A Global Torelli Theorem for Certain Calabi-Yau Threefolds

Abstract

We establish a global Torelli theorem for the complete family of Calabi-Yau threefolds arising from cyclic triple covers of ${{\mathbb {P}}}^3$ branched along six stable hyperplanes.

Comparison with Rohde’s Example

Given distinct $a,b,c\in {{\mathbb {C}}}\backslash \{0,1\}$, Rohde [14] constructed a singular Calabi-Yau threefold $X^{'}$ in the following way:

Let W be the surface in the weighted projective space ${{\mathbb {P}}}(2,2,1,1)$ defined by the equation $y_1^3+y_2^3+x_0x_1(x_1-x_0)(x_1-ax_0)(x_1-bx_0)(x_1-cx_0)=0$. Let F be the Fermat curve in ${{\mathbb {P}}}^2$ defined by the homogeneous equation $z_0^3+z_1^3+z_2^3=0$. Then the cyclic group $G={{\mathbb {Z}}}/3{{\mathbb {Z}}}$ acts on W and F. Fixing a generator $\sigma$ of G and let $\omega$ be a fixed primitive cubic root of unity. We define these actions explicitly:

$$\begin{aligned} \begin{aligned} \sigma : W&\rightarrow W, \\ [x_0:x_1:y_1:y_2]&\mapsto [x_0:x_1:\omega y_1:\omega y_2],\\ \sigma :&F\rightarrow F,\\ [z_0:z_1:z_2]&\mapsto [\omega z_0:z_1:z_2]. \end{aligned} \end{aligned}$$

Rohde [14] constructed the Calabi-Yau threefold as a crepant resolution of the quotient threefold $X^{'}:=W\times F/G$, where G acts on $W\times F$ in the diagonal way. Moreover, varying the parameters a, b, c in ${{\mathbb {C}}}\backslash \{0,1\}$, Rohde obtains a family of Calabi-Yau threefolds ${\mathcal {X}}^{'}\rightarrow {\mathfrak {M}}_{1,6}$, where we recall that ${\mathfrak {M}}_{1,6}$ is the moduli space of ordered six distinct points in ${{\mathbb {P}}}^1$. The main goal of this section is to show Rohde’s family is birationally equivalent to the family ${\mathcal {X}}_{AR}\rightarrow {\mathfrak {M}}_{3, 6}$, which is the universal family of cyclic triple covers of ${{\mathbb {P}}}^3$ branched along six hyperplanes in general position.

We first analyze the structure of the singular surface W. In general, if $X_1$ and $X_2$ are two varieties with G-action, we say $X_1$ and $X_2$ are G-birationally equivalent if there exists a birational map from $X_1$ to $X_2$ compatible with the G-action.

Let C be the cyclic triple cover of ${{\mathbb {P}}}^1$ branched along the six points $\{0, 1, \infty , a, b, c\}$ whose affine model is the curve in ${{\mathbb {C}}}^2$ defined by the equation $y^3-x(x-1)(x-a)(x-b)(x-c)=0$. Then G acts on C in the following way:

$$\begin{aligned} \begin{aligned} \sigma : C&\rightarrow C,\\ (x,y)&\mapsto (x, \omega y). \end{aligned} \end{aligned}$$

Let G act on the product $C\times F$ diagonally, then G acts on the quotient $C\times F/ G$ through the G-action on the first factor C.

Lemma A.1

W is G-birationally equivalent to the quotient $C\times F/ G$.

Proof

Let $F_0$ be the affine curve $\{(z_1,z_2)\in {{\mathbb {C}}}^2| 1+z_1^3+z_2^3=0\}$, and define a G-action on $F_0$ by $\sigma (z_1,z_2)=(\omega z_1,\omega z_2)$. Then $F_0$ is G-birational to the Fermat curve F. Define the following morphism:

$$\begin{aligned} \begin{aligned} C\times F_0&\rightarrow W, \\ (x,y,z_1,z_2)&\mapsto (x, z_1 y, z_2 y). \end{aligned} \end{aligned}$$

It is easy to see this morphism induces a G-birational equivalent between $C\times F_0/ G$ and W. So W is G-birationally equivalent to $C\times F/ G$. $\square$

Now we consider the following six hyperplanes in ${{\mathbb {P}}}^3$ which are in general position

$$\begin{aligned} H_i: X_i=0 \ (0\le i\le 3), \ \ H_4: \sum _{i=0}^3X_i=0, \ \ H_5: X_0+a X_1+b X_2+c X_3=0, \end{aligned}$$

where $[X_0:\cdots :X_3]$ is the homogeneous coordinates on ${{\mathbb {P}}}^3$. Let X be the cyclic triple cover of ${{\mathbb {P}}}^3$ branched along $\sum _{i=0}^5 H_i$.

In order to analyze the structure of X, we define some auxiliary varieties.

Let $u_1$, $v_1$ be linear functions of u, v defined by the following relations:

$$\begin{aligned} \left\{ \begin{array}{ll} u_1 &{} =1+u+v, \\ v_1 &{} =a+bu+cv. \end{array} \right. \end{aligned}$$

(A.1)

We define Y as the following affine variety:

$$\begin{aligned} Y=\left\{ (t_1,u,v,y_1)\in {{\mathbb {C}}}^4| y_1^3=\frac{uvt_1(t_1+1)}{u_1v_1(v_1-u_1)}\right\} . \end{aligned}$$

Let S be the following affine surface:

$$\begin{aligned} S=\Bigg \{(w, u, v)\in {{\mathbb {C}}}^3| w^3=\frac{uv}{u_1v_1(v_1-u_1)}\Bigg \}. \end{aligned}$$

Let G acts on S by $\sigma (w, u, v)=(\omega ^2 w, u, v)$.

Lemma A.2

We have the following birational isomorphsims:

• (1) X is birationally equivalent to Y.

• (2) Y is birationally equivalent to $S\times F /G$, where G acts on $S\times F$ diagonally.

• (3) S is G-birationally equivalent to $C\times F /G$.

Proof

(1) We take the following affine model of X:

$$\begin{aligned} X_1=\{(x_1,x_2,x_3,y)\in {{\mathbb {C}}}^4| y^3=x_1x_2x_3(1+x_1+x_2+x_3)(1+ax_1+bx_2+cx_3)\}. \end{aligned}$$

Under the coordinate transformation

$$\begin{aligned} \left\{ \begin{array}{ll} x_1 &{} =t, \\ x_2 &{} =tu,\\ x_3 &{} =tv,\\ y &{} =y. \end{array} \right. \end{aligned}$$

the hypersurface $X_1$ is birationally equivalent to the following hypersurface in ${{\mathbb {C}}}^4$:

$$\begin{aligned} X_2=\{(t,u,v,y)\in {{\mathbb {C}}}^4| y^3=t^3uv(1+u_1t)(1+v_1t)\}, \end{aligned}$$

where $u_1$, $v_1$ are defined by the equations (A.1). Then we can see $X_2$ is birational to Y under the following coordinate transformation:

$$\begin{aligned} \left\{ \begin{array}{ll} u &{} =u, \\ v &{} =v,\\ t &{} =(\frac{1}{v_1}-\frac{1}{u_1})t_1-\frac{1}{u_1},\\ y &{} =tu_1v_1(\frac{1}{u_1}-\frac{1}{v_1})y_1. \end{array} \right. \end{aligned}$$

(2) It is direct to see the following affine curve $F_1$ is G-birationally equivalent to the Fermat curve F:

$$\begin{aligned} F_1=\{(t_1,x)\in {{\mathbb {C}}}^2| x^3=t_1(t_1+1)\}, \end{aligned}$$

where G acts on $F_1$ by $\sigma (t_1, x)=(t_1, \omega x)$.

Then the following rational map induces the desired G-birationally equivalence between $S\times F /G$ and Y:

$$\begin{aligned} \begin{aligned} S\times F_1&\rightarrow Y, \\ (w, u, v, t_1, x)&\mapsto (t_1, u, v, wx). \end{aligned} \end{aligned}$$

(3) From the equations (A.1), we can view $(w, u_1, v_1)$ as a coordinate system on ${{\mathbb {C}}}^3$, and we make the following coordinate transformation:

$$\begin{aligned} \left\{ \begin{array}{ll} w &{} =w, \\ u_1 &{} =t_2,\\ v_1 &{} =t_2z. \end{array} \right. \end{aligned}$$

Under this coordinate transformation, we see S is G-birationally equivalent to the following surface:

$$\begin{aligned} S_1=\{(w,t_2,z)\in {{\mathbb {C}}}^3| w^3=\frac{(A_1t_2+a_1)(B_1t_2+b_1)}{t_2^3z(z-1)}\}, \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{array}{ll} A_1 &{} =\frac{b-z}{b-c}, \\ B_1 &{} =\frac{c-z}{c-b},\\ a_1 &{} =\frac{a-b}{b-c},\\ b_1 &{} =\frac{a-c}{c-b}, \end{array} \right. \end{aligned}$$

and G acts on $S_1$ by $\sigma (w, t_2, z)=(\omega ^2 w, t_2, z)$.

A further coordinate transformation:

$$\begin{aligned} \left\{ \begin{array}{ll} z&{}= z,\\ t_3 &{} =\frac{t_2+\frac{a_1}{A_1}}{\frac{b_1}{B_1}-\frac{a_1}{A_1}}, \\ w_1 &{} =\frac{wt_2}{A_1B_1(\frac{b_1}{B_1}-\frac{a_1}{A_1})}, \end{array} \right. \end{aligned}$$

shows that $S_1$ is G-birational to the surface:

$$\begin{aligned} S_2=\{(z,t_3,w_1)\in {{\mathbb {C}}}^3|w_1^3=\frac{t_3(t_3+1)}{z(z-1)A_1B_1(A_1b_1-B_1a_1)} \}, \end{aligned}$$

where G acts on $S_2$ by $\sigma (z, t_3, w_1)=(z, t_3, \omega ^2 w_1)$.

Now let $C_1$ be the following affine curve:

$$\begin{aligned} C_1:=\{(z,w_2)\in {{\mathbb {C}}}^2| w_2^3=z(z-1)A_1B_1(A_1b_1-B_1a_1)\}, \end{aligned}$$

and let G act on $C_1$ by $\sigma (z, w_2)=(z, \omega ^2 w_2)$. We consider the affine model

$$\begin{aligned} F_1=\{(t_3,x)\in {{\mathbb {C}}}^2| x^3=t_3(t_3+1)\} \end{aligned}$$

of the Fermat curve F as before. The rational map

$$\begin{aligned} \begin{aligned} C_1\times F_1&\rightarrow S_2, \\ (z, w_2, t_3, x)&\mapsto (z, t_3, \frac{x}{w_2}). \end{aligned} \end{aligned}$$

gives a G-birationally equivalence between $C_1\times F_1 /G$ and $S_2$. Moreover, we see the smooth projective model of $C_1$ is isomorphic to C, the cyclic triple cover of ${{\mathbb {P}}}^1$ branched along the six points $\{0,1,\infty , a, b, c\}$. By combining all of the birational equivalences above, we obtain S is G-birationally equivalent to $C\times F/ G$. $\square$

Now it is direct to see that, by combining Lemma A.1 and Lemma A.2, we obtain the following birational equivalence.

Proposition A.3

Given distinct $a,b,c\in {{\mathbb {C}}}\backslash \{0,1\}$, Rohde’s singular Calabi-Yau threefold $X^{'}=W\times F /G$ is birational to X, which is the cyclic triple cover of ${{\mathbb {P}}}^3$ branched along $\sum _{i=0}^5 H_i$ with $H_i$ defined by

$$\begin{aligned} H_i: X_i=0 \ (0\le i\le 3), \ \ H_4: \sum _{i=0}^3X_i=0, \ \ H_5: X_0+a X_1+b X_2+c X_3=0. \end{aligned}$$

Note that to give distinct $a,b,c\in {{\mathbb {C}}}\backslash \{0,1\}$ is equivalent to give six distinct points $\{0, 1, \infty , a, b, c\}$ in ${{\mathbb {P}}}^1$. Since the moduli space ${\mathfrak {M}}_{1,6}$ of ordered six distinct points in ${{\mathbb {P}}}^1$ is isomorphic to the moduli space ${\mathfrak {M}}_{3, 6}$ of ordered six hyperplane arrangements in general in ${{\mathbb {P}}}^3$, the following birational equivalence is a direct consequence of Proposition A.3.

Proposition A.4

Rohde’s Calabi-Yau family ${\mathcal {X}}^{'}\rightarrow {\mathfrak {M}}_{1,6}$ is birationally equivalent to the universal family ${\mathcal {X}}_{AR}\rightarrow {\mathfrak {M}}_{3, 6}$ of cyclic triple covers of ${{\mathbb {P}}}^3$ branched along six hyperplanes in general position.