vinogradov's mean value theorem and its associated restriction theory via … · 2014. 10....
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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 .
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 2 / 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 .
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 2 / 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 .
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 2 / 34
(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
supa
(‖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).
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 3 / 34
(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
supa
(‖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).
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 3 / 34
(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
supa
(‖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).
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 3 / 34
(Even) 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).
Conjecture (Main Restriction Conjecture)
For each ε > 0, one has
∮|fk,a(α;X )|2s dα�
X ε
( ∑n≤X|an|2
)s
, when s ≤ 12k(k + 1),
X s−12k(k+1)
( ∑n≤X|an|2
)s
, when s > 12k(k + 1).
Here, we write∮
for∫
[0,1)k .
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 4 / 34
(Even) 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).
Conjecture (Main Restriction Conjecture)
For each ε > 0, one has
∮|fk,a(α;X )|2s dα�
X ε
( ∑n≤X|an|2
)s
, when s ≤ 12k(k + 1),
X s−12k(k+1)
( ∑n≤X|an|2
)s
, when s > 12k(k + 1).
Here, we write∮
for∫
[0,1)k .
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 4 / 34
Some observations, I:
fk,a(α;X ) =∑
1≤n≤Xane(nα1 + . . .+ nkαk).
Conjecture (Main Restriction Conjecture)
∮|fk,a(α;X )|2s dα�
X ε
( ∑n≤X|an|2
)s
, when s ≤ 12k(k + 1),
X s−12k(k+1)
( ∑n≤X|an|2
)s
, 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.
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 5 / 34
Some observations, I:
fk,a(α;X ) =∑
1≤n≤Xane(nα1 + . . .+ nkαk).
Conjecture (Main Restriction Conjecture)
∮|fk,a(α;X )|2s dα�
X ε
( ∑n≤X|an|2
)s
, when s ≤ 12k(k + 1),
X s−12k(k+1)
( ∑n≤X|an|2
)s
, 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.
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 5 / 34
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
s
� 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
((u4
i + v4i )j − (u4
s+i + v4s+i )
j)
= 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: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
s
� 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
((u4
i + v4i )j − (u4
s+i + v4s+i )
j)
= 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: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
s
� 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
((u4
i + v4i )j − (u4
s+i + v4s+i )
j)
= 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
((u4
i + v4i )j − (u4
s+i + v4s+i )
j)
= 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!).
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 7 / 34
Some observations, II:
So the number N(X ) of integral solutions of the system of equations
s∑i=1
((u4
i + v4i )j − (u4
s+i + v4s+i )
j)
= 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!).
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 7 / 34
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
(s < 3),
∮ ∣∣∣ ∑1≤n≤X
ane(n2α + nβ)∣∣∣6 dα dβ � X ε
∑n≤X|an|2
3
,
∮ ∣∣∣ ∑1≤n≤X
ane(n2α + nβ)∣∣∣2s dα dβ � X s−3
∑n≤X|an|2
s
(s > 3).
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 8 / 34
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
6
n1 + n2 + n3 = n4 + n5 + n6
},
with each solution counted with weight
an1an2an3an4an5an6 .
Trevor D. Wooley (University of Bristol) Efficient congruencing Oxford, 29th September 2014 9 / 34
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
6
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 /
34
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
)2
≤∑
h
∑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
3)
= 2(n1 − n3)(n2 − n3),
and this is O(X ε) unless n1 = n3 or n2 = n3.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 11 /
34
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≤Xane(n2α + nβ)
∣∣∣6 dα dβ =∑
|hi |≤2X i (i=1,2)
( ∑(n1,n2,n3)∈B(h)
an1an2an3
)2
≤∑
h
∑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
3)
= 2(n1 − n3)(n2 − n3),
and this is O(X ε) unless n1 = n3 or n2 = n3.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 11 /
34
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≤Xane(n2α + nβ)
∣∣∣6 dα dβ =∑
|hi |≤2X i (i=1,2)
( ∑(n1,n2,n3)∈B(h)
an1an2an3
)2
≤∑
h
∑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
3)
= 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∮ ∣∣∣ ∑
1≤n≤Xane(n2α + nβ)
∣∣∣6 dα dβ � X ε∑
n1,n2,n3
|an1an2an3 |2
� X ε(∑
n
|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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 12 /
34
One should remove the special solutions with n1 = n3 or n2 = n3 inadvance, and for the remaining solutions one finds that∮ ∣∣∣ ∑
1≤n≤Xane(n2α + nβ)
∣∣∣6 dα dβ � X ε∑
n1,n2,n3
|an1an2an3 |2
� X ε(∑
n
|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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 12 /
34
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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 13 /
34
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
s
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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 14 /
34
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
s
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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 14 /
34
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
s
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) conclusionholds for s ≤ 2k − 1 in place of s ≤ k + 1.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 15 /
34
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
s
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) conclusionholds for s ≤ 2k − 1 in place of s ≤ k + 1.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 15 /
34
3. Efficient congruencing
Recent techniques applied in the context of Vinogradov’s mean valuetheorem allow one to establish:
Theorem (W. 2014)
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
s
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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 16 /
34
We now aim to sketch the ideas underlying a slightly simpler result:
Theorem
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
s
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 17 /
34
Consider an auxiliary prime number p (for now, think of p as being a verysmall power of X ).
Write
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 18 /
34
Consider an auxiliary prime number p (for now, think of p as being a verysmall power of X ).
Write
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 18 /
34
Observe that by Cauchy’s inequality, one has
|fa(α;X )| =∣∣∣ ∑1≤n≤X
ane(nα1 + . . .+ nkαk)∣∣∣
≤ X 1/2(∑n≤X|an|2
)1/2,
whence|fa(α;X )| ≤ X 1/2.
Thus
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 19 /
34
Observe that by Cauchy’s inequality, one has
|fa(α;X )| =∣∣∣ ∑1≤n≤X
ane(nα1 + . . .+ nkαk)∣∣∣
≤ X 1/2(∑n≤X|an|2
)1/2,
whence|fa(α;X )| ≤ X 1/2.
Thus
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 19 /
34
Observe that by Cauchy’s inequality, one has
|fa(α;X )| =∣∣∣ ∑1≤n≤X
ane(nα1 + . . .+ nkαk)∣∣∣
≤ X 1/2(∑n≤X|an|2
)1/2,
whence|fa(α;X )| ≤ X 1/2.
Thus
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 19 /
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Define
λs = lim supX→∞
sup(an)∈C[X ]
|an|≤1
logUs,k(X ; a)
logX.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 20 /
34
Define
λs = lim supX→∞
sup(an)∈C[X ]
|an|≤1
logUs,k(X ; a)
logX.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 20 /
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Define
λs = lim supX→∞
sup(an)∈C[X ]
|an|≤1
logUs,k(X ; a)
logX.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 20 /
34
Define
λs = lim supX→∞
sup(an)∈C[X ]
|an|≤1
logUs,k(X ; a)
logX.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 20 /
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Aim: Prove that Λ ≤ 0 for s ≥ k2.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 21 /
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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
.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 22 /
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[[Ka,b(X )]] =Ka,b(X )
(X/pa)k−12k(k+1)(X/pb)s
.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 23 /
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[[Ka,b(X )]] =Ka,b(X )
(X/pa)k−12k(k+1)(X/pb)s
.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 23 /
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(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.
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 24 /
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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
(j
l
)aj−l
s∑i=1
((xi − a)j − (yi − a)j) =s∑
i=1
((xi − a + a)j − (yi − a + a)j).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 25 /
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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
(j
l
)aj−l
s∑i=1
((xi − a)j − (yi − a)j) =s∑
i=1
((xi − a + a)j − (yi − a + a)j).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 25 /
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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∑
l=1
((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∑
l=1
(ujl − v jl ) (1 ≤ j ≤ k),
whencek∑
i=1
(xi − η)j ≡k∑
i=1
(yi − η)j (mod pjb) (1 ≤ j ≤ k).
In this way, we obtain a system of congruence conditions modulo pjb for1 ≤ j ≤ k .
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 26 /
34
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∑
l=1
((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∑
l=1
(ujl − v jl ) (1 ≤ j ≤ k),
whencek∑
i=1
(xi − η)j ≡k∑
i=1
(yi − η)j (mod pjb) (1 ≤ j ≤ k).
In this way, we obtain a system of congruence conditions modulo pjb for1 ≤ j ≤ k .
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 26 /
34
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∑
l=1
((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∑
l=1
(ujl − v jl ) (1 ≤ j ≤ k),
whencek∑
i=1
(xi − η)j ≡k∑
i=1
(yi − η)j (mod pjb) (1 ≤ j ≤ k).
In this way, we obtain a system of congruence conditions modulo pjb for1 ≤ j ≤ k .
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 26 /
34
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∑
l=1
((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∑
l=1
(ujl − v jl ) (1 ≤ j ≤ k),
whencek∑
i=1
(xi − η)j ≡k∑
i=1
(yi − η)j (mod pjb) (1 ≤ j ≤ k).
In this way, we obtain a system of congruence conditions modulo pjb for1 ≤ j ≤ k .
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 26 /
34
k∑i=1
(xi − η)j ≡k∑
i=1
(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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 27 /
34
k∑i=1
(xi − η)j ≡k∑
i=1
(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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 27 /
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k∑i=1
(xi − η)j ≡k∑
i=1
(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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 27 /
34
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,
where
Ξ =
∮ ∑1≤ξ′≤pkb
ξ′≡ξ (mod pa)
ρkb(ξ′)2
ρa(ξ)2|fkb(α; ξ′)|2
k
|fb(α; η)|2s dα.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 28 /
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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,
where
Ξ =
∮ ∑1≤ξ′≤pkb
ξ′≡ξ (mod pa)
ρkb(ξ′)2
ρa(ξ)2|fkb(α; ξ′)|2
k
|fb(α; η)|2s dα.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 28 /
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Ξ =
∮ ∑1≤ξ′≤pkb
ξ′≡ξ (mod pa)
ρkb(ξ′)2
ρa(ξ)2|fkb(α; ξ′)|2
k
|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∑
1≤ξ′≤pkbξ′≡ξ (mod pa)
ρkb(ξ′)2|fkb(α; ξ′)|2s
k/s
.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 29 /
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Ξ =
∮ ∑1≤ξ′≤pkb
ξ′≡ξ (mod pa)
ρkb(ξ′)2
ρa(ξ)2|fkb(α; ξ′)|2
k
|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∑
1≤ξ′≤pkbξ′≡ξ (mod pa)
ρkb(ξ′)2|fkb(α; ξ′)|2s
k/s
.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 29 /
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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 ,
where
Ξ1 = ρa(ξ)−2∑ξ′
ρkb(ξ′)2
∮|fb(α; η)2k fkb(α; ξ′)2s | dα
and
Ξ2 =
∮|fb(α; η)|2s+2k dα.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 30 /
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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 ,
where
Ξ3 = ρ0(1)−4pb∑η=1
pkb∑ξ′=1
ρb(η)2ρkb(ξ′)2
∮|fb(α; η)2k fkb(α; ξ′)2s |dα,
and
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 31 /
34
Then one can check that
[[Ka,b(X )]]� [[Kb,kb(X )]]k/s(X/Mb)(1−k/s)(Λ+ε).
Given the hypothesis that
[[Ka,b(X )]]� XΛ(pψ)Λ,
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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 32 /
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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.
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 33 /
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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).
Trevor D. Wooley (University of Bristol) Efficient congruencingOxford, 29th September 2014 34 /
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