Moonshine: Lecture 3. Moonshine: Lecture 3. Ken Ono (Emory University)
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1 Ken Ono (Emory University)
2 I m going to talk about...
3 I m going to talk about... I. History of Moonshine
4 I m going to talk about... I. History of Moonshine II. Distribution of Monstrous Moonshine
5 I m going to talk about... I. History of Moonshine II. Distribution of Monstrous Moonshine III. Umbral Moonshine
6 I. History of Moonshine The Monster
7 I. History of Moonshine The Monster Conjecture (Fischer and Griess (1973)) There is a huge simple group (containing a double cover of Fischer s B) with order
8 I. History of Moonshine The Monster Conjecture (Fischer and Griess (1973)) There is a huge simple group (containing a double cover of Fischer s B) with order Theorem (Griess (1982)) The Monster group M exists.
9 I. History of Moonshine Classification of Finite Simple Groups
10 I. History of Moonshine Classification of Finite Simple Groups Theorem (Classification of Finite Simple Groups) Finite simple groups live in natural infinite families, apart from the sporadic groups.
11 I. History of Moonshine Classification of Finite Simple Groups Theorem (Classification of Finite Simple Groups) Finite simple groups live in natural infinite families, apart from the sporadic groups.
12 I. History of Moonshine Ogg s Theorem
13 I. History of Moonshine Ogg s Theorem Theorem (Ogg, 1974) X 0 (N) est hyperelliptique pour exactement dix-neuf valuers de N.
14 I. History of Moonshine Ogg s Theorem Theorem (Ogg, 1974) X 0 (N) est hyperelliptique pour exactement dix-neuf valuers de N. Corollary (Ogg, 1974) Toutes les valuers supersingulières de j sont F p si, et seulement si, g + = 0,
15 I. History of Moonshine Ogg s Theorem Theorem (Ogg, 1974) X 0 (N) est hyperelliptique pour exactement dix-neuf valuers de N. Corollary (Ogg, 1974) Toutes les valuers supersingulières de j sont F p si, et seulement si, g + = 0, i.e. p Ogg ss := {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 47, 59, 71}.
16 I. History of Moonshine Ogg s Jack Daniels Problem
17 I. History of Moonshine Ogg s Jack Daniels Problem Remark This is the first hint of Moonshine.
18 I. History of Moonshine Second hint of moonshine John McKay observed that =
19 I. History of Moonshine John Thompson s generalizations Thompson further observed: = = = }{{} 1 } {{ } Coefficients of j(τ) Dimensions of irreducible representations of the Monster M
20 I. History of Moonshine Klein s j-function Definition Klein s j-function j(τ) 744 = n= 1 c(n)q n = q q q q satisfies ( ) aτ + b j = j(τ) for every matrix cτ + d ( ) a b SL c d 2 (Z).
21 I. History of Moonshine The Monster characters The character table for M (ordered by size) gives dimensions: χ 1 (e) = 1 χ 2 (e) = χ 3 (e) = χ 4 (e) = χ 194 (e) =
22 I. History of Moonshine Monster module Conjecture (Thompson) There is an infinite-dimensional graded module V = n= 1 V n with dim(v n) = c(n).
23 I. History of Moonshine The McKay-Thompson Series Definition (Thompson) Assuming the conjecture, if g M, then define the McKay Thompson series T g (τ) := tr(g Vn)q n. n= 1
24 I. History of Moonshine Conway and Norton Conjecture (Monstrous Moonshine) For each g M there is an explicit genus 0 discrete subgroup Γ g SL 2 (R)
25 I. History of Moonshine Conway and Norton Conjecture (Monstrous Moonshine) For each g M there is an explicit genus 0 discrete subgroup Γ g SL 2 (R) for which T g (τ) is the unique modular function with T g (τ) = q 1 + O(q).
26 I. History of Moonshine Borcherds work Theorem (Frenkel Lepowsky Meurman) The moonshine module V = n= 1 V n is a vertex operator algebra whose graded dimension is given by j(τ) 744, and whose automorphism group is M.
27 I. History of Moonshine Borcherds work Theorem (Frenkel Lepowsky Meurman) The moonshine module V = n= 1 V n is a vertex operator algebra whose graded dimension is given by j(τ) 744, and whose automorphism group is M. Theorem (Borcherds) The Monstrous Moonshine Conjecture is true.
28 I. History of Moonshine Borcherds work Theorem (Frenkel Lepowsky Meurman) The moonshine module V = n= 1 V n is a vertex operator algebra whose graded dimension is given by j(τ) 744, and whose automorphism group is M. Theorem (Borcherds) The Monstrous Moonshine Conjecture is true. Remark Earlier work of Atkin, Fong and Smith numerically confirmed Monstrous moonshine.
29 The Jack Daniels Problem Ogg s Problem
30 The Jack Daniels Problem Ogg s Problem Question A Do order p elements in M know the F p supersingular j-invariants?
31 The Jack Daniels Problem Ogg s Problem Question A Do order p elements in M know the F p supersingular j-invariants? Theorem (Dwork s Generating Function) If p 5 is prime, then (j(τ) 744) U(p) A p (α) j(τ) α α SS p g(x) SS p B p (g)j(τ) + C p (g) g(j(τ)) (mod p).
32 The Jack Daniels Problem Answer to Question A
33 The Jack Daniels Problem Answer to Question A If g M and p is prime, then Moonshine implies that T g + pt g U(p) = T g p.
34 The Jack Daniels Problem Answer to Question A If g M and p is prime, then Moonshine implies that T g + pt g U(p) = T g p. And so if g has order p, then T g + pt g U(p) = j 744.
35 The Jack Daniels Problem Answer to Question A If g M and p is prime, then Moonshine implies that T g + pt g U(p) = T g p. And so if g has order p, then T g + pt g U(p) = j 744. Which implies that T g j 744 (mod p).
36 The Jack Daniels Problem Answer to Question A If g M and p is prime, then Moonshine implies that T g + pt g U(p) = T g p. And so if g has order p, then T g + pt g U(p) = j 744. Which implies that T g j 744 (mod p)....giving us Dwork s generating function T g U(p) (j 744) U(p) (mod p).
37 The Jack Daniels Problem Ogg s Problem Question B If p Ogg ss, then why do we expect p #M?
38 The Jack Daniels Problem Ogg s Problem Question B If p Ogg ss, then why do we expect p #M? Answer By Moonshine, if g M has order p, then Γ g Γ + 0 (p) has genus 0.
39 The Jack Daniels Problem Ogg s Problem Question B If p Ogg ss, then why do we expect p #M? Answer By Moonshine, if g M has order p, then Γ g Γ + 0 (p) has genus 0. By Ogg, if p Ogg ss, then X 0 + (p) has positive genus.
40 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M?
41 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine implies j (h p U(p)) comes from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
42 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine implies j (h p U(p)) comes from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
43 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine implies j (h p U(p)) comes from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion lag s suffice iff p Ogg ss.
44 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine implies j (h p U(p)) comes from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
45 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine expects j (h p U(p)) to come from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
46 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine expects j (h p U(p)) to come from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
47 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine expects j (h p U(p)) to come from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
48 The Jack Daniels Problem Ogg s Problem Question C If p Ogg ss, then why do we expect (a priori) that p #M? Heuristic Argument? Let h p (τ) be the hauptmodul for Γ + 0 (p). Hecke implies that h p U(p) (j 744) U(p) (mod p). Deligne for E p 1 gives h p U(p) S p 1 (1) (mod p). Implies j (h p U(p)) S p+1 (1) (mod p). Moonshine expects j (h p U(p)) to come from Θ s. But Serre implies j (h p U(p)) S 2 (p) (mod p). We expect S 2 (p) (mod p) to be spanned by Θ s. Pizer proved Θ s from quaternion alg s suffice iff p Ogg ss.
49 The Jack Daniels Problem Witten s Conjecture (2007) Conjecture (Witten, Li-Song-Strominger) The vertex operator algebra V is dual to a 3d quantum gravity theory. Thus, there are 194 black hole states.
50 The Jack Daniels Problem Witten s Conjecture (2007) Conjecture (Witten, Li-Song-Strominger) The vertex operator algebra V is dual to a 3d quantum gravity theory. Thus, there are 194 black hole states. Question (Witten) How are these different kinds of black hole states distributed?
51 II. Distribution of Monstrous Moonshine Distribution of Monstrous Moonshine
52 II. Distribution of Monstrous Moonshine Open Problem Question Consider the moonshine expressions = = = c(n) = 194 m i (n)χ i (e) i=1 How many 1 s, s, etc. show up in these equations?
53 II. Distribution of Monstrous Moonshine Some Proportions n δ (m 1 (n)) δ (m 2 (n)) δ (m 194 (n)) /2 1/
54 II. Distribution of Monstrous Moonshine Some Proportions n δ (m 1 (n)) δ (m 2 (n)) δ (m 194 (n)) /2 1/
55 II. Distribution of Monstrous Moonshine Some Proportions n δ (m 1 (n)) δ (m 2 (n)) δ (m 194 (n)) /2 1/
56 II. Distribution of Monstrous Moonshine Some Proportions n δ (m 1 (n)) δ (m 2 (n)) δ (m 194 (n)) /2 1/
57 II. Distribution of Monstrous Moonshine Some Proportions n δ (m 1 (n)) δ (m 2 (n)) δ (m 194 (n)) /2 1/
58 II. Distribution of Monstrous Moonshine Distribution of Moonshine Theorem 1 (Duncan, Griffin, O) We have Rademacher style exact formulas for m i (n) of the form m i (n) = Kloosterman sums I -Bessel fcns χ i c
59 II. Distribution of Monstrous Moonshine Distribution of Moonshine Theorem 1 (Duncan, Griffin, O) We have Rademacher style exact formulas for m i (n) of the form m i (n) = Kloosterman sums I -Bessel fcns χ i c Remark The dominant term gives m i (n) dim(χ i) 2 n 3/4 M e4π n
60 II. Distribution of Monstrous Moonshine Distribution Remark We have that is well defined δ (m i ) := lim n + m i (n) 194 i=1 m i(n)
61 II. Distribution of Monstrous Moonshine Distribution Remark We have that is well defined, and δ (m i ) = δ (m i ) := lim n + m i (n) 194 i=1 m i(n) dim(χ i ) 194 j=1 dim(χ j) = dim(χ i )
62 II. Distribution of Monstrous Moonshine Proof Orthogonality Fact If G is a group and g, h G, then { C G (g) If g and h are conjugate χ i (g)χ i (h) = χ i 0 otherwise, where C G (g) is the centralizer of g in G.
63 II. Distribution of Monstrous Moonshine Proof T χ (τ) Define T χi (τ) = 1 χ i (g)t g (τ). M g M
64 II. Distribution of Monstrous Moonshine Proof T χ (τ) Define T χi (τ) = 1 χ i (g)t g (τ). M g M The orthogonality of characters gives the inverse relation 194 T g (τ) = χ i (g)t χi (τ). i=1
65 II. Distribution of Monstrous Moonshine Proof T χ (τ) Define T χi (τ) = 1 χ i (g)t g (τ). M g M The orthogonality of characters gives the inverse relation 194 T g (τ) = χ i (g)t χi (τ). i=1 From this we can work out that T χi (τ) = m i (n)q n. n= 1
66 II. Distribution of Monstrous Moonshine Proof Outline of the proof
67 II. Distribution of Monstrous Moonshine Proof Outline of the proof Theorem 2 (Duncan, Griffin,O) We have exact formulas for the coeffs of the T g (τ) and T χi (τ).
68 II. Distribution of Monstrous Moonshine Proof Outline of the proof Theorem 2 (Duncan, Griffin,O) We have exact formulas for the coeffs of the T g (τ) and T χi (τ). Sketch of Proof 1 Each Γ g contains some congruence subgroup Γ 0 (N g ).
69 II. Distribution of Monstrous Moonshine Proof Outline of the proof Theorem 2 (Duncan, Griffin,O) We have exact formulas for the coeffs of the T g (τ) and T χi (τ). Sketch of Proof 1 Each Γ g contains some congruence subgroup Γ 0 (N g ). 2 We first find the poles of T g (τ).
70 II. Distribution of Monstrous Moonshine Proof Outline of the proof Theorem 2 (Duncan, Griffin,O) We have exact formulas for the coeffs of the T g (τ) and T χi (τ). Sketch of Proof 1 Each Γ g contains some congruence subgroup Γ 0 (N g ). 2 We first find the poles of T g (τ). 3 We can then build a form with matching poles via Poincaré series.
71 II. Distribution of Monstrous Moonshine Proof Outline of the proof Theorem 2 (Duncan, Griffin,O) We have exact formulas for the coeffs of the T g (τ) and T χi (τ). Sketch of Proof 1 Each Γ g contains some congruence subgroup Γ 0 (N g ). 2 We first find the poles of T g (τ). 3 We can then build a form with matching poles via Poincaré series. 4 The Poincaré series give exact formulas for coefficients.
72 III. Umbral Moonshine Umbral (shadow) Moonshine
73 III. Umbral Moonshine Present day moonshine Observation (Eguchi, Ooguri, Tachikawa (2010)) There is a mock modular form H(τ) = q ( q + 231q q q q ).
74 III. Umbral Moonshine Present day moonshine Observation (Eguchi, Ooguri, Tachikawa (2010)) There is a mock modular form H(τ) = q ( q + 231q q q q ). The degrees of the irreducible repn s of the Mathieu group M 24 are: 1, 23,45, 231, 252, 253, 483, 770, 990, 1035, 1265, 1771, 2024, 2277, 3312, 3520, 5313, 5544, 5796,
75 III. Umbral Moonshine Mathieu Moonshine Theorem (Gannon (2013)) There is an infinite dimensional graded M 24 -module whose McKay-Thompson series are specific mock modular forms.
76 III. Umbral Moonshine Mathieu Moonshine Theorem (Gannon (2013)) There is an infinite dimensional graded M 24 -module whose McKay-Thompson series are specific mock modular forms. Remark Alleged multiplicities must be integral and non-negative.
77 III. Umbral Moonshine Mathieu Moonshine Theorem (Gannon (2013)) There is an infinite dimensional graded M 24 -module whose McKay-Thompson series are specific mock modular forms. Remark Alleged multiplicities must be integral and non-negative. Computed using wgt 1/2 weakly holomorphic modular forms.
78 III. Umbral Moonshine Mathieu Moonshine Theorem (Gannon (2013)) There is an infinite dimensional graded M 24 -module whose McKay-Thompson series are specific mock modular forms. Remark Alleged multiplicities must be integral and non-negative. Computed using wgt 1/2 weakly holomorphic modular forms. Integrality follows from theory of modular forms mod p.
79 III. Umbral Moonshine Mathieu Moonshine Theorem (Gannon (2013)) There is an infinite dimensional graded M 24 -module whose McKay-Thompson series are specific mock modular forms. Remark Alleged multiplicities must be integral and non-negative. Computed using wgt 1/2 weakly holomorphic modular forms. Integrality follows from theory of modular forms mod p. Non-negativity follows from effectivizing argument of Bringmann-O on Ramanujan s f (q) mock theta function.
80 III. Umbral Moonshine Mock modular forms What are mock modular forms? Notation. Throughout, let τ = x + iy H with x, y R.
81 III. Umbral Moonshine Mock modular forms What are mock modular forms? Notation. Throughout, let τ = x + iy H with x, y R. Hyperbolic Laplacian. ( ) ( ) k := y 2 2 x y 2 + iky x + i. y
82 III. Umbral Moonshine Mock modular forms Harmonic Maass forms Definition A harmonic Maass form of weight k on a subgroup Γ SL 2 (Z) is any smooth function M : H C satisfying:
83 III. Umbral Moonshine Mock modular forms Harmonic Maass forms Definition A harmonic Maass form of weight k on a subgroup Γ SL 2 (Z) is any smooth function M : H C satisfying: 1 For all A = ( ) a b c d Γ and z H, we have ( ) aτ + b M = (cz + d) k M(τ). cτ + d
84 III. Umbral Moonshine Mock modular forms Harmonic Maass forms Definition A harmonic Maass form of weight k on a subgroup Γ SL 2 (Z) is any smooth function M : H C satisfying: 1 For all A = ( ) a b c d Γ and z H, we have ( ) aτ + b M = (cz + d) k M(τ). cτ + d 2 We have that k M = 0.
85 III. Umbral Moonshine Mock modular forms Fourier expansions
86 III. Umbral Moonshine Mock modular forms Fourier expansions Fundamental Lemma If M H 2 k and Γ(a, x) is the incomplete Γ-function, then M(τ) = c + (n)q n + c (n)γ(k 1, 4π n y)q n. n n<0 Holomorphic part M + Nonholomorphic part M
87 III. Umbral Moonshine Mock modular forms Fourier expansions Fundamental Lemma If M H 2 k and Γ(a, x) is the incomplete Γ-function, then M(τ) = c + (n)q n + c (n)γ(k 1, 4π n y)q n. n n<0 Holomorphic part M + Nonholomorphic part M Remark We call M + a mock modular form.
88 III. Umbral Moonshine Mock modular forms Fourier expansions Fundamental Lemma If M H 2 k and Γ(a, x) is the incomplete Γ-function, then M(τ) = c + (n)q n + c (n)γ(k 1, 4π n y)q n. n n<0 Holomorphic part M + Nonholomorphic part M Remark We call M + a mock modular form. If ξ 2 k := 2iy 2 k τ, then the shadow of M is ξ 2 k(m ).
89 III. Umbral Moonshine Mock modular forms Shadows are modular forms Fundamental Lemma The operator ξ 2 k := 2iy 2 k τ defines a surjective map ξ 2 k : H 2 k S k.
90 III. Umbral Moonshine Mock modular forms Shadows are modular forms Fundamental Lemma The operator ξ 2 k := 2iy 2 k τ defines a surjective map ξ 2 k : H 2 k S k. Remark In M 24 Moonshine, the McKay-Thompson series are mock modular forms with classical Jacobi theta series shadows!
91 III. Umbral Moonshine Mock modular forms Larger Framework of Moonshine? Remark There are well known connections with even unimodular positive definite rank 24 lattices: M Leech lattice M 24 A 24 1 lattice.
92 III. Umbral Moonshine Mock modular forms Umbral Moonshine Conjecture Conjecture (Cheng, Duncan, Harvey (2013)) Let L X (up to isomorphism) be an even unimodular positive-definite rank 24 lattice, and let : X be the corresponding ADE-type root system.
93 III. Umbral Moonshine Mock modular forms Umbral Moonshine Conjecture Conjecture (Cheng, Duncan, Harvey (2013)) Let L X (up to isomorphism) be an even unimodular positive-definite rank 24 lattice, and let : X be the corresponding ADE-type root system. W X the Weyl group of X.
94 III. Umbral Moonshine Mock modular forms Umbral Moonshine Conjecture Conjecture (Cheng, Duncan, Harvey (2013)) Let L X (up to isomorphism) be an even unimodular positive-definite rank 24 lattice, and let : X be the corresponding ADE-type root system. W X the Weyl group of X. The umbral group G X := Aut(L X )/W X.
95 III. Umbral Moonshine Mock modular forms Umbral Moonshine Conjecture Conjecture (Cheng, Duncan, Harvey (2013)) Let L X (up to isomorphism) be an even unimodular positive-definite rank 24 lattice, and let : X be the corresponding ADE-type root system. W X the Weyl group of X. The umbral group G X := Aut(L X )/W X. For each g G X let H X g (τ) be a specific automorphic form with minimal principal parts.
96 III. Umbral Moonshine Mock modular forms Umbral Moonshine Conjecture Conjecture (Cheng, Duncan, Harvey (2013)) Let L X (up to isomorphism) be an even unimodular positive-definite rank 24 lattice, and let : X be the corresponding ADE-type root system. W X the Weyl group of X. The umbral group G X := Aut(L X )/W X. For each g G X let H X g (τ) be a specific automorphic form with minimal principal parts. Then there is an infinite dimensional graded G X module K X for which H X g (τ) is the McKay-Thompson series for g.
97 III. Umbral Moonshine Mock modular forms Remarks
98 III. Umbral Moonshine Mock modular forms Remarks Cheng, Duncan and Harvey constructed their mock modular forms using Rademacher sums.
99 III. Umbral Moonshine Mock modular forms Remarks Cheng, Duncan and Harvey constructed their mock modular forms using Rademacher sums. For X = A 12 2 we have G X = M 24 and Gannon s Theorem.
100 III. Umbral Moonshine Mock modular forms Remarks Cheng, Duncan and Harvey constructed their mock modular forms using Rademacher sums. For X = A 12 2 we have G X = M 24 and Gannon s Theorem. There are 22 other isomorphism classes of X, the H X g (τ) constructed from X and its Coxeter number m(x ).
101 III. Umbral Moonshine Mock modular forms Remarks Cheng, Duncan and Harvey constructed their mock modular forms using Rademacher sums. For X = A 12 2 we have G X = M 24 and Gannon s Theorem. There are 22 other isomorphism classes of X, the H X g (τ) constructed from X and its Coxeter number m(x ). Remark Apart from the Leech case, the H X g (τ) are always weight 1/2 mock modular forms whose shadows are weight 3/2 cuspidal theta series with level m(x ).
102 III. Umbral Moonshine Mock modular forms Our results...
103 III. Umbral Moonshine Mock modular forms Our results... Theorem 3 (Duncan, Griffin, Ono) The Umbral Moonshine Conjecture is true.
104 III. Umbral Moonshine Mock modular forms Our results... Theorem 3 (Duncan, Griffin, Ono) The Umbral Moonshine Conjecture is true. Remark This result is a numerical proof of Umbral moonshine. It is analogous to the work of Atkin, Fong and Smith in the case of monstrous moonshine.
105 III. Umbral Moonshine Mock modular forms Beautiful examples
106 III. Umbral Moonshine Mock modular forms Beautiful examples Example For M 12 the MT series include Ramanujan s mock thetas: f (q) = 1 + φ(q) = 1 + χ(q) = 1 + n=1 n=1 n=1 q n2 (1 + q) 2 (1 + q 2 ) 2 (1 + q n ) 2, q n2 (1 + q 2 )(1 + q 4 ) (1 + q 2n ), q n2 (1 q + q 2 )(1 q 2 + q 4 ) (1 q n + q 2n )
107 III. Umbral Moonshine Sketch of the proof Strategy of Proof
108 III. Umbral Moonshine Sketch of the proof Strategy of Proof For each X we compute non-negative integers m X i (n) for which K X = m X i (n)v χi. χ i n= 1
109 III. Umbral Moonshine Sketch of the proof Orthogonality Fact If G is a group and g, h G, then { C G (g) If g and h are conjugate χ i (g)χ i (h) = χ i 0 otherwise, where C G (g) is the centralizer of g in G.
110 III. Umbral Moonshine Sketch of the proof T X χ (τ) Define the weight 1/2 harmonic Maass form Tχ X i (τ) := 1 χ i (g)hg X (τ). M g G X
111 III. Umbral Moonshine Sketch of the proof T X χ (τ) Define the weight 1/2 harmonic Maass form T X χ i (τ) := 1 M g G X χ i (g)h X g (τ). The orthogonality of characters gives the inverse relation H X g (τ) = χ i (g)t X χ i (τ).
112 III. Umbral Moonshine Sketch of the proof T X χ (τ) Define the weight 1/2 harmonic Maass form T X χ i (τ) := 1 M g G X χ i (g)h X g (τ). The orthogonality of characters gives the inverse relation H X g (τ) = χ i (g)t X χ i (τ). It would be great if Tχ X i (τ) = m X i (n)q n. n= 1
113 III. Umbral Moonshine Sketch of the proof T X χ (τ) Define the weight 1/2 harmonic Maass form T X χ i (τ) := 1 M g G X χ i (g)h X g (τ). The orthogonality of characters gives the inverse relation H X g (τ) = χ i (g)t X χ i (τ). It would be great if Tχ X i (τ) = m X i (n)q n. n= 1 We only need to establish integrality and non-negativity!
114 III. Umbral Moonshine Sketch of the proof Difficulties
115 III. Umbral Moonshine Sketch of the proof Difficulties We have that T X χ i (τ) = period integral of a Θ-function + m X i (n)q n. n= 1
116 III. Umbral Moonshine Sketch of the proof Difficulties We have that T X χ i (τ) = period integral of a Θ-function + n= 1 m X i (n)q n. Method of holomorphic projection gives: π hol : H 1 2 M 2 = {wgt 2 quasimodular forms}.
117 III. Umbral Moonshine Sketch of the proof Holomorphic projection
118 III. Umbral Moonshine Sketch of the proof Holomorphic projection Definition Let f be a wgt k 2 (not necessarily holomorphic) modular form f (τ) = n Z a f (n, y)q n.
119 III. Umbral Moonshine Sketch of the proof Holomorphic projection Definition Let f be a wgt k 2 (not necessarily holomorphic) modular form f (τ) = n Z a f (n, y)q n. We say that f is admissible if the following hold:
120 III. Umbral Moonshine Sketch of the proof Holomorphic projection Definition Let f be a wgt k 2 (not necessarily holomorphic) modular form f (τ) = n Z a f (n, y)q n. We say that f is admissible if the following hold: 1 At cusps we have f (γ 1 w) ( d where w = γτ, dw τ) k 2 = c 0 + O(Im(w) ɛ ),
121 III. Umbral Moonshine Sketch of the proof Holomorphic projection Definition Let f be a wgt k 2 (not necessarily holomorphic) modular form f (τ) = n Z a f (n, y)q n. We say that f is admissible if the following hold: 1 At cusps we have f (γ 1 w) ( d where w = γτ, dw τ) k 2 a f (n, y) = O(y 2 k ) as y 0 for all n > 0. 2 = c 0 + O(Im(w) ɛ ),
122 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Definition If f is an admissible wgt k 2 nonholomorphic modular form, then its holomorphic projection is
123 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Definition If f is an admissible wgt k 2 nonholomorphic modular form, then its holomorphic projection is (π hol f )(τ) := (π k hol f )(τ) := c 0 + c(n)q n, n=1
124 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Definition If f is an admissible wgt k 2 nonholomorphic modular form, then its holomorphic projection is (π hol f )(τ) := (π k hol f )(τ) := c 0 + where for n > 0 we have c(n) = (4πn)k 1 (k 2)! 0 c(n)q n, n=1 a f (n, y)e 4πny y k 2 dy.
125 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued
126 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Fundamental Lemma If f is an admissible wgt k 2 nonholomorphic modular form on Γ 0 (N), then the following are true.
127 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Fundamental Lemma If f is an admissible wgt k 2 nonholomorphic modular form on Γ 0 (N), then the following are true. 1 If f is holomorphic, then π hol (f ) = f (τ).
128 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Fundamental Lemma If f is an admissible wgt k 2 nonholomorphic modular form on Γ 0 (N), then the following are true. 1 If f is holomorphic, then π hol (f ) = f (τ). 2 The function π hol (f ) lies in the space M k (Γ 0 (N)).
129 III. Umbral Moonshine Sketch of the proof Holomorphic projection continued Fundamental Lemma If f is an admissible wgt k 2 nonholomorphic modular form on Γ 0 (N), then the following are true. 1 If f is holomorphic, then π hol (f ) = f (τ). 2 The function π hol (f ) lies in the space M k (Γ 0 (N)). Remark Holomorphic projections appeared earlier in works of Sturm, and Gross-Zagier, and work of Imamoglu, Raum, and Richter, Mertens, and Zwegers in connection with mock modular forms.
130 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ).
131 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ). Up to the nonholomorphic part, they are the multiplicity gen fcns for V X χ i of the alleged moonshine modules K X.
132 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ). Up to the nonholomorphic part, they are the multiplicity gen fcns for V X χ i of the alleged moonshine modules K X. Compute holomorphic projections of products with shadows.
133 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ). Up to the nonholomorphic part, they are the multiplicity gen fcns for V X χ i of the alleged moonshine modules K X. Compute holomorphic projections of products with shadows. The m X χ i (n) are integers iff these holomorphic projections satisfy certain congruences.
134 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ). Up to the nonholomorphic part, they are the multiplicity gen fcns for V X χ i of the alleged moonshine modules K X. Compute holomorphic projections of products with shadows. The m X χ i (n) are integers iff these holomorphic projections satisfy certain congruences. The m X χ i (n) can be estimated using infinite sums of Kloosterman sums weighted by I -Bessel functions.
135 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ). Up to the nonholomorphic part, they are the multiplicity gen fcns for V X χ i of the alleged moonshine modules K X. Compute holomorphic projections of products with shadows. The m X χ i (n) are integers iff these holomorphic projections satisfy certain congruences. The m X χ i (n) can be estimated using infinite sums of Kloosterman sums weighted by I -Bessel functions. For sufficiently large n this establishes non-negativity.
136 III. Umbral Moonshine Sketch of the proof Sketch of the proof of umbral moonshine Compute each wgt 1/2 harmonic Maass form T X χ i (τ). Up to the nonholomorphic part, they are the multiplicity gen fcns for V X χ i of the alleged moonshine modules K X. Compute holomorphic projections of products with shadows. The m X χ i (n) are integers iff these holomorphic projections satisfy certain congruences. The m X χ i (n) can be estimated using infinite sums of Kloosterman sums weighted by I -Bessel functions. For sufficiently large n this establishes non-negativity. Check the finitely many (less than 400) cases directly.
137 Executive Summary Distribution of Monstrous Moonshine Theorem 1 (Duncan, Griffin, O) If 1 i 194, then we have Rademacher style exact formulas for m i (n).
138 Executive Summary Distribution of Monstrous Moonshine Theorem 1 (Duncan, Griffin, O) If 1 i 194, then we have Rademacher style exact formulas for m i (n). Remark We have that δ (m i ) = dim(χ i ) 194 j=1 dim(χ j) = dim(χ i )
139 Executive Summary Umbral Moonshine Theorem 3 (Duncan, Griffin, Ono) The Umbral Moonshine Conjecture is true.
140 Executive Summary Umbral Moonshine Theorem 3 (Duncan, Griffin, Ono) The Umbral Moonshine Conjecture is true. Question Do the infinite dimensional graded G X modules K X exhibit deep structure?
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