2.1 A representation of the split odd special orthogonal group

Let $W$ be the $\mathbb {Z}$-lattice of rank $2n+1$ equipped with a nondegenerate symmetric bilinear form $[-,-] \colon W \times W \to \mathbb {Z}$ that has signature $(n+1,n)$ after extending scalars to $\mathbb {R}$. By [Reference SerreSer73, Chapter V]), the lattice $W$ is unique up to isomorphism over $\mathbb {Z}$, and there is a $\mathbb {Z}$-basis

(3)\begin{equation} W = \mathbb{Z}\langle e_1, \ldots, e_n,u, e_1', \ldots, e_n' \rangle, \end{equation}

such that $[e_i, e_j] = [e_i',e_j'] = [e_i,u]= [e_i',u]= 0$, $[e_i, e_j'] = \delta _{ij}$, and $[u,u] = 1$ for all relevant pairs $(i,j)$. We denote by $A_0$ the matrix of $[-,-]$ with respect to the basis (3).

For a $\mathbb {Z}$-algebra $R$, let $W_R := W \otimes _\mathbb {Z} R$. For a field $k$ of characteristic not equal to $2$, the $k$-vector space $W_k$ equipped with the bilinear form $[-,-]$ is called the split orthogonal space of dimension $2n+1$ and determinant $(-1)^n$ over $k$ and is unique up to $k$-isomorphism [Reference Milnor and HusemollerMH73, § 6]. The space $W_k$ is called ‘split’ because it is a nondegenerate quadratic space containing a maximal isotropic subspace (defined over $k$) for the bilinear form $[-,-]$.

Let $T \in \operatorname {End}_R(W_R)$. Recall that the adjoint transformation $T^{\dagger} \in \operatorname {End}_R(W_R)$ of $T$ with respect to the form $[-,-]$ is determined by the formula $[Tv, w] = [v, T^{\dagger} w]$ for all $v,w \in W_R$, and that $T$ is said to be self-adjoint if $T = T^{\dagger}$. If $T$ is expressed as a matrix with respect to the basis (3), then $T$ is self-adjoint with respect to the form $[-,-]$ if and only if $T^TA_0 = A_0T$.

Let $V$ be the affine space defined over $\operatorname {Spec} \mathbb {Z}$ whose $R$-points are given by

\[ V(R) = \{T \in \operatorname{End}_R(W_R) : T = T^{\dagger} \} \]

for any $\mathbb {Z}$-algebra $R$. An $R$-automorphism $g \in \operatorname {Aut}_R(W_R)$ is called orthogonal with respect to the form $[-,-]$ if $[g(v),g(w)] = [v,w]$ for all $v,w \in W_R$. If $g$ is expressed as a matrix with respect to the basis (3), then $g$ is orthogonal with respect to the form $[-,-]$ if and only if $g^TA_0g = A_0$. Let $G := \operatorname {SO}(W)$ be the group scheme defined over $\operatorname {Spec} \mathbb {Z}$ whose $R$-points are given by

\[ G(R) = \{g \in \operatorname{Aut}_R(W_R) : g\text{ is orthogonal with respect to $[-,-]$ and }\det g = 1\} \]

for any $\mathbb {Z}$-algebra $R$. The group scheme $G$ is known as the split odd special orthogonal group of the lattice $W$, and it acts on $V$ by conjugation: for $g \in G(R)$ and $T \in V(R)$, the map $gTg^{-1} \in \operatorname {End}_R(W_R)$ is readily checked to be self-adjoint with respect to $[-,-]$ and is therefore an element of $V(R)$. We thus obtain a linear representation $G \to \operatorname {GL}(V)$ of dimension $\dim V = 2n^2 + 3n + 1$.

Since $G$ acts on $V$ by conjugation, the characteristic polynomial

\[ \operatorname{ch}(T) := \det(x \cdot \operatorname{id} - T) = x^{2n+1} + \sum_{i = 1}^{2n+1} c_i x^{2n+1-i} \in R[x] \]

is invariant under the action of $G(R)$ for each $T \in V(R)$. In fact, by [Reference BourbakiBou75, § 8.3, part (VI) of § 13.2], the coefficients $c_1, \ldots, c_{2n+1}$, which have respective degrees $1, \ldots, 2n+1$, freely generate the ring of $G$-invariant functions on $V$.

2.2 Rings associated to binary forms

For the rest of § 2, let $R$ be a principal ideal domain (we typically take $R$ to be $\mathbb {Z}$, $\mathbb {Q}$, $\mathbb {Z}_p$, $\mathbb {Z}/p\mathbb {Z}$, or $\mathbb {Q}_p$, where $p \in \mathbb {Z}$ is a prime), and let $k$ be the fraction field of $R$. Let $n \geq 1$, and let

\[ F(x,z) = \sum_{i = 0}^{2n+1} f_i x^{2n+1-i}z^i \in R[x,z] \]

be a separable binary form of degree $2n+1$ with leading coefficient $f_0 \in R\smallsetminus \{0\}$.

Consider the étale $k$-algebra

\[ K_F := k[x]/(F(x,1)). \]

Let $\theta$ denote the image of $x$ in $K_F$. For each $i \in \{1, \ldots, 2n\}$, let $p_i$ be the polynomial defined by

\[ p_i(t) := \sum_{j = 0}^{i-1} f_j t^{i-j}, \]

and let $\zeta _i := p_i(\theta )$. To the binary form $F$, there is a naturally associated free $R$-submodule $R_F \subset K_F$ having rank $2n+1$ and $R$-basis given by

(4)\begin{equation} R_F := R \langle 1, \zeta_1, \zeta_2, \ldots, \zeta_{2n} \rangle. \end{equation}

The module $R_F$ is of significant interest in its own right and has been studied extensively in the literature. In [Reference Birch and MerrimanBM72, proof of Lemma 3], Birch and Merriman proved that the discriminant of $F$ is equal to the discriminant of $R_F$, and in [Reference NakagawaNak89, Proposition 1.1], Nakagawa proved that $R_F$ is actually a ring (and, hence, an order in $K_F\!$) having multiplication table

(5)\begin{equation} \zeta_i\zeta_j = \sum_{k = j+1}^{\min\{i+j,2n+1\}} f_{i+j-k}\zeta_k - \sum_{k = \max\{i+j-(2n+1),1\}}^i f_{i+j-k}\zeta_k, \end{equation}

where $1 \leq i \leq j \leq 2n$ and we take $\zeta _0 = 1$ and $\zeta _{2n+1} = -f_{2n+1}$ for the sake of convenience. (To be clear, these results of Nakagawa are stated for the case of irreducible $F$, but as noted in [Reference WoodWoo11, § 2.1], their proofs continue to hold when $F$ is not irreducible.)

Also contained in $K_F$ is a natural family of free $R$-submodules $I_F^j$ of rank $2n+1$ for each $j \in \{0, \ldots, 2n\}$, having $R$-basis given by

(6)\begin{equation} I_F^j := R\langle1,\theta, \ldots, \theta^j,\zeta_{j+1}, \ldots, \zeta_{2n} \rangle. \end{equation}

Note that $I_F^0 = R_F$ is the unit ideal. By [Reference WoodWoo11, Proposition A.1], each $I_F^j$ is an $R_F$-module and, hence, a fractional ideal of $R_F$; moreover, the notation $I_F^j$ makes sense, because $I_F^j$ is equal to the $j$th power of $I_F^1$. By [Reference WoodWoo11, Proposition A.4], the fractional ideals $I_F^j$ are invertible precisely when the form $F$ is primitive, in the sense that $\sum _{i = 0}^{2n+1}Rf_i = R$. It is a result of Simon (see [Reference SimonSim08, Proposition 14]; cf. [Reference WoodWoo11, Theorem 2.4]) that

\[ J_F := I_F^{2n-1} \]

represents the ideal class of the inverse different of $R_F$.

Given a fractional ideal $I$ of $R_F$ having a specified basis (i.e. a based fractional ideal), the norm of $I$, denoted by $\operatorname {N}(I)$, is defined to be the determinant of the $k$-linear transformation taking the basis of $I$ to the basis of $R_F$ in (4). It is easy to check that $\operatorname {N}(I_F^j) = f_0^{-j}$ for each $j$ with respect to the basis in (6). The norm of $\kappa \in K_F^\times$ is $\operatorname {N}(\kappa ) := \operatorname {N}(\kappa \cdot I_F^0)$ with respect to the basis $\langle \kappa, \kappa \cdot \zeta _1, \ldots, \kappa \cdot \zeta _{2n}\rangle$ of $\kappa \cdot I_F^0$. Note that we have the multiplicativity relation

(7)\begin{equation} \operatorname{N}(\kappa \cdot I) = \operatorname{N}(\kappa) \cdot \operatorname{N}(I) \end{equation}

for any $\kappa \in K_F^\times$ and fractional ideal $I$ of $R_F$ with a specified basis.

We now explain how the objects $K_F$, $R_F$, and $I_F^j$ transform under the action of $\gamma \in \operatorname {SL}_2(R)$ on binary forms of degree $2n+1$ defined by $\gamma \cdot F = F((x,z) \cdot \gamma )$. If $F' = \gamma \cdot F$, then $\gamma$ induces an isomorphism $K_F \simeq K_{F'}$, under which the rings $R_F$ and $R_{F'}$ happen to be identified with each other (see [Reference NakagawaNak89, Proposition 1.2] for a direct proof and [Reference WoodWoo11, § 2.3] for a geometric argument). On the other hand, the ideals $I_F^j$ and $I_{F'}^j$ are isomorphic as $R_F$-modules but may not necessarily be identified under the isomorphism $K_F \simeq K_F'$. Indeed, as explained in [Reference BhargavaBha13, (7)], these ideals are related by the following explicit rule: if $\gamma = [\begin{smallmatrix} a & b \\ c & d\end{smallmatrix}]$, then for each $j \in \{0, \ldots, 2n\}$, the composition

(8)

is an injective map of $R_F$-modules having image equal to $I_{F'}^{j}$, where $\phi _{j,\gamma }$ sends each $\delta \in I_F^j$ to $(-b\theta +a)^{-j} \cdot \delta \in K_F$. When $j = 0$, we recover the identification of $R_F = I_F^0$ with $R_{F'} = I_{F'}^0$ from (

).

2.3 Parametrization of square roots of the class of $J_F$

We say that a based fractional ideal $I$ of $R_F$ is a square root of the class of the inverse different if there exists $\delta \in K_F^\times$ such that

(9)\begin{equation} I^2 \subset \delta \cdot J_F\quad \text{and}\quad \operatorname{N}(I)^2 = \operatorname{N}(\delta) \cdot \operatorname{N}(J_F\!). \end{equation}

We now recall an orbit parametrization for such square roots that we introduced in [Reference SwaminathanSwa23]. To do this, let $\widetilde {H}_F$ be the set of pairs $(I, \delta )$ satisfying (9). Let $H_F = \widetilde {H}_F/ \sim$, where the equivalence relation $\sim$ is defined as follows: $(I_1, \delta _1) \sim (I_2, \delta _2)$ if and only if there exists $\kappa \in K_F^\times$ such that the based fractional ideals $I_1$ and $\kappa \cdot I_2$ are equal up to an $\operatorname {SL}_{2n+1}(R)$-change-of-basis and such that $\alpha _1 = \kappa ^2 \cdot \alpha _2$. Now, take $(I, \delta ) \in H_F$, and consider the symmetric bilinear form:

(10)\begin{equation} \langle-,-\rangle \colon I \times I \to K_F, \quad (\alpha, \beta) \mapsto \langle \alpha, \beta \rangle =\delta^{-1} \cdot \alpha\beta. \end{equation}

Let $\pi _{2n-1},\pi _{2n} \in \operatorname {Hom}_{R}(J_F,R)$ be the maps defined on the $R$-basis (6) of $J_F$ by

\begin{align*} \pi_{2n-1}(\theta^{2n-1}) - 1 &= \pi_{2n-1}(\zeta_{2n}) = \pi_{2n-1}(\theta^i) = 0 \quad\text{for each } i \in \{0, \ldots, 2n-2\} ,\\ \pi_{2n}(\zeta_{2n}) + 1 &= \pi_{2n}(\theta^i) = 0 \quad\text{for each } i \in \{0, \ldots, 2n-1\}. \end{align*}

Let $A$ and $B$ respectively denote the matrices representing the symmetric bilinear forms $\pi _{2n} \circ \langle -, - \rangle \colon I \times I \to R$ and $\pi _{2n-1} \circ \langle -, - \rangle \colon I \times I \to R$ with respect to the $R$-basis of $I$, and observe that $A$ and $B$ are symmetric of dimension $(2n+1) \times (2n+1)$ and have entries in $R$. We have, thus, constructed a map of sets

\[ \mathsf{orb}_F \colon H_F \longrightarrow \{\text{$\operatorname{SL}_{2n+1}(R)$-orbits on $R^2 \otimes_R \operatorname{Sym}_2 R^{2n+1}$}\}, \]

where by $R^2 \otimes _R \operatorname {Sym}_2 R^{2n+1}$ we mean the space of pairs of $(2n+1) \times (2n+1)$ symmetric matrices with entries in $R$. The following result characterizes the map $\mathsf {orb}_F$.

For the purposes of this paper, it turns out (see Lemma 11) that we only need the restriction of the parametrization in Theorem 3 to those pairs $(A,B)$ such that $A$ defines a split quadratic form over $k$. This restricted parametrization can be expressed in terms of the action of $G(R)$ on $V(R)$. Indeed, suppose we are given a pair $(A,B) \in R^2 \otimes _{R} \operatorname {Sym}_2 R^{2n+1}$ such that the invariant form $\det (x \cdot A + z \cdot B)$ of $(A,B)$ is equal to $(-1)^n \cdot F_{\operatorname {mon}}(x,z)$ and such that $A$ defines a quadratic form that is split over $k$. Then there exists $g \in \operatorname {GL}_{2n+1}(R)$ satisfying $gAg^T = A_0$ (by the uniqueness statement in the first paragraph of § 2.1), and the matrix $T = (g^T)^{-1}\cdot -A^{-1}B \cdot g^T$ is self-adjoint with respect to the matrix $A_0$ and has characteristic polynomial $F_{\operatorname {mon}}(x,1)$. We thus obtain the following correspondence, the proof of which is immediate.

2.4 Orbits over fields

When $R = k$ is a field, the parametrization in Theorem 3 simplifies considerably. Indeed, every fractional ideal of $R_F = K_F$ is equal to $K_F$, so every element of $H_F$ is of the form $(K_F,\delta )$, where $f_0 \cdot \delta \in (K_F^\times /K_F^{\times 2})_{\operatorname {N}\equiv 1} := \{ \rho \in K_F^\times : \operatorname {N}(\rho ) \in k^{\times 2}\}/K_F^{\times 2}$. Moreover, condition $\mathrm {(b)}$ in Theorem 3 holds trivially in this case, implying that every $\operatorname {SL}_{2n+1}(k)$-orbit on $k^2 \otimes _k \operatorname {Sym}_2 k^{2n+1}$ with invariant form $F_{\operatorname {mon}}(x,1)$ arises from an element of $H_F$. We thus obtain the following result, which gives a convenient description of the orbits of $\operatorname {SL}_{2n+1}(k)$ on $k^2 \otimes _k \operatorname {Sym}_2 k^{2n+1}$ that arise via the parametrization.

In what follows, we say that an orbit of $\operatorname {SL}_{2n+1}(k)$ on $k^2 \otimes _k \operatorname {Sym}_2 k^{2n+1}$ (or of $G(k)$ on $V(k)$) is distinguished if it corresponds to $1 \in (K_F^\times /K_F^{\times 2})_{\operatorname {N}\equiv 1}$ under the bijection of Proposition 6. In geometric terms, the orbit of a pair $(A,B) \in k^2 \otimes _k \operatorname {Sym}_2 k^{2n+1}$ is distinguished if and only if $A$ and $B$ share a maximal isotropic space defined over $k$ (see [Reference Bhargava and GrossBG13, § 4]).

For ease of notation, write $f = F_{\operatorname {mon}}$. Given any $(A,B) \in k^2 \otimes _k \operatorname {Sym}_2 k^{2n+1}$ with invariant form $f$, we may regard $A$ and $B$ as quadratic forms cutting out a pair of quadric hypersurfaces $Q_A$ and $Q_B$ in $\mathbb {P}_k^{2n}$. Let $\mathscr {F}_{(A,B)}$ denote the Fano scheme parametrizing $(n-1)$-dimensional linear spaces contained in the base locus $Q_A \cap Q_B$ of the pencil of quadric hypersurfaces spanned by $Q_A$ and $Q_B$. The following proposition states that $\mathscr {F}_{(A,B)}$ is a torsor of the Jacobian of $\mathscr {S}_F$.

Proof. The Jacobian of a stacky curve $\mathscr {X}$ is defined to be the $0$th-degree component $\operatorname {Pic}^0(\mathscr {X})$ of the group $\operatorname {Pic}(\mathscr {X})$ of divisors on $\mathscr {X}$ modulo principal divisors. Let $k_{\operatorname {sep}}$ be a separable closure of $k$, and let $\theta _1, \ldots, \theta _{2n+1} \in \mathbb {P}^1_k(k_{\operatorname {sep}})$ be the roots of $F$. Applying the correspondence between isomorphism classes of line bundles and divisor classes to [Reference Candelori and CameronCC17, proof of Proposition 2.2] yields that $\operatorname {Pic}^0(\mathscr {S}_F\!)$ is the group generated by the classes of the divisors $(\tfrac {1}{2}\theta _i - \tfrac {1}{2}\theta _j)$ subject to the relations $2 \cdot (\tfrac {1}{2}\theta _i - \tfrac {1}{2}\theta _j) = 0$ for $1 \leq i < j \leq 2n+1$; i.e. we have

\[ \operatorname{Pic}^0(\mathscr{S}_F\!) = (\mathbb{Z}/2\mathbb{Z}) \langle (\tfrac{1}{2}\theta_i - \tfrac{1}{2}\theta_j) : 1 \leq i < j \leq 2n+1 \rangle. \]

There is a natural right action of the absolute Galois group $G_k := \operatorname {Gal}(k_{\operatorname {sep}}/k)$ on $\operatorname {Pic}^0(\mathscr {S}_F\!)$: for $\sigma \in G_k$, we have $(\tfrac {1}{2} \theta _i - \tfrac {1}{2} \theta _i) \cdot \sigma = (\tfrac {1}{2} (\sigma ^{-1} \cdot \theta _i) - \tfrac {1}{2} (\sigma ^{-1} \cdot \theta _j))$.

A result of Wang (see [Reference WangWan18, Corollary 2.5]) implies that, as long as $\#k$ is sufficiently large, the Fano scheme $\mathscr {F}_{(A,B)}$ is a torsor over $k$ of the $2$-torsion subgroup $J[2]$ of the Jacobian $J$ of the monic odd-degree hyperelliptic curve $y^2 = f(x,1)$. Thus, to prove that $\mathscr {F}_{(A,B)}$ is a torsor of $\operatorname {Pic}^0(\mathscr {S}_F\!)$, it suffices to show that there is a $G_k$-equivariant isomorphism $\operatorname {Pic}^0(\mathscr {S}_F\!) \simeq J[2]$. But this is clear: the divisor classes $(\theta _i - \theta _j)$ for $1 \leq i < j \leq 2n+1$ generate $J[2](k_{\operatorname {sep}})$ over $\mathbb {Z}/2\mathbb {Z}$.

Note, in particular, that the $\operatorname {SL}_{2n+1}(k)$-orbit of $(A,B)$ is distinguished if and only if $\mathscr {F}_{(A,B)}$ is the trivial torsor of the Jacobian of $\mathscr {S}_F$ (which happens if and only if $\mathscr {F}_{(A,B)}(k) \neq \varnothing$).

We conclude this section by explaining how to visualize the simply transitive action of $\operatorname {Pic}^0(\mathscr {S}_F\!)$ on $\mathscr {F}_{(A,B)}(k_{\operatorname {sep}})$ from Proposition 7. For each $i \in \{1, \ldots, 2n+1\}$, let $Q_i$ be the singular fiber lying over the point $\theta _i$ in the pencil of quadrics spanned by $Q_A$ and $Q_B$. Suppose that $T$ is generic, so that each of the quadrics $Q_i$ is a simple cone with cone point $q_i$. Let $\mathscr {F}_{(A,B)}(k_{\operatorname {sep}}) = \{p_1, \ldots, p_{2^{2n}}\}$, and for each $\ell \in \{1, \ldots, 2^{2n}\}$, let $\overline {p}_\ell$ denote the corresponding linear subspace of $\mathbb {P}_{k_{\operatorname {sep}}}^{2n}$. We illustrate this setup in the case where $n = 1$ in Figure 1.

Figure 1.

The pencil of conics spanned by $Q_A$ and $Q_B$ in $\mathbb {P}_{k_{\operatorname {sep}}}^2$, with singular fibers $Q_1$, $Q_2$, and $Q_3$ having cone points $q_1$, $q_2$, and $q_3$, respectively. In this case, $\mathscr {F}_{(A,B)}(k_{\operatorname {sep}}) = Q_A \cap Q_B = \{p_1, p_2, p_3, p_4\}$. The dashed arrows display the action of the degree-$0$ divisor $(\tfrac {1}{2}\theta _1 - \tfrac {1}{2}\theta _2)$ on $\mathscr {F}_{(A,B)}$.

We define an action of $\operatorname {Pic}^0(\mathscr {S}_F\!)$ on $\mathscr {F}_{(A,B)}(k_{\operatorname {sep}})$ by first specifying how each $(\tfrac {1}{2}\theta _i)$ acts on $\mathscr {F}_{(A,B)}(k_{\operatorname {sep}})$. Let $L_{i\ell } \subset \mathbb {P}_{k_{\operatorname {sep}}}^{2n}$ be the linear span of the $(n-1)$-plane $\overline {p}_\ell$ and the point $q_i$ for each pair $(i,\ell )$. By [Reference ReidRei72, Theorem 3.8], there exists precisely one element $p_{i(\ell )} \in \mathscr {F}_{(A,B)}(k_{\operatorname {sep}}) \smallsetminus \{p_\ell \}$ such that $\overline {p}_{i(\ell )} \subset L_{i\ell }$. For each $i$, let $(\tfrac {1}{2}\theta _i)$ act on $\mathscr {F}_{(A,B)}$ by swapping $p_\ell$ and $p_{i(\ell )}$ for each $\ell$. Then for each pair $(i,j)$, the divisor $(\tfrac {1}{2}\theta _i - \tfrac {1}{2}\theta _j)$ acts on $\mathscr {F}_{(A,B)}$ by swapping $p_\ell$ with $p_{i(j(\ell ))} = p_{j(i(\ell ))}$ for each $\ell$. When $n = 1$, for example, the divisor $(\tfrac {1}{2}\theta _1 - \tfrac {1}{2}\theta _2)$ acts on $\mathscr {F}_{(A,B)}$ by sending $(p_1, p_2, p_3, p_4)$ to $(p_3, p_4, p_1, p_2)$, as we demonstrate in Figure 1.

3.1 The construction

Let $(x_0,y_0,z_0) \in \mathscr {S}_F(R)$. For convenience, we say that the point $(x_0, y_0, z_0)$ is a Weierstrass point if $y_0 = 0$ and a non-Weierstrass point otherwise. Note that if $y_0 = 0$, then $z_0 \neq 0$ because $f_0 \neq 0$, so the binary form $F$ factors uniquely as

(11)\begin{equation} F(x,z) = \bigg(x - \frac{x_0}{z_0} \cdot z \bigg) \cdot \widetilde{F}(x,z) \end{equation}

where $\widetilde {F} \in R[x,z]$ is a binary form of degree $2n$ with leading coefficient equal to $f_0$.

We start by transforming the pair $(x_0, z_0)$ into a pair with simpler coordinates by constructing a suitable $\gamma \in \operatorname {SL}_2(R)$ such that $(x_0, z_0) \cdot \gamma = (0,1)$. Let $b_0, d_0 \in R$ be any elements such that $b_0x_0 + d_0z_0 = 1$ (such $b_0, d_0$ exist since $Rx_0 + Rz_0 = R$). We claim that there exists $t \in R$ such that $(d_0 -t x_0) + \theta (b_0 + t z_0) \in K_F^\times$. Indeed, if we were to have $(d_0 -t x_0) + \theta (b_0 + t z_0) \in K_F \smallsetminus K_F^\times$, then the pair $(x,z) = (d_0 - tx_0,-b_0 - t z_0)$ is a root of the equation $F(x,z) = 0$, which has finitely many roots up to scaling, implying that we can choose $t$ to avoid this outcome (since $k$ is of characteristic $0$). Take any such $t \in R$, let $b = b_0 + t z_0$ and $d = d_0 - t x_0$, and let $\gamma = [\begin{smallmatrix} z_0 & b\\ -x_0 & d\end{smallmatrix}]$. Now, let $F'$ be defined by

\[ F'(x,z) = F ((x,z) \cdot \gamma^{-1}) = \sum_{i = 0}^{2n+1} f_i' x^{2n+1-i}z^i \in R[x,z]. \]

Note that $f_{2n+1}' = y_0^2$ and that $f_0' \neq 0$ since $f_0' = f_0 \cdot \operatorname {N}(d + \theta b)$. If $y_0 = 0$, the binary form $F'$ factors uniquely as

\[ F'(x,z) = x \cdot \widetilde{F}'(x,z), \]

where $\widetilde {F}' \in R[x,z]$ is a binary form of degree $2n$ with leading coefficient equal to $f_0'$; further observe that $f_{2n}' = \widetilde {F}(x_0,z_0) \neq 0$ because $F$ is separable. Let $\theta ' = (-x_0 + \theta z_0)/(d+\theta b)$ be the root of $F'(x,1)$ in $K_{F'} = K_F$, and consider the free rank-$(2n+1)$ $R$-submodule $I' \subset K_{F'}$ defined by

(12)\begin{equation} I' := R\langle \xi, \theta', \ldots, {\theta'}^n, \zeta_{n+1}', \ldots, \zeta_{2n}'\rangle, \quad \text{where } \xi :=\begin{cases} y_0 & \text{if $y_0 \neq 0$,} \\ \zeta_{2n}'+f_{2n}' & \text{if $y_0 = 0$.} \end{cases} \end{equation}

Observe that the $R$-module $I'$ is a fractional ideal of $R_{F'}$ that resembles the fractional ideal $I_{F'}^n$, the only difference between them being that the basis element $1 \in I_{F'}^n$ is replaced by $\xi \in I'$.

Lemma 9 We have that

\[ {I'}^2 \subset \delta' \cdot J_{F'}, \quad \text{where } \delta' := \begin{cases} -\theta' & \text{if $y_0 \neq 0$,}\\ \widetilde{F}'(\theta',1)-\theta' & \text{if $y_0 = 0$}\end{cases} \]

and that $\operatorname {N}(I')^2 = \operatorname {N}(\delta ') \cdot \operatorname {N}(J_{F'})$.

Proof. Recall from (6) that the fractional ideal $J_{F'}$ has the $\mathbb {Z}$-basis

\[ J_{F'} = R\langle1 , \theta', \ldots, \theta'^{2n-1}, \zeta_{2n}' \rangle. \]

To prove the claimed containment, it suffices to check that when each of the pairwise products of the basis elements of $I'$ in (12) is divided by $\delta '$, what results is an element of $J_{F'}$. We have the following computations:

\begin{align*} \xi^2/\delta' & = \zeta_{2n}' + f_{2n}', & & \\ (\xi \cdot {\theta'}^i)/\delta' & = -y_0 {\theta'}^{i-1} & & \text{for $i \in \{1, \ldots, n\}$} ,\\ (\xi \cdot \zeta_j')/\delta' & = -y_0 \zeta_{j-1}' - y_0 f_{j-1}' & & \text{for }j \in \{n+1, \ldots, 2n\}, \\ ({\theta'}^i \cdot {\theta'}^j)/\delta' & = -{\theta'}^{i+j-1} & & \text{for }i,j \in \{1, \ldots, n\} ,\\ ({\theta'}^i \cdot \zeta_j')/\delta' & = -{\theta}'^{i-1} \zeta_j' & & \text{for }i \in \{1, \ldots, n\},\,j \in \{n+1, \ldots, 2n\}, \\ (\zeta_i'\cdot \zeta_j')/\delta' & = -\zeta_i'\zeta_{j-1}'-f_{j-1}'\zeta_i' & & \text{for }i,j \in \{n+1, \ldots, 2n\}. \end{align*}

Using the fact that $\zeta _i' \in R_{F'}$ for each $i \in \{0, \ldots, 2n+1\}$ together with the fact that $J_{F'}$ is closed under multiplication by elements in $R_{F'}$, one readily verifies that each of the expressions obtained on the right-hand side above is an element of $J_{F'}$. Crucially, these expressions do not depend in piecewise fashion on the value of $y_0$, even though $\xi$ and $\delta '$ do (see Remark 12).

As for the second part of the lemma, using $\operatorname {N}(I_{F'}^n) = {f_0'}^{-n}$, we find that $\operatorname {N}(I') = \widetilde {\xi } \cdot {f_0'}^{-n}$, where $\widetilde {\xi } = \xi = y_0$ if $y_0 \neq 0$ and $\widetilde {\xi } = \xi -\zeta _{2n}' = f_{2n}'$ if $y_0 = 0$. Moreover, a calculation reveals that $\operatorname {N}(\delta ') = \widetilde {\xi }^2 \cdot {f_0'}^{-1}$. Combining these norm calculations, we deduce that

\[ \operatorname{N}(I')^2 = (\widetilde{\xi} \cdot {f_0'}^{-n})^2 = (\widetilde{\xi}^2 \cdot {f_0'}^{-1}) \cdot ({f_0'}^{-(2n-1)}) = \operatorname{N}(\delta') \cdot \operatorname{N}(J_{F'}) \]

We now transform $I'$ and $\delta '$ back from $R_{F'}$ to $R_F$. Recall from (8) that

\[ I_F^{n} = \phi_{n,\gamma}(I_{F'}^n) = (-b \theta' + z_0)^{-n} \cdot I_{F'}^n, \]

where we can invert $-b\theta ' + z_0$ because its inverse is equal to $d + \theta b$. Since $I'$ resembles $I_{F'}^n$, it makes sense to consider the fractional ideal

(13)\begin{equation} I := \phi_{n,\gamma}(I') = (-b \theta' + z_0)^{-n} \cdot I' = (d + \theta b)^n \cdot I', \end{equation}

which has norm given by

(14)\begin{equation} \operatorname{N}(I) = \operatorname{N}((d + \theta b)^n \cdot I') = \operatorname{N}(d + \theta b)^n \cdot \operatorname{N}(I') = \widetilde{\xi}\cdot f_0^{-n}, \end{equation}

where we used the multiplicativity relation (7). Similarly, let $\delta \in K_F^\times$ be defined as follows:

(15)\begin{equation} \delta := \phi_{1,\gamma}(\delta') = \frac{\delta'}{-b\theta'+z_0} = \begin{cases} x_0 - \theta z_0 & \text{if $y_0 \neq 0$,} \\ \widetilde{F}(z_0\theta, z_0) + (x_0 - \theta z_0) & \text{if $y_0 = 0$.} \end{cases} \end{equation}

It then follows from Lemma 9 that

(16)\begin{equation} I^2 \subset \frac{\delta'}{-b\theta'+z_0}\cdot \frac{J_{F'}}{(-b \theta'+z_0)^{2n-1}} = \delta\cdot J_F, \end{equation}

where in the last step above, we used the fact that

\[ J_F = \phi_{2n-1,\gamma}(J_{F'}) = (-b \theta'+z_0)^{-(2n-1)} \cdot J_{F'}. \]

Note that $\operatorname {N}(I)^2 = \operatorname {N}(\delta ) \cdot \operatorname {N}(J_{F}\!)$ (this can be verified directly from (14) and (15), but it also follows from the fact that $\operatorname {N}(I')^2 = \operatorname {N}(\delta ') \cdot \operatorname {N}(J_{F'})$ by Lemma 9).

Proof. Let $\gamma = [\begin{smallmatrix} z_0 & b\\ -x_0 & d\end{smallmatrix}]\in \operatorname {SL}_2(R)$ satisfying $(x_0, z_0) \cdot \gamma = (0,1)$ and $d + \theta b \in K_F^\times$ as above. Any other $\check {\gamma } \in \operatorname {SL}_2(R)$ satisfying $(x_0,z_0) \cdot \check {\gamma }$ to $(0,1)$ is a right-translate of $\gamma$ by an element of the stabilizer of $(0,1)$, which is the group of unipotent matrices $M_t = [\begin{smallmatrix} 1 & t \\ 0 & 1 \end{smallmatrix} ]$ for $t \in R$. Thus, take $t \in R$, and let

\[ \check{\gamma} = \gamma \cdot M_t = \bigg[\begin{array}{cc} z_0 & b + t z_0 \\ -x_0 & d - t x_0 \end{array} \bigg]. \]

Let $\check {F}(x,z) = F((x,z) \cdot \check {\gamma }^{-1}) = \sum _{i = 0}^{2n+1} \check {f}_i x^{2n+1-i}z^i$, and let $\check {\theta }$ be the root of $\check {F}(x,1)$ in $K_F$. Assume that $t$ has been chosen so that $(d-t x_0) + \theta (b+tz_0) \in K_F^\times$, and note that

(17)\begin{equation} \check{\theta} = \frac{-x_0 + \theta z_0}{(d-t x_0) + \theta (b+tz_0)} = \frac{\theta'}{1 + \theta' t}. \end{equation}

Just as we associated a pair $(I', \delta ')$ to $\gamma$, we can associate a pair $(\check {I}, \check {\delta })$ to $\check {\gamma }$ satisfying the analogous properties $\check {I}^2 \subset \check {\delta } \cdot J_{\check {F}}$ and $\operatorname {N}(\check {I})^2 = \operatorname {N}(\check {\delta }) \cdot \operatorname {N}(J_{\check {F}})$. To prove the lemma, it suffices to show that

(18)\begin{equation} \phi_{n,\gamma}(I') = \phi_{n,\check{\gamma}}(\check{I}) \quad \text{and} \quad \phi_{1,\gamma}(\delta') = \phi_{1, \check{\gamma}}(\check{\delta}). \end{equation}

The second equality in (18) is readily deduced by inspecting (15); as for the first equality, note that it is equivalent to the following:

(19)\begin{equation} \bigg(\frac{-b \theta'+z_0}{-(b+tz_0) \check{\theta}+ z_0}\bigg)^n \cdot \check{I} = I'. \end{equation}

Using (17), we find that the fraction in parentheses in (19) is equal to $1+t \theta '$. Now, observe that we can express the ideals $I'$ and $\check {I}$ as

(20)\begin{equation} I' = R \langle \widetilde{\xi}, \theta' \cdot I_{F'}^{n-1} \rangle \quad \text{and} \quad \check{I} = R \langle \widetilde{\xi}, \check{\theta} \cdot I_{\check{F}}^{n-1}\rangle, \end{equation}

where $\widetilde {\xi } = y_0$ if $y_0 \neq 0$ and $\widetilde {\xi } = f_{2n}' = \check {f}_{2n}$ if $y_0 = 0$. Upon combining (19) and (20), we find that

\begin{align*} (1 + t \theta')^n \cdot \widetilde{I} & = R\langle (1+t \theta')^n \cdot \widetilde{\xi} , ((1+t\theta') \widetilde{\theta}) \cdot ((1+t\theta')I_{\widetilde{F}})^{n-1}\rangle \\ & = R\langle(1+t\theta')^n \cdot \widetilde{\xi}, \theta' \cdot I_{F'}^{n-1} \rangle = I', \end{align*}

where the last step above follows because ${\theta '}^i \in I'$ for each $i \in \{0, \ldots, n\}$.

By applying the correspondence in Theorem 3 to the pair $(I, \delta )$ constructed above, we see that the point $(x_0, y_0, z_0) \in \mathscr {S}_F(R)$ gives rise to a pair $(A,B) \in R^2 \otimes _R \operatorname {Sym}_2 R^{2n+1}$ with invariant form $(-1)^n \cdot F_{\operatorname {mon}}$. To complete the construction, it remains to verify that $A$ is split.

Proof. It suffices to exhibit an $n$-dimensional isotropic space $X$ over $k$ for the quadratic form defined by $A$. We claim that

\[ X = k\langle x_0 - \theta z_0, \theta(x_0 - \theta z_0), \ldots, \theta^{n-1}(x_0 - \theta z_0)\rangle \subset K_F \]

does the job. Indeed, since $F$ is separable, the elements $\theta ^i(x_0 - \theta z_0)$ for $i \in \{0, \ldots, n-1\}$ are linearly independent over $k$: if $y_0 \neq 0$, then this follows because the elements $\theta ^i \in K_F$ for $i \in \{0, \ldots, n-1\}$ are linearly independent and $x_0 - \theta z_0$ is an invertible element of $K_F$; if $y_0 = 0$, then this follows because the elements $\theta ^i \in K_{\widetilde {F}}$ for $i \in \{0, \ldots, n-1\}$ are linearly independent and $x_0 - \theta z_0$ is an invertible element of $K_{\widetilde {F}}$. Thus, we have that $\dim _k X = n$. Moreover, note that

\[ \pi_{2n}(\langle \theta^i(x_0 - \theta z_0) , \theta^j (x_0 - \theta z_0) \rangle) = \pi_{2n}(x_0\theta^{i+j} - z_0\theta^{i+j+1}) = 0 \]

for any $i,j \in \{0, \ldots, n-1\}$.

3.2 Distinguished points

We say that a point $(x_0, y_0, z_0) \in \mathscr {S}_F(R)$ is distinguished if its associated $G(k)$-orbit is distinguished. Note that Proposition 6 implies that $(x_0, y_0, z_0)$ is distinguished if and only if $f_0 \cdot (x_0 - \theta z_0)$ is a square in $K_F^\times$.

When $f_0$ is a multiple of a perfect square by a unit $u \in R^\times$, $\mathscr {S}_F$ has a trivial $R$-point, namely $(u,\sqrt {u^{2n+1} \cdot f_0},0)$. Since $f_0\cdot (u - \theta \cdot 0) = u \cdot f_0 \in K_F^{\times 2}$, the orbits associated to these trivial solutions are always distinguished. Whether a primitive integer solution is distinguished is not necessarily invariant under replacing $F$ with an $\operatorname {SL}_2(R)$-translate. Indeed, suppose $(x_0, y_0, z_0) \in \mathscr {S}_F(R)$, take $\gamma \in \operatorname {SL}_2(R)$ such that $(x_0, z_0) \cdot \gamma = (1,0)$, and let $F'(x,z) = F((x,z) \cdot \gamma ^{-1})$. Then the point $(1,y_0,0) \in \mathscr {S}_{F'}(R)$ is distinguished, regardless of whether the point $(x_0,y_0,z_0) \in \mathscr {S}_F(R)$ is distinguished. Nevertheless, whether a solution is distinguished is always preserved under replacing $F$ with a translate by a unipotent matrix of the form $M_t^T$ for $t \in R$.

Now take $R = \mathbb {Z}$. When $|f_0|$ is not a perfect square, we show in [Reference SwaminathanSwa23, § 3.3] that the non-square factors of $|f_0|$ tend to obstruct the distinguished $G(\mathbb {Q})$-orbit from arising via the map $\mathsf {orb}_F$ defined in § 2.3. More precisely, we have the following result, which states that there are strong limitations on the forms $F \in \mathscr {F}_{2n+1}(f_0)$ for which some point of $\mathscr {S}_F(\mathbb {Z})$ is distinguished:

Theorem 14 implies that points of $\mathscr {S}_F(\mathbb {Z})$ often fail to be distinguished, a fact that is crucial for the proofs of Theorems 1 and 2, because the geometry-of-numbers results from [Reference Bhargava and GrossBG13, § 9] that we apply in § 4 to count orbits of $G(\mathbb {Z})$ on $V(\mathbb {Z})$ only give us a count of the non-distinguished orbits.

4.1 Outline of the proof

Let $\delta$ be the upper density of forms $F \in \mathscr {F}_{2n+1}(f_0)$, enumerated by height, such that $\mathscr {S}_F(\mathbb {Z}) \neq \varnothing$. Proving Theorem 1 amounts to obtaining an upper bound on $\delta$.

To this end, let $\mathscr {F}_{2n+1}^*(f_0)$ be the subset of all $F \in \mathscr {F}_{2n+1}(f_0)$ satisfying both conditions in Theorem 14 for every prime $p \mid \kappa$; note that the complement of $\mathscr {F}_{2n+1}^*(f_0)$ consists entirely of binary forms for which the distinguished orbit does not arise from a primitive integer solution. Moreover, let $\mathscr {P}$ denote the ($G(\mathbb {Z})$-invariant) subset of integral elements of $V(\mathbb {Z})$ that arise via the construction in § 3.1 from separable forms $F \in \mathscr {F}_{2n+1}(f_0)$. By partitioning $\mathscr {F}_{2n+1}(f_0)$ into $\mathscr {F}_{2n+1}^*(f_0)$ and its complement, we see that $\delta$ can be bounded as follows:

(21)\begin{align} \delta \leq \lim_{X \to \infty} \frac{\#\{F \in \mathscr{F}_{2n+1}^*(f_0) : H(F) < X\} + \#\big(G(\mathbb{Q})\backslash \{\text{$T \in \mathscr{P}$ : $T$ irreducible, $H(T) < X$}\}\big)}{\#\{F \in \mathscr{F}_{2n+1}(f_0) : H(F) < X\}}, \end{align}

where we say that $T$ is irreducible if $\operatorname {ch}(T)$ is separable and the $G(\mathbb {Q})$-orbit of $T$ is non-distinguished.

The first term on the right-hand side of (21) constitutes the trivial bound for the density of forms $F \in \mathscr {F}_{2n+1}^*(f_0)$ such that $\mathscr {S}_F(\mathbb {Z}) \neq \varnothing$. Computing this first term is straightforward. Indeed, the first term in the numerator and the denominator are respectively given by

(22)\begin{align} \#\{F \in \mathscr{F}_{2n+1}^*(f_0) : H(F) < X\} & = \mu_{f_0} \cdot f_0^{-2n^2 - n} \cdot 2^{2n+1} \cdot X^{({n+1})/{2n}} + o(X^{({n+1})/{2n}}), \end{align}

(23)\begin{align} \#\{F \in \mathscr{F}_{2n+1}(f_0) : H(F) < X\} & = f_0^{-2n^2 - n} \cdot 2^{2n+1} \cdot X^{({n+1})/{2n}} + o(X^{({n+1})/{2n}}). \end{align}

The second term in the numerator on the right-hand side of (21) means ‘the number of $G(\mathbb {Q})$-equivalence classes of irreducible elements $T \in \mathscr {P}$ of height less than $X$’, and this constitutes a bound on the number forms $F \in \mathscr {F}_{2n+1}(f_0) \smallsetminus \mathscr {F}_{2n+1}^*(f_0)$ such that $\mathscr {S}_F(\mathbb {Z}) \neq \varnothing$. It is hard to describe $\mathscr {P}$ explicitly and, hence, to obtain a precise asymptotic for this second term. But $\mathscr {P}$ is contained within a subset of $V(\mathbb {Z})$ defined by local conditions: indeed, we have

\[ \mathscr{P} \subset V(\mathbb{Z}) \cap \bigcap_v \mathscr{P}_v, \]

where for each place $v$ of $\mathbb {Q}$, we write $\mathscr {P}_v$ for the subset of $V(\mathbb {Z}_v)$ that arises via the construction in § 3.1 from separable forms $F \in \mathscr {F}_{2n+1}(f_0, \mathbb {Z}_v)$. Thus, the second term in the numerator on the right-hand side of (21) is bounded by the following quantity:

(24)\begin{equation} \#\bigg(G(\mathbb{Q})\bigg\backslash \bigg\{\text{$T \in V(\mathbb{Z}) \cap \bigcap_v \mathscr{P}_v$ : $T$ is irreducible, $H(T) < X$}\bigg\}\bigg) \end{equation}

The rest of this section is devoted to bounding (24). We desire a bound of order $o(f_0^{-2n^2-n} \cdot 2^{-n})$ when $2 \nmid f_0$ and a bound of order $O(f_0^{-2n^2-n} \cdot 2^{-\varepsilon _1 n^{\varepsilon _2}})$ when $2 \mid f_0$. We obtain bounds of these magnitudes by means of the following steps.

(A) In § 4.2, we adapt results of Bhargava–Gross [Reference Bhargava and GrossBG13] to obtain an asymptotic upper bound for the number of irreducible orbits of $G(\mathbb {Z})$ on $V(\mathbb {Z})$ of bounded height, where the orbits are counted with respect to a weight function $\phi$ defined by infinitely many local conditions (see Theorem 16). By choosing the weight function $\phi$ appropriately, so that it picks out orbits belonging to $\mathscr {P}_v$ for every place $v$ and counts each $G(\mathbb {Q})$-equivalence class of integral orbits just once, we obtain an upper bound for the quantity (24). This bound is expressed a product of local mass formulas, one for each place of $\mathbb {Q}$.

(B) Proving Theorem 1 then amounts to bounding each of the local mass formulas obtained in step (A) (in other words, we must control the ‘size’ of $\mathscr {P}_v$ for each $v$).

• – In § 4.3, we bound the mass at $\infty$ by working over $\mathbb {R}$ (see Proposition 18). The mass at $\infty$ contributes a factor of $O(2^n \cdot X^{({n+1})/{2n}})$ to the bound.

• – In § 4.4, we bound the mass at $p$ when $2 \neq p \nmid f_0$ by working over $\mathbb {Z}/p\mathbb {Z}$ (see Proposition 22). As we later show in § 4.7, the masses at these primes together contribute a factor of $O(2^{-\varepsilon _1 n^{\varepsilon _2}})$ to the bound. This constitutes a small part of the saving when $2 \nmid f_0$ and the entire saving when $2 \mid f_0$.

• – In § 4.5, we bound the mass at $p$ when $2 = p \nmid f_0$ by working over $\mathbb {Z}/8\mathbb {Z}$ (see Proposition 26). In this case, the mass at $2$ contributes a factor of $O(2^{-2n})$ to the bound, which constitutes most of the saving.

• – In § 4.6, we bound the mass at $p$ when $p \mid f_0$ by working over $\mathbb {Q}_p$ and $\mathbb {Z}_p$ (see Propositions 30 and 33). In this case, the mass at $p$ contributes to the bound a factor of $O(|f_0|_p^{2n^2+n})$ when $p \neq 2$ and $O(2^{-n} \cdot |f_0|_2^{2n^2+n})$ when $p = 2$.

(C) The final step is to combine the bounds obtained in steps (A) and (B) and to simplify the result. We do this in § 4.7.

4.2 Step (A): counting irreducible integral orbits of bounded height

For a monic polynomial $f = x^{2n+1} + \sum _{i = 1}^{2n+1} c_i \cdot x^{2n+1-i} \in \mathbb {Z}[x]$, let the height of $f$ be defined by $H(f) := H(c_1, \ldots, c_{2n+1}) := \max \{|c_i|^{2n(2n+1)/i} : i = 1, \ldots, 2n+1\}$. For an element $T \in V(\mathbb {Z})$, let the height of (the $G(\mathbb {Z})$-orbit of) $T$ be defined by $H(T) = H(\operatorname {ch}(T))$.

Fix $m \in \{0, \ldots, n\}$, and define $\mathscr {I}(m) \subset \mathscr {F}_{2n+1}(1,\mathbb {R})$ to be the set of separable monic degree-$(2n+1)$ polynomials over $\mathbb {R}$ having exactly $2m+1$ real roots. Let $\mathscr {B} \subset \operatorname {ch}^{-1}(\mathscr {I}(m)) \subset V(\mathbb {R})$ be a measurable subset closed under the action of $G(\mathbb {R})$. Let $N(V(\mathbb {Z}) \cap \mathscr {B}; X)$ denote the number of $G(\mathbb {Z})$-orbits of irreducible elements $T \in V(\mathbb {Z}) \cap \mathscr {B}$ satisfying $H(T) < X$. Then we have the following adaptation of [Reference Bhargava and GrossBG13, Theorem 10.1 and (10.27)], giving an asymptotic for $N(V(\mathbb {Z}) \cap \mathscr {B}; X)$.

Theorem 15 There exists an absolute constant $\mathscr {J} \in \mathbb {Q}^\times$ such that

\begin{align*} N(V(\mathbb{Z}) \cap \mathscr{B}; X) &= \frac{1}{2^{m+n}} \cdot |\mathscr{J}| \cdot \operatorname{Vol}(G(\mathbb{Z}) \backslash G(\mathbb{R})) \\ &\quad \cdot \int_{\substack{f \in \mathscr{I}(m) \\ H(f) < X}} \#\bigg(\frac{\mathscr{B} \cap \operatorname{ch}^{-1}(f)}{G(\mathbb{R})}\bigg)\, df + o(X^{({n+1})/{2n}}), \end{align*}

where the fundamental volume $\operatorname {Vol}(G(\mathbb {Z}) \backslash G(\mathbb {R}))$ is computed with respect to a differential $\omega$ that generates the rank-$1$ module of top-degree differentials of $G$ over $\mathbb {Z}$.

The following result generalizes Theorem 15 by giving an upper bound on the number of integral orbits of height up to $X$, where the orbits are weighted by a function defined by local conditions.

Theorem 16 (Cf. [Reference Bhargava and GrossBG13, Theorem 10.12])

Let $\phi \colon V(\mathbb {Z}) \to [0,1] \subset \mathbb {R}$ be a function such that there exists a function $\phi _p \colon V(\mathbb {Z}_p) \to [0,1]$ for each prime $p$ satisfying the following conditions:

• – for all $T \in V(\mathbb {Z})$, the product $\prod _{p} \phi _p(T)$ converges to $\phi (T)$; and

• – for each $p$, the function $\phi _p$ is locally constant outside a closed set $S_p \subset V(\mathbb {Z}_p)$ of measure $0$.

Let $N_\phi (V(\mathbb {Z}) \cap \mathscr {B}; X)$ denote the number of $G(\mathbb {Z})$-orbits of irreducible elements $T \in V(\mathbb {Z}) \cap \mathscr {B}$ satisfying $H(T) < X$, where each orbit $G(\mathbb {Z}) \cdot T$ is counted with weight $\phi (T) := \prod _p \phi _p(T)$. Then

\[ N_\phi(V(\mathbb{Z}) \cap \mathscr{B}; X) \leq N(V(\mathbb{Z}) \cap \mathscr{B}; X) \cdot \prod_{p} \int_{T \in V(\mathbb{Z}_p)} \phi_p(T)\,dT + o(X^{({n+1})/{2n}}). \]

To apply Theorem 16 to bound (24), we need to choose the local weight functions $\phi _p$ appropriately. Given $T \in V(\mathbb {Z})$, denote by $m(T)$ the quantity

\[ m(T) := \sum_{T' \in \mathscr{O}(T)} \frac{\#\operatorname{Stab}_\mathbb{Q}(T')}{\#\operatorname{Stab}_\mathbb{Z}(T')} = \sum_{T' \in \mathscr{O}(T)} \frac{\#\operatorname{Stab}_\mathbb{Q}(T)}{\#\operatorname{Stab}_\mathbb{Z}(T')}, \]

where $\mathscr {O}(T)$ denotes a set of representatives for the action of $G(\mathbb {Z})$ on the $G(\mathbb {Q})$-orbit of $T$, and where $\operatorname {Stab}_\mathbb {Q}(T')$ and $\operatorname {Stab}_\mathbb {Z}(T')$ respectively denote the stabilizers of $T'$ in $G(\mathbb {Q})$ and $G(\mathbb {Z})$. By [Reference Bhargava and GrossBG13, Proposition 10.8], counting $G(\mathbb {Q})$-orbits on $V(\mathbb {Z})$ is, up to a negligible error, no different from counting $G(\mathbb {Z})$-orbits on $V(\mathbb {Z})$, where the $G(\mathbb {Z})$-orbit of $T \in V(\mathbb {Z})$ is weighted by $1/m(T)$.

The weight $m(T)$ can be expressed as a product over primes $p$ of local weights $m_p(T)$. Indeed, as stated in [Reference Bhargava and GrossBG13, (11.3)], we have that

(25)\begin{equation} m(T) = \prod_{p} m_p(T), \quad \text{where } m_p(T) := \sum_{T' \in \mathscr{O}_p(T)} \frac{\#\operatorname{Stab}_{\mathbb{Q}_p}(T)}{\#\operatorname{Stab}_{\mathbb{Z}_p}(T')}, \end{equation}

where $\mathscr {O}_p(T)$ denotes a set of representatives for the action of $G(\mathbb {Z}_p)$ on the $G(\mathbb {Q}_p)$-orbit of $T$, and where $\operatorname {Stab}_{\mathbb {Q}_p}(T')$ and $\operatorname {Stab}_{\mathbb {Z}_p}(T')$ denote the stabilizers of $T'$ in $G(\mathbb {Q}_p)$ and $G(\mathbb {Z}_p)$, respectively.

For each prime $p$, let $\psi _p \colon V(\mathbb {Z}_p) \to \{0,1\} \subset \mathbb {R}$ be the indicator function of elements $T \in \mathscr {P}_p$ such that $\operatorname {ch}(T)$ is separable. Then upon taking $\phi _p = \psi _p/m_p$ for each $p$, Theorems 15 and 16 together imply the following.

Proposition 17 We have that (24) is bounded above by

(26)\begin{align} & \sum_{m = 0}^n \frac{1}{2^{m+n}} \cdot |\mathscr{J}| \cdot \operatorname{Vol}(G(\mathbb{Z}) \backslash G(\mathbb{R})) \cdot\int_{\substack{f \in \mathscr{I}(m) \nonumber\\ H(f) < X}} \#\bigg(\frac{\mathscr{P}_\infty \cap \operatorname{ch}^{-1}(f)}{G(\mathbb{R})}\bigg) df \nonumber\\ & \quad \cdot \prod_{p} \int_{T \in V(\mathbb{Z}_p)} \frac{\psi_p(T)}{m_p(T)}dT + o(X^{({n+1})/{2n}}). \end{align}

In what follows, we call the sum over $m$ on the first line of (26) the mass at $\infty$, and we call the integral over $V(\mathbb {Z}_p)$ on the second line of (26) the mass at $p$.

4.3 Step (B): bounding the mass at $\infty$

In this section, we prove the following result, which shows that the mass at $\infty$ in (26) is bounded by $O(2^n \cdot X^{({n+1})/{2n}})$.

Proposition 18 The mass at $\infty$ in (26) is

\[ \leq 3 \cdot 2^{n} \cdot \prod_{p > 2} \frac{p^{2n^2+n}}{\#G(\mathbb{Z}/p\mathbb{Z})} \cdot \int_{H(f) < X} df. \]

Lemma 19 We have that

\[ |\mathscr{J}|\cdot \operatorname{Vol}(G(\mathbb{Z}) \backslash G(\mathbb{R})) \leq 2^{2n+1} \cdot \prod_{p > 2} \frac{p^{2n^2+n}}{\#G(\mathbb{Z}/p\mathbb{Z})}. \]

Proof. Let $R = \mathbb {C}$, $\mathbb {R}$, or $\mathbb {Z}_p$ where $p$ is a prime. Let $\mathscr {R} \subset R^{2n+1}$ be any open subset, and let $s \colon \mathscr {R} \to V(R)$ be a continuous function such that the characteristic polynomial of $s(c_1, \ldots, c_{2n+1})$ is given by $x^{2n+1} + \sum _{i = 1}^{2n+1} c_ix^{2n+1-i}$ for every $(c_1, \ldots, c_{2n+1}) \in \mathscr {R}$. The constant $\mathscr {J}$ arises as a multiplicative factor in the following change-of-measure formula.

Lemma 20 For any measurable function $\phi$ on $V(R)$, we have

\[ \int_{T \in G(R) \cdot s(\mathscr{R})} \phi(T)\, dT = |\mathscr{J}| \cdot \int_\mathscr{R} \int_{G(R)} \phi(g \cdot s(c_1, \ldots, c_{2n+1})) \omega(g)\, dr, \]

where we regard $G(R) \cdot s(\mathscr {R})$ as a multiset, $|-|$ denotes the standard absolute value on $R$, $dT$ is the Euclidean measure on $V(R)$, and $dr$ is the restriction to $\mathscr {R}$ of the Euclidean measure on $R^{2n+1}$.

Proof of Lemma 20 The proof is identical to that of [Reference Bhargava and ShankarBS15, Proposition 3.11].

In particular, the constant $\mathscr {J}$ is independent of the choices of $R = \mathbb {C}$, $\mathbb {R}$, or $\mathbb {Z}_p$ and of the region $\mathscr {R}$ and the functions $s$ and $\phi$. Thus, to compute $|\mathscr {J}|$, we can make the following convenient choices: take $R = \mathbb {Z}_p$, so that $|-| = |-|_p$; take

\[ \mathscr{R} = \bigg\{(c_1, \ldots, c_{2n+1}) \in \mathbb{Z}_p^{2n+1} : x^{2n+1} + \sum_{i = 1}^{2n+1} c_i x^{2n+1-i} \equiv f\ ({\rm mod}\ p)\bigg\} \]

for a fixed monic irreducible degree-$(2n+1)$ polynomial $f \in (\mathbb {Z}/p\mathbb {Z})[x]$; letting $\Sigma _f := \{T \in V(\mathbb {Z}_p) : \operatorname {ch}(T) \equiv f \pmod p\}$, take $s$ to be any continuous right-inverse to the function that sends $T \in \Sigma _f$ to the list of coefficients of $\operatorname {ch}(T)$; and take $\phi$ to be the function that sends $T \in \Sigma _f$ to ${1}/{\#\operatorname {Stab}(T)}$, where $\operatorname {Stab}(T) \subset G(\mathbb {Z}_p)$ is the stabilizer of $T$, and sends $T \in V(\mathbb {Z}_p) \smallsetminus \Sigma _f$ to $0$. The existence of the right-inverse $s$ is well-known; see, e.g., [Reference Shankar, Siad, Swaminathan and VarmaSSSV21, § 3.3].

Because $\phi$ is $G(\mathbb {Z}_p)$-invariant, Lemma 20 yields that for the above convenient choices, we have, on the one hand, that the $p$-adic density of $\Sigma _f$ is

(27) \begin{align} \operatorname{Vol}(\Sigma_f) := \int_{T \in \Sigma_f} dT & = \int_{T \in G(\mathbb{Z}_p) \cdot s(\mathscr{R})} \frac{1}{\#\operatorname{Stab}(T)}\, dT \nonumber\\ & = |\mathscr{J}|_p \cdot \operatorname{Vol}(G(\mathbb{Z}_p)) \cdot \int_{f'\equiv f\,(\operatorname{mod} p)}\sum_{\substack{T \in G(\mathbb{Z}_p)\backslash\Sigma_f \nonumber\\ \operatorname{ch}(T) = f'}} \frac{1}{\#\operatorname{Stab}(T)}\, dr. \end{align}

First suppose $p \neq 2$. We now compute the sum in the integrand in (27). Because we chose $f$ to be irreducible over $\mathbb {Z}/p\mathbb {Z}$, the ring $R_{f'} := \mathbb {Z}_p[x]/(f')$ is the maximal order in its field of fractions $K_{f'}$, and it is, in fact, the unique local ring of rank $2n+1$ over $\mathbb {Z}_p$ having residue field $\mathbb {F}_{p^{2n+1}}$ (hence, $R_{f'}$ does not depend on the choice of $f'$). Then by [Reference SwaminathanSwa23, (63)], there is precisely one $G(\mathbb {Z}_p)$-orbit with characteristic polynomial $f'$, and the size of the stabilizer of this orbit is equal to $\#R_{f'}^\times [2]_{\operatorname {N}\equiv 1} = 1$. Thus, we find that

(28)\begin{equation} \sum_{\substack{T \in G(\mathbb{Z}_p)\backslash\Sigma_f \\ \operatorname{ch}(T) = f'}} \frac{1}{\#\operatorname{Stab}(T)} = 1. \end{equation}

Combining (27) and (28) yields that

(29)\begin{equation} \operatorname{Vol}(\Sigma_f) = |\mathscr{J}|_p \cdot \operatorname{Vol}(G(\mathbb{Z}_p)) \cdot \int_{f'\equiv f\,(\operatorname{mod} p)} dr = |\mathscr{J}|_p \cdot \frac{\operatorname{Vol}(G(\mathbb{Z}_p))}{p^{2n+1}}. \end{equation}

Let $\overline {\Sigma }_f := \{T \in V(\mathbb {Z}/p\mathbb {Z}) : \operatorname {ch}(T) = f\}$. Then the mod-$p$ reduction map $\Sigma _f \to \overline {\Sigma }_f$ is surjective, and we have by [Reference Bhargava and GrossBG13, § 6.1] that $\#\overline {\Sigma }_f = \#G(\mathbb {Z}/p\mathbb {Z})$. Thus, we have, on the other hand, that

(30)\begin{equation} \operatorname{Vol}(\Sigma_f) = \frac{\#\overline{\Sigma}_f}{p^{\dim V}} = \frac{\#G(\mathbb{Z}/p\mathbb{Z})}{p^{2n^2+3n+1}}.\end{equation}

Equating (29) and (30) yields that

(31)\begin{equation} |\mathscr{J}|_p = \frac{\#G(\mathbb{Z}/p\mathbb{Z})}{p^{2n^2+n} \cdot \operatorname{Vol}(G(\mathbb{Z}_p))} = 1 \end{equation}

when $p \neq 2$, where the last equality follows from the fact that the group $G$ is smooth over $\mathbb {Z}_p$ when $p \neq 2$.

Next, suppose $p = 2$. We now derive an upper bound on the sum in the integrand in (27). By [Reference SwaminathanSwa22, Proposition 59], the number of $G(\mathbb {Z}_2)$-orbits with characteristic polynomial $f'$ is equal to $2^{n-1}(2^n+1)$. The size of the stabilizer of any such orbit is equal to $\#R_{f'}^\times [2]_{\operatorname {N}\equiv 1} = 1$, so we find that

(32)\begin{equation} \sum_{\substack{T \in G(\mathbb{Z}_p)\backslash\Sigma_f \\ \operatorname{ch}(T) = f'}} \frac{1}{\#\operatorname{Stab}(T)} = 2^{n-1}(2^n+1). \end{equation}

Combining (27) and (32) yields that

(33)\begin{equation} \operatorname{Vol}(\Sigma_f) = |\mathscr{J}|_2 \cdot \operatorname{Vol}(G(\mathbb{Z}_2)) \cdot\frac{2^{n-1}(2^n+1)}{2^{2n+1}}. \end{equation}

Let $\overline {\Sigma }_f = \{T \in V(\mathbb {Z}/2\mathbb {Z}) : \operatorname {ch}(T) = f\}$. Then the mod-$2$ reduction map $\Sigma _f \to \overline {\Sigma }_f$ is surjective, and by Propositions 4 and 6, the action of $G(\mathbb {Z}/2\mathbb {Z})$ on $\overline {\Sigma }_f$ is simply transitive, so $\#\overline {\Sigma }_f = \#G(\mathbb {Z}/2\mathbb {Z})$. Thus, we have, on the other hand, that

(34)\begin{equation} \operatorname{Vol}(\Sigma_f) = \frac{\#\overline{\Sigma}_f}{2^{\dim V}} = \frac{\#G(\mathbb{Z}/2\mathbb{Z})}{2^{2n^2+3n+1}}. \end{equation}

Combining (33) and (34) yields that

(35)\begin{equation} |\mathscr{J}|_2^{-1} = 2^{2n} \cdot \operatorname{Vol}(G(\mathbb{Z}_2)) \cdot \frac{2^{2n^2-1}(2^n+1)}{\#G(\mathbb{Z}/2\mathbb{Z})}. \end{equation}

Since $G$ is not smooth over $\mathbb {Z}_2$, computing $\operatorname {Vol}(G(\mathbb {Z}_2))$ is far more complicated, but we do not need to know the value of $\operatorname {Vol}(G(\mathbb {Z}_2))$ for our purpose. The value of $\#G(\mathbb {Z}/2\mathbb {Z})$ is given in [Reference ChoCho15, § 6] to be $\#G(\mathbb {Z}/2\mathbb {Z}) = 2^{n^2} \cdot \prod _{i = 1}^n (2^{2i}-1)$, so it follows that

(36)\begin{equation} \frac{2^{2n^2-1}(2^n+1)}{\#G(\mathbb{Z}/2\mathbb{Z})} = (2^{-1} + 2^{-n-1}) \cdot \prod_{i = 1}^n (1 - 2^{-2i})^{-1} \leq 1. \end{equation}

Now, applying the identity

(37)\begin{equation} |\mathscr{J}| \cdot \prod_{p} |\mathscr{J}|_p = 1 \end{equation}

along with the Tamagawa number identity (see [Reference LanglandsLan66])

(38)\begin{equation} \operatorname{Vol}(G(\mathbb{Z}) \backslash G(\mathbb{R})) \cdot \prod_{p} \operatorname{Vol}(G(\mathbb{Z}_p)) = 2 \end{equation}

and combining them with (31), (35), and (36) completes the proof of Lemma 19.

Next, in the following lemma, we compute the integrand in the integral over $f$.

Lemma 21 Let $F \in \mathscr {F}_{2n+1}(f_0,\mathbb {R})$, and suppose that $F$ has exactly $2m+1$ real roots. Then there are exactly $2m+1$ orbits of $G(\mathbb {R})$ on $V(\mathbb {R})$ that arise from points in $\mathscr {S}_F(\mathbb {R})$. In particular, the integrand in the integral in the mass at $\infty$ in (26) is a constant with value $2m+1$.

Proof. Let $f(x) = F_{\operatorname {mon}}(x,1)$, and let the real roots of $f$ be given in increasing order by $\lambda _1 < \cdots < \lambda _{2m+1}$. We now compute the number of orbits of $G(\mathbb {R})$ on $V(\mathbb {R})$ having characteristic polynomial $f$ that arise from points in $\mathscr {S}_F(\mathbb {R})$ via the construction in § 3. The $G(\mathbb {R})$-orbit associated to a point $(x_0, y_0, z_0) \in \mathscr {S}_F(\mathbb {R})$ is identified via the correspondence in Proposition 6 with some $f_0 \cdot \delta \in (K_F^\times /K_F^{\times 2})_{\operatorname {N} \equiv 1}$. If $y_0 \neq 0$, this class is represented by a sequence of the form

(39)\begin{equation} (f_0(x_0 - \lambda_1 z_0), \ldots, f_0(x_0 - \lambda_{2m+1} z_0), b_1, \ldots, b_{n-m}) \in \mathbb{R}^{2m+1} \times \mathbb{C}^{n-m} \simeq K_F. \end{equation}

Otherwise, if $y_0 = 0$, then ${x_0}/{z_0} = \lambda _j$ for some $j$; letting $\widetilde {F}$ be as in (11), we see that the class $f_0 \cdot \delta \in (K_F^\times /K_F^{\times 2})_{\operatorname {N} \equiv 1}$ is represented by a sequence of the form

(40)\begin{equation} (f_0(x_0 - \lambda_1 z_0), \ldots,f_0 \cdot \widetilde{F}(x_0,z_0) ,\ldots, f_0(x_0 - \lambda_{2m+1} z_0), b_1, \ldots, b_{n-m}) \in \mathbb{R}^{2m+1} \times \mathbb{C}^{n-m}, \end{equation}

where the term $f_0 \cdot \widetilde {F}(x_0, z_0)$ in (40) replaces the term $f_0(x_0 - \lambda _j z_0)$ in (39). The class in $(K_F^{\times }/K_F^{\times 2})_{\operatorname {N} \equiv 1}$ of the sequence in (39) or in (40) is given by the sequence of signs of its first $2m+1$ terms. Because $\operatorname {N}(b_j) > 0$ for every $j$, the condition that $\operatorname {N}(f_0 \cdot \delta )$ is a square in $\mathbb {R}^\times$ is equivalent to the condition that $\prod _{i = 1}^{2m+1} f_0(x_0 - \lambda _i z_0) > 0$ if $y_0 \neq 0$ and $(f_0 \cdot \widetilde {F}(x_0,z_0)) \cdot \prod _{\substack {i = 1 \\ i \neq j}}^{2m+1} f_0(x_0 - \lambda _i z_0) > 0$ if $y_0 = 0$. We now split into cases based on the signs of $f_0$ and $z_0$.

Case 1a: $f_0 > 0$, $z_0 > 0$. First suppose $y_0 \neq 0$. In this case, the sign of $f_0(x_0 - \lambda _i z_0)$ is equal to the sign of $(x_0/z_0) - \lambda _i$. The possible sequences of signs of the $f_0(x_0 - \lambda _i z_0)$ are therefore

(41)\begin{equation} (+,-,-,-,\ldots,-)\text{ or }(+,+,+,-,\ldots,-)\text{ or } \cdots\text{ or } (+,+,+,+,\ldots,+), \end{equation}

giving a total of $m+1$ orbits. We label the sign sequences in (41) from left to right by an index $\tau$ that runs from $1$ up to $m+1$. If $y_0 = 0$ and ${x_0}/{z_0} = \lambda _j$, then because $\lambda _i < \lambda _j$ when $i < j$ and $\lambda _i > \lambda _j$ when $i > j$, we obtain the same sign sequences as those listed in (41).

Case 1b: $f_0 > 0$, $z_0 < 0$. First suppose $y_0 \neq 0$. In this case, the sign of $f_0(x_0 - \lambda _i z_0)$ is equal to the opposite of the sign of $(x_0/z_0) - \lambda _i$. The possible sequences of signs of the $f_0(x_0 - \lambda _i z_0)$ are, therefore,

(42)\begin{equation} (+,+,+,\ldots,+,+)\text{ or }(-,-,+,\ldots,+,+)\text{ or } \cdots\text{ or } (-,-,-,\ldots,-,+), \end{equation}

giving a total of $m+1$ orbits. We label the sign sequences in (42) from left to right by an index $\tau$ that runs from $m+1$ up to $2m+1$. (Note that the first sign sequence in (42) is the same as the last sign sequence in (41), which is why they share the same value of $\tau$.) If $y_0 = 0$, we again obtain the same sign sequences as those listed in (42).

Case 2a: $f_0 < 0$, $z_0 > 0$. The possible sign sequences are the same as those listed in (42), and we likewise label them from left to right by an index $\tau$ that runs from $m+1$ up to $2m+1$.

Case 2b: $f_0 < 0$, $z_0 < 0$. The possible sign sequences are the same as those listed in (41), and we likewise label them from left to right by an index $\tau$ that runs from $1$ up to $m+1$.

Substituting the results of Lemmas 19 and 21 into the mass at $\infty$ in (26), we deduce that this mass is bounded by

(43)\begin{align} \leq \sum_{m = 0}^n \frac{1}{2^{m+n}} \cdot 2^{2n+1} \cdot \prod_{p > 2} \frac{p^{2n^2+n}}{\#G(\mathbb{Z}/p\mathbb{Z})} \cdot \int_{\substack{f \in \mathscr{I}(m) \\ H(f) < X}} (2m+1)\, df \leq 3 \cdot 2^{n} \cdot \prod_{p > 2} \frac{p^{2n^2+n}}{\#G(\mathbb{Z}/p\mathbb{Z})} \cdot \int_{H(f) < X} df, \end{align}

as necessary. This completes the proof of Proposition 18.

4.4 Step (B): bounding the mass at $p$ when $2 \neq p \nmid f_0$

Let $\mathscr {I}_p(m)$ denote the set of monic degree-$(2n+1)$ polynomials over $\mathbb {Z}/p\mathbb {Z}$ (not necessarily separable) having $m$ distinct irreducible factors. When $2 \neq p \nmid f_0$, we obtain the following result, which bounds the mass at $p$ in (26).

Proposition 22 For each prime $p \nmid 2f_0$, we have that

\[ \int_{T \in V(\mathbb{Z}_p)} \frac{\psi_p(T)}{m_p(T)}\, dT \leq \min\bigg\{1,\sum_{m = 1}^{2n+1} \frac{p+1}{2^{m-1}} \cdot \frac{\#G(\mathbb{Z}/p\mathbb{Z})}{p^{2n^2+n}} \cdot \frac{\#\mathscr{I}_p(m)}{p^{2n+1}}\bigg\}. \]

Proof of Proposition 22 Given a monic degree-$(2n+1)$ polynomial $f$ over $\mathbb {Z}/p\mathbb {Z}$, let $\overline {\Sigma }_f^{\operatorname {sol}} \subset V(\mathbb {Z}/p\mathbb {Z})$ denote the mod-$p$ reduction of the subset $\{T \in \mathscr {P}_p : \operatorname {ch}(T) \equiv f \pmod p\}$. Because $G$ is smooth over $\mathbb {Z}_p$, the map $G(\mathbb {Z}_p) \to G(\mathbb {Z}/p\mathbb {Z})$ is surjective, implying that $\overline {\Sigma }_f^{\operatorname {sol}}$ is $G(\mathbb {Z}/p\mathbb {Z})$-invariant. Then, since $m_p(T) \geq 1$, we have

(44)\begin{align} \int_{T \in V(\mathbb{Z}_p)} \frac{\psi_p(T)}{m_p(T)}\, dT &\leq \int_{T \in \mathscr{P}_p} dT\nonumber\\ & \leq \sum_{m = 1}^{2n+1} \sum_{f \in \mathscr{I}_p(m)} \frac{\#\overline{\Sigma}_f^{\operatorname{sol}}}{p^{\dim V}} = \sum_{m = 1}^{2n+1} \sum_{f \in \mathscr{I}_p(m)} \sum_{T \in G(\mathbb{Z}/p\mathbb{Z})\backslash \overline{\Sigma}_f^{\operatorname{sol}}} \frac{\#\mathscr{O}(T)}{p^{2n^2+3n+1}}, \end{align}

where for each $T \in V(\mathbb {Z}/p\mathbb {Z})$, we denote by $\mathscr {O}(T)$ the $G(\mathbb {Z}/p\mathbb {Z})$-orbit of $T$. Let $\operatorname {Stab}(T) \subset G(\mathbb {Z}/p\mathbb {Z})$ denote the stabilizer of $T$. Substituting $\#\mathscr {O}(T) = \#G(\mathbb {Z}/p\mathbb {Z})/\#\operatorname {Stab}(T)$ into (44) yields that

(45)\begin{equation} \int_{T \in V(\mathbb{Z}_p)} \frac{\psi_p(T)}{m_p(T)}\, dT \leq \sum_{m = 1}^{2n+1} \sum_{f \in \mathscr{I}_p(m)} \sum_{T \in G(\mathbb{Z}/p\mathbb{Z})\backslash \overline{\Sigma}_f^{\operatorname{sol}}} \frac{1}{\#\operatorname{Stab}(T)}\cdot \frac{\#G(\mathbb{Z}/p\mathbb{Z})}{p^{2n^2+3n+1}}. \end{equation}

The following lemma gives a formula for $\#\operatorname {Stab}(T)$ that is independent of $f \in \mathscr {I}_p(m)$.

Lemma 24 For $T \in \overline {\Sigma }_f^{\operatorname {sol}}$, we have that $\#\operatorname {Stab}(T) = 2^{m-1}$.

Proof of Lemma 24 Let $K_f = (\mathbb {Z}/p\mathbb {Z})[x]/(f)$. By Proposition 4, $\#\operatorname {Stab}(T) = \#K_f^\times [2]_{\operatorname {N}\equiv 1}$. Letting $f$ split over $\mathbb {Z}/p\mathbb {Z}$ as $f(x) = \prod _{i = 1}^m f_i(x)^{n_i}$, where the $f_i$ are distinct and irreducible, we have that $K_f \simeq \prod _{i = 1}^m (\mathbb {Z}/p\mathbb {Z})[x]/(f_i^{n_i})$. Consequently, $K_f^\times [2]$ contains the $2^m$ elements of the form $(\pm 1, \ldots, \pm 1)$. Exactly half of these elements have norm $1$, so $\#K_f^\times [2]_{\operatorname {N}\equiv 1} = 2^{m-1}$.

To bound the right-hand side of (45), it remains to control the number of $G(\mathbb {Z}/p\mathbb {Z})$-orbits on $\overline {\Sigma }_f^{\operatorname {sol}}$. But there can only be as many orbits as there are points in $\mathbb {P}^1(\mathbb {Z}/p\mathbb {Z})$, which has size $p+1$. Combining this observation with (45) and Lemma 24 yields the proposition.

Remark 25 It is well-known (see [Reference Bhargava and GrossBG13, § 6.1]) that for each odd prime $p$, we have

(46)\begin{equation} \#G(\mathbb{Z}/p\mathbb{Z}) = p^{n^2} \cdot \prod_{i = 1}^n (p^{2i}-1) \leq p^{2n^2+n}. \end{equation}

Further note that $\operatorname {Vol}(G(\mathbb {Z}_p)) = \#G(\mathbb {Z}/p\mathbb {Z})/p^{2n^2+n}$.

4.5 Step (B): bounding the mass at $2$ when $2 \nmid f_0$

Here, it turns out to be ineffective to work over $\mathbb {Z}/2\mathbb {Z}$ or even $\mathbb {Z}/4\mathbb {Z}$; instead, we work over $\mathbb {Z}/8\mathbb {Z}$. Let $\mathscr {I}_8(m)$ denote the set of monic degree-$(2n+1)$ polynomials in $(\mathbb {Z}/8\mathbb {Z})[x]$ having $m$ distinct irreducible factors over $\mathbb {Z}/2\mathbb {Z}$. When $2 = p \nmid f_0$, we obtain the following result, which shows that the mass at $2$ in (26) contributes a saving of $O(2^{-2n})$.

Proposition 26 We have that

\[ \int_{T \in V(\mathbb{Z}_2)} \frac{\psi_2(T)}{m_2(T)}\, dT \leq \min\bigg\{1,\sum_{m = 1}^{2n+1} \frac{12}{2^{2n+m-1}} \cdot 2 \cdot \frac{\#\mathscr{I}_8(m)}{8^{2n+1}}\bigg\}. \]

Proof. Given a monic degree-$(2n+1)$ polynomial $f$ over $\mathbb {Z}/8\mathbb {Z}$, let $\overline {\Sigma }_f^{\operatorname {sol}} \subset V(\mathbb {Z}/8\mathbb {Z})$ denote the closure under the action of $G(\mathbb {Z}/8\mathbb {Z})$ of the mod-$8$ reduction of the subset $\{T \in \mathscr {P}_2 : \operatorname {ch}(T) \equiv f \pmod 8\}$ (note that this subset may not be a priori $G(\mathbb {Z}/8\mathbb {Z})$-invariant, because $G$ is not smooth over $\mathbb {Z}_2$ and the map $G(\mathbb {Z}_2) \to G(\mathbb {Z}/8\mathbb {Z})$ is not surjective). Then we have

(47)\begin{align} \int_{T \in V(\mathbb{Z}_2)} \frac{\psi_2(T)}{m_2(T)}\, dT & \leq \int_{T \in \mathscr{P}_2} dT \nonumber\\ & \leq \sum_{m = 1}^{2n+1} \sum_{f \in \mathscr{I}_8(m)} \frac{\#\overline{\Sigma}_f^{\operatorname{sol}}}{8^{\dim V}} = \sum_{m = 1}^{2n+1}\sum_{f \in \mathscr{I}_8(m)} \sum_{T \in G(\mathbb{Z}/8\mathbb{Z})\backslash \overline{\Sigma}_f^{\operatorname{sol}}} \frac{\#\mathscr{O}(T)}{8^{2n^2+3n+1}}, \end{align}

where for each $T \in V(\mathbb {Z}/8\mathbb {Z})$, we denote by $\mathscr {O}(T)$ the $G(\mathbb {Z}/8\mathbb {Z})$-orbit of $T$. Let $\operatorname {Stab}(T) \subset G(\mathbb {Z}/8\mathbb {Z})$ denote the stabilizer of $T$. Substituting $\#\mathscr {O}(T) = \#G(\mathbb {Z}/8\mathbb {Z})/\#\operatorname {Stab}(T)$ into (47) yields that

(48)\begin{equation} \int_{T \in V(\mathbb{Z}_2)} \frac{\psi_2(T)}{m_2(T)}\, dT \leq \sum_{m = 1}^{2n+1} \sum_{f \in \mathscr{I}_8(m)} \sum_{T \in G(\mathbb{Z}/8\mathbb{Z})\backslash \overline{\Sigma}_f^{\operatorname{sol}}} \frac{1}{\#\operatorname{Stab}(T)}\cdot \frac{\#G(\mathbb{Z}/8\mathbb{Z})}{8^{2n^2+3n+1}}. \end{equation}

The following lemma gives a lower bound for $\#\operatorname {Stab}(T)$ that is independent of $f \in \mathscr {I}_8(m)$.

Proof of Lemma 27 Let $R_f = (\mathbb {Z}/8\mathbb {Z})[x]/(f)$. By Remark 13, $\operatorname {Stab}(T)$ contains a subgroup isomorphic to $R_f^\times [2]_{\operatorname {N}\equiv 1} := \{\rho \in R_f^\times : \rho ^2 = 1 = \operatorname {N}(\rho )\}$. Let $f$ split over $\mathbb {Z}/8\mathbb {Z}$ as $f(x) = \prod _{i = 1}^m f_i(x)$, where the $f_i$ have the property that their mod-$2$ reductions are powers of distinct irreducible polynomials. Then it follows from [Reference Lei and PoulinLP20, Theorem 7] together with the Chinese remainder theorem that $R_f \simeq \prod _{i = 1}^m (\mathbb {Z}/8\mathbb {Z})[x]/(f_i)$, and $R_f^\times [2]$ contains the $2^{2n+m+1}$ elements of the form

\[ \bigg(\ldots, \rho_0 + \sum_{j = 1}^{d_i-1} \rho_j\theta_i^j, \ldots\bigg) \in \prod_{i = 1}^m (\mathbb{Z}/8\mathbb{Z})[x]/(f_i), \]

where $\rho _0$ is any element of $(\mathbb {Z}/8\mathbb {Z})^\times$ and $\rho _j$ is any element of $\{0,4\} \subset \mathbb {Z}/8\mathbb {Z}$ for each $j$, $\theta _i$ is the image of $x$ in $(\mathbb {Z}/8\mathbb {Z})[x]/(f_i)$, and $d_i = \deg f_i$ for each $i$. At least $\frac {1}{4}$ of these elements have norm $1$, so $\#\operatorname {Stab}(T) \geq \#R_F^\times [2]_{\operatorname {N}\equiv 1} \geq 2^{2n+m-1}$.