Proof. Fix any $T$ in $V_{f}(k)$ . Since $T$ has distinct eigenvalues, its stabilizer scheme in $\operatorname{GL}(U)$ is a maximal torus. It contains and hence is equal to the maximal torus $\operatorname{Res}_{L/k}\mathbb{G}_{m}$ . For any $k$ -algebra $K$ , we have

$$\begin{eqnarray}\operatorname{Stab}_{\text{O}(U)}(T)(K)=\{g\in (K[T]/f(T))^{\times }\mid g^{\ast }g=1\}.\end{eqnarray}$$

Since $T=T^{\ast }$ and $g$ is a polynomial in $T$ , we have $g=g^{\ast }$ . Thus,

$$\begin{eqnarray}\displaystyle \operatorname{Stab}_{\text{O}(U)}(T) & \simeq & \displaystyle \operatorname{Stab}_{\operatorname{GL}(U)}(T)[2]\simeq \operatorname{Res}_{L/k}\unicode[STIX]{x1D707}_{2},\nonumber\\ \displaystyle \operatorname{Stab}_{\text{SO}(U)}(T) & \simeq & \displaystyle (\operatorname{Res}_{L/k}\unicode[STIX]{x1D707}_{2})_{N=1},\nonumber\\ \displaystyle \operatorname{Stab}_{\text{PSO}(U)}(T) & \simeq & \displaystyle (\operatorname{Res}_{L/k}\unicode[STIX]{x1D707}_{2})_{N=1}/\unicode[STIX]{x1D707}_{2}.\nonumber\end{eqnarray}$$

Since $T$ is self-adjoint, there is an orthonormal basis $\{u_{1},\ldots ,u_{2n+2}\}$ for $U$ consisting of eigenvectors of $T$ with eigenvalues $\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{2n+2}$ . If $T^{\prime }$ is another element of $V_{f}(k^{s}),$ then there is an orthonormal basis $\{u_{1}^{\prime },\ldots ,u_{2n+2}^{\prime }\}$ of $U$ consisting of eigenvectors of $T^{\prime }$ with eigenvalues $\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{2n+2}$ . Let $g\in \operatorname{GL}(U)(k^{s})$ be an operator sending $u_{i}$ to $\pm u_{i}^{\prime }$ , where the signs are chosen so that $g\in \operatorname{SL}(U)(k^{s})$ . Then $g\in \operatorname{SO}(U)(k^{s})$ and the image of $g$ in $\operatorname{PSO}(U)(k^{s})$ sends $T$ to  $T^{\prime }$ .◻

Proof. We prove Theorem 10 in the case when $k$ is a local field. For a complete proof, see § 3. Combining the descent sequence (9) and the long exact sequence obtained by taking Galois cohomology of the short exact sequence

$$\begin{eqnarray}1\rightarrow J[2]\rightarrow \text{Res}_{L/k}\unicode[STIX]{x1D707}_{2}/\unicode[STIX]{x1D707}_{2}\xrightarrow[{}]{N}\unicode[STIX]{x1D707}_{2}\rightarrow 1,\end{eqnarray}$$

we get the following commutative diagram.

(11)

The map $\unicode[STIX]{x1D6FF}^{\prime }$ is defined in [Reference Poonen and SchaeferPS97] by evaluating $(x-\unicode[STIX]{x1D6FD})$ on a given divisor class. As shown in [Reference Poonen and SchaeferPS97], the first row is not exact: the map $\unicode[STIX]{x1D6FF}^{\prime }$ lands inside, generally not onto, $(L^{\times }/L^{\times 2}k^{\times })_{N=1}$ with kernel the subgroup generated by the class $(\infty ^{\prime })-(\infty )$ . Note that $(\infty ^{\prime })-(\infty )\in 2J(k)$ if and only if the norm map $N:\operatorname{Res}_{L/k}\unicode[STIX]{x1D707}_{2}/\unicode[STIX]{x1D707}_{2}(k)\rightarrow \unicode[STIX]{x1D707}_{2}(k)$ is surjective if and only if there is a unique distinguished orbit.

To prove Theorem 10, it suffices to show that if $\unicode[STIX]{x1D6FC}\in (L^{\times }/L^{\times 2}k^{\times })_{N=1}$ lies in the image of $\unicode[STIX]{x1D6FF}^{\prime }$ , then $\langle \,,\,\rangle _{\unicode[STIX]{x1D6FC}}$ is split. We will prove this by explicitly writing down a $k$ -rational ( $n+1$ )-dimensional isotropic subspace in the special case when $k$ is a local field. For a complete and more conceptual proof using pencils of quadrics, see § 3. Suppose $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FF}^{\prime }([D])$ for some $[D]\in J(k)/2J(k)$ of the form

$$\begin{eqnarray}[D]=(Q_{1})+\cdots +(Q_{m})-m(\infty )\hspace{0.6em}{\rm mod}\hspace{0.2em}2J(k)\cdot \langle (\infty ^{\prime })-(\infty )\rangle ,\end{eqnarray}$$

where $Q_{1},\ldots ,Q_{m}\in C(k^{s})$ are non-Weierstrass non-infinity points and $m\leqslant n+1$ . When $k$ is a local field, every $[D]\in J(k)/2J(k)$ can be written in this form [Reference WangWan13b, Lemma 3.8]. If we write $Q_{i}=(x_{i},y_{i})$ , then $\unicode[STIX]{x1D6FC}=(x_{1}-\unicode[STIX]{x1D6FD})\cdots (x_{m}-\unicode[STIX]{x1D6FD})$ and

$$\begin{eqnarray}\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D707}\rangle _{\unicode[STIX]{x1D6FC}}=\operatorname{Tr}_{L/k}((x_{1}-\unicode[STIX]{x1D6FD})\cdots (x_{m}-\unicode[STIX]{x1D6FD})\unicode[STIX]{x1D706}\unicode[STIX]{x1D707}/f^{\prime }(\unicode[STIX]{x1D6FD})).\end{eqnarray}$$

We may also assume that the $x_{i}$ are all distinct since the sum of all the $Q_{i}$ whose $x$ -coordinates appear more than once lies in $2J(k)\cdot \langle (\infty ^{\prime })-(\infty )\rangle$ . Write

$$\begin{eqnarray}\widetilde{V}=\mathop{\prod }_{1\leqslant i<j\leqslant m}(x_{i}-x_{j})\end{eqnarray}$$

for the Vandermonde polynomial, and, for each $i=1,\ldots ,m$ , define

$$\begin{eqnarray}q_{i}:=\mathop{\prod }_{1\leqslant j\leqslant m,j\neq i}(x_{j}-x_{i}),\quad a_{i}:=\widetilde{V}/q_{i},\quad h_{i}(t):=\frac{f(t)-f(x_{i})}{t-x_{i}}.\end{eqnarray}$$

For any $j\geqslant 0,$ we define

$$\begin{eqnarray}g_{j}(t)=\mathop{\sum }_{i=1}^{m}x_{i}^{j}a_{i}\frac{h_{i}(t)}{y_{i}}.\end{eqnarray}$$

Then the $(n+1)$ -plane $Y$ defined below is $k$ -rational and isotropic [Reference WangWan13b, Lemma 2.44]:

$$\begin{eqnarray}Y:=\operatorname{Span}\{1,\unicode[STIX]{x1D6FD},\ldots ,\unicode[STIX]{x1D6FD}^{n-m^{\prime }},g_{0}(\unicode[STIX]{x1D6FD}),\ldots ,g_{m^{\prime }-1}(\unicode[STIX]{x1D6FD})\}\quad \text{if}~m=2m^{\prime }~\text{or}~m=2m^{\prime }+1.\end{eqnarray}$$

This completes the proof of Theorem 10 when $k$ is a local field.◻

Proof. Given a pair $(I,\unicode[STIX]{x1D6FC})$ such that $\langle \,,\,\rangle _{\unicode[STIX]{x1D6FC}}$ is split, $\unicode[STIX]{x1D6FC}I^{2}\subset R$ and $N(I)^{2}=N(\unicode[STIX]{x1D6FC}^{-1})$ , there exists an isometry over $\mathbb{Q}_{p}$ from $(L,\langle \,,\,\rangle _{\unicode[STIX]{x1D6FC}})$ to $(U,\langle \,,\,\rangle _{Q})$ that sends $I$ to the self-dual lattice  $M_{0}$ . The image of the multiplication by $\unicode[STIX]{x1D6FD}$ operator lies in $V_{f}(\mathbb{Z}_{p})$ . Any two such isometries differ by an element in $\text{O}(U)(\mathbb{Z}_{p})$ . Hence we get a well-defined $\text{O}(U)(\mathbb{Z}_{p})$ -orbit in $V_{f}(\mathbb{Z}_{p})$ . Along with (12), we have proved the first statement.

For the second statement, from the sequence of isometries (7), we see that since $\langle \,,\,\rangle _{\unicode[STIX]{x1D6FC}}$ is split, there exists $g\in \text{O}(U)(\mathbb{Q}_{p}^{s})$ such that

$$\begin{eqnarray}\text{}^{\unicode[STIX]{x1D70E}\!}\sqrt{\unicode[STIX]{x1D6FC}}/\sqrt{\unicode[STIX]{x1D6FC}}=g^{-1}\,\text{}^{\unicode[STIX]{x1D70E}\!}g\quad \forall \unicode[STIX]{x1D70E}\in \operatorname{Gal}(k^{s}/k).\end{eqnarray}$$

Here, the left-hand side is viewed as an element of $\operatorname{Stab}_{\text{O}(U)}(T_{f})(k^{s})$ . The rational orbit corresponding the pair $(I,\unicode[STIX]{x1D6FC})$ is the rational orbit of $T=gT_{f}g^{-1}$ . The rest follows formally from unwinding definitions.◻

Proof of Proposition 12.

First note that it suffices to show that the $\operatorname{PO}(U)(\mathbb{Q}_{p})$ -orbit of $T$ contains an integral representative. Since $T$ is soluble, there exists some $[D]\in J(\mathbb{Q}_{p})/2J(\mathbb{Q}_{p})$ such that $\widetilde{\unicode[STIX]{x1D6FC}}=\unicode[STIX]{x1D6FF}^{\prime }([D])\in (L^{\times }/L^{\times 2}\mathbb{Q}_{p}^{\times })_{N=1}$ corresponds to the $\operatorname{PO}(U)(\mathbb{Q}_{p})$ -orbit of  $T$ . By [Reference WangWan13b, Lemma 3.8], there exist non-Weierstrass non-infinity points $Q_{1},\ldots ,Q_{m}\in C(\mathbb{Q}_{p}^{s})$ , with $m\leqslant n+1$ , such that

(14) $$\begin{eqnarray}[D]=(Q_{1})+\cdots +(Q_{m})-m(\infty )\hspace{0.6em}{\rm mod}\hspace{0.2em}2J(\mathbb{Q}_{p})\cdot \langle (\infty ^{\prime })-(\infty )\rangle .\end{eqnarray}$$

Write each $Q_{i}=(x_{i},y_{i})\in C({\mathcal{O}}_{\mathbb{Q}_{p}^{s}})$ . Then $\unicode[STIX]{x1D6FC}=(x_{1}-\unicode[STIX]{x1D6FD})\cdots (x_{m}-\unicode[STIX]{x1D6FD})$ is a lift of $\widetilde{\unicode[STIX]{x1D6FC}}$ to $L^{\times }$ . We claim that either the $\text{O}(U)(\mathbb{Q}_{p})$ -orbit of $T$ corresponding to the image of $\unicode[STIX]{x1D6FC}$ in $L^{\times }/L^{\times 2}$ has an integral representative, or $[D]$ can be expressed in the form (14) with $m$ replaced by $m-2$ . Applying induction on $m$ completes the proof.

The claim follows verbatim from the proof of [Reference Bhargava and GrossBG13, Proposition 8.5]. We give a quick sketch here. Let $r(x)\in \mathbb{Q}_{p}[x]$ be a polynomial of degree at most $m-1$ such that, for all  $i$ , $r(x_{i})=y_{i}$ , and let

$$\begin{eqnarray}p(x)=(x-x_{1})\cdots (x-x_{m})\in \mathbb{Z}_{p}[x].\end{eqnarray}$$

Now $p(x)$ divides $r(x)^{2}-f(x)$ in $\mathbb{Q}_{p}[x]$ and we denote the quotient by $q(x)$ . By definition, $\unicode[STIX]{x1D6FC}=(-1)^{m}P(\unicode[STIX]{x1D6FD})$ . If the polynomial $r(x)\in \mathbb{Z}_{p}[x]$ , then the ideal $I=(1,r(\unicode[STIX]{x1D6FD})/\unicode[STIX]{x1D6FC})$ does the job. Note that $\unicode[STIX]{x1D6FC}I^{2}=(\unicode[STIX]{x1D6FC},r(\unicode[STIX]{x1D6FD}),q(\unicode[STIX]{x1D6FD}))$ . The integrality assumption of $r(x)$ is used to show that $r(\unicode[STIX]{x1D6FD}),q(\unicode[STIX]{x1D6FD})\in R$ . A computation of ideal norms shows that $N(I)^{2}=N(\unicode[STIX]{x1D6FC})^{-1}$ .

When $r(x)$ is not integral, a Newton polygon analysis on $f(x)-r(x)^{2}$ shows that $\operatorname{div}(y-r(x))-[D]$ has the form $D^{\ast }+E$ with $D^{\ast }$ , $E\in J(\mathbb{Q}_{p})$ , where $D^{\ast }$ can be expressed in (14) with $m$ replaced by $m-2$ and the $x$ -coordinates of the non-infinity points in $E$ have negative valuation. The condition of divisibility on the coefficients of $f(x)$ ensures that $E\in 2J(\mathbb{Q}_{p})\cdot ((\infty ^{\prime })-(\infty ))$ , or equivalently $(x-\unicode[STIX]{x1D6FD})(E)\in L^{\times 2}\mathbb{Q}_{p}^{\times }$ .◻

Proof of Proposition 14.

Once again, it suffices to work with $\operatorname{PO}(U)$ -orbits instead of $\operatorname{PSO}(U)$ -orbits directly. The assumption on $\unicode[STIX]{x1D6E5}(f)$ implies that $R$ is the maximal order. Hence there is a bijection between $\text{O}(U)(\mathbb{Z}_{p})$ -orbits and $(R^{\times }/R^{\times 2})_{N=1}$ . Note that over non-archimedean local fields, the splitness of the quadratic form is automatic from the existence of a self-dual lattice. Taking flat cohomology over $\text{Spec}(\mathbb{Z}_{p})$ of the sequence

$$\begin{eqnarray}1\rightarrow \unicode[STIX]{x1D707}_{2}\rightarrow \text{O}(U)\rightarrow \operatorname{PO}(U)\rightarrow 1\end{eqnarray}$$

gives:

$$\begin{eqnarray}1\rightarrow \text{O}(U)(\mathbb{Z}_{p})/\pm 1\rightarrow \operatorname{PO}(U)(\mathbb{Z}_{p})\rightarrow \mathbb{Z}_{p}^{\times }/\mathbb{Z}_{p}^{\times 2}\rightarrow 1.\end{eqnarray}$$

Hence $\operatorname{PO}(U)(\mathbb{Z}_{p})$ -orbits correspond bijectively to $(R^{\times }/R^{\times 2}\mathbb{Z}_{p}^{\times })_{N=1}$ .

On the other hand, the assumption on $\unicode[STIX]{x1D6E5}(f)$ implies that the projective closure ${\mathcal{C}}$ (in weighted projective space) of the hyperelliptic curve $C$ defined by affine equation $y^{2}=f(x)$ over $\operatorname{Spec}(\mathbb{Z}_{p})$ is regular. Since the special fiber of ${\mathcal{C}}$ is geometrically reduced and irreducible, the Neron model ${\mathcal{J}}$ of its Jacobian $J_{\mathbb{Q}_{p}}$ is fiberwise connected [Reference Bosch, Lutkebohmert and RaynaudBLR90, § 9.5, Theorem 1] and its 2-torsion ${\mathcal{J}}[2]$ is isomorphic to $(\operatorname{Res}_{R/\mathbb{Z}_{p}}\unicode[STIX]{x1D707}_{2})_{N=1}/\unicode[STIX]{x1D707}_{2}$ . Using diagram (11) after replacing $L,k,J$ by $R,\mathbb{Z}_{p},{\mathcal{J}}$ , we see that the vertical maps are all isomorphisms and $\unicode[STIX]{x1D6FF}^{\prime }$ maps ${\mathcal{J}}(\mathbb{Z}_{p})/2{\mathcal{J}}(\mathbb{Z}_{p})$ surjectively to $(R^{\times }/R^{\times 2}\mathbb{Z}_{p}^{\times })_{N=1}$ . The Neron mapping property implies that ${\mathcal{J}}(\mathbb{Z}_{p})/2{\mathcal{J}}(\mathbb{Z}_{p})=J(\mathbb{Q}_{p})/2J(\mathbb{Q}_{p})$ .

Suppose the $\text{O}(U)(\mathbb{Z}_{p})$ -orbit of some $T\in V_{f}(\mathbb{Z}_{p})$ corresponds to an equivalence class of pair $(I,\unicode[STIX]{x1D6FC})$ . Since $R$ is maximal, we have $\operatorname{End}_{R}(I)=R$ . Since $R^{\times }[2]=L^{\times }[2]$ , we see from (13) that it remains to compare $(R^{\times 2}\cap \mathbb{Z}_{p}^{\times })/\mathbb{Z}_{p}^{\times 2}$ with $(L^{\times 2}\cap \mathbb{Q}_{p}^{\times })/\mathbb{Q}_{p}^{\times 2}$ . These two groups are only non-trivial when $L$ contains a quadratic extension $K^{\prime }$ of  $\mathbb{Q}_{p}$ . The condition $p^{2}\nmid \unicode[STIX]{x1D6E5}(f)$ implies that $K^{\prime }=\mathbb{Q}_{p}(\sqrt{u})$ can only be the unramified quadratic extension of  $\mathbb{Q}_{p}$ . In other words, $u$ can be chosen to be a unit in $\mathbb{Z}_{p}^{\times }$ . Hence in this case $(L^{\times 2}\cap \mathbb{Q}_{p}^{\times })/\mathbb{Q}_{p}^{\times 2}$ and $(R^{\times 2}\cap \mathbb{Z}_{p}^{\times })/\mathbb{Z}_{p}^{\times 2}$ are both equal to the group of order 2 generated by the class of  $u$ .◻

Proof. It will be convenient to use the following parameters for $T$ :

$$\begin{eqnarray}\displaystyle s_{i} & = & \displaystyle t_{i}/t_{i+1}\quad \text{for}~i=1,\ldots ,n;\nonumber\\ \displaystyle s_{n+1} & = & \displaystyle t_{n}t_{n+1}.\nonumber\end{eqnarray}$$

The condition for $t\in T^{\prime }$ translates to $s_{i}>c$ for all $i$ . We pick the following Haar measure $dh$ on $G(\mathbb{R})=NTK$ :

(20) $$\begin{eqnarray}\displaystyle dh & = & \displaystyle du\mathop{\prod }_{j=1}^{n-1}s_{j}^{j(j-2n-1)}\cdot (s_{n}s_{n+1})^{-n(n+1)/2}d^{\times }s_{j}\,dk\nonumber\\ \displaystyle & = & \displaystyle du\unicode[STIX]{x1D6FF}(s)d^{\times }s\,dk,\end{eqnarray}$$

where $du$ is a Haar measure on the unipotent group $N$ , $dk$ is the Haar measure on $K$ normalized so that $K$ has volume 1, $\unicode[STIX]{x1D6FF}(s)$ denotes $\prod _{j=1}^{n-1}s_{j}^{j(j-2n-1)}\cdot (s_{n}s_{n+1})^{-n(n+1)/2}$ , and $d^{\times }s$ denotes $\prod _{j=1}^{n+1}d^{\times }s_{j}$ in which each $d^{\times }s_{j}=ds_{j}/s_{j}$ is the standard Haar measure on  $\mathbb{R}^{\times }$ . The conjugation action of $T$ on $N$ breaks up into a direct sum of characters of  $T$ . The Haar measure character $\unicode[STIX]{x1D6FF}(s)$ is the product of the inverses of all the characters of $T$ arising in this decomposition, in order for the measure $dh$ above to be left-invariant.

Then, since $G_{0}$ is $K$ -invariant, (19) implies that

(21) $$\begin{eqnarray}\displaystyle N(V(\mathbb{Z})(b_{11}=0);X) & = & \displaystyle O\biggl(\int _{h\in {\mathcal{F}}}\#\{hG_{0}R^{(m)}(X)\cap V(\mathbb{Z})(b_{11}=0)\}\,dh\biggr)\nonumber\\ \displaystyle & = & \displaystyle O\biggl(\int _{u\in N^{\prime }}\int _{t\in T^{\prime }}\#\{utG_{0}R^{(m)}(X)\cap V(\mathbb{Z})(b_{11}=0)\}\,\unicode[STIX]{x1D6FF}(s)\,d^{\times }s\,du\biggr)\nonumber\\ \displaystyle & = & \displaystyle O\biggl(\int _{t\in T^{\prime }}\#\{tG_{0}R^{(m)}(X)\cap V(\mathbb{Z})(b_{11}=0)\}\,\unicode[STIX]{x1D6FF}(s)\,d^{\times }s\biggr),\end{eqnarray}$$

where the final equality follows because $N^{\prime }$ has finite measure,

$$\begin{eqnarray}utG_{0}R^{(m)}(X)=t(t^{-1}ut)G_{0}R^{(m)}(X),\end{eqnarray}$$

and the coefficients of $t^{-1}ut$ are bounded independent of $t\in T^{\prime }$ and $u\in N^{\prime }$ .

Let $b_{ij}$ , with $i\leqslant j$ and $(i,j)\neq (n+1,n+2)$ , be the system of coordinates on $V(\mathbb{R})$ , where $b_{ij}$ is the $(i,j)$ th entry of the symmetric matrix $B$ . To each coordinate $b_{ij}$ we associate the weight $w(i,j)$ , which records how an element $s\in T$ scales $b_{ij}$ . For example,

$$\begin{eqnarray}\displaystyle w(1,1) & = & \displaystyle s_{1}^{-2}\cdots s_{n-1}^{-2}s_{n}^{-1}s_{n+1}^{-1}\nonumber\\ \displaystyle w(i,2n+3-i) & = & \displaystyle 1,\quad i=1,\ldots ,2n+2,\quad \text{coordinates on the anti}\text{-}\text{diagonal}\nonumber\\ \displaystyle w(i,2n+2-i) & = & \displaystyle s_{i}^{-1},\quad i=1,\ldots ,2n+1,\quad \text{coordinates above the anti}\text{-}\text{diagonal}\nonumber\\ \displaystyle w(n+1,n+1) & = & \displaystyle s_{n}s_{n+1}^{-1}.\nonumber\end{eqnarray}$$

Let $C$ be an absolute constant such that $CX^{1/d}$ bounds the absolute value of all the coordinates of elements $B\in G_{0}R^{(m)}(X)$ . If, for $(s_{1},\ldots ,s_{n+1})\in T^{\prime }$ , we have $CX^{1/d}\,w(i_{0},2n+2-i_{0})<1$ for some $i_{0}\in \{1,\ldots ,n+1\}$ , then $CX^{1/d}\,w(i,j)<1$ for all $i\leqslant i_{0},j\leqslant 2n+2-i_{0}$ . Hence the top left $i_{0}\times (2n+2-i_{0})$ block of any integral $B\in tG_{0}R^{(m)}(X)$ is 0. Just as [Reference Bhargava and GrossBG13, Lemma 10.3] shows, any such $B$ has zero discriminant. Hence, to prove Proposition 21, we may assume

(22) $$\begin{eqnarray}s_{i}\leqslant CX^{1/d},~i=1,\ldots ,n;\quad s_{n+1}\leqslant C^{2}X^{2/d}.\end{eqnarray}$$

We use $T_{X}$ to denote the set of $t=(s_{1},\ldots ,s_{n+1})\in T^{\prime }$ satisfying these bounds.

Let $U$ denote the set of pairs of integers $(i,j)$ with $1\leqslant i,j,\leqslant 2n+2$ and $i\leqslant j$ . For any subset $U_{1}$ of  $U$ , let $V(\mathbb{R})(U_{1})$ denote the subset of $V(\mathbb{R})$ consisting of elements $B$ whose $(i,j)$ th entry is less than 1 in absolute value when $(i,j)\in U_{1}$ and at least 1 in absolute value when $(i,j)\notin U_{1}$ . Let $V(\mathbb{Z})(U_{1})$ denote the set of integral points in $V(\mathbb{R})(U_{1})$ . Then to prove Proposition 21, it suffices to show that

(23) $$\begin{eqnarray}N(V(\mathbb{Z})(U_{1});X)=O_{\unicode[STIX]{x1D716}}(X^{(\dim V-1)/d+\unicode[STIX]{x1D716}}),\end{eqnarray}$$

for every set $U_{1}$ containing $(1,1)$ .

Proposition 19, in conjunction with the argument used to justify (21), implies

$$\begin{eqnarray}\displaystyle N(V(\mathbb{Z})(U_{1});X) & = & \displaystyle O\biggl(\int _{t\in T_{X}}\operatorname{Vol}(tG_{0}R^{(m)}(X)\cap V(\mathbb{R})(U_{1}))\,\unicode[STIX]{x1D6FF}(s)\,d^{\times }s\biggr)\nonumber\\ \displaystyle & = & \displaystyle O\biggl(X^{(\dim V-\#U_{1})/d}\int _{t\in T_{X}}\mathop{\prod }_{(i,j)\not \in U_{1}}w(i,j)\,\unicode[STIX]{x1D6FF}(s)\,d^{\times }s\biggr).\nonumber\end{eqnarray}$$

Hence, to prove (23), we need to bound

(24) $$\begin{eqnarray}\widetilde{I}(U_{1},X):=X^{(\dim V-\#U_{1})/d}\int _{t\in T_{X}}\mathop{\prod }_{(i,j)\not \in U_{1}}w(i,j)\,\unicode[STIX]{x1D6FF}(s)\,d^{\times }s,\end{eqnarray}$$

for every set $U_{1}$ containing $(1,1)$ .

Note that if $i^{\prime }\leqslant i$ and $j^{\prime }\leqslant j$ , then $w(i^{\prime },j^{\prime })$ has smaller exponents in all the $s_{k}$ than $w(i,j)$ . Thus, if a set $U_{1}$ contains $(i,j)$ but not $(i^{\prime },j^{\prime })$ , then

$$\begin{eqnarray}\widetilde{I}(U_{1}\backslash \{(i,j)\}\cup \{(i^{\prime },j^{\prime })\},X)\geqslant \widetilde{I}(U_{1},X).\end{eqnarray}$$

Hence, for the purpose of obtaining an upper bound for $\widetilde{I}(U_{1},X)$ , we may assume that if $(i,j)\in U_{1}$ , then $(i^{\prime },j^{\prime })\in U_{1}$ for all $i^{\prime }\leqslant i$ and $j^{\prime }\leqslant j$ . We say that such a set $U_{1}$ is closed. If a closed set $U_{1}$ contains any element on, or to the right of, the off-anti-diagonal, then every element in $V(\mathbb{Z})(U_{1})$ has discriminant 0 and, by definition, $N(V(\mathbb{Z})(U_{1});X)=0$ . Let $U_{0}$ denote the set of coordinates $(i,j)$ such that $i\leqslant j$ and $i+j\leqslant 2n+1$ . In other words, $U_{0}$ contains every coordinate to the left of the off-anti-diagonal. Since every element in $V(\mathbb{Z})(U_{0})$ is distinguished (by Proposition 7), hence reducible, it suffices to consider $\widetilde{I}(U_{1},X)$ for all $U_{1}\subsetneq U_{0}$ .

To this end, as the product of the weights over all coordinates is 1, we define

(25) $$\begin{eqnarray}I(U_{1},X)=X^{-\#U_{1}/d}\int _{s_{1},\ldots ,s_{n}=c}^{CX^{1/d}}\int _{s_{n+1}=c}^{C^{2}X^{2/d}}\!\mathop{\prod }_{(i,j)\in U_{1}}w(i,j)^{-1}\,\mathop{\prod }_{k=1}^{n-1}s_{k}^{k(k-2n-1)}\cdot (s_{n}s_{n+1})^{-n(n+1)/2}\,d^{\times }s.\end{eqnarray}$$

To complete the proof of Proposition 21, it suffices to prove the following lemma.

Lemma 22. Let $U_{1}$ be non-empty proper closed subset of $U_{0}$ . Then

$$\begin{eqnarray}I(U_{1},X)=O_{\unicode[STIX]{x1D716}}(X^{-1/d+\unicode[STIX]{x1D716}}).\end{eqnarray}$$

If $U_{1}=U_{0}$ or $U_{1}=\emptyset$ , then $I(U_{1},X)=O(1)$ .

Proof. The proof of this lemma is a combinatorial argument using induction on $n\geqslant 2$ . We first compute

(26) $$\begin{eqnarray}I(U_{0},X)=X^{-n(n+1)/d}\int _{s_{1},\ldots ,s_{n}=c}^{CX^{1/d}}\int _{s_{n+1}=c}^{C^{2}X^{2/d}}s_{1}s_{2}^{3}\cdots s_{n-1}^{2n-3}s_{n}^{n-1}s_{n+1}^{n}\,d^{\times }s=O(1).\end{eqnarray}$$

This is expected since $V(\mathbb{Z})(U_{0})$ contains all but negligibly few distinguished orbits (see Proposition 23). It is also easy to see that $I(\emptyset ,X)=O(1)$ . Let $U_{1}^{\prime }$ denote $U_{0}\backslash U_{1}$ , and define $I_{n}^{\prime }(U_{1}^{\prime },X)$ to equal $I(U_{1},X)$ . Combining (25) with (26), we obtain

$$\begin{eqnarray}\displaystyle I_{n}^{\prime }(U_{1}^{\prime },X) & = & \displaystyle I(U_{1},X)\nonumber\\ \displaystyle & = & \displaystyle X^{(\#U_{1}^{\prime }-n(n+1))/d}\int _{s_{1},\ldots ,s_{n}=c}^{CX^{1/d}}\int _{s_{n+1}=c}^{C^{2}X^{2/d}}\mathop{\prod }_{(i,j)\in U_{1}^{\prime }}w(i,j)\cdot s_{1}s_{2}^{3}\cdots s_{n-1}^{2n-3}s_{n}^{n-1}s_{n+1}^{n}\,d^{\times }s.\nonumber\end{eqnarray}$$

Even though we only need the result when $n\geqslant 2$ , for the purpose of the induction it is also necessary to work out the case $n=1$ . When $n=1$ , we have $U_{0}=\{(1,1),(1,2)\}$ and

$$\begin{eqnarray}\displaystyle I_{1}(\emptyset ,X) & = & \displaystyle O(1),\nonumber\\ \displaystyle I_{1}(\{(1,1)\},X) & = & \displaystyle O_{\unicode[STIX]{x1D716}}(X^{-1/d+\unicode[STIX]{x1D716}}),\nonumber\\ \displaystyle I_{1}(U_{0},X) & = & \displaystyle O_{\unicode[STIX]{x1D716}}(X^{\unicode[STIX]{x1D716}}).\nonumber\end{eqnarray}$$

To establish the inductive step, we write $U_{1}^{\prime }=U_{2}^{\prime }\cup U_{3}^{\prime }$ where $U_{2}^{\prime }$ is the set of coordinates $(1,j)$ in $U_{1}^{\prime }$ and $U_{3}^{\prime }=U_{1}^{\prime }\backslash U_{2}^{\prime }$ . Since we have

$$\begin{eqnarray}\int _{c}^{CX^{1/d}}s^{k}\,d^{\times }s\ll _{c,C}\int _{c}^{CX^{1/d}}s^{k_{1}}\,d^{\times }s\int _{c}^{CX^{1/d}}s^{k_{2}}\,d^{\times }s\end{eqnarray}$$

for every $k_{1}+k_{2}=k$ , it follows that we may bound $I_{n}^{\prime }(U_{1}^{\prime },X)$ by the product

$$\begin{eqnarray}I_{n}^{\prime }(U_{1}^{\prime },X)\ll _{c,C}J_{n}(U_{2}^{\prime },X)K_{n}(U_{3}^{\prime },X),\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle J_{n}(U_{2}^{\prime },X) & := & \displaystyle X^{(\#U_{2}^{\prime }-2n)/d}\int _{s_{1}=c}^{CX^{1/d}}\int _{s_{2},\ldots ,s_{n}=c}^{CX^{1/d}}\int _{s_{n+1}=c}^{C^{2}X^{2/d}}\mathop{\prod }_{(1,j)\in U_{2}^{\prime }}w(1,j)\,s_{1}s_{2}^{2}\cdots s_{n-1}^{2}s_{n}s_{n+1}\,d^{\times }s,\nonumber\\ \displaystyle K_{n}(U_{3}^{\prime },X) & := & \displaystyle X^{(\#U_{3}^{\prime }-(n-1)n)/d}\int _{s_{2},\ldots ,s_{n}=c}^{CX^{1/d}}\int _{s_{n+1}=c}^{C^{2}X^{2/d}}\mathop{\prod }_{(i,j)\in U_{3}^{\prime }}w(i,j)\,s_{2}s_{3}^{3}\cdots s_{n-1}^{2n-5}s_{n}^{n-2}s_{n+1}^{n-1}\,d^{\times }s.\nonumber\end{eqnarray}$$

Note that $K_{n}(U_{3}^{\prime },X)=I_{n-1}^{\prime }(\{(i,j):(i+1,j+1)\in U_{3}^{\prime }\},X)$ (which we denote by $I_{n-1}^{\prime }(U_{3}^{\prime },X)$ ) and we may estimate it using induction. Since $U_{1}$ is closed and non-empty, the subset $U_{2}^{\prime }$ is either empty or of the form $\{(1,k),(1,k+1),\ldots ,(1,2n)\}$ with $k\geqslant 2$ . A direct calculation gives

$$\begin{eqnarray}J_{n}(U_{2}^{\prime },X)=\left\{\begin{array}{@{}ll@{}}O(1)\quad & \text{if}~U_{2}^{\prime }=\emptyset ,\\ O_{\unicode[STIX]{x1D716}}(X^{(-k+1)/d+\unicode[STIX]{x1D716}})\quad & \text{if}~2\leqslant k\leqslant n+1,\\ O_{\unicode[STIX]{x1D716}}(X^{(k-2n-1)/d+\unicode[STIX]{x1D716}})\quad & \text{if}~n+2\leqslant k\leqslant 2n.\end{array}\right.\end{eqnarray}$$

Hence we have

(27) $$\begin{eqnarray}J_{n}(U_{2}^{\prime },X)=O_{\unicode[STIX]{x1D716}}(X^{-1/d+\unicode[STIX]{x1D716}}),\end{eqnarray}$$

unless $U_{2}^{\prime }=\emptyset$ , in which case it is $O(1)$ .

Hence, if $U_{2}^{\prime }$ is not empty, then the lemma follows by induction on $n$ (used to bound $I_{n-1}^{\prime }(U_{3}^{\prime },X)$ by $O_{\unicode[STIX]{x1D716}}(X^{\unicode[STIX]{x1D716}})$ ). If $U_{2}^{\prime }$ is empty, then $U_{3}^{\prime }$ must be non-empty since $U_{1}^{\prime }$ is non-empty. If, further, $U_{3}^{\prime }\neq U_{0}\backslash \{(1,1),\ldots ,(1,2n)\}$ , then by induction we have $I_{n-1}^{\prime }(U_{3}^{\prime },X)=O_{\unicode[STIX]{x1D716}}(X^{-1/d+\unicode[STIX]{x1D716}})$ . The only remaining case is when $U_{1}=\{(1,1),\ldots ,(1,2n)\}$ , for which a direct computation yields the result.◻

This concludes the proof of Proposition 21. ◻

Proof. Observe that if $B\in V(\mathbb{Z})$ is reducible over $\mathbb{Z}$ , then the image of $B$ in $V(\mathbb{F}_{p})$ is reducible for all  $p$ . For any prime  $p$ , let $\unicode[STIX]{x1D719}_{p}$ denote the $p$ -adic density of the set of elements of $V(\mathbb{Z}_{p})$ that are reducible mod  $p$ . Then, to prove Proposition 23, it suffices to show

$$\begin{eqnarray}\mathop{\prod }_{p}\unicode[STIX]{x1D719}_{p}=0.\end{eqnarray}$$

We show this by proving that $\unicode[STIX]{x1D719}_{p}$ is bounded above by some constant less than 1 when $p$ is large enough. For large enough  $p$ , there is a positive proportion $r_{n}$ (depending only on  $n$ ) of polynomials of degree $2n+2$ over $\mathbb{F}_{p}$ that factor into two linear terms and an irreducible polynomial of degree $2n$ . Suppose $f(x)\in \mathbb{Z}_{p}[x]$ with this reduction type over  $\mathbb{F}_{p}$ . Since it has a linear factor, Proposition 5 implies that there is one distinguished orbit. Since $H^{1}(\mathbb{F}_{p},J)=0$ by Lang’s theorem, every orbit is soluble. The number of orbits $\#J(\mathbb{F}_{p})/2J(\mathbb{F}_{p})$ is equal to the size of the stabilizer $\#J[2](\mathbb{F}_{p})$ . Since $f(x)$ has a factor of degree 2, $\#J[2](\mathbb{F}_{p})\geqslant 2$ . Therefore at least half of the elements in $V_{f}(\mathbb{F}_{p})$ are not distinguished. Hence, for $p$ large enough, $\unicode[STIX]{x1D719}_{p}\leqslant 1-\frac{1}{2}r_{n}<1$ .

We use the same technique to prove the second claim in Proposition 23. For $p$ large enough, there is a positive proportion $r_{n}^{\prime }$ (depending only on  $n$ ) of polynomials of degree $2n+2$ over $\mathbb{F}_{p}$ that factors into a linear term and an irreducible polynomial of degree $2n+1$ . If $B\in V_{f}(\mathbb{Z}_{p})$ where $f(x)$ has this reduction type mod  $p$ , then $p$ does not divide the discriminant of $f(x)$ . As a consequence, the hyperelliptic curve $y^{2}=f(x)$ is smooth over $\operatorname{Spec}(\mathbb{Z}_{p})$ and the 2-torsion of its Jacobian $J[2]$ is a finite étale group scheme over $\operatorname{Spec}(\mathbb{Z}_{p})$ . From the reduction type of $f(x)$ over  $p$ , we see that $\#J[2](\mathbb{Q}_{p})=\#J[2](\mathbb{F}_{p})=1$ . Denote by $\unicode[STIX]{x1D719}_{p}$ the $p$ -adic density of the set of elements of $V(\mathbb{Z}_{p})$ with non-trivial stabilizer in $G(\mathbb{Q}_{p})$ . Then we have shown that $\unicode[STIX]{x1D719}_{p}\leqslant 1-r_{n}^{\prime }<1$ for $p$ sufficiently large. This completes the proof.◻

Proof. Let ${\mathcal{F}}^{\prime }\subset {\mathcal{F}}$ be the set consisting of $h\in {\mathcal{F}}$ such that the $b_{11}$ -coefficient of any $B\in hG_{0}R^{(m)}(X)$ is less than 1 in absolute value. From (19), we see that $N(V(\mathbb{Z})^{(m)};X)$ is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\int _{h\in {\mathcal{F}}}\#\{hG_{0}R^{(m)}(X)\cap V(\mathbb{Z})^{\text{irr}}\}\,dh\nonumber\\ \displaystyle & & \displaystyle \quad =\displaystyle \frac{1}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\biggl(\int _{h\in {\mathcal{F}}\backslash {\mathcal{F}}^{\prime }}\#\{hG_{0}R^{(m)}(X)\cap V(\mathbb{Z})^{\text{irr}}\}\,dh\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{h\in {\mathcal{F}}^{\prime }}\#\{hG_{0}R^{(m)}(X)\cap V(\mathbb{Z})^{\text{irr}}\}\,dh\biggr).\nonumber\end{eqnarray}$$

From Propositions 21 and 23, we obtain

(28) $$\begin{eqnarray}\displaystyle N(V(\mathbb{Z})^{(m)};X)=\frac{1}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\int _{h\in {\mathcal{F}}\backslash {\mathcal{F}}^{\prime }}\#\{hG_{0}R^{(m)}(X)\cap V(\mathbb{Z})\}\,dh+o(X^{(\dim V)/d}). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Note that $b_{11}$ has minimal weight among all the $b_{ij}$ , that is, the powers of the $s_{k}$ in $w(1,1)/w(i,j)$ are non-negative for each $i,j,k$ . Furthermore, the length of the projection of $hG_{0}R^{(m)}(X)$ onto the $b_{11}$ -line is greater than 1 for any $h\in {\mathcal{F}}\backslash {\mathcal{F}}^{\prime }$ (by the definition of  ${\mathcal{F}}^{\prime }$ ). Hence, for $h\in {\mathcal{F}}\backslash {\mathcal{F}}^{\prime }$ , the volumes of all smaller-dimensional projections of $hG_{0}R^{(m)}(X)$ are bounded by a constant times the volume of its projection onto the $b_{11}=0$ hyperplane. Proposition 19 then implies that

$$\begin{eqnarray}\displaystyle N(V(\mathbb{Z})^{(m)};X) & = & \displaystyle \frac{1}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\int _{h\in {\mathcal{F}}\backslash {\mathcal{F}}^{\prime }}\operatorname{Vol}(hG_{0}R^{(m)}(X))\nonumber\\ \displaystyle & & \displaystyle +\,O\biggl(\frac{\operatorname{Vol}(hG_{0}R^{(m)}(X))}{X^{1/d}w(1,1)}\biggr)\,dh+o(X^{(\dim V)/d}).\nonumber\end{eqnarray}$$

Recall that ${\mathcal{F}}^{\prime }$ is defined by the condition $CX^{1/d}w(1,1)<1$ . Hence, to be in  ${\mathcal{F}}^{\prime }$ , one of the $s_{i}$ must be at least $C^{1/2n}X^{1/2nd}$ , which implies that the volume of ${\mathcal{F}}^{\prime }$ is bounded by $o(1)$ . Moreover, since $\int _{h\in {\mathcal{F}}\backslash {\mathcal{F}}^{\prime }}1/w(1,1)\,dh=O(1)$ , we obtain

(29) $$\begin{eqnarray}\displaystyle N(V(\mathbb{Z})^{(m)};X) & = & \displaystyle \frac{1}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\int _{h\in {\mathcal{F}}}\operatorname{Vol}(hG_{0}R^{(m)}(X))\,dh+o(X^{(\dim V)/d})\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\int _{h\in G_{0}}\operatorname{Vol}({\mathcal{F}}h\cdot R^{(m)}(X))\,dh+o(X^{(\dim V)/d})\nonumber\\ \displaystyle & = & \displaystyle \frac{\operatorname{Vol}({\mathcal{F}}\cdot R^{(m)}(X))}{\#J^{(m)}[2](\mathbb{R})\operatorname{Vol}(G_{0})}\int _{h\in G_{0}}\,dh+o(X^{(\dim V)/d})\nonumber\\ \displaystyle & = & \displaystyle \frac{\operatorname{Vol}({\mathcal{F}}\cdot R^{(m)}(X))}{\#J^{(m)}[2](\mathbb{R})}+o(X^{(\dim V)/d}),\end{eqnarray}$$

where the third equality follows because the volume of ${\mathcal{F}}h\cdot R^{(m)}(X)$ is independent of  $h$ . This concludes the proof of Theorem 24.◻