By Ribet K.A., Stein W.A.
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Sample text
If g, h ∈ G and α ∈ h∗ , then {α, gh(α)} = {α, gα} + {gα, ghα} = {α, gα} + {α, hα}, so the map is a group homomorphism. To see that the map does not depend on the choice of α, suppose β ∈ h∗ . 2, we have {α, β} = {gα, gβ}. Thus {α, gα} + {gα, β} = {gα, β} + {β, gβ}, so cancelling {gα, β} from both sides proves the claim. The fact that the map is surjective follows from general facts from algebraic topology. Let h0 be the complement of Γi ∪ Γρ in h, fix α ∈ h0 , and let X(G)0 = π(h0 ). The map h0 → X(G)0 is an unramified covering of (noncompact) Riemann surfaces with automorphism group G.
Thus Γ0 (N )\h parameterizes isomorphism classes of pairs (E, C) where E is an elliptic curve and C is a cyclic subgroup of order N , and Γ1 (N )\H parameterizes isomorphism classes of pairs (E, P ) where E is an elliptic curve and P is a point of exact order N . We can specify a point of exact order N on an elliptic curve E by giving an injection Z/N Z → E[N ], or equivalently, an injection µN → E[N ] where µN denotes the N th roots of unity. Then Γ(N )\h parameterizes isomorphism classes of pairs (E, {α, β}), where {α, β} is a basis for E[N ] ≈ (Z/N Z)2 .
1. TR ⊗R C ∼ = TC , as C-vector spaces. Proof. 1 assures us that there is a C-basis of Sk (C) consisting of forms with integral coefficients, we see that Sk (R) ≈ Rd where d = dimC Sk (C). 2) we see that TR leaves Sk (R) invariant, thus TR = R[. . , Td , . ] ⊂ EndR Sk (R). 50 6. 2 we defined a perfect pairing TC × Sk (C) → C. By restricting to R we again obtain a perfect pairing, so we see that TR ≈ Sk (R) ≈ ∼ Rd which implies that TR ⊗R C − → TC . The above argument also shows that Sk (C) ∼ = Sk (R) ⊗R C, so complex conjugation is defined over R.