Vinogradov s mean value theorem and its associated restriction theory via efficient congruencing.

Transcription

1 Vinogradov s mean value theorem and its associated restriction theory via efficient congruencing. Trevor D. Wooley University of Bristol Oxford, 29th September 2014 Oxford, 29th September /

2 1. Introduction Let k 2 be an integer, and consider g : T k C (T = R/Z [0, 1)), with an associated Fourier series g(α 1,..., α k ) = ĝ(n 1,..., n k )e(n 1 α n k α k ), n Z k in which ĝ(n) C and e(z) = e 2πiz. Oxford, 29th September /

3 1. Introduction Let k 2 be an integer, and consider g : T k C (T = R/Z [0, 1)), with an associated Fourier series g(α 1,..., α k ) = ĝ(n 1,..., n k )e(n 1 α n k α k ), n Z k in which ĝ(n) C and e(z) = e 2πiz. Restriction operator: (E. Stein, J. Bourgain, K. Hughes, et al.) Rg := n Z k n=(n,n 2,...,n k ) ĝ(n)e(n α). [This is just one example of a restriction operator!] Oxford, 29th September /

4 1. Introduction Let k 2 be an integer, and consider g : T k C (T = R/Z [0, 1)), with an associated Fourier series g(α 1,..., α k ) = ĝ(n 1,..., n k )e(n 1 α n k α k ), n Z k in which ĝ(n) C and e(z) = e 2πiz. Restriction operator: (E. Stein, J. Bourgain, K. Hughes, et al.) Rg := n Z k n=(n,n 2,...,n k ) ĝ(n)e(n α). [This is just one example of a restriction operator!] We are interested in the norm of the operator g Rg. Oxford, 29th September /

5 (Slightly) more concretely (for analytic number theorists): Consider a sequence (a n ) n=1 of complex numbers, not all zero, and define the exponential sum f a = f k,a (α; X ) by putting f k,a (α; X ) = a n e(nα n k α k ). 1 n X Oxford, 29th September /

6 (Slightly) more concretely (for analytic number theorists): Consider a sequence (a n ) n=1 of complex numbers, not all zero, and define the exponential sum f a = f k,a (α; X ) by putting f k,a (α; X ) = a n e(nα n k α k ). Aim: Obtain a bound for in terms of p, k and X. 1 n X sup ( f a L p/ a l 2) a Oxford, 29th September /

7 (Slightly) more concretely (for analytic number theorists): Consider a sequence (a n ) n=1 of complex numbers, not all zero, and define the exponential sum f a = f k,a (α; X ) by putting f k,a (α; X ) = a n e(nα n k α k ). Aim: Obtain a bound for in terms of p, k and X. 1 n X sup ( f a L p/ a l 2) a Conjecture (Main Restriction Conjecture) For each ε > 0, one has f a L p X ε, when 0 < p k(k + 1), ε,p,k a l 2 X 1 2 k(k+1) 2p, when p > k(k + 1). Oxford, 29th September /

8 (Even) more concretely (for analytic number theorists): Consider a sequence (a n ) n=1 of complex numbers, not all zero, and define the exponential sum f a = f k,a (α; X ) by putting f k,a (α; X ) = a n e(nα n k α k ). 1 n X Oxford, 29th September /

9 (Even) more concretely (for analytic number theorists): Consider a sequence (a n ) n=1 of complex numbers, not all zero, and define the exponential sum f a = f k,a (α; X ) by putting f k,a (α; X ) = a n e(nα n k α k ). 1 n X Conjecture (Main Restriction Conjecture) For each ε > 0, one has ( X ε f k,a (α; X ) 2s n X dα ( X s 1 2 k(k+1) Here, we write for [0,1) k. ) s a n 2, when s 1 2k(k + 1), n X ) s a n 2, when s > 1 2k(k + 1). Oxford, 29th September /

10 Some observations, I: f k,a (α; X ) = a n e(nα n k α k ). 1 n X Conjecture (Main Restriction Conjecture) ( X ε f k,a (α; X ) 2s n X dα ( X s 1 2 k(k+1) ) s a n 2, when s 1 2k(k + 1), n X ) s a n 2, when s > 1 2k(k + 1). Oxford, 29th September /

11 Some observations, I: f k,a (α; X ) = a n e(nα n k α k ). 1 n X Conjecture (Main Restriction Conjecture) ( X ε f k,a (α; X ) 2s n X dα ( X s 1 2 k(k+1) ) s a n 2, when s 1 2k(k + 1), n X ) s a n 2, when s > 1 2k(k + 1). Consider the sequence (a n ) = 1. Then MRC implies that f k,1 (α; X ) 2s dα X ε (X s + X 2s 1 2 k(k+1) ), an assertion equivalent to the Main Conjecture in Vinogradov s Mean Value Theorem. Oxford, 29th September /

12 Some observations, II: Consider the situation in which (a n ) is supported on a thin sequence, say a n = card { (x, y) Z 2 : n = x 4 + y 4}. Oxford, 29th September /

13 Some observations, II: Consider the situation in which (a n ) is supported on a thin sequence, say a n = card { (x, y) Z 2 : n = x 4 + y 4}. Then MRC implies that for 1 s 1 2k(k + 1), one should have f k,a (α; X ) 2s dα X ε n X a s n 2 X ε ( X 1/2) s = X s/2+ε. Oxford, 29th September /

14 Some observations, II: Consider the situation in which (a n ) is supported on a thin sequence, say a n = card { (x, y) Z 2 : n = x 4 + y 4}. Then MRC implies that for 1 s 1 2k(k + 1), one should have f k,a (α; X ) 2s dα X ε n X a s n 2 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 ( (u 4 i + v 4 i ) j (u 4 s+i + v 4 s+i) j) = 0 (1 j k), with 1 u 4 i + v 4 i X (1 i 2s). Oxford, 29th September /

15 Some observations, II: So the number N(X ) of integral solutions of the system of equations s ( (u 4 i + v 4 i ) j (u 4 s+i + v 4 s+i) j) = 0 (1 j k), with 1 u 4 i + v 4 i X (1 i 2s), satisfies N(X ) X s/2+ε. Oxford, 29th September /

16 Some observations, II: So the number N(X ) of integral solutions of the system of equations s ( (u 4 i + v 4 i ) j (u 4 s+i + v 4 s+i) j) = 0 (1 j k), with 1 u 4 i + v 4 i X (1 i 2s), satisfies N(X ) X s/2+ε. But the number of diagonal solutions with u i = u s+i and v i = v s+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!). Oxford, 29th September /

17 2. Classical results (Bourgain, 1993) The Main restriction Conjecture holds for k = 2, and in particular: a n e(n 2 α + nβ) 2s dα dβ n X a s n 2 (s < 3), 1 n X 1 n X 1 n X a n e(n 2 α + nβ) 6 dα dβ X ε n X a n 2 a n e(n 2 α + nβ) 2s dα dβ X s 3 n X a n 2 3, s (s > 3). Oxford, 29th September /

18 Sketch proof for the case k = 2 and s = 3: By orthogonality, the integral 1 n X a n e(n 2 α + nβ) 6 dα dβ counts the number of solutions of the simultaneous equations } n1 2 + n2 2 + n3 2 = n4 2 + n5 2 + n6 2, n 1 + n 2 + n 3 = n 4 + n 5 + n 6 with each solution counted with weight a n1 a n2 a n3 a n4 a n5 a n6. Oxford, 29th September /

19 Sketch proof for the case k = 2 and s = 3: By orthogonality, the integral 1 n X a n e(n 2 α + nβ) 6 dα dβ counts the number of solutions of the simultaneous equations } n1 2 + n2 2 n3 2 = n4 2 + n5 2 n6 2, n 1 + n 2 n 3 = n 4 + n 5 n 6 with each solution counted with weight a n1 a n2 a n3 a n4 a n5 a n6. Oxford, 29th September /

20 Let B(h) denote the set of integral solutions of the equation } n1 2 + n2 2 n3 2 = h 2, n 1 + n 2 n 3 = h 1 with 1 n i X. Oxford, 29th September /

21 Let B(h) denote the set of integral solutions of the equation } n1 2 + n2 2 n3 2 = h 2, n 1 + n 2 n 3 = h 1 with 1 n i X. Then by Cauchy s inequality, a n e(n 2 α + nβ) 6 dα dβ = 1 n X h h i 2X i (,2) ( (n 1,n 2,n 3 ) B(h) n 1,n 2,n 3 B(h) a n1 a n2 a n3 2. a n1 a n2 a n3 ) 2 Oxford, 29th September /

22 Let B(h) denote the set of integral solutions of the equation } n1 2 + n2 2 n3 2 = h 2, n 1 + n 2 n 3 = h 1 with 1 n i X. Then by Cauchy s inequality, a n e(n 2 α + nβ) 6 dα dβ = 1 n X h h i 2X i (,2) ( (n 1,n 2,n 3 ) B(h) n 1,n 2,n 3 B(h) a n1 a n2 a n3 2. But B(h) is bounded above by the number of solutions of h 2 1 h 2 = (n 1 + n 2 n 3 ) 2 (n n 2 2 n 2 3) = 2(n 1 n 3 )(n 2 n 3 ), a n1 a n2 a n3 ) 2 and this is O(X ε ) unless n 1 = n 3 or n 2 = n 3. Oxford, 29th September /

23 One should remove the special solutions with n 1 = n 3 or n 2 = n 3 in advance, and for the remaining solutions one finds that a n e(n 2 α + nβ) 6 dα dβ X ε 1 n X n 1,n 2,n 3 a n1 a n2 a n3 2 X ε( a n 2) 3. n Oxford, 29th September /

24 One should remove the special solutions with n 1 = n 3 or n 2 = n 3 in advance, and for the remaining solutions one finds that a n e(n 2 α + nβ) 6 dα dβ X ε 1 n X n 1,n 2,n 3 a n1 a n2 a n3 2 X ε( a n 2) 3. Key observation: With B(h) the set of integral solutions of the equation } n1 2 + n2 2 n3 2 = h 2, n 1 + n 2 n 3 = h 1 with 1 n i X, one has B(h) X ε (Very strong control of the number of solutions of the associated Diophantine system). n Oxford, 29th September /

25 Now let B s,k (h) denote the set of integral solutions of the system s x j i = h j (1 j k), with 1 x i X. Then we have B s,k (h) 1 (1 s k), and (using estimates from Vinogradov s mean value theorem) B s,k (h) X s 1 2 k(k+1), for s > 2k(k 1) (uses W., 2014). Oxford, 29th September /

26 f k,a (α; X ) = a n e(nα n k α k ). 1 n X Theorem (Bourgain, 1993; K. Hughes, 2012) For each ε > 0, one has MRC in the shape f k,a (α; X ) 2s dα X ε (1 + X s 1 2 k(k+1) ) a n 2 n X 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). Oxford, 29th September /

27 f k,a (α; X ) = a n e(nα n k α k ). 1 n X Theorem (Bourgain, 1993; K. Hughes, 2012) For each ε > 0, one has MRC in the shape f k,a (α; X ) 2s dα X ε (1 + X s 1 2 k(k+1) ) a n 2 n X 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 efficient congruencing results in Vinogradov s mean value theorem (W., 2014). Oxford, 29th September /

28 Theorem (Bourgain, 1993; K. Hughes, 2012) For each ε > 0, one has MRC in the shape f k,a (α; X ) 2s dα X ε (1 + X s 1 2 k(k+1) ) a n 2 n X 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). Oxford, 29th September /

29 Theorem (Bourgain, 1993; K. Hughes, 2012) For each ε > 0, one has MRC in the shape f k,a (α; X ) 2s dα X ε (1 + X s 1 2 k(k+1) ) a n 2 n X 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) conclusion holds for s 2k 1 in place of s k + 1. Oxford, 29th September /

30 3. 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 f k,a (α; X ) 2s dα X ε (1 + X s 1 2 k(k+1) ) a n 2 n X 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,..., and D(k) = 1 2 k(k + 1) 1 3 k + 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). Oxford, 29th September /

31 We now aim to sketch the ideas underlying a slightly simpler result: Theorem For each ε > 0, one has MRC in the shape f k,a (α; X ) 2s dα X ε (1 + X s 1 2 k(k+1) ) a n 2 n X s 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 (a n ) = (1)) indirectly. Oxford, 29th September /

32 Consider an auxiliary prime number p (for now, think of p as being a very small power of X ). Write and then define ( ρ c (ξ) = ρ c (ξ; a) = fa (α; X ) = ρ 0 (1) 1 1 n X 1 n X n ξ (mod p c ) [Note: if a n = 0 for all n, then define f a = 0.] a n 2 ) 1/2, a n e(nα n k α k ). Oxford, 29th September /

33 Consider an auxiliary prime number p (for now, think of p as being a very small power of X ). Write and then define ( ρ c (ξ) = ρ c (ξ; a) = fa (α; X ) = ρ 0 (1) 1 1 n X 1 n X n ξ (mod p c ) [Note: if a n = 0 for all n, then define f a = 0.] We investigate U s,k (X ; a) = a n 2 ) 1/2, a n e(nα n k α k ). f a (α; X ) 2s dα. Oxford, 29th September /

34 Observe that by Cauchy s inequality, one has f a (α; X ) = a n e(nα n k α k ) 1 n X X 1/2( a n 2) 1/2, n X whence f a (α; X ) X 1/2. Oxford, 29th September /

35 Observe that by Cauchy s inequality, one has f a (α; X ) = a n e(nα n k α k ) 1 n X X 1/2( a n 2) 1/2, n X whence Thus f a (α; X ) X 1/2. U s,k (X ; a) = f a (α; X ) 2s dα X s. Oxford, 29th September /

36 Observe that by Cauchy s inequality, one has f a (α; X ) = a n e(nα n k α k ) 1 n X X 1/2( a n 2) 1/2, n X whence Thus f a (α; X ) X 1/2. U s,k (X ; a) = f a (α; X ) 2s dα X s. Moreover, one has that U s,k (X ; a) is scale-invariant, by which we mean that it is invariant on scaling (a n ) to (γa n ) for any γ > 0. Oxford, 29th September /

37 Define λ s = lim sup X sup (a n) C [X ] a n 1 log U s,k (X ; a). log X Oxford, 29th September /

38 Define λ s = lim sup X sup (a n) C [X ] a n 1 log U s,k (X ; a). log X Then there exists a sequence (X m ) m=1 with lim m X m = + such that, for some sequence (a n ) C [Xm] with a n 1, one has that for each ε > 0, U s,k (X m ; a) X λs ε, whilst whenever 1 Y Xm 1/2, and for all sequences (a n ), at the same time one has U s,k (Y ; a) Y λs+ε. Oxford, 29th September /

39 Define λ s = lim sup X sup (a n) C [X ] a n 1 log U s,k (X ; a). log X Then there exists a sequence (X m ) m=1 with lim m X m = + such that, for some sequence (a n ) C [Xm] with a n 1, one has that for each ε > 0, U s,k (X m ; a) X λs ε, whilst whenever 1 Y Xm 1/2, and for all sequences (a n ), at the same time one has U s,k (Y ; a) Y λs+ε. We now fix such a value X = X m sufficiently large, and put Λ = λ s+k (s + k 1 2k(k + 1)). Oxford, 29th September /

40 Define λ s = lim sup X sup (a n) C [X ] a n 1 log U s,k (X ; a). log X Then there exists a sequence (X m ) m=1 with lim m X m = + such that, for some sequence (a n ) C [Xm] with a n 1, one has that for each ε > 0, U s,k (X m ; a) X λs ε, whilst whenever 1 Y Xm 1/2, and for all sequences (a n ), at the same time one has U s,k (Y ; a) Y λs+ε. We now fix such a value X = X m sufficiently large, and put Λ = λ s+k (s + k 1 2k(k + 1)). Aim: Prove that Λ 0 for s k 2. Oxford, 29th September /

41 Aim: Prove that Λ 0 for s k 2. This implies that U s+k (X ; a) X s+k 1 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 f c (α; ξ) = ρ c (ξ) 1 a n e(nα n k α k ), and then put 1 n X n ξ (mod p c ) pa p b K a,b (X ) = ρ 0 (1) 4 ρ a (ξ) 2 ρ b (η) 2 ξ=1 η=1 f a (α; ξ) 2k f b (α; η) 2s dα. Oxford, 29th September /

42 pa p b K a,b (X ) = ρ 0 (1) 4 ρ a (ξ) 2 ρ b (η) 2 ξ=1 η=1 f a (α; ξ) 2k f b (α; η) 2s dα. One expects that K a,b (X ) X ε (X /p a ) k 1 2 k(k+1) (X /p b ) s, and motivated by this observation, we define [[K a,b (X )]] = K a,b (X ) (X /p a ) k 1 2 k(k+1) (X /p b ) s. Oxford, 29th September /

43 [[K a,b (X )]] = K a,b (X ) (X /p a ) k 1 2 k(k+1) (X /p b ) s. Oxford, 29th September /

44 [[K a,b (X )]] = K a,b (X ) (X /p a ) k 1 2 k(k+1) (X /p b ) s. Strategy: (i) Show that if then U s+k,k (X ; a) X s+k 1 2 k(k+1)+λ, [[K 0,1 (X )]] X Λ. (ii) Show that whenever [[K a,b (X )]] X Λ (p ψ ) Λ, then there is a small non-negative integer h with the property that [[K a,b (X )]] X Λ (p ψ ) Λ, where ψ = (s/k)ψ + (s/k 1)b, a = b, b = kb + h. Oxford, 29th September /

45 (ii) Show that whenever [[K a,b (X )]] X Λ (p ψ ) Λ, then there is a small non-negative integer h with the property that [[K a,b (X )]] X Λ (p ψ ) Λ, where ψ = (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 for which b (n) k n and ψ (n) nk n [[K a (n),b (n)(x )]] X Λ (p ψ(n) ) 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. Oxford, 29th September /

46 4. Translation invariance, and the congruencing idea Observe that the system of equations s (x j i y j i ) = 0 (1 j k) (1) has a solution x, y if and only if, for any integral shift a, the system of equations s ((x i a) j (y i a) j ) = 0 (1 j k) is also satisfied Oxford, 29th September /

47 4. Translation invariance, and the congruencing idea Observe that the system of equations s (x j i y j i ) = 0 (1 j k) (1) has a solution x, y if and only if, for any integral shift a, the system of equations s ((x i a) j (y i a) j ) = 0 (1 j k) is also satisfied To see this, note that j l=1 ( ) j a j l l s ((x i a) j (y i a) j ) = s ((x i a + a) j (y i a + a) j ). Oxford, 29th September /

48 The mean value f a (α; ξ) 2k f b (α; η) 2s dα counts (with weights) the number of integral solutions of the system k (x j i y j i ) = s ((p b u l + η) j (p b v l + η) j ) l=1 with 1 x, y X and (1 η)/p b u, v (X η)/p b. (1 j k), Oxford, 29th September /

49 The mean value f a (α; ξ) 2k f b (α; η) 2s dα counts (with weights) the number of integral solutions of the system k (x j i y j i ) = s ((p b u l + η) j (p b v l + η) j ) l=1 (1 j k), 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 ((x i η) j (y i η) j ) = p jb s l=1 (u j l v j l ) (1 j k), Oxford, 29th September /

50 The mean value f a (α; ξ) 2k f b (α; η) 2s dα counts (with weights) the number of integral solutions of the system k (x j i y j i ) = s ((p b u l + η) j (p b v l + η) j ) l=1 (1 j k), 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 whence k ((x i η) j (y i η) j ) = p jb k (x i η) j s l=1 (u j l v j l ) (1 j k), k (y i η) j (mod p jb ) (1 j k). Oxford, 29th September /

51 The mean value f a (α; ξ) 2k f b (α; η) 2s dα counts (with weights) the number of integral solutions of the system k (x j i y j i ) = s ((p b u l + η) j (p b v l + η) j ) l=1 (1 j k), 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 whence k ((x i η) j (y i η) j ) = p jb k (x i η) j s l=1 (u j l v j l ) (1 j k), k (y i η) j (mod p jb ) (1 j k). In this way, we obtain a system of congruence conditions modulo p jb for 1 j k. Oxford, 29th September /

52 k (x i η) j k (y i η) j (mod p jb ) (1 j k). Oxford, 29th September /

53 k (x i η) j k (y i η) j (mod p jb ) (1 j k). Suppose that x is well-conditioned, by which we mean that x 1,..., x k lie in distinct congruence classes modulo p. Then, given an integral k-tuple n, the solutions of the system k (x i η) j n j (mod p) (1 j k), with 1 x p, may be lifted uniquely to solutions of the system k (x i η) j n j (mod p kb ) (1 j k), with 1 x p kb. Oxford, 29th September /

54 k (x i η) j k (y i η) j (mod p jb ) (1 j k). Suppose that x is well-conditioned, by which we mean that x 1,..., x k lie in distinct congruence classes modulo p. Then, given an integral k-tuple n, the solutions of the system k (x i η) j n j (mod p) (1 j k), with 1 x p, may be lifted uniquely to solutions of the system k (x i η) j n j (mod p kb ) (1 j k), with 1 x p kb. In this way, the initial congruences essentially imply that x y (mod p kb ), provided that we inflate our estimates by k!p 1 2 k(k 1)b. Oxford, 29th September /

55 x y (mod p kb ) Oxford, 29th September /

56 x y (mod p kb ) Now we are counting solutions with weights, so we reinsert this congruence information back into the mean value K a,b (X ) to obtain the relation K a,b (X ) p 1 pa 2 k(k 1)(a+b) ρ 0 (1) 4 ρ a (ξ) 2 ρ b (η) 2 Ξ, p b ξ=1 η=1 where Ξ = 1 ξ p kb ξ ξ (mod p a ) ρ kb (ξ ) 2 ρ a (ξ) 2 f kb(α; ξ ) 2 k f b (α; η) 2s dα. Oxford, 29th September /

57 Ξ = 1 ξ p kb ξ ξ (mod p a ) ρ kb (ξ ) 2 ρ a (ξ) 2 f kb(α; ξ ) 2 k f b (α; η) 2s dα. Oxford, 29th September /

58 Ξ = 1 ξ p kb ξ ξ (mod p a ) ρ kb (ξ ) 2 ρ a (ξ) 2 f kb(α; ξ ) 2 k f b (α; η) 2s dα. But by Hölder s inequality, the term here raised to power k is bounded above by ρ a (ξ) 2k ( 1 ξ p kb ξ ξ (mod p a ) ρ a(ξ) 2 ρ kb (ξ ) 2 f kb (α; ξ ) 2s ) k/s ( 1 ξ p kb ξ ξ (mod p a ) 1 ξ p kb ξ ξ (mod p a ) ρ kb (ξ ) 2 f kb (α; ξ ) 2s k/s. ρ kb (ξ ) 2 ) k k/s Oxford, 29th September /

59 Then another application of Hölder s inequality yields Ξ ρ a (ξ) 2 ρ kb (ξ ) 2 f kb (α; ξ ) 2s Ξ k/s 1 Ξ 1 k/s 2, ξ k/s f b (α; η) 2s dα where Ξ 1 = ρ a (ξ) 2 ξ ρ kb (ξ ) 2 f b (α; η) 2k f kb (α; ξ ) 2s dα and Ξ 2 = f b (α; η) 2s+2k dα. Oxford, 29th September /

60 Recall that K a,b (X ) p 1 pa 2 k(k 1)(a+b) ρ 0 (1) 4 ρ a (ξ) 2 ρ b (η) 2 Ξ, p b ξ=1 η=1 From here, yet another application of Hölder s inequality gives where and pb 4 Ξ 3 = ρ 0 (1) K a,b (X ) p 1 2 k(k 1)(a+b) Ξ k/s 3 Ξ 1 k/s 4, p kb η=1 ξ =1 ρ b (η) 2 ρ kb (ξ ) 2 pb p a 4 Ξ 4 = ρ 0 (1) ρ b (η) 2 ρ a (ξ) 2 η=1 ξ=1 (X /M b ) s+k 1 2 k(k+1)+λ+ε. f b (α; η) 2k f kb (α; ξ ) 2s dα, f b (α; η) 2s+2k dα Oxford, 29th September /

61 Then one can check that [[K a,b (X )]] [[K b,kb (X )]] k/s (X /M b ) (1 k/s)(λ+ε). Given the hypothesis that [[K a,b (X )]] X Λ (p ψ ) Λ, this implies that [[K b,kb (X )]] X Λ (p ψ ) Λ, where ψ = (s/k)ψ + (s/k 1)b, which is a little stronger than we had claimed earlier. Oxford, 29th September /

62 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 with 1 x, y X. x j i y m i = 2s i=s+1 x j i y m i (0 j 3, 0 m 2), The number of equations is r = (3 + 1)(2 + 1) 1 = 11, the largest total degree is k = = 5, the sum of degrees is K = (3 + 1) 22(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. Oxford, 29th September /

63 Most recent work: the efficient congruencing methods apply also to systems that are only approximately translation-invariant. Consider, for example, integers 1 k 1 < k 2 <... < k t, and the number T (X ) of solutions of the system s (x k j i y k j i ) = 0 (1 j t), with 1 x, y X. Then (W. 2014) one has T (X ) X s+ε, whenever 1 s 1 2 t(t + 1) ( 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). Oxford, 29th September 2014 /