BILINEAR FORMS WITH KLOOSTERMAN SUMS

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1 BILINEAR FORMS WITH KLOOSTERMAN SUMS 1. Introduction For K an arithmetic function α = (α m ) 1, β = (β) 1 it is often useful to estimate bilinear forms of the shape B(K, α, β) = α m β n K(mn). m n complex coefficients, With applications to modular forms in mind, we restrict our attention to the situation in which K is a Kloosterman or hyper-kloosterman sum, i.e., for some k we have K = Kl k (, q) : (Z/qZ) C n q 1 k x 1,...,x k (Z/qZ) x 1...x k =n e q (x x k ) We can extend K to an arithmetic function by 0, or in other controlled ways. Since the prime q shall be somewhat large compared to the supports of α β, the precise nature of this extension does not affect our results. Our coefficients α shall be supported on [M] := {1,..., M}, while our coefficients β shall be supported on an interval N [1, q 1] of length N. Since K 1 (as a well-known consequence of Deligne s work), we can use Cauchy or Hölder to bound B(K, α, β) trivially, for instance, B(K, α, β) n α β (MN) 1/ (q 1/4 + M 1/ + N 1/ q 1/4 log q), an estimate that is nontrivial in the ranges M q δ, N q 1/+δ for some δ > 0, for instance. A fundamental challenge, when dealing with incomplete character sums, is to go beyond the Pòlya-Vinogradov range. )For Dirichlet Characters, Burgess bounds are the archetype [... ].) This was achieved in the present context by Kowalski, Michel, Sawin [?]. Theorem 1.1 ([?]). Let q be a prime, let M, N R satisfy Then for any ɛ > 0 we have 1 M Nq 1/4, q 1/4 < MN < q 5/4. B(K, α, β) n,ɛ q ɛ α β (MN) 1/ (M 1/ + (MN) 3/16 q 11/64 ). This is nontrivial when M = N q 11/4 + δ, for instance. We offer the following bound, which goes further beyond the Pòlya-Vinogradov range: 1

2 BILINEAR FORMS WITH KLOOSTERMAN SUMS Theorem 1.. Assume Then 1 M Nq, q 7/8 MN q 64. B(K, α, β) n,ɛ q ɛ α β (MN) 1/ (M 1/ + (q 7 (MN) 8 ) 1/7 ). This beats the trivial bound when M = N q δ, for instance. In applications, it is often beneficial to have specific bounds tailored to the scenario in which β = 1 N. This is the Type I scenario arising in the Vaughan [?] Heath-Brown [?] identities, the more general situation addressed in Theorem 1.1 is known as Type II. Kowalski, Michel, Sawin obtained the following Type I estimate: Theorem 1.3 ([?]). Assume α 1, that Then 1 M N, N < q, MN < q 3. ( B(K, α, 1 N ) q ɛ α 1/ 1 α 1/ M M 1/4 N 5 ) 1/1 N. q 3 Note that Cauchy gives α 1 M 1/ α, so a trivial bound is B(K, α, 1 N ) N α 1 α 1/ 1 α 1/ M 1/4 N. Theorem 1.3 beats this when M = N q 3/7+δ, for instance. Theorem 1.4. Assume that α 1 1 M N 3, MN q. Then ( B(K, α, 1 N ) q ɛ α /3 1 α 1/3 q M 1/6 4 ) 1/4 N. M 3 N 7 This defeats the trivial estimate as soon as M = N q /5+δ, say. B(K, α, 1 N ) N α 1 α /3 1 α 1/3 M 1/6 N.. Proof of Theorem 1.4 To prove Theorem 1.4, we begin as in [?, ] with the +ab-shifting trick. Given parameters A, B 1 such that we have B(K, α, N ) qɛ AB AB N, AM < q, s AM ν(r, s)µ(r, s) ν(r, s) =... (note the N here should be N, an interval of length N) µ(r, s) = η B K(s(r + b)). B<b B

3 BILINEAR FORMS WITH KLOOSTERMAN SUMS 3 For ν, we have the moment estimates ν 1 AN α 1 ν q ɛ AN α from [?]. Now we apply Hölder s inequality with exponent 6: s AM ν(r, s)µ(r, s) = νµ 1 ν /3 3/ ν 1/3 6 µ 6 (AN α 1 ) /3 (q ɛ AN α ) 1/6 µ 6 We adapt the stard notational convention that ɛ denotes an arbitrarily small positive number, whose value may differ between instances. After a small amount of bookkeeping, we now have (1) B(K, α, N ) qɛ AB (AN) 5 /3 6 α 1 α 1/3 µ 6. By the triangle inequality, we have Here K(x) = K(x). S(K, b, AM) = µ 6 6 b B S(K, b; AM) B = { b Z 6 : B < b i B, 1 i 6} i=1 K(s(r + b i ))K(s(r + b i+3 )). Definition.1. Let V be the affine variety of sextuples b = (b1,..., b 6 ) A 6 F q defined by the conditions (1) If k is even, then for any i {1,..., 6} the cardinality #{j : b j = b i } is even. () If k is odd not equal to 3, then {{b 1, b, b 3 }} = {{b 4, b 5, b 6 }} is an equality of multisets. (3) If k = 3, then either {{b 1, b, b 3 }} = {{b 4, b 5, b 6 }} or b = (b, b, b, b, b, b ) for some b, b. The role of the diagonal set is played by B = { b B : b mod q V }. Observe that the contribution from the vectors b B to µ 6 6 satisfies b B S(K, b; AM) qab 3 M := x 1.

4 4 BILINEAR FORMS WITH KLOOSTERMAN SUMS For b B, we can exploit averaging over r: Lemma.. For b B\B s F q, we have i=1 K(s(r + b i ))K(s(r + b i+3 )) q 1/. In particular, for any B B\B we have S(K, b, AM) AMq 1/ B. b B We refer to 3 for the proof. Generically we ll need to save more than q 1/. An application of the Plancherel formula this is the Pòlya-Vinogradov method from 4 of our course notes yields with S(K, b; AM) (log q) max λ mod q Ŝ(K, b, λ) Ŝ(K, b, λ) = s mod q R(K, r, λ, b) R(K, r, λ, b) = R(K, r, λ, b) = e q (λs) K(s(r + b i ))K(s(r + b i+3 )). i=1 By following the proof of [?, Theorem.3], we obtain the following generic estimate. Theorem.3. There exists a codimension one subvariety V bad A 6 F q containing V, with degree bounded independently of q, such that if λ F q b V bad (F q ) then Ŝ(K, b, λ) q therefore S(K, b, AM) q log q. This uses the full power of Deligne-Katz [?], but an improvement could still be sought on the codimension. Write B bad = { b B : b mod q V bad (F q )} B gen = B\B bad. By Schwartz-Zippel uniformity of the degree bound, we have #B bad (deg V bad )B 5 B 5. Thus by Lemma. we have b B bad \B S(K, b; AM) q 1/ AB 5 M := x. By Theorem.3 we have S(K, b; AM) (log q)qb 6 := (log q)x 3. Thus b B gen µ 6 6 (x 1 + x + x 3 ) log q x 1 = qab 3 M, x = q 1/ AB 5 M, x 3 = qb 6.

5 BILINEAR FORMS WITH KLOOSTERMAN SUMS 5 Choosing A = M 1/4 N 3/4, B = M 1/4 N 1/4 ensures that AB = N x 1 = x 3. We note that the hypotheses of our theorem ensure that A 1, AM < q as our parameters are acceptable. Moreover, the hypothesis M N q ensures that x x 3 = q(mn) 3/. Now from (1) we have B(K, α, N ) qɛ N (AN)5/6 α /3 1 α 1/3 q 1/6 (MN) 1/4 = qɛ+1/6 N N 5/6( N 15 ) /3 1/4 α M 5 1 α 1/3 (MN) 1/4 = q ɛ+1/6 M 1/4 N 17/4 α /3 1 α 1/3 = q ɛ M 1/6 N α /3 1 α 1/3 ( q 4 ) 1/4 M 3 N 7 We use a similar strategy to prove Theorem 1.. This time Cauchy-Scwhartz, the +ab-shifiting trick, Hölder-6 give Here S (K, b; AM) = B(K, α, β) α β (N + qɛ AB M /3 (AN) 5/6 µ 6 ), i=1 s,s AM s1 s mod q µ 6 6 = b B S (K, b; AM). K(s 1 (r+b i ))K(s (r+b i ))K(s 1 (r+b i+3 ))K(s (r+b i+3 )). In 3, we shall also confirm the following analogue of Lemma.: Lemma.4. For any subset B B\B we have S (K, b, AM) (AM) q 1/ B. b B For B, we have the trivial bound () S (K, b, AM) qa B 3 M := y 1 b B We WHAT the condition s 1 s mod q by the indicator function expression, 1 1 e q (λ(s 1 s )). q λ mod q The Pólya-Vinogradov method then gives S (K, b, AM) (log q) + (log q) max Ŝ(K, b, λ 1, λ ). λ 1 λ λ 1,λ mod q

6 6 BILINEAR FORMS WITH KLOOSTERMAN SUMS Here Ŝ(K, b, λ 1, λ ) = ζ(λ 1, λ, b) 1 q ζ(λ 1, λ, b) = λ mod q ζ(λ 1 + λ, λ + λ, b), R(r, λ 1, b)r(r, λ, b). Mimicking the proof of [KMS, Theorem.5] gives Theorem.5. There exists a codimension one subvariety V bad A 6 F q containing V, with degree bounded independently of q, such that for every b V bad (F q ) every distinct λ 1, λ F q, we have Ŝ(K, λ 1, λ, b) q 3/. In fact, V bad is the same in Theorem.3 Theorem.5. Using Lemma.4 for b B bad \B Theorem.5 for B gen () for B gives Choosing we have µ 6 6 (log q) (y 1 + y + y 3 ) y 1 = qa B 3 M, y = q 1/ A B 5 M, y 3 = q 3/ B 6. A = q 1/3 M /3 N 1/3, B = q 1/3 M /3 N /3, AB = N, y = y 3. Moreover, the hypothesis MN q 7/8 implies that y 1 y 3 = q 1/ M 4 N 4. Now Thus B(K, α, β) α β N + q ɛ( M /3 (AN) 5/6 ) 1/ ( M 4 N 4 ) 1/1 AB q 1/ = N + q ɛ 1/4 M 1/3 (qm N 4 ) 5/36 M 1/3 N 1/6 = N + q ɛ+7/7 M 7/18 N 7/18 = N + q ɛ (MN) 1/ (q 7 (MN) 8 ) 1/7. B(K, α, β) q ɛ α β (MN) 1/ (M 1/ + q 7 (MN) 8 ) 1/7. 3. Proof of lemmas Proof of Lemma.. We appeal directly to [?, Corollary 1.6]. The relevant vector is γ = (γ s,1,..., γ s,6 ), for i = 1,..., 6. γ s,i = ( ) s sbi, 0 1 Case 1: k even. If b B then there exists an i such that #{j : b j = b i } is odd.

7 BILINEAR FORMS WITH KLOOSTERMAN SUMS 7 Case : k > 3 odd. Here σ = (1, 1, 1, c, c, c), c denotes complex conjugation. If b B, then {{b 1, b, b }} {{b 4, b 5, b 6 }}. In particular, there exists an i such that #{j : b i = b j, j 3} #{j : b i = b j, j 4}. As known from [?, Remark 1.9], the special involution is ( ) 1 0 ξ =. 0 1 Since q s F q, we can never have ξγ i = γ j. The conditions () (3) of [?, Definition 1.3] are thus equivalent. Case 3: k = 3 This is almost the same as Case. The only thing that could go wrong is if β = (b, b, b, b, b, b ), but we have explicitly excluded this. Proof of Lemma.4. We have γ = (γ s1,1,..., γ s1,6, γ s,1,..., γ s,6 σ = (1, 1, 1, c, c, c, c, c, c, 1, 1, 1). Recall that s 1 s mod q. Case 1: k even. If b B then there exists an i such that #{j : b j = b i } is odd. Thus #{j 1 : γ j = γ i } is odd; here γ i = γ s1,i, since i < b. Case : k > 3 odd. If b B then #{j : b i = b j, j 3} #{j : b i = b j, j 4}, so there exists i 6 such that #{j 3 : b j = b i } #{j 4 : b j = b i }. note that γ j γ i for j > 6 (as s 1 s mod q). Since k > 3, this also means that the two expressions are incongruent modulo k. Also, k-normality of ( γ, σ) is the same with or without respect to the special involution ξ, since we can never have ξγ i = γ j. To see this, note that if ξγ i = γ j then q s i or q (s 1 + s ), for some i = 1,, both of which are impossible since q, s i F q, 4AM = 4(qMN) 1/3 q 64MN q the latter given by our hypothesis. Case 3: k = 3. Again this is basically the same as Case, since we ve explicitly forbidden vectors b of the shape (b, b, b, b b b ).