Andrey Dymov
email: dymov@mi-ras.ru

Sergei Kuksin

Alberto Maiocchi
email: alberto.maiocchi@unimib.it

Sergei Vlȃdut
email: serge.vladuts@univ-amu.fr

A refinement of Heath-Brown's theorem on quadratic forms

In his paper from 1996 on quadratic forms Heath-Brown developed a version of the circle method to count points in the intersection of an unbounded quadric with a lattice of short period, if each point is given a weight, and approximated this quantity by the integral of the weight function against a measure on the quadric. The weight function is assumed to be C ∞ 0 -smooth and vanish near the singularity of the quadric. In our work we allow the weight function to be finitely smooth, not vanish at the singularity and have an explicit decay at infinity.

1 Introduction

Setting and result

Let us consider a non-degenerate quadratic form with integer coefficients on R d , d ≥ 4,

F (z) = 1 2 Az • z , (1.1)
which implies that A can be chosen as a non-degenerate symmetric matrix with integer elements whose diagonal elements are even. If F is sign-definite, then for t ∈ R the quadric

Σ t = {z : F t (z) = 0}, F t := F -t, (1.2)
is either an ellipsoid or an empty set, while in the non sign-definite case Σ t is an unbounded hyper-surface in R d , which is smooth if t = 0, while Σ 0 is a cone and has a singular point at zero. Let Z d L be the lattice of a small period L -1 ,

Z d L = L -1 Z d , L ≥ 1,
and let w be a regular real function on R d which means that w and its Fourier transform ŵ(ξ) are continuous functions which decay at infinity sufficiently fast:

|w(z)| ≤ C|z| -d-γ , ∀z ∈ R d , | ŵ(ξ)| ≤ C|ξ| -d-γ , ∀ξ ∈ R d , (1.3)
for some γ, C > 0, where | • | denotes the Euclidean norm. Our goal is to study the behaviour of series

N L (w; A, m) := z∈Σm∩Z d L w(z) ,
where m ∈ R is such that L 2 m is an integer. 1 Let w L (z) := w(z/L).

Then, obviously, N L (w; A, m) = N 1 (w L ; A, L 2 m) =: N (w L ; A, L 2 m) .

(1.4)

We will also write N L (w; A) := N L (w; A, 0) and N (w L ; A) := N (w L ; A, 0).

To study N L (w; A, m) we closely follow the circle method in the form, given to it by Heath-Brown in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. Our notation differs a bit from that in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. Namely, under the scaling z = z /L, z ∈ Z d , we count (with weights) solutions of equation F (z ) = mL2 , z ∈ Z d , while Heath-Brown writes the equation as F (z ) = m, z ∈ Z d , so that his m corresponds to our L 2 m. We start with a key theorem which expresses the analogue of Dirac's delta function on integers, i.e. the function δ : Z → R such that δ(n) := 1 for n = 0 0 for n = 0 , through a sort of Fourier representation. This result goes back at least to Duke, Friedlander and Iwaniec [START_REF] Duke | Bounds for automorphic Lfunction[END_REF] (cf. also [START_REF] Iwaniec | The circle method and the Fourier coefficients of modular forms[END_REF]) , and we state it in the form, given in [10, Theorem 1]; basically, it replaces (a major arc decomposition of) the trivial identity δ(n) = 1 0 e 2πiαn dα employed in the usual circle method. In the theorem below for q ∈ N we denote by e q the exponential function e q (x) := e 2πix q , and denote by a(mod q) * the summation over residues a with (a, q) = 1, i.e., over all integers a ∈ [1, q -1], relatively prime with q.

Theorem 1.1. For any Q ≥ 1, there exists c Q > 0 and a smooth function h(x, y) : R >0 × R → R, such that

δ(n) = c Q Q -2 ∞ q=1 a(mod q) * e q (an)h q Q , n Q 2 .
(1.5)

The constant c Q satisfies c Q = 1 + O N (Q -N ) for any N > 0, while h is such that h(x, y) ≤ c/x and h(x, y) = 0 for x > max(1, 2|y|) (so for each n the sum in (1.5) contains finitely many non zero terms).

Since for any function w on R d the quantity N ( w; A, t) may be written as z∈Z d w(z)δ(F t (z)), then Theorem 1.1 allows to represent the series N ( w; A, t) as an iterated sum. Transforming that sum further using the Poisson summation formula as in [10, Theorem 2] we arrive at the following result: 2 Theorem 1.2 (Theorem 2 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). For any regular function w, any t and any Q ≥ 1 we have the expression

N ( w; A, t) = c Q Q -2 c∈Z d ∞ q=1 q -d S q (c)I 0 q (c) , (1.6)
with S q (c) = S q (c; A, t) := a(mod q) * b(mod q) e q (aF t (b) + c • b) (1.7)

and

I 0 q (c) = I 0 q (c; A, t, Q) := R d w(z)h q Q , F t (z) Q 2 e q (-z • c) dz .
(1.8)

We will apply Theorem 1.2 to examine for large L the sum N (w L ; A, L 2 m) = N L (w; A, m), choosing w = w L , t = L 2 m and Q = L ≥ 1 and estimating explicitly the leading terms in L of S q (c) and I 0 q (c) as well as the remainders. The answer will be given in terms of the integral σ ∞ (w) = σ ∞ (w; A, t) = Σt w(z) µ Σt (dz)

(1.9)

(which is singular if t = 0). Here µ Σt (dz) = |∇F (z)| -1 dz| Σt = |Az| -1 dz| Σt , with dz| Σt representing the volume element over Σ t , induced from the standard euclidean structure on R d , and A the symmetric matrix in (1.1). For regular functions w this integral converges (see Section 7).

To write down the asymptotic for N L (w; A, m) we will need the following quantities, where p ranges over all primes and c ∈ Z d : (1.11)

σ c p = σ c p (A, L 2
The products in the formulas above are taken over all primes. In the asymptotics, where these quantities are used, they are bounded uniformly in L (see Theorems 1.3 and 1.4, as well as Proposition 1.5).

Everywhere below for a function f ∈ C k (R d ) we denote

f n 1 ,n 2 = sup z∈R d max |α| 1 ≤n 1 |∂ α f (z)| z n 2 ,
where n 1 ∈ N ∪ {0}, n 1 ≤ k, and n 2 ∈ R. Here

x := max{1, |x|} for x ∈ R l , l ∈ N, and |α| 1 ≡ α j for any integer vector α ∈ (N ∪ {0}) d . By C n 1 ,n 2 (R d ) we denote a linear space of C n 1 -smooth functions f : R d → R, satisfying

f n 1 ,n 2 < ∞.
Note that if w ∈ C d+1,d+1 (R d ) then the function w is regular, so Theorem 1.2 applies. Indeed, the first relation in (1.3) is obvious. To prove the second note that for any integer vector α ∈ (N ∪ {0}) d , ξ α ŵ(ξ) = i 2π

|α| 1 ∂ α x w(ξ). But if |α| 1 ≤ d + 1, then |∂ α x w| ≤ C x -d-1
, so ∂ α x w is an L 1 -function. Thus its Fourier transform ∂ α

x w is a bounded continuous function for each |α| 1 ≤ d + 1 and the second relation in (1.3) also holds. Now we formulate our main results. First we treat the case d ≥ 5.

Theorem 1.3. Assume that d ≥ 5. Then for any 0 < ε ≤ 1 there exist positive constants K 1 (d, ε), K 2 (d, ε) and

K 3 (d, ε), with K 2 (d, ε) ≤ K 3 (d, ε), such that if w ∈ C K 1 ,K 2 (R d ) ∩ C 0,K 3 (R d ) and a real number m satisfies L 2 m ∈ Z, then N L (w; A, m) -σ ∞ (w)σ(A, L 2 m)L d-2 ≤ CL d/2+ε ( w K 1 ,K 2 + w 0,K 3 ) , (1.12
) where the constant C depends on d, ε, m and A. The constant σ(A, L 2 m) is bounded uniformly in L and m. In particular if ε = 1/2, then one can take

K 1 = 2d(d 2 + d -1), K 2 = 4(d + 1) 2 + 3d + 1 and K 3 = K 1 + 3d + 4.
Next we study the case d = 4, restricting ourselves for the situation when m = 0. Theorem 1.4. Assume that d = 4 and m = 0. Then for any 0 < ε < 1/5 there exist positive constants K 1 (ε) and K 2 (ε), such that for w

∈ C K 1 ,K 2 (R d ) N L (w; A, 0) -η(0)σ ∞ (w)σ * (A)L d-2 log L -σ 1 (w; A, L)L d-2 ≤ C 0 L d-2-ε w K 1 ,K 2 , (1.13)
where the constant C 0 depends on ε and A. The constant η(0) is 1 if the determinant det A is a square of an integer and is 0 otherwise. The Lindependent constant σ * (A) is finite while the constant σ 1 satisfies

|σ 1 (w; A, L)| ≤ C 0 w K 1 ,K 2
uniformly in L. In the case of a square determinant det A, when η(0) = 1, it is given by (1.24). In the case of a non-square determinant det A, when η(0) = 0 and the term σ 1 (w; A, L)L d-2 gives the asymptotic of the sum N L , the constant σ 1 (w; A, L) does not depend on L and has the form

σ 1 (w; A) = σ ∞ (w)L(1, χ) p (1 -χ(p)p -1 )σ p (A, 0) , (1.14)
where χ is the Jacobi symbol ( det(A) * ) and L(1, χ) is the Dirichlet L-function. 3If η(0)σ * (A) = 0, then the asymptotic (1.13) degenerates. Similar (1.12) also degenerates to an upper bound on N L , unless we know that σ(A, L 2 m) admits a suitable positive lower bound, for all L. Luckily enough, the required lower bounds often exist, see below Proposition 1.5.

Theorems 1.3 and 1.4 refine Theorems 5, 6 and 7 from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] in three respects: firstly, now the weight function w has finite smoothness and sufficiently fast decays at infinity, while in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] w ∈ C ∞ 0 . Secondly, we specify how the remainder depends on w. Thirdly and the most importantly, we remove the restriction that the support of w does not contain the origin, imposed in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] in a number of crucial statements. These improvements are crucial for us since in our work [START_REF] Dymov | The long space-period limit for equations of discrete turbulence[END_REF], dedicated to the problem of wave turbulence, the two theorems above are used in the situation when w(0) = 0 and the support of w is not compact. A similar specification of the Heath-Brown method was obtained in [1, Section 5] to study an averaging problem, related to the questions, considered in [START_REF] Dymov | The long space-period limit for equations of discrete turbulence[END_REF]. Apart from wave turbulence and averaging, the replacement of sums over integer points of a quadric by integrals, with careful estimating the remainders, is needed in Kolmogorov Arnold Moser theory for partial differential equations, e.g. see (C.2) in [START_REF] Eliasson | KAM for the nonlinear beam equation[END_REF]. The publications [START_REF] Eliasson | KAM for the nonlinear beam equation[END_REF][START_REF] Buckmaster | Effective dynamics of the nonlinear Schrödinger equation on large domains[END_REF][START_REF] Dymov | The long space-period limit for equations of discrete turbulence[END_REF] are recent. We are certain that these days, when people, working in PDEs and dynamical systems, treat complicated non-linear phenomena with resonances more and more often, there will be increasing demand for the instrumental asymptotics (1.12), (1.13) and their variations. Our paper uses only basic results from the number theory and is well available to readers from Analysis.

We note that the papers [START_REF] Getz | Secondary terms in asymptotics for the number of zeros of quadratic forms over number fields[END_REF] and [START_REF] Tran | Secondary terms in asymptotics for the number of zeros of quadratic forms[END_REF] treat the sums N L (w; A, m) for even and odd dimensions d correspondingly, without the restriction that w(0) = 0, in a more general context than our Theorems 1.3 and 1.4. However, due to this generality the corresponding constants in the asymptotical in L formulas in [START_REF] Getz | Secondary terms in asymptotics for the number of zeros of quadratic forms over number fields[END_REF] and [START_REF] Tran | Secondary terms in asymptotics for the number of zeros of quadratic forms[END_REF] are very implicit (e.g., the question whether they vanish or not is highly non-trivial). The connection of the constants with singular integrals like (1.9) and the dependence of the remainders in the asymptotics on the weight function w, crucial for application in analysis, is not clear. Another feature of [START_REF] Getz | Secondary terms in asymptotics for the number of zeros of quadratic forms over number fields[END_REF][START_REF] Tran | Secondary terms in asymptotics for the number of zeros of quadratic forms[END_REF] is the use of rather advanced adelic technique, which makes it difficult for readers without serious number-theoretical background to use the result and the method of the work. Remarks. 1) Theorem 1.3 is a refinement of Theorem 5 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], while Theorem 1.4 refines Theorems 6 and 7 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. In [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] also is available some asymptotic in L information about behaviour of the sums N L (w; A, m) when d = 4, m = 0 and d = 3, m = 0. Since our proof of Theorems 1.3 and 1.4 is based on ideas from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], strengthened by Theorem 7.3, which is valid for d ≥ 3, then most likely our approach allows to generalise the abovementioned results of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]

for d = 3, 4 to the case when w ∈ C K 1 ,K 2 (R d ) with suitable K 1 , K 2 .
2) In our work the dependence of constants in estimates on m is uniform on compact intervals, while the dependence on the operator A is only via the norms of A and A -1 .

3) The values of constants K j (d, ε) in (1.12), given in Theorem 1.3, are far from optimal since it was not our goal to optimise them. 4) As the theorems' proof are based on the representation (1.6), then the function w should be regular (see (1.3)). But this holds true if w ∈ C d+1,d+1 and so is valid if the constants K 1 , K 2 are sufficiently big. E.g. if K 1 , K 2 are as big as in the last line of the assertion of Theorem 1.3.

Brief discussion of the proofs. We present in full only a proof of Theorem 1.3, which resembles that of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]Theorem 5] with an additional control of how the constants depend on w. The significant difference from the argument of Heath-Brown comes in Sections 3 and 4, where we do not assume that the function w vanishes near the origin, the last assumption being crucial in the analysis of integrals in Sections 6 and 7 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. To cope with this difficulty, which becomes apparent e.g. in Proposition 3.8 below, we have to examine the smoothness at zero of function

t → σ ∞ (w; A, t) (1.15)
and its decay at infinity. The corresponding analysis is performed in Section 7. There, using the techniques, developed in [START_REF] Dymov | Formal expansions in stochastic model for wave turbulence 1: kinetic limit[END_REF] to study integrals (1.9), we prove that function (1.15) is ( d/2 -2)-smooth, but in general for even d its derivative of order (d/2 -1) may have a logarithmic singularity at zero. There we also estimate the rate of decay of function (1.15) at infinity. The proof of Theorem 1.4 resembles that of Theorems 6 and 7 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] with a new addition given by Proposition 3.8, based on the result of Section 7.

We thus limit ourselves to a sketch of the theorem's demonstration, given in Section 1.3 in parallel to that of Theorem 1.3, and point out the main differences between the two proofs. Establishing Theorem 1.4 we use certain results from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] (namely, Lemmas 30 and 31) without proof.

Lower bounds for constant from the asymptotics. Let us now discuss lower bounds for the constants σ(A, L 2 m) and σ * (A) from Theorems 1.3 and 1.4. (ii) If d = 4 and m = 0 we have σ * (A) > 0 for any non-degenerate matrix A such that the corresponding equation 2F (z) = Az • z = 0 has nontrivial solutions in every p-adic field (in particular this holds if the equation has a non-trivial solution in Z4 ).

See Theorems 4, 6 and 7 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. We do not prove this result, but just note that its demonstration uses a refinement of the calculation in the second part of the proof of Lemma 2.3. Namely, while the lemma gives an upper bound for the desired quantity, a more thorough analysis permits also to establish the claimed lower bounds.

In Appendix B we give essentially a complete calculation, proving Proposition 1.5 in the case of the simplest quadratic form F = Σ d/2 i=1 x i y i , d = 2s ≥ 4, and m = 0. A proof of the proposition for any A may follow the same lines, replacing explicit formulas by some general results (e.g. Hensel's Lemma).

Non-homogeneous quadratic polynomials. Now consider a non-homogeneous quadratic polynomial F with the second order part, equal to F in (1.1):

F(z) = 1 2 Az • z + z * • z + τ, z * ∈ R d , τ ∈ R,
and the corresponding set Σ F = {z :

F(z) = 0}, N L (w; F) = z∈Σ F ∩Z d L w(z). Denote z = A -1 z * , z = z + z, m = 1 2 z • Az -τ, and assume that z ∈ Z d L 4 and L 2 τ ∈ Z. Then L 2 m ∈ Z, z ∈ Z d L if and only if z ∈ Z d L , and
F(z) = F (z ) -m. So setting w z (z ) = w(z -z) we have N L (w; F) = N L (w z ; A, m). Since σ ∞ (w z ; A, m) = Σm w z (z ) dz | Σm |∇F (z )| = Σ F w(z) dz | Σ F |∇F(z)| =: σ ∞ (w; F),
then we arrive at the following corollary from Theorem 1.3:

Corollary 1.6. If d ≥ 5, the quadratic form F is as in Theorem 1.3, F is a non-homogeneous quadratic form as above and L is such that z

:= A -1 z * ∈ Z d L , τ L 2 ∈ Z, then for any 0 < ε ≤ 1 and w ∈ C K 1 ,K 2 (R d ) ∩ C 0,K 3 (R d ) we have N L (w; F) -σ ∞ (w; F) σ(A, L 2 m)L d-2 ≤ CL d/2+ε ( w K 1 ,K 2 + w 0,K 3 ) .
Here the constants K 1 , K We always assume that function w belongs to the space C m,n (R d ) with sufficiently large m, n. If in the statement of an assertion we employ the norm w a,b then we assume that w ∈ C a,b (R d ).

We denote e q (x) = e 2πix/q and abbreviate e 1 (x) =: e(x). By • we denote the ceiling function, x = min n∈Z {n ≥ x}. By N we denote the set of positive integers.

Acknowledgements. The authors thank Professor Heath-Brown for advising them concerning the paper [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]

Scheme of the proof of Theorem 1.3

Let d ≥ 5. As it has been already discussed, if w satisfies assumptions of the theorem with sufficiently large constants K i then w is regular in the sense of Section 1.1, so Theorem 1.2 applies. Then, according to (1.6) and (1.4),

N L (w; A, m) = c L L -2 c∈Z d ∞ q=1 q -d S q (c)I q (c) , (1.16)
where the sum S q (c) = S q (c; A, L 2 m) is given by (1.7) with t = L 2 m and the integral I q (c) -by (1.8) with w = w L , Q = L and t = L 2 m,

I q (c; A, m, L) := R d w z L h q L , F L 2 m (z) L 2 e q (-z • c) dz . (1.17) Denoting n(c; A, m, L) = ∞ q=1 q -d S q (c)I q (c) , we have N L (w; A, m) = c L L -2 c∈Z d
n(c). Then for any γ 1 ∈ (0, 1/2) we write

N L as N L (w; A, m) = c L L -2 J 0 + J γ 1 < + J γ 1 > , (1.18)
where

J 0 := n(0) , J γ 1 < := c =0, |c|≤L γ 1 n(c) , J γ 1 > := |c|>L γ 1 n(c) . (1.19)
Proposition 5.1 (which is a modification of Lemmas 19 and 25 from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]) implies that |J γ 1 > | γ 1 ,m w N 0 ,2N 0 +d+1 with N 0 := d + (d + 1)/γ 1 (see Corollary 5.2). In Proposition 6.1, following Lemmas 22 and 28 from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], we show that

|J γ 1 < | γ 1 ,m L d/2+2+γ 1 (d+1) w N ,d+5 + w 0, N +3d+4 , (1.20) N = d 2 /γ 1 -2d.
To analyse J 0 we write it as J 0 = J + 0 + J - 0 , where

J + 0 := q>ρL q -d S q (0)I q (0) , J - 0 := q≤ρL q -d S q (0)I q (0) , (1.21)
with ρ = L -γ 2 for some 0 < γ 2 < 1 to be determined. Lemma 4.2, which is a combination of Lemmas 16 and 25 from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], modified using the results from Section 7, implies that

J + 0 L d/2+2+γ 2 (d/2-1) |w| L 1 L d/2+2+γ 2 (d/2-1) w 0,d+1 .
Finally Lemma 4.3, which is a combination of Lemma 13 and simplified Lemma 31 from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] with the results from Section 7, establishes that J -

0 equals L d σ ∞ (w)σ(A, L 2 m) + O γ 2 ,m w d/2-2,d-1 + w 0,d+1 L d/2+2+γ 2 (d/2-2)
(see (1.9) and (1.11)). Identity (1.18) together with the estimates above implies the desired result if we choose

γ 2 = ε/(d/2 -1) and γ 1 = ε/(d + 1).
Uniform in L and m boundedness of the product σ(A, L 2 m) follows from Lemma 2.3.

Scheme of the proof of Theorem 1.4

In this section we assume that d = 4 and m = 0. The proof proceeds exactly as in the previous section up to formula (1.20), which is not sharp enough for the case d = 4 and should be replaced by

J γ 1 < -L d c =0 η(c)σ * c (A)σ c ∞ (w; A, L) γ 1 L 7/2+(d+4)γ 1 w K1 , K2 (1.22)
for appropriate constants K1 , K2 , where the terms σ * c (A) are introduced in (1.10), terms σ c ∞ (w; A) are given by

σ c ∞ (w; A, L) := L -d ∞ q=1 q -1 I q (c; A, 0, L) , (1.23)
and the constants η(c) = ±1 are defined in Lemma A.1. In particular, η(0) = 1 if the determinant det A is a square of an integer and η(0) = 0 otherwise. The proof of the bound (1.22) makes use of Lemma A.1 (Lemma 30 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]), involving only minor modifications of the argument in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] and is left to the reader.

The bound on J 0 must be refined too and this is done in Appendix A. We consider only the case when the determinant det A is a square of an integer, so in particular η(0) = 1. The opposite case can be obtained by minor modification of the latter, following [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] (see Appendix A for a discussion). In Proposition A.3, which is a combination of Lemmas 13, 16 and 31 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], modified using Proposition 3.8, we prove that in the case of square determinant det

A J 0 =σ ∞ (w)σ * (A)L d log L + K(0)L d + O ε L d-ε w d/2-2,d-1 + w 0,d+1 ,
where a constant K(0) = K(0; w, A) is defined in Section A.1. Again, identity (1.18) together with the estimates above implies the desired result if we choose γ 1 = ( 12 -ε)/(d + 4) and put

σ 1 (w; A, L) := K(0) + c =0 η(c)σ * c (A)σ c ∞ (w; A, L) . (1.24)
Finiteness of the products σ * c (A) follow from Lemma A.2 while the claimed in the theorem estimate for the constant σ 1 (w; A, L) is established in Section A.3.

Series S q

Now we start to prove Theorem 1.3, following the scheme presented in Section 1.2. Part of the assertions, forming the proof, do not use that d ≥ 5. So below in all assertion involving the dimension d, we indicate the real requirements on d. We recall that the constants in estimates may depend on d and A, but this dependence is not indicated (see Notation and agreements).

In the present section we analyse the sums S q (c) = S q (c; A, L 2 m) entering, in particular, the definitions of the singular series σ(A, L 2 m) and σ p (A, L 2 m).

Lemma 2.1 (Lemma 25 in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). For any d ≥ 1 we have

|S q (c; A, L 2 m)| q d/2+1 , uniformly in c ∈ Z d .
Proof. According to (1.7), an application of the Cauchy-Schwarz inequality shows that

|S q (c)| 2 ≤ φ(q) a(mod q) * b(mod q) e q (aF L 2 m (b) + c • b) 2 = φ(q) a(mod q) * u,v(mod q) e q a(F L 2 m (u) -F L 2 m (v)) + c • (u -v) , (2.1) where φ(q) is the Euler totient function. Since F t (z) = 1 2 Az • z -t, then F L 2 m (u) -F L 2 m (v) = (Av) • w + F (w) = v • Aw + F (w). So e q a(F L 2 m (u) -F L 2 m (v)) + c • (u -v) = e q aF (w) + c • w e q (av • Aw).
Now we see that the summation over v in (2.1) produces a zero contribution, unless each component of the vector Aw is divisible by q. This property holds for at most a finite number N of vectors w, where the constant N depends only on det A. Thus,

|S q (c)| 2 φ(q) a(mod q) * v(mod q) 1 ≤ φ 2 (q) q d .
The lemma's assertion shows that the sums σ c p , defined in (1.10), are finite:

Corollary 2.2. If d ≥ 5, for any prime p we have σ c p (A, L 2 m) 1. Recall that σ(A, L 2 m) = p σ p (A, L 2 m) (see (1.11)). Lemma 2.3. For any d ≥ 5 and 1 ≤ X ≤ ∞ we have q≤X q -d S q (0) = σ(A, L 2 m) + O(X -d/2+2 ).
In particular, σ(A,

L 2 m) = ∞ q=1 q -d S q (0). So |σ(A, L 2 m)| 1 in view Lemma 2.1.
Proof. We start by showing the multiplicative property of trigonometric sums S qq (0) = S q (0)S q (0) ,

whenever (q, q ) = 1 (cf. Lemma 23 from [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). By definition

S qq (0) = a(mod qq ) * v(mod qq ) e qq (aF L 2 m (v)) .
When (q, q ) = 1 we can replace the summation on a (mod qq ) by a double summation on a q modulo q and a q modulo q by writing a = qa q + q a q . Thus

S qq (0) = aq(mod q) * a q (mod q ) * v(mod qq ) e q (a q F L 2 m (v))e q (a q F L 2 m (v)) .
Then we replace the summation on v (mod qq ) with the double summation on v q modulo q and v q modulo q by writing v = q qv q + q q v q , where q and q are defined through relations q q = 1 ( mod q ) and q q = 1 ( mod q). We observe that

F L 2 m (v) = q 2 q2 F (v q ) + q 2 q 2 F (v q ) + q qq q Av q • v q -L 2 m , so that e q (a q F L 2 m (v)) = e q (a q q 2 q 2 F (v q ) -a q L 2 m) = e q (a q F L 2 m (v q )),
by the definition of q and since e q (qN ) = 1 for any integer N . Similar,

e q (a q F L 2 m (v)) = e q (a q F L 2 m (v q )) .
This gives (2.2).

Next we note that, due to Lemma 2.1,

q≥X q -d |S q (0)| q≥X q -d/2+1 X -d/2+2 . (2.3) By (2.
2) and the definition of σ,

σ = lim n→∞ σ n , σ n = p≤n n l=0 p -dl S p l (0) = q∈Pn q -d S q (0),
where p are primes and P n denotes the set of natural numbers q with prime factorization of the form

q = p k 1 1 • • • p km m , where 2 ≤ p 1 < p 2 • • • < p m ≤ n, k j ≤ n
and m ≥ 0 (m = 0 corresponds to q = 1). Since any q ≤ n belongs to P n , then according to (2.3),

q∈P N q -d S q (0) - q≤X q -d S q (0) X -d/2+2 ∀ N ≥ X,
for any finite X > 0. Passing in this estimate to a limit as N → ∞ we recover the assertion if X < ∞. Then the result with X = ∞ follows in an obvious way.

3 Singular integrals I 0 q 3.1 Properties of h(x, y)

Following [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], Section 3, we construct the function h(x, y) ∈ C ∞ (R > , R), entering Theorem 1.1, starting from the weight function w 0 ∈ C ∞ 0 (R), defined as

w 0 (x) = exp 1 x 2 -1 for |x| < 1 0 for |x| ≥ 1 . (3.1)
We denote c 0 := ∞ -∞ w 0 (x) dx and introduce the shifted weight function

ω(x) = 4 c 0 w 0 (4x -3) , which of course belongs to C ∞ 0 (R). Obviously, 0 ≤ ω ≤ 4e -1 /c 0 , ω is supported on (1/2, 1), and ∞ -∞ ω(x) dx = 1 . The required function h : R >0 × R → R is defined in terms of ω as h(x, y) := h 1 (x) -h 2 (x, y) with h 1 (x) := ∞ j=1 1 xj ω(xj) , h 2 (x, y) := ∞ j=1 1 xj ω |y| xj . (3.2)
For any fixed pair (x, y) each of the two sum in j contains a finite number of nonzero terms. So h is a smooth function.

In [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], Section 3, it is shown how to derive Theorem 1.1 from the definition (3.2). 5 Here we limit ourselves to providing some relevant properties of h, proved in Section 4 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. In particular these properties imply that for small x, h(x, y) behaves as the Dirac delta function in y Lemma 3.1 (Lemma 4 in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). We have:

1. h(x, y) = 0 if x ≥ 1 and |y| ≤ x/2.

If

x ≤ 1 and |y| ≤ x/2, then h(x, y) = h 1 (x), and for any m ≥ 0

∂ m h(x, y) ∂x m m 1 x m+1 . 3. If |y| ≥ x/2, then for any m, n ≥ 0 ∂ m+n h(x, y) ∂x m ∂y n m,n 1 x m+1 |y| n .
Lemma 3.2 (Lemma 5 in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). Let m, n, N ≥ 0. Then for any x, y

∂ m+n h(x, y) ∂x m ∂y n N,m,n 1 x 1+m+n δ(n)x N + min 1, (x/|y|) N . Lemma 3.2 with m = n = N = 0 immediately implies Corollary 3.3. For any x, y ∈ R > × R we have |h(x, y)| 1/x.
Lemma 3.4 (Lemma 6 in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). Fix X ∈ R >0 and 0 < x < C min {1, X}, for some C > 0. Then for any N ≥ 0,

X -X h(x, y) dy = 1 + O N,C Xx N -1 + O N,C x N X N . Lemma 3.5 (Lemma 8 in [10]). Fix X ∈ R >0 and n ∈ N. Let x < C min {1, X} for C > 0. Then X -X y n h(x, y) dy N,C X n Xx N -1 + x N X N .
The previous results are used to prove the key Lemma 9 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], which can be extended to the following

Lemma 3.6. Let a function f ∈ C M -1,0 (R) ∩ L 1 (R), M ≥ 1, be such that its (M -1)-st derivative f (M -1) is absolutely continuous on [-1, 1], and let 0 < x ≤ C for some C > 0. Then for any 0 < β ≤ 1 and any N ≥ 0, R f (y)h(x, y) dy =f (0) + O M x M β M +1 1 X X -X |f (M ) (y)| dy + O N,C (x N + β N ) f M -1,0 + x -1 |f | L 1 , (3.3)
where X := min {1, x/β}.

Proof. By Lemma 3.2 with m = n = 0, for any N ≥ 0 we have |h(x, y)| N (x N + β N )x -1 if |y| ≥ X. So the tail-integral for f h dy may be bounded as

|y|≥X f (y)h(x, y) dy N (x N +β N )x -1 |y|≥X |f (y)| dy N (x N +β N )x -1 |f | L 1 .
(3.4) For the integral in |y| < X, instead we take the Taylor expansion of f (y) around zero and get that

X -X f (y)h(x, y) dy = M -1 j=0 f (j) (0) j! X -X y j h(x, y) dy + O M X M x X -X |f (M ) (y)| dy , (3.5
f (0) X -X h(x, y) dy = f (0) + O N,C f 0,0 Xx N + x N +1 X N +1 , (3.6)
while by Lemma 3.5, for any j > 0 we have

f (j) (0) j! X -X y j h(x, y) dy N,j,C f j,0 X j Xx N + x N +1 X N +1 . ( 3
< δ < 1, R f (y)h(x, y) dy = f (0) + O M,C,δ x M -δ ( f M,0 + |f | L 1 ) .
Proof. The assertion follows from Lemma 3.6 by choosing for any 0 < δ < 1, β = x δ/(M +1) if x ≤ 1 and β = 1 if x > 1. Indeed, then for 0 < x ≤ 1 we have that x M β -(M +1) = x M -δ , and that

(x N + β N )x -1 ≤ 2β N x -1 ≤ 2x M -δ if N ≥ N δ = (M -δ + 1)(M + 1)/δ. While if 1 ≤ x ≤ C, then x M ≤ C δ x M -δ
, and choosing N = 0 we get that (x N + 1) = 2 ≤ 2x M -δ . The obtained relations imply the assertion.

Approximation for I q (0)

In what follows it is convenient to write the integrals I q (c; A, L 2 m) as

I q (c) = L d Ĩq (c), (3.8)
where

Ĩq (c) = Ĩq (c; A, m, L) = R d w(z) h q L , F m (z) e q (-z • cL) dz . (3.9)
The proposition below replaces Lemmas 11, 13 and Theorem 3 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. In difference with those results we do not assume that 0 / ∈ supp w. Since for c = 0 the exponent e q in the definition of the integral I q (c) equals one, we can consider I q (0) as a function of a real argument q ∈ R, and we do so in the proposition below; we will use this in Appendix A. Proposition 3.8. Let q ∈ R, q ≤ CL with some C > 0.

a) If d ≥ 5 and N M < d/2 -1, then for any δ > 0,

I q (0; A, m, L) = L d σ ∞ (w; A, m) + O m,M,C,δ q M -δ L d-M +δ w M,d+1 . (3.10) b) If d = 4, N M ≤ d/2
-1 and m = 0, then for any 0 < β ≤ 1 and N ≥ 0,

I q (0; A, 0, L) =L d σ ∞ (w; A, 0) + O β -M -1 q M L d-M log q Lβ w M,d+1 + O C,N (q N L d-N + β N )( w M -1,d+1 + Lq -1 w 0,d+1 ) . (3.11)
Proof. For d ≥ 4, applying the co-area formula (see [START_REF] Chavel | Riemannian Geometry: a Modern Introduction[END_REF], Theorem 6.3) we re-write the integral in (3.9) with c = 0 in terms of integrals over hypersurfaces Σ t as follows:

Ĩq (0) = R I(m+t)h(q/L, t) dt , I(t) = Σt w(z) µ Σt (dz) , (3.12)
where the measure µ Σt is the same as in (1.9). By Theorem 7.3,

I k, K k,K, K w k,K if K < K + 2 -d 2 , K > d, (3.13)
and

k < d/2 -1. Denote f m (y) = I(m + y). Then f m k, K m, K I k, K , and by (3.13) |f m | L 1 = |I| L 1 I 0,4/3 w 0,d+1 . (3.14)
To prove a) we apply Corollary 3.7 with f = f m and x = q/L to the first integral in (3.12). Note that f m (0) = I(m) = σ ∞ (w; A, m). Then, using (3.13) with K = 0, K = d + 1 and k = M jointly with (3.14) we get that

Ĩq (0) = σ ∞ (w) + O M,m,C,δ q M -δ L -M +δ w M,d+1 .
So (3.10) follows.

To establish (3.11), we apply Lemma 3.6 to write the integral in (3.12) with m = 0 as

R I(t)h(x, t) dt = I(0) + O M β -M -1 x M 1 X X -X |I (M ) (t)| dt + O C,N (x N + β N )( I M -1,0 + x -1 |I| L 1 )
, where x = q/L and X = min{1, x/β}. By applying Theorem 7.3, with k = M and M = d + 1, we get

X -X |I (M ) (t)| dt X log X w M,d+1 .
Using this estimate jointly with (3.13) and (3.14) we arrive at (3.11).

4 The J 0 term

In this section we prove the following proposition concerning the term J 0 defined in (1.19):

Proposition 4.1. Let d ≥ 5. Then for any 0 < γ 2 < 1,

J 0 -L d σ ∞ (w)σ(A, L 2 m) γ 2 ,m L d 2 +2+γ 2 ( d 2 -1) w d/2 -2,d+1 .
Proof. To establish the result we write J 0 in the form (1.21). Then the assertion follows from Lemmas 4.2 and 4.3 below which estimate the terms J + 0 and J - 0 , noting that |w| L 1 w 0,d+1 .

Lemma 4.2. Assume that w ∈ L 1 (R d ) and d ≥ 3. Then we have the bound 1) |w| L 1 , for any γ 2 ∈ (0, 1).

|J + 0 | L d/2+2+γ 2 (d/2-
Proof. Since according to Lemma 2.1 |S q (0)| q d/2+1 , then

|J + 0 | q>L 1-γ 2
q -d/2+1 I q (0).

Writing integral I q as in (3.8), by Corollary 3.

3 we get |I q (0)| L d+1 q |w| L 1 .
Therefore,

|J + 0 | L d+1 |w| L 1 q>L 1-γ 2 q -d/2 L d+1 |w| L 1 L (-d/2+1)(1-γ 2 ) = L d/2+2+γ 2 (d/2-1) |w| L 1 .
Lemma 4.3. Let d ≥ 5. Then for any γ 2 ∈ (0, 1),

J - 0 = L d σ ∞ (w)σ(A, L 2 m) + O γ 2 ,m L d/2+2+γ 2 (d/2-2) w d/2 -2,d+1 .
Proof. Inserting (3.10) with C = 1 into the definition of the term J - 0 , we get J - 0 = I A + I B , where

I A := L d σ ∞ (w) q≤L 1-γ 2
q -d S q (0),

|I B | M,δ,m L d-M +δ w M,d+1 q≤L 1-γ 2
S q (0)q -d+M , for M < d/2 -1 and any δ > 0. Lemma 2.3 implies that

q≤L 1-γ 2 q -d S q (0) = σ(A, L 2 m) + O(L (-d/2+2)(1-γ 2 ) ), so I A = L d σ ∞ (w)σ(A, L 2 m) + O(σ ∞ (w)L d/2+2+γ 2 (d/2-2) ) ,
whereas |σ ∞ (w)| = |I(m)| ≤ I 0,0 ≤ w 0,d+1 on account of (3.13). As for the term I B , Lemma 2.1 implies that

|I B | M,δ,m L d-M +δ w M,d+1 q≤L 1-γ 2 q -d/2+1+M .
Choosing M = d/2 -2 and δ = γ 2 /2, we get

|I B | δ,m w d/2 -2,d+1 L d/2+2+δ ln L γ 2 ,m w d/2 -2,d+1 L d/2+2+γ 2 .
5 The J γ 1 > term

We provide here an estimate of the term J γ 1 > defined in (1.19). The key point of the proof is an adaptation of Lemma 19 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] to our case. We recall the notation (3.8).

Proof. Let f q (z) := w (z) h q L , F m (z) . Since i 2π q L |c| -2 (c • ∇ z ) e q (-z • cL) = e q (-z • cL) ,
then integrating by parts N times the integral (3.9) we get that

Ĩq (c) ≤ q 2πL |c| -2 N R d (c • ∇ z ) N f q (z) dz N q L N |c| -N 0≤n≤N R d max 0≤l≤n/2 ∂ n-l ∂y n-l h q L , F m (z) × |z| n-2l ∇ N -n z w(z) dz ,
where ∂ ∂y h stands for the derivative of h with respect to the second argument.

Assume first that q ≤ L. Then, by Lemma 3.2 with N = 0,

max 0≤l≤n/2 ∂ n-l ∂y n-l h q L , F m (z) |z| n-2l ∇ N -n z w(z) ≤ (L/q) n+1 z -d-1 w N -n,n+d+1 .
This implies (5.1) since n ≤ N . Let now q > L. Then, due to item 1 of Lemma 3.1, h is different from zero only if

2|F m (z)| > q L . (5.2)
Then for such z and for l ≤ n, item 3 of Lemma 3.1 implies that

∂ n-l ∂y n-l h q L , F m (z) n-l L q 1 |F m (z)| n-l n-l L q n-l+1 . So max 0≤l≤n/2 ∂ n-l ∂y n-l h q L , F m (z) |z| n-2l ∇ N -n z w(z) max 0≤l≤n (L/q) n-l+1 z 2(N -n+l)
w N -n,2N -n+d+1 z d+1 .

Since from (5.2) we have that q/L m z 2 , then the first fraction above is bounded by (L/q) N +1 , and again (5.1) follows.

As a corollary we get an estimate for J γ 1 > :

Corollary 5.2. For J γ 1 > defined in (1.19) with γ 1 ∈ (0, 1) and d ≥ 3 we have |J γ 1 > | γ 1 ,m w N 0 ,2N 0 +d+1 , where N 0 := d + (d + 1)/γ 1 .
Proof. Denoting by | • | 1 the l 1 -norm, by the definition of J γ 1 > we have

|J γ 1 > | s≥L γ 1 s d-1 ∞ q=1 q -d sup |c| 1 =s |S q (c)||I q (c)| s≥L γ 1 s d-1 ∞ q=1 q 1-d/2 L d sup |c| 1 =s | Ĩq (c)| N ,m s≥L γ 1 s d-1 ∞ q=1 q -d/2 s -N L d+1 w N,2N +d+1 ,
where the second line follows from Lemma 2.1, while the third one -from Proposition 5.1. The sum in q is bounded by a constant. Choosing N = N 0 we get that

L d+1 s≥L γ 1 s d-1 s -N ≤ L d+1 s≥L γ 1 s -1-(d+1)/γ 1 1 .
This concludes the proof.

6 The J γ 1 < term

The estimate

Our next (and final) goal is to estimate the term J γ 1 < from (1.18). Proposition 6.1. For any d ≥ 3 and γ 1 ∈ (0, 1/2),

|J γ 1 < | γ 1 ,m L d/2+2+γ 1 (d+1) w N ,d+5 + w 0, N +3d+4 ,
where N = N (d, γ 1 ) := d 2 /γ 1 -2d. Proposition 6.1 will follow from the next lemma which is a modification of Lemma 22 in [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] and is proved in the next subsection: Lemma 6.2. For any d ≥ 3 and c = 0,

|I q (c)| γ 1 ,m L d/2+1+γ 1 q/|c| d/2-1-γ 1 w N ,d+5 + w 0, N +3d+4 ,
where N and γ 1 are the same as above.

Proof of Proposition 6.1. Accordingly to Lemma 2.1,

|J γ 1 < | c =0, |c|≤L γ 1 ∞ q=1 q -d q d/2+1 |I q (c)| L dγ 1 max c =0: |c|≤L γ 1 |I q (c)| ∞ q=1 q -d/2+1 = L dγ 1 q<L + q≥L q -d/2+1 max c =0: |c|≤L γ 1 |I q (c)| = J -+ J + , with
J -:= L dγ 1 q<L q -d/2+1 max c =0: |c|≤L γ 1 |I q (c)| , J + := L dγ 1 q≥L q -d/2+1 max c =0: |c|≤L γ 1 |I q (c)| .
Corollary 3.3 together with (3.8), (3.9) implies

|I q (c)| L d+1 q |w| L 1 , (6.1)
so that

J + L dγ 1 L d+1 |w| L 1 q≥L q -d/2 L dγ 1 +d/2+2 |w| L 1 L dγ 1 +d/2+2 w 0,d+1 .
On the other hand, since |c| ≥ 1, from Lemma 6.2 we get

J -γ 1 ,m L dγ 1 L d/2+1+γ 1 w N ,d+5 + w 0, N +3d+4 q<L q -γ 1 ≤ w N ,d+5 + w 0, N +3d+4 L γ 1 (d+1)+d/2+2 .

Proof of Lemma 6.2

We begin with

Application of the inverse Fourier transform

Note that the proof is nontrivial only for q L|c|: indeed, for any α > 0 the bound (6.1) implies that

|I q (c)| α L d |w| L 1 α L d L|c|/q -d/2+1+γ 1 |w| L 1 if q ≥ αL|c|,
since |c| ≥ 1 and -d/2+1+γ 1 < 0. So, it remains to use again the inequality |w| L 1 w 0,d+1 . Let us take a small enough α = α(d, γ 1 , A) ∈ (0, 1) and assume that q < αL|c|. Consider the function w 2 (x) = 1/(1 + x 2 ) and set

w(z) := w(z) w 2 (F m (z)) = w(z)(1 + F m (z) 2 ). (6.2) Let p(t) := +∞ -∞ w 2 (v)h(q/L, v)e(-tv) dv, e(x) := e 1 (x) = e 2πix . (6.3)
This is the Fourier transform of function w 2 (•)h(q/L, •). Then, expressing w 2 h via p by the inverse Fourier transform and writing w(z) = w(z)w 2 (F m (z)), we find that

w(z)h(q/L, F m (z)) = w(z) +∞ -∞ p(t)e(tF m (z)) dt.
Inserting this representation into (3.9) we get Ĩq (c) = Note that |u| = |c|L/q > α -1 > 1 since q < α|c|L. Now let us denote

W 0 (x) = c -d 0 d i=1 w 0 (x i ) (see (3.1)). Then W 0 ∈ C ∞ 0 (R d ), W 0 ≥ 0 and supp W 0 = [-1, 1] d ⊂ {x ∈ R d : |x| ≤ √ d}, R d W 0 (x) dx = 1. (6.4)
Let us set δ = |u| -1/2 < √ α and write w as

w(z) = δ -d d R W 0 z -a δ w(z) da.
Then setting b := za δ we get that

| Ĩq (c)| ≤ R d +∞ -∞ |p(t)||I a,t | dt da,
where in view of (6.4),

I a,t := {|b|≤ √ d} W 0 (b) w(z) e(tF (z) -u • z) db, z := a + δb.
Consider the exponent in the integral I a,t :

f (b) = f a,t (b) := tF (a + δb) -u • (a + δb).
At the next step we will estimate integral I a,t , regarding (a, t) as a parameter. Consider another parameter R, satisfying 1 ≤ R ≤ |u| 1/3 ; its value will be chosen later. Below we distinguish two cases: 1. (a, t) belongs to the "good" domain S R , where

S R = (a, t) : |∇f (0)| = δ|tAa -u| ≥ R t/|u| = R δ 2 t ; 2. (a, t) belongs to the "bad" set S R c = (R d × R) \ S R .

Integral over S R .

We consider first the integral over the good set S R :

Lemma 6.3. For any d ≥ 1, N ≥ 0 and R ≥ 2 A √ d we have S R |p(t)| |I a,t | da dt N,m L q R -N w N,d+5 . (6.5)
Proof. Let l := ∇f (0)/|∇f (0)| and L = l • ∇ b . Then for (a, t) ∈ S R and |b| ≤ √ d (see (6.4)),

|Lf (b)| = Lf (0) + δ 2 t∇f (0) • Ab/|∇f (0)| ≥ |∇f (0)| -δ 2 |t||Ab| ≥ R δ 2 t -δ 2 |t| A R 2 A ≥ 1 2 R δ 2 t ≥ R/2.
|I a,t | N max |b i |≤1 ∀i max 0≤k≤N L N -k w(δb + a) L 2 f (b) k Lf (b) N +k ,
where we have used that

L m f (b) = 0 for m ≥ 3. Since |L 2 f (b)| ≤ δ 2 |t||l • Al| ≤ δ 2 |t| A
, then in view of (6.6)

L 2 f (b) Lf (b) ≤ δ 2 |t| A 1 2 R δ 2 t = 2 A R ≤ 1 √ d . So using that 1 Lf (b) ≤ 2 R by (6.6), we find |I a,t | N R -N max |b i |≤1 ∀i max 0≤k≤N L k w(δb + a) .
Thus, denoting by 1 S R the indicator function of the set S R , we have

R d |I a,t |1 S R da N R -N R d a d+1 max |b i |≤1 ∀i max 0≤k≤N L k w(δb + a) da a d+1 N R -N w N,d+1 N,m R -N w N,d+5 ,
for every t. Then l.h.s. of (6.5

) N,m R -N w N,d+5 +∞ -∞ |p(t)| dt. (6.7)
To prove (6.5) it remains to show that

∂ k ∂v k h(x, v) k x -k-1 min{1, x 2 /v 2 } , k ≥ 1,
and by Corollary 3.3, |h(x, v)| x -1 . Then an integration by parts in (6.3) shows that, for any M ≥ 0,

|p(t)| M |t -M | ∞ -∞ |w (M ) 2 (v)|x -1 dv + max 1≤k≤M ∞ -∞ |w (M -k) 2 (v)| x -k-1 min 1, x 2 v 2 dv ,
where x := q/L. Writing the latter integral as a sum |v|≤x + |v|>x we see that

|v|≤x = x -k-1 |v|≤x |w (M -k) 2 (v)| dv M x -k and |v|>x = x -k+1 |v|>x |w (M -k) 2 (v)| v 2 dv M x -k .
Then, for any M ≥ 0

|p(t)| M q L |t| -M if q L < 1 and |p(t)| M q L -1 |t| -M if q L ≥ 1.
(6.9) Choosing M = 2 when |t| > L/q and M = 0 when |t| ≤ L/q we get (6.8). The fact that (a, t) ∈ S R c implies that the integration in da for a fixed t is restricted to the region, where Aa -t -1 u ≤ (R/δ|t|) t/|u| , or

Integral over

a - A -1 u t ≤ A -1 R δ|t| t/|u| . (6.11)
We first consider the case |t| ≥ |u| In view of (6.10) -(6.12),

R d |I a,t |1 S R c (a, t)da R d |u| -d/2+β w 0,0 .
Since |F m (z)| m z 2 , by definition (6.2) of the function w we have w 0,0 m w 0,4 . Then the r.h.s. above is m R d |u| -d/2+β w 0,4 . Taking into account that, by (6.8),

|t|≥|u| 1-β/d |p(t)| dt L q ≤ |u| , we get |t|≥|u| 1-β/d R d |p(t)||I a,t |1 S R c (a, t) da dt m R d |u| -d/2+1+β w 0,4 .
(6.13) Now let |t| ≤ |u| 1-β/d . Then the r.h.s. of (6.11) is bounded by the quantity

A -1 R/(δ|t|), so that |a| |A -1 u|/|t| -A -1 R/(δ|t|). Since |A -1 u| ≥ C A |u| and R ≤ |u| 1/3 , then |a| A |u| -RC A |u| |t| ≥ (1 -C A |u| -1/6 ) |u| |t| ≥ 1 2 |u| |t| ≥ 1 2 |u| β/d with C A = C -1 A A -1 , since |u| -1 ≤ α, if α is so small that 1-C A α 1/6 ≥ 1/2. Then 1 |a|/|u| β/d on S R c , so that 1 S R c (a, t) |u| -d/2+β/d |a| d 2 /2β-1
, and we deduce from (6.10) that for such values of t

R d |I a,t |1 S R c (a, t)da |u| -d/2+β/d R d |a| d 2 /2β-1 max |b i |≤1 ∀i | w(δb + a)| da m |u| -d/2+β/d w 0,K(d,β)
, where K(d, β) = d + d 2 /2β + 4. On the other hand, by (6.9) with M = 0, (6.14) Putting together (6.13) and (6.14) we get the assertion.

End of the proof

In order to complete the proof of Lemma 6.2 we combine Lemmas 6.3 and 6.4 to get that

| Ĩq (c)| N ,m L q R -N + R d |u| -d/2+1+β w N,d+5 + w 0,K(d,β) .
We fix here

γ 1 ∈ (0, 1/2), β = γ 1 /2, R = |u| γ 1 2d ≤ |u| 1 3 and pick N = d 2 γ 1 - 2d > 0 (notice that R ≥ α -γ 1 /2d ≥ 2 A √ d if α is small enough, so that assumption of Lemma 6.3 is satisfied). Then K(d, β) = N + 3d + 4, R -N ≤ |u| -d/2+γ 1 ≤ |c| (L|c|/q) -d/2+γ 1 since |c| ≥ 1. Moreover, R d |u| -d/2+1+β = |u| -d/2+1+γ 1 = (L|c|/q) -d/2+1+γ
1 . This concludes the proof.

Integrals over quadrics

Our goal in this section is to study integrals I(t; w) over the quadrics Σ t . We start with a case of quadratic forms F , written in a convenient normal form (Theorem 7.1), and show later in Section 7.4 (Theorem 7.3) how to reduce general integrals I(t; w) to those, corresponding to the quadratic forms like that. In this section we assume that d ≥ 3 and not use the bold font to denote vectors since most of variables we use are vectors.

Quadratic forms in normal form

On R d = R n u × R d 1 x × R d 1 y = {z = (u, x, y)}, where d ≥ 3, n ≥ 0 and d 1 ≥ 1, consider the quadratic form F (z) = 1 2 |u| 2 + x • y = 1 2 Az • z , A(u, x, y) = (u, y, x) . (7.1)
Note that A is an orthogonal operator, |Az| = |z|. As in Section 1.1 we define the quadrics Σ t = {z : F (z) = t}, t ∈ R. Note that for t = 0 Σ t is a smooth hypersurface, while Σ 0 is a cone with a singularity at the origin. We denote the volume element on Σ t (on Σ 0 \{0} if t = 0), induced from R d , as dz | Σt and set

µ Σt (dz) = |Az| -1 dz | Σt (7.2)
(see below concerning this measure when t = 0). For a k * ∈ N ∪ {0} and a function f on R d satisfying

f ∈ C k * ,M (R d ) , M > d , (7.3)
we will study the integrals

I(t) = I(t; f ) = Σt f (z)µ Σt (dz) . (7.4)
Our first goal is to demonstrate the following result:

Theorem 7.1. For the quadratic form F (z) as in (7.1) and a function f ∈ C k * ,M (R d ), M > d, consider integral I(t; f ), defined in (7.4). Then the function I(t), defined by

(7.4), is C k -smooth if k < d/2 -1, k ≤ k * , and is C k -smooth outside zero if k ≤ min(d/2 -1, k * ). For 0 < |t| ≤ 1 we have ∂ k I(t) k,M f k,M if k < d/2 -1, ∂ k I(t) k,M f k,M (1 -ln |t|) if k ≤ d/2 -1. (7.5) While for |t| ≥ 1, denoting κ = M +2-d 2
, we have

∂ k I(t) k,M f k,M t -κ if 1 ≤ k ≤ d/2 -1, k ≤ k * , |I(t)| M,κ f 0,M t -κ ∀ κ < κ. (7.6)
An example, see [START_REF] Dymov | Some remarks on Heath-Brown's theorem on quadratic forms[END_REF]Example A.3], shows that in general the log-factor cannot be removed from the r.h.s. in (7.5).

The theorem is proved below in number of steps. In the proof for a given vector x ∈ R d 1 we consider its orthogonal complement in R d 1 -the hyperspace x ⊥ . We denote its elements x, and provide x ⊥ with the Lebesgue measure dx. If d 1 = 1, then x ⊥ degenerates to the space R 0 = {0}, and dx -to the δ-measure at 0. Practically it means that when d 1 = 1, the spaces x ⊥ and y ⊥ (and integrals over them) disappear from our construction. It makes the case d 1 = 1 easier, but notationally different from d 1 ≥ 2. For example, in formula (7.8) with d 1 = 1 the affine space σ x t (u , x ) becomes the point (u , x , (t -

1 2 |u | 2 )|x | -2
x ), the measure dµ Σt | Σ x t in (7.14) becomes du |x| -1 dx, etc. Accordingly, below we write the proof only for d 1 ≥ 2, leaving the case d 1 = 1 as an easy exercise for the reader.

Disintegration of the two measures

Our goal in this subsection is to find a convenient disintegration of the measures dz | Σt and µ Σt , following the proof of Theorem 3.6 in [START_REF] Dymov | Formal expansions in stochastic model for wave turbulence 1: kinetic limit[END_REF].

Recall that we write elements z ∈ R d as z = (u, x, y), where u ∈ R d and

x, y ∈ R d 1 . Let us denote Σ x t = {(u, x, y) ∈ Σ t : x = 0} (if t < 0, then Σ x t = Σ t ).
Then for any t Σ x t is a smooth hypersurface in R d , and the mapping

Π x t : Σ x t → R n × R d 1 \{0} , (u, x, y) → (u, x) , (7.7)
is a smooth affine euclidean vector bundle. Its fibers are

σ x t (u , x ) := Π x t -1 (u , x ) = u , x , x ⊥ + t -1 2 |u | 2 |x | 2 x , (7.8)
where x ⊥ is the orthogonal complement to x in R d 1 . For any x = 0 denote

U x = x : |x -x | ≤ 1 2 |x | , U = R n × U x × R d 1 .
Now we construct a trivialisation of the bundle Π x t over U . To do this we fix in R d 1 any orthonormal frame (e 1 , . . . , e d 1 ) such that the ray R + e 1 intersects U x . Then

x 1 > 0 ∀x = (x 1 , . . . , x d 1 ) =: (x 1 , x) ∈ U x .
We wish to construct an affine in the third argument diffeomorphism

Φ t : R n × U x × R d 1 -1 → U ∩ Σ t of the form Φ t (u, x, η) = (u, x, Φ u,x t (η)) , Φ u,x t (η) = (ϕ t (u, x, η), η) ∈ R d 1 , η ∈ R d 1 -1 . (7.9) We easily see that Φ t (u, x, η) ∈ Σ t if and only if ϕ t (u, x, η) = t -1 2 |u| 2 -x • η x 1 . (7.10)
The mapping η → Φ u,x t (η) with this function ϕ t is affine, and the range of Φ t equals U ∩ Σ t .

In the coordinates (u, x, η

1 , η) ∈ R n × U x × R × R d 1 -1 on the domain U ⊂ R d the hypersurface Σ x t is embedded in R d as a graph of the function (u, x, η) → η 1 = ϕ t .
Accordingly, in the coordinates (u, x, η) on U ∩ Σ t the volume element on Σ t reads as ρt (u, x, η)du dx dη, where

ρt = 1 + |∇ϕ t | 2 1/2 = 1 + |u| 2 + |η| 2 + |x| 2 + x -2 1 (t -1 2 |u| 2 -x • η) 2 x 2 1 1/2 . Passing from the variable η ∈ R d 1 -1 to y = Φ u,x t (η) ∈ σ x t (u, x) we replace dη by | det Φ u,x t (η)|d σ x t (u,x) y.
Here d σ x t (u,x) y is the Lebesgue measure on the (d 1 -1)-dimensional affine euclidean space σ x t (u, x) while by det Φ u,x t we denote the determinant of the linear mapping Φ u,x t , viewed as a linear isomorphism of the euclidean space R d 1 -1 = {η} and the tangent space to σ x t (u, x), identified with the euclidean space x ⊥ ⊂ R d 1 . Accordingly we write the volume element on Σ t ∩ U as ρ t (u, x, y)du dx d σ x t (u,x) y with ρ t (u, x, y) = ρt (u, x, η)| det Φ u,x t (η)| , (u, x, y) ∈ Σ t , where Φ u,x t (η) = y. Now we will calculate the density ρ t . Let us take any point z * = (u * , x * , y * ) ∈ U ∩ Σ t and choose a frame (e 1 , . . . , e d 1 ) such that e 1 = x * /|x * |. Then

x * = (|x * |, 0) , y * = y * 1 , ȳ * , y * 1 = t -1 2 |u * | 2 |x * | , ȳ * ∈ R d 1 -1 .
So (see (7.9)-(7.10)) the mapping Φ t is such that Φ

u * ,x * t (η) = y * 1 , η = ỹ ∈ σ x t (u * , x * ) (i.e. ϕ t (z * ) = y * 1 ). In these coordinates ρ t (u * , x * , y * 1 , ȳ * ) = ρt (u * , x * , ȳ * ), which equals 1 + |x * | -2 |u * | 2 + |ȳ * | 2 + |y * 1 | 2 1/2 = |x * | 2 + |u * | 2 + |ȳ * | 2 + |y * 1 | 2 1/2 |x * | . That is, ρ t (z * ) = |z * | |x * | . Since z * is any point in U ∩ Σ t ,
dz | Σ x t = du |x| -1 dx |z|d σ x t (u,x) y . (7.11)
That is, for any function f ∈ C 0 0 (Σ x t ),

f (z)dz | Σ x t = R n R d 1 |x| -1 σ x t (u,x) |z|f (z) d σ x t (u,x) y dx du .
Similarly, if we set Σ y t = {(u, x, y) ∈ Σ t : y = 0} and consider the projection

Π y t : Σ y t → R n × R d 1 \{0} , (u, x, y) → (u, y) , then dz | Σ y t = du |y| -1 dy |z|d σ y t (u,y) x . (7.12)
Let us denote Σ 0 t = {(u, x, y) ∈ Σ t : x = y = 0}. Then Σ t \Σ 0 t is a smooth manifold and dz | Σt defines on it a smooth measure. By (7.11) and (7.12), for any t the volume of the set {z ∈ Σ t \Σ 0 t : 0 < |x| 2 + |y| 2 ≤ ε} goes to zero with ε. So assigning to Σ 0 t zero measure we extend dz | Σt to a Borel measure on Σ t such that each set {z ∈ Σ t : |z| ≤ R} has a finite measure and

(dz | Σt ) (Σ x t ∪ Σ y t ) c = 0 . (7.13)
By (7.11) and (7.12) function |z| -1 is locally integrable Σ t with respect to the measure dz | Σt . So µ Σt (see (7.2)) is a well defined Borel measure on Σ t . Since |Az| = |z|, then, in view of (7.11) and (7.12),

dµ Σt | Σ x t = du |x| -1 dx d σ x t (u,x) y , dµ Σt | Σ y t = du |y| -1 dy d σ y t (u,y) x . (7.
14) The measure µ Σt defines on R d a Borel measure, supported by Σ t . It will also be denoted µ Σt .

Analysis of the integral I(t; f )

Note that for any t the mapping

L t : Σ x 0 → Σ x t , (u, x, y) → (u, x, y + t|x| -2 x)
defines an affine isomorphism of the bundles Π 0 | Σ x 0 and Π t | Σ x t . Since L t preserves the Lebesgue measure on the fibers, then in view of (7.11) it sends the measure µ Σ 0 to µ Σt . Using (7.14) we get that for any t the integral I(t), defined in (7.4), may be written as

I(t; f ) Σ 0 f (L t (z))µ Σ 0 (dz) = R n ×R d 1 |x| -1 σ(u,x)
f (u, x, y + t|x| -2 x)d σ x (u,x) y du dx . (7.15) Here σ(u, x)

:= σ x 0 (u, x) = x ⊥ -1 2 |u| 2 |x| -2
x . We recall that f (u, x, y) satisfies (7.3). Taking any smooth function ϕ(t) ≥ 0 on R which vanishes for |t| ≥ 2 and equals one for |t| ≤ 1 we write f = f 00 + f 1 , where f 00 = ϕ(|(x, y)| 2 )f and

f 1 = (1 -ϕ(|(x, y)| 2 ))f . Denoting B r (R m ) = {ξ ∈ R m : |ξ| ≤ r} and B r (R m ) = {ξ ∈ R m : |ξ| ≥ r} we see that supp f 00 ⊂ R n × B √ 2 (R 2d 1 ) , supp f 1 ⊂ R n × B 1 (R 2d 1 ) . (7.16) Setting next f 11 (z) = f 1 (z)(1 -ϕ(4|x| 2 )), f 10 (z) = f 1 (z)ϕ(4|x| 2 ) we write f = f 00 + f 11 + f 10 . Since (x, y) ∈ B 1 (R 2d 1 ) implies that |x| ≥ 1/ √ 2 or |y| ≥ 1/ √ 2, then in view of (7.16), supp f 11 ⊂ R n × B 1/2 (R d 1 x ) × R d 1 y , supp f 10 ⊂ R n × R d 1 x × B 1/ √ 2 (R d 1 y ) . (7.17)
Obviously, for i, j = 0, 1 we have

f ij k,m ≤ C k,m f k,m , for all k ≤ k * , m ≤ M .
Setting I ij (t) = I(t; f ij ) we have:

I(t; f ) = I 00 (t
∂ k I 00 (t) = R n B √ 2 (R d 1 )
|x| -1 dx du y∈σ(u,x)

d k /dt k f 00 (u, x, y + t|x| -2 x) d σ(u,x) y = R n B √ 2 (R d 1 ) |x| -1 y∈σ(u,x) d k y f 00 (u, x, y + t|x| -2 x) |x| -2 x d σ(u,x) y dxdu, (7.18)
where by d k y f 00 |x| -2 x we denote the action of the differential d k y f 00 on the set of k vectors, each of which equals to |x| -2 x. Setting τ = t -1 2 |u| 2 , for y ∈ σ(u, x) we have

y + t|x| -2 x = ȳ + τ |x| -2 x,
for some ȳ ∈ x ⊥ . (

Then we write the integral over y in (7.18) as

x ⊥ d k y f 00 (u, x, ȳ + τ |x| -2 x) |x| -2 x dȳ. (7.20) Since |ȳ + τ x|x| -2 | 2 = |ȳ| 2 + τ 2 |x| -2 , then on the support of the integrand |x| ≤ √ 2, |ȳ| 2 + τ 2 |x| -2 ≤ 2. ( 7.21)
In particular,

|τ | = t - 1 2 |u| 2 ≤ √ 2|x| ≤ 2 in (7.20). ( 7
k,m |x| -k u -m f k,m . Denoting R = |u|, r = |x| we get that |∂ k I 00 (t)| k,M f k,M √ 2 0 r d 1 -k-2 ∞ 0 R n-1 R -M χ |τ |≤ √ 2r dR dr . (7.23)
If n = 0, then the integral over R should be removed from the r.h.s. Below we estimate the integral ∂ k I 00 (t) separately for the cases n = 0 and n ≥ 1. a) If n = 0, then τ = t, we get from (7.22) that |x| ≥ t/ √ 2 and see from (7.16) that, for t = 0, I 00 (t) is C k * -smooth (since f ∈ C k * ). Then from (7.23) we obtain

|∂ k I 00 (t)| k,M f k,M √ 2 |t|/ √ 2 r d 1 -k-2 χ |t|≤2 dr . (7.24)
From here it follows that

|∂ k I 00 (t)| k f k,M if k ≤ min(d 1 -2, k * ), |∂ k I 00 (t)| k f k,M 1 + ln |t| if k ≤ min(d 1 -1, k * ) , (7.25)
while I 00 (t) = 0 for |t| ≥ 2. b) If n ≥ 1, then to estimate ∂ k I 00 (t) we split the integral for I 00 (t) in a sum of two. Namely, for a fixed t = 0 we write f 00 as f 00 = f 00< +f 00> , with f 00< = f 00 ϕ(8|x| 2 /t 2 ), where ϕ is the function, used to define the functions

f ij , 0 ≤ i, j ≤ 1. Then supp f 00< ⊂ {2|x| ≤ |t|}, supp f 00> ⊂ {2 √ 2|x| ≥ |t|}. (7.26)
With an obvious notation we have I 00 (t) = I 00< (t) + I 00> (t), where

I 00< (t) = R n B √ 2 (R d 1 )∩B |t|/2 (R d 1 ) |x| -1 y∈σ(u,x) |x| 2 +|y+t|x| -2 x| 2 ≤2 f 00< (u, x, y + t|x| -2 x) d σ(u,x) y dxdu , I 00> (t) = R n B √ 2 (R d 1 )∩B |t|/2 √ 2 (R d 1 ) |x| -1 y∈σ(u,x) |x| 2 +|y+t|x| -2 x| 2 ≤2 f 00> (u, x, y + t|x| -2 x) d σ(u,x) y dxdu .
Consider first function I 00< (t). We observe that, by (7.19), for y ∈ σ(u, x) and |x| ≤ |t|/2 (cf. (7.26))

|y + t|x| -2 x| ≥ |τ ||x| -1 = t - 1 2 |u| 2 |x| -1 ≥ -t|x| -1 > √ 2 , for t < 0 ,
so that I 00< (t) = 0 for t < 0. For t > 0, performing the change of variables √ tu = u, tx = x, we get

I 00< (t) =t d/2-1 R n B √ 2/t (R d 1 )∩B 1/2 (R d 1 ) |x | -1 ϕ(8|x | 2 ) y∈σ(u ,x ) |x | 2 t 2 +|y+|x | -2 x | 2 ≤2 f 00 ( √ tu , tx , y + |x | -2 x ) d σ(u ,x ) y dx du ,
where we notice that σ(u , x ) = σ(u, x). We differentiate with respect to t, observing that, by induction in k, for any l and k we have

d k dt k t l g( √ tu , tx ) = l 1 +l 2 +l 3 =k c l 1 ,l 2 ,l 3 t l-l 1 -l 2 /2 u l 2 • ∇ u l 2 x l 3 • ∇ x l 3 g( √ tu , tx ) ,
for any sufficiently regular function g and suitable constants c l 1 ,l 2 ,l 3 . From this we get

∂ k I 00< (t) k,M max l 1 +l 2 +l 3 =k t d/2-1-l 1 -l 2 /2 f k,M R n |u | l 2 u √ t -M B √ 2/t (R d 1 )∩B 1/2 (R d 1 ) |x | l 3 -1 y∈σ(u ,x ) |x | 2 t 2 +|y+|x | -2 x | 2 ≤2 d σ(u ,x ) y dx du .
Denoting points of the space x ⊥ as ȳ, we see that the integral over d σ(u ,x ) y is bounded by

ȳ∈x ⊥ |x | 2 t 2 +|ȳ+τ |x | -2 x | 2 ≤2 1 dȳ, τ = 1 - 1 2 |u | 2 . ( 7
- √ 2|x | ≤ |u | 2 2 ≤ 1 + √ 2|x | . (7.28)
As the the domain of integration in ȳ is bounded, then integral (7.27) is bounded by a constant. So putting |x | = r , |u | = R and using (7.28) we have

∂ k I 00< (t) k,M max l 1 +l 2 +l 3 =k f k,M t d/2-l 1 -l 2 /2-1 1/2 0 r d 1 -2+l 3 √ 2 √ 1+ √ 2r √ 2 √ 1- √ 2r R n-1+l 2 R 2 t -M/2 dR dr .
Since r ≤ 1/2, then on the domain of integration

2 - √ 2 ≤ R ≤ 2 + √ 2, while √ 2 1 + √ 2r - √ 2 1 - √ 2r
r . So the integral in dR is bounded by C t -M/2 r . Therefore

∂ k I 00< (t) k,M max l 1 +l 2 +l 3 =k f k,M t d/2-l 1 -l 2 /2-1 t -M/2 1/2 0 r d 1 -1+l 3 dr .
This implies that for 0 < t ≤ 4, for any k ≤ k * and any d 1 ≥ 1 we have

|∂ k I 00< (t)| k f k,0 t d/2-k-1 . (7.29)
While for any t ≥ 4 and any k ≤ k * ,

|∂ k I 00< (t)| k,M max l 1 +l 2 +l 3 =k f k,M,d t d/2-M/2-l 1 -l 2 /2-1 × √ 2/t 0 r d 1 -1+l 3 dr k,M f k,M t -(M +2+k+2d 1 -d)/2 .
(7.30)

We recall that I 00< (t) vanishes for t < 0.

For I 00> (t) we first note that by (7.21) and (7.26) function I 00> (t) vanishes if |t| > 4. Next, by induction in k, we observe that

d k dt k g(tx|x| -2 )(1 -ϕ(8|x| 2 /t 2 )) = l 1 +l 2 +l 3 =k c l 1 ,l 2 ,l 3 |x| 2(l 2 -l 1 ) t -3l 2 -l 3 × (x • ∇) l 1 g d l 2 dy l 2 (1 -ϕ) , (7.31)
where c l 1 ,l 2 ,l 3 = 0 if l 3 > 0 and l 2 = 0. Since ϕ = 0 only for

|t|/2 √ 2 ≤ |x| ≤ |t|/2, then d l 2 dy l 2 (1 -ϕ)t -3l 2 -l 3 l 2 ,l 3 |x| -3l 2 -l 3 , l 2 > 0 , so that d k dt k g(tx|x| -2 )(1 -ϕ(8|x| 2 /t 2 )) k |x| -k g k,0 .
From here, in a way analogous to (7.23), putting again |x| = r and |u| = R, we get that

|∂ k I 00> (t)| k,M f k,M √ 2 |t|/2 √ 2 r d 1 -k-2 ∞ 0 R n-1 R -M χ |τ |≤
|∂ k I 00> (t)| k,M,n f k,M √ 2 |t|/2 √ 2 dr r d/2-k-2 k,M f k,M , k < d/2 -1, f k,M (1 + | ln |t||) , k ≤ d/2 -1. (7.32) If k < d/2 -1,
then by the above ∂ k I 00 (t) is bounded for all t. In this case, modifying the integrand in (7.18) by the factor χ |x|≥ε , we see that thus obtained functions I ε 00> , I ε 00< satisfy the same estimates as the functions I 00> , I 00< above, so the function I ε 00 also does. The functions ∂ k I ε 00 (t) with ε > 0 obviously are continuous in t and converge to ∂ k I 00 (t) uniformly on bounded intervals. So the latter function also is continuous. Similar ∂ k I 00 (t) with k = d/2 -1 is continuous on any set |t| ≥ ε > 0, so is continuous for t = 0.

Integral I 11 (t).

Due to (7.17) and similar to (7.18), (7.20), for any k ≤ k * we have

∂ k I 11 (t) = R n |x|≥1/2 |x| -1 x ⊥ d k y f 11 (u, x, ȳ + τ x|x| -2
∂ k I 11 (t) = R n |x|≥1/2 |x| -k-1 x ⊥ Φ k (z) dȳdxdu , (7.34)
where z = (u, x, ȳ), ȳ ∈ x ⊥ , and

|Φ k (z)| k f k,M ẑ -M , ẑ = (u, x, ȳ + τ x|x| -2 ). ( 7
K := C k f k,M O |x|≥1/2 |x| -k-1 x ⊥ |t| + |x| 2 + |ȳ| 2 -M/2 dȳdxdu .
Since O 1 du ≤ Ct n/2 , then by putting r = |x|, |t| + r 2 = T 2 and R = |ȳ|/T we find that

K k f k,M t n/2 ∞ 1/2 r d 1 -2-k T d 1 -1-M ∞ 0 R d 1 -2 1 + R 2 -M/2 dRdr .
The integral in dR is bounded since M > d 1 , so that

K k,M f k,M t n/2 ∞ 1/2 r d 1 -2-k |t| + r 2 (d 1 -1-M )/2 dr .
Recalling that we are considering the case t ≥ 1, we put r = √ t l. Then

K kM f k,M t n+1+d 1 -2-k+d 1 -1-M 2 ∞ t -1/2 /2 l d 1 -2-k (1 + l 2 ) d 1 -1-M 2
dl .

Since M > 2d 1 , the integral over l converges and we get

K k,M f k,M |t| -(M +2-d+k)/2 |t| max(0,k+1-d 1 )/2 Y (t) , with Y = ln t if k = d 1 -1 and Y = 1 otherwise. Then, in the case Y = 1 the component of (7.34), corresponding to u ∈ O, is bounded by C(k, M, d) f k,M |t| -κ , κ = M + 2 -d 2 , ( 7
|Φ k (z)| k (|t||x| -1 + |x|) -M . Let M = M 1 + M 2 , M j ≥ 0.
Then the part of the integral (7.34) for u ∈ O c is bounded by

C f k,M |x|≥1/2 |x| -1-k t|x| -1 + |x| -M 1 R n x ⊥ (u, ȳ) -M 2 dȳdu dx Choosing M 2 = n + d 1 -1 + γ with 0 < γ < 1 (then M 1 , M 2 > 0 since M > d)
we achieve that the integral over du dȳ is bounded by C(γ), for any γ. Since by Young's inequality6

(A + B) -1 ≤ C a A -a B a-1 , 0 < a < 1 , for any A, B > 0, then t|x| -1 + |x| -M 1 ≤ C a |x| (2a-1)M 1 |t| -aM 1 (0 < a < 1
). So the integral above is bounded by

C(γ) f k,M |t| -aM 1 |x|≥1/2 |x| -1-k+bM 1 dx , b = 2a -1 ∈ (-1, 1) . Denote b * = 1+k-d 1 M 1 .
* > -1 if γ is sufficiently small, since M > d. Noting that a(b * )M 1 = b * + 1 2 M 1 = M + 2 + k -d -γ 2 = κ + k 2 - γ 2
(κ was defined in (7.37)), we see that the part of integral (7.34), corresponding to u ∈ O c , is bounded by (7.37) if k ≥ 1, while for k = 0 it is bounded by (7.37) with κ replaced by any κ < κ.

(7.38)

2) Now let n = 0. Then

∂ k I 11 (t) ≤ |x|≥1/2 |x| -1-k x ⊥ Φ k (z) dȳdx , z = (x, ȳ) , (7.39)
where |Φ k (z)| k ẑ -M with ẑ = (x, ȳ + tx|x| -2 ). Repeating literally the step 1b) above with n = 0 we get that for |t| ≥ 1 the integral in (7.39) may also be bounded by (7.37). We recall that for |t| ≤ 1 the derivative ∂ k I 11 (t) was estimated in (7.33).

Integral I 10 (t).

Now we use the second disintegration in (7.14) instead of the first. Since by (7.17) on the support of the integrand |y| ≥ 1/ √ 2, then repeating the argument above with x and y swapped we get that I 10 (t) meets the same estimates as I 11 (t).

7.3.4

End of the proof of Theorem 7.1

Finally, -combining together relations (7.25), (7.29), (7.32) and ( 7.33) we estimate ∂ k I(t) for 0 < |t| ≤ 4, while -combining together (7.30), (7.37), (7.38) and using the fact that ∂ k I 00> (t) and ∂ k I 00 (t) vanish for |t| ≥ 4 when n = 0, we estimate ∂ k I(t) for t ≥ 4.

For the reason, explained at the end of Section 7.3.1, the involved derivatives are continuous functions. This proves the theorem.

Linear transformations of quadrics

In this subsection we denote by C 0 spaces of continuous functions with compact support.

In R d = {z} let us consider a quadratic form with real coefficients7 F (z) = 1 2 Az • z of signature (n 0 , n + , n -) such that n 0 = 0, n + ≥ n -=:

d 1 ≥ 1. Denote n = n + -n -.
Using the standard diagonal normal form for a symmetric quadratic form, we construct a linear transformation

L : R d → R d , z → Z = (u, x, y), u ∈ R n , x, y ∈ R d 1 , such that Q(L(z)) = F (z), where Q(Z) = 1 2 |u| 2 + x • y. Consider the cor- responding quadrics Σ Q t = {Z : Q(Z) = t}, Σ F t = {z : F (z) = t},
µ Q t , f Q = lim ε→0 1 2ε t-ε≤Q(Z)≤t+ε f Q (Z) dZ, (7.40) µ F t , f F = lim ε→0 1 2ε t-ε≤F (z)≤t+ε f F (z) dz, where f Q , f F ∈ C 0 (R d )
Q ∈ C 0 Σ Q t \ {0} and f F ∈ C 0 Σ F t \ {0} we have µ Q t , f Q = Σ Q t f Q (Z) |∇Q(Z)| dZ | Σ Q t , µ F t , f F = Σ F t f F (z) |∇F (z)| dz | Σ F t , where dZ | Σ Q(or F ) t is the volume element on Σ Q(or F ) t \ {0}, induced from R d , see [13]. Now let f F = f Q • L. Then the integral in (7.40) equals t-ε≤Q(Z)≤t+ε f Q (Z) dZ = | det(L)| t-ε≤F (z)≤t+ε
f F (z) dz, so passing to the limit we get that

L • | det(L)|µ F t = µ Q t . (7.41)
Thus, to examine the function

t → I F (t; f ) = µ F t , f , µ F t = |∇F (z)| -1 dz | Σ F t , (7.42)
we are free to use any linear coordinate system in R d since changing the coordinates we only modify function I F by a constant factor.

Sign definite forms

Finally let us consider the case when n 0 = 0 and min(n + , n -) = 0, i.e. when the form F (z) = 1 2 Az • z is sign-definite and non degenerate. Suppose for definitenes that n -= 0. Then there exists a linear transformation L such that F (z) = Q(L(z)), where Q(Z) = 1 2 |Z| 2 , Z ∈ R d . The quadric Σ t reduces to the empty set for t < 0, so function I F (t) (see (7.42)) vanishes for t < 0. The calculation of previous subsection remains true in this case, so (7.41) and the change of coordinates Z = √ 2t Z show that

I F (t; f ) = C(d, L)t -1 |Z|= √ 2t f Q (Z) µ S d-1 √ 2t (dZ) = C(d, L)t d/2-1 |Z |=1 f Q ( √ 2tZ ) µ S d-1 1 (dZ ), t > 0, f Q = f • L -1 ,
Then, setting R l = 2 l ρL we get

J + 0 = ∞ l=0 R l <q≤R l+1 q -d I q S q =η(0)σ * (A) q>ρL q -1 I q + O γ ρ α+γ-d-1 L α+γ |w| L 1 ∞ l=0 2 -l(d+1-α-γ) =η(0)σ * (A) q>ρL q -1 I q + O γ ρ α+γ-d-1 L α+γ |w| L 1 .
It remains to compare the sum A := q>ρL q -1 I q with the integral B := L d ∞ ρ r -1 I * (r) dr. Since L d I * (r) = I rL , then changing the variable of integration r to q = rL, B takes the form ∞ ρL q -1 I q dq. Then,

|A -B| ≤ q>ρL q -1 I q - ∞ ρL +1
q -1 I q dq + ρL +1 ρL q -1 I q dq . (A.11) Due to (A.8), |q -1 I q | q -2 L d+1 |w| L 1 and |∂ q (q -1 I q )| q -3 L d+1 |w| L 1 . Thus,the both terms in the r.h.s. of (A.11) are bounded by (ρL

) -2 L d+1 |w| L 1 = ρ -2 L d-1 |w| L 1 .
Recall that ĈA is a constant arising in Lemma A.2.

I A := L d σ ∞ (w)
q≤ρL q -d S q (0), I B := q≤ρL S q (0)q -d (f q + g q ) , with |f q | qL d-1 log( q L ) w d/2-1,d+1 , |g q | N q N L d-N + 1 Lq -1 w 0,d+1 .

In view of the definitions (1.10)- (1.11), our first aim is to compute the constants σ p (A, 0). For a prime p and k ∈ N let consider the set Note that p -dt S p t (0) = 1 for t = 0, while for t = 1: We proceed now by induction, supposing that, for k ≥ 1,

  m) := ∞ l=0 p -dl S p l (c; A, L 2 m), σ p := σ 0 p ,(1.10)whereS 1 ≡ 1, σ * c (A) := p (1 -p -1 )σ c p (A, 0), σ * (A) := σ * 0 (A) = p (1 -p -1 )σ p (A, 0), and σ(A, L 2 m) = p σ 0 p (A, L 2 m) = p σ p (A, L 2 m).

Proposition 1 . 5 .

 15 (i) If d ≥ 5 then there exist positive constants c(A) < C(A) such that 0 < c(A) ≤ σ(A, L 2 m) ≤ C(A) < ∞ for any non-degenerate matrix A, uniformly in L and m.

  . The work of AD was funded by a grant from the Russian Science Foundation (Project 20-41-09009) [Sections 1-7] and by the Grant of the President of the Russian Federation (Project MK-1999.2021.1.1) [Sections A-C]. Research of SK was equally supported by the Ministry of Science and Higher Education of the Russian Federation (megagrant No. 075-15-2022-1115) and by l'Agence Nationale de la Recherche (France), grant 17-CE40-0006.

) by Corollary 3 . 3 .

 33 Next we use Lemma 3.4 with N replaced by N + 1 to get that

Corollary 3 . 7 .

 37 Let an integrable function f belong to the class C M,0 (R), M ∈ N, and 0 < x ≤ C for some C > 0. Then, for any 0

Proposition 5 . 1 .

 51 For any d ≥ 1, N > 0 and c = 0, | Ĩq (c)| N ,m L q |c| -N w N,2N +d+1 (5.1)

  e tF (z) -u • z dz dt, u := c L/q.

(6. 6 )

 6 Since (2πiLf (b)) -1 Le(f (b)) = e(f (b)), then integrating by parts N times integral I a,t we get

8 )

 8 In virtue of Lemma 3.2 with N = 2,

Lemma 6 . 4 .

 64 S R c . Now we study the integral over the bad set S R c . For any d ≥ 1, 1 ≤ R ≤ |u| 1/3 and 0 < β < 1 we have S c R |p(t)||I a,t | da dt m R d |u| -d/2+1+β w 0,K(d,β) , where K(d, β) = d + d 2 /2β + 4. Proof. On S R c we use for I a,t the easy upper bound |I a,t | max |b i |≤1 ∀i | w(δb + a)| ≤ w 0,0 . (6.10)

|t|≤|u| 1 -

 1 β/d |p(t)|dt |u| 1-β/d , from which we obtain |t|≤|u| 1-β/d R d |p(t)||I a,t |1 S R c (a, t) da dt m |u| -d/2+1 w 0,K(d,β) .

  .27) By (7.22), on the support of the integrand |τ | ≤ √ 2|x |. So there 1

√ 6 √

 6 2r dR dr (here and below b a dr = 0 if b ≤ a). Since on the integration domain, due to (7.26) and the indicator function χ |τ |≤ √ 2r , we have R 2 ≤ 2r, then

1 2

 1 .37) for all |t| ≥ 1, since max(0, k + 1 -d 1 ) ≤ k. If Y = ln t the same estimate holds in the case d 1 ≥ 2 since max(0, k + 1 -d 1 ) < k. In the case d 1 = 1 and Y = ln t (i.e. k = 0) we get (7.37) with κ replaced by any κ < κ (and constant C depending on κ ). b) Now consider the integral for u ∈ O c = R n \O. There |τ | = |t -|u| 2 | ≥ 1 2 |t|. So, by inequalities (7.35) and (7.36), |Φ k (z)| k (u, ȳ) -M and

  Then for b = b * the exponent for |x| in the formula above equals -d 1 , and b

Lemma A. 5 .

 5 Assume that the determinant det A is a square of an integer. Then for any γ > 0, N > 1, any ρ ≤ 1 and L satisfying ρL > 1,J - 0 =L d σ ∞ (w; A, 0) σ * (A) log(ρL) + ĈA + O γ,N ρ α+γ-d L α+γ + L d ρ log L + ρ N -1 + L 1-d w d/2-1,d+1 .Proof. Inserting Proposition 3.8 b) with M = d/2 -1 = 1 and β = 1 into the definition of the term J - 0 , we get J - 0 = I A + I B , where

Sp 1 + p - 1 .

 11 p (k) = {(x, y) mod p k : F d (x, y) = 0 mod p k } and denote N p (k) := S p (k). Note that the set S p (k) and the constant N p (k) depend on d. Then the constants σ p can be rewritten asσ p (d) := σ p (A, 0) = lim k→∞ N p (k) p (d-1)k . (B.2)This relation is mentioned in[START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], p. 199, without a proof; we sketch its rigorous derivation at the end of this appendix.Let N p (d) := N p (1) be the number of F p -rational points on {F d = 0 mod p}.Lemma B.1. For any prime p,σ p (d) = N p (d) -1 p d-1 -p 1-d . (B.3)Proof. For j = 0, 1, . . . , k we define S p (k, j) as a set of (x, y) ∈ S p (k) such that (x, y) = p j (x , y ) mod p k , where p (x , y ).So S p (k, 0) = {(x, y) ∈ S p (k) : p (x, y)} and S p (k, k) = {(0, 0)}. Sets S p (k, j) and S p (k, j ) with j = j do not intersect, and denoting N p (k, j) = S p (k, j) we haveS p (k) = k j=0 S p (k, j), N p (k) = k j=0 N p (k, j) .In particular, N p (1, 0) = N p -1 since N p (1, 1) = 1. We claim thatN p (k, 0) = N p (k -1, 0)p (d-1) , and thusN p (k, 0) = N p (1, 0)p (d-1)(k-1) = (N p -1) p (d-1)(k-1) . (B.4)Indeed, we argue by induction in k. Let k = 2 and (x, y) ∈ S p (2, 0). Let us write (x, y) as (x 0 + pa, y 0 + pb) with (x 0 , y 0 ), (a, b) ∈ F d p . Then p (x 0 , y 0 ), so (x 0 , y 0 ) ∈ S p (1, 0). Let us now fix any (x 0 , y 0 ) ∈ S p (1, 0) and look for (a, b) ∈ F d p such that (x 0 + pa, y 0 + pb) ∈ S p (2, 0). Since p 2 F (a, b) = 0 mod and thusσ p (d) = 1 + p 1-s -p -s -p 1-d 1 -p 2-2d = (1 + p 1-s )(1 -p -s ) 1 -p 2-2d .Since by Euler's formula p (1 -p -l ) = 1/ζ(l) for any l > 1, then in the case d = 4 we get from (1.11) and the obtained formula for σ p (d) thatσ(A, 0; d = 4This does not converge, butσ * (A; d = 4) = p (1 -p -1 )σ p (4) = ζ(A, 0; d) = ζ(s -1)ζ(2d -2) ζ(s)ζ(d -2) = (1 + 2 1-s )(1 + 2 2-4s ) (1 + 2 -s )(1 + 22-2s ) +o(1) = 1+o(1) tends to 1 when d = 2s ≥ 10 grows. It remains to prove (B.2). By definition (1.10), σ p = ∞ t=0 p -dt S p t (0), where S p t (0) = a mod p t * b mod p t e p t (aF (b)).

p

  -d S p (0) = p -d p, p F (b) e p (aF (b)) = p -d (p -1)N p (d) + p -d (-1)(p d -N p (d)) = p 1-d N p (d) -1 , n, m = 0 such that (m, n) = 1. Therefore 1 t=0 p -dt S p t (0) = p 1-d N p (1).

p 1 s=0F

 1 -dt S p t (0) = p(1-d)k N p (k) .Then we writeS p k+1 (0) = a mod p k+1 * b mod p k+1 e p k+1 (aF (b)) = Σ 1 + Σ 2 + Σ 3 ,where we have definedΣ 1 := a mod p k+1 * p k+1 |F (b) 1 = p k (p -1)N p (k + 1) , Σ 2 := a mod p k+1 * F (b)=lp k e p (al) = -p k (p d N p (k) -N p (k + 1)) , Σ 3 := a mod p k+1 * k-(b)=lp s e p k+1-s (al) = 0 ,with a non-zero l = l(b) such that p l. The equalities above essentially follow by a repeated application of (B.5). This way we have gotS p k+1 (0) p d(k+1) = p k+1 N p (k + 1) -p d+k N p (k) p d(k+1) = N p (k + 1) p (d-1)(k+1) -N p (k) p (d-1)k ,which completes the induction step, thus proving (B.2).

  2 , K 3 depend on d and ε, while C depends on d, ε, A and τ, |z * |. Notation and agreements. We write A a,b B if A ≤ CB, where the constant C depends on a and b. Similar, O a,b ( w m 1 ,m 2 ) stands for a quantity, bounded in absolute value by C(a, b) w m 1 ,m 2 . We do not indicate the dependence on the matrix norms A , A -1 and on the dimension d since most of our estimates depend on these quantities.

  1-β/d . Since |u| > 1, then considering separately the cases |t| ≤ |u| and |t| ≥ |u| we see that

	R δ|t|	t/|u| ≤ R|u| -1/2+β/d .	(6.12)

  ) + I 10 (t) + I 11 (t) . I 00 (t) is a continuous function, and for 1 ≤ k ≤ k * ,

	7.3.1 Integral I 00 (t).
	By (7.15)

  )[x|x| -2 ] dȳ dxdu .

	We easily see that I 11 (t) is a C k -smooth function and, since M > d and
	ȳ + τ x|x| -2 ≥ |ȳ|, then
	|∂ k I 11 (t) k,M f k,M ∀t.	(7.33)
	Now let |t| ≥ 1. Let us write ∂

k I 11 as

  It is empty if t < 0, while for t ≥ 0, O = {u : t ≤ |u| 2 ≤ 3t} . By(7.35) and the first relation in (7.36), for t ≥ 0 the part of the integral in (7.34) with u ∈ O is bounded by

	1) Let n ≥ 1.
	a) We first integrate in (7.34) over u in the spherical layer
	O := {u : |τ | = t -1 2 |u| 2 ≤ 1 2 t} .
		.35)
	Obviously,
	|ẑ| ≥ |z| , |ẑ| ≥ 2 -1/2 |z| + |τ ||x| -1 .	(7.36)
	Below we separate the cases n ≥ 1 and n = 0.

E.g., m = 0 -this case is the most important for us.

In[START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF] the result below is stated for w ∈ C ∞ 0 . But the argument there, based on the Poisson summation, applies as well to regular functions w.

Concerning the classical notion of the Jacobi symbol and the Dirichlet L-function we refer a reader without number-theoretical background e.g. to[START_REF] Serre | A Course in Arithmetic[END_REF] and[START_REF] Karatsuba | Basic Analytic Number Theory[END_REF].

This holds e.g. if det A = ±1 and z * ∈ Z d L .

Actually it is proved there that any function h defined through (3.2) with arbitrary weight function ω ∈ C ∞ 0 (R), supported on [1/2, 1], may provide a representation of δ(n).

Indeed, by Young's inequality with p = 1/a, q = 1/(1 -a) we have that A a B (1-a) ≤ aA + (1 -a)B ≤ Ca(A + B). This proves the assertion.

 is the only part of our work, where quadratic forms are allowed to have non-rational coefficients.

where µ S d-1 r is the volume element on the d -1 sphere of radius r. From this relation we immediately get that for any k ≤ min(d/2 -1, k * ),

and

General result

We sum up the obtained results in the following Theorem 7.3. Consider any nondegenerate quadratic form F (z) = 1 2 Az • z on R d , d ≥ 3, and a function f ∈ C k * ,M (R d ), M > d. Then the corresponding integral I F (t; f ) = µ F t , f (see (7.42)) meets the assertions of Theorem 7.1.

Proof. i) If n + ≥ n -, then by means of a linear change of variable F may be put to the normal form (7.1), where d 1 ≥ 0. Now the assertion follows from the argument in Subsections 7.4, 7.5 and Theorem 7.1.

ii) If n -> n + , then the quadratic form -F is as in i), and the assertion follows again since obviously I -F (t; f ) = I F (-t; f ).

A The J 0 term: case d = 4

In this section we find asymptotic for the term J 0 from (1.19) in the case d = 4 and m = 0.

(A. [START_REF] Buckmaster | Effective dynamics of the nonlinear Schrödinger equation on large domains[END_REF] Below in this section we always assume (A.1).

A.1 Preliminary results and definitions

We will need Lemmas 30 and 31 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF], restricted for the case m = 0 and d = 4, which we state below without a proof. Recall that constants σ * c (A) are defined in (1.10) and σ * (A) = σ * 0 (A). Set α := 7/2 and recall (A.1). Lemma A.1 (Lemma 30 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). For any ε > 0 and X ∈ N,

where η(c) = 1 if c • A -1 c = 0 and at the same time det A is a square of an integer, and

Lemma A.2 (Lemma 31 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]). Assume that the determinant det A is a square of an integer. Then for any ε > 0 and X ∈ N,

where ĈA is a constant depending only on A. Otherwise, if det A is not a square of an integer, then for any ε > 0 and

where χ is the Jacobi symbol ( det(A) * ) and L(1, χ) is the Dirichlet L-function. We will also need the following construction. Let us define for r ∈ R >0

Consider a function K(ρ; w, A), ρ ∈ R >0 , given by

) where constant η(0) is defined according to Lemma A.1 and ĈA -according to Lemma A.2. Note that functions I * (r) and K(ρ) do not depend on L.

We claim that function K(ρ), ρ > 0, can be extended at ρ = 0 by continuity. Indeed, for 0 < ρ 1 < ρ 2 ≤ 1

(A.5) Using that I * (r) = L -d I rL (0) (see (3.8)), we write the term I * (r) from (A.5) in the form, given by Proposition 3.8 b). Then I * (r) takes the form of the r.h.s. of (3.11), divided by L d , with q = rL. The leading term in the obtained formula for I * (r) is σ ∞ (w; A, 0) and the corresponding integral [START_REF] Dymov | Formal expansions in stochastic model for wave turbulence 1: kinetic limit[END_REF]) cancels the first term in the brackets of (A.5). Then, setting M = d/2 -1, β = r γ , γ = γ/d and 0 < γ < 1 in the just discussed formula for I * (r), obtained from (3.11), we get the estimate

The last inequality here is obtained by choosing N = N (γ) to be sufficiently large and writing r d/2(1-γ)-2 log r γ r d/2(1-γ)-2-γ/2 = r d/2-2-γ . Therefore K(ρ) extends at ρ = 0 by continuity and

-Hölder continuous at zero, for any γ > 0.

A.2 Estimate for J 0

Argument in this section is related to Section 13 of [START_REF] Heath-Brown | A new form of the circle method, and its application to quadratic forms[END_REF]. Here we restrict ourselves for the case when the determinant det A is a square of an integer, so in particular η(0) = 1. We use this specification only in the proof of Lemma A.5, when applying Lemma A.2. The case of non-square determinant is easier and can be obtained similarly, using the second assertion of Lemma A.2.

Proposition A.3. Assume that the determinant det A is a square of an integer. Then for any 0 < ε < 1/5,

Proof. To establish Proposition A.3 we write J 0 in the form (1.21), J 0 = J + 0 + J - 0 , where J + 0 := q>ρL q -d S q (0)I q (0) and J - 0 := q≤ρL q -d S q (0)I q (0) , with ρ ≤ 1. Then the assertion follows from Lemmas A.4 and A.5 below.

Recall that α = 7/2.

Then for any γ > 0, any ρ ≤ 1 and L satisfying ρL > 1,

Proof. To simplify the notation, in this proof we denote I q := I q (0) and S q := S q (0). Let us recall the summation by parts formula for sequences (f q ) and (g q ):

We take arbitrary R ∈ N and apply the latter with m = R, n = 2R, f q = q -d I q and g q = R<q ≤q S q , so that g R = 0 and S q = g q -g q-1 for q > R. We find R<q≤2R q -d S q I q =(2R)

where for a sequence (a q ) we denote ∂q a q := a q+1 -a q . By (3.8)-(3.9),

where the first estimate above follows from Corollary 3.3 while the second one -from Lemma 3.2 with m = 1, n = N = 0. Then, | ∂q (q

where we recall that σ * 0 (A) = σ * (A). Let us view the r.h.s. of (A.7) as a linear functional G (S q ) on the space of sequences (S q ). Then, inserting formula (A.9) in the r.h.s. of (A.7), we get R<q≤2R q -d S q I q = η(0)σ * (A)G (q d-1 )

where the O γ term is obtained by applying (A.8) together with the estimate for ∂q (q -d I q ) above and replacing the sums S q , S q in the r.h.s. of (A.7) by O γ (R α+γ ). According to the summation by parts formula (A.7) with S q replaced by q d-1 , we have R<q≤2R q -d q d-1 I q = G (q d-1 ) . Thus, by (A.10),

So,

(A.12) on account of (3.13). As for the term I B , since d = 4, Lemma 2.1 implies that

for N ≥ 2. The obtained estimates on I A and I B imply the assertion. Now we conclude the proof of Proposition A.3. The leading term of J 0 is given by the sum of leading terms from formulas for J + 0 and J - 0 in Lemmas A.4 and A.5. Since η(0) = 1, it takes the form

where in the last equality we used (A.4) and (A.6). Then we find

w 0,d+1 . We now pick ρ = L -1/5 and N = 2, and, using that d = 4, get the assertion of proposition.

A.3 Estimate for σ 1 (w; A, L)

In this section we get an upper bound for the subleading order term σ 1 of the asymptotics from Theorem 1.4.

In the case when the determinant det A is not a square of an integer, σ 1 is given by (1.14) and the task is not complicated. Indeed, according to Lemma A.2, the product p (1 -χ(p)p -1 )σ p (A, 0) is finite (and independent from L). On the other hand, by (A.12), |σ ∞ (w; A, 0)| w 0,d+1 . Thus,

In the case when det A is a square, σ 1 is given by (1.24) and the required estimate is less trivial.

Proposition A.6. Assume that det A is a square of an integer. Then

Proof. Since η(c) takes values 0 or 1, then according to the definition (1.24) of σ 1 , we have

Let us first estimate the term K(0). According to (A.6),

On the other hand, σ * (A) is independent from L and, in view of Lemma A.2 is finite. Then, by the definition (A.4) of K(ρ),

Due to the definition (A.3) of the integral I * (r) and Corollary 3.3, |I * (r)| r -1 |w| L 1 r -1 w 0,d+1 . Then, in view of (A.12), |K(1)| w 0,d+1 , so that, by (A. [START_REF] Birch | Forms in Many Variables[END_REF],

Let us now estimate the terms σ c ∞ (w), which are given by (1.23):

where

) and M ∈ N will be chosen later. Using that d = 4, according to Lemma 6.2,

where we denoted C(w) := w N ,d+5 + w 0, N +3d+4 . On the other hand, by Proposition 5.1, B Constants σ(A, 0) and σ * (A)

It is clear that our result provides an approximation to the series N L (w; A, m) through the singular integral σ ∞ (w) only if the singular series σ(A, m) or σ * (A) are strictly positive. In fact, the singular series is known to be strictly positive under a very general condition, namely, for non-singular forms of any degree that have non-singular solutions in R and in every p-adic field (provided the singular series is absolutely convergent), see, e.g., Section 7 of [START_REF] Birch | Forms in Many Variables[END_REF]. However, since the most interesting case in applications to mathematical physics is the case of the quadratic form F d (x, y) below, we give in this Appendix a direct elementary treatment of the evaluation of the constants σ(A, 0) for d ≥ 5 and σ * (A) for d = 4 in this case, independent of the general theory.

In this section we consider the case when the quadratic form reads as So each (x 0 , y 0 ) ∈ S p (1, 0) generates exactly p d-1 vectors (x, y) ∈ S p (2, 0), which proves the formula for k = 2. This argument remains valid for any k ≥ 2, by representing (x, y) mod p k in the form (x 0 + p k-1 a, y 0 + p k-1 b) with (x 0 , y 0 ) ∈ F d p k-1 and (a, b) ∈ F d p . Let now (x, y) ∈ S p (k, j) with j ≥ 1. Then (x, y) = p j (x , y ) mod p k , where p (x , y ) and (x , y ) satisfies p 2j F (x , y ) = 0 mod p k . Thus (x , y

=: j k . The correspondence (x, y) → (x , y ) is a well defined mapping from S p (k, j) to S p (k -2j, 0). Indeed, if (x 1 , y 1 ) ∼ (x, y) in S p (k, j), then p k-j | (x 1 , y 1 ) -(x , y ) , so (x 1 , y 1 ) ∼ (x , y ) in S p (k -2j, 0). Since this map is obviously surjective, then it is a bijection of S p (k, j) onto S p (k -2j, 0), which in view of (B.4) implies N p (k, j) = N p (k -2j, 0) = (N p -1) p (d-1)(k-2j-1) .

By (B.4) this formula as well holds for j = 0. Any (x, y) such that p j |(x, y) with j ≥ j k + 1 satisfies F (x, y) = 0 mod p k . Thus