θ − |mn|
(1 − v 2 )2 dsdvdθ, 4v 2
where the last expression is derived by taking x = λv sin θ,
y = λv cos θ
and x =
2syv − x . v2
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If mn > 0, by the change v = w + π
2 ∞ ∞
877
√ w2 + 1 we can write the above integral as
cos 4πλ s cos θ − w2 + 1 sin θ k s2 + w2 dsdwdθ
−∞ −∞ −π 2
1 = 2
2π ∞ ∞
cos 4πλ s2 + w2 + 1 cos θ k s2 + w2 dsdwdθ.
0 −∞ −∞
A similar analysis applies to the case mn < 0, which leads to 1 2
2π ∞ ∞
cos 4πλ s2 + w2 cos θ k s2 + w2 dsdwdθ.
0 −∞ −∞
Now, it is sufficient to use polar coordinates and the integral representation of J0 in (16) to obtain G(x). 2 3. The reversed Kuznetsov formula It is an elementary fact that the Fourier transform of f ( x2 + y 2 ) is given by ∞ 2 2 2π 0 rf (r)J0 (2πr ξ + η ) dr. Renaming the variables and using that the Fourier transform is involutive for even functions, one gets at once the inversion formula for the Hankel transform ∞ F (x) =
∞ rf (r)J0 (rx) dr
implies
0
f (r) =
xF (x)J0 (rx) dx.
(18)
0
Note that G in Theorem 2.1 can be written as a transform of this kind ∞ G(x) = 8πx
rk r2 − J0 (rx) dr.
The asymmetry of the reversed formula for mn > 0 comes from the fact that the interval [0, 1] is missing for = 1 and we have to compensate for this part. This appears in Kuznetsov’s original approach [12, (6.19)] but is hidden under involved integral transforms. With our formulation it becomes more transparent and natural. In this asymmetric case mn > 0 of the reversed formula, for the purpose of applying Petersson’s formulas, it is convenient to consider Bessel functions of integral orders Jn(t) that, according to (13), are defined as the Fourier coefficients of eit sin(2πθ) ; i.e., 1 eit sin(2πθ) e(−nθ) dθ.
Jn (t) = 0
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To ensure convergence in the reversed formula, one needs some regularity conditions on the test function f , and that f (0) = 0. Here we use strong regularity conditions in our proof for simplicity, although these are certainly stronger than strictly necessary; specifically that f (k) (0) = 0 for all k even and hence f can be extended to a regular odd function on R. Kuznetsov imposes f (0) = 0 but this seems to be unnecessary [7]. In general, regularity matters are less important for applications. Theorem 3.1. Let f ∈ C0∞ (R) be odd and as in Theorem 2.1. Then ∞ 1 c=1
c
S(m, n; c)f
4π
∞ |mn| δmn = Snm − f (t)J0 (t) dt c 2π 0
+
∞ 1 c=1
c
S(m, n; c)V
4π
|mn| , c
where Snm is as in (8), with h the Selberg transform of the function k(r) = ∞ √ 1 8π 0 f (x)J0 (x r + ) dx, and V , which only appears when mn > 0, admits the representations ∞ 1 V (x) = x
rf (t)J0 (rt)J0 (rx) drdt 0
0
and V (x) = 2
∞ j=1
∞ (2j − 1)J2j−1 (x)
t−1 f (t)J2j−1 (t) dt.
0
Remark 3.2. By a formula due to H. Petersson [7], for m, n > 0 √ ∞ (2j − 2)! 1 4π mn j √ S(m, n; c)J2j−1 , alj (m)alj (n) = δmn + 2π(−1) c c (4π mn)2j−1 c=1 l
where alj (n) is the n-th Fourier coefficient of the l-th element of an orthonormal basis (with respect to the Petersson inner product) of the classical cusp forms of weight 2j. Using this formula, one can completely rule out Kloosterman sums in the term containing V . If we think of classical modular forms as harmonic functions (corresponding to the zero eigenvalue of Laplacian), one obtains a truly reversed formula containing Kloosterman sums in one side and spectral information in the other side. In practice the term involving V gives a minor contribution. ∞ √ 1 Proof. Define k(r) = 8π f (x)J0 (x r + ) dx. Recall that k(r) a constant) 0 is (up to the radial component of the Fourier transform of the function f ( x2 + y 2 )/ x2 + y 2 ∈
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√ C0∞ evaluated at r + . Then we have that k is a rapidly decreasing function, and consequently g in (5) is also rapidly decreasing, in fact with exponential decay, and by partial integration we conclude that RC holds true for the corresponding Selberg ∞ transform h. Note that 8πk(0) = 0 f (x)J0 (x) dx for = 1. In any case, by (18) f can be recovered as ∞ f (x) = 4πx
√ k(r)J0 (x r + ) dr,
−
and by Theorem 2.1 it only remains to prove that, with = 1, 0 V (x) = 4πx
√ k(r)J0 (x r + 1) dr
−1
for both expressions of V . The definition of k gives, after a change of variables, 0 4πx
∞ 1
√
k(r)J0 (x r + 1) dr = x
−1
rf (y)J0 (ry)J0 (rx) drdy, 0
0
which is the first expression. The second one would follow from the following Bessel function identity 1 xy
rJ0 (rx)J0 (ry) dr = 2
∞
(2j − 1)J2j−1 (x)J2j−1 (y).
(19)
j=1
0
Indeed, the definition of Jn (t) implies eiy sin θ = Jn (y)e(nθ), and using this Fourier expansion as a generating function, one easily deduces Jn−1 (y) − Jn+1 (y) = 2Jn (y)
and Jn−1 (y) + Jn+1 (y) =
2n Jn (y), y
which leads to the relation Jn−1 (z)Jn−1 (w) − Jn+1 (z)Jn+1 (w) =
2n 2n Jn (w)Jn (z) + Jn (z)Jn (w) w z
and gives a telescoping series when summing over (positive) odd numbers n. Namely zwJ0 (z)J0 (w) = 2
n zJn (w)Jn (z) − wJn (z)Jn (w) .
n odd
Substituting z = ry, w = rx, and integrating against r−1 in the interval [0, 1], we obtain (19). 2
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4. Equivalence with the classic Kuznetsov formula This section is devoted to the proof of the equivalence between the classical Kuznetsov formula and Theorem 2.1. To see this it suffices to prove that H equals G, by the very definition of H we are forced to use some special functions. Theorem 4.1. We have G(x) = H(x), for all x > 0. ∞ Proof. Define the sine transform SF (ξ) = 0 F (x) sin(ξx) dx. By Fourier inversion it is enough to prove SG = SH . It will be convenient to consider the formula d SG f (ξ) f (ξ) = − dξ
∞
dx . G(x) cos xf (ξ) x
(20)
0
If mn < 0, using ∞ 0
⎧ 1 , ⎪ ⎨ β 2 −α2 J0 (xβ) cos(αx) dx = ∞, ⎪ ⎩ 0,
0 < α2 < β 2 , β 2 = α2 , 0 < β 2 < α2 ,
and (5), we can manipulate (20) with f (ξ) = sinh ξ, changing the order of integration to get
SG (sinh ξ) cosh ξ = −4π
d dξ
∞ sinh2 ξ
k(r) r − sinh ξ 2
dr = −4πg (2ξ).
(21)
If mn > 0, we consider two cases. First, if ξ > 1 then f (ξ) = cosh ξ in (20) also leads to −4πg (2ξ), as in (21), and a similar procedure for ξ < 1 and f (ξ) = sin ξ yields d SG (sin ξ) cos ξ = −4π dξ
∞ 0
k(r) dr. r + cos2 ξ
(22)
Once we have simplified SG , we are going to check that coincides with SH . For mn < 0, using the definition (11), 4 SH (sinh ξ) = π
∞
th(t) sinh(πt)
−∞
∞
K2it (x) sin(x sinh ξ) dx dt. 0
By (12), the term inside the parenthesis is equal to
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∞ 2 sinh ξ
π sin(2tξ) cos(2tv) sinh(πt) dv = , cosh(2v) + cosh(2ξ) 2 cosh ξ
881
(23)
0
where the last equality follows after the change w = e2v and using the residue theorem over a keyhole contour. The case mn < 0 is then settled because ∞ SH (sinh ξ) cosh ξ = 2
th(t) sin(2tξ) dt = −4πg (2ξ).
−∞
If mn > 0, the definition (10) gives ∞ SH (η) = −∞
i th(t) cosh(πt)
∞
J2it (x) − J−2it (x) sin(xη) dx dt
0
and by (13), we may write the term inside the parentheses as π ∞
d 2 π cosh(πt) dη
sinh(2tθ) sin(x sin θ) cos(xη) 0
+
4η π
∞ 0
dx dθ x
0
cos(2tθ) sinh(πt) dθ. sinh2 θ + η 2
(24)
If η > 1, the first integral vanishes and writing η = cosh ξ, the second integral is the same as the one appearing in (23). Hence SH (cosh ξ) sinh ξ = −4πg (2ξ). If 0 < η < 1, taking η = sin ξ, (24) equals d 2 cosh(πt) cos ξ dξ
π−ξ
∞
ξ
0
4 sinh(πt) sinh(2tθ) dθ + π
cos(2tθ) dθ. sinh2 θ + sin2 ξ
Now, the first integral is immediate and the second one can be computed following the same steps as in (23). Combining these results we get ∞ SH (sin ξ) cos ξ = − −∞
By (5), the previous integral is
d sinh(2tξ) dt = − th(t) cosh(πt) dξ
∞ h(t) −∞
cosh(2tξ) dt. cosh(πt)
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∞
∞ g(r)
−∞
−∞
cos(rt) cosh(2tξ) dtdr = 8 cosh(πt)
∞
−∞
q(sinh2 r) cos ξ cosh r dr cosh(2r) + cos(2ξ)
∞ ∞ =4 0 w
k(r) cos ξ drdw . (w + cos2 ξ) w(r − w)
Exchanging the order of integration, the inner integral can be computed using residue theory by integration over a dumbbell contour around [0, r], which leads to (22), and this finishes the proof. 2 5. Some estimates and examples Our new statement for the Kuznetsov formula not only shows advantages in its proof, it also eases estimations in practice and the search of explicit examples. The following result shows that the Selberg transform can be estimated as a Fourier transform evaluated in the hyperbolic distance. Lemma 5.1. Let h be a function satisfying the regularity conditions RC and assume ∞ j t h(t)eirt dt Bj (r) for j = 0, 1, where B0 (r) and B1 (r)/ sinh r are non-increasing −∞ functions for r > 0, then k2 (u) B0 (ρ)B1 (ρ)/ sinh ρ, where ρ and u are linked by (2). Proof. By (6), we have q(sinh2 v2 ) B0 (v) and q (sinh2 v2 ) B1 (v)/ sinh v. For ξ > u, write ∞ k(u) = u
−q (v) dv √ = π v−u
∞ ξ + = I0 (ξ) + I1 (ξ). ξ
u
By the second mean value theorem 1 I0 (ξ) √ ξ−u
ξ ξ
B0 (ρ) q (v) dv √ ξ−u
and B1 (ρ) I1 (ξ) sinh ρ
ξ √ u
Choosing ξ = u +
B0 B1
B1 (ρ) dv ξ − u. sinh ρ v−u
sinh ρ the result follows. 2
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883
For instance, for the Gaussian choice h(t) = e−t /T we can take B0 (r) = T e−T r /4 2 2 and B1 (r) = rT 3 e−T r /4 , getting 2 2 u cosh r − 1 T 2 e−T r /4 , (25) k 2 sinh u 2
2
2 2
which is in fact sharp and plays a role in the large sieve inequalities of [2]. In general, using the Weil bound [8, Corollary 11.12], the relation (7) and J0 (t) (1 + |t|)−1/2 , we deduce at once Proposition 5.2. For Snm as in (8) and with the notation of Lemma 5.1, we have δmn Snm − π
∞ t tanh(πt)h(t) dt |mn|
1/2
∞ τ (c) c=1
−∞
with
∞ I(λ) =
c3/2
1/2
(m, n, c)
mn I c2
B0 (v)B1 (v) sinh v dv, 1 + (|λ| cosh v + λ)1/2
0
where τ (c) is the number of divisors of c and (m, n, c) represents the greatest common divisor. To illustrate its applicability we give an estimate for smoothed sums of Fourier coefficients. Corollary 5.3. Given δ > 0, we have ∞ 2 −t2 /T 2 1 νj (n) e j ηt (n)2 e−t2 /T 2 dt = π −1 T 2 + O |n|1/2+δ T , + 4π j −∞
for any n = 0 and T > 1. This improves (16.56) in [8]. We think that the argument employed there does not match the exponent in (16.55), that can be lowered, but our application of Proposition 5.2 gives anyway a better result. Proof of Corollary 5.3. Choose h(t) = e−t /T and n = m in Proposition 5.2. As in the 2 2 2 2 derivation of (25), we can take B0 (r) = T e−T r /4 and B1 (r) = rT 3 e−T r /4 . Then we have 2 ∞ 1/2 −T 2 v2 +v v e n 2 I 2 T dv T 1/2 min 1, c1/2 n−1/2 c 1 + ev/2 (n/c)1/2 2
0
that gives the expected error term.
2
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Finally, note that tanh(πt) = ±1 + O(e−π|t| ) where the sign of ±1 coincides with that of t and consequently the main term is π −1 T 2 + O(1). 2 Another advantage provided by the new expression is to get some explicit instances of Kuznetsov formula. That is, given k and its Selberg transform, finding the corresponding G function. There are some pairs of k and h having closed formulas, this is employed in [3] to produce some unusual approximate formulas. We use some of these pairs in our statement with special emphasis in giving simple and self-contained proofs. Forexample, taking k(r) = e−μr with μ > 0, the corresponding Selberg transform is μ/2 4e π/μKit (μ/2) and using the Taylor expansion of the cosine in (16) we have ∞
x2n G(x) = 2x (−1) (2n)! n=0
∞
n
e
−μr n
2π cos2n θ dθ.
r dr
0
0
Both integrals are simple, and the resulting series is the Taylor expansion of 2 4πxμ−1 e−x /4μ . If we consider functions of the type √ αβ
k(r) = e− 4 (α + β)2 + 4αβr
(α+β)2 +4αβr
,
where α, β > 0, we have h(t) = Kit (α)Kit (β) and after a change of variables we can write G as πx(α + β) √ 2 αβ
∞ e
−β(α+β)y
J0
x(α + β) 2 √ y − 1 dy. 2 αβ
1
It follows from (16) that 2π 1 J0 γ y 2 − 1 = eiγy cos θ−γ sin θ dθ 2π 0
so, after integrating in y, it is sufficient to apply the residue theorem to obtain G(x) =
πx 4αβ + x2
α+β √ αβ
e− 2
4αβ+x2
.
In other examples, like functions of the type k(r) = (1 + r)−μ with μ > 1, it may be helpful to use the Hankel transform. In this case, the Selberg transform is h(t) =
4π Γ (μ − 1/2 + it)Γ (μ − 1/2 − it). Γ 2 (μ)
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On the other hand, observe that the generalized binomial theorem gives
1 + r2
−μ
=
∞ Γ (μ + n) (−1)n r2n . n!Γ (μ) n=0
Furthermore, by (12) we have ∞
∞ x
μ+2n
Kμ−1 (x) dx = Γ (μ + 2n + 1)
0
cosh((μ − 1)v) dv (cosh v)μ+2n+1
0 μ+2n−1
=2
n!Γ (μ + n).
To see this, just write the hyperbolic cosine as a sum of exponential and apply the change v = ± 12 log 1−y y in the resulting integrals. Substituting this expression into the above, gives the expansion of J0 (or the Taylor series of the cosine as was done in the first example) and finally we obtain
1+r
2 −μ
∞ =
xμ Kμ−1 (x)J0 (xr) dx. 2μ−1 Γ (μ)
0
Now, the inversion formula for the Hankel transform gives G(x) = In fact, if μ − the above as
1 2
πxμ Kμ−1 (x). 2μ−4 Γ (μ)
∈ Z+ we can take the series expansion of the Bessel function to write
μ−3/2 πe−x (μ − 1/2)! (μ − 3/2 + k)! (2x)μ−1/2−k . 8(2μ − 1)! k!(μ − 3/2 − k)! k=0
Other instances of “exact” Hankel transforms can be found in mathematical tables but rarely do they correspond to an explicit function h. Acknowledgments This paper is part of the Ph.D. Thesis of the second author to be defended at Universidad Autónoma de Madrid. We acknowledge the research environment provided by the Mathematics Department and the ICMAT. We are indebted to N. Pitt and A. Ubis for some comments and suggestions. We are also very grateful to the anonymous referee for pointing out the unpublished proof by Zagier and its extension to GL(n).
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