Subconvexity for $GL(3)\times GL(2)$ $L$-functions in $t$-aspect

Let $\pi$ be a Hecke-Maass cusp form for $SL(3,\mathbb Z)$ and $f$ be a holomorphic (or Maass) Hecke form for $SL(2,\mathbb{Z})$. In this paper we prove the following subconvex bound $$ L\left(\tfrac{1}{2}+it,\pi\times f\right)\ll_{\pi,f,\varepsilon} (1+|t|)^{\frac{3}{2}-\frac{1}{42}+\varepsilon}. $$


Introduction
For π a Hecke-Maass cusp form for SL (3, Z), and f a holomorphic Hecke cusp form for SL(2, Z) the associated Rankin-Selberg L-series is given by λ π (n, r)λ f (n) (nr 2 ) s , in the half plane σ > 1. (Here λ π and λ f are the normalized Fourier coefficients of the forms.) This series extends to an entire function and satisfies a functional equation of the Riemann type s → 1 − s with a gamma factor of 'degree six'. This particular L-function plays a crucial role in quantum chaos (see [11]), and hence it is important to study its deeper analytic properties. In particular one seeks to understand the size of these functions inside the critical strip. A standard consequence of the functional equation is the easy convexity bound L 1 2 + it, π × f ≪ π,f,ε (1 + |t|) 3 2 +ε . The Lindelöf hypothesis predicts that such a bound holds with any positive exponent in place of 3/2 + ε. But even breaking the convexity barrier is hard and has remained open so far. The purpose of this paper is to prove the following subconvex bound. Theorem 1. Let π be a Hecke-Maass cusp form for SL (3, Z), and f a holomorphic Hecke cusp form for SL(2, Z). Then we have Subconvex bounds in the t-aspect are known for L-functions of degree upto three over the field of rationals (see [12], [3] and [10]). Similar bounds are also known for the Rankin-Selberg L-function L(s, f × g) for two GL(2) forms f and g. The t-aspect subconvexity for genuine GL(4) L-functions remains an important open problem. Our method of proof is similar to the one given in [10] and is based on the separation

The set-up
Let λ π (n, m) denote the normalised Fourier coefficients of the form π (see Chapter 6 of [2]) and let λ f (n) denote the normalised Fourier coefficients of the form f (see [4]). Suppose t > 2, then by approximate functional equation (see [4]) we have for some smooth function V supported in [1,2] and satisfying V (j) (x) ≪ j 1.
Remark 1 (Notation). In this paper the notation α ≪ A will mean that for any ε > 0, there is a constant c such that |α| ≤ cAt ε . The dependence of the constant on π, f and ε, when occurring, will be ignored.
Using the Ramanujan bound on average , i.e.
we further conclude that Hence to establish subconvexity we need to show cancellation in the sum S r (N) for N roughly of size t 3 and r small. We can and shall further normalize V , for convenience, so that V (y)dy = 1.
2.1. The delta method. There are three oscillatory factors contributing to the sum S r (N). Our method is based on separating these oscillations using the circle method.
In the present situation we will use a version of the delta method of Duke, Friedlander and Iwaniec. More specifically we will use the expansion (20.157) given in Chapter 20 of [4]. Let δ : Z → {0, 1} be defined by We seek a Fourier expansion which matches with δ in the range [−2M, 2M]. For this we pick Q = 2M 1/2 . Then we have for n ∈ Z ∩ [−2M, 2M] (and e(z) = e 2πiz ). The ⋆ on the sum indicates that the sum over a is restricted by the condition (a, q) = 1. The function g is the only part in the formula which is not explicitly given. We only need the following two properties (see (20.158) and (20.159) of [4]) g(q, x) ≪ |x| −A for any A > 1. In particular the second property implies that the effective range of the integral in (3) 2.2. Separation of oscillation. We apply (3) directly to S r (N) as a device to separate the oscillations of λ(n, r) and λ f (n)n −it . This by itself does not suffice, and as in [9] and [10] we need a 'conductor lowering mechanism'. For this purpose we introduce an extra integral namely where t ε < K < t 1−ε is a parameter which will be chosen optimally later, and U is a smooth function supported in [1/2, 5/2], with U(x) = 1 for x ∈ [1, 2] and U (j) ≪ j 1.
For n, m ≍ N, the integral 1 and we get that upto a negligible error term S r (N) is given by where W is a smooth bump function with support [−t ε , t ε ] 2.3. Sketch of proof. We end this section with a brief sketch of the proof. For simplicity let us focus on the generic case, i.e. N = t 3 , r = 1 and q ∼ Q = t 3/2 /K 1/2 , so that the main object of study is given by Our aim is to save N plus a 'little more'. First we apply the Voronoi summation formulae to both the m and n sums. In the GL(2) (resp. GL(3)) Voronoi we save (NK) 1/2 /t (resp. N 1/4 /K 3/4 ) and the dual length becomes m ⋆ ∼ t 2 /K (resp. n ⋆ ∼ K 3/2 N 1/2 ). Also we save √ Q in the a sum and √ K in the v integral. Hence in total we have saved N/t, and it remains to save t plus a little extra in a sum of the form where I is an integral transform which oscillates like n iK with respect to n, and the character sum is given by Next applying the Cauchy inequality we arrive at where we seek to save t 2 plus extra. Opening the absolute value square we apply the Poisson summation formula on the sum over n. We save enough in the zero frequency (diagonal contribution) if t 2 Q/K > t 2 i.e. if K < t. On the other hand we save enough in the non-zero frequencies if K 3/2 N 1/2 /K 1/2 > t 2 which boils down to K > t 1/2 .

Remark 2.
Notice that since the character sum boils down to an additive character we are saving more than the usual. In the usual case we would have saved K 3/2 N 1/2 /QK 1/2 , which would be larger than t 2 only if we had K > t 4/3 . This would contradict the upper bound K < t.

Voronoi summation formulae
Consider the sum over m in (6). Applying the Voronoi summation formula this transforms into where k is the weight of the form f . Extracting the oscillation of the Bessel function we see that the above sum is essentially given by a sum of two terms of the form By repeated integration by parts it follows that the integral is negligibly small if m ≫ t ε max{K, t 2 q 2 /N} =: M 0 . In the complementary range the size of the integral is given by the second derivative bound. However we need a more precise analysis of the integral based on the stationary phase expansion. In particular we note that when Nx/qQ ≪ t 1−ε then m ≍ (qt) 2 /N, otherwise the integral is negligibly small.
In the present case we have g(n) = e (nx/qQ) n iv V (n/N). Extracting the oscillation of the integral transform (see e.g. Lemma 2.1 of [5]), as in the case of GL(2) above, we essentially arrive at the following expression By repeated integration by parts we see that the integral is negligibly small if n 2 1 n 2 ≫ t ε ((qK) 3 r/N + K 3/2 N 1/2 rx 3 ) =: N 0 . We now substitute (7) in place of the third line and (8) in place of the second line of (6), to get the object of focus.

Reduction of integrals
4.1. Simplifying the integrals. We have transformed the sum in (6) into a new object with four integrals, which we need to simplify. Consider the integral over x which boils down to R W (x) g(q, x)e Nx(z − y) qQ dx.
Using (4) this splits as the sum of two integrals where in the second integral the weight function h has smaller size. In the first integral by repeated integration by parts we see that it is negligibly small unless |z − y| ≪ t ε q/QK. (We will continue our analysis with the first integral. For the second integral, apart from the fact that the weight function h is of size 1/qQ, we are able to get a weaker restriction |z − y| ≪ t ε /K by considering the v integral. As such we obtain much better final bound in this case.) Writing z = y + u with |u| ≪ t ε q/QK we arrive at the y integral I(m, n 2 1 n 2 , q) := ∞ 0 U(y)y −it e ± 2 √ mNy q ± 3(Nn 2 1 n 2 (y + u)) 1/3 qr 1/3 dy.
where W is a bump function. Then we have L ≪ 1/t.
Proof. To prove this assertion we make a change of variable z = y 1/2 , so that the phase function in (9) reduces to For this to be smaller than t in magnitude one at least needs a negative sign in the second term and 3(NN 0 ) 1/3 w/qr 1/3 ≍ t. Except this case we have I(. . . ) ≪ t −1/2 by the second derivative bound. In the special situation we have N 0 ≍ (tq) 3 r/N. Opening the absolute value square we arrive at The lemma follows. Splitting q in dyadic blocks q ∼ C, and writing q = q 1 q 2 with q 1 |(n 1 r) ∞ , (n 1 r, q 2 ) = 1, we see that the contribution of the C-block to the above sum is dominated by To analyse the sum in (10) further we break the sum over m into dyadic blocks. Then applying Cauchy's inequality and using the Ramanujan bound on average we see that the expression in (10)

Poisson summation.
Smoothing out the outer sum in (12), opening the absolute value square and applying the Poisson summation formula we arrive at Ω ≪ N 0 By repeated integration by parts we see that the integral is negligibly small if Moreover from our analysis in Subsection 4.2 it follows that I ≪ t −1 .

5.3.
The zero frequency. The zero frequency n 2 = 0 has to be treated differently.
Let Ω 0 denotes the contribution of the zero frequency to Ω, and let Σ 0 be its contribution to (11). and Proof. In the case n 2 = 0 it follows from the congruence conditions that q 2 = q ′ 2 and α = α ′ . So the character sum is bounded as and hence we get Trivially executing the remaining sums we get the first part of the lemma. This bound when substituted in place of Ω in (11) yields the bound Here if we substitute N/K in place of C and use the fact that K = t 1−η , then we get O(r 1/3 N 1/2 t 1+η/2 ) as the final bound to (11). This takes care of all the terms in (14) except the single term which has C 1/2 in the denominator. This occurs only when M 1 ∼ K, which is possible only if N|x|/CQ ∼ t (as otherwise the integral in (7) is negligibly small). In this case we get The integral over x takes care of the x 1/2 in the denominator, and we see that the total contribution of this term to (11) is dominated by O(r 1/3 N 1/2 t 3/2−3η/2 ). The lemma follows.
6. Analysis of non-zero frequencies 6.1. The character sum. Our next lemma gives a bound for C.
Proof. The 'character sum' C can be dominated by a product of two sums C ≪ C 1 C 2 where In the second sum since (n 1 , q 2 q ′ 2 ) = 1, we get α ≡ −mn 1 mod d 2 and α ′ ≡ −m ′n 1 mod d ′ 2 . Then using the congruence modulo q 2 q ′ 2 we are able to conclude that In the first sum C 1 the congruence condition determines α ′ uniquely in terms of α, and hence This completes the proof of the lemma.
We now substitute these bounds in (13). Writing q 2 d 2 in place of q 2 and q ′ 2 d ′ 2 in place of q ′ 2 we get that the contribution of the non-zero frequencies to Ω is We denote by Σ =0 the term we get by substituting this for Ω in (11).

6.2.
The case of small modulus. In this section we will consider the case where q ∼ C ≪ t 1+ε . Recall that we have I ≪ 1/t and n 2 = 0. Lemma 4. The contribution of q ∼ C ≪ t 1+ε , and n 2 = 0 to (11) is bounded by Proof. We use the congruences to count the number of (m, m ′ ) in (15). This comes out to be dominated by ). It follows that the contribution of this case to Ω =0 is dominated by Summing over n 2 and q 2 we arrive at Next summing over d 2 we get Executing the remaining sums we get N 0 q 1 rC 2 N 2 Suppose M 1 ≍ (tC) 2 /N or M 1 ≫ C/q 1 , then when the above bound is substituted for Ω in (11) we get the bound In the complementary range when M 1 ≪ C/q 1 and M 1 is not of size (tC) 2 /N, then N 0 ≍ (Ct) 3 r/N. In this case we adopt a different strategy for counting.
) In this case q 2 d 2 n 1 + m ′ n 2 ≪ Cn 1 /q 1 + M 1 N 2 ≪ Cn 1 /q 1 + N/n 1 q 2 1 t 2 . Writing q 2 d 2 n 1 + m ′ n 2 = −d ′ 2 h we see that h ≪ Cn 1 /q 1 D ′ + N/n 1 q 2 1 t 2 D ′ := H. With this we transform (15) to Using the second congruence we count the number of d ′ 2 which comes out to be O((d 2 , m ′ n 2 )D ′ /D). The first congruence gives us the number of m which comes out to be O((n 2 , d 2 )(1 + M 1 /D)). It follows that (17) is dominated by Then summing over n 2 , m ′ and d 2 we arrive at which is dominated by Now we substitute D ≪ C/q 1 , M 1 ≪ C/q 1 and C ≪ t 1+ε . When the above bound is substituted in place of Ω in (11) we get the bound This is dominated by the previous bound. The lemma follows.
6.3. The generic case. It now remains to tackle the case where C ≫ t 1+ε and n 2 = 0.
Proof. In this case we need a better bound for I. To this end we seek to apply stationary phase analysis to the integral I(. . . ) in (9), namely where A = 2 √ mN /q and B = 3(Nn 2 1 n 2 ) 1/3 /qr 1/3 . Since C ≫ t 1+ε , from (7) we conclude that we have plus sign with A and that A ≍ t. From (8) we conclude that B ≪ t 1−η/2 . (Otherwise the integrals in (7) and (8) are negligibly small.) As such the stationary point can be written as y 0 + y 1 + y 2 + . . . with y i ≪ (B/t) i . Explicit calculation yields t πA 8/3 , and in general y k = f k (t, A)(B/t) k for some function f k . It follows that I(m, n 2 1 n 2 , q) is essentially given by where g 1 (A) = ∓t 2/3 /3(πA) 2/3 ≪ 1 and g 2 (A) ≪ 1/t. Also note that B ≍ (NN 0 ) 1/3 /qr 1/3 . It follows that the integral I is given by × e − N 0 n 1 n 2 y q 2 q ′ 2 q 1 r dy where in B, B ′ we replace n 2 1 n 2 by N 0 y. Since n 2 = 0 we get N 0 n 1 n 2 y q 2 q ′ 2 q 1 r ≫ N 0 n 1 C 2 r ≫ t ε NN 0 C 3 rt 2 as C ≫ t 1+ε and N ≪ t 3+ε . Making a change of variable y = z 3 and using the third derivative bound for the exponential integral we get I ≪ 1 t q 2 q ′ 2 q 1 r N 0 n 1 n 2 1/3 ≪ Cr 1/3 t 2/3 t(NN 0 ) 1/3 .
In our bounds for Ω (see (16) and (18)), we had the factor N 0 N 2 which boils down to C(NN 0 ) 1/3 r 2/3 /n 1 q 1 by substituting the value of N 2 . Now when we incorporate the new bound for the integral, this factor is replaced by C 2 r/n 1 q 1 t 1/3 . Making this replacement in the proof of Lemma 4, we get Lemma 5.