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.


  1. 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.

  2. 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.

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

  • $\begingroup$ I'm having trouble making the matrix display correctly in TeX (it should be diagonal with entries $\omega$ and $\omega^2$) -- feel free to correct it if you know how. $\endgroup$
    – macbeth
    Nov 15, 2011 at 19:58
  • $\begingroup$ I fixed the matrix by using \cr as the line separator. $\endgroup$ Nov 15, 2011 at 20:14

1 Answer 1


The complex manifold $X$ is a $K3$ surface with $9$ singular points of type $\frac{1}{3}(1,2)$, i.e. rational double points of type $A_2$.

In fact, let us denote by $A$ the abelian surface $\mathbb{C}^2 / \Lambda ^2$ and by $\pi \colon A \longrightarrow X$ the natural projection. Then the action of $\mathbb{Z}/3 \mathbb{Z}$ on $A$ is locally given by $$ \omega \cdot (x, y)=(\omega x, \omega^2 y).$$ Therefore the $1$-forms $dx$, $dy$, which give a basis of $H^0(\Omega^1_A)$, are not invariant, whereas the $2$-form $dx \wedge dy$, which gives a basis of $H^0(\Omega^2_A)$, is invariant. This means that $$p_g(X)=1, \quad q(X)=0.$$ Moreover, since $X$ has only rational double points, one can write $$K_A=\pi^*K_X,$$ which implies $K_X^2=0$. Moreover $K_A$ nef implies $K_X$ nef, hence $X$ is a minimal model.

By Enriques-Kodaira classification it follows that $X$ is a $K3$ surface.

The case with group $\mathbb{Z}/4 \mathbb{Z}$ is similar: the quotient is a $K3$ surface whose singularities are $4$ points of type $\frac{1}{4}(1, 3)$, i.e. rational double points of type $A_3$. The minimal resolution of each of them is a tree made of $3$ smooth rational curves, each having self-intersection $(-2)$.

Remark. I called $X$ the quotient of $A$ by the finite group, without blowing up the singularities. Since $X$ has only rational double points, this does not make any difference in the computation of invariants.


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