Cool theorems proved by undergraduates

Transcription

1 Cool theorems proved by undergraduates Ken Ono Emory University

2 Child s play...

3 Child s play... Thanks to the,

4 Child s play... Thanks to the, undergrads and I play by...

5 Child s play... Thanks to the, undergrads and I play by...shooting off

6 Child s play... Thanks to the, undergrads and I play by...shooting off and getting in trouble,

7 Child s play... Thanks to the, undergrads and I play by...shooting off and getting in trouble,...and by proving theorems in number theory.

8 Child s play... Our toys include...

9 Child s play... Our toys include... (Prime numbers)

10 Child s play... Our toys include... (Prime numbers) (Partitions)

11 Child s play... Our toys include... (Prime numbers) (Partitions) (Numbers and Number Fields)

12 Prime Numbers Some theorems on primes

13 Prime Numbers Some theorems on primes Theorem (Euclid) There are infinitely many prime numbers.

14 Prime Numbers Some theorems on primes Theorem (Euclid) There are infinitely many prime numbers. Theorem (Prime Number Theorem) If π(x) := #{p x : prime}, then π(x) x ln x.

15 Prime Numbers

16 Prime Numbers Theorem (Dirichlet) If 0 r < t are integers for which gcd(r, t) = 1, then let π r,t (x) := #{p x : prime and p r (mod t)}.

17 Prime Numbers Theorem (Dirichlet) If 0 r < t are integers for which gcd(r, t) = 1, then let π r,t (x) := #{p x : prime and p r (mod t)}. Then in terms of Euler s phi-function, we have π r,t (x) 1 φ(t) x ln x.

18 Prime Numbers Theorem (Dirichlet) If 0 r < t are integers for which gcd(r, t) = 1, then let π r,t (x) := #{p x : prime and p r (mod t)}. Then in terms of Euler s phi-function, we have π r,t (x) 1 φ(t) x ln x. Example (Half of the primes are of the following forms:) 3n + 1, 3n + 2 4n + 1, 4n + 3 6n + 1, 6n + 5

19 Prime Numbers Arithmetic progressions of primes

20 Prime Numbers Arithmetic progressions of primes Theorem (van der Corput (1939)) There are ly many length 3 arithmetic progressions of primes.

21 Prime Numbers Arithmetic progressions of primes Theorem (van der Corput (1939)) There are ly many length 3 arithmetic progressions of primes. Example Here are examples of arithmetic progressions of length 3: (3, 3 + 2, ) = (3, 5, 7), (5, , ) = (5, 47, 89), (43, , ) = (43, 73, 103).

22 Prime Numbers How long can these get?

23 Prime Numbers How long can these get? Theorem (Green-Tao (2000s)) For every k there are ly many length k AP of primes.

24 Prime Numbers How long can these get? Theorem (Green-Tao (2000s)) For every k there are ly many length k AP of primes. Example Length 10 example: 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089.

25 Prime Numbers Different problem Question Can you find k consecutive primes which end with the digit 1?

26 Prime Numbers Different problem Question Can you find k consecutive primes which end with the digit 1? Example We have the following consecutive primes 181, 191, 241, 251,. 4831, 4861, 4871.

27 Prime Numbers A great theorem...

28 Prime Numbers A great theorem... Theorem (Shiu (2000)) Let p 1 = 2, p 2 = 3,... be the primes in order. If gcd(r, t) = 1, then for every positive integer k there is an n for which p n p n+1 p n+2 p n+k r (mod t).

29 Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression.

30 Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression. Corollary Define the sequence of integers N π := { π, 2π, 3π,... }.

31 Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression. Corollary Define the sequence of integers N π := { π, 2π, 3π,... }. Let p π (n) be the nth prime in this sequence, and so p π (1) = 3, p π (2) = 31, p π (3) = 37,....

32 Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression. Corollary Define the sequence of integers N π := { π, 2π, 3π,... }. Let p π (n) be the nth prime in this sequence, and so p π (1) = 3, p π (2) = 31, p π (3) = 37,.... Then for every gcd(r, t) = 1 and every k there exists an n for which

33 Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression. Corollary Define the sequence of integers N π := { π, 2π, 3π,... }. Let p π (n) be the nth prime in this sequence, and so p π (1) = 3, p π (2) = 31, p π (3) = 37,.... Then for every gcd(r, t) = 1 and every k there exists an n for which p π (n) p π (n + 1) p π (n + 2) p π (n + k) r (mod t).

34 Prime Numbers More on strings of prime

35 Prime Numbers More on strings of prime Example The first 6 consecutive primes 5 (mod 7) in N π is , , , , ,

36 Prime Numbers More on strings of prime Example The first 6 consecutive primes 5 (mod 7) in N π is , , , , , Remark Special sets includes primes in the sets N α := { α, 2α, 3α,... } α irrational real alg. int., N := { n log log n : n = 1, 2, 3,... }.

37 Prime Numbers

38 Recurrence sequences The Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377

39 Recurrence sequences The Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377

40 Recurrence sequences The Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377 Conjecture (Folklore) These are the only perfect powers!

41 Recurrence sequences Perfect Powers in the Fibonacci Sequence Theorem (Siksek, Bugeaud, Mignotte, Annals of Math 2006) F 0 = 0 p, F 1 = 1 p, F 6 = 2 3, and F 12 = 12 2 are the only perfect powers in the Fibonacci sequence.

42 Recurrence sequences Perfect Powers in the Fibonacci Sequence Theorem (Siksek, Bugeaud, Mignotte, Annals of Math 2006) F 0 = 0 p, F 1 = 1 p, F 6 = 2 3, and F 12 = 12 2 are the only perfect powers in the Fibonacci sequence. Proof. Modularity of elliptic curves :-)

43 Recurrence sequences Lucas Sequences Definition (Lucas Sequence) A Lucas sequence u n is a nondegenerate integral linear binary recurrence relation defined by with u 0 = 0 and u 1 = 1. u n+2 = bu n+1 + cu n

44 Recurrence sequences Lucas Sequences Definition (Lucas Sequence) A Lucas sequence u n is a nondegenerate integral linear binary recurrence relation defined by with u 0 = 0 and u 1 = 1. u n+2 = bu n+1 + cu n Definition (Companion Sequence) The companion sequence v n is defined by v n+2 = bv n+1 + cv n with v 0 = 2 and v 1 = b.

45 Recurrence sequences Lucas Sequences A Lucas sequence (b, c) has characteristic polynomial and roots g(z) = z 2 bz c, α, β = b ± b 2 + 4c. 2

46 Recurrence sequences Lucas Sequences A Lucas sequence (b, c) has characteristic polynomial and roots g(z) = z 2 bz c, α, β = b ± b 2 + 4c. 2 Fact 1. u n = αn β n α β v n = α n + β n 2. u 2k = u k v k 3. (b 2 + 4c)u 2 n = v 2 n 4( c) n

47 Recurrence sequences Specific Examples Theorem (Silliman, Vogt) For the following values of b and c: (b, c) = (3, 2), (5, 6), (7, 12), (17, 72), (9, 20) the Lucas sequence u n has no nontrivial pth powers, except u 2 = 3 2 in (9, 20).

48 Recurrence sequences General Bound Conjecture (Frey-Mazur) Let E, E be two elliptic curves defined over Q. If E[p] = E [p] for some p > 17, then E and E are isogenous.

49 Recurrence sequences General Bound Conjecture (Frey-Mazur) Let E, E be two elliptic curves defined over Q. If E[p] = E [p] for some p > 17, then E and E are isogenous. Theorem (Silliman, Vogt) Assume the Frey Mazur conjecture. Then a pth power u n = y p satisfies p Ψ(b, c). where Ψ(b, c) is an explicit constant.

50 Recurrence sequences More Examples Example The sequence (3, 1): 0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, , We haven t found any nontrivial perfect powers...

51 Recurrence sequences More Examples Example The sequence (3, 1): 0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, , We haven t found any nontrivial perfect powers... Theorem (Silliman, Vogt) Assume the Frey-Mazur Conjecture. There are no nontrivial powers in the sequences (b, c) = (3, 1), (5, 1), and (7, 1).

52 Recurrence sequences

53 Partitions Beautiful identities

54 Partitions Beautiful identities Euler proved that q (1 q 24n ) = q q 25 q 49 + q q 169 q 289. n=1

55 Partitions Beautiful identities Euler proved that q (1 q 24n ) = q q 25 q 49 + q q 169 q 289. n=1 Jacobi proved that q (1 q 8n ) 3 = q 3q 9 + 5q 25 7q q 81 11q n=1

56 Partitions Beautiful identities Euler proved that q (1 q 24n ) = q q 25 q 49 + q q 169 q 289. n=1 Jacobi proved that q (1 q 8n ) 3 = q 3q 9 + 5q 25 7q q 81 11q n=1 Gauss proved that q n=1 (1 q 16n ) 2 (1 q 8n ) = q+q 9 +q 25 +q 49 +q 81 +q 121 +q 169 +q 225 +

57 Partitions Such rare identities have been discovered by:

58 Partitions Such rare identities have been discovered by: Crazy combinatorial manipulation of power series. Higher identities such as Jacobi s n=1(1 q 2n )(1 + z 2 q 2n 1 )(1 + z 2 q 2n 1 ) = m Z z 2m q m2. Modular forms.

59 Partitions Nekrasov-Okounkov Theory

60 Partitions Nekrasov-Okounkov Theory Deeper structure for such identities.

61 Partitions Nekrasov-Okounkov Theory Deeper structure for such identities. One doesn t have to multiply out and combine terms.

62 Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n.

63 Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n. p(n) := # {partitions of n}.

64 Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n. p(n) := # {partitions of n}. Example {Partitions of 4} = {4, 3 + 1, 2 + 2, , }

65 Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n. p(n) := # {partitions of n}. Example {Partitions of 4} = {4, 3 + 1, 2 + 2, , } = p(4) = 5.

66 Nekrasov-Okounkov Generating function for p(n)

67 Nekrasov-Okounkov Generating function for p(n) Lemma We have that p(n)q n 1 = 1 q n. n=0 n=1

68 Nekrasov-Okounkov Wishful thinking

69 Nekrasov-Okounkov Wishful thinking Question Is there a combinatorial theory of infinite products where coefficient of q n = formula in partitions of n?

70 Nekrasov-Okounkov Hooklengths of partitions

71 Nekrasov-Okounkov Hooklengths of partitions Definition Hooks are the maximal. in the Ferrers board of λ.

72 Nekrasov-Okounkov Hooklengths of partitions Definition Hooks are the maximal. in the Ferrers board of λ. Hooklengths are their lengths, and H(λ) = { multiset of hooklengths of λ}

73 Nekrasov-Okounkov An example Example (λ = )

74 Nekrasov-Okounkov An example Example (λ = ) An so H(λ) = {1, 1, 1, 2, 2, 3, 4, 4, 6, 7}.

75 Nekrasov-Okounkov Nekrasov-Okounkov q-series Definition For z C, let O z (q) := λ q λ h H(λ) (1 z h 2 ).

76 Nekrasov-Okounkov Example (λ = )

77 Nekrasov-Okounkov Example (λ = ) The λ-contribution to O z (q) is q 10 (1 z ) h 2 h H(λ) = q 10 (1 z) 3 ( 1 z 4 ) 2 ( 1 z ) ( 1 z ) 2 ( 1 z ) ( 1 z )

78 Nekrasov-Okounkov A famous identity revisited

79 Nekrasov-Okounkov A famous identity revisited Letting z = 4, we consider O 4 (q) = λ q λ h H(λ) (1 4h 2 ).

80 Nekrasov-Okounkov A famous identity revisited Letting z = 4, we consider O 4 (q) = λ q λ h H(λ) (1 4h 2 ). We only need λ where every hook h 2.

81 Nekrasov-Okounkov A famous identity revisited Letting z = 4, we consider O 4 (q) = λ q λ h H(λ) (1 4h 2 ). We only need λ where every hook h 2. = {Triangular partitions k}

82 Nekrasov-Okounkov A famous identity revisited.

83 Nekrasov-Okounkov A famous identity revisited. λ 8 λ + 1 H(λ) (1 4/h 2 ) 0 1 φ {1} {1, 1, 3} {1,..., 5} 7

84 Nekrasov-Okounkov A famous identity revisited. λ 8 λ + 1 H(λ) (1 4/h 2 ) 0 1 φ {1} {1, 1, 3} {1,..., 5} 7 And so we have = qo 4 (q 8 ) = q 3q 9 + 5q 25 7q 49 +

85 Nekrasov-Okounkov A famous identity revisited. λ 8 λ + 1 H(λ) (1 4/h 2 ) 0 1 φ {1} {1, 1, 3} {1,..., 5} 7 And so we have = qo 4 (q 8 ) = q 3q 9 + 5q 25 7q 49 + Jacobi? = q (1 q 8n ) 3 n=1

86 Nekrasov-Okounkov Big Identity

87 Nekrasov-Okounkov Big Identity Theorem (Nekrasov-Okounkov) If z is complex, then O z (q) := q λ (1 z ) h 2 = (1 q n ) z 1. λ n=1 h H(λ)

88 Nekrasov-Okounkov Big Identity Theorem (Nekrasov-Okounkov) If z is complex, then O z (q) := λ q λ (1 z ) h 2 = (1 q n ) z 1. h H(λ) n=1 Remark Letting z = 0 gives the generating function for p(n).

89 Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities

90 Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities Question 1 When do the sums below vanish? A(a, b; n) := λ =n h H a(λ) ( 1 ab ) h 2.

91 Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities Question 1 When do the sums below vanish? A(a, b; n) := λ =n h H a(λ) ( 1 ab ) h 2. 2 Are there more identities of Euler, Gauss and Jacobi-type?

92 Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities Question 1 When do the sums below vanish? A(a, b; n) := λ =n h H a(λ) ( 1 ab ) h 2. 2 Are there more identities of Euler, Gauss and Jacobi-type? 3 If so, find them all.

93 Nekrasov-Okounkov A cool theorem

94 Nekrasov-Okounkov A cool theorem Theorem (Clader, Kemper, Wage) The list of all pairs (a, b) for which almost all the A(a, b; n) vanish are (1, 2), (1, 3), (1, 4), (1, 5), (1, 7), (1, 9), (1, 11), (1, 15), (1, 27), (2, 2), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 9), (4, 5), (4, 7), (7, 9), (7, 15).

95 Nekrasov-Okounkov A cool theorem Theorem (Clader, Kemper, Wage) The list of all pairs (a, b) for which almost all the A(a, b; n) vanish are (1, 2), (1, 3), (1, 4), (1, 5), (1, 7), (1, 9), (1, 11), (1, 15), (1, 27), (2, 2), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 9), (4, 5), (4, 7), (7, 9), (7, 15). Remark These pairs are the Euler, Gauss and Jacobi identities for A(a, b; n)q n := n=1 (1 q an ) b (1 q n ).

96 Nekrasov-Okounkov

97 Number Fields Number fields Wrong Definition This is a number field.

98 Number Fields Number fields Wrong Definition This is a number field..

99 Number Fields Number fields Wrong Definition This is a number field.. Definition A finite dimensional field extension of Q is called a number field.

100 Number Fields An invariant Remark The discriminant K Z \ {0} of a number field K does:

101 Number Fields An invariant Remark The discriminant K Z \ {0} of a number field K does: Measures the volume of algebraic integers.

102 Number Fields An invariant Remark The discriminant K Z \ {0} of a number field K does: Measures the volume of algebraic integers. Controls some properties of primes (ramification).

103 Number Fields Quadratic fields If D is square-free and K := Q( D), then { D if D 1 (mod 4), K := 4D otherwise.

104 Number Fields Quadratic fields If D is square-free and K := Q( D), then { D if D 1 (mod 4), K := 4D otherwise. Example For example, we have Q(i) = = 4 Q( 2) = = 8.

105 Number Fields Distribution of quadratic fields Notation. Let N 2 (X ) := #{quad fields with X }.

106 Number Fields Distribution of quadratic fields Notation. Let N 2 (X ) := #{quad fields with X }. X N 2 (X ) N 2 (X )/X

107 Number Fields Distribution of quadratic fields Notation. Let N 2 (X ) := #{quad fields with X }. X N 2 (X ) N 2 (X )/X Theorem (Easy) We have that N 2 (X ) lim = 6 X + X π 2 =

108 Number Fields General Number Fields Conjecture (Linnik) If we let N n (X ) := #{degree n fields with X }, then N n (X ) c n X.

109 Number Fields General Number Fields Conjecture (Linnik) If we let N n (X ) := #{degree n fields with X }, then N n (X ) c n X. Theorem (Easy, Davenport-Heilbronn (70s), Bhargava (2000s)) Linnik s Conjecture is true for n = 2, 3, 4, 5.

110 Number Fields General results

111 Number Fields General results Theorem (Schmidt (1995), Ellenberg-Venkatesh (2006)) We have that X n+2 4 if 6 n 84393, N n (X ) n X exp(c log n) if n >

112 Number Fields Related Functions Notation. If G is a finite group, then let N n (G; X ) := number of deg n fields with X whose Galois closure has Galois group G

113 Number Fields Related Functions Notation. If G is a finite group, then let N n (G; X ) := number of deg n fields with X whose Galois closure has Galois group G Remark Estimating N n (G; X ) is easy only for n = 2 and G = Z/2Z.

114 Number Fields Prime cyclic cases Theorem (Wright (1989)) If p is an odd prime, then there is a constant c(p) for which N p (Z/pZ; X ) = c(p) X 1 p 1 + O(X 1/p ).

115 Number Fields Prime cyclic cases Theorem (Wright (1989)) If p is an odd prime, then there is a constant c(p) for which N p (Z/pZ; X ) = c(p) X 1 p 1 + O(X 1/p ). Conjecture (Cohen, Diaz y Diaz, Olivier (2006)) We have that N 3 (Z/3Z; X ) = c(3) X O(X 1/6 ).

116 Number Fields A cool theorem...

117 Number Fields A cool theorem... Theorem (Lee-Oh) Assuming the GRH, if p is an odd prime, then N p (Z/pZ; X ) = c(p) X 1 p 1 + X 1 3(p 1) R p (log X ) + O(X 1 4(p 1) ), where R p (x) is a polynomial of degree p(p 2)/3 1.

118 Number Fields A cool theorem... Theorem (Lee-Oh) Assuming the GRH, if p is an odd prime, then N p (Z/pZ; X ) = c(p) X 1 p 1 + X 1 3(p 1) R p (log X ) + O(X 1 4(p 1) ), where R p (x) is a polynomial of degree p(p 2)/3 1. Corollary (Lee-Oh) Assuming GRH, the Cohen, Diaz y Diaz, Olivier Conjecture is true.

119 Number Fields Linear regression when p = 3: Slope 1 6

120 Number Fields Linear regression when p = 3: Slope x axis = log 10 (X ) y axis = log 10 (N 3 (Z/3Z; X ) c(3) X 1 2 )

121 In conclusion... Our summer toys include...

122 In conclusion... Our summer toys include... (Prime numbers) (Partitions) (Numbers and Number Fields)

123 In conclusion... So, we basically...

124 In conclusion... So, we basically......prove theorems, and...explode stuff...