Bounds for local Kloosterman sums

Then

Kl_p n, ψ, ψ⁰

=

p^` 1−p⁻¹−r X

x∈T \X(n)

N(x)S_w(θ_x;`), whereN(x) :=|T ∗x|is the size of T-orbit of x∈X(n).

Proof. Rewrite the Kloosterman sum Kl_p n, ψ, ψ⁰

= X

x∈T \X(n)

X

y∈T ∗x

ψ(u(y))ψ⁰ u⁰(y)

= X

x∈T \X(n)

X

y∈T ∗x r

Y

i=1

e(niκi(y))

r

Y

i=1 w(αi)<0

e n⁰_iκ⁰_i(y)

=|V_w(`)|⁻¹ X

x∈T \X(n)

X

y∈T ∗x

X

λ×λ⁰∈Vw(`) r

Y

i=1

e(λiniκi(y))

r

Y

i=1 w(αi)<0

e λ⁰_in⁰_iκ⁰_i(y)

=|V_w(`)|⁻¹ X

x∈T \X(n)

N(x) X

λ×λ⁰∈Vw(`) r

Y

i=1

e(λiniκi(y))

r

Y

i=1 w(αi)<0

e λ⁰_in⁰_iκ⁰_i(y)

=

p^` 1−p⁻¹−r X

x∈T \X(n)

N(x)Sw(θx;`).

alone is in general insufficient to give a non-trivial bound for Kl_p(n_w,r,s, ψ, ψ⁰). To obtain non-trivial bounds, we use two different approaches. Note that Klp(nw,r,s, ψ, ψ⁰) is in general an exponential sum of the form

X

x∈S

e f(x)

p^k

for some k∈N. The approach we use then depends on the value of k:

(i) whenk≥2, we use thep-adic stationary phase method [DF97];

(ii) when k = 1, the stationary phase method fails, and we instead apply known results for exponential sums, which are derived using algebro-geometric arguments.

We now give an overview of the p-adic stationary phase method, following [DF97]. Let us first consider a simple case. Letf be a polynomial with coefficients in Z. Form∈Nwe consider the exponential sum

S_m(f) := X

x∈Z/p^mZ

e f(x)

p^m

. Consider the Taylor expansion off

f(x+p^m−jy) =f(x) +p^m−jf⁰(x)y+1

2p^2(m−j)f⁰⁰(x)y²+· · ·. If2(m−j)≥m (or2(m−j)−1≥m if p= 2), then we see that

Sm(f) =p^−j X

x∈Z/p^mZ

X

y∈Z/p^jZ

e

f(x+p^m−jy) p^m

= X

x∈Z/p^mZ

e f(x)

p^m

·p^−j X

y∈Z/p^jZ

e

f⁰(x)y p^j

.

The inner sum vanishes unlessf⁰(x)≡0 (mod p^j), hence the sum becomes Sm(f) = X

x∈Z/p^mZ f⁰(x)≡0 (modp^j)

e f(x)

p^m

.

This generalises easily to higher-dimensional cases. Let V be a smooth scheme of dimension n, and f :V →A¹=A¹_Zp aZp-morphism. We consider the exponential sum

S =Sm(f) := X

x∈V(Z/p^mZ)

e f(x)

p^m

. (3.26)

Letj≤mbe a positive integer. We write D(Z/p^jZ) :=

x∈V(Z/p^jZ)

∇f(x)≡0 (mod p^j) (3.27) to denote the “approximate critical points” off. For x∈(Z/p^jZ)ⁿ, we define

S_x = X

x∈V(Z/p^mZ) x≡x(modp^j)

e f(x)

p^m

.

Clearly we have

S= X

x∈(Z/p^jZ)ⁿ

S_x.

Theorem 3.8. [DF97, Theorem 1.8(a)] If 2j ≤ m, then S_x = 0 unless x ∈ D(Z/p^jZ). Now suppose m = 2j or 2j+ 1, and let x ∈ (Z/p^mZ)ⁿ map to x ∈D(Z/p^jZ). If m = 2j, then we have

S_x=p^mn/2e f(x)

p^m

. Ifm= 2j+ 1, then we have

Sx=p^(m−1)n/2e f(x)

p^m

X

y∈(Z/pZ)ⁿ

e

1

2y^THxy+p^−j∇f(x)·y p

! ,

whereH_x is the Hessian matrix of f at x. In particular, if we let tdenote the maximum value of n−rank_FpH_x for x∈D(Z/p^jZ), then|S| ≤

D(Z/p^jZ)

p^(mn+t)/2.

Proof. We give a proof to the special case where V = Aⁿ is the affine space. Then f is a polynomial with coefficients in Zp. The general case follows from a reduction lemma [DF97, Lemma 1.18], which reduces the general case into this special case.

Consider the Taylor expansion off

f(x+p^m−jy) =f(x) +p^m−j∇f(x)·y+1

2p^2(m−j)y^TH_xy+· · · Since2j≤m, we have

f(x+p^m−jy) =f(x) +p^m−j∇f(x)·y ∈Z/p^mZ.

This is obvious whenpis odd, and when p= 2, the diagonal entries of the HessianH_x are even, so the second-order term vanishes as well. Hence

S_x=p^−nj X

x∈(Z/p^mZ)ⁿ x≡x(modp^j)

X

y∈(Z/p^jZ)ⁿ

e f(x+p^(m−j)y) p^m

!

= X

x∈(Z/p^mZ)ⁿ x≡x(modp^j)

e f(x)

p^m

·p^−nj X

y∈(Z/p^jZ)ⁿ

e

∇f(x)·y p^j

.

The inner sum vanishes unless ∇f(x) ≡0 (modp)^j, that is, x ∈D(Z/p^jZ). Assuming this is the case, we continue

S_x = X

y∈(Z/p^m−jZ)ⁿ

e

f(x+p^jy) p^m

. Ifm= 2j, thenf(x+p^jy) =f(x) +p^j∇f(x)·y =f(x)∈Z/p^mZ, so

S_x=p^mn/2e f(x)

p^m

. Ifm= 2j+ 1, then we have

f(x+p^jy) =f(x) +p^j∇f(x)·y+1

2p^2jy^TH_xy∈Z/p^mZ. Hence

S_x=e f(x)

p^m

X

y∈(Z/p^m−jZ)ⁿ

e

1

2p^2jy^TH_xy+p^j∇f(x)·y p^m

!

=p^(m−1)n/2e f(x)

p^m

X

y∈(Z/pZ)ⁿ

e

1

2y^THxy+p^−j∇f(x)·y p

! .

Finally, we observe that the inner sum is an n-dimensional Gauß sum, and it follows from straightforward computations that the Gauß sum is bounded byp^{n−rankFpHx/2}. The bound for S then follows.

Theorem 3.9. Let 0 ≤ s ≤ r be integers, and ψ = ψm1,m2, ψ⁰ = ψn1,n2 characters of U(Qp)/U(Zp). Then

Klp nsαsβ,r,s, ψ, ψ⁰

min

p^2s

|m₁|⁻¹_p , p^r−s

, p^r

|m₂|⁻¹_p , p^s 1/2

|n₂|⁻¹_p , p^s 1/2

. Proof. We may assumev_p(m₁)≤r−s, and v_p(m₂), v_p(n₂)≤s. Observe that

Kl_p n_sαsβ,r,s, ψ_m1,m2, ψ_n1,n2

=p^k+2lKl_p n_sαsβ,r−k−l,s−lψ_m1p^−k_,m2p^−l, ψ_n1,n2p^−l

wheneverp^k|(m1, p^r−s)andp^l |(m2, n2, p^s). So we may assumes= 0,r =s, orp-m1(m2, n2).

Ifs= 0, then

Klp nsαsβ,r,0, ψ, ψ⁰ =

X

v3(modp^r) (v3,p^r)=1

e

m1v3

p^r

≤p^vp(m1).

Ifr =s, then

Klp nsαs_β,r,r, ψ, ψ⁰ =

X

v4(modp^r) (v4,p^r)=1

X

v3(modp^r)

e

m2v4v₃²+n2v4

p^r

≤p^{r+vp(m2 2)+vp(n22)}

is just a summation of quadratic Gauss sums, and is easily evaluated.

Now supposep-m1(m2, n2). Ifp|m2 and s >1, then Kl_p n_sαsβ,r,s, ψ, ψ⁰

= X

v⁰₄(modp^s−1) (v₄⁰,p)=1

X

v3(modp^r) (v3,p^r−s)=1

p−1

X

k=0

e

m₁v₃ p^r−s

e m₂v₄⁰v₃²+n₂ v₄⁰ +kp^s−1 p^s

!

=p

p−1

X

k=0

e n2k

p

Klp nsαsβ,r−1,s−1, ψ_m1,m2/p, ψ⁰

= 0.

Ifp|m2 and s= 1, the same argument shows that the sum is either 0 orp. Similarly, ifp|n2, the sum is also either 0 orp. So we may assume p-m₁m₂n₂.

Ifr >2s, we writer = 2s+l, for l >0. Then Kl_p n_{sαsβ,2s+l,s}, ψ, ψ⁰

= X

v4(modp^s) (v4,p)=1

p^s+l−1−1

X

v⁰₃=0 (v⁰₃,p)=1

p−1

X

k=0

e m1(v₃⁰ +kp^s+l−1) +p^lm2v4(v₃⁰ +kp^s+l−1)²+p^ln2v4

p^s+l

! ,

where k (modp) is chosen such that (v⁰₃+kp^s+l−1)(v₃⁰ +kp^s+l−1) ≡ 1 (modp^s+l). Then the sum becomes

Kl_p n_{sαsβ,2s+l,s}, ψ, ψ⁰

= X

v4(modp^s) (v4,p)=1

p^s+l−1−1

X

v⁰₃=0 (v₃⁰,p)=1

e m₁v⁰₃+p^lm₂v₄v₃⁰²+p^ln₂v₄ p^s+l

!_p−1 X

k=0

e m₁k

p

= 0.

Ifr <2s, we writer = 2s−l, for 0< l < s. Then Klp n_sαsβ,2s−l,s, ψ, ψ⁰

= X

v4(modp^s) (v4,p)=1

X

v3(modp^2s−l) (v3,p)=1

e

p^lm1v3+m2v4v₃²+n2v4

p^s

=p^s−l X

v4(modp^s) (v4,p)=1

X

v3(modp^s) (v3,p)=1

e

p^lm₁v₃+m₂v₄v²₃+n₂v₄ p^s

.

Whenp is odd, we apply the same argument and see that Kl_p n_sαsβ,2s−l,s, ψ, ψ⁰

=p^s−l X

v4(modp^s) (v4,p)=1

p^s−1−1

X

v⁰₃=0 (v₃⁰,p)=1

p−1

X

k=0

e p^lm₁(v₃⁰ +kp^s−1) +m₂v₄(v₃⁰ +kp^s−1)²+n₂v₄ p^s

!

=p^s−l X

v4(modp^s) (v4,p)=1

p^s−1−1

X

v⁰₃=0 (v₃⁰,p)=1

e p^lm₁v₃⁰ +m₂v₄v₃⁰²+n₂v₄ p^s

!p−1

X

k=0

e

2m₂v₄v⁰₃k p

= 0.

Whenp= 2, if we further assumel≥2, then we have Kl_p n_sαsβ,2s−l,s, ψ, ψ⁰

= p^s−l X

v4(modp^s) (v4,p)=1

p^s−2−1

X

v⁰₃=0 (v₃⁰,p)=1

p²−1

X

k=0

e p^lm₁(v₃⁰ +kp^s−2) +m₂v₄(v⁰₃+kp^s−2)²+n₂v₄ p^s

! ,

where nowk (mod p²) is chosen such that(v⁰₃+kp^s+l−2)(v⁰₃+kp^s+l−2)≡1 (mod p^s+l). Then the sum becomes

Kl_p n_{sαsβ,2s−l,s}, ψ, ψ⁰

= p^s−l X

v4(modp^s) (v4,p)=1

p^s−2−1

X

v₃⁰=0 (v⁰₃,p)=1

e p^lm₁v⁰₃+m₂v₄v₃⁰²+n₂v₄ p^s

!_p²₋₁ X

k=0

e

2m₂v₄v₃⁰k p²

= 0.

Therefore, it remains to consider the case r= 2s, and, ifp= 2, the caser= 2s−1.

Now supposer = 2s. Whens= 1, we have Klp nsαs_β,2,1, ψ, ψ⁰

=p X

v4(modp) (v4,p)=1

X

v3(modp) (v3,p)=1

e

m1v3+m2v4v₃²+n2v4

p

.

When p = 2, there is nothing to prove. When p is odd, this exponential sum is estimated by Adolphson and Sperber [AS89, Corollary 4.3] to be of O(p²)as well. So we conclude that

Klp nsαs_β,2,1, ψ, ψ⁰ p². So the theorem holds for this case.

If s >1, we apply the stationary phase method. Let f(x, y) = ^m_x¹ +^m2_y^x2 +n₂y. Consider the sum

S= X

x,y∈(Z/p^sZ)^×

e

f(x, y) p^s

=p^−sKlp nsαsβ,2s,s, ψ, ψ⁰ . Letj≥1be such that2j≤s. Define as in (3.27)

D Z/p^jZ

= n

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

∇f(x, y)≡0 (modp^j) o

=

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

2m₂x³ ≡m₁y (mod p^j), m2x² ≡n2y² (modp^j)

.

It is straightforward to check that

D Z/p^jZ

≤ 4, and H_x,y is invertible over Fp for all (x, y)∈D Z/p^jZ

, sorank_FpHx,y= 2. So we deduce from Theorem 3.8 that

Klp nsαsβ,r,s, ψ, ψ⁰

≤4p^2s.

Now supposep= 2, andr = 2s−1. It suffices to prove the bound for sufficiently larges, so we can always use stationary phase method. Letf(x, y) = ^2m_x¹ +^m2_y^x2 +n₂y. Consider the sum

S = X

x,y∈(Z/p^sZ)^×

e

f(x, y) p^s

=p^−s+1Klp(nsαs_β,2s−1,s, ψ, ψ⁰).

Letj≥1be such that2j≤s. Define as in (3.27) D Z/p^jZ

= n

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

∇f(x, y)≡0 (modp^j) o

=

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

2m₂x³ ≡2m₁y (modp^j), m2x²≡n2y² (modp^j)

. Then we have

D Z/p^jZ

≤16. The Hessian Hx,y is not invertible, but nevertheless we have from Theorem 3.8 that

Klp nsαsβ,2s−1,s, ψ, ψ⁰

≤64p^2s−1. This finishes the proof of the theorem.

Theorem 3.10. Let 0 ≤ 2r ≤ s be integers, and ψ = ψ_m1,m2, ψ⁰ = ψ_n1,n2 characters of U(Qp)/U(Zp). Then

Kl_p n_sβsα,r,s, ψ, ψ⁰

minn p^3r

|m₂|⁻¹_p , p^s−2r , p^s

|m₁|⁻¹_p ,|n₁|⁻¹_p , p^ro .

Remark. Up to multiplication by a constant, this Kloosterman sum can also be considered as a GL(3)Kloosterman sum. Precisely, following the notation in [BFG88, (4.3)], we have

Kl_p n_sβsα,r,s, ψ, ψ⁰

=p^rS n₁, m₁, m₂;p^r, p^s−r . A non-trivial bound for Kl_p n_sβsα,r,s, ψ, ψ⁰

then follows from Larsen [BFG88, Appendix]. For sake of completeness, we still give a proof below.

Proof. We may assume that v_p(m₂)≤s−2r, and v_p(m₁), v_p(n₁)≤r. Observe that Kl_p n_sβsα,r,s, ψ_m1,m2, ψ_n1,n2

=p^3k+lKl_p n_sβsα,r−k,s−2k−l, ψ_m1p−k,m2p^−l, ψ_n1p−k,n2

whenever p^k | (m₁, n₁, p^r) and p^l | (m₂, p^s−2r). So we may assume r = 0, s = 2r, or p -m2(m1, n1).

Ifr = 0, then

Klp nsβsα,0,s, ψ, ψ⁰ =

X

v34(modp^s) (v34,p^s)=1

e

m2v34

p^s

≤p^vp(m2).

Ifs= 2r, then

Kl_p n_sβsα,r,2r, ψ, ψ⁰ =

X

v24(modp^r) (v24,p^r)=1

X

v34(modp^2r)

e

m₁v₂₄v₃₄+n₁v₂₄ p^r

≤p^{2r+min{vp(m1),vp(n1)}}.

Now supposep-m₂(m₁, n₁). Ifp|m₁ andr >1, then Kl_p(n_sβsα,r,s, ψ, ψ⁰) = X

v₂₄⁰ (modp^r−1) (v⁰₂₄,p)=1

X

v34(modp^s) (v34,p^s−2r)=1

p−1

X

k=0

e m₁v⁰₂₄v₃₄+n₁(v⁰₂₄+kp^r−1) p^r

! e

m₂v₃₄ p^s−2r

=p²

p−1

X

k=0

e n₁k

p

Kl_p n_sβsα,r−1,s−2, ψ_m1/p,m2, ψ⁰

= 0.

Ifp|m₁ and r= 1, the same argument shows that the sum is either 0 orp. Similarly, ifp|n₁, the sum is also either 0 orp. So we may assume p-m1m2n1.

Ifs >3r, we writes= 3r+l, for l >0. Then Kl_p n_{sβsα,r,3r+l}, ψ, ψ⁰

=p^2r X

v24(modp^r) (v24,p)=1

p^r+l−1−1

X

v⁰₃₄=0 (v⁰₃₄,p)=1

p−1

X

k=0

e p^lm₁v₂₄(v₃₄⁰ +kp^r+l−1) +p^ln₁v₂₄+m₂(v₃₄⁰ +kp^r+l−1) p^r+l

! ,

wherek (mod p) is chosen such that(v⁰₃₄+kp^r+l−1)(v⁰₃₄+kp^r+l−1)≡1 (mod p^r+l). Then the sum becomes

Kl_p n_{sβsα,r,3r+l}, ψ, ψ⁰

=p^2r X

v24(modp^r) (v24,p)=1

p^r+l−1−1

X

v⁰₃₄=0 (v⁰₃₄,p)=1

e p^lm₁v₂₄v⁰₃₄+p^ln₁v₂₄+m₂v⁰₃₄ p^r+l

!_p−1 X

k=0 p−1

X

k=0

e m₂k

p

= 0.

Ifs <3r, we writes= 3r−l, for 0< l < r. We apply the same argument, and obtain Kl_p n_sβsα,r,3r−l, ψ, ψ⁰

=p^2r−l X

v24(modp^r) (v24,p)=1

p^r−1−1

X

v⁰₃₄=0 (v₃₄⁰ ,p)=1

p−1

X

k=0

e m₁v₂₄(v⁰₃₄+kp^r−1) +n₁v₂₄+p^lm₂(v₃₄⁰ +kp^r−1) p^r

!

=p^2r−l X

v24(modp^r) (v24,p)=1

p^r−1−1

X

v⁰₃₄=0 (v₃₄⁰ ,p)=1

e m₁v₂₄v₃₄⁰ +n₁v₂₄+p^lm₂v₃₄⁰ p^r

!_p−1 X

k=0

e

m₁v₂₄k p

= 0.

So it remains to consider the cases= 3r. Whenr= 1, we have Klp nsβsα,1,3, ψ, ψ⁰

=p² X

v24(modp) (v24,p)=1

X

v34(modp) (v34,p)=1

e

m1v24v34+n1v24+m2v34

p

.

Letx=m1v24v34,y =n1v24, and z=m2v34. After this change of variables, the sum becomes

p² X

x,y,z∈Fp

xyz=m2m1n1

e

x+y+z p

,

which is known as a generalised Kloosterman sum in the sense of Deligne [Del77]. By a theorem of Deligne [Del77, Sommes. trig., 7.1.3], this sum is bounded by 3p³. So the theorem holds for this case.

Forr >1, we apply the stationary phase method. Let f(x, y) = ^m_x^1y +n₁x+^m_y². Consider the sum

S = X

x,y∈(Z/p^rZ)^×

e

f(x, y) p^r

=p^−2rKlp nsβsα,r,3r, ψ, ψ⁰ .

Letj≥1be such that2j≤r. Define as in (3.27) D Z/p^jZ

=n

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

∇f(x, y)≡0 (mod p^j)o

=

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

m₁y≡n₁x² (modp^j), m₁y² ≡m₂x (modp^j)

. We have

D(Z/p^jZ)

≤ 3. The Hessian H_x,y is invertible unless p = 3. So we conclude from Theorem 3.8 that

Klp(nsβsα,r,3r, ψ, ψ⁰)

p^3r. This finishes the proof of the theorem.

Theorem 3.11. Let 0 ≤ s ≤ 2r be integers, and ψ = ψ_m1,m2, ψ⁰ = ψ_n1,n2 characters of U(Qp)/U(Zp). Then

Klp ns_αs_βs_α,r,s, ψ, ψ⁰







p^{r3+2s3+23min{vp(m1)+s,vp(n1)+r}+13vp(m2)} if s≤r, p^{r+min{vp(m2),r+vp(n1)}}+p^r+min{^s₂^+vp(m₁),r−^s₂+v_p(n₁)} if r < s <2r,

p^{r+min{vp(m2),r+vp(n1)}} if s= 2r.

Proof. We make use of the stratification of Kloosterman sums in Section 3.2. Forw=sαsβsα, we have∆_w ={α}. Hence, for`∈N, we have

A_w(`) = (Z/p<sup>`</sup>Z)²×(Z/p^`Z).

Lett= diag a₁, a₂, ca⁻¹₁ , ca⁻¹₂

∈ T. Thens:=n⁻¹tn= diag ca⁻¹₁ , a₂, a₁, ca⁻¹₂

. We compute κ⁰₁(t∗x) =ca⁻¹₁ a⁻¹₂ κ⁰₁(x).

So

Vw(`) =

λ×λ⁰ ∈Aw(`)

λ₁, λ₂, λ⁰₁ ∈(Z/p^`Z)^×, λ1λ2λ⁰₁= 1

. Letθ:A_w(`)→C^× be a character given by

θ(λ×λ⁰) =e

n1λ1+n2λ2

p^`

e

n⁰₁λ⁰₁ p^`

for n1, n2, n⁰₁ ∈Z, then

S_w(θ, `) = X

λ2∈(Z/p^`Z)^×

e n₂λ₂

p^`

S(n₁λ₂, n⁰₁;p^`). (3.28)

Let n= nsαsβsα,r,s. In terms of Plücker coordinates (see Section 2.2.4), this says v1 =p^r and v14=p^s. Suppose x^v_a,b³ ∈X(n) has coordinates

(v₁, v₂, v₃, v₄;v₁₄) = (p^r, p^r−a, v₃, p^r−b;p^s).

Let δ⁰ = (p^r−a, p^av₃ +p^r−b). Then v₁₄ = p^r+a/δ⁰. This says s−r ≤ a ≤ s/2, b ≤ r, and δ⁰ =p^r+a−s. From Bruhat decomposition, we have

u⁰

x^v_a,b³

=









1 p^−a v₃p^−r p^−b 1 p^−b

1

−p^−a 1









(modU(Zp)).

LetX_a,b^v3(n) =T ∗x^v_a,b³, and define S_a,b^v3 n, ψ, ψ⁰

= X

x∈X_a,b^v3(n)

ψ(u(x))ψ⁰ u⁰(x) .

We also set

X_a,b(n) = [

v3 (modp^r)

(^pr−a,pav3+p^r−b)^=pr+a−s

X_a,b^v3(n),

and

S_a,b n, ψ, ψ⁰

= X

x∈X_a,b(n)

ψ(u(x))ψ⁰ u⁰(x) . It is easy to see that

X(n) = a

s−r≤a≤s/2 0≤b≤r

X_a,b(n).

Asr ≥s/2≥a,r ≥b, we see thatu(x), u⁰(x) have entries inp^−2rZp/Zp for all x∈X(n). Let S_a,b be a finite subset of Zp such that

Xa,b(n) = a

v3∈S_a,b

X_a,b^v3(n).

By Theorem 3.7, we have Sa,b n, ψ, ψ⁰

=p^−4r 1−p⁻¹−2 X

v3∈S_a,b

X_a,b^v3(n) Sw

θ_a,b^v3; 2r

, where

θ^v_a,b³ λ×λ⁰

=e

m₂uλ₂ p^s

e

m₁ˆv₂λ₁+n₁p^r−aλ⁰₁ p^r

, withvˆ2 andu given as in (3.6) and (3.7). By (3.28), we have

Sw

θ^v_a,b³; 2r

= X

x,y∈(Z/p^2rZ)^×

e

m2ux p^s

e

m1vˆ2xy+n1p^r−ay p^r

, (3.29)

and we easily deduce that X

v3∈S_a,b

X_a,b^v3(n)

≤ |S_a,b|p^a+b ≤p^r+a+b. (3.30)

We estimate the size of Sw

θ_a,b^v3; 2r

below. We start by computing vp(ˆv2) and vp(u). From (3.6), it is clear that v_p(ˆv₂) = s−a. Now we consider v_p(u). If a 6= s/2, then we have (after putting v⁰₂=v⁰₂= 1)

u=p^a+r−s(−p^av₃+v₄) +V⁰v₃²p^2a

=p^a+r−s(p^av3+v4)−2v3p^2a+r−s+V⁰v²₃p^2a

=p^2a+2r−2sV⁰−2v3p^2a+r−s+V⁰v²₃p^2a

=p^2aV⁰ p^2r−2sV⁰²−2p^r−sv₃V⁰+v₃²

=p^2aV⁰ p^r−sV⁰−v₃2

=p^2aV⁰ p^−av₄2

=v²₄V⁰.

So v_p(u) = 2 (r−b). Ifa=s/2, then (again we setv⁰₂=v₂⁰ = 1)

u=−v3p^2a+r−s+v4p^a+r−s=p^a+r−s(2v4−(p^av3+v4)). (3.31) These expressions will be useful in computing vp(u), when more conditions are given.

Case I: Suppose s < r. We deduce from (3.6) that v_p(v₃) = 0, v_p(v₄) = a, so only terms with r = a+b contribute. When a6=s/2, we have v_p(u) = 2 (r−b) = 2a. When a=s/2, we can still takevp(u) =s= 2a. So vp(u) = 2aalways holds.

(i) Supposea≤ ^2s−r₃ . Writeu=p^2au⁰. Let

t= min{v_p(m₂), v_p(m₁) + 2s−r−3a, v_p(n₁) +s−3a}, and

f(x, y) =p^−t

m2u⁰y+m1vˆ2p^s−r−2ax

y +n1p^s−3a x

=m⁰₂y+m⁰₁x y +n⁰₁

x , wherem⁰₁ =m₁ˆv₂p^{s−r−2a−t},m⁰₂ =m₂u⁰p^−t,n⁰₁ =n₁p^s−3a−t. Consider the sum

S = X

x,y∈(Z/p^s−2a−tZ)^×

e

f(x, y) p^s−2a−t

=p2s−4a−4r−2t

Sw

θ^v_a,b³; 2r

.

Whens−2a−t >1, letj≥1be such that2j≤s−2a−t. Define as in (3.27) D Z/p^jZ

=n

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

∇f(x, y)≡0 (modp^j)o

=

(x, y)∈ Z/p^jZ×

× Z/p^jZ×

m⁰₁x²≡n⁰₁y (modp^j), m⁰₂y² ≡m⁰₁x (modp^j)

. Note that at least one ofm⁰₁,m⁰₂andn⁰₁is not divisible byp. It then follows thatD Z/p^jZ is empty unless vp(m2) =vp(m1) + 2s−r−3a=vp(n1) +s−3a. Then this reduces to the situation seen in the proof of Theorem 3.10, and we obtain a bound

S_w

θ^v_a,b³; 2r

p^4r+2a−s+t. (3.32)

Now supposes−2a−t= 1. Ifp-m⁰₁m⁰₂n⁰₁, then it again follows by the theorem of Deligne [Del77, Sommes. trig., 7.1.3] that S p. When p divides some (but not all) of m⁰₁, m⁰₂, n⁰₁, then the sum reduces to a Ramanujan sum, and is easily evaluated thatS pas well.

So the bound (3.32) also holds for this case.

The bounds for S_w

θ^v_a,b³; 2r

in other cases are obtained analogously, and we shall omit the repetitive computations thereafter.

(ii) Suppose a > ^2s−r₃ . Writevˆ₂=p^s−avˆ₂⁰. Let

t= min{v_p(m₂) +r+ 3a−2s, v_p(m₁), v_p(n₁) +r−s}, and

f(x, y) =p^−t

m₂up^r+a−2sy+m₁ˆv⁰₂x

y + n₁p^r−s x

=m⁰₂y+m⁰₁x y +n⁰₁

x , wherem⁰₁vˆ₂⁰p^−t,m⁰₂ =m2up^r+a−2s−t,n⁰₁ =n1p^r−s−t. Then we have

S = X

x,y∈(Z/p^r+a−s−tZ)^×

e

f(x, y) p^r+a−s−t

=p2a−2r−2s−2tS_w

θ^v_a,b³; 2r . Then we obtain analogously

S_w

θ_a,b^v3; 2r

p^3r−a+s+t.

Recall that we have δ⁰ = (p^r−a, p^a(v₃+ 1)) =p^r+a−s. A necessary condition for this to hold is thatp^r−s |v3+ 1. So|S_a,b| ≤p^s. So, from (3.30) we actually have

X

v3∈S_a,b

X_a,b^v3(n)

≤p^s+a+b. Hence

Kl_p n, ψ, ψ⁰

≤ X

0≤a≤s/2 b=r−a

S_a,b n, ψ, ψ⁰

X

0≤a≤s/2 b=r−a

p^−4rp^s+a+bSw

θ^v_a,b³; 2r

X

0≤a≤s/2

minn

p^r+2a+vp(m2), ps−a+min{s+vp(m1),r+vp(n1)}o p^{r3+2s3+23min{vp(m1)+s,vp(n1)+r}+13vp(m2)}.

Case II: Supposes=r. We deduce from (3.6) that when a6= 0, thenv_p(v₃) = 0, v_p(v₄)≥a. So, only terms withr≥a+bcontribute. Whena6=s/2, we have vp(u) = 2 (r−b). Whena=s/2, we can still take v_p(u) =s= 2 (r−b). Sov_p(u) = 2 (r−b) always holds. We compute

S_w

θ_a,b^v3; 2r

p^2rminn

p^{3r−2b+vp(m2)}, p^{2r−a+min{vp(m1),vp(n1)}}o . Hence

Klp n, ψ, ψ⁰

≤ X

0≤a≤r/2 b≤r−a

S_a,b n, ψ, ψ⁰

X

0≤a≤s/2 b≤r−a

p^−4rp^r+a+b

p^2rminn

p^{3r−2b+vp(m2)}, p^{2r−a+min{vp(m1),vp(n1)}}o

X

0≤a≤s/2 b≤r−a

p^−r+a+bminn

p^{3r−2b+vp(m2)}, p^{2r−a+min{vp(m1),vp(n1)}}o

p^{5r3+23min{vp(m1),vp(n1)}+13vp(m2)}. Case III:2r > s > r. We consider the following subcases:

(a) Supposea=s−r. Then the condition p^r−a, p^av₃+p^r−b

= 1impliesb=r. Sov_p(u) = 0.

We deduce from (3.6) that ˆv2= 0. So

S_w

θ_a,b^v3; 2r

p^3r−sminn

p^r+vp(m2), p^2r+vp(n1)o .

(b) Suppose s−r < a < s/2. Then we deduce from (3.6) that v_p(v₃) = 0, v_p(v₄) ≥ a. So a+b≤r. Meanwhile, as r+a−s < a, the condition p^r−a, p^av3+p^r−b

=p^r+a−s says r−b=r+a−s, which impliesa+b=s > r, a contradiction. So there is no contribution from this case.

(c) Supposea=s/2. Again, we deduce from (3.6) thatvp(v3) = 0,vp(v4)≥a. So, only terms withr≥a+b contribute. In this case, we do not have a good bound for v_p(u). So

S_w

θ^v_a,b³; 2r

p^3r+min{^s₂+vp(m1),r−^s₂+vp(n1)}.

Hence

Kl_p n, ψ, ψ⁰

≤ X

s−r≤a≤s/2 b≤r−a

S_a,b n, ψ, ψ⁰

X

a=s−r b=r

p^−4rp^r+a+b

p^3r−sminn

p^r+vp(m2), p^2r+vp(n1)o

+ X

a=s/2 b≤r−s/2

p^−4rp^r+a+b

p^3r+min{^s₂+vp(m1),r−^s

2+vp(n1)}

p^{r+min{vp(m2),r+vp(n1)}}+p^r+min{₂^s+vp(m1),r−^s

2+vp(n1)}.

Case IV: s = 2r. In this case, we have a=r, and v₃, v₄ = p^r−b is arbitrary. We deduce from (3.6) thatvˆ2 = 0. We consider the following subcases:

(a) Suppose b= 0. We may assumev₄ = 0. Thenv_p(u) =r+v_p(v₃). We compute

Sw

θ^v_a,b³; 2r

p^rmin n

p^{2r+vp(v3)+vp(m2)}, p^2r+vp(n1) o

. Fixc≤r. Then

|{v₃ ∈ S_a,b|vp(v3) =c}| ≤p^r−c. (b) Suppose b >0. Thenv_p(u) =r−b. We compute

Sw

θ_a,b^v3; 2r

p^rmin n

p^{2r−b+vp(m2)}, p^2r+vp(n1) o

. Hence

Kl_p n, ψ, ψ⁰

≤ X

a=r/2 b≤r

S_a,b n, ψ, ψ⁰

X

a=r/2 b=0c≤r

p^−4rp^r−c+a+b

p^rminn

p^2r+c+vp(m2), p^2r+vp(n1)o

+ X

a=r/2 b>0

p^−4rp^r+a+b

p^rminn

p^{2r−b+vp(m2)}, p^2r+vp(n1)o

p^{r+min{vp(m2),r+vp(n1)}}. This finishes the proof of the theorem.

Theorem 3.12. Let 0 ≤ r ≤ s be integers, and ψ = ψ_m1,m2, ψ⁰ = ψ_n1,n2 characters of U(Qp)/U(Zp). Then

Klp ns_βsαs_β,r,s, ψ, ψ⁰







p^{s2+r2+12vp(m1)+12min{2r+vp(m2),s+vp(n2)}} if r ≤s/2, p^{s−r2+12vp(m1)+12min{2r+vp(m2),s+vp(n2)}} if s/2< r < s, p^{s+min{vp(m1),vp(n2)}}. if r =s.

Proof. We make use of the stratification of Kloosterman sums in Section 3.2. For w=s_βs_αs_β, we have∆w ={β}. Hence, for`∈N, we have

A_w(`) = Z/p<sup>`</sup>Z

2

× Z/p^`Z

.

Lett= diag a₁, a₂, ca⁻¹₁ , ca⁻¹₂

∈ T. Thens=n⁻¹tn= diag ca⁻¹₂ , ca⁻¹₁ , a₂, a₁

. We compute κ⁰₂(t∗x) =ca⁻²₁ κ⁰₂(x).

So

V_w(`) =

λ×λ⁰ ∈A_w(`)

λ₁, λ₂, λ⁰₂ ∈ Z/p^`Z×

, λ²₁λ2λ⁰₂ = 1

. Letθ:A_w(`)→C^× be a character given by

θ λ×λ⁰

=e

n₁λ₁+n₂λ₂ p^`

e

n⁰₂λ⁰₂ p^`

for n1, n2, n⁰₂ ∈Z, then

S_w(θ, `) = X

λ1∈(Z/p^`Z)^× e

n₁λ₁ p^`

S

n₂λ⁻²₁ , n⁰₂;p^`

. (3.33)

Let n=ns_βsαs_β,r,s. In terms of Plücker coordinates (see Section 2.2.4), this saysv2 =p^r, and v₁₂=p^s. Suppose x^v_a,b²³ ∈X(n)has coordinates

(v12, v13, v14, v23) =

p^s, p^s−a, p^s−b, v23

.

The condition(v12, v14)|v²₁₃sayss−b≤2 (s−a), that is,2a−b≤s. We also havemax{a, b}= r. From Bruhat decomposition, we have

u⁰ x^v_a,b²³

=









1 −v₂₃p^−s p^−a 1 p^−a p^−b

1

1









(modU(Zp)).

LetX_a,b^v23(n) =T ∗x^v_a,b²³, and define S_a,b^v23 n, ψ, ψ⁰

= X

x∈X_a,b^v23(n)

ψ(u(x))ψ⁰ u⁰(x) .

We also set

X_a,b(n) = a

v23 (modp^s)

(^ps−r,v23,p^−bv23−p^s−2a)⁼¹

X_a,b^v23(n),

and

S_a,b n, ψ, ψ⁰

= X

x∈X_a,b(n)

ψ(u(x))ψ⁰ u⁰(x) . It is easy to see that

X(n) = a

0≤a,b≤r max{a,b}=r

2a−b≤s

X_a,b(n).

It is clear thatu(x), u⁰(x) have entries inp^−sZp/Zp for all x∈X(n). Let S_a,b be a finite subset of Zp such that

Xa,b(n) = a

v23∈S_a,b

X_a,b^v23(n).

By Theorem 3.7, we have S_a,b n, ψ, ψ⁰

=p^−2s 1−p⁻¹−2 X

v23∈S_a,b

X_a,b^v23(n) Sw

θ_a,b^v23;s

, where

θ^v_a,b²³ λ×λ⁰

=e

m₁uλ₁ p^r

e

m₂ˆv₁₄λ₂+n₂p^s−bλ⁰₂ p^s

. withvˆ₁₄ and ugiven as in (3.13) and (3.14). By (3.33), we have

Sw

θ^v_a,b²³;s

= X

x,y∈(Z/p^sZ)^×

e

m1ux p^r

e

m2vˆ14x²y+n2p^s−by p^s

, (3.34)

and we easily deduce that

X

v23∈S_a,b

X_a,b^v23(n)

≤ |S_a,b|p^a+b ≤p^s+a. (3.35)

We estimate the size ofSw

θ_a,b^v23;s

. We start by computing vp(ˆv14)and vp(u) in (3.34). From (3.13), we see that

up^r−a≡v₂₃ (modp^r), up^r−b ≡ −p^s−a (mod p^r). (3.36) So, if a = r, then u ≡ v23 (modp^r), and if b = r, then u ≡ −p^s−a (mod p^r). (Recall that max{a, b}=r.) Also, we know that

v₂₃=−p^s−2a+b+βp^b (3.37)

for some β ∈Z such that β, p^s−2r+b

= 1 (see Section 2.2.4). Meanwhile, from (3.14), we see that unlessr =s, we havev_p(ˆv₁₄) = 2r−b.

Case I: Suppose r < s/2. We deduce from (3.37) that v_p(v₂₃) = b. From (3.36), we deduce a≥b. So we actually have a=r, and thenv_p(u) =b.

(i) Supposeb≤ ^3r−s₂ . Writeu=p^bu⁰. Let

t= min{v_p(m1), vp(m2) + 3r−2b−s, vp(n2) +r−2b}

and

f(x, y) =p^−t

m1u⁰

x +m2vˆ14p^r−b−sx²

y +n2p^r−2by

= m⁰₁

x +m⁰₂x²

y +n⁰₂y, wherem⁰₁ =m1u⁰p^−t,m⁰₂ =m2vˆ14p^{r−b−s−t},n⁰₂ =n2p^r−2b−t. Consider the sum

S = X

x,y∈(Z/p^r−b−tZ)^×

e

f(x, y) p^r−b−t

=p2r−2s−2b−2tS_w

θ_a,b^v23;s .

Whenr−b−t >1, let j≥1 be such that2j≤r−b−t. Define as in (3.27) D(Z/p^jZ) =

(x, y)∈(Z/p^jZ)^××(Z/p^jZ)^×

∇f(x, y)≡0 (modp^j)

=

(x, y)∈(Z/p^jZ)^××(Z/p^jZ)^×

2m⁰₂x³ ≡m⁰₁y (mod p^j) m⁰₂x² ≡n⁰₂y² (modp^j)

.

Note that at least one ofm⁰₁,m⁰₂ andn⁰₂ is not divisible byp. It then follows that whenp is odd,D(Z/p^jZ)is empty unless vp(m1) =vp(m2) + 3r−2b−s=vp(n2) +r−2b. Then this reduces to the situation seen in the proof of Theorem 3.9 (see the caser = 2s). When p = 2, D(Z/p^jZ) is empty unless v_p(m₁)−1 =v_p(m₂) + 3r−2b−s =v_p(n₂) +r−2b.

This is also dealt with in the proof of Theorem 3.9 (see the case r = 2s−1). In either case, we obtain a bound

Sw

θ^v_a,b²³;s

p^2s−r+b+t. (3.38)

Now supposer−b−t= 1. Ifp-m⁰₁m⁰₂n⁰₂, then it again follows from the argument in the proof of Theorem 3.9 that|S| p. Whenp divides some (but not all) ofm⁰₁, m⁰₂, n⁰₂, then the sum reduces to Gauß sums or Ramanujan sums, and is easily evaluated that|S| p as well. So the bound (3.38) also holds for this case.

The bounds forS_w

θ^v_a,b²³;s

in other cases are obtained analogously, and we shall omit the repetitive computations thereafter.

(ii) Suppose b > ^3r−s₂ . Writevˆ14=p^2r−bvˆ₁₄⁰ . Let

t= min{v_p(m₁) +s+ 2b−3r, v_p(m₂), v_p(n₂) +s−2r}, and

f(x, y) =p^−t

m₁up^s+b−3r

x +m₂vˆ₁₄⁰ x²

y +n2p^s−2ry

= m⁰₁

x + m⁰₂x²

y +n⁰₂y, wherem⁰₁ =m1up^s+b−3r−t,m⁰₂ =m2ˆv⁰₁₄p^−t,n⁰₂ =n2p^s−2r−t. Then we have

S = X

x,y∈(Z/p^s+b−2r−tZ)^×

e

f(x, y) p^s+b−2r−t

=p^2b−4r−2tSw

θ_a,b^v23;s

.

Then we obtain analogously S_w

θ^v_a,b²³;s

p^s+2r−b+t. Hence

Kl_p n, ψ, ψ⁰

≤ X

0≤b≤ra=r

S_a,b n, ψ, ψ⁰

X

0≤b≤ra=r

p^−2sp^s+a Sw

θ^v_a,b²³;s

X

0≤b≤ra=r

p^−2sp^s+a

p^s−rminn

p^s+b+vp(m1), pr−b+min{2r+vp(m2),s+vp(n2)}o

p^{s2+r2+12min{2r+vp(m2),s+vp(n2)}+12vp(m1)}. Case II: Supposer=s/2. We consider the following subcases:

(a) Suppose b=r. From (3.36), we may assume u= 0. We compute

S_w

θ^v_a,b²³;s

p^{3s2+min{vp(m2),vp(n2)}}.

(b) Suppose b < r. Then a= r. From (3.37), we see that v₂₃ = (β−1)p^b for some β ∈ Z such that β, p^b

= 1. So vp(v23) ≥b. And from (3.36), we deduce that vp(u) =vp(v23).

We compute S_w

θ_a,b^v23;s

p^s/2minn

p^{s+vp(v23)+vp(m1)}, p^{3s2−b+min{vp(m2),vp(n2)}}o . Fixc≥b. Then

|{v₂₃∈ S_a,b|v_p(v₂₃) =c}| ≤p^s−c. Hence

Kl_p n, ψ, ψ⁰

≤ X

a,b≤r max{a,b}=r

S_a,b n, ψ, ψ⁰

X

a≤rb=r

p^−2sp^s+a

p^{3s2+min{vp(m2),vp(n2)}}

+ X

a=r b<r b≤c≤r

p^−2sp^s−c+a+b

p^s/2minn

p^{s+vp(v23)+vp(m1)}, p^{3s2−b+min{vp(m2),vp(n2)}}o

p^{5s4+12vp(m1)+12min{vp(m2),vp(n2)}}.

Case III: Supposes > r > s/2. We consider the following subcases:

(a) Suppose b=r. Then vp(u) =s−a, and vp(ˆv14) =r. We compute

S_w

θ_a,b^v23;s

p^s−rminn

p^{2s−a+vp(m1)}, p^{r+min{r+vp(m2)},s−r+vp(n2)}o .

(b) Suppose b < r. Then a = r. Then from (3.37) we deduce that vp(v23) = p^s−2r+b, and hencev_p(u) =p^s−2r+b. We compute

S_w

θ^v_a,b²³;s

p^s−rminn

p^{2s−2r+b+vp(m1)}, pr−b+min{2r+vp(m2),s+vp(n2)}o . Hence

Klp n, ψ, ψ⁰

≤ X

a,b≤r max{a,b}=r

2a−b≤s

S_a,b n, ψ, ψ⁰

X

a≤rb=r

p^−2sp^s+a

p^s−rminn

p^{2s−a+vp(m1)}, p^{r+min{r+vp(m2)},s−r+vp(n2)}o

+ X

2r−s≤b<ra=r

p^−2sp^s+a

p^s−rmin n

p^{2s−2r+b+vp(m1)}, pr−b+min{2r+v_p(m2),s+vp(n2)}o

p^{s−r2+12vp(m1)+12min{2r+vp(m2),s+vp(n2)}}.

Case IV: r =s. In this case we only have to consider terms with b=r. Indeed, if b < r, then a=r, and then by (3.36), we see that up^r−b ≡ −1 (mod p^r), which saysb=r, a contradiction.

Whenb=r, we have vp(u) =s−a, and from (3.14) we may assumevˆ14= 0. We compute S_w

θ_a,b^v23;s

minn

p^{2s−a+vp(m1)}, p^s+vp(n2)o .

Hence

Kl_p n, ψ, ψ⁰ ≤ X

b=sa≤s

S_a,b n, ψ, ψ⁰

X

b=sa≤s

p^−2sp^s+a

min n

p^{2s−a+vp(m1)}, p^s+vp(n2) o

p^{s+min{vp(m1),vp(n2)}}. This finishes the proof of the theorem.

Theorem 3.13. Letψ=ψ_m1,m2,ψ⁰ =ψ_n1,n2 be characters ofU(Qp)/U(Zp). Then

Klp nw0,r,s, ψ, ψ⁰

|m₁m2|⁻¹_p ,|n₁n2|⁻¹_p 1/2

(s+ 1)p^{r2+3s4+12min{r,s}}.

Proof. We make use of the stratification of Kloosterman sums in Section 3.2. For w =w0, we have ∆_w0 = ∆. Hence, for`∈N, we have

Aw0(`) =

Z/p^`Z 2

× Z/p^`Z

2

. Lett= diag a1, a2, ca⁻¹₁ , ca⁻¹₂

∈ T. Thens=n⁻¹tn= diag ca⁻¹₁ , ca⁻¹₂ , a1, a2

. We compute κ⁰₁(t∗x) =a2a⁻¹₁ κ⁰₁(x), κ⁰₂(t∗x) =ca⁻²₂ κ⁰₂(x).

So

V_w0(`) =

λ×λ⁰∈A_w0(`)

λ₁λ⁰₁ = 1, λ₂λ⁰₂ = 1 . Letθ:Aw0(`)→C^× be a character given by

θ λ×λ⁰

=

2

Y

i=1

e n_iλ_i

p^` 2

Y

i=1

e n⁰_iλ⁰_i

p^`

for n₁, n₂, n⁰₁, n⁰₂ ∈Z, then

S_w0(θ;`) =S

n₁, n⁰₁;p^` S

n₂, n⁰₂;p^`

. (3.39)

Let n = nw0,r,s. In terms of Plücker coordinates (see Section 2.2.4), this says v1 = p^r, and v12=p^s. Suppose x^v_a,b^3,v4,v13 ∈X(n)has coordinates

(v₁, v₂, v₃, v₄;v₁₂, v₁₃, v₁₄) =

p^r, p^r−a, v₃, v₄;p^s, v₁₃, p^s−b . Note that this also saysr ≥a, s≥b. From Bruhat decomposition, we have

u⁰

x^v_a,b^3,v4,v13

=









1 p^−a v₃p^−r v₄p^−r 1 v13p^−s p^−b

1

−p^−a 1









(modU(Zp)).

LetX_a,b^v3,v4,v13(n) =T ∗x^v_a,b^3,v4,v13, and define S_a,b^v3,v4,v13 n, ψ, ψ⁰

= X

x∈X_a,b^v3,v4,v13(n)

ψ(u(x))ψ⁰ u⁰(x) .

We also set

X_a,b(n) = [

v3,v4(modp^r) v13(modp^s)

conditions

X_a,b^v3,v4,v13(n),

and

S_a,b n, ψ, ψ⁰

= X

x∈X_a,b(n)

ψ(u(x))ψ⁰ u⁰(x) . It is easy to see that

X(n) = a

0≤a≤r 0≤b≤s

X_a,b(n).

Now we consider casesr ≥sand r < sseparately.

(i) Supposer > s. As r ≥a, r ≥s≥b, we see that u(x), u⁰(x) have entries inp^−rZp/Zp for all x∈X(n). Let S_a,b be a finite subset of Z³p such that

X_a,b(n) = a

(v3,v4,v13)∈S_a,b

X_a,b^v3,v4,v13(n).

By Theorem 3.7, we have Sa,b n, ψ, ψ⁰

=p^−2r 1−p⁻¹−2 X

(v3,v4,v13)∈S_a,b

X_a,b^v3,v4,v13(n) Sw0

θ^v_a,b^3,v4,v13;r

, where

θ^v_a,b^3,v4,v13 λ×λ⁰

=e

m₁vˆ₂λ₁+n₁p^r−aλ⁰₁ p^r

e

m₂vˆ₁₄+n₂p^s−b p^s

. By (3.39), we have

S_w0

θ^v_a,b^3,v4,v13;r

=S m₁vˆ₂, n₁pˆ^r−a;p^r S

m₂vˆ₁₄p^r−s, n₂p^r−b;p^r . And we obtain a bound by applying (3.25):

S_w0

θ_a,b^v3,v4,v13;r

≤4p^r

gcd m₁vˆ₂, n₁p^r−a, p^r gcd

m₂vˆ₁₄p^r−s, n₂p^r−b, p^r1/2

. (ii) Suppose s ≥ r. Then u(x), u⁰(x) has entries in p^−sZp/Zp for all x ∈ X(n). Again, by

Theorem 3.7 we have S_a,b n, ψ, ψ⁰

=p^−2s 1−p⁻¹−2 X

(v3,v4,v13)∈S_a,b

X_a,b^v3,v4,v13(n) S_w0

θ^v_a,b^3,v4,v13;s ,

where

θ_a,b^v3,v4,v13 λ×λ⁰

=e (m₁ˆv₂p^s−r)λ₁+ (m₂vˆ₁₄)λ₂+ (n₁p^s−a)λ⁰₁+ n₂p^s−b λ⁰₂ p^s

! . By (3.39), we have

Sw0

θ_a,b^v3,v4,v13;s

=S m1vˆ2p^s−r, n1p^s−a;p^s S

m2vˆ14, n2p^s−b;p^s

. Applying (3.25) gives

S_w0

θ_a,b^v3,v4,v13;s

≤4p^s

gcd m₁vˆ₂p^s−r, n₁p^s−a, p^s ,gcd

m₂ˆv₁₄, n₂p^s−b, p^s1/2

.

Now we give a bound to the size of Kl_p(n, ψ, ψ⁰). To ease computations, we consider a relaxed bound by ignoringvˆ2 andvˆ14.

Suppose r > s. Then the bound says

S_w0

θ^v_a,b^3,v4,v13;r

≤4p^r

gcd m₁ˆv₂, n₁p^r−a, p^r gcd

m₂ˆv₁₄p^r−s, n₂p^r−b, p^r1/2

≤4p^r

|n₁n2|⁻¹_p p^2r−a−b 1/2

= 4p^2r−a+b2 |n₁n2|^−1/2_p . Note that

X

(v3,v4,v13)∈S_a,b

X_a,b^v3,v4,v13(n)

≤ |S_a,b|p^a+b. Hence

Klp n, ψ, ψ⁰ ≤X

a≤r b≤s

Sa,b n, ψ, ψ⁰

≤X

a≤r b≤s

p^−2r 1−p⁻¹−2

4|n₁n2|^−1/2_p |S_a,b|p^2r+a+b2

|n₁n₂|^−1/2_p X

a≤r b≤s

|S_a,b|p^a+b2 .

So it suffices to give an upper bound to|S_a,b|. Such bounds were computed in Section 2.4. Note that we require r≥a+b in order to have S_a,b nonempty.

Case I: Suppose s−r+a≥0.

(a) Ifs−2r+ 2a+b≥0, then|S_a,b| ≤p^r+s−a−b.

(b) Ifs−2r+ 2a+b <0, then|S_a,b| ≤p^{2s−b−ds−b2 e}≤p^3s/2−b/2. Case II: Supposes−r+a <0. Then|S_a,b| ≤p^{2s−b−ds−b2 e}≤p^3s/2−b/2. Combining the cases, we obtain

X

a≤r b≤s

|S_a,b|p^a+b2 ≤ X

r−s≤a≤r 2r−2a−s≤b≤r−a

p^{r+s−a2−b2} + X

r−s≤a≤r b<2r−2a−s

p^3s2+a2 + X

a<r−s b≤s

p^3s2+a2

(s+ 1)p^r2+5s4. Hence, we have forr > s

Klp n, ψ, ψ⁰

|n₁n2|^−1/2_p (s+ 1)p^r2+5s4. (3.40) Forr ≤s, applying the same argument gives

Klp n, ψ, ψ⁰

|n₁n2|^−1/2_p (s−r+ 1)p^r+3s4 . (3.41) Combining (3.40) and (3.41), we get

Kl_p n, ψ, ψ⁰

|n₁n₂|^−1/2_p (s+ 1)p^{r2+3s4+12min{r,s}}. (3.42) By Proposition 3.2, we can swap the characters, so

Kl_p n, ψ, ψ⁰

|m₁m₂|^−1/2_p (s+ 1)p^{r2+3s4+12min{r,s}} (3.43) as well. Combining (3.42) and (3.43) yields the theorem.