THE CIRCLE METHOD AND BOUNDS FOR L-FUNCTIONS-III: t ...

THE CIRCLE METHOD AND BOUNDS FOR L-FUNCTIONS - III: t-ASPECT SUBCONVEXITY FOR GL(3) L-FUNCTIONS

arXiv:1301.1007v2 [math.NT] 19 Jan 2013

RITABRATA MUNSHI Abstract. Let π be a Hecke-Maass cusp form for SL(3, Z). In this paper we will prove the following subconvex bound
3 1 L 21 + it, π ≪π,ε (1 + |t|) 4 − 32 +ε .

1. Introduction Let π be a Hecke-Maass cusp form of type (ν1 , ν2 ) for SL(3, Z). Let the normalized Fourier coefficients of π be given by λ(m1 , m2 ) (so that λ(1, 1) = 1). The Langlands parameters (α1 , α2 , α3 ) associated with π are defined as α1 = −ν1 − 2ν2 + 1, α2 = −ν1 + ν2 and α3 = 2ν1 + ν2 − 1. The Ramanujan-Selberg conjecture predicts that Re(αi ) = 0. From the work of Jacquet and Shalika [6], we (at least) know that |Re(αi )| < 12 . The L-series associated with π is given by L(s, π) =

∞ X

λ(1, n)n−s

n=1

in the domain σ = Re(s) > 1. This extends to an entire function and satisfies a functional equation. More precisely there is an associated gamma factor given by   3 Y s − αi − s2 , π Γ γ(s, π) = 2 i=1 so that

γ(s, π)L(s, π) = γ(s, π ˜ )L(1 − s, π ˜ ).

Here π ˜ is the dual form having Langlands parameters (−α3 , −α2 , −α1 ). The convexity principle implies that L(1/2 + it, π) ≪π (1 + |t|)3/4 - the convexity bound. The purpose of this paper is to prove the following. Theorem 1. Let π be a Hecke-Maass cusp form for SL(3, Z). Then we have
1 3 L 21 + it, π ≪π,ε (1 + |t|) 4 − 32 +ε .

1 A similar subconvex bound (with exponent 43 − 16 + ε) is known for the symmetric square lifts of SL(2, Z) forms (or self dual forms for SL(3, Z)) due to the work of Li [7]. (Other subconvexity results in the case of degree three L-functions in different aspects can be found in [1], [9], [10], [11] and [13].) Subconvex bound in the t-aspect was first established by Weyl [15] for degree one L-functions, and by Good [3] for degree two L-functions. This paper settles the problem for degree three L-functions.

Like the two previous papers [12] and [13], with the same title, we will yet again demonstrate the power of the circle method in the context of subconvexity. In the present situation Kloosterman’s version of the circle method works best. Let ( 1 if n = 0; δ(n) = 0 otherwise. 1991 Mathematics Subject Classification. 11F66, 11M41. Key words and phrases. subconvexity, GL(3) Maass forms, twists. 1

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RITABRATA MUNSHI

Then for any real number Q, we have Z (1) δ(n) = 2 Re

1

X X⋆

0 1≤q≤Q Kt1/4 or K > t1/4 . In particular by choosing K in the range t1/4 < K < t1/2 we can get a bound which breaks the convexity barrier. Acknowledgements. The author is partly supported by SwarnaJayanti Fellowship, Department of Science and Technology, Government of India. The author also gratefully acknowledges the hospitality of TIFR CAM Bangalore, where a part of the work was done.

2. GL(3) Voronoi summation formula and stationary phase method 2.1. Voronoi type summation formula for SL(3, Z). Suppose π is a Maass form of type (ν1 , ν2 ) for SL3 (Z), which is an eigenfunction of all the Hecke operators with Fourier coefficients λ(n1 , n2 ), normalized so that λ(1, 1) = 1. Since we shall work entirely at the level of L-functions, we simply refer to Goldfeld’s book [2] for details regarding automorphic forms on higher rank groups. In this subsection we recall two important results - a summation formula for the Fourier coefficients twisted by additive characters and a bound on the average size of the Fourier coefficients - which will play vital role in our analysis. R∞ Let g be a compactly supported smooth function on (0, ∞), and let g˜(s) = 0 g(x)xs−1 dx be its Mellin transform. For σ > −1 + max{−Re(α1 ), −Re(α2 ), −Re(α3 )} and ℓ = 0, 1 define


3 1 +ℓ 2 +ℓ 3 +ℓ Γ 1+s+α Γ 1+s+α π −3s− 2 Γ 1+s+α 2 2 2


, γℓ (s) := 2 +ℓ 3 +ℓ 1 +ℓ 2 Γ −s−α Γ −s−α Γ −s−α 2 2 2

set γ± (s) = γ0 (s) ∓ iγ1 (s) and let (7)

G± (y) =

1 2πi

Z

(σ)

y −s γ± (s)˜ g (−s)ds.

The following Voronoi type summation formula (see [7], [8]) will play a crucial role in our analysis. Recall the definition of the Kloosterman sum X⋆  aα + bα  e S(a, b; c) = , c α mod c

where α ¯ denotes the multiplicative inverse of α mod c.

t-ASPECT SUBCONVEXITY FOR GL(3) L-FUNCTIONS

5

Lemma 1. Let g be a compactly supported smooth function on (0, ∞), we have  2    ∞ ∞ XX X X λ(n2 , n1 ) n1 n2 an g(n) =q (8) , S(¯ a, ±n2 ; q/n1 )G± λ(1, n)e q n1 n2 q3 ± n =1 n=1 n1 |q

2

where (a, q) = 1 and a ¯ denotes the multiplicative inverse of a mod q. We need to study the behaviour of the gamma factor γ± (s) more closely, especially for s restricted in vertical strips. Using Stirling formula we can pull out the oscillatory part, and the remaining part satisfies a ‘scaling property’. Indeed for s = − 21 + iτ with |τ | ≫ tε , we apply Stirling’s formula to get    3iτ 1 1 |τ | γ± − + iτ = Φ± (τ ), where Φ′± (τ ) ≪ (9) . 2 eπ |τ | The following lemma, which gives Ramanujan conjecture on average, is also well-known. It follows from standard properties of the Rankin-Selberg L-function. Lemma 2. We have XX n21 n2 ≤x

|λ(n1 , n2 )|2 ≪ x1+ε ,

where the implied constant depends on the form π and ε. 2.2. Stationary phase method. We will need estimates for exponential integrals of the form Z b (10) I= g(x)e(f (x))dx, a

where f and g are smooth real valued functions. First we recall an easy estimate. Suppose in the range of the integral we have |f ′ (x)| ≥ B and f (j) (x) ≪ B 1+ε for j ≥ 2. Suppose Supp(g) ⊂ (a, b) and g (j) (x) ≪a,b,j 1. Then by making the change of variable u = f (x) we get Z f (b) g(f −1 (u)) I= e(u)du. ′ −1 (u)) f (a) f (f By applying integration by parts j times we get I ≪a,b,j,ε B −j+ε . This will be used at several places to show that certain exponential integrals are negligibly small in the absence of the stationary phase. In case there is a single stationary phase point then the integral has an asymptotic expansion. A sharp version of this stationary phase method, which can be found in [4], will be useful for our purpose. The estimates are in terms of parameters Θf , Ωf ≫ (b − a) and Ωg , for which the derivatives satisfy (11)

f (i) (x) ≪ Θf /Ωif ,

g (j) (x) ≪ 1/Ωjg .

For the second assertion we will moreover require (12)

f (2) (x) ≫ Θf /Ω2f .

Lemma 3. Suppose f and g are smooth real valued satisfying (11) for i = 2, 3 and j = 0, 1, 2. Suppose g(a) = g(b) = 0. (1) Suppose f ′ and f ′′ do not vanish in [a, b]. Let Λ = min[a,b] |f ′ (x)|. Then we have ! Ω2f Ωf Λ Θf . + 2 I ≪ 2 3 1+ Ωf Λ Ωg Ωg Θf /Ωf

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RITABRATA MUNSHI

(2) Suppose f ′ changes sign from negative to positive at the unique point x0 ∈ (a, b). Let κ = min{b − x0 , x0 − a}. Further suppose (11) holds for i = 4 and (12) holds. Then we have ! Ω3f Ω4f Ωf g(x0 )e(f (x0 ) + 1/8) p + 3/2 + 3/2 I= . +O Θ2f κ3 f ′′ (x0 ) Θ Θ Ω2g f

f

Finally we recall the second derivative bound for exponential integrals in two variables. Let Z bZ d I(2) = (13) g(x, y)e(f (x, y))dydx a

c

where f and g are smooth real valued functions. First suppose g = 1, and we have positive parameters r1 and r2 such that in the rectangle [a, b] × [c, d] we have h i2 f (2,0) (x, y) ≫ r12 , f (0,2) (x, y) ≫ r22 , f (2,0) (x, y)f (0,2) (x, y) − f (1,1) (x, y) ≫ r12 r22 , (14) where f (i,j) (x, y) =

∂ i+j ∂xi ∂y j f (x, y)

and the implied constants are absolute. Then we have (see [14]) I(2) ≪

1 . r1 r2

To extend this result to smooth g with Supp(g) ⊂ (a, b) × (c, d), we apply integration by parts once in each variable. To state the result we define the total variation of g to be Z bZ d 2 ∂ dydx. var(g) := g(x, y) ∂x∂y a c Lemma 4. Suppose f , g, r1 and r2 are as above satisfying the condition (14). Then we have I(2) ≪

var(g) r1 r2

with an absolute implied constant. 2.3. An integral. Let W be a smooth real valued function with Supp(W ) ⊂ [a, b] ⊂ (0, ∞). Suppose W (j) (x) ≪a,b,j 1. Consider the exponential integral Z ∞ (15) W † (r, s) = W (x)e(−rx)xs−1 dx 0

where r ∈ R and s = σ + iβ ∈ R. In particular W † (r, 1) is the Fourier transform of W and W † (0, s) is the Mellin transform of W . The integral is of the form (10) with g(x) = W (x)xσ−1

and

f (x) = −rx +

1 β log x. 2π

Then

1 β 1 β and f (j) (x) = (−1)j−1 (j − 1)! 2π x 2π xj for j ≥ 2. The unique stationary point is given by f ′ (x) = −r +

x0 = and we can write

β , 2πr

  x  β 1 1 0 =r f (x) = − −1 . 2π x x0 x Suppose a, b and σ are fixed and we are interested in the dependence of the integral on β and r. Suppose |β|, |r| ≥ 1. If x0 ∈ / [a/2, 2b] then in the support of the integral, i.e. W (x) 6= 0, we have |f ′ (x)| ≫ max{|β|, |r|} and |f (j) (x)| ≪j |β|, where the implied constants depend on a, b and σ. So in this case W † (r, s) ≪j min{|β|−j , |r|−j }, where again the implied constant depends on a, b and σ. ′

t-ASPECT SUBCONVEXITY FOR GL(3) L-FUNCTIONS

7

On the other hand if x0 ∈ [a/2, 2b] then using the second statement of Lemma 3 (with Θf = |β| and Ωf = Ωg = 1) we get   g(x0 )e(f (x0 ) + 1/8) p + O |β|−3/2 . W † (r, s) = f ′′ (x0 )

The error term can also be written as O(|r|−3/2 ), as x0 ∈ [a/2, 2b] implies that |r| ≍ |β|. Note that for β > 0 we need to take conjugate so that the conditions of the lemma are satisfied. Also we note that above asymptotic holds regardless the location of x0 . For the following statement we take √ the πi/2 −1 = e . Lemma 5. Let W , r and s be as above. We have √   σ  iβ   2πe(1/8) β β β † √ W (r, s) = W + O min{|β|−3/2 , |r|−3/2 } , 2πr 2πr 2πer −β where the implied constant depends on a, b and σ. We also have ( )! β j r j † W (r, s) = Oa,b,σ,j min , . r β

3. Application of summation formula

3.1. Applying Poisson summation. For simplicity let us assume that t > 2. First we will apply the Poisson summation formula on the sum over m in (4), i.e.     ∞ m X mx m¯ a −i(t+v) e V⋆ . m e − q aq N m=1 Breaking the sum into congruence classes modulo q we get       X α + mq (α + mq)x α¯ a X ⋆ −i(t+v) V . (α + mq) e e − q aq N m∈Z

α mod q

Then applying Poisson we obtain       Z X α¯ a X α + yq (α + yq)x e − V⋆ e(−my)dy. (α + yq)−i(t+v) e q aq N R α mod q

m∈Z

Making the change of variable (α + yq)/N → 7 u and executing the resulting complete character sum we arrive at   Z X N (x − ma) 1−i(t+v) ⋆ −i(t+v) N (16) u du. V (u) u e aq R m∈Z m≡¯ a mod q

The above integral, in the notation of Subsection 2.3 is (17)


V ⋆† N (aq)−1 (ma − x), 1 − i(t + v) .

p 2−ε −1 −1 Recall that a ≍ N/K, and by our choice, √ see (5), K < t /N . So |N (aq) (ma − x)| ≍ q N |m| −1 −1 if m 6= 0, and |N (aq) (ma − x)| ≪ q N K if m = 0. Applying the second statement of Lemma 5 it follows that the contribution of the zero frequency m = 0 (which occurs only for q = 1 due to the condition (m, q) = 1) in (16) is negligibly small, and also the contribution of the tail |m| ≫ qt1+ε /N is negligibly small. We only need to consider m with 1 ≤ |m| ≪ qt1+ε /N , which in turn implies that we only need to focus on q which are in the range p N/t1+ε ≪ q ≤ Q = N/K. Taking a dyadic subdivision we conclude the following.

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RITABRATA MUNSHI

Lemma 6. Suppose N and K satisfy (5), then we have X N (18) S(N, C) + O(t−2012 ) S + (N ) = K √ 1+ε N/t

≪C≤ N/K C dyadic

where (19)

S(N, C) =

Z

0

1

Z

N

×

∞ X

−i(t+v)

R

v V K

λ(1, n)e

n=1



nm q

X

CK 1/3 −i(t+Kv)  i(Kv−τ )  (Kv − τ )aq (t + Kv)aq dvdx × 2πeN (x − ma) 2πeN x     1 10K ε + O √ 4/3 min 1, t . |τ | tK Observe that the new error term absorbs the contribution of the earlier error term in (22). Now we can eliminate the restriction on the integral over v. Indeed in the complementary range the weight p function is non-vanishing only for τ ≤ Kv and x ≪ K −2/3 (recall that a, q ≪ N/K). We get r     Z Z 1 aq 10K dx (Kv − τ )aq V (v) V dv √ ≪ √ 3/2 min 1, . tN 0≤x≪K −2/3 0≤v− Kτ ≤K −2/3 2πN x |τ | x tK So we conclude that (23)

⋆⋆

I (q, m, τ ) = c2

√     (Kv − τ )aq (t + Kv)aq t + Kv ⋆ V V (v) √ V 2πN (x − ma) 2πN x x(x − ma) 0 R  −i(t+Kv)  i(Kv−τ ) (t + Kv)aq (Kv − τ )aq × dvdx 2πeN (x − ma) 2πeN x     1 10K ε t + O √ 4/3 min 1, |τ | tK

 aq 3/2 Z N

for some absolute constant c2 .

1

Z

4.2. Integral over v. Now we will study the integral over v in (23). This term vanishes unless m < 0. For x < 1/K we bound the integral trivially. Indeed, in this case the weight function restricts the integral over v to a range of length N/K 2 aq. So estimating trivially we get √      aq 3/2 Z 1/K Z (Kv − τ )aq (t + Kv)aq t + Kv ⋆ (24) V dvdx V (v) √ V N x(x − ma) 2πN (x − ma) 2πN x 0 R s 1 N ≪√ . 5/2 aq tK Let us now take x ∈ [1/K, 1]. Temporarily we set     Kv − τ t + Kv (t + Kv)aq (Kv − τ )aq + f (v) = − log log 2π 2πeN (x − ma) 2π 2πeN x

and

√√     t t + Kv aq (Kv − τ )aq (t + Kv)aq ⋆ g(v) = V . V (v) V N (x − ma) 2πN (x − ma) 2πN x √ We are multiplying by an extra t to balance the size of the function. Then   (j − 1)!(−K)j (j − 1)!(−K)j (t + Kv)x K ′ , f (j) (v) = − + , log f (v) = − j−1 2π (Kv − τ )(x − ma) 2π(t + Kv) 2π(Kv − τ )j−1 for j ≥ 2. The stationary phase is given by

v0 = −

(t + τ )x − τ ma . Kma

In the support of the integral we have f

(j)

Nx ≍ aq



Kaq Nx

j

t-ASPECT SUBCONVEXITY FOR GL(3) L-FUNCTIONS

for j ≥ 2, and g

(j)

11

j  Kaq . (v) ≪ 1 + Nx

for j ≥ 0. Moreover we can write

    K K K(v0 − v) K(v0 − v) − . f (v) = log 1 + log 1 + 2π (t + Kv) 2π (Kv − τ ) √ 1+ε √ In the support of the integral we have 0 ≤ Kv − τ ≤ N/aq ≤ Kt / N ≪ t3/4+ε (recall that √ 1+ε < q). It follows that if v0 ∈ / [.5, 3] then in the support of the integral we t < N , K < t and N/t have   Kaq . |f ′ (v)| ≫ K 1−ε min 1, Nx ′

Applying the first statement of Lemma 3 with Nx Θf = , aq

Nx Ωf = , Kaq

    Kaq Nx 1−ε , and Λ = K min 1, Ωg = min 1, Kaq Nx

we obtain the bound Θf g(v)e(f (v))dv ≪ 2 3 Ωf Λ R

Z

(25)

Ω2f Ωf Λ 1+ + 2 Ωg Ωg Θf /Ωf

!

tε .

On the other hand if v0 ∈ [.5, 3] then treating the integral as a finite integral over the range [.1, 4] and applying the second part of Lemma 3 it follows that ! ! Z Ω3f Ω4f Ωf g(v0 )e(f (v0 ) + 1/8) p (26) + 3/2 tε . g(v)e(f (v))dv = +O 2 + 3/2 ′′ (v ) 2 Θ f R Θf Θ f Ωg 0 f Notice that we have κ > .4. The bound from (25) and the error term of (26) together make a total contribution of size   1 N ε O √ t t aqK 3

in (23). We arrive at this through explicit calculation (estimating the integral over x trivially) of all the six factors and using the lower bound on K from (5) to determine the dominating contribution. As a demonstration let us compute the contribution of the last term in (26). If x ≤ Kaq/N then Ωg = Ωf . Hence r Ω3f 1 aq . = 3/2 2 K N x Θ f Ωg On the other hand if x > Kaq/N then Ωg = 1, and we get 3/2  Ω3f Nx 1 . = 3/2 K 3 aq Θf Ω2g

We then plug in the other factors from (23), drop the condition that v0 ∈ [.5, 3] (which is a restriction on the range of the x integral) and get r Z max{1/K,Kaq/N } Z 1 Z Ω3f aq dx aq 1 dx N √ √ √ ≪ xdx + tN 1/K Θ3/2 Ω2g x x KN t 1/K K 3 taq max{1/K,Kaq/N } f

≪

N tε √ . K 3 taq

Similarly we can compute the contribution of the other terms. But it turns out, given the restriction (5) on K, that the other terms make a smaller contribution. Also the above term dominates the

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RITABRATA MUNSHI

bound obtained in (24). We conclude that  3/2 Z 1    −i(t+τ )  (t + τ )q (t + τ )q (t + τ )q q t+τ ⋆ ⋆⋆ V − − − dx V I (q, m, τ ) = c3 K −mN 2πN m 2πN m 2πeN m 1/K     10K ε 1 N ε 1 t +√ t + O √ 4/3 min 1, |τ | tK t K 3 aq

for some absolute constant c3 . Recall that V ⋆ (y)V (y) = V (y). Set √   N 1 10K 1 B(C, τ ) = √ 4/3 min 1, +√ (27) . |τ | tK t K 5/2 C Note that   Z + √NC K tε 1 N 1 1 √ √ (28) + B(C, τ )dτ ≪ tε ≪ √ 1/3 tε . √ 2 1/3 2 K ε tK tK C tK t − N C

The last inequality follows as C > N/t1+ε and as we are assuming the lower bound (5) for K. Lemma 8. Suppose C ≤ q ≤ 2C, with N/t1+ε ≪ C ≤

p N/K and N , K satisfy (5). We have

I⋆⋆ (q, m, τ ) = J1 (q, m, τ ) + J2 (q, m, τ )

where c4 J1 (q, m, τ ) = √ t+τ K for some absolute constant c4 and

 3 −i(t+τ )    (t + τ )q 2 (t + τ )q − V − 2πeN m 2πN m

J2 (q, m, τ ) := I⋆⋆ (q, m, τ ) − J1 (q, m, τ ) = O (B(C, τ )tε ) , with B(C, τ ) as defined in (27). Consequently from Lemma 7 we derive the following decomposition for S(N, C). Lemma 9. We have S(N, C) =

X

J∈J

where Sℓ,J (N, C) =

1 N 2 −it K X 2π ±

{S1,J (N, C) + S2,J (N, C)} + O(t−2012 )

XX

√ n21 n2 ≪ N K 3/2 tε

×

XX

λ(n2 , n1 ) √ n2 S(m, ¯ ±n2 ; q/n1 ) 3

C