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I am being stuck by the proof of the existence of Poincare line bundle of complex torus in Griffiths-Harris. Here is the question:

Let $M$ be a complex torus and $M'$ be the complex torus dual to $M$ (the one consists of all holomorphic line bundle whose chern class is zero). By Kunneth's formula and the definition of $M',$ one knows that the identity element $I$ of $$ Hom(H^1(M,\mathbb{Z}),H^1(M,\mathbb{Z}))\equiv H^1(M,\mathbb{Z})\otimes H^1(M',\mathbb{Z})$$ belongs to $H^2(M\times M',\mathbb{Z})$ and is of type (1,1). Hence, there exists a line bundle $P$ over $M\times M'$ having $I$ as its chern class. For each $\xi$ in $M'$, let $P_{\xi}$ be the restriction of $P$ on $M\times \{\xi\}$, clearly it is in $M'$. Hence we get a mapping $\Phi: M' \rightarrow M'$ that send $\xi$ to $P_{\xi}.$

Why the induced mapping of $\Phi$ on the first homology is the identity?

Thank you so much.

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1 Answer 1

[Not exactly an answer. Still, it may be helpful, I hope.]

I think there are two obscure points in this proof: apart from the homology question, why the map is holomorphic? (Is it really obvious?). Of course, all this can be settled. However, I prefer a more explicit construction of the Poincare bundle, which can be found, e.g., in Hindry&Silverman, "Diophantine Geometry", Exercise A.5.6. (I like it, but even if you won't, it is definitely worth reading).

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did you check in Birkenhacke-Lange's "Abelian varieties"? –  IMeasy Jan 2 '13 at 22:23

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