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On page 21 of the first volume of Geometry of Algebraic Curves, there is stated the isomorphism \begin{equation} \textrm{Hom}(\Lambda^2H_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z), \end{equation} where $A$ is a complex torus defined by a lattice $\Lambda=H_1(A,\mathbb Z)\subset \mathbb C^g$. It is not explained, so it should be obvious, but I find no way to prove it.

Can anyone give me a clue to see why this isomorphism holds? Is it a special case of something more general (that I do not know)?

Thanks in advance.

3 Answers 3

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By the characteric property of the exterior product, there is a canonical isomorphism of $\mathbb Z$- modules \begin{equation} \textrm{Hom}_\mathbb Z(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= Alt^2 _\mathbb Z(H_1(A,\mathbb Z),\mathbb Z) \quad (1)\end{equation} For a complex torus $A= V/\Lambda$ (where $V$ is a complex vector space and $\Lambda \subset V$ a full lattice) there is a canonical isomorphism $H_1(A,\mathbb Z)=\Lambda \quad (0)$ so that $(1)$ becomes \begin{equation} \textrm{Hom}_\mathbb Z(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= Alt^2_\mathbb Z (\Lambda,\mathbb Z) \quad (2)\end{equation}

and multilinear algebra furnishes an isomorphism $Alt^2 (\Lambda ,\mathbb Z)=\wedge ^2 \check{\Lambda} \quad (3)$ where we have used the notation $\check {\Lambda }=Hom_\mathbb Z(\Lambda,\mathbb Z)$
Hence $(2)$ becomes \begin{equation} \textrm{Hom}(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= \wedge ^2 \check{\Lambda} \quad (4)\end{equation}
From algebraic topology we have an isomorphism $ H^1(A,\mathbb Z)\stackrel {algtop} {=}Hom_\mathbb Z(H_1(A,\mathbb Z),\mathbb Z)\stackrel {(0)} {=} Hom_\mathbb Z(\Lambda,\mathbb Z)\stackrel {def} {=}\check {\Lambda } \quad (5)$ which joined to Künneth's theorem permits to prove that the cup product $\Lambda ^2_\mathbb Z \check {\Lambda } \stackrel {(5)}{=}\Lambda ^2_\mathbb Z H^1(A,\mathbb Z) \stackrel {cup}{\to }H^2(A,\mathbb Z) \quad (6)$ is an isomorphism.
From $(4)$ and $(6)$ we obtain the required canonical isomorphism of $\mathbb Z$- modules \begin{equation} \textrm{Hom}(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z)\quad \text {(FINAL)}\end{equation}

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    My pleasure, atricolf!2012-12-24
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For a complex torus one can calculate explicitly cohomology as exterior algebra (over integers) on the lattice $\Lambda\subset \mathbb C^g$. You can find this result, for example, in Griffths and Harris Principles of Algebraic geometry. Than your isomorphism is a connection between homology and cohomology in this situation.

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Just to add a little to Alex's answer:

Topologically, a $g$-dimensional abelian variety is the product of $2g$ circles.
Using the Kunneth Theorem, you can prove that the cohomology ring of a product of circles is generated by the $H^1$, and indeed is isomorphic to the exterior algebra on $H^1$ (with any coefficients).

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    Thank you @Matt E! I see that Kunneth formula might help, but for now I miss some details. Thank you very much for the clue.2012-12-21