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Compactification of the moduli space of abelian varieties, Kyoto ...

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COMPACTIFICATION OF THE MODULI SPACE OF ABELIAN VARIETIES 34. over C, any Hesse cubic is <strong>the</strong> image <strong>of</strong> E(ω) :=C/Z + Zω, a complextorus by <strong>the</strong>tasx k = θ k (q, w) = ∑ e 2πi(3m+k)2ω/6 e 2πi(3m+k)zm∈Z= ∑ q (3m+k)2 w 3m+km∈Zwhere q = e 2πiω/6 , w = e 2πiz .Then K is <strong>the</strong> image <strong>of</strong> ker(3 : E(ω) → E(ω)) = 〈 1 3 , ω 3 〉.2.2. The <strong>moduli</strong> <strong>space</strong> <strong>of</strong> Hesse cubics — <strong>the</strong> Stone-age (Neolithic)level structure. Consider <strong>the</strong> <strong>moduli</strong> <strong>space</strong> <strong>of</strong> Hesse cubics.1. <strong>the</strong> <strong>moduli</strong> <strong>space</strong> SQ 1,3 :=<strong>the</strong> set <strong>of</strong> isom. classes <strong>of</strong> (C(μ),K),where <strong>the</strong> definition <strong>of</strong> an isom.(C(μ),K) ≃ (C(μ ′ ),K) : isom. iff∃ f : C(μ) → C(μ ′ ) : an isom. with f |K =id K ,This extra condition f |K =id K for isom. is <strong>the</strong> classical level str.,2. if (C(μ),K) ≃ (C(μ ′ ),K), <strong>the</strong>n μ = μ ′ , because <strong>the</strong> isom is given by amatrix A, whose eigenvectors are K with |K| = 9, hence easy to proveA is scalar.3. SQ 1,3 ≃ P 1 , in fact, SQ 1,3 ≃ X(3) modular curve over Z[ζ 3 , 1/3], Thiscompatifies A 1,3 := {(C(μ),K); C(μ)smooth} = P 1 \{4 points}.2.3. The <strong>moduli</strong> <strong>space</strong> <strong>of</strong> smooth cubics — classical level structure.Consider <strong>the</strong> <strong>moduli</strong> <strong>space</strong> <strong>of</strong> Hesse cubics.1. <strong>the</strong> <strong>moduli</strong> <strong>space</strong> A 1,3 :=<strong>the</strong> set <strong>of</strong> isom. classes <strong>of</strong> (C, C[3],ι),where C a smooth cubic, C[3] <strong>the</strong> 3-division points,ι :(C[3],e C ) → (K, e K )a symplectic isom,where e C Weil pairing <strong>of</strong> C, that is,e C : C[3] × C[3] → μ 3alternating nondeg.and K =(Z/3Z) ⊕ , e K (e 1 ,e 2 )=ζ 3 , e i stand. basis <strong>of</strong> K, e K alt.,2. <strong>the</strong> definition <strong>of</strong> an isom.(C, C[3],ι) ≃ (C ′ ,C ′ [3],ι ′ ) : isom. iff∃ f : C → C ′ : isom. , f |C[3] : C[3] → C ′ [3] isom. ι ′ · f = ι,This extra condition f |C[3] : C[3] → C ′ [3] isom such that ι ′ · f = ιfor isom. is <strong>the</strong> classical level str.,3. Note that (C(μ),C(μ)[3], id K ) ∈ A 1,3 because C(μ)[3] = K,4. any (C, C[3],ι) ≃ (C(μ),C(μ)[3], id) for some μ,5. if (C(μ),C(μ)[3], id) ≃ (C(μ ′ ),C(μ ′ )[3], id), <strong>the</strong>n μ = μ ′ , because f anisom satisfies id K ·f =id K , hence f =id K ,Neolithic isom, μ = μ ′ ,6. This proves A 1,3 := {(C(μ),K); C(μ)smooth} = P 1 \{4 points}, hencewhat to add to A 1,3 are 3-gons.

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