Vinogradov’s mean value theorem and its associated
restriction 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
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1. Introduction
Let k ≥ 2 be an integer, and consider
g : Tk → C
(T = R/Z ' [0, 1)),
with an associated Fourier series
X
gb(n1 , . . . , nk )e(n1 α1 + . . . + nk αk ),
ge(α1 , . . . , αk ) =
n∈Zk
in which gb(n) ∈ C and e(z) = e 2πiz .
Trevor D. Wooley (University of Bristol)
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1. Introduction
Let k ≥ 2 be an integer, and consider
g : Tk → C
(T = R/Z ' [0, 1)),
with an associated Fourier series
X
gb(n1 , . . . , nk )e(n1 α1 + . . . + nk αk ),
ge(α1 , . . . , αk ) =
n∈Zk
in which gb(n) ∈ C and e(z) = e 2πiz .
Restriction operator: (E. Stein, J. Bourgain, K. Hughes, et al.)
X
Rg :=
gb(n)e(n · α).
n∈Zk
n=(n,n2 ,...,nk )
[This is just one example of a restriction operator!]
Trevor D. Wooley (University of Bristol)
Efficient congruencing
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1. Introduction
Let k ≥ 2 be an integer, and consider
g : Tk → C
(T = R/Z ' [0, 1)),
with an associated Fourier series
X
gb(n1 , . . . , nk )e(n1 α1 + . . . + nk αk ),
ge(α1 , . . . , αk ) =
n∈Zk
in which gb(n) ∈ C and e(z) = e 2πiz .
Restriction operator: (E. Stein, J. Bourgain, K. Hughes, et al.)
X
Rg :=
gb(n)e(n · α).
n∈Zk
n=(n,n2 ,...,nk )
[This is just one example of a restriction operator!]
We are interested in the norm of the operator g 7→ Rg .
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(Slightly) more concretely (for analytic number theorists):
Consider a sequence (an )∞
n=1 of complex numbers, not all zero, and define
the exponential sum fa = fk,a (α; X ) by putting
X
fk,a (α; X ) =
an e(nα1 + . . . + nk αk ).
1≤n≤X
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(Slightly) more concretely (for analytic number theorists):
Consider a sequence (an )∞
n=1 of complex numbers, not all zero, and define
the exponential sum fa = fk,a (α; X ) by putting
X
fk,a (α; X ) =
an e(nα1 + . . . + nk αk ).
1≤n≤X
Aim: Obtain a bound for
sup (kfa kLp /kak`2 )
a
in terms of p, k and X .
Trevor D. Wooley (University of Bristol)
Efficient congruencing
Oxford, 29th September 2014
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(Slightly) more concretely (for analytic number theorists):
Consider a sequence (an )∞
n=1 of complex numbers, not all zero, and define
the exponential sum fa = fk,a (α; X ) by putting
X
fk,a (α; X ) =
an e(nα1 + . . . + nk αk ).
1≤n≤X
Aim: Obtain a bound for
sup (kfa kLp /kak`2 )
a
in terms of p, k and X .
Conjecture (Main Restriction Conjecture)
For each ε > 0, one has

X ε ,
when 0 < p ≤ k(k + 1),
kfa kLp
ε,p,k
1 k(k+1)
X 2 − 2p , when p > k(k + 1).
kak`2
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(Even) more concretely (for analytic number theorists):
Consider a sequence (an )∞
n=1 of complex numbers, not all zero, and define
the exponential sum fa = fk,a (α; X ) by putting
fk,a (α; X ) =
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
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(Even) more concretely (for analytic number theorists):
Consider a sequence (an )∞
n=1 of complex numbers, not all zero, and define
the exponential sum fa = fk,a (α; X ) by putting
fk,a (α; X ) =
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
Conjecture (Main Restriction Conjecture)
For each ε > 0, one has

!s

P

ε

|an |2 ,
when s ≤ 21 k(k + 1),

I
X
n≤X
!s
|fk,a (α; X )|2s dα 
1
P

s− 2 k(k+1)
2

|an |
, when s > 21 k(k + 1).

X
n≤X
Here, we write
H
for
R
[0,1)k .
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Some observations, I:
fk,a (α; X ) =
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
Conjecture (Main Restriction Conjecture)
I
|fk,a (α; X )|2s dα Trevor D. Wooley (University of Bristol)



ε


X
!s
P
|an
|2
when s ≤ 21 k(k + 1),
,
n≤X

1

s− k(k+1)


X 2
!s
P
|an |2
n≤X
Efficient congruencing
, when s > 21 k(k + 1).
Oxford, 29th September 2014
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Some observations, I:
fk,a (α; X ) =
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
Conjecture (Main Restriction Conjecture)
I
|fk,a (α; X )|2s dα 


ε


X
!s
P
|an
|2
when s ≤ 21 k(k + 1),
,
n≤X

1

s− k(k+1)


X 2
!s
P
|an |2
n≤X
, when s > 21 k(k + 1).
Consider the sequence (an ) = 1. Then MRC implies that
I
1
|fk,1 (α; X )|2s dα X ε (X s + X 2s− 2 k(k+1) ),
an assertion equivalent to the Main Conjecture in Vinogradov’s Mean
Value Theorem.
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Some observations, II:
Consider the situation in which (an ) is supported on a thin sequence, say
an = card (x, y ) ∈ Z2 : n = x 4 + y 4 .
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Some observations, II:
Consider the situation in which (an ) is supported on a thin sequence, say
an = card (x, y ) ∈ Z2 : n = x 4 + y 4 .
Then MRC implies that for 1 ≤ s ≤ 12 k(k + 1), one should have

s
I
X
|fk,a (α; X )|2s dα X ε 
|an |2 
n≤X
X ε X 1/2
Trevor D. Wooley (University of Bristol)
Efficient congruencing
s
= X s/2+ε .
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Some observations, II:
Consider the situation in which (an ) is supported on a thin sequence, say
an = card (x, y ) ∈ Z2 : n = x 4 + y 4 .
Then MRC implies that for 1 ≤ s ≤ 12 k(k + 1), one should have

s
I
X
|fk,a (α; X )|2s dα X ε 
|an |2 
n≤X
X ε X 1/2
s
= X s/2+ε .
But by orthogonality, when s is a positive integer, this integral counts the
number of solutions of the system of equations
s
X
4
4
(ui4 + vi4 )j − (us+i
+ vs+i
)j = 0
(1 ≤ j ≤ k),
i=1
with 1 ≤ ui4 + vi4 ≤ X (1 ≤ i ≤ 2s).
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Some observations, II:
So the number N(X ) of integral solutions of the system of equations
s
X
4
4
(ui4 + vi4 )j − (us+i
+ vs+i
)j = 0
(1 ≤ j ≤ k),
i=1
with 1 ≤ ui4 + vi4 ≤ X (1 ≤ i ≤ 2s), satisfies
N(X ) X s/2+ε .
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Some observations, II:
So the number N(X ) of integral solutions of the system of equations
s
X
4
4
(ui4 + vi4 )j − (us+i
+ vs+i
)j = 0
(1 ≤ j ≤ k),
i=1
with 1 ≤ ui4 + vi4 ≤ 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 all
i, has order of growth X s/2 .
So this shows that “on average”, the solutions are diagonal. This is not a
conclusion that follows from the Main Conjecture in Vinogradov’s mean
value theorem (by any method known to me!).
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2. Classical results (Bourgain, 1993)
The Main restriction Conjecture holds for k = 2, and in particular:

s
I X
2s
X
|an |2 
(s < 3),
an e(n2 α + nβ) dα dβ 
1≤n≤X
n≤X

3
I X
6
X
an e(n2 α + nβ) dα dβ X ε 
|an |2  ,
1≤n≤X
n≤X

s
I X
2s
X
an e(n2 α + nβ) dα dβ X s−3 
|an |2 
1≤n≤X
Trevor D. Wooley (University of Bristol)
(s > 3).
n≤X
Efficient congruencing
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Sketch proof for the case k = 2 and s = 3:
By orthogonality, the integral
I X
6
an e(n2 α + nβ) dα dβ
1≤n≤X
counts the number of solutions of the simultaneous equations
)
n12 + n22 + n32 = n42 + n52 + n62
,
n1 + n2 + n3 = n4 + n5 + n6
with each solution counted with weight
an1 an2 an3 an4 an5 an6 .
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Sketch proof for the case k = 2 and s = 3:
By orthogonality, the integral
I X
6
an e(n2 α + nβ) dα dβ
1≤n≤X
counts the number of solutions of the simultaneous equations
)
n12 + n22 − n32 = n42 + n52 − n62
,
n1 + n2 − n3 = n4 + n5 − n6
with each solution counted with weight
an1 an2 an3 an4 an5 an6 .
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34
Let B(h) denote the set of integral solutions of the equation
)
n12 + n22 − n32 = h2
,
n1 + n2 − n3 = h1
with 1 ≤ ni ≤ X .
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34
Let B(h) denote the set of integral solutions of the equation
)
n12 + n22 − n32 = h2
,
n1 + n2 − n3 = h1
with 1 ≤ ni ≤ X .
Then by Cauchy’s inequality,
I X
6
an e(n2 α + nβ) dα dβ =
X
X
an1 an2 an3
2
|hi |≤2X i (i=1,2) (n1 ,n2 ,n3 )∈B(h)
1≤n≤X
≤
X X
|B(h)||an1 an2 an3 |2 .
h n1 ,n2 ,n3
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34
Let B(h) denote the set of integral solutions of the equation
)
n12 + n22 − n32 = h2
,
n1 + n2 − n3 = h1
with 1 ≤ ni ≤ X .
Then by Cauchy’s inequality,
I X
6
an e(n2 α + nβ) dα dβ =
X
X
an1 an2 an3
2
|hi |≤2X i (i=1,2) (n1 ,n2 ,n3 )∈B(h)
1≤n≤X
≤
X X
|B(h)||an1 an2 an3 |2 .
h n1 ,n2 ,n3
But |B(h)| is bounded above by the number of solutions of
h12 − h2 = (n1 + n2 − n3 )2 − (n12 + n22 − n32 )
= 2(n1 − n3 )(n2 − n3 ),
and this is O(X ε ) unless n1 = n3 or n2 = n3 .
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34
One should remove the special solutions with n1 = n3 or n2 = n3 in
advance, and for the remaining solutions one finds that
I X
6
X
2
an e(n α + nβ) dα dβ X ε
|an1 an2 an3 |2
n1 ,n2 ,n3
1≤n≤X
Xε
X
|an |2
3
.
n
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One should remove the special solutions with n1 = n3 or n2 = n3 in
advance, and for the remaining solutions one finds that
I X
6
X
2
an e(n α + nβ) dα dβ X ε
|an1 an2 an3 |2
n1 ,n2 ,n3
1≤n≤X
Xε
X
|an |2
3
.
n
Key observation: With B(h) the set of integral solutions of the equation
)
n12 + n22 − n32 = h2
,
n1 + n2 − n3 = h1
with 1 ≤ ni ≤ X , one has |B(h)| X ε (Very strong control of the
number of solutions of the associated Diophantine system).
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Now let Bs,k (h) denote the set of integral solutions of the system
s
X
xij = hj
(1 ≤ j ≤ k),
i=1
with 1 ≤ xi ≤ X . Then we have
|Bs,k (h)| 1
(1 ≤ s ≤ k),
and (using estimates from Vinogradov’s mean value theorem)
1
|Bs,k (h)| X s− 2 k(k+1) ,
for s > 2k(k − 1) (uses W., 2014).
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34
fk,a (α; X ) =
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
Theorem (Bourgain, 1993; K. Hughes, 2012)
For each ε > 0, one has MRC in the shape
s

I
1
|fk,a (α; X )|2s dα X ε (1 + X s− 2 k(k+1) ) 
X
|an |2 
n≤X
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).
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fk,a (α; X ) =
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
Theorem (Bourgain, 1993; K. Hughes, 2012)
For each ε > 0, one has MRC in the shape
s

I
1
|fk,a (α; X )|2s dα X ε (1 + X s− 2 k(k+1) ) 
X
|an |2 
n≤X
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 “efficient
congruencing” results in Vinogradov’s mean value theorem (W., 2014).
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Theorem (Bourgain, 1993; K. Hughes, 2012)
For each ε > 0, one has MRC in the shape
s

I
|fk,a (α; X )|2s dα X ε (1 +
X
1
X s− 2 k(k+1) ) 
|an |2 
n≤X
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).
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Theorem (Bourgain, 1993; K. Hughes, 2012)
For each ε > 0, one has MRC in the shape
s

I
|fk,a (α; X )|2s dα X ε (1 +
X
1
X s− 2 k(k+1) ) 
|an |2 
n≤X
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).
Very recently: Bourgain and Demeter, 2014: The above (MRC) conclusion
holds for s ≤ 2k − 1 in place of s ≤ k + 1.
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3. Efficient congruencing
Recent techniques applied in the context of Vinogradov’s mean value
theorem allow one to establish:
Theorem (W. 2014)
For each ε > 0, one has MRC in the shape
s

I
|fk,a (α; X )|2s dα X ε (1 +
X
1
X s− 2 k(k+1) ) 
|an |2 
n≤X
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, ... , and
D(k) = 21 k(k + 1) − 13 k + O(k 2/3 ) (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).
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We now aim to sketch the ideas underlying a slightly simpler result:
Theorem
For each ε > 0, one has MRC in the shape

I
|fk,a (α; X )|2s dα X ε (1 +
X
1
X s− 2 k(k+1) ) 
s
|an |2 
n≤X
whenever s ≥ k(k + 1).
It is worth noting that we tackle the mean value directly, rather than using
results about Vinogradov’s mean value theorem (the special case
(an ) = (1)) indirectly.
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Consider an auxiliary prime number p (for now, think of p as being a very
small power of X ).
Write
X
ρc (ξ) = ρc (ξ; a) =
2
|an |
1/2
,
1≤n≤X
n≡ξ (mod p c )
and then define
e
fa (α; X ) = ρ0 (1)−1
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
[Note: if an = 0 for all n, then define e
fa = 0.]
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Consider an auxiliary prime number p (for now, think of p as being a very
small power of X ).
Write
X
ρc (ξ) = ρc (ξ; a) =
2
|an |
1/2
,
1≤n≤X
n≡ξ (mod p c )
and then define
e
fa (α; X ) = ρ0 (1)−1
X
an e(nα1 + . . . + nk αk ).
1≤n≤X
[Note: if an = 0 for all n, then define e
fa = 0.]
We investigate
I
Us,k (X ; a) =
|e
fa (α; X )|2s dα.
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Observe that by Cauchy’s inequality, one has
X
|fa (α; X )| = an e(nα1 + . . . + nk αk )
1≤n≤X
≤ X 1/2
X
|an |2
1/2
,
n≤X
whence
|e
fa (α; X )| ≤ X 1/2 .
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Observe that by Cauchy’s inequality, one has
X
|fa (α; X )| = an e(nα1 + . . . + nk αk )
1≤n≤X
≤ X 1/2
X
|an |2
1/2
,
n≤X
whence
|e
fa (α; X )| ≤ X 1/2 .
Thus
I
Us,k (X ; a) =
|e
fa (α; X )|2s dα X s .
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Observe that by Cauchy’s inequality, one has
X
|fa (α; X )| = an e(nα1 + . . . + nk αk )
1≤n≤X
≤ X 1/2
X
|an |2
1/2
,
n≤X
whence
|e
fa (α; X )| ≤ X 1/2 .
Thus
I
Us,k (X ; a) =
|e
fa (α; X )|2s dα X s .
Moreover, one has that Us,k (X ; a) is scale-invariant, by which we mean
that it is invariant on scaling (an ) to (γan ) for any γ > 0.
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Define
λs = lim sup
sup
X →∞ (an )∈C[X ]
|an |≤1
log Us,k (X ; a)
.
log X
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Define
λs = lim sup
sup
X →∞ (an )∈C[X ]
|an |≤1
log Us,k (X ; a)
.
log X
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) X λs −ε ,
1/2
whilst whenever 1 ≤ Y ≤ Xm , and for all sequences (an ), at the same
time one has
Us,k (Y ; a) Y λs +ε .
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Define
λs = lim sup
sup
X →∞ (an )∈C[X ]
|an |≤1
log Us,k (X ; a)
.
log X
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) X λs −ε ,
1/2
whilst whenever 1 ≤ Y ≤ Xm , and for all sequences (an ), at the same
time one has
Us,k (Y ; a) Y λs +ε .
We now fix such a value X = Xm sufficiently large, and put
Λ = λs+k − (s + k − 12 k(k + 1)).
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Define
λs = lim sup
sup
X →∞ (an )∈C[X ]
|an |≤1
log Us,k (X ; a)
.
log X
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) X λs −ε ,
1/2
whilst whenever 1 ≤ Y ≤ Xm , and for all sequences (an ), at the same
time one has
Us,k (Y ; a) Y λs +ε .
We now fix such a value X = Xm sufficiently large, and put
Λ = λs+k − (s + k − 12 k(k + 1)).
Aim: Prove that Λ ≤ 0 for s ≥ k 2 .
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Aim: Prove that Λ ≤ 0 for s ≥ k 2 .
This implies that
1
Us+k (X ; a) X s+k− 2 k(k+1)+ε ,
for s + k ≥ k(k + 1), thereby confirming MRC under the same condition
on s.
Approach this problem through an auxiliary mean value. Define
X
fc (α; ξ) = ρc (ξ)−1
an e(nα1 + . . . + nk αk ),
1≤n≤X
n≡ξ (mod p c )
and then put
a
−4
Ka,b (X ) = ρ0 (1)
b
p X
p
X
2
ρa (ξ) ρb (η)
2
I
|fa (α; ξ)2k fb (α; η)2s | dα.
ξ=1 η=1
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a
Ka,b (X ) = ρ0 (1)
−4
b
p X
p
X
2
2
I
ρa (ξ) ρb (η)
|fa (α; ξ)2k fb (α; η)2s | dα.
ξ=1 η=1
One “expects” that
1
Ka,b (X ) X ε (X /p a )k− 2 k(k+1) (X /p b )s ,
and motivated by this observation, we define
[[Ka,b (X )]] =
Ka,b (X )
1
.
(X /p a )k− 2 k(k+1) (X /p b )s
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[[Ka,b (X )]] =
Ka,b (X )
1
.
(X /p a )k− 2 k(k+1) (X /p b )s
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[[Ka,b (X )]] =
Ka,b (X )
.
1
(X /p a )k− 2 k(k+1) (X /p b )s
Strategy:
(i) Show that if
1
Us+k,k (X ; a) X s+k− 2 k(k+1)+Λ ,
then
[[K0,1 (X )]] X Λ .
(ii) Show that whenever
[[Ka,b (X )]] X Λ (p ψ )Λ ,
then there is a small non-negative integer h with the property that
0
[[Ka0 ,b0 (X )]] X Λ (p ψ )Λ ,
where
ψ 0 = (s/k)ψ + (s/k − 1)b,
a0 = b,
b 0 = kb + h.
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(ii) Show that whenever
[[Ka,b (X )]] X Λ (p ψ )Λ ,
then there is a small non-negative integer h with the property that
0
[[Ka0 ,b0 (X )]] X Λ (p ψ )Λ ,
where
ψ 0 = (s/k)ψ + (s/k − 1)b,
a0 = b,
b 0 = kb + h.
By iterating this process, we obtain sequences (a(n) ), (b (n) ), (ψ (n) ) with
b (n) ≈ k n
and ψ (n) ≈ nk n
for which
(n)
[[Ka(n) ,b(n) (X )]] X Λ (p ψ )Λ .
Suppose that Λ > 0. Then the right hand side here increases so rapidly
that, for large enough values of n, it is larger than the trivial estimate for
the left hand side. This gives a contradiction, so that Λ ≤ 0.
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4. Translation invariance, and the congruencing idea
Observe that the system of equations
s
X
(xij − yij ) = 0
(1 ≤ j ≤ k)
(1)
i=1
has a solution x, y if and only if, for any integral shift a, the system of
equations
s
X
((xi − a)j − (yi − a)j ) = 0 (1 ≤ j ≤ k)
i=1
is also satisfied
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4. Translation invariance, and the congruencing idea
Observe that the system of equations
s
X
(xij − yij ) = 0
(1 ≤ j ≤ k)
(1)
i=1
has a solution x, y if and only if, for any integral shift a, the system of
equations
s
X
((xi − a)j − (yi − a)j ) = 0 (1 ≤ j ≤ k)
i=1
is also satisfied
To see this, note that
j s
s
X
X
j j−l X
a
((xi − a)j − (yi − a)j ) =
((xi − a + a)j − (yi − a + a)j ).
l
l=1
i=1
i=1
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The mean value
I
|fa (α; ξ)2k fb (α; η)2s | dα
counts (with weights) the number of integral solutions of the system
k
X
i=1
(xij − yij ) =
s
X
((p b ul + η)j − (p b vl + η)j )
(1 ≤ j ≤ k),
l=1
with 1 ≤ x, y ≤ X and (1 − η)/p b ≤ u, v ≤ (X − η)/p b .
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The mean value
I
|fa (α; ξ)2k fb (α; η)2s | dα
counts (with weights) the number of integral solutions of the system
k
X
(xij − yij ) =
i=1
s
X
((p b ul + η)j − (p b vl + η)j )
(1 ≤ j ≤ k),
l=1
with 1 ≤ x, y ≤ X and (1 − η)/p b ≤ u, v ≤ (X − η)/p b .
By translation invariance (Binomial Theorem), this system is equivalent to
k
s
X
X
j
j
jb
((xi − η) − (yi − η) ) = p
(ulj − vlj )
i=1
(1 ≤ j ≤ k),
l=1
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The mean value
I
|fa (α; ξ)2k fb (α; η)2s | dα
counts (with weights) the number of integral solutions of the system
k
X
(xij − yij ) =
i=1
s
X
((p b ul + η)j − (p b vl + η)j )
(1 ≤ j ≤ k),
l=1
with 1 ≤ x, y ≤ X and (1 − η)/p b ≤ u, v ≤ (X − η)/p b .
By translation invariance (Binomial Theorem), this system is equivalent to
k
s
X
X
j
j
jb
((xi − η) − (yi − η) ) = p
(ulj − vlj )
i=1
(1 ≤ j ≤ k),
l=1
whence
k
X
k
X
(xi − η) ≡
(yi − η)j
j
i=1
(mod p jb )
(1 ≤ j ≤ k).
i=1
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The mean value
I
|fa (α; ξ)2k fb (α; η)2s | dα
counts (with weights) the number of integral solutions of the system
k
X
(xij − yij ) =
i=1
s
X
((p b ul + η)j − (p b vl + η)j )
(1 ≤ j ≤ k),
l=1
with 1 ≤ x, y ≤ X and (1 − η)/p b ≤ u, v ≤ (X − η)/p b .
By translation invariance (Binomial Theorem), this system is equivalent to
k
s
X
X
j
j
jb
((xi − η) − (yi − η) ) = p
(ulj − vlj )
i=1
(1 ≤ j ≤ k),
l=1
whence
k
X
k
X
(xi − η) ≡
(yi − η)j
j
i=1
(mod p jb )
(1 ≤ j ≤ k).
i=1
In this way, we obtain a system of congruence conditions modulo p jb for
1 ≤ j ≤ k.
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k
X
(xi − η)j ≡
i=1
k
X
(yi − η)j
(mod p jb )
(1 ≤ j ≤ k).
i=1
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k
X
(xi − η)j ≡
i=1
k
X
(yi − η)j
(mod p jb )
(1 ≤ j ≤ k).
i=1
Suppose that x is well-conditioned, by which we mean that x1 , . . . , xk lie in
distinct congruence classes modulo p. Then, given an integral k-tuple n,
the solutions of the system
k
X
(xi − η)j ≡ nj
(mod p)
(1 ≤ j ≤ k),
i=1
with 1 ≤ x ≤ p, may be lifted uniquely to solutions of the system
k
X
(xi − η)j ≡ nj
(mod p kb )
(1 ≤ j ≤ k),
i=1
with 1 ≤ x ≤
p kb .
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k
X
(xi − η)j ≡
i=1
k
X
(yi − η)j
(mod p jb )
(1 ≤ j ≤ k).
i=1
Suppose that x is well-conditioned, by which we mean that x1 , . . . , xk lie in
distinct congruence classes modulo p. Then, given an integral k-tuple n,
the solutions of the system
k
X
(xi − η)j ≡ nj
(1 ≤ j ≤ k),
(mod p)
i=1
with 1 ≤ x ≤ p, may be lifted uniquely to solutions of the system
k
X
(xi − η)j ≡ nj
(mod p kb )
(1 ≤ j ≤ k),
i=1
with 1 ≤ x ≤
p kb .
In this way, the initial congruences essentially imply that
x≡y
(mod p kb ),
1
provided that we inflate our estimates by k!p 2 k(k−1)bOxford,
.
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x≡y
(mod p kb )
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x≡y
(mod p kb )
Now we are counting solutions with weights, so we reinsert this congruence
information back into the mean value Ka,b (X ) to obtain the relation
Ka,b (X ) pa X
pb
X
1
k(k−1)(a+b)
−4
p2
ρ0 (1)
ρa (ξ)2 ρb (η)2 Ξ,
ξ=1 η=1
where
k

I 

Ξ= 

X
1≤ξ 0 ≤p kb
ξ 0 ≡ξ (mod p a )

ρkb (ξ 0 )2
0 2
|f
(α;
ξ
)|
 |fb (α; η)|2s dα.
kb
ρa (ξ)2

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k

Ξ=
I 



X
1≤ξ 0 ≤p kb
ξ 0 ≡ξ (mod p a )

ρkb (ξ 0 )2
0 2
|f
(α;
ξ
)|
 |fb (α; η)|2s dα.
kb
ρa (ξ)2

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k

Ξ=
I 



X
1≤ξ 0 ≤p kb
ξ 0 ≡ξ (mod p a )

ρkb (ξ 0 )2
0 2
|f
(α;
ξ
)|
 |fb (α; η)|2s dα.
kb
ρa (ξ)2

But by Hölder’s inequality, the term here raised to power k is bounded
above by
!k/s
X
ρa (ξ)−2k
ρkb (ξ 0 )2 |fkb (α; ξ 0 )|2s
1≤ξ 0 ≤p kb
ξ 0 ≡ξ (mod p a )
!k−k/s
X
ρkb (ξ 0 )2
1≤ξ 0 ≤p kb
ξ 0 ≡ξ (mod p a )
k/s



ρa (ξ)−2

X
1≤ξ 0 ≤p kb


ρkb (ξ 0 )2 |fkb (α; ξ 0 )|2s 

.
ξ 0 ≡ξ (mod p a )
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Then another application of Hölder’s inequality yields
k/s

I
Ξ
ρa (ξ)−2
X
ρkb (ξ 0 )2 |fkb (α; ξ 0 )|2s 
|fb (α; η)|2s dα
ξ0
k/s
1−k/s
Ξ1 Ξ2
,
where
Ξ1 = ρa (ξ)
−2
X
0 2
ρkb (ξ )
I
|fb (α; η)2k fkb (α; ξ 0 )2s | dα
ξ0
and
I
Ξ2 =
|fb (α; η)|2s+2k dα.
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Recall that
pa X
pb
X
1
k(k−1)(a+b)
−4
ρ0 (1)
p2
ρa (ξ)2 ρb (η)2 Ξ,
Ka,b (X ) ξ=1 η=1
From here, yet another application of Hölder’s inequality gives
1
k/s
1−k/s
Ka,b (X ) p 2 k(k−1)(a+b) Ξ3 Ξ4
,
where
b
Ξ3 = ρ0 (1)
−4
kb
p X
p
X
ρb (η)2 ρkb (ξ 0 )2
I
|fb (α; η)2k fkb (α; ξ 0 )2s | dα,
η=1 ξ 0 =1
and
b
−4
Ξ4 = ρ0 (1)
a
p X
p
X
2
ρb (η) ρa (ξ)
2
I
|fb (α; η)|2s+2k dα
η=1 ξ=1
1
(X /M b )s+k− 2 k(k+1)+Λ+ε .
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Then one can check that
[[Ka,b (X )]] [[Kb,kb (X )]]k/s (X /M b )(1−k/s)(Λ+ε) .
Given the hypothesis that
[[Ka,b (X )]] X Λ (p ψ )Λ ,
this implies that
0
[[Kb,kb (X )]] X Λ (p ψ )Λ ,
where
ψ 0 = (s/k)ψ + (s/k − 1)b,
which is a little stronger than we had claimed earlier.
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5. Further restriction ideas
Parsell, Prendiville and W., 2013 consider general translation invariant
systems (cf. Arkhipov, Karatsuba and Chubarikov, 1980, 2000’s). For
example, consider the number J(X ) of solutions of the system
s
X
xij yim =
i=1
2s
X
xij yim
(0 ≤ j ≤ 3, 0 ≤ m ≤ 2),
i=s+1
with 1 ≤ x, y ≤ X .
The number of equations is r = (3 + 1)(2 + 1) − 1 = 11, the largest total
degree is k = 3 + 2 = 5, the sum of degrees is
K = 21 3(3 + 1) · 21 2(2 + 1) = 18,
and the number of variables in a block is 2.
(General) theorem shows that whenever s > r (k + 1), then
J(X ) X 2sd−K . Can develop a restriction variant of this work.
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Most recent work: the “efficient congruencing” methods apply also to
systems that are only approximately translation-invariant. Consider, for
example, integers 1 ≤ k1 < k2 < . . . < kt , and the number T (X ) of
solutions of the system
s
X
k
k
(xi j − yi j ) = 0
(1 ≤ j ≤ t),
i=1
with 1 ≤ x, y ≤ X . Then (W. 2014) one has
T (X ) X s+ε ,
whenever 1 ≤ s ≤ 12 t(t + 1) − ( 31 + 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).
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