An abelian subfield of the dyadic division field of a hyperelliptic Jacobian

TL;DR

This paper studies the structure of abelian subfields within the dyadic division field of a hyperelliptic Jacobian, providing explicit descriptions and bounds for these subextensions in relation to torsion points.

Contribution

It explicitly characterizes the maximal abelian subextension of the dyadic division field of a hyperelliptic Jacobian and shows its containment within specific torsion point fields.

Findings

Maximal abelian subextension is contained in $K(J[8])$ for genus ≥ 2.

Explicit description of the abelian subextension $K(J[4])$.

Action of a specific automorphism on these subfields is described.

Abstract

Given a field $k$ of characteristic different from $2$ and an integer $d \geq 3$, let $J$ be the Jacobian of the "generic" hyperelliptic curve given by $y^2 = \prod_{i = 1}^d (x - \alpha_i)$, where the $\alpha_i$'s are transcendental and independent over $k$; it is defined over the transcendental extension $K / k$ generated by the symmetric functions of the $\alpha_i$'s. We investigate certain subfields of the field $K_{\infty}$ obtained by adjoining all points of $2$-power order of $J(\bar{K})$. In particular, we explicitly describe the maximal abelian subextension of $K_{\infty} / K(J[2])$ and show that it is contained in $K(J[8])$ (resp. $K(J[16])$) if $g \geq 2$ (resp. if $g = 1$). On the way we obtain an explicit description of the abelian subextension $K(J[4])$, and we describe the action of a particular automorphism in $\mathrm{Gal}(K_{\infty} / K)$ on these subfields.

Equations

$$y^2=\prod _{i=1}^d(x-{\alpha }_i)$$

$$\mathrm{GSp}(T_2(J)):=\{\sigma \in {\mathrm{Aut}}_{{\mathbb{Z}}_2}(T_2(J))|e_2(P^{\sigma },Q^{\sigma })={\chi }_2(\sigma )e_2(P,Q)\forall P,Q\in T_2(J)\}$$

$$\mathrm{Sp}(T_2(J)):=\{\sigma \in {\mathrm{Aut}}_{{\mathbb{Z}}_2}(T_2(J))|e_2(P^{\sigma },Q^{\sigma })=e_2(P,Q)\forall P,Q\in T_2(J)\}$$

$$K_2({\mu }_2)=K_1({\mu }_2,\{\sqrt{{\gamma }_{i,j}}{\}}_{1\le i<j\le 2g+1})⊊K_{\infty }^{\mathrm{ab}}=K_2({\mu }_2,\{\sqrt[4]{{\prod }_{j\ne i}{\gamma }_{i,j}}{\}}_{i=1}^{2g+1})⊊K_3({\mu }_2)$$

$$K_2({\mu }_2)=K_1({\mu }_2,\{\sqrt{{\gamma }_{i,j}}{\}}_{1\le i<j\le 3})⊊K_{\infty }^{\mathrm{ab}}=K_2({\mu }_2,\sqrt[8]{{\gamma }_{1,2}{\gamma }_{1,3}{\gamma }_{2,3}^2},\sqrt[8]{{\gamma }_{2,3}{\gamma }_{2,1}{\gamma }_{3,1}^2},\sqrt[8]{{\gamma }_{3,1}{\gamma }_{3,2}{\gamma }_{1,2}^2})⊊K_4({\mu }_2)$$

$$K_2({\mu }_2)=K_1({\mu }_2,\{\sqrt{{\gamma }_{i,j}}{\}}_{1\le i<j\le 3})⊊K_{\infty }^{\mathrm{ab}}=K_2({\mu }_2,\sqrt[4]{{\gamma }_{1,2}{\gamma }_{1,3}},\sqrt[4]{{\gamma }_{2,3}{\gamma }_{2,1}},\sqrt[4]{{\gamma }_{3,1}{\gamma }_{3,2}})⊊K_3({\mu }_2)$$

$$k(\bar{J}[2])(\{\sqrt{{\bar{\gamma }}_{i,j}}{\}}_{1\le i<j\le 2g+1},\{\sqrt[4]{{\prod }_{j\ne i}{\bar{\gamma }}_{i,j}}{\}}_{i=1}^{2g+1},{\zeta }_8).$$

$$k(\bar{J}[2])(\{\sqrt{{\bar{\gamma }}_{i,j}}{\}}_{1\le i<j\le 2g+1},\sqrt[8]{{\bar{\gamma }}_{1,2}{\bar{\gamma }}_{1,3}{\bar{\gamma }}_{2,3}^2},\sqrt[8]{{\bar{\gamma }}_{2,3}{\bar{\gamma }}_{2,1}{\bar{\gamma }}_{3,1}^2},\sqrt[8]{{\bar{\gamma }}_{3,1}{\bar{\gamma }}_{3,2}{\bar{\gamma }}_{1,2}^2},{\zeta }_{16}).$$

$$\Psi ({\Lambda }_3),\Psi ({\Lambda }_3^{\prime })=-\Psi ({\Lambda }_2)-(\Psi ({\Lambda }_2)-\Psi ({\Lambda }_2^{\prime }))\pm 4(\sqrt{{\gamma }_{1,2}}+\sqrt{{\gamma }_{1,3}})\sqrt[4]{{\gamma }_{1,2}}\sqrt[4]{{\gamma }_{1,3}};$$

$$\Psi ({\Lambda }_4)=-\Psi ({\Lambda }_3)-(\Psi ({\Lambda }_3)-\Psi ({\Lambda }_3^{\prime }))+4\sqrt{2}(\sqrt{{\gamma }_{1,2}}+\sqrt{{\gamma }_{1,3}}+2\sqrt[4]{{\gamma }_{1,2}}\sqrt[4]{{\gamma }_{1,3}})\sqrt{\sqrt{{\gamma }_{1,2}}+\sqrt{{\gamma }_{1,3}}}\sqrt[8]{{\gamma }_{1,2}}\sqrt[8]{{\gamma }_{1,3}}.$$

$$\sqrt{\sqrt{{\gamma }_{1,2}}+\sqrt{{\gamma }_{1,3}}}\sqrt[8]{{\gamma }_{1,2}}\sqrt[8]{{\gamma }_{1,3}}\in K_4.$$

$$\sqrt[4]{-{\gamma }_{2,3}}\sqrt[8]{{\gamma }_{1,2}}\sqrt[8]{{\gamma }_{1,3}}=\sqrt{-{\gamma }_{2,3}}\sqrt{\sqrt{{\gamma }_{1,2}}+\sqrt{{\gamma }_{1,3}}}\sqrt[8]{{\gamma }_{1,2}}\sqrt[8]{{\gamma }_{1,3}}/B_1\in K_4.$$

$${\Sigma }^{\prime }=A_{1,2}(A_{1,3}{A}_{2,3})...(A_{1,d-1}{A}_{2,d-1}...A_{d-2,d-1})\in P_d.$$

$${\pi }_1({\mathbb{P}}_{\mathbb{C}}^1(\mathbb{C})\setminus \{z_i{\}}_{1\le i\le d},{\bar{P}}_0)⊳{\pi }_1({\mathcal{C}}_{{\underline{z}}_0}(\mathbb{C})\setminus \{(z_i,0){\}}_{1\le i\le d},P_0)\twoheadrightarrow {\pi }_1({\mathcal{C}}_{{\underline{z}}_0}(\mathbb{C}),P_0)$$

$$(x^{\prime },y^{\prime })=(1/({\alpha }_{2g+2}-x),y/(\beta ({\alpha }_{2g+2}-x{)}^{g+1})),$$

$$K_2(\beta )=K_1(\beta ,\{\sqrt{({\alpha }_{2g+2}-{\alpha }_j{)}^{-1}-({\alpha }_{2g+2}-{\alpha }_i{)}^{-1}}{\}}_{1\le i<j\le 2g+1})=K_1(\beta ,\{\beta \sqrt{{\gamma }_{i,j}}{\}}_{1\le i<j\le 2g+1}).$$

$${\sigma }^m{\tau }^n{\sigma }^{-m}{\tau }^{-n}=\left[\begin{array}{cc}1-2^{m+n}& 2^{2m+n}\\ -2^{m+2n}& 1+2^{m+n}+2^{2m+2n}\end{array}\right]\in [\Gamma (2),\Gamma (2)].$$

$$T_{a_1}{T}_{b_1}^4{T}_{a_1}^{-1}\equiv T_{b_1}^4{T}_{a_1}^4(\text{mod}[\Gamma (2),\Gamma (2)]).$$

$$\{[1,2{]}^{\prime }+[1,3{]}^{\prime }+2[2,3{]}^{\prime },[2,3{]}^{\prime }+[2,1{]}^{\prime }+2[3,1{]}^{\prime },[3,1{]}^{\prime }+[3,2{]}^{\prime }+2[1,2{]}^{\prime }\}\subset \tilde{V}.$$

$$2[1,2{]}^{\prime }+2[1,3{]}^{\prime },2[2,3{]}^{\prime }+2[2,1{]}^{\prime },2[3,1{]}^{\prime }+2[3,2{]}^{\prime }\in V.$$

$$S:=\mathrm{Spec}(\mathbb{Z}[\frac{1}{2},\{{\alpha }_i{\}}_{1\le i\le d},\{({\alpha }_i-{\alpha }_j{)}^{-1}{\}}_{1\le i<j\le 2g+1}{]}^{S_d}),$$

$$\mathrm{Gal}((K_{\mathbb{Q}}{)}_2/(K_{\mathbb{Q}}{)}_1)\cong \tilde{\Gamma }(2)/\tilde{\Gamma }(4)=\Gamma (2)/\Gamma (4)\times ⟨\iota ⟩\cong (\mathbb{Z}/2\mathbb{Z}{)}^{2g^2+g}\times ⟨\iota ⟩,$$

Full text

1. Introduction

Let $k$ be any field of characteristic different from $2$; let $\alpha_{1},...,\alpha_{d}$ be transcendental and independent over $k$ for some integer $d\geq 3$; and let $K$ denote the extension of $k$ obtained by adjoining the symmetric functions of the $\alpha_{i}$’s with separable closure denoted $\bar{K}$. The equation given by

[TABLE]

defines a hyperelliptic curve $C$ of genus $g:=\lfloor(d-1)/2\rfloor$ over $K$. Its Jacobian, denoted by $J$, is a principally polarized abelian variety over $K$ of dimension $g$. For each integer $n\geq 1$, we write $J[2^{n}]\subset J(\bar{K})$ for the $2^{n}$-torsion subgroup of $J$ and $K_{n}:=K(J[2^{n}])$ for the (finite algebraic) extension of $K$ obtained by adjoining the coordinates of the points in $J[2^{n}]$ to $K$; we denote the (infinite algebraic) extension $\bigcup_{n=1}^{\infty}K_{n}$ by $K_{\infty}=K(J[2^{\infty}])$. Let $T_{2}(J)$ denote the $2$-adic Tate module of $J$; it is a free $\mathbb{Z}_{2}$-module of rank $2g$ given by the inverse limit of rank-$2g$ $\mathbb{Z}/2^{n}\mathbb{Z}$-modules $J[2^{n}]$ with respect to the multiplication-by-$2$ map. The canonical principal polarization on $J$ defines the Weil pairing $e_{2}:T_{2}(J)\times T_{2}(J)\to\mathbb{Z}_{2}$; it is a nondegenerate, skew-symmetric, $\mathbb{Z}_{2}$-bilinear pairing on $T_{2}(J)$.

We have the natural action of the absolute Galois group $G_{K}=\mathrm{Gal}(\bar{K}/K)$ on each $J[2^{n}]$. It is well known that this action respects the Weil pairing $e_{2}$ up to multiplication by the cyclotomic character $\chi_{2}:G_{K}\to\mathbb{Z}_{2}^{\times}$ (in particular, this implies that $K_{\infty}$ contains the multiplicative subgroup $\mu_{2}$ of $2$-power roots of unity in the separable closure of $k$). Each element $\sigma\in G_{K}$ therefore acts as an automorphism in the group

[TABLE]

of symplectic similitudes. We denote this natural Galois action by $\rho_{2}:G_{K}\to\mathrm{GSp}(T_{2}(J))$ and each modulo-$2^{n}$ action by $\bar{\rho}_{2^{n}}:G_{K}\to\mathrm{GSp}(J[2^{n}])$. For any field $F$, we write $F(\mu_{2})$ for the algebraic extension of $F$ obtained by adjoining all $2$-power roots of unity. Clearly the image of $\rho_{2}$ is contained in the symplectic group

[TABLE]

if and only if $K=K(\mu_{2})$. For each $n\geq 0$, we write $\Gamma(2^{n})\lhd\mathrm{Sp}(T_{2}(J))$ for the level-$2^{n}$ principal congruence subgroup consisting of automorphisms whose images modulo $2^{n}$ are trivial.

It is well known that we always have $K_{1}\subseteq k(\alpha_{1},...,\alpha_{d})$ and that equality holds except when $d=4$ (see [16]). The main purpose of this paper is to provide an explicit description of the maximal abelian subextension of $K_{\infty}/K_{1}$, which we denote by $K_{\infty}^{\mathrm{ab}}$. (Below for any integer $n\geq 1$, we write $\zeta_{2^{n}}\in\bar{k}$ to denote a $2^{n}$th root of unity.)

Theorem 1.1**.**

For $1\leq i,j\leq 2g+1$, let $\gamma_{i,j}=\alpha_{j}-\alpha_{i}$ (resp. $\gamma_{i,j}=(\alpha_{j}-\alpha_{i})\prod_{l\neq i,j}(\alpha_{d}-\alpha_{l})$) if $d=2g+1$ (resp. if $d=2g+2$).

a) Suppose that $g\geq 2$. If $\zeta_{4}\in k$, we have

[TABLE]

and $\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{1}(\mu_{2}))\cong(\mathbb{Z}/2\mathbb{Z})^{2g^{2}-g}\times(\mathbb{Z}/4\mathbb{Z})^{2g}$. If $\zeta_{4}\notin k$, we instead have $K_{\infty}^{\mathrm{ab}}=K_{2}(\mu_{2})$.

b) Suppose that $g=1$. If $\zeta_{8}\in k$, we have

[TABLE]

and $\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{1}(\mu_{2}))\cong\mathbb{Z}/2\mathbb{Z}\times(\mathbb{Z}/8\mathbb{Z})^{2}$.

If $\zeta_{8}\notin k$ but $\zeta_{4}\in k$, we instead have

[TABLE]

and $\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{1}(\mu_{2}))\cong\mathbb{Z}/2\mathbb{Z}\times(\mathbb{Z}/4\mathbb{Z})^{2}$.

Finally, if $\zeta_{4}\notin k$, we instead have $K_{\infty}^{\mathrm{ab}}=K_{2}(\mu_{2})=K_{1}(\mu_{2},\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 3})$ and $\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{1}(\mu_{2}))\cong(\mathbb{Z}/2\mathbb{Z})^{3}$.

c) Let $\sigma\in G_{K}$ be any Galois automorphism such that $\rho_{2}(\sigma)=-1\in\mathrm{GSp}(T_{2}(J))$. Then $\sigma$ acts on $K_{\infty}^{\mathrm{ab}}$ by fixing $K_{1}(\mu_{2})$, changing the signs of all generators of the form $\sqrt{\gamma_{i,j}}$, and fixing (resp. changing the signs of) the remaining generators given in (2) and (3) if $g$ is even (resp. if $g$ is odd).

The following corollary is proven via the argument in Step 4 of the proof of [16, Lemma 3].

Corollary 1.2**.**

Let $a_{0},...,a_{d-1}\in k$ be elements such that $f(x):=x^{d}+\sum_{i=0}^{d-1}a_{i}x^{i}\in k[x]$ is separable, and let $\bar{J}$ be the Jacobian of the hyperelliptic curve defined over $k$ by the equation $y^{2}=f(x)$. Let $\bar{\alpha}_{1},...,\bar{\alpha}_{d}\in\bar{k}$ denote the roots of the polynomial $f(x)\in k[x]$ and let $\bar{\gamma}_{i,j}\in\bar{k}$ be given by formulas in terms of the $\bar{\alpha}_{i}$’s analogous to those used to define the $\gamma_{i,j}$’s in the statement of Theorem 1.1.

a) If $g\geq 2$, the extension $k(\bar{J}[8])/k(\bar{J}[2])$ contains the subextension

[TABLE]

b) If $g=1$, the extension $k(\bar{J}[16])/k(\bar{J}[2])$ contains the subextension

[TABLE]

c) Let $\sigma$ be any automorphism in the absolute Galois group of $k$ which acts on $k(\bar{J}[16])$ as multiplication by $-1$. Then $\sigma$ acts on the subfields described above by changing the signs of all generators of the form $\sqrt{\gamma_{i,j}}$ and by fixing (resp. changing the signs of) all remaining generators given in (5) and (6) if $g$ is even (resp. if $g$ is odd).

Remark 1.3**.**

We can also verify part (b) of Theorem 1.1 for the $d=3$ case by combined use of the formulas given in [14] and [16]. We illustrate how to see that $K_{4}(\mu_{2})$ contains an element whose $8$th power is $\gamma_{1,2}\gamma_{1,3}\gamma_{2,3}^{2}$ as follows (one may use a similar argument for the other generators). For $n=1,2,3$ and $1\leq i<j\leq 3$, we fix elements $\sqrt[2^{n}]{\gamma_{i,j}}\in\bar{K}$ whose $2^{n}$th powers are $\gamma_{i,j}\in\bar{K}$ and which are compatible in the obvious way, and we fix a square root of $\sqrt{\gamma_{1,2}}+\sqrt{\gamma_{1,3}}$. (Note that due to the equivariance of the Weil pairing, $-1$ has a $2^{n}$th root in $K(J[2^{n+1}])$ for each $n\geq 1$.) Let $\mathcal{L}$ be the $3$-regular tree defined in [14], and assume the notation used throughout that paper. Let $\{\Lambda_{0},\Lambda_{1},\Lambda_{2},\Lambda_{3},\Lambda_{4}\}$ be a non-backtracking path in $\mathcal{L}$, where $\Lambda_{0}$ is the root and $\Lambda_{1}=\Lambda(\alpha_{1})$. Then it is tedious but straightforward to verify that there exists a decoration $\Psi:\mathcal{L}\setminus\{\Lambda_{0}\}\to\bar{K}$ (see [14, Definition 1.2]) such that $\Psi(\Lambda_{2}),\Psi(\Lambda_{2}^{\prime})=\gamma_{1,2}+\gamma_{1,3}\pm 2\sqrt{\gamma_{1,2}}\sqrt{\gamma_{1,3}}$;

[TABLE]

and

[TABLE]

By [14, Proposition 2.5(b)], we have $\Psi(\Lambda_{2}),\Psi(\Lambda_{2}^{\prime}),\Psi(\Lambda_{3}),\Psi(\Lambda_{3}^{\prime}),\Psi(\Lambda_{4})\in K_{4}$. Moreover, from [16, Theorem 1, Remark 11(b)], we see that $\sqrt{2}(\sqrt{\gamma_{1,2}}+\sqrt{\gamma_{1,3}}+2\sqrt[4]{\gamma_{1,2}}\sqrt[4]{\gamma_{1,3}})\in K_{3}$. It follows that

[TABLE]

By [16, Theorem 1], we have $\sqrt{\pm\gamma_{i,j}},B_{1}:=\sqrt[4]{-\gamma_{2,3}}\sqrt{\sqrt{\gamma_{1,2}}+\sqrt{\gamma_{1,3}}}\in K_{3}$. Therefore, we have

[TABLE]

The rest of this paper is dedicated to a proof of Theorem 1.1; our plan is as follows. We will first assume that $k=\mathbb{C}$ and prove Theorem 1.1 in that case by viewing the situation in a topological setting similar to the author’s strategy in [15]; we will retain this assumption throughout §2 and §3. In §2, we determine generators for the $4$-torsion field $K_{2}$, which is contained in $K_{\infty}^{\mathrm{ab}}$. Then in §3, we determine generators for $K_{\infty}^{\mathrm{ab}}$ over $K_{2}$, treating the $g\geq 2$ case and the $g=1$ case separately. Finally, in §4 we generalize these results to the situation where $k$ is any field of characteristic different from $2$.

The author would like to thank the referee for a number of corrections and suggestions which have improved this text.

2. The $4$-division field over $\mathbb{C}$

We assume for this section as well as in §3 that $k=\mathbb{C}$, so that $K$ is generated over $\mathbb{C}$ by the symmetric functions of the transcendental elements $\alpha_{i}$. We will consider $K$ as a subfield of the function field of the ordered configuration space $Y_{d}$ of $d$-element ordered subsets of $\mathbb{C}$; we view $Y_{d}(\mathbb{C})$ as a topological space. The fundamental group of $Y_{d}$ is well known to be the pure braid group on $d$ strands, which we denote by $P_{d}$. This has a well-known presentation (see [3, Lemma 1.8.2]) with generators $A_{i,j}$ for $1\leq i<j\leq d$. It is known (see [3, Corollary 1.8.4] and its proof) that the center of $P_{d}$ is cyclically generated by the element $\Sigma:=A_{1,2}(A_{1,3}A_{2,3})...(A_{1,d}A_{2,d}...A_{d-1,d})\in P_{d}$. The profinite completion $\widehat{P}_{d}$ of $P_{d}$ is the étale fundamental group of $Y_{d}$ and may be identified with the Galois group of $K^{\mathrm{unr}}/K(\{\alpha_{i}\}_{1\leq i\leq d})$, where $K^{\mathrm{unr}}$ is the maximal extension of $K$ unramified away from the primes $(\alpha_{j}-\alpha_{i})$ for $1\leq i<j\leq d$. The criterion of Néron-Ogg-Shafarevich ([12, Theorem 1]) implies that the natural $\ell$-adic representation $\rho_{\ell}:G_{K}\to\mathrm{GSp}(T_{\ell}(J))$, restricted to the subgroup fixing the Galois extension $K(\{\alpha_{i}\}_{1\leq i\leq d})$, factors through the restriction map $\mathrm{Gal}(\bar{K}/K(\{\alpha_{i}\}_{1\leq i\leq d}))\twoheadrightarrow\widehat{P}_{d}$; we denote the induced representation of $\widehat{P}_{d}$ also by $\rho_{\ell}$.

There is a “universal” family of hyperelliptic curves $\mathcal{C}\to Y_{d}$ whose fiber $\mathcal{C}_{\underline{z}}$ over each point $\underline{z}=(z_{1},...,z_{d})\in Y_{d}(\mathbb{C})$ is the hyperelliptic curve given by the monic polynomial in $\mathbb{C}[x]$ whose roots are the elements of the $d$-element ordered set $\underline{z}$; this family has $C$ as its generic fiber. We write $\rho^{\mathrm{top}}:P_{d}\to\mathrm{Aut}(H_{1}(C_{\underline{z}_{0}},\mathbb{Z}))$ for the representation induced by the monodromy representation $P_{d}\cong\pi_{1}(Y_{d},\underline{z}_{0})\to\mathrm{Aut}(\pi_{1}(\mathcal{C}_{\underline{z}_{0}},P_{0}))$ associated to the family $\mathcal{C}\to Y_{d}$, where $\underline{z}_{0}:=(1,...,d)\in Y_{d}(\mathbb{C})$ and $P_{0}\in\mathcal{C}_{\underline{z}}$ are basepoints). The monodromy action respects the intersection pairing on $\mathcal{C}_{\underline{z}_{0}}$, and therefore, the image of $\rho^{\mathrm{top}}$ is contained in the group of symplectic automorphisms $\mathrm{Sp}(H_{1}(C_{\underline{z}_{0}},\mathbb{Z}))$. (See [15, §2] for more details of this construction.)

As both $P_{d}$ and $\mathrm{Sp}(H_{1}(C_{\underline{z}_{0}},\mathbb{Z}))$ are residually finite, the representation $\rho^{\mathrm{top}}$ induces a representation of the profinite completion $\widehat{P}_{d}$ on each pro-$\ell$ completion $H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z})\otimes\mathbb{Z}_{\ell}$ of $H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z})$. For each prime $\ell$, we denote this representation by $\rho_{\ell}^{\mathrm{top}}:\widehat{P}_{d}\to\mathrm{Sp}(H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z})\otimes\mathbb{Z}_{\ell})$. Our technique is to study $\rho_{2}$ by relating it to the topologically-defined representation $\rho_{2}^{\mathrm{top}}$ using a key comparison result proved by the author as [15, Proposition 2.2].

Lemma 2.1**.**

For any prime $\ell$, there is an isomorphism of $\mathbb{Z}_{\ell}$-modules $H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z})\otimes\mathbb{Z}_{\ell}\stackrel{{\scriptstyle\sim}}{{\to}}T_{\ell}(J)$ making the representations $\rho_{\ell}^{\mathrm{top}}$ and $\rho_{\ell}$ isomorphic.

We now state and prove some properties of the representation $\rho^{\mathrm{top}}$ that we will need below.

Proposition 2.2**.**

a) The image of $P_{d}$ under $\rho^{\mathrm{top}}$ coincides with the principal congruence subgroup $\Gamma(2)\lhd\mathrm{Sp}(H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z}))$.

b) If $d$ is odd, we have $\rho^{\mathrm{top}}(\Sigma)=-1\in\Gamma(2)$.

c) If $d$ is even, we have $\rho^{\mathrm{top}}(\Sigma)=1\in\Gamma(2)$ and $\rho^{\mathrm{top}}(\Sigma^{\prime})=-1\in\Gamma(2)$, where

[TABLE]

Proof.

The statement of (a) has been shown in several works: see [1, Théorème 1], [10, Lemma 8.12], or [17, Theorem 7.3(ii)].

Assume that $d$ is odd, so $d=2g+1$, where $g$ is the genus of $\mathcal{C}_{\underline{z}_{0}}$. Then one deduces directly from the presentation of $P_{d}$ given by [3, Lemma 1.8.2] that the abelianization of $P_{d}$ is a free $\mathbb{Z}$-module of rank $2g^{2}+g$ whose generators are the images of the elements $A_{i,j}$, $1\leq i<j\leq 2g+1$; its maximal abelian quotient of exponent $2$ is therefore a $(2g^{2}+g)$-dimensional $\mathbb{F}_{2}$-vector space generated by the images of the $A_{i,j}$’s. Meanwhile, it follows directly from [11, Corollary 2.2] that the maximal abelian exponent-$2$ quotient of $\Gamma(2)$ also has rank $2g^{2}+g$. It follows that $\rho^{\mathrm{top}}$ induces an isomorphism between the exponent-$2$ abelianizations of $P_{d}$ and $\Gamma(2)$. Now since $\Sigma$ is a product of each of the elements $A_{i,j}\in P_{d}$, it has nontrivial image in the exponent-$2$ abelianization of $P_{d}$ and therefore has nontrivial image in $\Gamma(2)$. Meanwhile, as $\Sigma$ lies in the center of $P_{d}$ and $\rho^{\mathrm{top}}(P_{d})=\Gamma(2)$ by (a), the image $\rho^{\mathrm{top}}(\Sigma)$ lies in the center of $\Gamma(2)$. The only nontrivial central element of $\Gamma(2)$ is the scalar $-1$, proving part (b).

Now assume that $d$ is even. Note that the family $\mathcal{C}\to Y_{d}$ is an unramified degree-$2$ cover of the family $Y_{d+1}^{\prime}\to Y_{d}$ whose fiber over each $\underline{z}=(z_{1},...,z_{d})\in Y_{d}$ is $\mathbb{P}_{\mathbb{C}}^{1}\smallsetminus\{z_{1},...,z_{d}\}$ ($Y_{d+1}^{\prime}$ is essentially the ordered configuration space of cardinality-$(d+1)$ subsets of $\mathbb{P}_{\mathbb{C}}^{1}$ whose first $d$ elements lie in $\mathbb{C}$). This implies that the monodromy action $\rho^{\mathrm{top}}$ is induced by the monodromy action associated to the family $Y_{d+1}^{\prime}\to Y_{d}$ via the inclusion of and quotients by characteristic subgroups

[TABLE]

(here $\bar{P}_{0}$ is the projection of the baspoint $P_{0}\in\mathcal{C}_{\underline{z}_{0}}$). In fact, if we let $x_{1},...,x_{d}$ denote the generators of $\pi_{1}(\mathbb{P}_{\mathbb{C}}^{1}(\mathbb{C})\smallsetminus\{z_{i}\}_{1\leq i\leq d},\bar{P}_{0})$ given in [7, §4], then $\pi_{1}(\mathcal{C}_{\underline{z}_{0}}(\mathbb{C})\smallsetminus\{(z_{i},0)\}_{1\leq i\leq d},P_{0})$ is the subgroup generated by the elements $x_{i}x_{i+1}$ for $1\leq i\leq d-1$ and $x_{j}^{2}$ for $1\leq j\leq d$, and $\pi_{1}(\mathcal{C}_{\underline{z}_{0}}(\mathbb{C}),P_{0})$ is the quotient of this by the elements $x_{j}^{2}$. Using the fact that the images of the elements $x_{i}x_{i+1}$ for $1\leq i\leq d-2$ form a $\mathbb{Z}$-basis of the abelianization $H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z})$, one may then explicitly compute the automorphisms of $H_{1}(\mathcal{C}_{\underline{z}_{0}},\mathbb{Z})$ induced by $\Sigma$ and $\Sigma^{\prime}$ using the statement and proof of [7, Lemma 4.1], thus verifying part (c). (One can also prove that $\rho^{\mathrm{top}}(\Sigma)=1$ using [7, Lemma 4.2] and the well-known fact that the kernel of Birman’s surjection onto the mapping class group coincides with the center of $P_{d}$.)

∎

Below, for each integer $N\geq 1$, we fix $\zeta_{N}\in\mathbb{C}$ to be the $N$th root of unity given by $e^{2\pi\sqrt{-1}/N}$.

Lemma 2.3**.**

For any integer $N\geq 1$, the maximal abelian exponent-$N$ subextension of $K^{\mathrm{unr}}/K(\{\alpha_{i}\}_{1\leq i\leq d})$ coincides with $K_{1}(\{\sqrt[N]{\alpha_{j}-\alpha_{i}}\}_{1\leq i<j\leq d})$. Each standard generator $A_{i,j}$ of $P_{d}\subset\widehat{P}_{d}=\mathrm{Gal}(K^{\mathrm{unr}}/K(\{\alpha_{i}\}_{1\leq i\leq d})$ acts on this subextension by sending $\sqrt[N]{\alpha_{j}-\alpha_{i}}$ to $\zeta_{N}\sqrt[N]{\alpha_{j}-\alpha_{i}}$ and fixing all of the other generators.

Proof.

The first statement results from a standard application of Kummer theory.

To prove the second statement, we fix some $N\geq 1$ and assume without loss of generality that $(i,j)=(d-1,d)$. Let $Y_{d}^{(N)}\to Y_{d}$ be the covering corresponding to the maximal abelian exponent-$N$ subextension of $K^{\mathrm{unr}}/K_{1}$. We choose as a topological representative of $A_{i,j}$ the loop given by $t\mapsto(1,...,d-1,d-1+e^{2\pi\sqrt{-1}t}\varepsilon/2)\in Y_{d}(\mathbb{C})$ for $t\in[0,1]$, where $\varepsilon$ is a real number satisfying $\min_{1\leq i\leq d-2}|z_{d-1}-z_{i}|>\varepsilon>0$ (we are allowed to move the basepoint to $(1,...,d-1,d-1+\varepsilon/2)\in Y_{d}(\mathbb{C})$ because the monodromy action we are considering factors through the abelianization of the fundamental group). We have a closed embedding of the punctured disk $B^{*}:=\{z\in\mathbb{C}\ |\ 0<|z|<1\}$ into $Y_{d}(\mathbb{C})$ given by $z\mapsto(1,...,d-1,d-1+\varepsilon z)$ which takes a loop representing the standard generator of $\pi_{1}(B^{*},1/2)$ to the loop representing $A_{d-1,d}$ defined above. The pullback of the cover $Y_{d}^{(N)}(\mathbb{C})\to Y_{d}(\mathbb{C})$ via $B^{*}\hookrightarrow Y_{d}(\mathbb{C})$ is clearly homeomorphic to the cover $B^{*}\to B^{*}$ given by $z\mapsto z^{N}$. Locally, the standard generator of $\pi_{1}(B^{*},1/2)$ acts on the ring of holomorphic functions defined on the covering space as $\sqrt[N]{z}\mapsto\zeta_{N}\sqrt[N]{z}$, and the claim follows.

∎

We are now ready to state and prove the main result of this section which explicitly describes the extension $K_{2}/K$ and shows in particular that it is abelian over $K_{1}$ and therefore contained in $K_{\infty}^{\mathrm{ab}}$.

Theorem 2.4**.**

For $1\leq i,j\leq 2g+1$, let $\gamma_{i,j}$ be defined as in the statement of Theorem 1.1. Then we have $K_{2}=K_{1}(\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1})$. Moreover, any Galois automorphism in $G_{K}$ which acts as multiplication by $-1$ on the subgroup $J[4]$ acts on $K_{2}$ by fixing $K_{1}$ and changing the signs of all generators $\sqrt{\gamma_{i,j}}$.

Proof.

For $g=1$, this was already shown by the author as [16, Proposition 6(a),(b)], so we assume that $g\geq 2$. In particular, this means that $K_{1}=\mathbb{C}(\{\alpha_{i}\}_{1\leq i\leq d})$. The first statement of the theorem was proved for odd $d$ as [15, Proposition 3.1], but the following argument proves the full theorem for general $d$.

By [15, Corollary 1.2(c)], we have that $\rho_{2}$ induces an isomorphism $\bar{\rho}_{4}:\mathrm{Gal}(K_{2}/K_{1})\stackrel{{\scriptstyle\sim}}{{\to}}\Gamma(2)/\Gamma(4)$. We note from the proof of [11, Corollary 2.2] that the largest abelian quotient of $\Gamma(2)$ of exponent $2$ is in fact $\Gamma(2)/\Gamma(4)\cong(\mathbb{Z}/2\mathbb{Z})^{2g^{2}+g}$; therefore, $K_{2}$ is the maximal abelian subextension of $K_{\infty}/K_{1}$ of exponent $2$. Since the extension $K_{\infty}/K_{1}$ is unramified over all primes of the form $(\alpha_{j}-\alpha_{i})$, such a subextension must be a subfield of $\widetilde{K}_{2}:=K_{1}(\{\sqrt{\alpha_{j}-\alpha_{i}}\}_{1\leq i<j\leq d})$. If $d=2g+1$, then $\mathrm{Gal}(\widetilde{K}_{2}/K_{1})$ already has rank $2g^{2}+g$ and therefore $K_{2}=\widetilde{K}_{2}$; moreover, Proposition 2.2(b) combined with Lemma 2.3 implies that any Galois element whose image under $\bar{\rho}_{4}$ is $-1\in\Gamma(2)/\Gamma(4)$ changes the sign of each $\sqrt{\alpha_{j}-\alpha_{i}}=\sqrt{\gamma_{i,j}}$. This proves the statement in the $d=2g+1$ case.

Now suppose that $d=2g+2$. Then $\mathrm{Gal}(\widetilde{K}_{2}/K_{1})$ has rank $d(d-1)/2>2g^{2}+g$, which implies that $K_{2}\subsetneq\widetilde{K}_{2}$. Kummer theory then tells us that $\mathrm{Gal}(K_{2}/K_{1})$ is canonically identified with the dual of some subgroup $H\subset K_{1}^{\times}/(K_{1}^{\times})^{2}$, where both are considered as vector spaces over $\mathbb{F}_{2}$ of dimension $2g^{2}+g$. Clearly $H$ is a subspace of the space $\widetilde{H}\subset K_{1}^{\times}/(K_{1}^{\times})^{2}$ (which itself is the dual of $\mathrm{Gal}(\widetilde{K}_{2}/K_{1})$) generated by images of the elements $\alpha_{j}-\alpha_{i}\in K_{1}^{\times}$; for $1\leq i<j\leq d$; we write $[\alpha_{j}-\alpha_{i}]=[\alpha_{i}-\alpha_{j}]\in\widetilde{H}$ for each such image and use additive notation for elements of $\widetilde{H}$.

We now identify $\widetilde{H}$ with its dual $\mathrm{Gal}(\widetilde{K}_{2}/K_{1})$ via the basis $\{[\alpha_{j}-\alpha_{i}]\}_{1\leq i<j\leq d}$. Lemma 2.3 implies that the image of each $A_{i,j}\in P_{d}\subset\widehat{P}_{d}=\mathrm{Gal}(K^{\mathrm{unr}}/K_{1})$ under $\rho_{2}$ composed with reduction modulo $4$ is the one induced by $[\alpha_{j}-\alpha_{i}]\in\mathrm{Gal}(\tilde{K}_{2}/K_{1})=\tilde{H}$. It follows from Proposition 2.2(c) that each element of $H$ must be the sum of an even number of generators of $\widetilde{H}$.

We note that the degree-$(d-1)$ curve $C^{\prime}:y^{\prime 2}=\prod_{i=1}^{2g+1}(x^{\prime}-1/(\alpha_{2g+2}-\alpha_{i}))$ is isomorphic to the degree-$d$ curve $C:y^{2}=\prod_{i=1}^{2g+2}(x-\alpha_{i})$ over the quadratic extension $K(\beta)/K$ via the change of variables

[TABLE]

where $\beta\in\bar{K}$ is a square root of the element $\prod_{i=1}^{2g+1}(\alpha_{i}-\alpha_{2g+2})$. (This is just the isomorphism of hyperelliptic curves induced from an automorphism of the projective line which moves $\alpha_{2g+2}$ to $\infty$.) From what was shown above for the odd-degree case, we have

[TABLE]

Thus, $K_{2}(\beta)$ is generated over $K_{1}(\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1})$ by the element $\beta$. Since each $\gamma_{i,j}$ corresponds to the element $[\alpha_{j}-\alpha_{i}]+\sum_{l\neq i,j}[\alpha_{2g+2}-\alpha_{l}]\in H$, which is the sum of an even number of generators while $\beta^{2}$ corresponds to $\sum_{i\neq 2g+2}[\alpha_{2g+2}-\alpha_{i}]\in H$, which is not, the extension $K_{1}(\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1})/K_{1}$ must be the fixed field corresponding to $H\subset\widetilde{H}$ and therefore coincides with $K_{2}$. Moreover, if $\Sigma^{\prime}\in P_{2g+2}$ is the braid defined as in the statement of Proposition 2.2(c), then that proposition says that $\Sigma^{\prime}$ corresponds to a Galois automorphism whose restriction to $K_{\infty}$ is $-1\in\Gamma(2)\cong\mathrm{Gal}(K_{\infty}/K_{1})$. Then Lemma 2.3 implies that $\Sigma^{\prime}$ changes the sign of each $\sqrt{\gamma_{i,j}}$, thus implying that any Galois element whose image under $\bar{\rho}_{4}$ is $-1\in\Gamma(2)/\Gamma(4)$ acts in this way, hence the statement in the $d=2g+2$ case.

∎

3. The maximal abelian subfield over $\mathbb{C}$

3.1. The abelianization of the Galois group

We retain our assumption from the last section that $k=\mathbb{C}$. Having found a particular abelian subextension of $K_{\infty}/K_{1}$, namely $K_{2}/K_{1}$, we shall now determine the maximal abelian subextension. In order to do this, we first need to know what its Galois group over $K_{1}$ looks like.

Lemma 3.1**.**

The abelianization $\Gamma(2)^{\mathrm{ab}}$ of the principal congruence subgroup $\Gamma(2)\lhd\mathrm{Sp}_{2g}(\mathbb{Z}_{2})$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{2g^{2}-g}\times(\mathbb{Z}/4\mathbb{Z})^{2g}$ (resp. $\mathbb{Z}/2\mathbb{Z}\times(\mathbb{Z}/8\mathbb{Z})^{2}$), and the abelianization map $\pi:\Gamma(2)\twoheadrightarrow\Gamma(2)^{\mathrm{ab}}$ factors through $\Gamma(2)/\Gamma(8)$ (resp. $\Gamma(2)/\Gamma(16)$) if $g\geq 2$ (resp. if $g=1$).

Proof.

The description of $\Gamma(2)^{\mathrm{ab}}$ for the $g\geq 2$ case is given by [11, Corollary 2.2]. We therefore assume that $g=1$ and proceed to compute the commutator subgroup $[\Gamma(2),\Gamma(2)]\lhd\Gamma(2)$.

We first claim that $[\Gamma(2),\Gamma(2)]$ contains $\Gamma(16)$. Write $\sigma=\begin{bmatrix}1&-2\\ 0&1\end{bmatrix}$ and $\tau=\begin{bmatrix}1&0\\ 2&1\end{bmatrix}$. We verify by straightforward computation that for any integers $m,n\geq 1$, we have the formula

[TABLE]

Using this formula, we compute $(\sigma^{2}\tau\sigma^{-2}\tau^{-1})(\sigma\tau\sigma^{-1}\tau^{-1})^{2}\equiv\tau^{8}$, $(\sigma\tau^{2}\sigma^{-1}\tau^{-2})(\sigma\tau\sigma^{-1}\tau^{-1})^{2}\equiv\sigma^{8}$, and $(\sigma\tau\sigma^{-1}\tau^{-1})^{4}\equiv 17$ modulo $32$. It is easy to show that for $n\geq 1$, the images modulo $2^{n+1}$ of $\sigma^{2^{n-1}}$, $\tau^{2^{n-1}}$, and the scalar matrix $1+2^{n}$ generate $\Gamma(2^{n})/\Gamma(2^{n+1})\cong(\mathbb{Z}/2\mathbb{Z})^{3}$; thus, in particular, $[\Gamma(2),\Gamma(2)]$ contains $\Gamma(16)$ modulo $32$. Now we show by induction that for each $n\geq 5$, $[\Gamma(2),\Gamma(2)]$ contains $\Gamma(2^{n})$ modulo $2^{n+1}$, which suffices to prove that $[\Gamma(2),\Gamma(2)]\supset\Gamma(16)$. Assume this is the case for $n-1$; then in particular, $[\Gamma(2),\Gamma(2)]$ contains elements which are equivalent modulo $2^{n}$ to $\sigma^{2^{n-2}}$, $\tau^{2^{n-2}}$, and $1+2^{n-1}$. On computing that the squares of such elements must be equivalent modulo $2^{n+1}$ to $\sigma^{2^{n-1}}$, $\tau^{2^{n-1}}$, and $1+2^{n}$ respectively, the claim is proven.

We next claim that the image of $[\Gamma(2),\Gamma(2)]$ modulo $16$ is cyclically generated by the image modulo $16$ of $\sigma\tau\sigma^{-1}\tau^{-1}$. To see this, we recall the well-known fact that $\Gamma(2)$ decomposes as the direct product of $\{\pm 1\}$ and the subgroup generated by $\sigma$ and $\tau$ and therefore, $[\Gamma(2),\Gamma(2)]$ coincides with the commutator subgroup of $\langle\sigma,\tau\rangle\lhd\Gamma(2)$. On checking that $\sigma\tau\sigma^{-1}\tau^{-1}$ commutes with both $\sigma$ and $\tau$ modulo $16$, we deduce as an easy exercise in group theory that the commutator of any two elements in $\Gamma(2)/\Gamma(16)$ is a power of the image of $\sigma\tau\sigma^{-1}\tau^{-1}$. Since the smallest normal subgroup of $\Gamma(2)/\Gamma(16)$ containing these powers is simply the cyclic subgroup generated by the image of $\sigma\tau\sigma^{-1}\tau^{-1}$, we have proven the claim.

Now it follows from the fact that $\Gamma(2^{n})/\Gamma(2^{n+1})\cong(\mathbb{Z}/2\mathbb{Z})^{3}$ for $n\geq 1$ that $\Gamma(2)/\Gamma(16)$ has order $2^{9}=512$; meanwhile, we see from what we have computed above that the image of $\sigma\tau\sigma^{-1}\tau^{-1}$ modulo $16$, which generates the image of $[\Gamma(2),\Gamma(2)]$, has order $4$. Therefore, $\Gamma(2)^{\mathrm{ab}}$ has order $128$. Since $\Gamma(2)^{\mathrm{ab}}$ is generated by the images of $-1$, $\sigma$, and $\tau$ modulo $[\Gamma(2),\Gamma(2)]$, the first statement of the lemma follows from an easy verification (by considering the image of $[\Gamma(2),\Gamma(2)]$ modulo $16$) that the images of $\sigma$ and $\tau$ each have order $8$ in $\Gamma(2)^{\mathrm{ab}}$.

∎

In order to find the extension of $K_{1}$ corresponding to the Galois quotient described by the lemma, we consider the $g\geq 2$ and $g=1$ cases separately.

3.2. The $g\geq 2$ case

Lemma 3.1, together with the results of §2 and the fact that $K_{\infty}\subset K^{\mathrm{unr}}$, implies that $K_{\infty}^{\mathrm{ab}}$ is an extension of $K_{2}=K_{1}(\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1})$ obtained by adjoining $2g$ independent $4$th roots of products of the elements $(\alpha_{j}-\alpha_{i})$. Similarly to what we saw in the proof of Theorem 2.4, Kummer theory tells us that $\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{2})$ is canonically identified with the dual of some subgroup $V\subset K_{2}^{\times}/(K_{2}^{\times})^{2}$, where both are considered as vector spaces over $\mathbb{F}_{2}$ of dimension $2g$. In fact, since $K_{\infty}\subset K^{\mathrm{unr}}$, the first statement of Lemma 2.3 implies that $V$ is also a subspace of the space $\widetilde{V}$ generated by the images in $\widetilde{K}_{2}^{\times}/(\widetilde{K}_{2}^{\times})^{2}$ of elements of the form $\sqrt{\alpha_{j}-\alpha_{i}}$, where $\widetilde{K}_{2}=K_{2}(\{\sqrt{\alpha_{j}-\alpha_{i}}\}_{1\leq i<j\leq d})$ as in the proof of Theorem 2.4. For ease of notation, we denote each of these images by $[i,j]=[j,i]\in\widetilde{V}$. We now proceed to explicitly determine the $2g$-dimensional subspace $V\subset K_{2}^{\times}/(K_{2}^{\times})^{2}\cap\widetilde{V}$.

Note that since $[\Gamma(2),\Gamma(2)]$ is a characteristic subgroup of $\Gamma(2)$, which in turn is normal in $\mathrm{Sp}(T_{2}(J))$, we have $[\Gamma(2),\Gamma(2)]\lhd\mathrm{Sp}(T_{2}(J))$. In particular, $[\Gamma(2),\Gamma(2)]$ is a normal subgroup of the image $G_{2}\subset\mathrm{Sp}(T_{2}(J))$ of $\rho_{2}$, so $K_{\infty}^{\mathrm{ab}}$ is Galois over $K$ and the action of $G_{2}$ on $\Gamma(2)$ by conjugation induces an action of $G_{2}$ on $\Gamma(2)^{\mathrm{ab}}$. This induces an $\mathbb{F}_{2}$-linear action of $G_{2}$ on $V$. It follows from a straightforward calculation that this action sends an element $\sigma\in G_{2}$ to the automorphism of $V$ determined by $[i,j]\mapsto[\bar{\sigma}(i),\bar{\sigma}(j)]$ for $1\leq i<j\leq d$, where the permuataion $\bar{\sigma}$ is the image of $\sigma$ in $G_{2}/\Gamma(2)\cong\mathrm{Gal}(K_{1}/K)=S_{d}$. We therefore have an action of $S_{d}$ on $V$, which we denote by $\psi:S_{d}\to\mathrm{Aut}(V)$.

As $V$ is a vector space over $\mathbb{F}_{2}$ of dimension $2g$, one candidate for the action $\psi$ is the well-known standard representation of $S_{d}$ over $\mathbb{F}_{2}$ (see [13, §2.2] for the construction of the standard representations over characteristic $2$ of dimension $d-1$ if $d$ is odd and of dimension $d-2$ if $d$ is even). In our situation, this turns out to be the case.

Lemma 3.2**.**

The $2g$-dimensional representation $(V,\psi)$ of $S_{d}$ defined above is isomorphic to the standard representation.

Proof.

We fix a symplectic ordered basis $\{a_{1},...,a_{g},b_{1},...,b_{g}\}$ of $T_{2}(J)$, i.e. a basis satisfying $e_{2}(a_{i},b_{i})=-1\in\mathbb{Z}_{2}$ for $1\leq i\leq g$ and $e_{2}(a_{i},b_{j})=0$ for $1\leq i<j\leq g$ and such that the image of each $a_{i}$ (resp. each $b_{i}$) in $J[2]$ is represented by the even-cardinality subset of roots given by $\{\alpha_{2i-1},\alpha_{2i}\}$ (resp. $\{\alpha_{2i},...,\alpha_{2g+1}\}$) (see the statement and proof of [10, Corollary 2.11] for a description of the elements in $J[2]$ in terms of the roots $\alpha_{i}$). In the following argument, we use the description of $[\Gamma(2),\Gamma(2)]$ given by [11, Proposition 2.1] as the subgroup of matrices (with respect to our symplectic basis) in $\mathrm{Sp}(T_{2}(J))$ which lie in $\Gamma(4)$ and whose $(i,i+g)$th and $(i+g,i)$th elements are divisible by $8$ for $1\leq i\leq g$. Note in particular that the automorphisms given by $v\mapsto v+4e_{2}(v,a_{i})a_{i}$ and $v\mapsto v+4e_{2}(v,b_{i})b_{i}$ each have $\pm 4$ as their $(i,i+g)$th or $(i+g,i)$th entries respectively, so their images in $\Gamma(4)/[\Gamma(2),\Gamma(2)]$ are nontrivial and distinct and in fact form a basis for the $\mathbb{F}_{2}$-space $\Gamma(4)/[\Gamma(2),\Gamma(2)]$.

We claim that the representation $(V,\psi)$ is faithful. Indeed, suppose that $\psi$ has nontrivial kernel. Since $d\geq 5$, this implies that the kernel of $\psi$ contains $A_{d}\lhd S_{d}$. Consider any element of $V$, written as a linear combination $\sum_{1\leq i<j\leq 2g+1}c_{i,j}[i,j]\in V$ with $c_{i,j}\in\mathbb{F}_{2}$; by our assumptions this element must be fixed by $A_{d}$. But then the $2$-transitivity of $A_{d}$ implies that the $c_{i,j}$’s are all equal, so that $V$ is spanned by the element $\sum_{1\leq i<j\leq d}[i,j]$, which contradicts the fact that $V$ is $2g$-dimensional.

If $g\geq 3$, then [13, Theorem 1.1] implies that $(V,\psi)$ is isomorphic to the standard representation, and we are done. We therefore assume that $g=2$. It follows from the statement and proof of [13, Lemma 3.2(iii),(iv)] that $(V,\psi)$ is isomorphic to the standard representation if and only if the transpositions in $S_{d}$ map to transvections in $\mathrm{Aut}(V)$ (in this context a transvection is defined to be any operator $A\in\mathrm{Aut}(V)$ such that $A-1$ has rank $1$). It therefore suffices to show that the transposition $(12)\in S_{d}$ acts on $V$ as a transvection. This is equivalent to the claim that any element of $G_{2}$ whose image modulo $2$ is $(12)\in S_{d}=\mathrm{Gal}(K_{1}/K)$ acts by conjugation on $\Gamma(4)/[\Gamma(2),\Gamma(2)]\cong\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{2})$ (which is the dual of $V$) as a transvection. For any $a\in T_{2}(J)$, let $T_{a}\in\mathrm{Aut}(V)$ denote the automorphism given by $v\mapsto v+e_{2}(v,a)a$. Then we see from the description of elements of $J[2]$ in terms of subsets of the set of $\alpha_{i}$’s which was mentioned above that the image of $T_{a_{1}}\in\mathrm{Sp}(T_{2}(J))$ modulo $2$ is $(12)$; since (as noted above) $\{T_{a_{1}}^{4},T_{a_{2}}^{4},T_{b_{1}}^{4},T_{b_{2}}^{4}\}$ is an $\mathbb{F}_{2}$-basis of $\Gamma(4)/[\Gamma(2),\Gamma(2)]$, we only need to calculate the conjugates of each of these basis elements by $T_{a_{1}}$. We compute that $T_{a_{1}}$ commutes with each of them except for $T_{b_{1}}^{4}$, and that

[TABLE]

Thus, we have seen that $T_{a_{1}}$ minus the identity operator acts on the $\mathbb{F}_{2}$-space $\Gamma(4)/[\Gamma(2),\Gamma(2)]$ by sending all basis elements to [math] except for $T_{b_{1}}^{4}$, which it sends to $T_{a_{1}}^{4}$. Therefore, $T_{a_{1}}$ acts as a transvection, as desired.

∎

Now the following proposition suffices to prove Theorem 1.1(a) when $k=\mathbb{C}$.

Proposition 3.3**.**

The subspace $V$ defined above is generated by the images in $\widetilde{V}$ of the elements $\prod_{j\neq i}\sqrt{\gamma_{i,j}}\in K_{2}^{\times}$ for $1\leq i\leq 2g+1$.

Proof.

For $1\leq i<j\leq 2g+1$, we write $[i,j]^{\prime}=[j,i]^{\prime}$ for the elements of $K_{2}^{\times}/(K_{2}^{\times})^{2}$ represented by $\sqrt{\gamma_{i,j}}$ (note that $[i,j]^{\prime}=[i,j]$ in the $d=2g+1$ case); we need to show that $V$ is spanned as an $\mathbb{F}_{2}$-space by the set $\{\sum_{j\neq i}[i,j]^{\prime}\}_{1\leq i\leq 2g+1}$.

We know from Lemma 3.2 that the action $\psi:S_{d}\to\mathrm{Aut}(V)$ defines the standard representation of $S_{d}$ of dimension $d-1$ (resp. $d-2$) if $d$ is odd (resp. even). We first assume that $d=2g+1$. By the construction of the standard representation, there exist elements $v_{i}\in V$ for $1\leq i\leq 2g+1$ which span $V$ and satisfy the unique linear relation $\sum_{1\leq i\leq 2g+1}v_{i}=0$, and such that $S_{2g+1}$ acts on the set of $v_{i}$’s by $v_{i}^{\sigma}=v_{\sigma(i)}$ for each permutation $\sigma$. We claim that $v_{1}=\sum_{j\neq 1}[1,j]$, from which it follows by acting on $v_{1}$ by any transposition $(1i)$ that $v_{i}=\sum_{j\neq i}[i,j]$ for $2\leq i\leq 2g+1$, and we get the desired spanning set for $V$.

As $v_{1}$ is obviously nontrivial, some $[s,t]$ appears in its expansion as a linear combination of basis elements of $\widetilde{V}$. Suppose that $1\in\{s,t\}$. Then the elements $[1,j]$ appear in the expansion of $v_{i}$ for all $2\leq j\leq 2g+1$, due to the fact that $v_{1}$ is fixed by every permutation in $S_{2g+1}$ which fixes $1$. If, on the other hand, $1\notin\{s,t\}$, then by a similar argument, all elements $[s,t]$ with $1\notin\{s,t\}$ appear in the expansion of $v_{1}$. It follows that either (i) $v_{1}=\sum_{j\neq 1}[1,j]$, (ii) $v_{1}=\sum_{s,t\neq 1}[s,t]$, or (iii) $v_{1}=\sum_{1\leq s<t\leq 2g+1}[s,t]$. In case (ii), we see that $\sum_{1\leq i\leq 2g+1}v_{i}=\sum_{1\leq s<t\leq 2g+1}[s,t]\neq 0$, a contradiction. In case (iii), we see that $v_{1}$ is fixed by all elements of $S_{2g+1}$ and therefore, all the elements $v_{i}$ are equal, which contradicts the fact that $V$ has dimension $2g$. Therefore, (i) holds, and we are done.

Now assume that $d=2g+2$. By the construction of the standard representation, there exist elements $v_{i,j}=v_{j,i}\in V$ for $1\leq i<j\leq 2g+2$ such that for any $i$, $\{v_{i,j}\}_{j\neq i}$ spans $V$ and satisfies the unique linear relation $\sum_{j\neq i}v_{i,j}=0$; such that $v_{s,t}+v_{s,j}=v_{t,j}$ for distinct $s,t,j$; and such that $S_{2g+2}$ acts on the set of $v_{i,j}$’s by $v_{i,j}^{\sigma}=v_{\sigma(i),\sigma(j)}$ for each permutation $\sigma$. We claim that $v_{1,2g+2}=\sum_{2\leq j\leq 2g+1}[1,j]^{\prime}$, from which we can again see by acting on $v_{1,2g+2}$ by any transposition $(1i)$ that $v_{i,2g+2}=\sum_{j\neq i}[i,j]^{\prime}$ for $2\leq i\leq 2g+1$, and we again get the desired spanning set for $V$.

By a similar analysis to what was done for the $d=2g+1$ case, we deduce that $v_{1,2g+2}$ is some linear combination of the elements $[i,2g+2]$, $\sum_{2\leq j\leq 2g+1}([1,j]+[2g+2,j])$, and $\sum_{2\leq s<t\leq 2g+1}[s,t]$. We first note that $v_{1,2g+2}$ cannot be the sum of an odd number of terms $[s,t]$, because then that would be the case for each other $v_{i,2g+2}$, and then the $2g+1$ terms $v_{i,2g+2}$ could not sum to [math]. We also note that, as in the $d=2g+1$ case, the element $v_{1,2g+1}$ can neither be trivial nor equal to the sum $\sum_{1\leq s<t\leq 2g+2}[s,t]$. Finally, it is straightforward to check that if $\sum_{2\leq s<t\leq 2g+1}[s,t]$ appears in the expansion of $v_{1,2g+2}$, then the $v_{s,t}+v_{s,j}=v_{t,j}$ property does not hold. Our only remaining choice is that $v_{1,2g+2}=\sum_{2\leq j\leq 2g+1}([1,j]+[2g+2,j])=\sum_{2\leq j\leq 2g+1}[1,j]^{\prime}$, and we are done.

∎

3.3. The $g=1$ case

Lemma 3.1, together with the results of §2 and the fact that $K_{\infty}\subset K^{\mathrm{unr}}$, imply that $K_{\infty}^{\mathrm{ab}}$ is an extension of $K_{2}=K_{1}(\sqrt{\gamma_{1,2}},\sqrt{\gamma_{1,3}},\sqrt{\gamma_{2,3}})$ obtained by adjoining $2$ independent $4$th roots of products of the elements $\gamma_{i,j}\in K_{1}$ (recall from [16] that $K_{1}$ is generated over $K$ by the $\gamma_{i,j}$’s both when $d=3$ and when $d=4$). Therefore, in this case, we get via Kummer theory a canonical identification of $\mathrm{Gal}(K_{\infty}^{\mathrm{ab}}/K_{2})$ with some subgroup $V\subset K_{2}^{\times}/(K_{2}^{\times})^{4}$; the submodule $V$ of the $\mathbb{Z}/4\mathbb{Z}$-module $K_{2}^{\times}/(K_{2}^{\times})^{4}$ is free of rank $2$. In fact, $V$ is contained in the rank-$3$ free submodule $\widetilde{V}\subset K_{2}^{\times}/(K_{2}^{\times})^{4}$ generated by images of the elements $\sqrt{\gamma_{i,j}}\in K_{2}^{\times}$. We denote each of these images by $[i,j]^{\prime}=[j,i]^{\prime}\in\widetilde{V}$. We now proceed to explicitly determine the rank-$2$ submodule $V\subset\widetilde{V}$.

As with the $g\geq 2$ case, we have an action of $G_{2}$ on $V$ which factors through the quotient $\mathrm{Gal}(K_{1}/K)$; this quotient is isomorphic to $S_{3}$ both when $d=3$ and when $d=4$. This action sends a permutation $\bar{\sigma}\in S_{3}$ to the automorphism of $V$ determined by $[i,j]^{\prime}=[\bar{\sigma}(i),\bar{\sigma}(j)]^{\prime}$ for $1\leq i<j\leq 3$. We again write $\psi:S_{3}\to\mathrm{Aut}(V)$ for this action. The following proposition suffices to prove Theorem 1.1(b) when $k=\mathbb{C}$.

Proposition 3.4**.**

The submodule $V\subset\widetilde{V}$ is generated by

[TABLE]

Proof.

We first note that the action $\psi$ has trivial kernel, by the same argument as was used in the proof of Proposition 3.3. Since $\mathrm{Aut}(V/2V)\cong\mathrm{SL}_{2}(\mathbb{F}_{2})\cong S_{3}$, this implies that the induced action of $\psi$ modulo $2V$ is isomorphic to the standard representation of $S_{3}$. By essentially the same argument that was used in the proof of Proposition 3.3 for the odd-degree case, this shows that $V/2V$ is spanned by the images modulo $2$ of the elements $\sigma_{j\neq i}[i,j]^{\prime}$ for $i=1,2,3$, implying that

[TABLE]

Therefore, the element $2[1,2]^{\prime}+2[1,3]^{\prime}+2[2,3]^{\prime}\in\widetilde{V}$ cannot lie in $V$, because otherwise $V$ would contain the subgroup generated by $2[1,2]^{\prime},2[1,3]^{\prime},2[2,3]^{\prime}$, which is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{3}$, contradicting the fact that $V$ has rank $2$.

Let $\Phi:\widetilde{V}\to\mathbb{Z}/4\mathbb{Z}$ be the functional sending $c_{1}[2,3]^{\prime}+c_{2}[3,1]^{\prime}+c_{3}[1,2]^{\prime}$ to $c_{1}+c_{2}+c_{3}\in\mathbb{Z}/4\mathbb{Z}$. We claim that $V$ coincides with the kernel of $\Phi$. Suppose that for some $v\in V$, we have $\Phi(v)\neq 0$. Since $\Phi(\psi(\bar{\sigma})(v))=\Phi(v)$ for all $\bar{\sigma}\in S_{3}$, we have $\Phi(\sum_{\bar{\sigma}\in A_{3}}\psi(\bar{\sigma})(v))=3\Phi(v)\neq 0$. But $\sum_{\bar{\sigma}\in A_{3}}\psi(\bar{\sigma})(v)\in V$ is fixed by $A_{3}\lhd S_{3}$ and is therefore some nontrivial multiple of $[1,2]^{\prime}+[1,3]^{\prime}+[2,3]^{\prime}\in\widetilde{V}$, which contradicts the fact that $2[1,2]^{\prime}+2[1,3]^{\prime}+2[2,3]^{\prime}\notin V$. Since the image of $\Phi$ has order $4$, its kernel has order $64/4=16$, hence the claim. Meanwhile, the generators given in the statement of the proposition lie in the kernel of $\Phi$, and the statement now follows from the elementary verification that the group they generate also has order $16$.

∎

3.4. The action of $-1$

We have shown that parts (a) and (b) of Theorem 1.1 hold when $k=\mathbb{C}$; we will now prove that the element $-1\in\Gamma(2)\cong\mathrm{Gal}(K_{\infty}/K_{1})$ acts as stated in Theorem 1.1(c), and then the full theorem will be proved over the complex numbers. In the odd-degree case, it follows immediately from the fact that $\rho^{\mathrm{top}}(\Sigma)=-1$ by Proposition 2.2(b) combined with Lemma 2.3 that since $\gamma_{i,j}=\alpha_{j}-\alpha_{i}$, the element $-1$ acts trivially (resp. by sign change) on the generators of $K_{\infty}^{\mathrm{ab}}/K_{2}$ listed in the statement of the theorem if $g$ is even (resp. if $g$ is odd). Similarly in the even-degree case, it follows immediately from the fact that $\rho^{\mathrm{top}}(\Sigma^{\prime})=-1$ by Proposition 2.2(c) combined with Lemma 2.3 that since $\gamma_{i,j}=(\alpha_{j}-\alpha_{i})\prod_{l\neq i,j}(\alpha_{2g+2}-\alpha_{l})$, the element $-1$ acts trivially (resp. by sign change) on the generators of $K_{\infty}^{\mathrm{ab}}/K_{2}$ listed in the statement of the theorem if $g$ is even (resp. if $g$ is odd).

4. Proof of the theorem in the general case

We have shown that Theorem 1.1 holds when $k=\mathbb{C}$; we will now prove that this suffices to show that Theorem 1.1 holds in general. In this section, whenever the ground field $k$ is specified (e.g. $k=\mathbb{Q}$), we will write $K_{k}$ (e.g. $K_{\mathbb{Q}}$) for the extension of $k$ obtained by adjoining the symmetric functions of the independent transcendental elements $\alpha_{1},...,\alpha_{d}$; our goal is to show that the particular generators of $(K_{\mathbb{C}})_{\infty}^{\mathrm{ab}}/K_{\mathbb{C}}$ that we found in §2 and §3 can also be used to generate $K_{\infty}^{\mathrm{ab}}/K$ (and that any Galois element mapping to $-1\in\mathrm{GSp}(T_{2}(J))$ acts in the same way on them), where $K=K_{k}$ for any field $k$ of characteristic different from $2$. The rest of this section will be devoted to proving the following proposition.

Proposition 4.1**.**

To prove Theorem 1.1, it suffices to prove the statement when $k=\mathbb{C}$.

We assume that $g\geq 2$ and only prove part (a) of the theorem as well as part (c) for the $g\geq 2$ case, noting that the claims for the $g=1$ case result from very similar arguments. In what follows, we will freely use the obvious fact that given an abelian variety $A$ over a field $F$ and an extension $F^{\prime}/F$, the finite algebraic extension $F^{\prime}(A[2^{n}])$ coincides with the compositum $F^{\prime}F(A[2^{n}])$ for any $n\geq 1$. We first need the following lemmas.

Lemma 4.2**.**

Let $k$ be any field of characteristic different from $2$ with separable closure $\bar{k}$. Then we have

a) $\mathrm{Gal}(K_{n}(\mu_{2})/K_{1}(\mu_{2}))\cong\Gamma(2)/\Gamma(2^{n})$ for $n\geq 1$ and thus $\mathrm{Gal}(K_{\infty}(\mu_{2})/K_{1}(\mu_{2}))\cong\Gamma(2)$; and

b) $K_{n}\cap\bar{k}=k(\zeta_{2^{n}})$ for $n\geq 1$ and thus $K_{\infty}\cap\bar{k}=k(\mu_{2})$.

Proof.

The author has shown (a) for $k$ of characteristic [math] (as [15, Proposition 4.1]) but the following argument proves (a) in the case of positive characteristic also. We first claim that the image of $\rho_{2}$ in $\mathrm{GSp}(T_{2}(J))$ contains a transvection given by $v\mapsto v+e_{2}(v,a)a$ for some $a\in T_{2}(J)\smallsetminus 2T_{2}(J)$. This follows from the discussion in [2, §2.3] (see also [5, §3.v] and [6, §9-10] and note that the argument holds in positive characteristic as well). Meanwhile, since the polynomial defining the hyperelliptic curve $C$ has full Galois group, the image of $\bar{\rho}_{2}$ is isomorphic to $S_{2g+1}$ or $S_{2g+2}$. It now follows from [4, Theorem 2.1.1, §2.2] that the image $G_{2}$ of $\rho_{2}$ in $\mathrm{GSp}(T_{2}(J))$ contains $\Gamma(2)\lhd\mathrm{Sp}(T_{2}(J))$. After restricting to the absolute Galois group of $K_{1}(\mu_{2})$, this image coincides with $\Gamma(2)$, and (a) immediately follows.

The linear disjointness of $K_{n}(\mu_{2})$ and $\bar{k}K_{1}$ over $K_{1}(\mu_{2})$ follows immediately from the fact that $\mathrm{Gal}(\bar{k}K_{n}/\bar{k}K_{1})\cong\mathrm{Gal}(K_{n}(\mu_{2})/K_{1}(\mu_{2}))\cong\Gamma(2)/\Gamma(2^{n})$ by part (a). Moreover, it is clear from the well-known description of $2$-torsion fields discussed above that $K_{1}(\mu_{2})\cap\bar{k}=k(\mu_{2})$, so we get $K_{n}(\mu_{2})\cap\bar{k}=(K_{n}(\mu_{2})\cap\bar{k}K_{1})\cap\bar{k}=k(\mu_{2})$. It follows that we have $K_{\infty}\cap\bar{k}=k(\mu_{2})$, so to prove part (b) it suffices to show that $K_{n}\cap k(\mu_{2})=k(\zeta_{2^{n}})$ for $n\geq 1$.

Let $\widetilde{\Gamma}(2^{n})\lhd\mathrm{GSp}(T_{2}(J))$ denote the kernel of reduction modulo $2^{n}$ for each $n\geq 1$, and write $G_{2}\subset\mathrm{GSp}(T_{2}(J))$ for the image of $\rho_{2}$. It is clear that $G_{2}\cap\widetilde{\Gamma}(2^{n})$ is isomorphic to the subgroup of $\mathrm{Gal}(K_{\infty}/K)$ fixing $K_{n}$; meanwhile, part (a) says that $G_{2}\cap\Gamma(2)=\Gamma(2)$ is isomorphic to the Galois subgroup fixing $k(\mu_{2})$. Therefore, $K_{n}\cap k(\mu_{2})$ is the fixed field of the subgroup of $\mathrm{Gal}(K_{\infty}/K)$ generated by $G_{2}\cap\widetilde{\Gamma}(2^{n})$ and $\Gamma(2)$, which is easily seen to coincide with the kernel of the mod-$2^{n}$ determinant map $G_{2}\to(\mathbb{Z}/2^{n}\mathbb{Z})^{\times}$. But by equivariance of the Weil pairing, the fixed field of this subgroup coincides with $k(\zeta_{2^{n}})$, and we are done.

∎

Let $J_{0}$ denote the Jacobian of the hyperelliptic curve $C_{0}$ defined over $K_{\mathbb{Q}}$ given by the equation in (1). Note that $C_{0}$ admits a smooth model $\mathfrak{C}$ over

[TABLE]

where the superscript “$S_{d}$” indicates taking the subring of invariants under the obvious permutation action on the $\alpha_{i}$’s. Define $\mathfrak{J}\to S$ to be the abelian scheme representing the Picard functor of the scheme $\mathfrak{C}\to S$ (see [9, Theorem 8.1]). Note that the ring $\mathbb{Z}[\textstyle\frac{1}{2},\{\alpha_{i}\}_{1\leq i\leq d},\{(\alpha_{i}-\alpha_{j})^{-1}\}_{1\leq i<j\leq 2g+1}]$, along with all subrings of invariants under finite groups of automorphisms, is integrally closed; in particular, $\mathcal{O}_{S}:=\mathbb{Z}[\textstyle\frac{1}{2},\{\alpha_{i}\}_{1\leq i\leq d},\{(\alpha_{i}-\alpha_{j})^{-1}\}_{1\leq i<j\leq 2g+1}]^{S_{d}}$ is integrally closed.

For each $n\geq 1$, [8, Proposition 20.7] implies that the kernel of the multiplication-by-$2^{n}$ map on $\mathfrak{J}\to S$, which we denote by $\mathfrak{J}[2^{n}]\to S$, is a finite étale group scheme over $S$. Since the morphism $\mathfrak{J}[2^{n}]\to S$ is finite, $\mathfrak{J}[2^{n}]$ is an affine scheme; we write $\mathcal{O}_{S,n}\supset\mathcal{O}_{S}$ for the minimal extension of scalars under which $\mathfrak{J}[2^{n}]$ becomes constant. It follows from the fact that $\mathcal{O}_{S}$ is integrally closed and from the finite étaleness of $\mathfrak{J}[2^{n}]$ that $\mathcal{O}_{S,n}$ is also integrally closed; its fraction field coincides with $K_{0}(J_{0}[2^{n}])$.

Let $\mathcal{O}_{S,\infty}^{\mathrm{ab}}\supset\mathcal{O}_{S}$ denote the integrally closed extension whose fraction field coincides with the maximal abelian subextension $(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}$ of $(K_{\mathbb{Q}})_{\infty}/(K_{\mathbb{Q}})_{1}(\mu_{2})$. Lemmas 3.1 and 4.2 together imply that $(K_{\mathbb{Q}})_{2}(\mu_{2})\subsetneq(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}\subsetneq(K_{\mathbb{Q}})_{3}(\mu_{2})$; that the extension $(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}/(K_{\mathbb{Q}})_{1}(\mu_{2})$ has Galois group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{2g^{2}-g}\times(\mathbb{Z}/4\mathbb{Z})^{2g}$; and that the analogous statements hold over each $\mathbb{F}_{p}$. Thus it is clear that for each prime $p\neq 2$, the fraction field of $\mathcal{O}_{S,\infty}^{\mathrm{ab}}/(p)$ coincides with the subfield of $(K_{\mathbb{F}_{p}})_{3}(\mu_{2})$ fixed by the kernel of the map $\bar{\pi}:\Gamma(2)/\Gamma(8)\twoheadrightarrow(\mathbb{Z}/2\mathbb{Z})^{2g^{2}-g}\times(\mathbb{Z}/4\mathbb{Z})^{2g}$ induced by the abelianization map $\pi$ and is therefore the maximal abelian subextension of $(K_{\mathbb{F}_{p}})_{\infty}/(K_{\mathbb{F}_{p}})_{1}(\mu_{2})$. Moreover, if $k$ is any field of characteristic $p\neq 2$, then it similarly follows from Lemmas 3.1 and 4.2 that $(K_{k(\mu_{2})})_{\infty}^{\mathrm{ab}}$ coincides with the subfield of $K_{3}(\mu_{2})$ fixed by the kernel of $\bar{\pi}$.

Note that $k(\mu_{2})$ contains $\mathfrak{f}(\mu_{2})$, where the prime subfield $\mathfrak{f}$ is $\mathbb{Q}$ (resp. $\mathbb{F}_{p}$) if the characteristic of $k$ is [math] (resp. $p\geq 3$). It then follows from the linear disjointness of $K_{3}(\mu_{2})$ and $\bar{\mathfrak{f}}$ over $\mathfrak{f}(\mu_{2})$ given by Lemma 4.2(b) that the subfield of $K_{3}(\mu_{2})$ fixed by the kernel of $\bar{\pi}$ coincides with the compositum of $K_{1}(\mu_{2})$ with the subfield of $(K_{\mathfrak{f}})_{3}(\mu_{2})$ fixed by the kernel of $\bar{\pi}$. The extension $(K_{k(\mu_{2})})_{\infty}^{\mathrm{ab}}$ is therefore generated over $K_{1}(\mu_{2})$ by the generators of $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ over $\mathcal{O}_{S,1}[\mu_{2}]$ (resp. by the images of these generators modulo $(p)$) if $k$ has characteristic [math] (resp. if $k$ has characteristic $p\geq 3$).

It remains to show that these generators are the same ones appearing in Theorem 1.1, for which we need another lemma.

Lemma 4.3**.**

The fields $(K_{\mathbb{C}})_{n}$ for $n\geq 1$, $(K_{\mathbb{C}})_{\infty}$, and $(K_{\mathbb{C}})_{\infty}^{\mathrm{ab}}$ coincide with the compositums of $\mathbb{C}$ with $(K_{\mathbb{Q}(\mu_{2})})_{n}$, $(K_{\mathbb{Q}(\mu_{2})})_{\infty}$, and $(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}$ respectively.

Proof.

For $n\geq 1$, let $\theta_{n}:\mathrm{Gal}((K_{\mathbb{C}})_{\infty}/(K_{\mathbb{C}})_{n})\to\mathrm{Gal}((K_{\mathbb{Q}})_{\infty}/(K_{\mathbb{Q}})_{n}(\mu_{2}))$ be the map given by the composition of the obvious inclusion map with the obvious restriction map. It is shown in the proof of [15, Proposition 4.1] that each $\theta_{n}$ is an isomorphism (this can also be deduced from Lemma 4.2(a)). Since $\theta_{1}$ and $\theta_{3}$ are isomorphisms, they induce an isomorphism $\Gamma(2)/\Gamma(8)\cong\mathrm{Gal}((K_{\mathbb{C}})_{3}/(K_{\mathbb{C}})_{1})\stackrel{{\scriptstyle\sim}}{{\to}}\mathrm{Gal}((K_{\mathbb{Q}})_{3}(\mu_{2})/(K_{\mathbb{Q}})_{1}(\mu_{2}))$, the image of whose restriction to $\mathrm{Gal}((K_{\mathbb{C}})_{3}/(K_{\mathbb{C}})_{\infty}^{\mathrm{ab}})$ fixes the subfield $(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}$. It follows from the definition of $\theta_{3}$ that $K_{\infty}^{\mathrm{ab}}=\mathbb{C}(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}$.

∎

We now claim that $\{\gamma_{i,j}\}_{1\leq i<j\leq 2g+1}$ is a set of generators for $\mathcal{O}_{S,2}[\zeta_{8}]$ over $\mathcal{O}_{S,1}[\zeta_{8}]$. Indeed, we see that $(K_{\mathbb{C}})_{1}(\{\gamma_{i,j}\}_{1\leq i<j\leq 2g+1})=(K_{\mathbb{C}})_{2}=\mathbb{C}(K_{\mathbb{Q}})_{2}$ and

[TABLE]

using Lemma 4.2(a) and (b) and Lemma 3.1, where $\iota\in\mathrm{Gal}((K_{\mathbb{Q}})_{2}/(K_{\mathbb{Q}})_{1})$ is any automorphism that acts on $\mathbb{Q}(\zeta_{4})$ by complex conjugation. It follows that $\mathcal{O}_{S,2}$ is generated over $\mathcal{O}_{S,1}$ by $\zeta_{4}$ and the square roots of integral elements $a_{i,j}\gamma_{i,j}$ for some $a_{i,j}\in\mathbb{Z}$ for $1\leq i<j\leq 2g+1$. Since the extension $\mathcal{O}_{S,2}$ is unramified over $\mathcal{O}_{S}$, we have $a_{i,j}\in\{\pm 1,\pm 2\}$. But $\sqrt{\pm 1},\sqrt{\pm 2}\in\mathbb{Z}[\zeta_{8}]\subset\mathcal{O}_{S,1}[\zeta_{8}]$, so we have $\mathcal{O}_{S,2}[\zeta_{8}]=\mathcal{O}_{S,1}[\zeta_{8},\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1}]$, as claimed.

Next we find formulas for generators of $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ over $\mathcal{O}_{S,1}(\mu_{2})$ using the ones we have shown for $k=\mathbb{C}$. We know that $\mathcal{O}_{S,\infty}^{\mathrm{ab}}\supsetneq\mathcal{O}_{S,2}(\mu_{2})=\mathcal{O}_{S,1}[\mu_{2},\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1}]$, and so, by Lemma 3.1, $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ is generated over $\mathcal{O}_{S,2}(\mu_{2})$ by square roots of $2g$ independent integral elements. Then it is clear from Lemma 4.3 that we may choose these $2g$ elements to be of the form $a_{i}\sqrt{\scriptstyle\prod_{j\neq i}\gamma_{i,j}}$ for some $a_{i}\in\mathbb{Z}[\mu_{2}]$ for $1\leq i\leq 2g$ and that the extension also contains a square root of $a_{2g+1}\sqrt{\scriptstyle\prod_{j\neq 2g+1}\gamma_{2g+1,j}}$ for some $a_{2g+1}\in\mathbb{Z}[\mu_{2}]$. Using the fact that $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ is Galois over $\mathcal{O}_{S}[\mu_{2}]$, from conjugating by Galois automorphisms that fix $\mathbb{Q}(\mu_{2})$ but permute the $\alpha_{i}$’s, we see that we may choose the elements $a_{1},...,a_{2g+1}$ to be the same element $a\in\mathbb{Z}[\mu_{2}]$. Note that the product of these $2g+1$ elements $a\sqrt{\scriptstyle\prod_{j\neq i}\gamma_{i,j}}$ can be written as $\pm a^{2g+1}\prod_{1\leq i<j\leq 2g+1}\gamma_{i,j}$, and this product must have a square root in $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$. But we already know that $\pm\prod_{1\leq i<j\leq 2g+1}\gamma_{i,j}$ has a square root in $\mathcal{O}_{S,2}[\mu_{2}]$, so we have $\sqrt{a}\in\mathcal{O}_{S,\infty}^{\mathrm{ab}}$. Since $\sqrt{a}$ is algebraic over $\mathbb{Q}(\mu_{2})$, we get $\sqrt{a}\in\mathbb{Q}(\mu_{2})$ by Lemma 4.2(b). Then $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ is generated over $\mathcal{O}_{S,2}[\mu_{2}]$ by the elements $\sqrt{a}\sqrt[4]{\scriptstyle\prod_{j\neq i}\gamma_{i,j}}$. Thus, the fraction field $(K_{\mathbb{Q}(\mu_{2})})_{\infty}^{\mathrm{ab}}$ of $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ (resp. the fraction field $(K_{\mathbb{F}_{p}(\mu_{2})})_{\infty}^{\mathrm{ab}}$ of $\mathcal{O}_{S,\infty}^{\mathrm{ab}}/(p)$ for each prime $p\neq 2$) is generated over $(K_{\mathbb{Q}(\mu_{2})})_{2}$ (resp. $(K_{\mathbb{F}_{p}(\mu_{2})})_{2}$) by the elements given in Theorem 1.1(a).

What we have shown above is that given any field $k$, the statement of Theorem 1.1 holds over $k(\mu_{2})$. It is now clear that $K_{1}(\mu_{2},\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1},\{\sqrt[4]{\scriptstyle\prod_{j\neq i}\gamma_{i,j}}\}_{1\leq i\leq 2g+1})$ is a subextension of $(K_{k(\mu_{2})})_{\infty}=K_{\infty}(\mu_{2})=K_{\infty}/K_{1}$. If $\zeta_{4}\in k$, then this subextension is clearly Kummer and therefore abelian, and it must be maximal abelian since there is no larger subextension which is abelian over $K_{1}(\mu_{2})$. If $\zeta_{4}\notin k$, then $K_{\infty}^{\mathrm{ab}}$ must be the largest subextension of $K_{1}(\mu_{2},\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1},\{\sqrt[4]{\scriptstyle\prod_{j\neq i}\gamma_{i,j}}\}_{1\leq i\leq 2g+1})$ which is abelian over $K_{1}$; this clearly coincides with $K_{2}(\mu_{2})=K_{1}(\mu_{2},\{\sqrt{\gamma_{i,j}}\}_{1\leq i<j\leq 2g+1})$. Thus, the statement of Theorem 1.1(a) is proved over $k$.

Finally, let $\sigma\in G_{K}$ be an element such that $\rho_{2}(\sigma)=-1\in\mathrm{GSp}(T_{2}(J))$. Then it is clear from tracing through the above arguments that $\sigma$ acts on $\mathcal{O}_{S,\infty}^{\mathrm{ab}}$ by changing the signs of the generators $\sqrt{\gamma_{i,j}}$ and by fixing (resp. changing the signs of) the remaining generators $\scriptstyle\sqrt{\prod_{j\neq i}\gamma_{i,j}}$ if $g$ is even (resp. if $g$ is odd), and that therefore, it acts this way on $K_{\infty}^{\mathrm{ab}}$, proving part (c) over $k$.