Vinogradov's mean value theorem and its associated restriction theory via … · 2014. 10. 8. ·...

Vinogradov’s mean value theorem and its associatedrestriction theory via efficient congruencing.

Trevor D. Wooley

University of Bristol

Oxford, 29th September 2014

Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 1 / 34

1. IntroductionLet k ≥ 2 be an integer, and consider

g : Tk → C (T = R/Z ' [0, 1)),

with an associated Fourier series

g(α1, . . . , αk) =∑n∈Zk

g(n1, . . . , nk)e(n1α1 + . . .+ nkαk),

in which g(n) ∈ C and e(z) = e2πiz .

Restriction operator: (E. Stein, J. Bourgain, K. Hughes, et al.)

Rg :=∑n∈Zk

n=(n,n2,...,nk )

g(n)e(n ·α).

[This is just one example of a restriction operator!]

We are interested in the norm of the operator g 7→ Rg .

g : Tk → C (T = R/Z ' [0, 1)),

g(α1, . . . , αk) =∑n∈Zk

g(n1, . . . , nk)e(n1α1 + . . .+ nkαk),

Rg :=∑n∈Zk

n=(n,n2,...,nk )

g(n)e(n ·α).

g : Tk → C (T = R/Z ' [0, 1)),

g(α1, . . . , αk) =∑n∈Zk

g(n1, . . . , nk)e(n1α1 + . . .+ nkαk),

Rg :=∑n∈Zk

n=(n,n2,...,nk )

g(n)e(n ·α).

(Slightly) more concretely (for analytic number theorists):

Consider a sequence (an)∞n=1 of complex numbers, not all zero, and definethe exponential sum fa = fk,a(α;X ) by putting

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

Aim: Obtain a bound for

(‖fa‖Lp/‖a‖`2)

in terms of p, k and X .

Conjecture (Main Restriction Conjecture)

For each ε > 0, one has

‖fa‖Lp‖a‖`2

�ε,p,k

X ε, when 0 < p ≤ k(k + 1),

X12−

k(k+1)2p , when p > k(k + 1).

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

(‖fa‖Lp/‖a‖`2)

‖fa‖Lp‖a‖`2

�ε,p,k

X ε, when 0 < p ≤ k(k + 1),

X12−

k(k+1)2p , when p > k(k + 1).

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

(‖fa‖Lp/‖a‖`2)

‖fa‖Lp‖a‖`2

�ε,p,k

X ε, when 0 < p ≤ k(k + 1),

X12−

k(k+1)2p , when p > k(k + 1).

(Even) more concretely (for analytic number theorists):

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

∮|fk,a(α;X )|2s dα�

( ∑n≤X|an|2

, when s ≤ 12k(k + 1),

X s−12k(k+1)

( ∑n≤X|an|2

, when s > 12k(k + 1).

Here, we write∮

for∫

[0,1)k .

(Even) more concretely (for analytic number theorists):

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

( ∑n≤X|an|2

, when s ≤ 12k(k + 1),

X s−12k(k+1)

( ∑n≤X|an|2

, when s > 12k(k + 1).

Here, we write∮

for∫

[0,1)k .

Some observations, I:

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

( ∑n≤X|an|2

, when s ≤ 12k(k + 1),

X s−12k(k+1)

( ∑n≤X|an|2

, when s > 12k(k + 1).

Consider the sequence (an) = 1. Then MRC implies that∮|fk,1(α;X )|2s dα� X ε(X s + X 2s−1

2k(k+1)),

an assertion equivalent to the Main Conjecture in Vinogradov’s MeanValue Theorem.

Some observations, I:

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

( ∑n≤X|an|2

, when s ≤ 12k(k + 1),

X s−12k(k+1)

( ∑n≤X|an|2

, when s > 12k(k + 1).

Consider the sequence (an) = 1. Then MRC implies that∮|fk,1(α;X )|2s dα� X ε(X s + X 2s−1

2k(k+1)),

an assertion equivalent to the Main Conjecture in Vinogradov’s MeanValue Theorem.

Some observations, II:Consider the situation in which (an) is supported on a thin sequence, say

an = card{

(x , y) ∈ Z2 : n = x4 + y4}.

Then MRC implies that for 1 ≤ s ≤ 12k(k + 1), one should have

∮|fk,a(α;X )|2s dα� X ε

∑n≤X|an|2

� X ε(X 1/2

)s= X s/2+ε.

But by orthogonality, when s is a positive integer, this integral counts thenumber of solutions of the system of equations

s∑i=1

i + v4i )j − (u4

s+i + v4s+i )

= 0 (1 ≤ j ≤ k),

with 1 ≤ u4i + v4

i ≤ X (1 ≤ i ≤ 2s).

an = card{

(x , y) ∈ Z2 : n = x4 + y4}.

∑n≤X|an|2

� X ε(X 1/2

)s= X s/2+ε.

s∑i=1

i + v4i )j − (u4

s+i + v4s+i )

= 0 (1 ≤ j ≤ k),

with 1 ≤ u4i + v4

i ≤ X (1 ≤ i ≤ 2s).

an = card{

(x , y) ∈ Z2 : n = x4 + y4}.

∑n≤X|an|2

� X ε(X 1/2

)s= X s/2+ε.

s∑i=1

i + v4i )j − (u4

s+i + v4s+i )

= 0 (1 ≤ j ≤ k),

with 1 ≤ u4i + v4

i ≤ X (1 ≤ i ≤ 2s).Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 6 / 34

Some observations, II:

So the number N(X ) of integral solutions of the system of equations

s∑i=1

i + v4i )j − (u4

s+i + v4s+i )

= 0 (1 ≤ j ≤ k),

with 1 ≤ u4i + v4

i ≤ X (1 ≤ i ≤ 2s), satisfies

N(X )� X s/2+ε.

But the number of diagonal solutions with ui = us+i and vi = vs+i , for alli , has order of growth X s/2.

So this shows that “on average”, the solutions are diagonal. This is not aconclusion that follows from the Main Conjecture in Vinogradov’s meanvalue theorem (by any method known to me!).

Some observations, II:

So the number N(X ) of integral solutions of the system of equations

s∑i=1

i + v4i )j − (u4

s+i + v4s+i )

= 0 (1 ≤ j ≤ k),

with 1 ≤ u4i + v4

i ≤ X (1 ≤ i ≤ 2s), satisfies

N(X )� X s/2+ε.

But the number of diagonal solutions with ui = us+i and vi = vs+i , for alli , has order of growth X s/2.

So this shows that “on average”, the solutions are diagonal. This is not aconclusion that follows from the Main Conjecture in Vinogradov’s meanvalue theorem (by any method known to me!).

2. Classical results (Bourgain, 1993)

The Main restriction Conjecture holds for k = 2, and in particular:

∮ ∣∣∣ ∑1≤n≤X

ane(n2α + nβ)∣∣∣2s dα dβ �

∑n≤X|an|2

(s < 3),

∮ ∣∣∣ ∑1≤n≤X

ane(n2α + nβ)∣∣∣6 dα dβ � X ε

∑n≤X|an|2

∮ ∣∣∣ ∑1≤n≤X

ane(n2α + nβ)∣∣∣2s dα dβ � X s−3

∑n≤X|an|2

(s > 3).

Sketch proof for the case k = 2 and s = 3:

By orthogonality, the integral∮ ∣∣∣ ∑1≤n≤X

ane(n2α + nβ)∣∣∣6 dα dβ

counts the number of solutions of the simultaneous equations

n21 + n2

2 + n23 = n2

4 + n25 + n2

n1 + n2 + n3 = n4 + n5 + n6

with each solution counted with weight

an1an2an3an4an5an6 .

Sketch proof for the case k = 2 and s = 3:

By orthogonality, the integral∮ ∣∣∣ ∑1≤n≤X

ane(n2α + nβ)∣∣∣6 dα dβ

counts the number of solutions of the simultaneous equations

n21 + n2

2 − n23 = n2

4 + n25 − n2

n1 + n2 − n3 = n4 + n5 − n6

with each solution counted with weight

an1an2an3an4an5an6 .

Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 10 /

Let B(h) denote the set of integral solutions of the equation

n21 + n2

2 − n23 = h2

n1 + n2 − n3 = h1

with 1 ≤ ni ≤ X .

Then by Cauchy’s inequality,∮ ∣∣∣ ∑1≤n≤X

ane(n2α + nβ)∣∣∣6 dα dβ =

∑|hi |≤2X i (i=1,2)

( ∑(n1,n2,n3)∈B(h)

an1an2an3

≤∑

∑n1,n2,n3

|B(h)||an1an2an3 |2.

But |B(h)| is bounded above by the number of solutions of

h21 − h2 = (n1 + n2 − n3)2 − (n2

1 + n22 − n2

= 2(n1 − n3)(n2 − n3),

and this is O(X ε) unless n1 = n3 or n2 = n3.

n21 + n2

2 − n23 = h2

n1 + n2 − n3 = h1

with 1 ≤ ni ≤ X .Then by Cauchy’s inequality,∮ ∣∣∣ ∑

1≤n≤Xane(n2α + nβ)

∣∣∣6 dα dβ =∑

|hi |≤2X i (i=1,2)

( ∑(n1,n2,n3)∈B(h)

an1an2an3

≤∑

∑n1,n2,n3

h21 − h2 = (n1 + n2 − n3)2 − (n2

1 + n22 − n2

= 2(n1 − n3)(n2 − n3),

and this is O(X ε) unless n1 = n3 or n2 = n3.

n21 + n2

2 − n23 = h2

n1 + n2 − n3 = h1

with 1 ≤ ni ≤ X .Then by Cauchy’s inequality,∮ ∣∣∣ ∑

∣∣∣6 dα dβ =∑

|hi |≤2X i (i=1,2)

( ∑(n1,n2,n3)∈B(h)

an1an2an3

≤∑

∑n1,n2,n3

h21 − h2 = (n1 + n2 − n3)2 − (n2

1 + n22 − n2

= 2(n1 − n3)(n2 − n3),

and this is O(X ε) unless n1 = n3 or n2 = n3.Trevor D. Wooley (University of Bristol) Efficient congruencing

Oxford, 29th September 2014 11 /34

One should remove the special solutions with n1 = n3 or n2 = n3 inadvance, and for the remaining solutions one finds that∮ ∣∣∣ ∑

∣∣∣6 dα dβ � X ε∑

n1,n2,n3

|an1an2an3 |2

� X ε(∑

|an|2)3.

Key observation: With B(h) the set of integral solutions of the equation

n21 + n2

2 − n23 = h2

n1 + n2 − n3 = h1

with 1 ≤ ni ≤ X , one has |B(h)| � X ε (Very strong control of thenumber of solutions of the associated Diophantine system).

One should remove the special solutions with n1 = n3 or n2 = n3 inadvance, and for the remaining solutions one finds that∮ ∣∣∣ ∑

∣∣∣6 dα dβ � X ε∑

n1,n2,n3

|an1an2an3 |2

� X ε(∑

|an|2)3.

Key observation: With B(h) the set of integral solutions of the equation

n21 + n2

2 − n23 = h2

n1 + n2 − n3 = h1

with 1 ≤ ni ≤ X , one has |B(h)| � X ε (Very strong control of thenumber of solutions of the associated Diophantine system).

Now let Bs,k(h) denote the set of integral solutions of the system

s∑i=1

x ji = hj (1 ≤ j ≤ k),

with 1 ≤ xi ≤ X . Then we have

|Bs,k(h)| � 1 (1 ≤ s ≤ k),

and (using estimates from Vinogradov’s mean value theorem)

|Bs,k(h)| � X s−12k(k+1),

for s > 2k(k − 1) (uses W., 2014).

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

Theorem (Bourgain, 1993; K. Hughes, 2012)

For each ε > 0, one has MRC in the shape

∮|fk,a(α;X )|2s dα� X ε(1 + X s−1

2k(k+1))

∑n≤X|an|2

whenever:(a) k = 2, or(b) s ≤ k + 1, or(c) s ≥ 2k(k − 1).Moroever, the factor X ε may be removed when s > 2k(k − 1).

The result (c) and its sequel depends on the latest “efficientcongruencing” results in Vinogradov’s mean value theorem (W., 2014).

fk,a(α;X ) =∑

1≤n≤Xane(nα1 + . . .+ nkαk).

∮|fk,a(α;X )|2s dα� X ε(1 + X s−1

2k(k+1))

∑n≤X|an|2

The result (c) and its sequel depends on the latest “efficientcongruencing” results in Vinogradov’s mean value theorem (W., 2014).

∮|fk,a(α;X )|2s dα� X ε(1 + X s−1

2k(k+1))

∑n≤X|an|2

Very recently: Bourgain and Demeter, 2014: The above (MRC) conclusionholds for s ≤ 2k − 1 in place of s ≤ k + 1.

∮|fk,a(α;X )|2s dα� X ε(1 + X s−1

2k(k+1))

∑n≤X|an|2

Very recently: Bourgain and Demeter, 2014: The above (MRC) conclusionholds for s ≤ 2k − 1 in place of s ≤ k + 1.

3. Efficient congruencing

Recent techniques applied in the context of Vinogradov’s mean valuetheorem allow one to establish:

Theorem (W. 2014)

∮|fk,a(α;X )|2s dα� X ε(1 + X s−1

2k(k+1))

∑n≤X|an|2

whenever:(a) k = 2, 3 (cf. classical k = 2), or

(b) 1 ≤ s ≤ D(k), where D(4) = 8, D(5) = 10, D(6) = 17, ... , andD(k) = 1

2k(k + 1)− 13k + O(k2/3) (cf. classical D(k) = k + 1), or

(c) s ≥ k(k − 1) (cf. classical s ≥ 2k(k − 1)).

Moroever, the factor X ε may be removed when s > k(k − 1).

We now aim to sketch the ideas underlying a slightly simpler result:

Theorem

∮|fk,a(α;X )|2s dα� X ε(1 + X s−1

2k(k+1))

∑n≤X|an|2

whenever s ≥ k(k + 1).

It is worth noting that we tackle the mean value directly, rather than usingresults about Vinogradov’s mean value theorem (the special case(an) = (1)) indirectly.

Consider an auxiliary prime number p (for now, think of p as being a verysmall power of X ).

ρc(ξ) = ρc(ξ; a) =

( ∑1≤n≤X

n≡ξ (mod pc )

|an|2)1/2

and then define

fa(α;X ) = ρ0(1)−1∑

1≤n≤Xane(nα1 + . . .+ nkαk).

[Note: if an = 0 for all n, then define fa = 0.]

We investigate

Us,k(X ; a) =

∮|fa(α;X )|2s dα.

Consider an auxiliary prime number p (for now, think of p as being a verysmall power of X ).

ρc(ξ) = ρc(ξ; a) =

( ∑1≤n≤X

n≡ξ (mod pc )

|an|2)1/2

and then define

fa(α;X ) = ρ0(1)−1∑

1≤n≤Xane(nα1 + . . .+ nkαk).

[Note: if an = 0 for all n, then define fa = 0.]

We investigate

Us,k(X ; a) =

∮|fa(α;X )|2s dα.

Observe that by Cauchy’s inequality, one has

|fa(α;X )| =∣∣∣ ∑1≤n≤X

ane(nα1 + . . .+ nkαk)∣∣∣

≤ X 1/2(∑n≤X|an|2

whence|fa(α;X )| ≤ X 1/2.

Us,k(X ; a) =

∮|fa(α;X )|2s dα� X s .

Moreover, one has that Us,k(X ; a) is scale-invariant, by which we meanthat it is invariant on scaling (an) to (γan) for any γ > 0.

|fa(α;X )| =∣∣∣ ∑1≤n≤X

ane(nα1 + . . .+ nkαk)∣∣∣

≤ X 1/2(∑n≤X|an|2

Us,k(X ; a) =

∮|fa(α;X )|2s dα� X s .

|fa(α;X )| =∣∣∣ ∑1≤n≤X

ane(nα1 + . . .+ nkαk)∣∣∣

≤ X 1/2(∑n≤X|an|2

Us,k(X ; a) =

∮|fa(α;X )|2s dα� X s .

Define

λs = lim supX→∞

sup(an)∈C[X ]

|an|≤1

logUs,k(X ; a)

Then there exists a sequence (Xm)∞m=1 with limm→∞ Xm = +∞ such that,for some sequence (an) ∈ C[Xm] with |an| ≤ 1, one has that for each ε > 0,

Us,k(Xm; a)� Xλs−ε,

whilst whenever 1 ≤ Y ≤ X1/2m , and for all sequences (an), at the same

time one hasUs,k(Y ; a)� Y λs+ε.

We now fix such a value X = Xm sufficiently large, and put

Λ = λs+k − (s + k − 12k(k + 1)).

Aim: Prove that Λ ≤ 0 for s ≥ k2.

Define

sup(an)∈C[X ]

|an|≤1

logUs,k(X ; a)

Λ = λs+k − (s + k − 12k(k + 1)).

Define

sup(an)∈C[X ]

|an|≤1

logUs,k(X ; a)

Λ = λs+k − (s + k − 12k(k + 1)).

Define

sup(an)∈C[X ]

|an|≤1

logUs,k(X ; a)

Λ = λs+k − (s + k − 12k(k + 1)).

This implies that

Us+k(X ; a)� X s+k−12k(k+1)+ε,

for s + k ≥ k(k + 1), thereby confirming MRC under the same conditionon s.

Approach this problem through an auxiliary mean value. Define

fc(α; ξ) = ρc(ξ)−1∑

1≤n≤Xn≡ξ (mod pc )

ane(nα1 + . . .+ nkαk),

and then put

Ka,b(X ) = ρ0(1)−4pa∑ξ=1

pb∑η=1

ρa(ξ)2ρb(η)2

∮|fa(α; ξ)2k fb(α; η)2s | dα.

Ka,b(X ) = ρ0(1)−4pa∑ξ=1

pb∑η=1

ρa(ξ)2ρb(η)2

∮|fa(α; ξ)2k fb(α; η)2s | dα.

One “expects” that

Ka,b(X )� X ε(X/pa)k−12k(k+1)(X/pb)s ,

and motivated by this observation, we define

[[Ka,b(X )]] =Ka,b(X )

(X/pa)k−12k(k+1)(X/pb)s

[[Ka,b(X )]] =Ka,b(X )

Strategy:

(i) Show that if

Us+k,k(X ; a)� X s+k−12k(k+1)+Λ,

then[[K0,1(X )]]� XΛ.

(ii) Show that whenever

[[Ka,b(X )]]� XΛ(pψ)Λ,

then there is a small non-negative integer h with the property that

[[Ka′,b′(X )]]� XΛ(pψ′)Λ,

whereψ′ = (s/k)ψ + (s/k − 1)b, a′ = b, b′ = kb + h.

[[Ka,b(X )]] =Ka,b(X )

Strategy:

(i) Show that if

Us+k,k(X ; a)� X s+k−12k(k+1)+Λ,

then[[K0,1(X )]]� XΛ.

[[Ka′,b′(X )]]� XΛ(pψ′)Λ,

By iterating this process, we obtain sequences (a(n)), (b(n)), (ψ(n)) with

b(n) ≈ kn and ψ(n) ≈ nkn

for which[[Ka(n),b(n)(X )]]� XΛ(pψ

(n))Λ.

Suppose that Λ > 0. Then the right hand side here increases so rapidlythat, for large enough values of n, it is larger than the trivial estimate forthe left hand side. This gives a contradiction, so that Λ ≤ 0.

4. Translation invariance, and the congruencing idea

Observe that the system of equations

s∑i=1

(x ji − y ji ) = 0 (1 ≤ j ≤ k) (1)

has a solution x, y if and only if, for any integral shift a, the system ofequations

s∑i=1

((xi − a)j − (yi − a)j) = 0 (1 ≤ j ≤ k)

is also satisfied

To see this, note that

j∑l=1

)aj−l

s∑i=1

((xi − a)j − (yi − a)j) =s∑

((xi − a + a)j − (yi − a + a)j).

4. Translation invariance, and the congruencing idea

Observe that the system of equations

s∑i=1

(x ji − y ji ) = 0 (1 ≤ j ≤ k) (1)

has a solution x, y if and only if, for any integral shift a, the system ofequations

s∑i=1

((xi − a)j − (yi − a)j) = 0 (1 ≤ j ≤ k)

is also satisfied

To see this, note that

j∑l=1

)aj−l

s∑i=1

((xi − a)j − (yi − a)j) =s∑

((xi − a + a)j − (yi − a + a)j).

The mean value ∮|fa(α; ξ)2k fb(α; η)2s |dα

counts (with weights) the number of integral solutions of the system

k∑i=1

(x ji − y ji ) =s∑

((pbul + η)j − (pbvl + η)j) (1 ≤ j ≤ k),

with 1 ≤ x, y ≤ X and (1− η)/pb ≤ u, v ≤ (X − η)/pb.

By translation invariance (Binomial Theorem), this system is equivalent to

k∑i=1

((xi − η)j − (yi − η)j) = pjbs∑

(ujl − v jl ) (1 ≤ j ≤ k),

whencek∑

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

In this way, we obtain a system of congruence conditions modulo pjb for1 ≤ j ≤ k .

k∑i=1

(x ji − y ji ) =s∑

((pbul + η)j − (pbvl + η)j) (1 ≤ j ≤ k),

with 1 ≤ x, y ≤ X and (1− η)/pb ≤ u, v ≤ (X − η)/pb.By translation invariance (Binomial Theorem), this system is equivalent to

k∑i=1

((xi − η)j − (yi − η)j) = pjbs∑

(ujl − v jl ) (1 ≤ j ≤ k),

whencek∑

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

k∑i=1

(x ji − y ji ) =s∑

((pbul + η)j − (pbvl + η)j) (1 ≤ j ≤ k),

k∑i=1

((xi − η)j − (yi − η)j) = pjbs∑

(ujl − v jl ) (1 ≤ j ≤ k),

whencek∑

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

k∑i=1

(x ji − y ji ) =s∑

((pbul + η)j − (pbvl + η)j) (1 ≤ j ≤ k),

k∑i=1

((xi − η)j − (yi − η)j) = pjbs∑

(ujl − v jl ) (1 ≤ j ≤ k),

whencek∑

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

k∑i=1

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

Suppose that x is well-conditioned, by which we mean that x1, . . . , xk lie indistinct congruence classes modulo p. Then, given an integral k-tuple n,the solutions of the system

k∑i=1

(xi − η)j ≡ nj (mod p) (1 ≤ j ≤ k),

with 1 ≤ x ≤ p, may be lifted uniquely to solutions of the system

k∑i=1

(xi − η)j ≡ nj (mod pkb) (1 ≤ j ≤ k),

with 1 ≤ x ≤ pkb.

In this way, the initial congruences essentially imply that

x ≡ y (mod pkb),

provided that we inflate our estimates by k!p12k(k−1)b.

k∑i=1

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

k∑i=1

(xi − η)j ≡ nj (mod p) (1 ≤ j ≤ k),

k∑i=1

(xi − η)j ≡ nj (mod pkb) (1 ≤ j ≤ k),

x ≡ y (mod pkb),

k∑i=1

(xi − η)j ≡k∑

(yi − η)j (mod pjb) (1 ≤ j ≤ k).

k∑i=1

(xi − η)j ≡ nj (mod p) (1 ≤ j ≤ k),

k∑i=1

(xi − η)j ≡ nj (mod pkb) (1 ≤ j ≤ k),

x ≡ y (mod pkb),

x ≡ y (mod pkb)

Now we are counting solutions with weights, so we reinsert this congruenceinformation back into the mean value Ka,b(X ) to obtain the relation

Ka,b(X )� p12k(k−1)(a+b)ρ0(1)−4

pa∑ξ=1

pb∑η=1

ρa(ξ)2ρb(η)2Ξ,

∮ ∑1≤ξ′≤pkb

ξ′≡ξ (mod pa)

ρkb(ξ′)2

ρa(ξ)2|fkb(α; ξ′)|2

|fb(α; η)|2s dα.

x ≡ y (mod pkb)

Now we are counting solutions with weights, so we reinsert this congruenceinformation back into the mean value Ka,b(X ) to obtain the relation

Ka,b(X )� p12k(k−1)(a+b)ρ0(1)−4

pa∑ξ=1

pb∑η=1

ρa(ξ)2ρb(η)2Ξ,

∮ ∑1≤ξ′≤pkb

ξ′≡ξ (mod pa)

ρkb(ξ′)2

|fb(α; η)|2s dα.

∮ ∑1≤ξ′≤pkb

ξ′≡ξ (mod pa)

ρkb(ξ′)2

|fb(α; η)|2s dα.

But by Holder’s inequality, the term here raised to power k is boundedabove by

ρa(ξ)−2k

( ∑1≤ξ′≤pkb

ξ′≡ξ (mod pa)

ρkb(ξ′)2|fkb(α; ξ′)|2s)k/s( ∑

1≤ξ′≤pkbξ′≡ξ (mod pa)

ρkb(ξ′)2

)k−k/s

ρa(ξ)−2∑

ρkb(ξ′)2|fkb(α; ξ′)|2s

∮ ∑1≤ξ′≤pkb

ξ′≡ξ (mod pa)

ρkb(ξ′)2

|fb(α; η)|2s dα.

But by Holder’s inequality, the term here raised to power k is boundedabove by

ρa(ξ)−2k

( ∑1≤ξ′≤pkb

ξ′≡ξ (mod pa)

ρkb(ξ′)2|fkb(α; ξ′)|2s)k/s( ∑

ρkb(ξ′)2

)k−k/s

ρa(ξ)−2∑

ρkb(ξ′)2|fkb(α; ξ′)|2s

Then another application of Holder’s inequality yields

Ξ�∮ ρa(ξ)−2

∑ξ′

ρkb(ξ′)2|fkb(α; ξ′)|2sk/s

|fb(α; η)|2s dα

� Ξk/s1 Ξ

1−k/s2 ,

Ξ1 = ρa(ξ)−2∑ξ′

ρkb(ξ′)2

∮|fb(α; η)2k fkb(α; ξ′)2s | dα

∮|fb(α; η)|2s+2k dα.

Recall that

Ka,b(X )� p12k(k−1)(a+b)ρ0(1)−4

pa∑ξ=1

pb∑η=1

ρa(ξ)2ρb(η)2Ξ,

From here, yet another application of Holder’s inequality gives

Ka,b(X )� p12k(k−1)(a+b)Ξ

k/s3 Ξ

1−k/s4 ,

Ξ3 = ρ0(1)−4pb∑η=1

pkb∑ξ′=1

ρb(η)2ρkb(ξ′)2

∮|fb(α; η)2k fkb(α; ξ′)2s |dα,

Ξ4 = ρ0(1)−4pb∑η=1

pa∑ξ=1

ρb(η)2ρa(ξ)2

∮|fb(α; η)|2s+2k dα

� (X/Mb)s+k−12k(k+1)+Λ+ε.

Then one can check that

[[Ka,b(X )]]� [[Kb,kb(X )]]k/s(X/Mb)(1−k/s)(Λ+ε).

Given the hypothesis that

this implies that[[Kb,kb(X )]]� XΛ(pψ

′)Λ,

whereψ′ = (s/k)ψ + (s/k − 1)b,

which is a little stronger than we had claimed earlier.

5. Further restriction ideas

Parsell, Prendiville and W., 2013 consider general translation invariantsystems (cf. Arkhipov, Karatsuba and Chubarikov, 1980, 2000’s). Forexample, consider the number J(X ) of solutions of the system

s∑i=1

x ji ymi =

2s∑i=s+1

x ji ymi (0 ≤ j ≤ 3, 0 ≤ m ≤ 2),

with 1 ≤ x, y ≤ X .

The number of equations is r = (3 + 1)(2 + 1)− 1 = 11, the largest totaldegree is k = 3 + 2 = 5, the sum of degrees is

K = 12 3(3 + 1) · 1

2 2(2 + 1) = 18,

and the number of variables in a block is 2.

(General) theorem shows that whenever s > r(k + 1), thenJ(X )� X 2sd−K . Can develop a restriction variant of this work.

Most recent work: the “efficient congruencing” methods apply also tosystems that are only approximately translation-invariant. Consider, forexample, integers 1 ≤ k1 < k2 < . . . < kt , and the number T (X ) ofsolutions of the system

s∑i=1

(xkji − y

kji ) = 0 (1 ≤ j ≤ t),

with 1 ≤ x, y ≤ X . Then (W. 2014) one has

T (X )� X s+ε,

whenever 1 ≤ s ≤ 12 t(t + 1)− ( 1

3 + o(1))t (t large).

Again, one can develop a restriction variant of these ideas.

(cf. classical s ≤ t + 1; and Bourgain and Bourgain-Demeter, 2014).

Vinogradov's mean value theorem and its associated restriction theory via … · 2014. 10. 8. ·...

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