Definitions Application Quaternions HMF Application II
Half integral weight modular forms
Ariel Pacetti
Universidad de Buenos Aires
Explicit Methods for Modular Forms March 20, 2013
Ariel Pacetti Half integral weight modular forms The Dedekind eta function ∞ πiz Y 2πinz η(z) = e 12 (1 − e ). n=1
It is well know that η(z)24 = ∆(z) a weight 12 cusp form, so η “should be” of weight 1/2. Actually η turns out to be weight 1/2 but with a character of order 24.
We can consider two classical examples:
Definitions Application Quaternions HMF Application II
Motivation
What is a half integral modular form?
Ariel Pacetti Half integral weight modular forms The Dedekind eta function ∞ πiz Y 2πinz η(z) = e 12 (1 − e ). n=1
It is well know that η(z)24 = ∆(z) a weight 12 cusp form, so η “should be” of weight 1/2. Actually η turns out to be weight 1/2 but with a character of order 24.
Definitions Application Quaternions HMF Application II
Motivation
What is a half integral modular form? We can consider two classical examples:
Ariel Pacetti Half integral weight modular forms It is well know that η(z)24 = ∆(z) a weight 12 cusp form, so η “should be” of weight 1/2. Actually η turns out to be weight 1/2 but with a character of order 24.
Definitions Application Quaternions HMF Application II
Motivation
What is a half integral modular form? We can consider two classical examples: The Dedekind eta function ∞ πiz Y 2πinz η(z) = e 12 (1 − e ). n=1
Ariel Pacetti Half integral weight modular forms Actually η turns out to be weight 1/2 but with a character of order 24.
Definitions Application Quaternions HMF Application II
Motivation
What is a half integral modular form? We can consider two classical examples: The Dedekind eta function ∞ πiz Y 2πinz η(z) = e 12 (1 − e ). n=1
It is well know that η(z)24 = ∆(z) a weight 12 cusp form, so η “should be” of weight 1/2.
Ariel Pacetti Half integral weight modular forms Definitions Application Quaternions HMF Application II
Motivation
What is a half integral modular form? We can consider two classical examples: The Dedekind eta function ∞ πiz Y 2πinz η(z) = e 12 (1 − e ). n=1
It is well know that η(z)24 = ∆(z) a weight 12 cusp form, so η “should be” of weight 1/2. Actually η turns out to be weight 1/2 but with a character of order 24.
Ariel Pacetti Half integral weight modular forms a b It is not hard to see that if γ = c d ∈ Γ0(4), then
θ(γz)2 −1 = (cz + d). θ(z) d
2 So θ(z) ∈ M1(Γ0(4), χ−1).
Definitions Application Quaternions HMF Application II
Motivation
The classical theta function ∞ X 2 θ(z) = e2πin z . n=−∞
Ariel Pacetti Half integral weight modular forms 2 So θ(z) ∈ M1(Γ0(4), χ−1).
Definitions Application Quaternions HMF Application II
Motivation
The classical theta function ∞ X 2 θ(z) = e2πin z . n=−∞