The classification of Hyperelliptic threefolds

TL;DR

This paper completes the classification of hyperelliptic threefolds, specifically describing those with group D_4, highlighting their algebraic nature and their place within a 2-dimensional family.

Contribution

It provides a complete and elementary classification of hyperelliptic threefolds with group D_4, including their algebraic properties and family structure.

Findings

Hyperelliptic threefolds with group D_4 form an irreducible 2-dimensional family.

The classification is elementary and self-contained.

These threefolds are algebraic.

Abstract

We complete the classification of hyperelliptic threefolds, describing in an elementary way the hyperelliptic threefolds with group $D_4$. These are algebraic and form an irreducible 2-dimensional family. Our paper is fully self-contained.

Equations

$$G:=D_4:=⟨r,s|r^4=1,s^2=1,(rs{)}^2=1⟩,$$

$$T^{\prime}:=E\times E\times E^{\prime}=:E_{1}\times E_{2}\times E_{3},\ \$$

$$H:=⟨\omega ⟩,\omega :=\left(h+k,h+k,0\right)\in T^{\prime }[2],$$

$$r(z)=(z_2,-z_1,z_3+h^{\prime })=R(z_1,z_2,z_3)+\left(0,0,h^{\prime }\right),$$

$$s(z)=\left(z_1+h,-z_2+k,-z_3\right)=S(z_1,z_2,z_3)+\left(h,k,0\right).$$

$$E=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau ),E^{\prime }=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}{\tau }^{\prime }),\tau ,{\tau }^{\prime }\in \mathcal{H}:=\{z\in \mathbb{C}|Im(z)>0\},$$

$$h=1/2,k=\tau /2,h^{\prime }=1/4$$

$$r(z_1,z_2,z_3):=(z_2,-z_1,z_3+1/4)$$

$$s(z_1,z_2,z_3):=(z_1+1/2,-z_2+\tau /2,-z_3).$$

$$G=⟨r,s|r^4=s^2=(rs{)}^2=1⟩\cong D_4,$$

$$V=V_1\oplus V_2,$$

$$\text{Case 1:}R=\left(\begin{array}{ccc}0& 1& \\ -1& 0& \\ & & 1\end{array}\right),S=\left(\begin{array}{ccc}1& & \\ & -1& \\ & & -1\end{array}\right),$$

$$\text{Case 2:}R=\left(\begin{array}{ccc}0& 1& \\ -1& 0& \\ & & 1\end{array}\right),S=\left(\begin{array}{ccc}1& & \\ & -1& \\ & & 1\end{array}\right).$$

$$(E_1\times E_1\times E_3)/H,H=\Lambda /({\Lambda }_1\oplus {\Lambda }_2\oplus {\Lambda }_3).$$

$$r(z_1,z_2,z_3)=(z_2,-z_1,z_3+c_3),$$

$$s(z_1,z_2,z_3):=(z_1+a_1,-z_2+a_2,\pm z_3+a_3),$$

$$(z_1,z_2)\mapsto (z_1-z_2,z_1+z_2)$$

$$s(z_1,z_2,z_3):=(z_1+a_1,-z_2+a_2,z_3+a_3),$$

$$(2a_1,0,2a_3)\in H.$$

$$rs(z)=(-z_2+a_2,-z_1-a_1,z_3+a_3+c_3).$$

$$(a_1+a_2,-(a_1+a_2),2(a_3+c_3))\in H.$$

$$(a_2-a_1,-(a_1+a_2),2c_3)\in H,$$

$$s(z_1,z_2,z_3):=(z_1+a_1,-z_2+a_2,-z_3).$$

$$s(z_1,z_2,z_3):=(z_1+a_1,-z_2+a_2,-z_3),$$

$$(2a_1,0,0)\in H$$

$$(a_1,w_2,w_3).$$

$$2\lambda ={\lambda }_1+{\lambda }^{\prime },2{\lambda }^{\prime }={\lambda }_2+{\lambda }_3$$

$$\Lambda \subset (1/2){\Lambda }_1+(1/2){\Lambda }_2+(1/4){\Lambda }_3.$$

$$2\lambda =\underset{=:{\lambda }_1\in {\Lambda }_1}{\underbrace{(I+S)\lambda }}+\underset{=:{\lambda }^{\prime }\in \Lambda }{\underbrace{(I-S)\lambda }}.$$

$$2{\lambda }^{\prime }=\underset{=:{\lambda }_3\in {\Lambda }_3}{\underbrace{(I+R^2){\lambda }^{\prime }}}+\underset{=:{\lambda }_2\in \Lambda \cap W_2={\Lambda }_2}{\underbrace{(I-R^2){\lambda }^{\prime }}}.$$

Full text

AMS Classification: 14K99, 14D99, 32Q15

The present work took place in the framework of the ERC Advanced grant n. 340258, ‘TADMICAMT’

Introduction

A Generalized Hyperelliptic Manifold $X$ is defined to be a quotient $X=T/G$ of a complex torus $T$ by the free action of a finite group $G$ which contains no translations. We say that $X$ is a Generalized Hyperelliptic Variety if moreover the torus $T$ is projective, i.e., it is an Abelian variety $A$.

The main purpose of the present paper is to complete the classification of the Generalized Hyperelliptic Manifolds of complex dimension three. The cases where the group $G$ is Abelian were classified by H. Lange in [La01], using work of Fujiki [Fu88] and the classification of the possible groups $G$ given by Uchida and Yoshihara in [UY76]: the latter authors showed that the only possible non Abelian group is the dihedral group $D_{4}$ of order $8$.

This case was first excluded but it was later found that it does indeed occur (see [CD18] for an account of the story and of the role of the paper [DHS08]). Our paper is fully self-contained and show that the family described in [CD18] gives all the possible hyperelliptic threefolds with group $D_{4}$.

Our main theorem is the following

Theorem 0.1**.**

Let $T$ be a complex torus of dimension $3$ admitting a fixed point free action of the dihedral group

[TABLE]

such that $G=D_{4}$ contains no translations.

Then $T$ is algebraic. More precisely, there are two elliptic curves $E,E^{\prime}$ such that:

(I) $T$ is a quotient $T:=T^{\prime}/H,\ H\cong\mathbb{Z}/2$, where

[TABLE]

[TABLE]

and $h,k$ are 2-torsion element $h,k\in E[2]$, such that $h,k\neq 0,h+k\neq 0$;

(II) there is an element $h^{\prime}\in E^{\prime}$ of order precisely $4$, such that, for $z=(z_{1},z_{2},z_{3})\in T^{\prime}$:

[TABLE]

Conversely, the above formulae give a fixed point free action of the dihedral group $G=D_{4}$ which contains no translations.

In particular, we have the following normal form:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

In particular, the Teichmüller space of hyperelliptic threefolds with group $D_{4}$ is isomorphic to the product $\mathcal{H}^{2}$ of two upper halfplanes.

1. Proof of the main theorem

We use the following notation: $T=V/\Lambda$ is a complex torus of dimension $3$, which admits a free action of the group

[TABLE]

such that the complex representation $\rho\colon G\to\operatorname{GL}(3,\mathbb{C})$ is faithful.

A first observation is that the complex representation $\rho$ of $G$ must contain the $2$-dimensional irreducible representation $V_{1}$ of $G$ (else, $\rho$ would be a direct sum of 1-dimensional representations: this, by the assumption on the faithfulness of $\rho$, would imply that $G$ is Abelian, a contradiction).

Hence we have a splitting

[TABLE]

where $V_{2}$ is 1-dimensional, and we can choose an appropriate basis so that, setting $R:=\rho(r),S:=\rho(s)$, we are left with the two cases

[TABLE]

which are distinguished by the multiplicity of the eigenvalue $1$ of $S$.

Indeed $R$ is necessarily of the form above, since the freeness of the $G$-action implies that $\rho(g)$ must have eigenvalue $1$ for every $g\in G$.

Lemma 1.1**.**

In both Cases 1 and 2, the complex torus $T=V/\Lambda$ is isogenous to a product of three elliptic curves, $T\sim_{\text{isog.}}E_{1}\times E_{2}\times E_{3}$, where $E_{i}\subset T$, for $i=1,2,3$ and $E_{1}$ and $E_{2}$ are isomorphic elliptic curves. In other words, writing $E_{j}=W_{j}/\Lambda_{j}$, the complex torus $T$ is isomorphic to

[TABLE]

Proof.

Let $I$ be the identity of $T$.

In Case 1, we set $E_{1}:=\ker(S-I)^{0}=\operatorname{im}(S+I)$, $E_{3}:=\ker(R-I)^{0}$ and $E_{2}:=R(E_{1})$ (here, the superscript zero denotes the connected component of the identity). Then it is clear that $E_{1}\cong E_{2}$, and that $T$ is isogenous to $E_{1}\times E_{2}\times E_{3}$.

In Case 2, we define similarly $E_{2}:=\ker(S+I)^{0}=\operatorname{im}(S-I)$, $E_{3}:=\ker(R-I)_{0}$ and $E_{1}:=R(E_{2})$. We obtain again $E_{1}\cong E_{2}$, and that $T$ is isogenous to $E_{1}\times E_{2}\times E_{3}$.

∎

Lemma 1.2**.**

Writing $E_{j}=W_{j}/\Lambda_{j}$, the following statements hold.

• (1)

In Case 1, the lattice $\Lambda_{2}$ is equal to $W_{2}\cap\Lambda$.

• (2)

In Case 2, the lattice $\Lambda_{1}$ is equal to $W_{1}\cap\Lambda$.

Proof.

(1) Obviously, $E_{2}=R(E_{1})=W_{2}/R(\Lambda_{1})$, i.e., $\Lambda_{2}=R(\Lambda_{1})\subset W_{2}\cap\Lambda$. On the other hand, $R(W_{2}\cap\Lambda)\subset W_{1}\cap\Lambda=\Lambda_{1}$, and applying the automorphism $R$ of $\Lambda$ gives $W_{2}\cap\Lambda\subset R(\Lambda_{1})=\Lambda_{2}$.

(2) Here, $E_{1}=R(E_{2})=W_{1}/R(\Lambda_{2})$, i.e., $\Lambda_{1}=R(\Lambda_{2})\subset W_{1}\cap\Lambda$. For the converse inclusion, observe $R(W_{1}\cap\Lambda)\subset W_{2}\cap\Lambda=\Lambda_{2}$, and applying $R$ yields again the result.

∎

We can now choose coordinates on $V$ such that $r$ is induced by a transformation of the form

[TABLE]

by choosing as the origin in $V_{1}$ a fixed point of the restriction of $r$ to $V_{1}$.

We can now view $r,s$ as affine self maps of $T$ induced by affine self maps of $E_{1}\times E_{2}\times E_{3}$ of the form

[TABLE]

[TABLE]

and sending the subgroup $H$ to itself.

Lemma 1.3**.**

The freeness of the action of the powers of $r$ is equivalent to: $H$ contains no element with last coordinate equal to $c_{3}$, or $2c_{3}$.

Moreover, $(0,0,4c_{3})\in H$.

Proof.

$r(z)=z$ is equivalent to $(z_{1}-z_{2},z_{1}+z_{2},-c_{3})\in H$. However, the endomorphism

[TABLE]

of $E_{1}\times E_{2}$ is surjective, hence $H$ cannot contain any element with last coordinate equal to $c_{3}$.

Since $r^{2}(z)=(-z_{1},-z_{2},z_{3}+2c_{3})$, $r^{2}(z)=z$ is equivalent to $(-2z_{1},-2z_{2},2c_{3})\in H$, and we reach the similar conclusion that $H$ cannot contain any element with last coordinate equal to $2c_{3}$.

Finally, the condition that $r^{4}$ is the identity is equivalent to $(0,0,4c_{3})\in H$.

∎

Proposition 1.1**.**

Case 2 does not occur.

Proof.

Since we assume that

[TABLE]

and that $s^{2}$ is the identity, it must be

[TABLE]

Consider now $rs$:

[TABLE]

The condition that $(rs)^{2}$ is the identity is equivalent to:

[TABLE]

This condition, plus the previous one, imply that

[TABLE]

contradicting Lemma 1.3.

∎

Henceforth we shall assume that we are in Case 1, and we can choose the origin in $E_{3}$ so that

[TABLE]

Lemma 1.4**.**

If

[TABLE]

then

[TABLE]

and $H$ contains no element of the form

[TABLE]

Proof.

The first condition is equivalent to $s^{2}$ being the identity, while the second is equivalent to the condition that $s$ acts freely, since $s(z)=z$ is equivalent to $(a_{1},-2z_{2}+a_{2},-2z_{3})\in H$.

∎

Proposition 1.2**.**

For each $\lambda\in\Lambda$ there exist $\lambda^{\prime}\in\Lambda,\lambda_{1}\in\Lambda_{1},\lambda_{2}\in\Lambda_{2},\lambda_{3}\in\Lambda_{3},$, such that

[TABLE]

More precisely, we even have:

[TABLE]

Proof.

Let $\lambda\in\Lambda$: we can write

[TABLE]

Furthermore, since $\lambda^{\prime}\in\operatorname{im}(I-S)$, we obtain

[TABLE]

Hence, $\lambda=\frac{\lambda_{1}}{2}+\frac{\lambda_{2}}{4}+\frac{\lambda_{3}}{4}$ for unique $\lambda_{j}\in\Lambda_{j}$.

Applying the automorphism $R$ of $\Lambda$ and the unicity of the $\lambda_{j}$ yields the result, since $R$ exchanges $\Lambda_{1}$ and $\Lambda_{2}$.

∎

Proposition 1.3**.**

We have

[TABLE]

Proof.

For $\lambda\in\Lambda$ we can write $\lambda=\frac{\lambda_{1}}{2}+\frac{\lambda_{2}}{2}+\frac{\lambda_{3}}{4}$ for unique $\lambda_{j}\in\Lambda_{j}$.

We now use the property

[TABLE]

Indeed, $2\lambda=\lambda_{1}+\lambda_{2}+\frac{\lambda_{3}}{2}$, hence $(0,0,[\frac{\lambda_{3}}{2}])\in H$ and $\frac{\lambda_{3}}{2}=0$ in $E_{3}$. Equivalently, there is an element $\lambda^{\prime}_{3}\in\Lambda_{3}$ with

[TABLE]

∎

Lemma 1.5**.**

Consider the transformation $rs$:

[TABLE]

The condition that its square is the identity amounts to

[TABLE]

while the freeness of its action is equivalent to the fact that $H$ contains no element of the form

[TABLE]

Proof.

The first condition is straighforward, while the freeness of the action is equivalent to the non existence of $(z_{1},z_{2},z_{3})$ such that

[TABLE]

As usual, we observe that for each $w_{1},w_{3}$ there exist $z_{1},z_{2},z_{3}$ with $z_{1}+z_{2}=w_{1},2z_{3}-c_{3}=w_{3}.$

∎

We put together the conclusions of Lemmas 1.3, 1.4, 1.5,

• •

(i) $(0,0,4c_{3})\in H$

• •

(ii) $(2a_{1},0,0)\in H$

• •

(iii) $(a_{1}+a_{2},-a_{1}-a_{2},0)\in H$, hence also $(a_{1}-a_{2},a_{1}+a_{2},0)\in H$.

1. (1)

$H$ contains no element of the form $(w_{1},w_{2},c_{3})$, 2. (2)

nor of the form $(w_{1},w_{2},2c_{3})$ 3. (3)

nor of the form $(a_{1},w_{2},w_{3})$ 4. (4)

nor of the form $(w_{1},w_{2},w_{3})$ with $w_{1}+a_{2}=w_{2}-a_{1}$.

It follows from (iii) and (3) that $a_{2}\neq 0$. While the condition that each element of $H$ which has two coordinates equal to zero is indeed zero (since $E_{i}$ embeds in $T$!) imply

[TABLE]

By conditions (1), (2), (3) the elements $a_{1}$, $c_{3}$ have respective orders exactly $2,4$. Moreover:

• •

(4) and (i) imply that $a_{1}+a_{2}\neq 0$

• •

(ii), (iii) and the fact that $H$ has exponent 2 implies $2a_{2}=2a_{1}=0$, $2a_{1}+2a_{2}=0$. Hence $a_{1}\neq a_{2}$ are nontrivial 2-torsion elements.

We have thus obtained the desired elements

[TABLE]

It suffices to show that $H$ is generated by $\omega:=(h+k,h+k,0)=(a_{1}+a_{2},a_{1}+a_{2},0)$.

Observe first that $\omega\in H$, by condition (iii).

Condition (4) implies that the first coordinate of an element of $H$ must be a multiple of $(a_{1}+a_{2})$: since it cannot equal $a_{1}$, by condition (3), and if it equals $a_{2}$, we can add $\omega$ and obtain an element of $H$ with first coordinate $a_{1}$. Using $R$, we infer that both coordinates must be a multiple of $(a_{1}+a_{2})$. Possibly adding $\omega$, we may assume that $w_{1}=0$: then by (4) we conclude that also $w_{2}=0$. Finally, the condition that each element of $H$ which has two coordinates equal to zero is indeed zero, show that $H$ is then generated by $\omega$, as we wanted to show.

The last assertions of the main theorem follow now in a straightforward way (see [CC17] concerning general properties of Teichmüller spaces of hyperelliptic manifolds).