# Hyperelliptic genus 3 curves with involutions and a Prym map

Paweł Borówka and Anatoli Shatsila

###### Abstract

We characterise genus 3 complex smooth hyperelliptic curves that contain two
additional involutions as curves that can be build from five points in
$\mathbb{P}^{1}$ with a distinguished triple. We are able to write down
explicit equations for the curves and all their quotient curves.

We show that, fixing one of the elliptic quotient curve, the Prym map becomes
a 2:1 map and therefore the hyperelliptic Klein Prym map, constructed recently
by the first author with A. Ortega, is also 2:1 in this case.

As a by-product we show an explicit family of $(1,d)$ polarised abelian
surfaces (for $d>1$), such that any surface in the family satisfying a certain
explicit condition is abstractly non-isomorphic to its dual abelian surface.

##  1 Introduction

Curves with involutions are interesting objects to study. The existence of
automorphisms provides a way to find models of curves (e.g. equations for
hyperelliptic curves) or periods of Jacobians (e.g. [BL04, Section 11.7]).
Moreover, involutions enable to consider (quotient) coverings that are central
constructions in the Prym theory.

In the paper, we consider hyperelliptic curves of genus 3 that contain two
additional involutions. The motivation to study them came from the Klein
covering construction that has been described in [BO18, BO20] and [BO23]. A
covering is called Klein if it is Galois 4:1 covering with the deck group
being isomorphic to the Klein four-group. A covering is called hyperelliptic
if both top and bottom curves are hyperelliptic. In [BO23] the authors show
that the Prym map of the hyperelliptic Klein covering is injective in all non-
trivial cases (i.e. when genera of curves in the construction increases).

The covering that has been left in [BO23] is the Klein covering of
hyperelliptic genus 3 curve over an elliptic curve, called baby mixed case. It
is because in such a case, one has a tower of curves $C\to E^{\prime}\to E$,
where $E,E^{\prime}$ are elliptic, so $P(C/E)=P(C/E^{\prime})$, i.e. the Klein
Prym variety equals the classical Prym variety. The main result of the paper
is the following theorem.

######  Theorem 1.1 (Theorem 3.10).

Let $\mathcal{R}^{H}_{1,4}$ be a moduli space of hyperelliptic Klein coverings
of elliptic curves branched in 4 points and let $\mathcal{E}_{d}(d,d)$ be the
moduli space of $(1,d)$ polarised abelian surfaces containing a pair of
complementary elliptic curves of exponent $d$.

Then the considered Prym map
$Pr:\mathcal{R}^{H}_{1,4}\longrightarrow\mathcal{E}_{2}(2,2)$ is finite,
dominant, not surjective map of generic degree two.

Such result is a little surprising because on the one hand the classical Prym
maps of small genus curves are usually not finite and on the other hand all
other hyperelliptic Klein Prym maps are injective.

The main idea of the proof is to show that there are precisely two smooth
hyperelliptic genus 3 curves on a general abelian surface that belongs to
$\mathcal{E}_{2}(2,2)$. In order to obtain the above result, we describe the
moduli space of genus 3 hyperelliptic curves with two additional commuting
involutions in Theorem 3.3. It turns out that they can be encoded as the
5-tuple of points in $\mathbb{P}^{1}$ with the distinguished triple (up to
projective equivalence).

As a by-product of some lemmas needed for the main result, we have been able
to show that most of abelian surfaces that belong to $\mathcal{E}_{d}(d,d)$
for $d>1$ are not isomorphic to their duals:

######  Theorem 1.2 (Theorem 2.5).

Let $E,F$ be non-isogenous elliptic curves with $P\in E[d]$ and $Q\in F[d]$
primitive $d$-torsion points. Assume that $E/P$ is not isomorphic to $E$ or
$F/Q$ is not isomorphic to $F$. Then $E\times F/\langle(P,Q)\rangle$ (of type
$(1,d)$) is not (abstractly) isomorphic to its dual.

The existence of the involutions implies that the Jacobian of the genus 3
curve is completely decomposable i.e., isogenous to the product of three
quotient elliptic curves. Such curves are classically interesting examples,
since the decomposition allows us to understand their periods in terms of
periods of curves of smaller genera, especially in terms of elliptic curves,
see [ES93, KR89, Pau08, PR17, LGRS23]. In particular, genus 3 curves with
completely decomposable Jacobians have recently gained a lot of attention
[HLP00, BO20, Kat21, MK22]. Therefore, in Proposition 4.3 we show explicit
equations of all the curves appearing in the construction. Since the Prym map
is generically 2:1, we are able to write an explicit (rational) deck
involution for the Prym map, see Proposition 4.6.

The structure of the paper is as follows. Section 2 is devoted to the theory
of non-simple abelian surfaces and the result of dual surfaces being non-
isomorphic is proved there. In Section 3 we describe the moduli space and
prove the main result concerning the Prym map. The last section of the paper
is devoted to finding equations of the curves that appear in the construction
and describing the deck involution. Moreover, we provide a construction of a
period matrix of the Prym variety in terms of periods of elliptic curves. We
conclude the paper with some problems for future work.

### Acknowledgements

Some results presented in this article are a part of the second author’s
Master Thesis at Jagiellonian University in Kraków. Both authors have been
supported by the Polish National Science Center project number
2019/35/D/ST1/02385. During the preparation of this article Anatoli Shatsila
was a participant of the tutoring programme under the Excellence Initiative at
the Jagiellonian University.

##  2 Non-simple abelian surfaces

We would like to start by recalling some results from the theory of abelian
surfaces. Let $\mathcal{A}_{2}(1,d)$ be the moduli of $(1,d)$ polarised
abelian surfaces and $\mathcal{A}_{2}=\mathcal{A}_{2}(1,1)$ be the moduli of
principally polarised ones. Recall from [BL04] and [Bor16] that for a
principally polarised abelian surface we have the following:

######  Proposition 2.1.

Let $A\in\mathcal{A}_{2}$ be a principally polarised abelian surface that
contains an elliptic curve $E$ of exponent 2. Then there exists a
complementary elliptic curve $F$ also of exponent 2 such that the addition map
$E\times F\to A$ is a polarised isogeny of degree 4. Note that $E\cap
F=E[2]=F[2]\subset A$.

Moreover, the locus of principally polarised abelian surfaces containing an
elliptic curve of exponent 2 is an irreducible surface in $\mathcal{A}_{2}$
that is an image of a so- called singular relation of discriminant $4$.

###### Proof.

The first part is proved in [BL04, Cor 12.1.2 and 12.1.4]. The second part is
essentially due to Humbert, see [Bor16, Thm 4]. ∎

Now, we would like to show a similar result for a $(1,2)$ polarised abelian
surfaces. Denote by

$\mathcal{E}_{2}(2,2)=\\{A\in\mathcal{A}_{2}(1,2):A\text{ contains a
complementary pair of elliptic curves of exponents }2,2\\}.$

Recall the notation from [BO18]: for an abelian variety $X$ and its abelian
subvarieties $M_{1},\ldots,M_{k}$ with associated symmetric idempotents
$\epsilon_{1},\ldots,\epsilon_{k}$, we write

$A=M_{1}\boxplus\ldots\boxplus M_{k}\ \text{ if }\
\epsilon_{1}+\ldots+\epsilon_{k}=1.$

In particular, for a pair of complementary elliptic curves $E,F$, we often
write $A=E\boxplus F$. As usual, we also abuse notation by identifying an
elliptic curve $E$ with its image under an embedding $E\to A$ unless we
consider the product $A=E\times F$.

The following lemma is a special case of [AB22, Thm 5.2].

######  Lemma 2.2.

The locus $\mathcal{E}_{2}(2,2)$ is a (non-empty) irreducible surface in
$\mathcal{A}_{2}(1,2)$.

###### Proof.

Let $X(2)$ be the moduli space of triples $(E,P,P^{\prime})$, where
$P,P^{\prime}\in E$ is a (symplectic) basis of 2-torsion points. Consider the
map

$\displaystyle\Phi:X(2)\times X(2)$ $\displaystyle\to\mathcal{A}_{2}(1,2)$
$\displaystyle((E,P,P^{\prime}),(F,Q,Q^{\prime}))$
$\displaystyle\mapsto((E\times F)/\langle(P,Q)\rangle,H),$

where $\Phi^{*}H=\mathcal{O}_{E}(2)\boxtimes\mathcal{O}_{F}(2).$ Since
$(P,Q)\in(E\times F)[2]$, the quotient $X=(E\times F)/\langle(P,Q)\rangle$ is
a $(1,2)$-polarised abelian surface. Moreover, it follows from [BL04, Cor.
6.3.5] that the quotient map $\pi$ is a polarised isogeny hence $\Phi$ is
well-defined. Since $X(2)$ is irreducible, it is enough to show that
$\operatorname{Im}(\Phi)=\mathcal{E}_{2}(2,2)$. Since $\pi_{|E}$ is injective,
$E$ has exponent 2 in $X$ so the image of $\Phi$ lies in
$\mathcal{E}_{2}(2,2).$ Now, take $(A,H)\in\mathcal{E}_{2}(2,2)$ and let
$(E,F)$ be a complementary pair of elliptic curves in $A$. Poincare’s
Reducibility Theorem yields an addition map $\mu$ that is a polarised isogeny
of degree 2

$\mu:(E\times F,\mathcal{O}_{E}(2)\boxtimes\mathcal{O}_{F}(2))\to(A,H).$

By [BL04, Prop. 5.3.11] we know that $\ker\mu=\langle(P,Q)\rangle$ for $P\in
E[2],Q\in F[2]$ hence $(A,H)\in\operatorname{Im}(\Phi)$. ∎

Since we want to show an auxiliary result (i.e. Theorem 2.5), from now on, let
$d>1$ and assume that $P\in E[d],\ Q\in F[d]$ are primitive $d$-torsions (i.e.
their orders are precisely $d$). In this case $A=(E\times
F)/\langle(P,Q)\rangle$ is $(1,d)$ polarised and $A\in\mathcal{E}_{d}(d,d)$.
We have the following.

######  Proposition 2.3.

If $E,F$ are non-isogenous then $A$ does not contain any other elliptic curve.

###### Proof.

Let $G$ be en elliptic curve contained in $A$ and consider the Norm map of
$E$. Note that either $\operatorname{Im}(\operatorname{Nm}_{E}|_{G})=\\{0\\}$
or $\operatorname{Im}(\operatorname{Nm}_{E}|_{G})=E$ and the latter occurs
when $E$ and $G$ are isogenous. Now, recall that
$m_{d}\operatorname{Nm}_{E}+m_{d}\operatorname{Nm}_{F}=m_{d^{2}}$ where
$m_{k}$ is multiplication by $k$ (see [BL04, p. 124]), so
$\operatorname{Im}(\operatorname{Nm}_{E}|_{G}+\operatorname{Nm}_{F}|_{G})=G$,
hence if one of the images is $0$ then $G$ is equal to the other curve.
However, if both maps are non-zero then $G$ is isogenous to both curves, hence
$E,F$ are isogenous. ∎

######  Lemma 2.4.

Let $E,F$ be complementary elliptic curves in $A=(E\times
F)/\langle(P,Q)\rangle$ of exponent d. Then $Q=-P\in A$ and $E\cap F=\langle
P\rangle\subset\ker\phi_{H}$. Moreover
$E\cap\ker\phi_{H}=F\cap\ker\phi_{H}=\langle P\rangle$.

###### Proof.

Recall that $\mu(P,0)=P\in E,\ \mu(0,Q)=Q\in F$. Now, the first equation
follows trivially from the fact that $(P,0)+(0,Q)=(P,Q)\in\ker\mu$.

To see that $E\cap F=\langle P\rangle$ one can check that
$\mu(x,0)=\mu(0,y)\iff(x,-y)\in\ker\mu=\langle(P,Q)\rangle$ and use the fact
that $P,Q$ are primitive $d$-torsions.

Now, in order to finish the proof, one needs to use the fact that
$\ker\phi_{H}=\mu((\ker\mu)^{\perp})$ where $\perp$ stands for the symplectic
complement (with respect to the symplectic pairing $\omega$), see [BL04, Prop
6.2.1 and 6.3.3]. Since $x\in E[d]$ is of the form $aP+bP^{\prime}$ (where
$a,b\in\mathbb{Z}_{d}$) with $\omega(P,P^{\prime})=1\in\mathbb{Z}_{d}$ we get
that $(aP+bP^{\prime},0)\in(\ker\mu)^{\perp}\iff b=0$. ∎

Now, we are ready to show a way to find an explicit condition that proves that
a surface is not abstractly isomorphic to its dual.

######  Theorem 2.5.

Let $E,F$ be non-isogenous elliptic curves with $P\in E[d]$ and $Q\in F[d]$
primitive $d$-torsion points. Assume that $E/P$ is not isomorphic to $E$ or
$F/Q$ is not isomorphic to $F$. Then $E\times F/\langle(P,Q)\rangle$ (of type
$(1,d)$) is not (abstractly) isomorphic to its dual.

###### Proof.

By symmetry we can assume $E/P$ is not isomorphic to $E$. Denote by $A=E\times
F/\langle(P,Q)\rangle$. Then its dual is $\hat{A}=A/\ker\phi_{H}$. By Lemma
2.4, we get that $E\cap\ker\phi_{H}=\langle P\rangle$, so
$E^{\prime}=E/\langle P\rangle$ is an elliptic curve embedded in $\hat{A}$. If
$\hat{A}$ was isomorphic to $A$ then the image of $E^{\prime}$ via the
isomorphism would be an elliptic curve embedded in $A$. By assumption it
cannot be $E$ and it cannot be $F$ since they are non-isogenous. Since this
contradicts Proposition 2.3, we get that $\hat{A}$ is not isomorphic to $A$. ∎

######  Remark 2.6.

It is well-known that a general non-principally polarised abelian variety is
not isomorphic to its dual. The advantage of Theorem 2.5 is that it provides a
family of examples that enables to write down explicit period matrices of
surfaces.

We would like to finish this section by describing better the connection
between $A$ and $\hat{A}$.

######  Proposition 2.7.

If $A=E\times F/\langle(P,Q)\rangle$, then $A/P=E^{\prime}\times F^{\prime}$
and $\hat{A}=E^{\prime}\times F^{\prime}/G$, where
$G\simeq\ker\phi_{H}/P\simeq\langle(P^{\prime},-Q^{\prime})\rangle$. In
particular, $A\in\mathcal{E}_{2}(2,2)\iff\hat{A}\in\mathcal{E}_{2}(2,2)$.

###### Proof.

Note that $A/P=(E\times F)/\langle(P,0),(0,Q)\rangle$ and
$\langle(P,0),(0,Q)\rangle$ is an isotropic subgroup of the form
$\mathbb{Z}_{d}\oplus\mathbb{Z}_{d}$ coming from
$\mathcal{O}_{E}(d)\boxtimes\mathcal{O}_{F}(d)$. Thus, $A/P=E^{\prime}\times
F^{\prime}$ (with a principal polarization) and the quotient map is a
polarised isogeny by [BL04, Cor. 6.3.5]. Since $\hat{A}=A/\ker\phi_{H}$, we
get the desired equality for $\hat{A}$. In the case $d=2$,
$(P^{\prime},-Q^{\prime})$ is a 2-torsion point of $E^{\prime}\times
F^{\prime}$, hence $\hat{A}$ lies in the image of $\Phi$, which finishes the
proof. ∎

##  3 Geometry of hyperelliptic genus 3 curves with involutions

Now, we are ready to set up the construction. We start by recalling a useful
result concerning involutions on hyperelliptic curves that is an easy
application of Accola’s Lemma [Acc69].

######  Lemma 3.1.

Let $X$ be a smooth hyperelliptic curve of genus $g=3$ with $\iota$ the
hyperelliptic involution. For any $\tau\in\operatorname{Aut}(X)$,
$\tau^{2}=\operatorname{id}$, $\tau\neq\iota$ we have

$\operatorname{Fix}(\tau)=\varnothing,\>|\operatorname{Fix}(\iota\tau)|=4,\>\>\>\text{or}\>\>\>|\operatorname{Fix}(\tau)|=4,\operatorname{Fix}(\iota\tau)=\varnothing.$

Now, let $\sigma\neq\tau$ be commuting involutions on $X$ such that
$|\operatorname{Fix}(\sigma)|=|\operatorname{Fix}(\tau)|=4$. Then
$|\operatorname{Fix}(\sigma\tau)|=4$ and $g(X/\langle\sigma,\tau\rangle)=0$.

###### Proof.

By Accola’s Lemma ([Acc69, Lemma 1]) applied to
$G_{0}=\langle\tau,\iota\rangle$ we get

$3=g(X/\tau)+g(X/\iota\tau).$

Then, Hurwitz formula yields $g(X/\tau)=2$ and $g(X/\iota\tau)=1$ or vice
versa.

For the second part, applying Accola’s Lemma ([Acc69]) to
$G_{0}=\langle\tau,\sigma\rangle$ we have

$3+2g(X/\langle\sigma,\tau\rangle)=g(X/\sigma)+g(X/\tau)+g(X/\sigma\tau).$

By Hurwitz formula, $g(X/\sigma)=g(X/\tau)=1$, hence $g(X/\sigma\tau)$ is odd.
The only possibility is $g(X/\sigma\tau)=1$, hence
$|\operatorname{Fix}(\sigma\tau)|=4$. Since the right hand side of the
equation equals 3, we get that $g(X/\langle\sigma,\tau\rangle)=0$. ∎

From now on, we will view elliptic curves both as one-dimensional abelian
varieties and genus one curves with a chosen point. We will also use the fact
that any elliptic curve is canonically isomorphic to its dual (as an abelian
variety). With such convention, we state the following result.

######  Lemma 3.2.

Let $C$ be a smooth curve of genus $g>1$ and $E$ an elliptic curve. Then the
following are equivalent:

  * i)

there exists a double covering $f:C\to E$;

  * ii)

$E$ can be embedded into $JC$ as an abelian subvariety of exponent $2$.

###### Proof.

The equivalence follows from the fact that $f$ does not factorise via a cyclic
étale covering (because it is branched of degree 2), hence $f^{*}\circ
Nm_{f}=Nm_{E}$, see [BL04, Prop 12.3.2]. ∎

In [BO23] the authors have investigated hyperelliptic curves with additional
commuting involutions and they have shown the injectivity of the hyperelliptic
Klein Prym map in a general case. The only case that has been left is a genus
3 hyperelliptic curve that is a Klein covering of an elliptic curve. Now, we
are ready to investigate such a case.

Let $\tilde{C}$ be a hyperelliptic genus 3 curve such that
$(\mathbb{Z}/2\mathbb{Z})^{3}\simeq\langle\iota,\sigma,\tau\rangle\subseteq\operatorname{Aut}(\tilde{C})$.
Here $\iota$ denotes the hyperelliptic involution on $\tilde{C}$. In addition,
we assume that $\sigma$ and $\tau$ each have 4 fixed points and hence
$\sigma\tau$ also has 4 fixed points (see Lemma 3.1). We also get
$|\operatorname{Fix}(\iota\tau)|=|\operatorname{Fix}(\iota\sigma)|=|\operatorname{Fix}(\iota\sigma\tau)|=0$.
We introduce the following notation for quotient curves:

$\displaystyle E_{\sigma}:=\tilde{C}/\langle\sigma\rangle,$ $\displaystyle\ \
\ \ E_{\tau}:=\tilde{C}/\langle\tau\rangle,$ $\displaystyle
E_{\sigma\tau}:=\tilde{C}/\langle\sigma\tau\rangle,$ $\displaystyle\ \ \ \
E_{\sigma,\iota\tau}:=\tilde{C}/\langle\sigma,\iota\tau\rangle,$
$\displaystyle
E_{\tau,\iota\sigma}:=\tilde{C}/\langle\tau,\iota\sigma\rangle,$
$\displaystyle\ \ \ \
E_{\iota\sigma,\iota\tau}:=\tilde{C}/\langle\iota\sigma,\iota\tau\rangle,$
$\displaystyle C_{\iota\sigma}:=\tilde{C}/\langle\iota\sigma\rangle,$
$\displaystyle\ \ \ \ C_{\iota\tau}:=\tilde{C}/\langle\iota\tau\rangle,$
$\displaystyle C_{\iota\sigma\tau}:=\tilde{C}/\langle\iota\sigma\tau\rangle.$

Note that Hurwitz formula yields that the six curves denoted by letter $E$ are
elliptic, while the other three denoted by $C$ have genus 2. We have the
following commutative diagram, where by +4 (resp. +2) we indicate that a
degree two map is branched at 4 points (resp. at 2 points).

${\tilde{C}}$${C_{\iota\tau}}$${C_{\iota\sigma}}$${C_{\iota\sigma\tau}}$${E_{\tau}}$${E_{\sigma}}$${E_{\sigma\tau}}$${{E_{\tau,\iota\sigma}}}$${{E_{\sigma,\iota\tau}}}$${{E_{\iota\sigma,\iota\tau}}}$$\scriptstyle{et}$$\scriptstyle{et}$$\scriptstyle{et}$$\scriptstyle{+4}$$\scriptstyle{+4}$$\scriptstyle{+4}$$\scriptstyle{+2}$$\scriptstyle{+2}$$\scriptstyle{et}$$\scriptstyle{et}$$\scriptstyle{et}$

Our aim is to investigate the diagram above. We start by showing how to
associate five points on $\mathbb{P}^{1}$ (with a distinguished triple) to a
curve $\tilde{C}$. Let
$w,\sigma(w),\tau(w),\sigma\tau(w),w^{\prime},\sigma(w^{\prime}),\tau(w^{\prime}),\sigma\tau(w^{\prime})$
be the eight Weierstrass points of $\tilde{C}$ and

$\displaystyle\operatorname{Fix}(\sigma)$ $\displaystyle=\\{p_{1},p_{2},\iota
p_{1},\iota p_{2}\\},$ $\displaystyle\operatorname{Fix}(\tau)$
$\displaystyle=\\{q_{1},q_{2},\iota q_{1},\iota q_{2}\\},$
$\displaystyle\operatorname{Fix}(\sigma\tau)$
$\displaystyle=\\{r_{1},r_{2},\iota r_{1},\iota r_{2}\\}.$

A proof that all points defined above are distinct from each other can be
found in [BO23].

Let $f:\tilde{C}\to\tilde{C}/\langle\sigma,\tau\rangle\simeq\mathbb{P}^{1}$ be
the quotient $4:1$ map and $[w]$ and $[w^{\prime}]$ be the images of
Weierstrass points under $f$. Let $g:\mathbb{P}^{1}\to\mathbb{P}^{1}$ be a 2:1
map ramified at $[w]$ and $[w^{\prime}]$ and $[p],[q],[r]$ be the images under
$g\circ f$ of fixed points of $\sigma,\tau$ and $\sigma\tau$ respectively.
Finally, let $h:\mathbb{P}^{1}\to\mathbb{P}^{1}$ be a 4:1 map branched at
$[p],[q],[r]$ with two simple ramifications at each fiber and
$\pi_{\tilde{C}}:\tilde{C}\to\mathbb{P}^{1}$ be the hyperelliptic $2:1$ map.
In this way we have constructed the following commutative diagram

${\tilde{C}}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$$\scriptstyle{\pi_{\tilde{C}}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{h}$

where

$h\circ\pi_{\tilde{C}}=g\circ
f:\tilde{C}\to\tilde{C}/\langle\sigma,\tau,\iota\rangle=\mathbb{P}^{1}$

is the $8:1$ quotient map. We have a 5-tuple of points lying in the bottom-
right $\mathbb{P}^{1}$, namely

$(g([w]),g([w^{\prime}]);[p],[q],[r]).$

This is the 5-tuple associated to $\tilde{C}$. Here, $[p],[q]$ and $[r]$ form
a distinguished triple.

On the other hand, starting with a 5-tuple with a distinguished triple

$(x,y;[p],[q],[r])$

we can reconstruct the curve $\tilde{C}$ as follows. Take a
$\mathbb{Z}_{2}^{2}$-Galois covering branched at $[p],[q],[r]$. The preimages
of $x,y$ give us 8 points $w_{1},\ldots,w_{8}$. Then, $\tilde{C}$ is a double
cover of $\mathbb{P}^{1}$ branched at these points. Since $w_{1},\ldots,w_{8}$
are invariant under the action of $\mathbb{Z}_{2}^{2}$, there are two
additional commuting involutions on $\tilde{C}$ lifted from $\mathbb{P}^{1}$.

######  Theorem 3.3.

The following are equivalent:

  1. 1.

the locus of 5 points in $\mathbb{P}^{1}$ with a chosen triple (up to
projective equivalence respecting the distinguished triple);

  2. 2.

the locus of hyperelliptic genus 3 curves having two additional commuting
involutions (different from the hyperelliptic one);

  3. 3.

the locus of genus 3 curves $\tilde{C}$ having $\mathbb{Z}^{3}_{2}\subset
Aut(\tilde{C})$;

  4. 4.

the locus of hyperelliptic curves given by
$y^{2}=(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1)$, $a,b\in\mathbb{C},\ a\neq b$
$a^{2},b^{2}\neq 4$.

###### Proof.

The equivalence $(1)\Leftrightarrow(2)$ follows from the construction. The
implication $(2)\Rightarrow(3)$ is obvious while $(3)\Rightarrow(2)$ comes
from Accola’s lemma ([Acc69]). The equivalence $(2)\Leftrightarrow(4)$ follows
from Proposition 4.2 that is proved independently. ∎

######  Remark 3.4.

Note that irreducibility of the moduli space and its dimension being equal to
two follows easily from point 4 of Theorem 3.3. One can also see Theorem 3.3
as a characterisation of the moduli space of hyperelliptic curves that have a
completely decomposed Richelot isogeny, see [Kat21].

###  3.1 From bottom to up and a Prym map

Now, we would like to describe the construction starting from one of elliptic
curves, say $E_{\sigma}$. We start with a characterisation lemma that is
similar to Theorem 3.3.

######  Lemma 3.5.

The following are equivalent:

  1. 1.

the locus of 5 points in $\mathbb{P}^{1}$ with a distinguished triple and a
point in the triple (up to projective equivalence respecting the triple and
the chosen point);

  2. 2.

the locus of pairs $(\tilde{C},\sigma)$, where
$\mathbb{Z}_{2}^{3}\subset\operatorname{Aut}(\tilde{C})$ and $\sigma$ is an
involution with $|\operatorname{Fix}(\sigma)|=4$;

  3. 3.

the locus of pairs $(E,B)$ where E is an elliptic curve and $B$ is a set of 4
points invariant under both $(-1)$ and a translation by a 2-torsion point (and
$B\cap E[2]=\emptyset$);

  4. 4.

the locus of genus 2 curves $C$ that are double coverings of elliptic curves
together with a pair of Weierstrass points on $C$ whose images under both
covering maps are equal.

###### Proof.

The equivalence $(1)\Leftrightarrow(2)$ follows from the construction, since
there is a one to one correspondence between involutions on $\tilde{C}$ with
four fixed points in the case $(2)$ and the points of the distinguished triple
in the case $(1)$.

To obtain $(2)\Rightarrow(3)$ we take $E:=E_{\sigma}$ and $B$ equal to the
branch locus of the covering defined by $\sigma$. To show $(3)\Rightarrow(2)$
one considers the covering branched at $B$ and defined by the line bundle
$\eta=\mathcal{O}_{E}(2O)$. Similarly, to get $(2)\Rightarrow(4)$ we take
$C=\tilde{C}/\iota\sigma$. We leave the details of the implications to the
reader since they are straightforward applications of the construction.

It remains to show $(4)\Rightarrow(2)$. Let $w_{1},w_{2}$ be two Weierstrass
points on $C$ whose images under the coverings of elliptic curves are equal.
We define $\tilde{C}\to C$ to be an étale hyperelliptic covering defined by
$\mathcal{O}_{C}(w_{1}-w_{2})$. Let $\sigma$ be an involution of $\tilde{C}$
exchanging the sheets of the covering. Then $(\tilde{C},\iota\sigma)$ is the
corresponding pair, because the covering involution of $C/E$ can be lifted to
an involution on $\tilde{C}$ that commutes with $\sigma$. ∎

######  Remark 3.6.

Note that changing a distinguished point within a triple means that one starts
with another elliptic curve (another involution) but the genus 3 curve
$\tilde{C}$ is the same.

Using the fact that $\tilde{C}/\langle\sigma,\tau\rangle\simeq\mathbb{P}^{1}$
we get that

$J\tilde{C}=E_{\sigma}\boxplus E_{\tau}\boxplus E_{\sigma\tau}.$

Recall that the Prym variety of the covering $f:X\to Y$ can be defined as the
complementary abelian subvariety to $f^{*}(JY)$ in $JX$ and is usually denoted
by $P(f)$ or $P(X/Y)$, see [BL04, Chapter 12]. With such notation, we can
write

$JX=f^{*}(JY)\boxplus P(X/Y).$

In our case, by choosing $\sigma$ we get that the Prym variety of the covering
$\tilde{C}\to E_{\sigma}$ is equal to
$P(\tilde{C}/E_{\sigma})=E_{\tau}\boxplus E_{\sigma\tau}$. Therefore, since
the Prym variety is complementary to the elliptic curve of exponent 2 and by
Lemma 3.2, we get that

$P(\tilde{C}/E_{\sigma})\in\mathcal{E}_{2}(2,2).$

######  Remark 3.7.

Note that $P(\tilde{C}/E_{\sigma})=P(\tilde{C}/E_{\sigma,\iota\tau})$, hence
we also study Pryms of hyperelliptic Klein coverings.

The main point of the Prym theory is to consider the Prym map that assigns to
a covering its Prym variety, see for example [Bea77, BO22]. Let

$\mathcal{R}^{H}_{1,4}=\\{(\tilde{C},\sigma)\>|\>g(\tilde{C})=3,(\mathbb{Z}/2\mathbb{Z})^{3}\subset\operatorname{Aut}(\tilde{C}),\sigma^{2}=\operatorname{id},|\operatorname{Fix}(\sigma)|=4\\}$

be the moduli space considered in Lemma 3.5. It follows that we have a well-
defined Prym map

$\Pr:\mathcal{R}^{H}_{1,4}\to\mathcal{E}_{2}(2,2).$

We want to find the generic degree of $\Pr$. We start by recalling some
results.

######  Proposition 3.8 ([Bar87], Duality Theorem 1.12).

Let $C$ be a hyperelliptic genus 3 curve and let $A$ be a $(1,2)$ polarised
abelian surface. Then the following are equivalent:

  1. 1.

$C$ can be embedded in $\hat{A}$;

  2. 2.

$A$ can be embedded in $JC$;

  3. 3.

$C$ is a double covering of an elliptic curve that is complementary to $A$ in
$JC$.

Note that by [Bor16, Prop 6], there are precisely three smooth hyperelliptic
curves on a general $(1,2)$ polarised abelian surface. Now, we are ready to
show a similar result in our case.

######  Proposition 3.9.

On $A\in\mathcal{E}_{2}(2,2)$ there are at most two smooth hyperelliptic genus
3 curves (up to a translation) and there is an open dense set
$U\subset\mathcal{E}_{2}(2,2)$ such that for $A\in U$ there are precisely two
smooth curves on $A$.

###### Proof.

Firstly, by [Bor16, Prop 6], there is a 1:1 correspondence between smooth
hyperelliptic curves on $A$ and polarised isogenies to Jacobians of smooth
curves and there are at most three of them.

Recall that a principally polarised abelian surface is either a Jacobian of a
smooth curve or a polarised product of two elliptic curves. Hence if $\pi:A\to
E^{\prime}\times F^{\prime}$ is a polarised isogeny of degree 2, then
$\pi^{-1}(E^{\prime}\times\\{0\\})$ and $\pi^{-1}(\\{0\\}\times F^{\prime})$
are elliptic curves in $A$. By Proposition 2.3 there are precisely two
elliptic curves on a general $A\in\mathcal{E}_{2}(2,2)$.

Moreover, if $A=E\times F/\langle(R,Q)\rangle$, then $A/\langle
R\rangle=E^{\prime}\times F^{\prime}$, so indeed there is at least one product
of elliptic curves and at most two Jacobians of smooth curves downstairs.

Let $U=\\{A:\exists P,Q\in A[2],A/P\text{ and }A/Q\text{ are Jacobians of
smooth curves}\\}$. Note that $U$ is open and dense because the set of
Jacobians is open and dense. ∎

For low genera, the Prym map is usually dominant and with positive dimensional
fibres for dimensional reasons. Since we are interested in special coverings,
both domain and codomain have the same dimension, so we can consider the
(generic) degree of the Prym map. Now we would like to compute the degree of
the Prym map.

######  Theorem 3.10.

The considered Prym map
$Pr:\mathcal{R}^{H}_{1,4}\longrightarrow\mathcal{E}_{2}(2,2)$ is finite,
dominant, not surjective map of generic degree two.

###### Proof.

Let $P\in\mathcal{E}_{2}(2,2)$. By Proposition 2.7, we get that
$\widehat{P}\in\mathcal{E}_{2}(2,2)$, so by Proposition 3.9 there exists at
most 2 smooth genus 3 hyperelliptic curves (precisely 2 on a general $P$),
called $C_{i}$ (that are coverings of $E_{i}$) that are embedded into
$\widehat{P}$. On the other hand, Proposition 3.8 says that this is a
necessary and sufficient condition for $P$ to be the Prym of a covering
$P(C_{i}/E_{i})$. The fact that $Pr$ is not surjective follows from Example
3.11. ∎

######  Example 3.11.

There exists $A\in\mathcal{E}_{2}(2,2)$ not in the image of $Pr$.

###### Proof.

Let $E$ be an elliptic curve given by the equation $y^{2}=x^{3}+x$. Then
$\operatorname{Aut}(E)=\langle\alpha\rangle\simeq\mathbb{Z}/4\mathbb{Z}$. Let
$E_{1}=E_{2}=E$. We denote non-zero two-torsion points on $E_{i}$ by
$e_{i},f_{i},e_{i}+f_{i}$ in such a way that

$\alpha(e_{i})=e_{i},\ \alpha(f_{i})=e_{i}+f_{i},\ \alpha(e_{i}+f_{i})=f_{i}.$

We denote four elliptic curves isomorphic to $E$ and lying in $E_{1}\times
E_{2}$ as follows:

$\displaystyle E_{\Delta}$ $\displaystyle:=\\{(s,s):s\in E\\};$ $\displaystyle
E_{-\Delta}$ $\displaystyle:=\\{(s,-s):s\in E\\};$ $\displaystyle E_{\alpha}$
$\displaystyle:=\\{(s,\alpha(s)):s\in E\\};$ $\displaystyle E_{-\alpha}$
$\displaystyle:=\\{(s,-\alpha(s)):s\in E\\}.$

Let $A=(E_{1}\times E_{2})/\langle(e_{1},e_{2})\rangle\in\mathcal{E}_{2}(2,2)$
and $\mu$ be the quotient map. We abuse the notation by writing
$\mu(x_{1})=\mu(x_{1},0)$ and $\mu(x_{2})=\mu(0,x_{2})$. In this way we can
write $\mu(e_{1})=\mu(e_{2})\in A$. In particular, we get that that
$\mu(E_{1})\cap\mu(E_{2})=\\{0,\mu(e_{1})\\}$.

Note that $\mu(E_{\Delta})$ and $\mu(E_{-\Delta})$ are isomorphic to
$E/\langle e\rangle$ and are complementary in $A$ of exponent 2, and
$\mu(E_{\Delta})\cap\mu(E_{-\Delta})=\\{0,\mu(f_{1},f_{2})\\}$. Analogously,
$\mu(E_{\alpha})$ and $\mu(E_{-\alpha})$ are isomorphic to $E/\langle
e\rangle$ and complementary in $A$ of exponent 2, and
$\mu(E_{\alpha})\cap\mu(E_{-\alpha})=\\{0,\mu(e_{1}+f_{1},f_{2})\\}$.

Recall that we compute the kernel of the polarising isogeny by
$\ker\phi_{A}=\mu((\ker\mu)^{\perp})$, hence
$\ker\phi_{A}=\\{0,\mu(e_{1}),\mu(f_{1},f_{2}),\mu(e_{1}+f_{1},f_{2})\\}$. But
we have

$\displaystyle A/\langle\mu(e_{1})\rangle$ $\displaystyle=E_{1}/\langle
e_{1}\rangle\times E_{2}/\langle e_{2}\rangle,$ $\displaystyle
A/\langle\mu(f_{1},f_{2})\rangle$
$\displaystyle=E_{\Delta}/E_{\Delta}[2]\times E_{-\Delta}/E_{-\Delta}[2],$
$\displaystyle A/\langle\mu(e_{1}+f_{1},f_{2})\rangle$
$\displaystyle=E_{3}/E_{3}[2]\times E_{4}/E_{4}[2],$

where all products are principally polarized. Since none of the quotients is a
Jacobian of a smooth genus 2 curve, it follows that $A$ is not in the image of
$\Pr$. ∎

######  Remark 3.12.

By [Bor16], a general $A\in\mathcal{A}_{(}1,2)$ contains three smooth
hyperelliptic genus 3 curves. By Proposition 3.9, a general $A=E\boxplus
F\in\mathcal{E}_{2}(2,2)$ contains two smooth hyperelliptic genus 3 curves and
a singular curve $E\cup F$ of arithmetic genus 3. One can use the proof of
Example 3.11 to show that on $E\boxplus E$ there are two singular curves
$E\cup E$ and $E_{\Delta}\cup E_{-\Delta}$ of arithmetic genus 3, hence the
degree of the Prym map at $E\times E/\langle(R,R)\rangle$ is at most one. In
the special case when $E$ has automorphism of order 4, we have got three
singular curves of arithmetic genus 3 and no smooth genus 3 hyperelliptic
curves, so the Prym map is not surjective.

##  4 Equations

In this section we find equations of curves that appear in the construction.
We start with a useful lemma.

######  Lemma 4.1.

Let $C$ be a smooth hyperelliptic curve of genus 2 with a non-hyperelliptic
involution $\sigma$. Then $C$ is isomorphic to a curve given by the equation

$y^{2}=(x^{2}-1)(x^{2}-t_{1}^{2})(x^{2}-t_{2}^{2})$

for some $t_{1},t_{2}\in\mathbb{C}$ satisfying $t_{1}\neq\pm t_{2},\
t_{1}^{2},t_{2}^{2}\neq 1$.

###### Proof.

Since $C$ is hyperelliptic, the involution $\sigma$ induces an involution
$\overline{\sigma}$ of $\mathbb{P}^{1}$ and we have a commutative diagram

${C}$${C}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$$\scriptstyle{\sigma}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\scriptstyle{\overline{\sigma}}$

By Hurwitz’s formula applied to
$\pi_{\overline{\sigma}}:\mathbb{P}^{1}\to\mathbb{P}^{1}/\overline{\sigma}\simeq\mathbb{P}^{1}$,
we see that $\overline{\sigma}$ has two fixed points $p,q\in\mathbb{P}^{1}$.
Let $r=\pi(w_{1})$ be an image of some Weierstrass point on $C$. By changing
the coordinates on $\mathbb{P}^{1}$ we can assume that $p=[1:0],q=[0:1]$ and
$r=[1:1]$. It follows that $\overline{\sigma}([x:y])=[-x:y]$. Without loss of
generality, we can assume that $\sigma(x,y)=(-x,y)$ and, since $\sigma$
preserves Weierstrass points, $C$ has the defining equation

$y^{2}=(x^{2}-1)(x^{2}-t_{1}^{2})(x^{2}-t_{2}^{2})$

for some $t_{1},t_{2}\in\mathbb{C}$ satisfying $1\neq t_{1}^{2}\neq
t_{2}^{2}\neq 1$. ∎

######  Proposition 4.2.

Let $\tilde{C}$ be a smooth hyperelliptic curve of genus $3$ such that
$\operatorname{Aut}(\tilde{C})=\langle\sigma,\tau,\iota\rangle$ with
$|\operatorname{Fix}(\sigma)|=|\operatorname{Fix}(\tau)|=4$ and $\iota$ the
hyperelliptic involution. Then $\tilde{C}$ is isomorphic to a smooth
hyperelliptic curve given by the equation

$y^{2}=(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1),$

where $(a,b)\in(\mathbb{C}\setminus\\{-2,2\\})^{2}\setminus\Delta$.

###### Proof.

Let $\overline{\sigma}$ and $\overline{\tau}$ be the involutions of
$\mathbb{P}^{1}$ induced by $\sigma$ and $\tau$ respectively. As before, we
can assume that $\overline{\sigma}([x:y])=[-x:y]$. It is easy to see that any
involution $j$ of $\mathbb{P}^{1}$ is given by $j([x:y])=[ax+by:cx-ay]$ with
$a^{2}+bc=1$. Thus, we can assume that $\overline{\tau}$ is of this form. Now,
since $\overline{\tau}$ and $\overline{\sigma}$ commute, we have
$\overline{\tau}([x:y])=[ky:x]$, for some $k\in\mathbb{C}$. By Lemma 3.1 and
Hurwitz formula we get $g(\tilde{C}/\iota\tau)=2$, and the following diagram
commutes:

${\tilde{C}}$${\tilde{C}}$${\tilde{C}/\iota\tau}$${\tilde{C}/\iota\tau}$${\mathbb{P}^{1}/\overline{\tau}}$${\mathbb{P}^{1}/\overline{\tau}}$$\scriptstyle{\sigma}$$\scriptstyle{\pi_{\iota\tau}}$$\scriptstyle{\pi_{\iota\tau}}$$\scriptstyle{\sigma^{\prime}}$$\scriptstyle{\pi_{\tilde{C}/\iota\tau}}$$\scriptstyle{\pi_{\tilde{C}/\iota\tau}}$$\scriptstyle{\overline{\sigma}(mod\overline{\tau})}$

By Lemma 4.1, we can assume that $\tilde{C}/\iota\tau$ has the defining
equation

$y^{2}=(x^{2}-1)(x^{2}-t_{1}^{2})(x^{2}-t_{2}^{2})$

over $\mathbb{P}^{1}/\overline{\tau}\simeq\mathbb{P}^{1}$ and, moreover, the
preimages $p_{1},p_{2}$ under $\pi_{\iota\tau}$ of the Weierstrass point
$(1,0)$ on $\tilde{C}/\iota\tau$ are not the Weierstrass points on
$\tilde{C}$. Then $\pi_{\tilde{C}}(p_{1})=\pi_{\tilde{C}}(p_{2})=[1:1]$ is a
fixed point of $\overline{\tau}$, hence $k=1$. It follows that the Weierstrass
points on $\tilde{C}$ are $(\pm t_{1},0),(\pm
t_{2},0),(\pm\frac{1}{t_{1}},0),(\pm\frac{1}{t_{2}},0)$ for some
$t_{1},t_{2}\in\mathbb{C}$, hence the defining equation of $\tilde{C}$ is

$y^{2}=(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1)$

with $a=-t_{1}^{2}-\frac{1}{t_{1}^{2}},b=-t_{2}^{2}-\frac{1}{t_{2}^{2}}$. In
order for $\tilde{C}$ to be smooth the Weierstrass points must be pairwise
distinct. Thus, we have
$(a,b)\in(\mathbb{C}\setminus\\{-2,2\\})^{2}\setminus\Delta$. ∎

The following proposition can be seen as a first step to see the construction
in coordinates.

######  Proposition 4.3.

Let $\tilde{C}$ be a hyperelliptic genus 3 curve given by the equation

$y^{2}=(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1)$

and $\sigma,\tau$ be the involutions of $\tilde{C}$ such that the involutions
$\overline{\sigma},\overline{\tau}$ of the projective line induced by $\sigma$
and $\tau$ are given by

$\overline{\sigma}([x:y])=[-x:y],\>\>\>\overline{\tau}([x:y])=[y:x].$

Then the quotient curves have the following defining equations:

$\displaystyle C_{\iota\tau}:\>\>\>y^{2}$
$\displaystyle=(x^{2}-4)(x^{2}+a-2)(x^{2}+b-2),$ $\displaystyle
C_{\iota\sigma}:\>\>\>y^{2}$ $\displaystyle=x(x^{2}+ax+1)(x^{2}+bx+1),$
$\displaystyle C_{\iota\sigma\tau}:\>\>\>y^{2}$
$\displaystyle=(x^{2}+4)(x^{2}+a+2)(x^{2}+b+2),$ $\displaystyle
E_{\tau}:\>\>\>y^{2}$ $\displaystyle=(x^{2}+a-2)(x^{2}+b-2),$ $\displaystyle
E_{\sigma}:\>\>\>y^{2}$ $\displaystyle=(x^{2}+ax+1)(x^{2}+bx+1),$
$\displaystyle E_{\sigma\tau}:\>\>\>y^{2}$
$\displaystyle=(x^{2}+a+2)(x^{2}+b+2),$ $\displaystyle
E_{\iota\sigma,\iota\tau}:\>\>\>y^{2}$ $\displaystyle=(x+a)(x+b)(x-2),$
$\displaystyle E_{\iota\tau,\sigma}:\>\>\>y^{2}$
$\displaystyle=(x+a)(x+b)(x-2)(x+2),$ $\displaystyle
E_{\iota\sigma,\tau}:\>\>\>y^{2}$ $\displaystyle=(x+a)(x+b)(x+2).$

###### Proof.

In the proof of Proposition 4.2 we found the equation of $C_{\iota\tau}$ over
$\mathbb{P}^{1}/{\overline{\tau}}$. We want to find its equation over
$\mathbb{P}^{1}$. The isomorphism
$\varphi:\mathbb{P}^{1}/{\overline{\tau}}\to\mathbb{P}^{1}$ is given by
$\varphi([x:1])=[x+\frac{1}{x}:1]$ for $x\neq 0$ and $\varphi([0:1])=[1:0]$,
hence the defining equation of $C_{\iota\tau}$ over $\mathbb{P}^{1}$ is given
by

$y^{2}=(x^{2}-4)\left(x^{2}-\left(t_{1}+\frac{1}{t_{1}}\right)^{2}\right)\left(x^{2}-\left(t_{2}+\frac{1}{t_{2}}\right)^{2}\right)=(x^{2}-4)(x^{2}+a-2)(x^{2}+b-2).$

From the commutative diagram below we see that the Weierstrass points on
$E_{\sigma}$ are the images under $\pi_{\sigma}$ of points in
$\operatorname{Fix}(\iota)\cup\operatorname{Fix}(\iota\sigma)=\operatorname{Fix}(\iota)$
since $\iota\sigma$ is fixed-point free.

${C}$${C}$${E_{\sigma}}$${E_{\sigma}}$$\scriptstyle{\iota}$$\scriptstyle{\pi_{\sigma}}$$\scriptstyle{\pi_{\sigma}}$$\scriptstyle{\iota_{C}}$

Thus, $E_{\sigma}$ has the following defining equation over
$\mathbb{P}^{1}/{\overline{\sigma}}:$

$y^{2}=(x-t_{1})\left(x-\frac{1}{t_{1}}\right)(x-t_{2})\left(x-\frac{1}{t_{2}}\right).$

Transforming the equations by
$\varphi:\mathbb{P}^{1}/{\overline{\sigma}}\to\mathbb{P}^{1}$ with
$\varphi([x:y])=[x^{2}:y]$ we get the equation

$y^{2}=(x-t_{1}^{2})\left(x-\frac{1}{t_{1}^{2}}\right)(x-t_{2}^{2})\left(x-\frac{1}{t_{2}^{2}}\right)=(x^{2}+ax+1)(x^{2}+bx+1).$

Analogously, the defining equations of $E_{\tau}$ and $E_{\sigma\tau}$ are
given by

$y^{2}=\left(x^{2}-\left(t_{1}+\frac{1}{t_{1}}\right)^{2}\right)\left(x^{2}-\left(t_{2}+\frac{1}{t_{2}}\right)^{2}\right)=(x^{2}+a-2)(x^{2}+b-2)$

and

$y^{2}=\left(x^{2}-\left(t_{1}-\frac{1}{t_{1}}\right)^{2}\right)\left(x^{2}-\left(t_{2}-\frac{1}{t_{2}}\right)^{2}\right)=(x^{2}+a+2)(x^{2}+b+2)$

respectively.

The Weierstrass points of $C_{\iota\sigma}$ are the images under
$\pi_{\iota\sigma}$ of points in
$\operatorname{Fix}(\iota)\cup\operatorname{Fix}(\sigma)$. That is,

$W(C_{\iota\sigma})=\left\\{\pi_{\iota\sigma}((t_{1},0)),\pi_{\iota\sigma}\left(\left(\frac{1}{t_{1}},0\right)\right),\pi_{\iota\sigma}((t_{2},0)),\pi_{\iota\sigma}\left(\left(\frac{1}{t_{2}},0\right)\right),\pi_{\iota\sigma}((0,1)),\pi_{\iota\sigma}(\infty)\right\\}.$

Hence, the defining equation of $C_{\iota\sigma}$ is

$y^{2}=x(x-t_{1}^{2})\left(x-\frac{1}{t_{1}^{2}}\right)(x-t_{2}^{2})\left(x-\frac{1}{t_{2}^{2}}\right)=x(x^{2}+ax+1)(x^{2}+bx+1).$

Analogously,

$W(C_{\sigma\tau})=\\{\pi_{\sigma\tau}((t_{1},0)),\pi_{\sigma\tau}((t_{2},0)),\pi_{\sigma\tau}((-t_{1},0)),\pi_{\sigma\tau}((-t_{2},0)),\pi_{\sigma\tau}((i,y_{0})),\pi_{\sigma\tau}((-i,y_{0}))\\},$

where $y_{0}$ is such that $y_{0}^{2}=(2-a)(2-b)$. It follows that the
defining equation of $C_{\sigma\tau}$ is

$y^{2}=(x^{2}+4)\left(x^{2}-\left(t_{1}-\frac{1}{t_{1}}\right)^{2}\right)\left(x^{2}-\left(t_{2}-\frac{1}{t_{2}}\right)^{2}\right)=(x^{2}+4)(x^{2}+a+2)(x^{2}+b+2).$

One can find equations of remaining elliptic curves using the fact that they
are quotients of $C_{\iota\tau}$ and $C_{\iota\sigma\tau}$. For instance,
since $E_{\iota\sigma,\iota\tau}=C_{\iota\tau}/\iota\sigma$, the Weierstrass
points of $E_{\iota\sigma,\iota\tau}$ are $-a+2,-b+2,\infty,4$. Applying the
projective transformation translating these points by $-2$ we get the defining
equation of $E_{\iota\sigma,\iota\tau}$

$y^{2}=(x+a)(x+b)(x+2).$

One finds equations of the remaining elliptic curves in a similar way.

∎

######  Remark 4.4.

The equation of $\tilde{C}$ can be found in [Sha03] and the equations of
$E_{\sigma},E_{\tau}$ and $E_{\sigma\tau}$ can be found in [LGRS23].

We can describe the correspondence established in Theorem 3.3 in terms of
equations. This will help us to describe an involution of the moduli space in
a convenient form (see Proposition 4.6).

######  Proposition 4.5.

Let $\tilde{C}$ be a hyperelliptic genus 3 curve defined by the equation

$y^{2}=(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1).$

Then the 5-tuple of points with a distinguished triple on the projective line
corresponding to $\tilde{C}$ is

$([-a:1],[-b:1];[1:0],[2:1],[-2:1]).$

Conversely, given a 5-tuple of points with a distinguished triple on the
projective line, we can find a projective transformation to a 5-tuple of the
form

$([-a:1],[-b:1];[1:0],[2:1],[-2:1])$

for some $a,b\in\mathbb{C}$. Then the curve $\tilde{C}$, corresponding to the
5-tuple is given by the equation

$y^{2}=(x^{4}+ax^{2}+1)(x^{4}+bx^{2}+1).$

###### Proof.

Recall that we have a commutative diagram

${\tilde{C}}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$${\mathbb{P}^{1}}$$\scriptstyle{\pi_{\tilde{C}}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{h}$

One can see from the fact that $\overline{\sigma}([x:y])=[-x:y]$ and
$\overline{\tau}([x:y])=[y:x]$ that the map $h$ is given by
$h([x:y])=[x^{4}+y^{4}:x^{2}y^{2}]$ (here we use notation from Proposition
4.2). Also note that

$\displaystyle\operatorname{Fix}(\overline{\sigma})$
$\displaystyle=\\{[1:0],[0:1]\\},$
$\displaystyle\operatorname{Fix}(\overline{\tau})$
$\displaystyle=\\{[1:1],[-1:1]\\},$
$\displaystyle\operatorname{Fix}(\overline{\sigma})$
$\displaystyle=\\{[i:1],[-i:1]\\}.$

Thus, the 5-tuple we get is

$(h([t_{1}:1]),h([t_{2}:1]);h([1:0]),h([1:1]),h([i:1])=([-a:1],[-b:1];[1:0],[2:1],[-2:1]).$

The second part of the proposition follows from the fact that the 5-tuple with
distinguished triple is defined up to projective equivalence. ∎

In the following proposition we find the involution on the moduli space
corresponding to the involution exchanging points in the fibres of the Prym
map in Theorem 3.10.

######  Proposition 4.6.

Let

$\varphi:\mathcal{R}^{H}_{1,4}\ni([-a:1],[-b:1];[1:0];[2:1],[-2:1])\mapstochar\dashrightarrow\\\
([-b:1],[-a:1];[1:0];[-2-a-b:1],[-2:1])\in\mathcal{R}^{H}_{1,4}$ (4.1)

be the rational involution on $\mathcal{R}^{H}_{1,4}$. Here we modify the
notation from Proposition 4.5 to indicate, that $[1:0]$ is the distinguished
point in the triple.

Then $\Pr(T)=\Pr(\varphi(T))$ whenever $\varphi$ is defined on
$T\in\mathcal{R}^{H}_{1,4}$.

###### Proof.

Let $\tilde{C}/E_{\sigma}$ and $\tilde{C}^{\prime}/E_{\sigma}^{\prime}$ be the
coverings associated to the 5-tuple

$([-a:1],[-b:1];[1:0];[2:1],[-2:1]),$

and

$([-b:1],[-a:1];[1:0];[-2-a-b:1],[-2:1])$

respectively. Note that the latter 5-tuple is projectively equivalent to

$([-a:1],[-b:1];[1:0];[2:1],[2-a-b:1]).$

By looking at 2-torsion points we obtain $E_{\iota\sigma,\tau}$ is isomorphic
to $E_{\iota\sigma,\tau}^{\prime}$ and $E_{\iota\sigma,\iota\tau}$ is
isomorphic to $E_{\iota\sigma,\iota\tau}^{\prime}$.

Applying the projective transformation given by $\begin{pmatrix}4&8+2a+2b\\\
0&-a-b\end{pmatrix}$ to the 5-tuple $([-b:1],[-a:1];[1:0];[-2-a-b:1],[-2:1])$
we have another description of $\varphi$

$\varphi([-a:1],[-b:1];[1:0],[2:1],[-2:1])=\\\
\left(\left[-\frac{2b-2a-8}{a+b}:1\right],\left[-\frac{2a-2b-8}{a+b}:1\right];[1:0];[2:1],[-2:1]\right).$
(4.2)

One can check that $j$-invariants of $E_{\tau}$ (resp. $E_{\sigma\tau}$) and
$E_{\tau}^{\prime}$ (resp. $E_{\sigma\tau}^{\prime}$) are equal, hence the
respective curves are isomorphic.

It follows that the two-torsion points $e_{1}$ and $e_{1}^{\prime}$ defining
the coverings $E_{\tau}\to E_{\iota\sigma,\tau}$ and $E_{\tau}^{\prime}\to
E_{\iota\sigma,\tau}^{\prime}$ are the same (similarly for another pair of
covers), hence

$P(\tilde{C}/E_{\sigma})=E_{\tau}\times E_{\sigma\tau}/\langle
e_{1}+f_{1}\rangle=E_{\tau}^{\prime}\times E_{\sigma\tau}^{\prime}/\langle
e_{1}^{\prime}+f_{1}^{\prime}\rangle=P(\tilde{C}^{\prime}/E_{\sigma}^{\prime})$

which shows the claim. ∎

######  Remark 4.7.

Note that if $b=-a$ the curves $E_{\tau}$ and $E_{\sigma\tau}$ are isomorphic
(see Prop 4.3), hence on such curves $\Pr$ has degree one (see Remark 3.12)
and $\varphi$ is not defined on such covers.

Moreover, one can compute that for $b\neq-a$ the 5-tuples $T$ and $\varphi(T)$
are not projectively equivalent, hence top curves are not isomorphic and the
degree of the Prym map is precisely two at such points.

###  4.1 Period matrix of $P(\tilde{C}/E_{\sigma})$

Let us compute a period matrix of $P(\tilde{C}/E_{\sigma})$. We know that
$P(\tilde{C}/E_{\sigma})=(E_{\tau}\times E_{\sigma\tau})/\langle
e_{1}+f_{1}\rangle$. Possibly after a symplectic transformation, we can assume
that small period matrices of $E_{\tau}$ and $E_{\sigma\tau}$ are given by
$[z_{1}]$ and $[z_{2}]$ respectively with $e_{1}+f_{1}$ given by
$\begin{bmatrix}1\\\ 1\end{bmatrix}$ (note that the polarization on the
product is $(2,2)$).

######  Proposition 4.8.

With the assumptions above, a (big) period matrix of $P(\tilde{C}/E_{\sigma})$
is

$\begin{bmatrix}z_{1}&z_{1}&1&0\\\ z_{1}&z_{1}+z_{2}&0&2\end{bmatrix}$

The embeddings of $E_{\tau}$ and $E_{\sigma\tau}$ are given by
$x\mapsto\begin{bmatrix}x\\\ x\end{bmatrix}$ and $x\mapsto\begin{bmatrix}0\\\
x\end{bmatrix}$ respectively.

###### Proof.

A big period matrix of a $(2,2)$-polarized product $E_{\tau}\times
E_{\sigma\tau}$ is

$M=\begin{bmatrix}z_{1}&0&2&0\\\ 0&z_{2}&0&2\end{bmatrix},$

and we denote the column vectors of this matrix as $f_{1},f_{2},e_{1},e_{2}$.
Note that $f_{1}^{\prime}=f_{1}$, $f_{2}^{\prime}=f_{2}-f_{1}$,
$e_{1}^{\prime}=e_{1}+e_{2}$, $e_{2}^{\prime}=e_{2}$ is a symplectic basis
with respect to $e^{\mathcal{O}_{E}(2)\boxtimes\mathcal{O}_{F}(2)}$. Writing
$M$ in the new basis and taking quotient by $e_{1}^{\prime}/2$ we get the
period matrix of $P(C/E_{\tau})$:

$M^{\prime}=\begin{bmatrix}z_{1}&z_{1}&1&0\\\ 0&z_{2}&-1&2\end{bmatrix}.$

Finally, applying symplectic transformation of the form $\begin{bmatrix}1&0\\\
1&1\end{bmatrix}$ we get

$\begin{bmatrix}z_{1}&z_{1}&1&0\\\ z_{1}&z_{1}+z_{2}&0&2\end{bmatrix}.$

One easily checks that embeddings of the curves $E_{\tau}$ and
$E_{\sigma\tau}$ are given by $x\mapsto\begin{bmatrix}x\\\ x\end{bmatrix}$ and
$x\mapsto\begin{bmatrix}0\\\ x\end{bmatrix}$. ∎

We would like to conclude the paper with the following problems.

  1. 1.

Is it possible to naturally compactify the domain, codomain and the Prym map
from Theorem 3.10?

  2. 2.

The investigated moduli space (Theorem 3.3) is of dimension 2. Therefore, one
can expect that if three elliptic curves satisfy some relation there should
exists a hyperelliptic genus 3 curve that is a covering of these three
elliptic curves. Can one find such an explicit relation?

  3. 3.

We have been able to compute period matrices of Prym varieties. Is it possible
to use them to construct explicit period matrices of completely decomposable
Jacobians of hyperelliptic genus 3 curves? Note that our construction lives in
the boundary of a family considered in [EF18].

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P. Borówka, Institute of Mathematics, Jagiellonian University in Kraków,
Poland
email address<EMAIL_ADDRESS>
A. Shatsila, Institute of Mathematics, Jagiellonian University in Kraków,
Poland
email address<EMAIL_ADDRESS>