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 $F$. $\endgroup$ Jul 3, 2014 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, 2014 at 14:24

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.