I'd like to know "what" (say, in the classification of complex surfaces) the following complex manifold $X$ is:

**Construction:** Let $\Lambda$ be the hexagonal lattice in $\mathbb{C}$; that is, the lattice generated by $(1, \omega)$ where $\omega=e^{2i\pi/3}$ is a third root of unity.

Observe that the lattice $\Lambda^2\subseteq \mathbb{C}^2$ is invariant under the $\mathbb{Z}/3\subseteq SU(2)$ action $\begin{pmatrix} \omega & 0 \cr 0 & \omega^2 \end{pmatrix}$.

I'm curious about the complex manifold $X$ obtained by quotienting $\mathbb{C}^2/\Lambda^2$ by $\mathbb{Z}/3$, and then blowing up at the $9=3^2$ singular points.

**Remarks:**

Doing this with $\mathbb{Z}/4\subseteq SU(2)$ instead of $\mathbb{Z}/3$, and the square (i.e., generated by $(1, i)$) lattice rather than the hexagonal lattice, and the resulting $4=2^2$ singular points, is something I'm equally curious about and equally unable to answer.

Doing this with $\mathbb{Z}/2$ instead of $\mathbb{Z}/3$, and any lattice in $\mathbb{C}^2$ at all (since all are invariant under this action), and the resulting $16=4^2$ singular points, gives a K3 surface; this was my motivation for the question.

The exceptional divisor at each of the 9 blowups is a pair of $\mathbb{P}^1$'s intersecting at a point.

`\cr`

as the line separator. $\endgroup$ – José Figueroa-O'Farrill Nov 15 '11 at 20:14