Determining the <PERSON> numbers of some orbifold examples
I'm currently reading about complex geometry in order to get a feeling of how to determine the <PERSON> numbers, e.g. of certain orbifold constructions. Since I'm a physicist with no deeper mathematical knowledge in algebraic geometry and complex geometry, I apologize for the following possibly trivial questions.
In the following I'm referring to two examples of the string theory book by Becker², <PERSON>.
Example 1 (p. 368):
$T^4/\mathbb{Z}_2$, i.e. $z_1 \sim z_1+1$ and $z_2 \sim z_2 + i$, where the $\mathbb{Z}_2$ isometry is generated by $\mathcal{I}:(z_1,z_2) \to (-z_1,-z_2)$. First of all, it is easy to see that there is no invariant $(1,2)-$form, hence $h^{1,2}=0$ and that $dz_1 \wedge dz_2$ is the only $(2,0)-$form, i.e. $h^{2,0}=1$. 364
Moreover, there are four invariant $(1,1)-$forms (they're obvious, so I'm not gonna write them down) which contribute to $h^{1,1}$. But there are in total also $16$ fixed points of the orbifold which can be blown up à la Eguchi-Hanson. It is claimed that $h^{1,1}=4+1 \times 16 =20$ because there is one $(1,1)-$cycle for every blown up singularity.
Sorry for this questions: How do I see that there is just one $(1,1)-$form for every singularity? And why aren't there cycles from the EH-Space that contribute to $h^{2,2}$ and the other Hodge-numbers?
Example 2 (pp. 372):
Now we consider $T^2 \times T^2/ \mathbb{Z}_3$ with the action $(z_1,z_2) \to (\omega z_1, \omega^{-1}z_2)$ with $\omega =\exp(2\pi i/3)$. One has then $9$ singularities.
In this case there are $2$ invariant $(1,1)-$forms. It is claimed that there are now $2$ $(1,1)-$cycles for every blow-up, i.e. $h^{1,1}=2+ 2 \times 9 =20$.
Why are there now two $(1,1)-$cycles for each singularity? And why aren't there other cycles contributing to the remaining Hodge-numbers?
I would be happy if someone could explain me the intuition and the maths behind this counting.
psm 610