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Let $X$ be an abelian variety of dimension $3$. I wish to compute the Euler characteristic of a certain subscheme $V$ of the Hilbert scheme $H=\textrm{Hilb}^3X$.

Definition of $V\subset H$: it consists of those $[Z]\in H$ supported on two distinct points, and whose cycle adds up to zero (according to the group structure on $X$).

Thus, $V$ can be described as the set of pairs $(z_1,z_2)\in X\times X$ such that $z_1+2z_2=0_X$, $z_1\neq z_2$, and one of the two points comes with a tangent vector. Moreover, there is an action of the group of order $2$, exchanging $z_1$ and $z_2$.

I just noted that instead of $$"(z_1,z_2) \,\textrm{ such that } \,z_1+2z_2=0_X,\,z_1\neq z_2"$$ we may write $$"(-2z,z) \,\textrm{ such that } \,z\in X\setminus X_3 "$$ where $X_3\subset X$ denotes the subgroup of $3$-torsion points of $X$. Moreover, the choice of a tangent vector amounts to the choice of a $\mathbb P^2$ (lines in $T_zX=\mathbb C^3$), for every couple as above. But I am not able to combine these data ($+$ the action of $\mathfrak S_2$) in order to write $V$ in a convenient form.

Can anyone help me describing $V$ in such a way that it becomes easy to compute $\chi(V)$?

Thank you.

Brenin
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    Dear Brenin, I'm probably just missing something, but since the equation $z_1 + 2 z_2$ is not symmetric in $z_1$ and $z_2$, how is there an action of $\mathfrak S_2$? Regards, – Matt E Jan 15 '14 at 16:39
  • @MattE: thanks for your comment. You might be right. What I had in mind, though, is that $(-2z,z)$ and $(z,-2z)$ should be identified (should they?). But you are right that there is no symmetry in the previous formulation. – Brenin Jan 15 '14 at 16:49
  • Here's my first thought; let me know if I'm missing something. The map $(z_1,z_2) \mapsto z_1$ exhibits the set of pairs as a $2^6$-fold covering space of $X\setminus X_3$, which is just $X$ with $3^6$ points deleted. Then the Hilbert scheme is a $P^2$-bundle over that. All those operations (deleting points, taking covering spaces, taking more general fibre bundles) change the Euler char. in known ways, so this should give the answer. –  Jan 15 '14 at 17:00
  • Dear @AsalBeagDubh, I do not see how to get a $2^6$-covering by using a projection. However, if you mean that the map $\textrm{Hilb}^3_2X\to X\setminus X_3$ (maybe restricted to $V$) given by $(-2z,z)\mapsto z$ is a $\mathbb P^2$-bundle, then we have $\chi(V)=\chi(\mathbb P^2)\cdot \chi(X\setminus X_3)=-3\cdot 3^6$. I am not convinced at all, though. Could you please clarify your argument? thank you very much! Regards – Brenin Jan 15 '14 at 18:11
  • Dear Brenin: what I had in mind was this. Let $W=\left{(z_1,z_2):z_1+2z_2=0 \right}$. Then $W \rightarrow X \setminus X_3$ given by $(z_1,z_2) \mapsto z_1$ is a covering space of degree $2^6$. Then your Hilbert scheme is a $P^2$ bundle over $W$. –  Jan 16 '14 at 09:28
  • @AsalBeagDubh: thanks for clarifying this. I am still asking myself the probably stupid question: why, for fixed $z_1\in X\setminus X_3$, do there exist $2^6$ elements $z_2\in X$ such that $z_1+2z_2=0$? – Brenin Jan 16 '14 at 15:52
  • Dear Brenin, given $z_1$, we want to find all elements $z_2$ such that $2z_2=z_1$. Abelian varieties (over $\mathbf{C}$ say) are divisible, so there is at least one such $z_2$; moreover, any two differ by a 2-torsion element of the group, and conversely adding a 2-torsion element gives another. There are $2^6$ 2-torsion elements. –  Jan 16 '14 at 16:15
  • Sorry, I meant $-z_1$ in place of $z_1$ in the previous comment. –  Jan 16 '14 at 17:08
  • Yes, but now it is very clear what you meant. Thank you so much! – Brenin Jan 16 '14 at 18:33
  • @Brenin: ok, hope it helps! –  Jan 17 '14 at 08:54

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