Systems of quadratic and cubic diagonal equations

Transcription

1 Systems of quadratic and cubic diagonal equations Julia Brandes Chalmers Institute of Technology / University of Gothenburg February 03, 2017 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

2 Introduction The question Consider systems of diagonal equation c i,1 x k i c i,sx k i s = 0 (1 i r). What is the number N(P) of integral solutions 1 x 1,..., x s P to such a system? Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

3 Introduction The question Consider systems of diagonal equation c i,1 x k i c i,sx k i s = 0 (1 i r). What is the number N(P) of integral solutions 1 x 1,..., x s P to such a system? Heuristics: Typically expect N(P) P s k i. if equations are suff. independent non-singularity conditions! possible obstructions to real or p-adic solubility Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

4 Introduction The question Consider systems of diagonal equation c i,1 x k i c i,sx k i s = 0 (1 i r). What is the number N(P) of integral solutions 1 x 1,..., x s P to such a system? Heuristics: Typically expect N(P) P s k i. if equations are suff. independent non-singularity conditions! possible obstructions to real or p-adic solubility square root barrier : without further info on coefficients, can get N(P) P s/2. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

5 Introduction The question Consider systems of diagonal equation c i,1 x k i c i,sx k i s = 0 (1 i r). What is the number N(P) of integral solutions 1 x 1,..., x s P to such a system? Heuristics: Typically expect N(P) P s k i. if equations are suff. independent non-singularity conditions! possible obstructions to real or p-adic solubility square root barrier : without further info on coefficients, can get N(P) P s/2. expect N(P) cp s k i for s 2 k i + 1. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

6 Introduction What do we know? Question For what systems c i,1 x k i c i,sx k i s = 0 (1 i r) can we show that N(P) cp s k i for s 2 k i + 1? Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

7 Introduction What do we know? Question For what systems c i,1 x k i c i,sx k i s = 0 (1 i r) can we show that N(P) cp s k i for s 2 k i + 1? Systems of linear and quadratic forms: classical. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

8 Introduction What do we know? Question For what systems c i,1 x k i c i,sx k i s = 0 (1 i r) can we show that N(P) cp s k i for s 2 k i + 1? Systems of linear and quadratic forms: classical. Vinogradov systems: k i = i for 1 i K. (follows from VMVT see BDG) Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

9 Introduction What do we know? Question For what systems c i,1 x k i c i,sx k i s = 0 (1 i r) can we show that N(P) cp s k i for s 2 k i + 1? Systems of linear and quadratic forms: classical. Vinogradov systems: k i = i for 1 i K. (follows from VMVT see BDG) One quadratic and one cubic equation Wooley 2015 (uses cubic VMVT) Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

10 Introduction What do we know? Question For what systems c i,1 x k i c i,sx k i s = 0 (1 i r) can we show that N(P) cp s k i for s 2 k i + 1? Systems of linear and quadratic forms: classical. Vinogradov systems: k i = i for 1 i K. (follows from VMVT see BDG) One quadratic and one cubic equation Wooley 2015 (uses cubic VMVT) Systems of r cubic equations: N(P) cp s 3r whenever s 6r + 1 (Brüdern Wooley 2016). Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

11 Introduction What do we know? Question For what systems c i,1 x k i c i,sx k i s = 0 (1 i r) can we show that N(P) cp s k i for s 2 k i + 1? Systems of linear and quadratic forms: classical. Vinogradov systems: k i = i for 1 i K. (follows from VMVT see BDG) One quadratic and one cubic equation Wooley 2015 (uses cubic VMVT) Systems of r cubic equations: N(P) cp s 3r whenever s 6r + 1 (Brüdern Wooley 2016). This talk: One cubic and r 2 quadratic equations, One quadratic and r 3 cubic equations. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

12 The circle method Warm-up: One quadratic and one cubic equation Wooley s theorem (r 2 = r 3 = 1) by the circle method: Count solutionns 1 x i P to c 1 x c s x 2 s = 0 d 1 x d s x 3 s = 0. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

13 The circle method Warm-up: One quadratic and one cubic equation Wooley s theorem (r 2 = r 3 = 1) by the circle method: Count solutionns 1 x i P to c 1 x c s x 2 s = 0 d 1 x d s x 3 s = 0. Let Then by orthogonality, N(P) = x 1,...x s 1 0 f (α 2, α 3 ) = e(α c i x 2 i )dα 1 0 P e(α 2 x 2 + α 3 x 3 ). x=1 e(β d i x 3 i )dβ Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

14 The circle method Warm-up: One quadratic and one cubic equation Wooley s theorem (r 2 = r 3 = 1) by the circle method: Count solutionns 1 x i P to c 1 x c s x 2 s = 0 d 1 x d s x 3 s = 0. Let Then by orthogonality, N(P) = x 1,...x s 1 0 f (α 2, α 3 ) = e(α c i x 2 i )dα 1 0 P e(α 2 x 2 + α 3 x 3 ). x=1 e(β d i xi 3 )dβ = [0,1] 2 s f (c i α, d i β)dαdβ. i=1 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

15 Need to understand The circle method s N(P) = f (c i α, d i β)dαdβ. [0,1] 2 i=1 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

16 The circle method Need to understand s N(P) = f (c i α, d i β)dαdβ. [0,1] 2 i=1 Idea: dissect unit interval in major arcs M, where the integrand is large minor arcs m, where the integrand is small main term, error term. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

17 The circle method Need to understand s N(P) = f (c i α, d i β)dαdβ. [0,1] 2 i=1 Idea: dissect unit interval in major arcs M, where the integrand is large minor arcs m, where the integrand is small main term, error term. Take [0, 1] 2 = M m, where M = {α [0, 1] 2 : qα i a i P 9/4, q P 3/4, a i, q i N}. By fairly standard methods (s 11): s N(P; M) = f (c i α 2, d i α 3 )dα cp s 5. M i=1 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

18 The circle method Need to understand contribution from minor arcs. Let s 11, then N(P; m) = s m i=1 f (c i α 2, d i α 3 )dα sup f (α) s 10 f (α) α m [0,1] 10 dα. 2 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

19 The circle method Need to understand contribution from minor arcs. Let s 11, then N(P; m) = s m i=1 Classical: sup α m f (α) P 3/4+ε. f (c i α 2, d i α 3 )dα sup f (α) s 10 f (α) α m [0,1] 10 dα. 2 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

20 The circle method Need to understand contribution from minor arcs. Let s 11, then N(P; m) = s m i=1 Classical: sup α m f (α) P 3/4+ε. Theorem (Wooley 2015) f (c i α 2, d i α 3 )dα sup f (α) s 10 f (α) α m [0,1] 10 dα. 2 [0,1] 2 f (α) 10 dα P 5+1/6+ε. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

21 The circle method Need to understand contribution from minor arcs. Let s 11, then N(P; m) = s m i=1 Classical: sup α m f (α) P 3/4+ε. Theorem (Wooley 2015) Altogether, when s 11: f (c i α 2, d i α 3 )dα sup f (α) s 10 f (α) α m [0,1] 10 dα. 2 [0,1] 2 f (α) 10 dα P 5+1/6+ε. N(P) = N(P; M) + O(N(P; m)) cp s 5 + O(P (3/4)(s 10)+ε P 5+1/6+ε ) = cp s 5 + O(P s 5 δ ). Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

22 General systems of diagonal equations What happens for larger systems? We have s N(P) = f (γ i,2, γ i,3 )dα, [0,1] r 2 +r 3 i=1 where P f (α 2, α 3 ) = e(α 2 x 2 + α 3 x 3 ) and γ i,k = r k x=1 j=1 c i,j α j,k (k = 2, 3). Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

23 General systems of diagonal equations What happens for larger systems? We have s N(P) = f (γ i,2, γ i,3 )dα, [0,1] r 2 +r 3 i=1 where f (α 2, α 3 ) = P e(α 2 x 2 + α 3 x 3 ) and γ i,k = x=1 r k j=1 c i,j α j,k (k = 2, 3). Major arcs dissection as before. Need to understand minor arcs contribution. We have (morally) N(P; m) sup f (α) s 4r2 6r3 α m [0,1] r 2 +r 3 r 3 i=1 f (γ i,2, γ i,3 ) 10 r 2 i=r 3+1 f (γ i,2, γ i,3 ) 4 dα. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

24 General systems of diagonal equations Consider the case r 2 = 2, r 3 = 1. The mean value I (P) = f (γ 1,2, γ 1,3 ) 10 f (γ 2,2, γ 2,3 ) 4 dα [0,1] 3 counts solutions to the system d 1,1 (x1 3 + y 5 3) +d 1,2(x6 3 + y 7 3) = 0 c 1,1 (x1 2 + y 5 2) +c 1,2(x6 2 + y 7 2) = 0 c 2,1 (x1 2 + y 5 2) +c 2,2(x6 2 + y 7 2) = 0. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

25 General systems of diagonal equations Consider the case r 2 = 2, r 3 = 1. The mean value I (P) = f (γ 1,2, γ 1,3 ) 10 f (γ 2,2, γ 2,3 ) 4 dα [0,1] 3 counts solutions to the system d 1,1 (x1 3 + y 5 3) +d 1,2(x6 3 + y 7 3) = 0 c 1,1 (x1 2 + y 5 2) +c 1,2(x6 2 + y 7 2) = 0 c 2,1 (x1 2 + y 5 2) +c 2,2(x6 2 + y 7 2) = 0. We diagonalise the bottom 2 2 matrix: Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

26 General systems of diagonal equations Consider the case r 2 = 2, r 3 = 1. The mean value I (P) = f (γ 1,2, γ 1,3 ) 10 f (γ 2,2, γ 2,3 ) 4 dα [0,1] 3 counts solutions to the system d 1,1 (x1 3 + y 5 3) +d 1,2(x6 3 + y 7 3) = 0 c 1,1 (x1 2 + y 5 2) = 0 +c 2,2 (x6 2 + y 7 2) = 0. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

27 General systems of diagonal equations Consider the case r 2 = 2, r 3 = 1. The mean value I (P) = f (γ 1,2, γ 1,3 ) 10 f (γ 2,2, γ 2,3 ) 4 dα [0,1] 3 counts solutions to the system d 1,1 (x1 3 + y 5 3) +d 1,2(x6 3 + y 7 3) = 0 c 1,1 (x1 2 + y 5 2) = 0 +c 2,2 (x6 2 + y 7 2) = 0. The last equation is a quadratic form in four variables and has P 2+ε solutions. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

28 General systems of diagonal equations Consider the case r 2 = 2, r 3 = 1. The mean value I (P) = f (γ 1,2, γ 1,3 ) 10 f (γ 2,2, γ 2,3 ) 4 dα [0,1] 3 counts solutions to the system d 1,1 (x1 3 + y 5 3) +d 1,2(x6 3 + y 7 3) = 0 c 1,1 (x1 2 + y 5 2) = 0 +c 2,2 (x6 2 + y 7 2) = 0. The last equation is a quadratic form in four variables and has P 2+ε solutions. This fixes the variables x 6, x 7, y 6, y 7, and we need to solve d 1,1 (x y 3 5 ) = w 1 c 1,1 (x y 2 5 ) = 0. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

29 General systems of diagonal equations Consider the case r 2 = 2, r 3 = 1. The mean value I (P) = f (γ 1,2, γ 1,3 ) 10 f (γ 2,2, γ 2,3 ) 4 dα [0,1] 3 counts solutions to the system d 1,1 (x1 3 + y 5 3) +d 1,2(x6 3 + y 7 3) = 0 c 1,1 (x1 2 + y 5 2) = 0 +c 2,2 (x6 2 + y 7 2) = 0. The last equation is a quadratic form in four variables and has P 2+ε solutions. This fixes the variables x 6, x 7, y 6, y 7, and we need to solve d 1,1 (x y 3 5 ) = w 1 c 1,1 (x y 2 5 ) = 0. This system has P 5+1/6+ε solutions by Wooley s theorem, and we get I (P) P 2+ε P 5+1/6+ε P 7+1/6+ε. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

30 General systems of diagonal equations The results, part I This argument leads to Theorem 1 (JB & S. T. Parsell 2015, to appear) Suppose that the quadratic and cubic systems are suitably non-singular. We have N(P) cp s 2r2 3r3, with c 0, provided that s 4r 2 + (20/3)r if r 2 r 3 Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

31 General systems of diagonal equations The results, part I This argument leads to Theorem 1 (JB & S. T. Parsell 2015, to appear) Suppose that the quadratic and cubic systems are suitably non-singular. We have N(P) cp s 2r2 3r3, with c 0, provided that s 4r 2 + (20/3)r if r 2 r 3, s 8r 3 + (8/3)r if r 3 > r 2. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

32 General systems of diagonal equations The results, part I This argument leads to Theorem 1 (JB & S. T. Parsell 2015, to appear) Suppose that the quadratic and cubic systems are suitably non-singular. We have N(P) cp s 2r2 3r3, with c 0, provided that s 4r 2 + (20/3)r if r 2 r 3, s 8r 3 + (8/3)r if r 3 > r 2. The argument generalises to arbitrary systems of arbitrary degrees. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

33 General systems of diagonal equations The results, part I This argument leads to Theorem 1 (JB & S. T. Parsell 2015, to appear) Suppose that the quadratic and cubic systems are suitably non-singular. We have N(P) cp s 2r2 3r3, with c 0, provided that s 4r 2 + (20/3)r if r 2 r 3, s 8r 3 + (8/3)r if r 3 > r 2. The argument generalises to arbitrary systems of arbitrary degrees. For r 3 = 1 we obtain essentially square root cancellation. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

34 General systems of diagonal equations The results, part I This argument leads to Theorem 1 (JB & S. T. Parsell 2015, to appear) Suppose that the quadratic and cubic systems are suitably non-singular. We have N(P) cp s 2r2 3r3, with c 0, provided that s 4r 2 + (20/3)r if r 2 r 3, s 8r 3 + (8/3)r if r 3 > r 2. The argument generalises to arbitrary systems of arbitrary degrees. For r 3 = 1 we obtain essentially square root cancellation. The case r 3 > r 2 is clearly unsatisfactory... Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

35 The case r 3 2r 2 : Complification The case r 3 2r 2 : Complification Recall s N(P) = f (γ i,2, γ i,3 )dα, [0,1] r 2 +r 3 i=1 where P f (α 2, α 3 ) = e(α 2 x 2 + α 3 x 3 ) and γ i,k = r k x=1 j=1 c i,j α 3,j (k = 2, 3). Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

36 The case r 3 2r 2 : Complification The case r 3 2r 2 : Complification Recall where f (α 2, α 3 ) = We have where J 1 (P) = N(P) = [0,1] r 2 +r 3 i=1 s f (γ i,2, γ i,3 )dα, P e(α 2 x 2 + α 3 x 3 ) and γ i,k = x=1 [0,1] r 2 +r 3 i=1 r k j=1 N(P; m) sup f (α 2, α 3 ) s 4r2 6r3 J 1 (P), α m r 3 r 3+r 2 f (γ i ) 2 i=r 3+1 2r 3 2r 2 f (γ i ) 8 i=r 3+r 2+1 c i,j α 3,j (k = 2, 3). 2r 3 r 2 f (γ i ) 4 i=2r 3 2r 2+1 f (γ i ) 8 dα Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

37 The case r 3 2r 2 : Complification Recall J 1 (P) = r 3 [0,1] r 2 +r 3 i=1 r 3+r 2 f (γ i ) 2 i=r 3+1 2r 3 2r 2 f (γ i ) 8 i=r 3+r 2+1 Idea: Do the counter-intuitive thing blow the system up! 2r 3 r 2 f (γ i ) 4 i=2r 3 2r 2+1 f (γ i ) 8 dα. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

38 The case r 3 2r 2 : Complification Recall J 1 (P) = r 3 [0,1] r 2 +r 3 i=1 r 3+r 2 f (γ i ) 2 i=r 3+1 2r 3 2r 2 f (γ i ) 8 i=r 3+r 2+1 Idea: Do the counter-intuitive thing blow the system up! Cauchy-Schwarz gives where the cubic system is of the shape J n (P) (P (5+1/6)r2+ε ) 1/2 J 2n (P) 1/2, 2r 3 r 2 f (γ i ) 4 i=2r 3 2r 2+1 f (γ i ) 8 dα. Figure: J 1(P) Figure: J 2(P) Figure: J 4(P) Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

39 The case r 3 2r 2 : Complification After m inductive steps: J 1 (P) (P (5+1/6)r2+ε ) 1 2 m J 2 m(p) 2 m The mean value J n (P) contains R n = n(r 3 r 2 ) + r 2 cubic and r 2 quadratic polynomials in 6R n + 4r 2 variables. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

40 The case r 3 2r 2 : Complification After m inductive steps: J 1 (P) (P (5+1/6)r2+ε ) 1 2 m J 2 m(p) 2 m The mean value J n (P) contains R n = n(r 3 r 2 ) + r 2 cubic and r 2 quadratic polynomials in 6R n + 4r 2 variables. Now a miracle happens: J n (P) P 3n(r3 r2)+9r2+ε P 3Rn+6r2+ε. We miss by a constant exponent regardless of the value of n!!! Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

41 The case r 3 2r 2 : Complification After m inductive steps: J 1 (P) (P (5+1/6)r2+ε ) 1 2 m J 2 m(p) 2 m The mean value J n (P) contains R n = n(r 3 r 2 ) + r 2 cubic and r 2 quadratic polynomials in 6R n + 4r 2 variables. Now a miracle happens: J n (P) P 3n(r3 r2)+9r2+ε P 3Rn+6r2+ε. We miss by a constant exponent regardless of the value of n!!! This gives J 1 (P) (P (5+1/6)r2+ε ) 1 2 m (P 3 2m (r 3 r 2)+9r 2+ε ) 2 m P 3r3+(2+1/6)r2+2 m 23 6 r2+ε. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

42 The case r 3 2r 2 : Complification The results, part II This leads to Theorem 2 (JB 2016, to appear) Suppose that the quadratic and cubic systems are suitably non-singular, and r 3 2r 2 > 1. We have with c 0, provided that N(P) cp s 2r2 3r3 s 6r 3 + (14/3)r Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

43 The case r 3 2r 2 : Complification The results, part II This leads to Theorem 2 (JB 2016, to appear) Suppose that the quadratic and cubic systems are suitably non-singular, and r 3 2r 2 > 1. We have with c 0, provided that N(P) cp s 2r2 3r3 s 6r 3 + (14/3)r For r 3 (r 2, 2r 2 ), we have the bound s 4r 2 + 6r 3 + (2/3) min{r 2, r 3 } + 1. Unfortunately, the middle range for r 3 seems to be harder. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

44 The case r 3 2r 2 : Complification The results, part II This leads to Theorem 2 (JB 2016, to appear) Suppose that the quadratic and cubic systems are suitably non-singular, and r 3 2r 2 > 1. We have with c 0, provided that N(P) cp s 2r2 3r3 s 6r 3 + (14/3)r For r 3 (r 2, 2r 2 ), we have the bound s 4r 2 + 6r 3 + (2/3) min{r 2, r 3 } + 1. Unfortunately, the middle range for r 3 seems to be harder. We have square root cancellation for min{r 2, r 3 } = 1. Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15

45 The end The end Thank you for your attention! Julia Brandes (Gothenburg) Systems of quadratic and cubic diagonal equations February 03, / 15