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Download A First Course in Modular Forms (Graduate Texts in by Fred Diamond, Jerry Shurman PDF

By Fred Diamond, Jerry Shurman

This e-book introduces the idea of modular types, from which all rational elliptic curves come up, with an eye fixed towards the Modularity Theorem. dialogue covers elliptic curves as advanced tori and as algebraic curves; modular curves as Riemann surfaces and as algebraic curves; Hecke operators and Atkin-Lehner idea; Hecke eigenforms and their mathematics homes; the Jacobians of modular curves and the Abelian kinds linked to Hecke eigenforms. because it provides those rules, the ebook states the Modularity Theorem in a variety of varieties, touching on them to one another and relating their purposes to quantity concept. The authors imagine no heritage in algebraic quantity conception and algebraic geometry. routines are incorporated.

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Let N be a positive integer. (a) The moduli space for Γ0 (N ) is S0 (N ) = {[Eτ , 1/N + Λτ ] : τ ∈ H}. 5 Modular curves and moduli spaces 39 Two points [Eτ , 1/N + Λτ ] and [Eτ , 1/N + Λτ ] are equal if and only if Γ0 (N )τ = Γ0 (N )τ . Thus there is a bijection ∼ ψ0 : S0 (N ) −→ Y0 (N ), [C/Λτ , 1/N + Λτ ] → Γ0 (N )τ. (b) The moduli space for Γ1 (N ) is S1 (N ) = {[Eτ , 1/N + Λτ ] : τ ∈ H}. Two points [Eτ , 1/N + Λτ ] and [Eτ , 1/N + Λτ ] are equal if and only if Γ1 (N )τ = Γ1 (N )τ . Thus there is a bijection ∼ ψ1 : S1 (N ) −→ Y1 (N ), [C/Λτ , 1/N + Λτ ] → Γ1 (N )τ.

If N is the order of K as a subgroup then K ⊂ E[N ] ∼ = Z/N Z × Z/N Z, and so by the theory of finite Abelian groups K ∼ = Z/nZ × Z/nn Z for some positive integers n and n . The multiply-by-n isogeny [n] of C/Λ takes K to a cyclic subgroup nK isomorphic to Z/n Z, and then the quotient isogeny π from C/Λ to C/nK has kernel nK. Follow this by the map C/nK −→ C/Λ given by z + nK → (m/n)z + (m/n)nK, now viewing nK as a lattice in C. This map makes sense and is an isomorphism since (m/n)nK = mK = Λ .

N∈Z To show that the Laurent series truncates from the left to a power series it suffices to show that lim ((f [α]k )(τ ) · qN ) = 0. qN →0 If α fixes ∞ then this is immediate from the Fourier series of f itself. 2 Congruence subgroups 23 lim |(f [α]k )(τ ) · qN | ≤ C lim (y r−k |qN |). qN →0 qN →0 Recalling that qN = e , show that y = C log(1/|qN |), and use the fact that polynomials dominate logarithms to complete the proof. 7. 4. 8. (a) Verify the Fourier expansion of G2 . 4) for two particular matrices γ1 , γ2 ∈ SL2 (Z).

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