i am reading the article "Counter-example to global Torelli problem for irreducible symplectic manifolds" by Yoshinori Namikawa and i have two questions i can't answer:

1) he takes $T$ a complex torus, $Sym^3(T)$ the symmetric 3-product and $\alpha:Sym^3(T)\rightarrow T$ the sum map. Then he writes $K^2(T)$ for the generalized Kummer variety of dimension $4$, $\overline{K}^2(T)$ for $\alpha^{-1}(0)$ and $\sum$ for the singular locus of $\overline{K}^2(T)$. Of course $K^2(T)$ and $\overline{K}^2(T)$ are bimeromorphic and the exceptional locus is an irreducible divisor $E$ (which i believe is the irreducible divisor of $Hilb^3(T)\rightarrow Sym^3(T)$ intersected with $K^2(T)$). Finally, let $F$ be a resolution of $E$ and so there is a holomorphic surjective map from $F$ to $\sum$: Namikawa claims this is the Albanese map of $F$. For sure my knowledge about the Albanese map is very limited, but i can't see why this is the Albanese map of $F$.. could you give me a hint?

2) Given $T$ a complex torus and $T^*$ its dual, we know that there is a class $\delta\in H^2(K^2(T),\mathbb{Z})$ such that $H^2(K^2(T),\mathbb{Z})=H^2(T,\mathbb{Z})\oplus\mathbb{Z}\delta$, $H^{1,1}(K^2(T))=H^{1,1}(T)\oplus\mathbb{C}\delta$ and $2\delta=E$ where $E$ is the class of the exceptional divisor of $Hilb^3(T)\rightarrow Sym^3(T)$ intersected with $K^2(T)$. The class $\delta^*$ is the class in $H^2(K^2(T^*))$ with the same properties.

Namikawa takes $T$ complex torus with $NS(T)=0$ and dual torus $T^*$ not isomorphic to $T$. Then proves that if $f:K^2(T)--\rightarrow K^2(T^*)$ is a meromorphic map, then it is an isomorphism in codimension 1 and it must be $f^*(\delta^*)=\pm \delta$ (and i'm ok with that) but then says that "since $2\delta$ (resp. $2\delta^*$) is represented by $E$ (resp. $E^*$) the case $f^*(\delta^*)=- \delta$ does not occur because $K^2(T)$ and $K^2(T^*)$ are kahler manifolds". This sentence is what i'm not understanding: i don't know ho to use the kahler hypothesis and the fact that $2\delta =E$ to exclude $f^*(\delta^*)=- \delta$. Also, he wants to prove $f^*(\delta^*)= \delta$ because then $f$ induces a meromorphic map between $E$ and $E^*$, so i don't understand how the hypothesis $2\delta=E$ could be useful..

Thank you very much, i'm very sorry because i understand these are quite specific questions, but i would be very grateful to you if you could give me a hand because i'm out of ideas..

  • $\begingroup$ Regarding (1),it looks to me like $\Sigma$ is the image of an embedding $T\to \overline{K}^2(T)$ sending $p$ to $2(\underline{-p}) + \underline{2p}$. Also, $F$ seems to be bimeromorphic to a projective space bundle over $\Sigma$. Since the Albanese variety is a bimeromorphic invariant, that would imply that the map from $F$ to $\Sigma$ is (equivalent to) the Albanese morphism for $F$. $\endgroup$ – Jason Starr Jul 3 '14 at 14:16
  • $\begingroup$ yes! thak you, i think you're right, because given the resolution of singularities $Hilb^3(T)\rightarrow Sym^3(T)$ we have that generally the fiber on the singular locus is a $\mathbb{P}^1$, so intersecting with $Ker(\alpha)$ we can say $F$ is bimeromorphic to a $\mathbb{P}^1$-bundle on $T$, which of course has $T$ as Albanese variety! $\endgroup$ – igor guedz Jul 3 '14 at 14:24

Question 1. It's not hard to see that $F$ is $\Sigma \times {\Bbb C} P^1$ and $\Sigma$ is a torus. Therefore the Albanese map is a projection to a torus.

Question 2. If $f^* \delta=-\delta$, then the effective cycle $f^* \delta+\delta$ is homologous to 0. On a Kahler manifold this is impossible, because an integral of a Kahler form over an effective cycle is always positive.

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