$([h_{1}(a_{1}):h_{2}(a_{2})]_{\mathcal{Q}})^{8}=[h_{1}(a_{1}^{8}):h_{2}(a_{2})]_{\mathcal{Q}}=[h_{1}(d_{1}):h_{2}(d_{2})]_{\mathcal{Q}}.$

We can assume that $\mathcal{P}_{K^{*}}(I)=\mathcal{Q}$. Lemma 4.5.13 implies
that there exist an $(O^{*},I,L^{2})$-approximation $(b_{1},b_{2})$ of
$(a_{1}^{8},a_{2})$ such that $b_{2}\in({\bf O}^{*})^{\times}$ and for every
$P\in\mathcal{R}:=\operatorname{supp}_{\mathcal{P}_{K^{*}}\smallsetminus\mathcal{P}_{K}^{*}(IL)}(b_{1})$,
$\operatorname{val}_{P}(b_{1})=-1$, and a
$(O^{*},\prod_{P\in\mathcal{R}}P,L^{2})$-approximation $(c_{1},c_{2})$ of
$(h_{1}(b_{1}),h_{2}(b_{2}^{-2}))$ such that
$h_{1}(c_{1})_{\mathcal{R}^{c}}\in\operatorname{Spin}_{f_{s}}({\bf O}^{*})$
and $c_{2}\in({\bf O}^{*})^{\times}$. Note that $\mathcal{R}$ and
$\mathcal{P}_{K^{*}}(L)$ are disjoint so
$h_{1}(c_{2})_{\mathcal{R}}\in\operatorname{Spin}_{f_{s}}({\bf
O}^{*})[L^{2}]$.

By the choice of $I$ and since $b_{2}\in({\bf O}^{*})^{\times}$, Lemma 4.5.3
implies that

$([h_{1}(a_{1}),h_{2}(a_{2})]_{\mathcal{Q}})^{8}=[h_{1}(a_{1})^{8},h_{2}(a_{2})]_{\mathcal{Q}}=[h_{1}(b_{1}),h_{2}(b_{2}^{-1})]_{\mathcal{Q}}.$

Lemma 4.5.4, the existence of
$(O^{*},\prod_{P\in\mathcal{R}}P,J)$-approximation $(c_{1},c_{2})$ of
$(b_{1},b_{2}^{-2})$ such that
$h_{1}(c_{1})_{\mathcal{R}^{c}}\in\operatorname{Spin}_{f_{s}}({\bf O}^{*})$
and $c_{2}\in({\bf O}^{*})^{\times}$ imply that

$([h_{1}(a_{1}),h_{2}(a_{2})]_{\mathcal{Q}})^{16}=[h_{1}(b_{1}),h_{2}(b_{2}^{-2})]_{\mathcal{Q}}=1.$

∎

######  Corollary 4.5.15.

Under Setting 4.4.2, for every $a_{1},a_{2}\in({\bf A}^{*})^{\times}$,
$[h_{1}(a_{1}):h_{2}(a_{2})]=1$.

###### Proof.

Follows from Proposition 4.5.12 and Corollary 4.5.14.

∎

###  4.6 Words have finite width

######  Lemma 4.6.1.

For every $I\in\mathcal{I}_{K^{*}}$, the commutator subgroup of
$\operatorname{SL}_{3}(O^{*})[I]$ contains
$\operatorname{SL}_{3}(O^{*})[I^{2}]$.

###### Proof.

It is enough to show that there is a constant $C$ such that, for any ideal
$I\in\mathcal{I}_{K}$, any element of $\operatorname{SL}_{3}(O)[I^{2}]$ is a
product of $C$ commutators of elements of $\operatorname{SL}_{3}(O)[I]$. Let
$e_{i,j}(t)$ be the elementary matrix with $t$ in the $(i,j)$th entry. For an
ideal $I\in\mathcal{I}_{K}$, denote $X_{I}=\left\\{g^{-1}e_{1,3}(t)g\mid t\in
I,g\in\operatorname{SL}_{3}(O)\right\\}$, $Y_{I}=\left\\{[g,h]\mid
g,h\in\operatorname{SL}_{3}(O)[I]\right\\}$, and let $E(K,I)=\langle
X_{I}\rangle$.

The identity $[e_{1,2}(t),e_{2,3}(s)]=e_{1,3}(st)$ implies that, if $t\in
I^{2}$, then $e_{1,3}(t)\in Y_{I}$. By [52], there is a constant $C$ such
that, for every $I\in\mathcal{I}_{K}$, $E(K,I)=X_{I}^{C}$. Thus, for every
$I\in\mathcal{I}_{K}$,

$E(K,I^{2})\subseteq Y_{I}^{C}.$

By the Congruence Subgroup Property,
$E(K,I^{2})=\operatorname{SL}_{3}(O)[I^{2}]$, proving the claim. ∎

######  Proposition 4.6.2.

For every $I\in\mathcal{I}_{K^{*}}$, the commutator subgroup of
$\operatorname{Spin}_{f_{s}}(O^{*}))[I]$,
$\operatorname{Spin}_{f_{s}}(O^{*})[I]^{\prime}$, contains
$\operatorname{Spin}_{f_{s}}(O^{*})[I^{4}]$.

###### Proof.

Choosing a Chevalley basis, for any root $\alpha$ we get a 1-parameter
subgroup $x_{\alpha}:\mathbb{G}_{a}\rightarrow\operatorname{Spin}_{f_{s}}$.
Denote by $E(K^{*},I)$ the normal subgroup of
$\operatorname{Spin}_{f_{s}}(O^{*})$ generated by the root elements
$\left\\{x_{\alpha}(t)\mid t\in I,\alpha\in\Phi\right\\}$. We use the
following claims:

  1. (1)

If $\alpha,\beta$ are roots such that $\alpha+\beta$ is also a root, then
there is $\epsilon_{\alpha,\beta}\in\left\\{\pm 1\right\\}$ such that
$[x_{\alpha}(t),x_{\beta}(s)]=x_{\alpha+\beta}(\epsilon_{\alpha,\beta}ts)$.
Thus, $E(K^{*},I^{2})\subseteq\operatorname{Spin}_{f_{s}}(O^{*})[I]^{\prime}$.

  2. (2)

By [47, Theorem 4.8], there is a $O^{*}$-embedding
$\rho:\operatorname{SL}_{3}\rightarrow\operatorname{Spin}_{f_{s}}$ such that
$\operatorname{Spin}_{f_{s}}(O^{*})[I^{2}]\subseteq
E(K^{*},I)\cdot\rho(\operatorname{SL}_{3}(O^{*},I))$.

  3. (3)

By Lemma 4.6.1,
$\operatorname{SL}_{3}(O^{*},I^{2})\subseteq\operatorname{SL}_{3}(O^{*},I)^{\prime}$.

Thus,

$\operatorname{Spin}_{f_{s}}(O^{*})[I^{4}]\overset{\eqref{cond:CKP.CKP}}{\subseteq}E(K,I^{2})\cdot\rho\left(\operatorname{SL}_{3}(O^{*},I^{2})\right)\overset{\eqref{cond:CKP.roots},\eqref{cond:CKP.lemma}}{\subseteq}\operatorname{Spin}_{f_{s}}(O^{*})[I]^{\prime}.$

∎

######  Theorem 4.6.3.

Let $w$ be a non-trivial word. There exists a constant $N$ such that for every
$I\in\mathcal{I}_{n}$ $\operatorname{wid}_{w}(G(O)[I])\leq N$.

###### Proof assuming the results of section 6 and appendix A..

Assume otherwise. Corollary 3.2.15 implies that there exists a discrete
$\operatorname{Spin}_{q}(K^{*})$-metaplectic extension

$1\longrightarrow D\longrightarrow
E\overset{\rho}{\longrightarrow}\operatorname{Spin}_{f_{a}}(\mathbb{K})\longrightarrow
1$

which does not have a continuous splitting. Thus, there is a
$\psi(\operatorname{Spin}_{q}(K^{*}))$-metaplectic extension

$1\longrightarrow D\longrightarrow
E\overset{\psi\circ\rho}{\longrightarrow}\operatorname{Spin}_{f_{s}}(\mathbb{K})\longrightarrow
1$ (21)

which does not have a continuous splitting.

Theorem 4.3.2 and Corollary 4.5.15 imply that $\psi\circ\rho$ has a section
$\alpha$ over $\operatorname{Spin}_{f_{s}}(K^{*})$. By the definition of a
discrete $\psi(\operatorname{Spin}_{f_{a}}(K^{*}))$-metaplectic extension,
there exist $L\in\mathcal{I}_{K^{*}}$ and a continuous section $\beta$ over
$\operatorname{Spin}_{f_{s}}(\mathbb{O})[L]$.

Since
$\operatorname{Spin}_{f_{s}}(\mathbb{O})[L]\cap\operatorname{Spin}_{f_{s}}(K^{*})=\operatorname{Spin}_{f_{s}}(O^{*})[L]$,
Proposition 4.6.2 implies that
$\alpha\restriction_{\operatorname{Spin}_{f_{s}}(O^{*})[L^{4}]}=\beta\restriction_{\operatorname{Spin}_{f_{s}}(O^{*})[L^{4}]}$.
Lemma 2.6.4 applied to $\alpha$ and
$\beta\restriction_{\operatorname{Spin}_{f_{s}}(\mathbb{O})[L^{4}]}$ implies
that there is a continuous homomorphism
$\gamma:\operatorname{Spin}_{f_{s}}(\mathbb{K})\rightarrow E$ such that
$\gamma\restriction_{\operatorname{Spin}_{f_{s}}(K^{*})}=\alpha$ and
$\gamma\restriction_{\operatorname{Spin}_{f_{s}}(\mathbb{O})[L^{4}]}=\beta\restriction_{\operatorname{Spin}_{f_{s}}(\mathbb{O})[L^{4}]}$.
Since
$\operatorname{Spin}_{f_{s}}G(\mathbb{K})=\operatorname{Spin}_{f_{s}}(K^{*})\operatorname{Spin}_{f_{s}}(\mathbb{O})[L^{4}]$,
$\gamma$ must also be a section of Extension (21), a contradiction. ∎

###  4.7 Proof of Theorem 1.1.1

###### Proof of Theorem 1.1.1 assuming the results of section 6 and appendix A
.

For every non-central $g\in\operatorname{Spin}_{f_{a}}(O)$, denote

$d_{g}:=\operatorname{wid}_{\operatorname{Spin}_{f_{a}}(\mathbb{R})}(\operatorname{gcl}_{\operatorname{Spin}_{f_{a}}(\mathbb{R}}(g)).$

Corollary A.3.6 implies that there exists a constant $C_{1}$ such that for
every non-central $g\in\operatorname{Spin}_{f_{a}}(O)$, there exists
$I\in\mathcal{I}_{K}$ such that

$\\{[g_{1},g_{2}]\mid
g_{1},g_{2}\in\operatorname{Spin}_{f_{a}}(O)[I]\\}\subseteq\operatorname{gcl}_{\operatorname{Spin}_{f_{a}}(O)}(g)^{C_{1}d_{g}}.$

Let $w$ be the commutator word $xyx^{-1}y^{-1}$. Theorem 4.6.3 applied to the
$w$ implies that there exists a constant $C_{2}$, such that for every for
every $I\in\mathcal{I}_{K}$,
$\operatorname{wid}_{w}(\operatorname{Spin}_{f_{a}}(O))\leq C_{2}$.

Denote $C_{3}:=C_{1}C_{2}$, by the Congruence Subgroup Property, for every
$I\in\mathcal{I}_{K}$, there exists $J\in\mathcal{I}_{K}$ such that
$\operatorname{Spin}_{f_{a}}(O)[J]\subseteq\big{(}\operatorname{Spin}_{f_{a}}(O)\big{)}_{w}$.
We get that for every non-central $g\in\operatorname{Spin}_{f_{a}}(O)$, there
exists $J\in\mathcal{I}_{K}$ such that
$\operatorname{Spin}_{f_{a}}(O)[J]\subseteq
w(\operatorname{gcl}_{\operatorname{Spin}_{f_{a}}(O)}(g))^{C_{3}d_{g}}$.

[2, Lemma 3.20] and [2, Lemma 3.22] (see the proof of Lemma A.3.4) imply that
there exists a constant $C_{4}$ such that for every
$g\in\operatorname{Spin}_{f_{a}}(O)$ and every $J\in\mathcal{I}_{K}$,

$\operatorname{wid}_{\operatorname{Spin}_{f_{a}}(O)/\operatorname{Spin}_{f_{a}}(O[J])}(\operatorname{gcl}_{\operatorname{Spin}_{f_{a}}(O)/\operatorname{Spin}_{f_{a}}(O)[J]}(g\operatorname{Spin}_{f_{a}}(O)[J])\leq
C_{4}.$

Since for every $g\in\operatorname{Spin}_{f_{a}}(O)$, $d_{g}\geq 1$, we can
take $C:=C_{3}+C_{4}$. ∎

##  5 Existence of approximations

###  5.1 One-dimensional tori

######  Definition 5.1.1.

The standard form $\mathbb{G}_{m}$ of the split 1-dimensional $O$-torus is the
zero locus of the po-klynomial $xy-1$ in $\mathbb{A}_{O}^{2}$. Given a prime
$P\in\mathcal{P}$ and an element $(x,x^{-1})\in\mathbb{G}_{m}(K_{P})$, let
$\operatorname{val}_{P}((x,x^{-1}))=\min\left\\{\operatorname{val}_{P}(x),\operatorname{val}_{P}(x^{-1})\right\\}$.

######  Definition 5.1.2.

Given $t\in O\smallsetminus\left\\{0\right\\}$, let
$O_{K\left(\sqrt{-t}\right)}$ be the ring of integers of
$K\left(\sqrt{-t}\right)$, let $\widetilde{T_{t}}$ be the restriction of
scalars of $\mathbb{G}_{m}$ from $O_{K\left(\sqrt{-t}\right)}$ to $O$, and let
$T_{t}$ be the kernel of the norm homomorphism of $O$-group schemes
$\widetilde{T_{t}}\rightarrow\mathbb{G}_{m}$.

We will use the following properties of $T_{t}$:

######  Proposition 5.1.3.

  1. (1)

$T_{t}$ is $K$-isomorphic to the zero locus of $x^{2}+ty^{2}-1$ in
$\mathbb{A}_{K}^{2}$. In particular, $T_{t}$ has weak approximation and
$T_{t}(K)=\left\\{x\in K(\sqrt{-t})\mid N_{L/K}(x)=1\right\\}$.

  2. (2)

$T_{t}(O)$ is the set of norm-one elements in the ring of integers of
$K(\sqrt{-t})$.

  3. (3)

If $P\in\mathcal{P}_{K}$ is ramified or inert in $K(\sqrt{-t})$, then
$T_{t}(K_{P})=T_{t}(O_{P})$.

  4. (4)

If $P\in\mathcal{P}_{K}$ is split in $K(\sqrt{-t})$, then there is an
$O_{P}$-isomorphism $T_{t}\cong\mathbb{G}_{m}$. Such an isomorphism is unique
up to composition with the flip $(x,y)\mapsto(y,x)$. Identifying $T_{t}(K)$
with the elements of norm 1 in $K(\sqrt{-t})$, we have
$\operatorname{val}_{P}(z)=\min\left\\{\operatorname{val}_{\mathfrak{p}}(z),-\operatorname{val}_{\mathfrak{p}}(z)\right\\}$,
for any $z\in T_{t}(K)$, where $\mathfrak{p}$ is a prime of $K(\sqrt{-t})$
lying over $P$.

###### Proof.

Recall that $K=\mathbb{Q}$ and $O=\mathbb{Z}[\frac{1}{2}]$. Write
$t=s^{2}\cdot r$, where $s,r\in O$ and $r$ is square-free. Since
$O=\mathbb{Z}[\frac{1}{2}]$ and $O_{K\left(\sqrt{-t}\right)}=O\cdot 1\oplus
O\cdot\sqrt{-r}$, we get that $O_{K\left(\sqrt{-t}\right)}\otimes R=R\cdot
1\oplus R\cdot\sqrt{-r}$, for every $O$-algebra $R$. It follows that
$T_{t}(R)=\left\\{(x,y)\mid x^{2}+ry^{2}=1\right\\}$. All statements follow
from this. ∎

######  Definition 5.1.4.

Let $t\in O\smallsetminus\left\\{0\right\\}$.

  1. (1)

If $P\in\mathcal{P}_{K}$ splits in $K(\sqrt{-t})$, a trivialization of $T_{t}$
over $P$ is a choice of a $K_{P}$-isomorphism
$\rho_{P}:T_{t}\rightarrow\mathbb{G}_{m}$. Abusing notation, we also denote
$\rho_{P}:T_{t}(K_{P})\rightarrow K_{P}^{\times}$ the composition of
$\rho_{P}$ and the projection to the first coordinate.

  2. (2)

A trivialization of $T_{t}$ over $P$ is the same information as a choice of a
square root of $-t$ in $O_{P}$. Abusing notation, we denote this square root
by $\rho_{P}(\sqrt{-t})$.

  3. (3)

Let $\mathcal{Q}\subseteq\mathcal{P}_{K}$. A trivialization of $T_{t}$ over
$\mathcal{Q}$ is a set of trivializations $\left\\{\rho_{P}\mid
P\in\mathcal{Q}\right\\}$, where $\rho_{P}$ is a trivialization of $T_{t}$
over $P$.

######  Remark 5.1.5.

Let $t\in O\smallsetminus\left\\{0\right\\}$ and let
$L=K\left(\sqrt{-t}\right)$. A trivialization $\rho_{P}$ of $T_{t}$ defines an
embedding $L\rightarrow K_{P}$ and, thus, a prime $\mathfrak{p}\triangleleft
O_{L}$.

###  5.2 Approximation in tori

######  Definition 5.2.1.

Recall that $K=\mathbb{Q}$. Let $K_{+}\subseteq K$ be the set of positive
rationals and let $O_{+}=O\cap K_{+}$.

######  Lemma 5.2.2.

Let $a_{1},a_{2}\in K^{\times}$ and let $I\in\mathcal{I}_{K}$. Then there are

  1. (1)

An element $t\in O_{+}$,

  2. (2)

Elements $z_{1},z_{2}\in T_{t}(K)$,

  3. (3)

Finite sets $\mathcal{R}_{1},\mathcal{R}_{2}\subseteq\mathcal{P}_{K}$,

  4. (4)

A trivialization $\rho$ of $T_{t}$ over
$\mathcal{P}_{K}(I)\cup\mathcal{R}_{1}\cup\mathcal{R}_{2}$,

such that

  1. (5)

$\rho_{P}(z_{i})a_{i}^{-1}\in 1+P^{\operatorname{val}_{P}(I)}O_{P}$, for every
$1\leq i\leq 2$ and every $P\in\mathcal{P}_{K}(I)$.

  2. (6)

The sets
$\mathcal{P}_{K}(I),\mathcal{R}_{1},\mathcal{R}_{2},\mathcal{P}_{K}(tO)$ are
pairwise disjoint.

  3. (7)

$z_{i}\in T_{t}(O_{P})$, for every
$P\notin\mathcal{P}_{K}(I)\cup\mathcal{R}_{i}$.

###### Proof.

Choose an element $t\in O_{+}$ prime to $I$ such that for every
$P\in\mathcal{P}_{K}(I)$, $-t$ has a square root in $O_{P}$. (for example,
choose $t$ such that $t\equiv-1\text{ (mod I)}$). Choosing such a root for
every $P\in\mathcal{P}_{K}(I)$ gives a trivialization $\rho$ of $T_{t}$ over
$\mathcal{P}(I)$.

Since $\mathcal{P}_{K}(I)$ and $\mathcal{P}_{K}(tO)$ are disjoint, the weak
approximation theorem in $T_{t}$ implies that there is an element $z_{1}\in
T_{t}(K)$ such that:

  * •

For every $P\in\mathcal{P}_{K}(I)$, $\rho_{P}(z_{1})a_{1}^{-1}\in
1+P^{\operatorname{val}_{P}(I)}O_{P}$

  * •

For every $P\in\mathcal{P}_{K}(tO)$, $z_{1}\in T_{t}(O_{P})$.

Let
$\mathcal{R}_{1}=\left\\{P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(I)\mid
z_{1}\notin T_{t}(O_{P})\right\\}$. If $P\in\mathcal{R}_{1}$, then $P$ splits
in $K(\sqrt{-t})$ by 5.1.33. It follows that $\rho$ can be extended to a
trivialization of $T_{t}$ over $\mathcal{P}(I)\cup\mathcal{R}_{1}$.

The finite sets $\mathcal{P}_{K}(I)$ and
$\mathcal{R}_{1}\cup\mathcal{P}_{K}(t)$ are disjoint, so the weak
approximation theorem in $T_{t}$ implies that there is an element $z_{2}\in
T_{t}(K)$ such that:

  * •

For every $P\in\mathcal{P}(I)$, $\rho_{P}(z_{2})a_{2}^{-1}\in
1+P^{\operatorname{val}_{P}(I)}O_{P}$.

  * •

For every $P\in\mathcal{P}(tO)\cup\mathcal{R}_{1}$, $z_{2}\in T_{t}(O_{P})$.

As before, let
$\mathcal{R}_{2}=\left\\{P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(I)\mid
z_{2}\notin T_{t}(O_{P})\right\\}$ and extend $\rho$ to a trivialization over
$\mathcal{P}_{K}(I)\cup\mathcal{R}_{1}\cup\mathcal{R}_{2}$. The claims of the
lemma are satisfied. ∎

######  Lemma 5.2.3.

Let $s_{1},\ldots,s_{r}\in\mathbb{Z}$ elements which are not prefect squares,
$x\in\mathbb{Z}$ and $m\geq 1$ a positive integer. Then there exists $k$ such
that:

  1. (1)

$t_{k}:=2^{2mk}-x$ belongs to $O_{+}$ and is not a prefect square.

  2. (2)

For every $1\leq i\leq r$, $\sqrt{s_{i}}\notin K(\sqrt{-t_{k}})$.

###### Proof.

If $k$ is large enough then $t_{k}$ is a positive integer so $t_{k}\in O_{+}$
. Assume that $2^{k}>|x|$ then

$(2^{k}-1)^{2}=2^{2k}-2\cdot 2^{k}+1<t_{k}=2^{2k}-x<2^{2k}+2\cdot
2^{k}+1=(2^{k}+1)^{2}$

so $t_{k}$ is not a perfect square.

Let $s\in\mathbb{Z}$ be an element which is not a perfect square. Then
$\sqrt{s}\in K(-\sqrt{t_{k}})$ if and only if $-st_{k}$ is a prefect square in
$\mathbb{Z}$.

Choose $k_{0}$ such that,

$|\sqrt{st_{k_{0}}}|>(2^{mr}+1)|sx|+1.$

Let $k\geq k_{0}$ and assume that, $-st_{k}$ is a perfect square, say,
$b^{2}=-st_{k}$ where $b$ is a positive integer. For every $1\leq i\leq r$,

$(2^{mi}b)^{2}\neq st_{k+i}=2^{2mi}st_{k}+2^{2mi}sx-sx.$

so, by the choice of $k_{0}$ and the fact that $t_{k}\geq t_{k_{0}}$,

$(2^{mi}b-1)^{2}=2^{2mi}b^{2}-2^{mi}b+1<st_{k+i}<2^{2mi}b^{2}+2^{mi}b+1=(2^{mk}b+1)^{2}.$

It follows that if $-st_{k}$ is a perfect square for some $k\geq k_{0}$, then
for every $1\leq i\leq r$, $-st_{k+i}$ is not a perfect square.

Choose $k_{1}$ such that for every $1\leq j\leq r$,

$|\sqrt{s_{j}t_{k_{1}}}|>(2^{mr}+1)|s_{j}x|+1.$

Then there exists $0\leq i\leq r$ such that for every $1\leq j\leq r$,
$-s_{j}t_{k_{1}+i}$ is not a perfect square.

∎

######  Lemma 5.2.4.

Let $I\in\mathcal{I}_{K}$, $a\in K^{\times}$ be such that $a\in
O_{P}^{\times}$, for all $P\in\mathcal{P}_{K}(I)$, and $s_{1},\ldots,s_{r}\in
O$ elements which are not prefect squares . Then there are

  1. (1)

An element $t\in O_{+}$ which is not a square,

  2. (2)

A trivialization $\rho$ of $T_{t}$ over $\mathcal{P}_{K}(I)$,

  3. (3)

An element $z\in T_{t}(O)$,

such that for every $1\leq i\leq r$, $\sqrt{s_{i}}\notin K(\sqrt{-t})$ and for
all $P\in\mathcal{P}(I)$, $\rho_{P}(z)a^{-1}\in
1+P^{\operatorname{val}_{P}(I)}O_{P}$.

###### Proof.

If $a=\pm 1$, take any $t\in O_{+}$ such that $-t$ has a square root modulo
$I$, $-s_{1}t,\ldots,-s_{r}t$ are not prefect squares and take $z:=\pm 1\in
T_{t}(O)$.

In the rest of the proof, assume that $a\neq\pm 1$. We will find $t,x\in O$
and $k\in\mathbb{N}$ such that $t$ and
$z=\frac{x}{2^{k}}+\frac{1}{2^{k}}\sqrt{-t}$ satisfy the requirements of the
lemma.

Since $a\neq\pm 1$, there is $N\in\mathbb{N}$ such that for every
$P\in\mathcal{P}_{K}(I)$, $a\not\equiv\pm 1\text{ (mod $P^{N}$)}$ and
$N>\operatorname{val}_{P}(I)$. Since $a$ is invertible in $O_{P}$ and
$O=\mathbb{Z}[\frac{1}{2}]$, there exists $x\in\mathbb{Z}$ such that
$x\equiv\frac{a+a^{-1}}{2}\text{ (mod $I^{2N}$)}$. By Lemma 5.2.3, there
exists $k\geq 1$ such that:

  * •

$2^{k}=1\text{(mod $I^{2N}$)}$.

  * •

$t:=2^{2k}-x^{2}$ belongs to $O_{+}$ and is not a prefect square.

  * •

For every $1\leq i\leq r$, $\sqrt{s_{i}}\notin K(\sqrt{-t})$.

We have
$\left(\frac{x}{2^{k}}\right)^{2}+\left(\frac{1}{2^{k}}\right)^{2}t=\frac{x^{2}}{2^{2k}}+\frac{t}{2^{2k}}=1$,
by the definition of $t$. Since $O=\mathbb{Z}[\frac{1}{2}]$,
$z=\frac{x}{2^{k}}+\frac{1}{2^{k}}\sqrt{-t}\in T_{t}(O)$.

Let $P\in\mathcal{P}_{K}(I)$. Then

$\left(\frac{a-a^{-1}}{2}\right)^{2}=\left(\frac{a+a^{-1}}{2}\right)^{2}-1\equiv
x^{2}-2^{2k}=-t\text{ (mod $P^{2N}$)}.$

Since $a\not\equiv\pm 1\text{ (mod $P^{N}$)}$, the following properties hold:

  * •

$x\not\equiv\pm 1\text{ (mod $P^{N}$)}$ so $t\equiv 1-x^{2}\not\equiv 0\text{
(mod $P^{N}$)}$.

  * •

$\operatorname{val}_{P}\left(\frac{a-a^{-1}}{2}\right)<N$

It follows that $\operatorname{val}_{P}(-t)$ is even and smaller than $2N$,
and that the leading coefficient of $-t$ is a square in $O/P$. Therefore, $-t$
has a square root $\zeta_{P}$ in $O_{P}$ such that
$\operatorname{val}_{P}(\zeta_{P})<N$ and

$\left(\frac{a-a^{-1}}{2\zeta_{P}}\right)^{2}\equiv 1\text{ (mod
$P^{2N-2\operatorname{val}_{P}(\zeta_{P})}$)}.$

Since $O=\mathbb{Z}[\frac{1}{2}]$, $P$ is odd so

$\left(\frac{a-a^{-1}}{2\zeta_{P}}\right)\equiv\pm 1\text{ (mod
$P^{2N-2\operatorname{val}_{P}(\zeta_{P})}$)}$

and

$\left(\frac{a-a^{-1}}{2}\right)\equiv\pm\zeta_{P}\text{ (mod
$P^{2N-\operatorname{val}_{P}(\zeta_{P})}$)}$

Since $2N-\operatorname{val}_{P}(\zeta_{P})\geq N$, by replacing $\zeta_{P}$
with $-\zeta_{P}$ if necessary, we can assume that

$\zeta_{P}\equiv\frac{a-a^{-1}}{2}\text{ (mod $P^{N}$)}.$

Define a trivialization $\rho$ of $T_{t}$ over $\mathcal{P}_{K}(I)$ by for
every $P\in\mathcal{P}_{K}(I)$, $\rho_{P}\left(\sqrt{-t}\right):=\zeta_{P}$.
Since $z=\frac{x}{2^{k}}+\frac{1}{2^{k}}\sqrt{-t})$,

$\rho_{P}(z)-a\equiv\frac{1}{2^{k}}\left(x+\frac{a-a^{-1}}{2}\right)-a\equiv
x+\frac{a-a^{-1}}{2}-a=x-\frac{a+a^{-1}}{2}\equiv 0\text{ (mod $I$)}.$

Since $a\in O_{P}^{\times}$, for every $P\in\mathcal{P}_{K}(I)$,

$\operatorname{val}_{P}(\rho_{P}(z)a^{-1}-1)=\operatorname{val}_{P}(\rho_{P}(z)-a)\geq\operatorname{val}_{P}(I)$

and the result follows. ∎

######  Lemma 5.2.5.

Let $\mathcal{R}\subseteq\mathcal{P}_{K}$ be a finite set of primes, let $t\in
O\smallsetminus\left\\{0\right\\}$, let $\rho$ be a trivialization of $T_{t}$
over $\mathcal{R}$, let $a\in K^{\times}$, and let
$\mathcal{Q}\subseteq\mathcal{P}_{k}$. Assume that, for each cofinite subset
$\mathcal{Q}^{\prime}\subseteq\mathcal{Q}$, the set
$\left\\{Q\in\mathcal{P}_{K\left(\sqrt{-t}\right)}\mid Q\cap
O\in\mathcal{Q}^{\prime}\right\\}$ generates the ideal class group
$\operatorname{Cl}_{K\left(\sqrt{-t}\right)}$ of $K\left(\sqrt{-t}\right)$.
Then there is an element $z\in T_{t}(K)$ such that

  1. (1)

$\rho_{P}(z)a^{-1}\in O_{P}^{\times}$, for all $P\in\mathcal{R}$.

  2. (2)

If $Q\in\mathcal{P}_{K}\smallsetminus\mathcal{R}$ and $z\notin T_{t}(O_{Q})$,
then $Q\in\mathcal{Q}$ and $\operatorname{val}_{Q}(z)=-1$.

###### Proof.

Denote $L=K\left(\sqrt{-t}\right)$, denote the ring of integers of $L$ by
$O_{L}$, and denote the non-trivial element in $\operatorname{Gal}(L/K)$ by
$\sigma$. By assumption, every $P\in\mathcal{R}$ splits in $L$. We identify
$T_{t}(K)$ with the group of elements of norm 1 in $L$ as in Proposition
5.1.31.

We divide the argument into the following steps:

Step 1: We prove the claim assuming $\mathcal{R}=\left\\{P\right\\}$ is a
singleton and $\operatorname{val}_{P}(a)=1$. The trivialization $\rho_{P}$
determines an embedding $\rho_{P}:L\hookrightarrow K_{P}$ and, thus, a prime
$\mathfrak{p}\triangleleft O_{L}$. Since $P$ splits in $L$,
$\mathfrak{p}\neq\sigma(\mathfrak{p})$.

By the assumption on $\mathcal{Q}$ and the finiteness of
$\operatorname{Cl}_{K\left(\sqrt{-t}\right)}$, there are prime ideals
$\mathfrak{q}_{1},\ldots,\mathfrak{q}_{m}\triangleleft O_{L}$ such that

  1. (1)

$\mathfrak{q}_{i}\cap O\in\mathcal{Q}$ and $\mathfrak{q}_{i}\cap O\neq P$, for
every $i\in\left\\{1,\ldots,m\right\\}$.

  2. (2)

$\mathfrak{q}_{i}\cap O\neq\mathfrak{q}_{j}\cap O$, for every
$i,j\in\left\\{1,\ldots,m\right\\},i\neq j$.

  3. (3)

There is an element $b\in L$ such that
$\mathfrak{p}\cdot\mathfrak{q}_{1}\cdots\mathfrak{q}_{m}=bO_{L}$.

Denote $z=\frac{b}{\sigma(b)}\in L$. Clearly, $N_{L/K}(z)=1$, so $z\in
T_{t}(K)$. By Item 2, $\operatorname{val}_{\sigma(\mathfrak{p})}(b)=0$. We
have

$\operatorname{val}_{P}(\rho_{P}(z))=\operatorname{val}_{\mathfrak{p}}(z)=\operatorname{val}_{\mathfrak{p}}(b)-\operatorname{val}_{\mathfrak{p}}(\sigma(b))=\operatorname{val}_{\mathfrak{p}}(b)-\operatorname{val}_{\sigma(\mathfrak{p})}(b)=1,$

so $\rho_{P}(z)a^{-1}\in O_{P}^{\times}$.

Let $Q\in\mathcal{P}$ be a prime different from $P$, and assume that $z\notin
T_{t}(O_{Q})$. By Item 3 of Proposition 5.1.3, $Q$ splits in $L$. Assume first
that $Q$ is different from any of the ideals $\mathfrak{q}_{i}\cap O$ and let
$\mathfrak{q}\in\mathcal{P}_{L}$ be an ideal above $Q$. By Item 4 of
Proposition 5.1.3,

$\operatorname{val}_{Q}(z)=\min\left\\{\operatorname{val}_{\mathfrak{q}}(z),\operatorname{val}_{\mathfrak{q}}(z^{-1})\right\\}=0$

and $z\in T_{t}(O_{Q})$, a contradiction.

Now suppose that there is an index $i$ such that $Q=\mathfrak{q}_{i}\cap O$.
By Item 4 of Proposition 5.1.3 and Item 2,
$\operatorname{val}_{Q}(z)=\min\left\\{\operatorname{val}_{\mathfrak{q}_{i}}(z),\operatorname{val}_{\mathfrak{q}_{i}}(z^{-1})\right\\}=-1$,
and the lemma follows.

Step 2: We prove the claim assuming $\mathcal{R}=\left\\{P\right\\}$ is a
singleton. Denote $m=|\operatorname{val}_{P}(a)|$. Inductively find elements
$z_{1},\ldots,z_{m}\in T_{t}(K)$ such that
$\operatorname{val}_{P}(\rho_{P}(z_{i}))=1$, such that Condition (2) of the
lemma holds, and such that the sets
$\mathcal{Q}_{i}:=\left\\{Q\in\mathcal{P}\mid z_{i}\notin
T_{t}(O_{Q})\right\\}$ are disjoint. To find the $i$-th element, apply Step 1
to the set $\mathcal{Q}\smallsetminus\cup_{j<i}\mathcal{Q}_{j}$. Finally, take
$z=z_{1}\cdots z_{m}$ or $z=z_{1}^{-1}\cdots z_{m}^{-1}$.

Step 3: We prove the claim in its full generality. Let
$\mathcal{R}=\left\\{P_{1},\ldots,P_{m}\right\\}$. By replacing $\mathcal{Q}$
with $\mathcal{Q}\smallsetminus\mathcal{R}$, we can assume that
$\mathcal{Q}\cap\mathcal{R}=\emptyset$. As before, using Step 2, we can
inductively find elements $z_{1},\ldots,z_{m}\in T_{t}(K)$ such that
$\operatorname{val}_{P}(\rho_{P}(z_{i}))=\operatorname{val}_{P}(\rho_{P}(a))$,
such that Condition (2) of the lemma holds, and such that the sets
$\mathcal{Q}_{i}:=\left\\{Q\in\mathcal{P}\mid z_{i}\notin
T_{t}(O_{Q})\right\\}$ are disjoint. The element $z=z_{1}\cdots z_{m}$
satisfies the conditions of the lemma. ∎

The proof of the next lemma is almost identical to the proof of Lemma 5.2.5
and is omitted. For a field $L$, denote by $\operatorname{Cl}_{L}^{2}$ the
subgroup of $\operatorname{Cl}_{L}$ consisting of all squares.

######  Lemma 5.2.6.

Let $\mathcal{R}\subseteq\mathcal{P}$ be a finite set of primes, let $t\in
O\smallsetminus\left\\{0\right\\}$, let $\rho$ be a trivialization of $T_{t}$
over $\mathcal{R}$, let $a\in K^{\times}$, and let
$\mathcal{Q}\subseteq\mathcal{P}$ be an infinite set of primes. Assume that,
for each cofinite subset $\mathcal{Q}^{\prime}\subseteq\mathcal{Q}$, the set
$\left\\{Q\in\mathcal{P}_{K\left(\sqrt{-t}\right)}\mid Q\cap
O\in\mathcal{Q}^{\prime}\right\\}$ generates a subgroup which contains the
subgroup $\\{x^{8}\mid x\in\operatorname{Cl}_{K\left(\sqrt{-t}\right)}\\}$ of
$\operatorname{Cl}_{K\left(\sqrt{-t}\right)}$. Then there is an element $z\in
T_{t}(K)$ such that

  1. (1)

$\rho_{P}(z)a^{-8}\in O_{P}^{\times}$, for all $P\in\mathcal{R}$.

  2. (2)

If $Q\in\mathcal{P}\smallsetminus\mathcal{R}$ and $z\notin T_{t}(O_{Q})$, then
$Q\in\mathcal{Q}$ and $\operatorname{val}_{Q}(z)=-1$.

###  5.3 From tori to $\operatorname{Spin}$

######  Lemma 5.3.1.

Let $L$ be a non-Archimedean local field of characteristic zero. let $(V,q)$
be a non-degenerate quadratic space of dimension at least 5 over $L$, and let
$v_{1},v_{2}\in V$. Assume that $q(v_{1})=q(v_{2})\neq 0$. Then there is an
element $g\in\operatorname{Spin}_{q}(L)$ such that $g(v_{1})=v_{2}$.

###### Proof.

By Witt’s lemma there exists $\sigma\in\mathrm{O}_{q}(L)$ such that
$\sigma(v_{1})=v_{2}$. Let $\tau=1$ if $\sigma\in\operatorname{SO}_{q}(L)$ and
otherwise choose arbitrary
$\tau\in\mathrm{O}_{q\restriction_{(Lv_{2})^{\perp}}}(L)\smallsetminus\operatorname{SO}_{q\restriction_{(Lv_{2})^{\perp}}}(L)$.
By replacing $\sigma$ with $\tau\sigma$ we can assume that
$\sigma\in\operatorname{SO}_{q}(L)$. Let $\alpha$ be the spinor norm of
$\sigma$. Since every regular quadratic space of dimension at least 4 over $L$
is universal, there exists
$\tau^{\prime}\in\operatorname{SO}_{q\restriction_{(Lv_{2})^{\perp}}}(L)$ such
that the spinor norm of $\tau^{\prime}$ is equal to $\alpha$. By replacing
$\sigma$ with $\tau^{\prime}\sigma$, we can assume that
$\sigma\in\operatorname{SO}_{q}(L)$ has spinor norm 1 and thus lifts to an
element of $\operatorname{Spin}_{q}(L)$. ∎

######  Lemma 5.3.2.

Let $t\in O_{+}$, let $P\in\mathcal{P}_{K}$, $U\subseteq K^{2n}$ a regular
five dimensional subspace of $(K^{2n},f_{s})$ and let $v_{1},v_{2}\in
O_{P}^{n}\cap U^{\perp}$ be orthogonal of norms
$f_{s}(v_{1})=1,f_{s}(v_{2})=t$. Let
$\varphi:T_{t}\rightarrow\operatorname{Spin}_{f_{s}}$ be the $K_{P}$ morphism
defined by $\varphi\left(x+y\sqrt{-t}\right)=x+yv_{1}v_{2}$. Assume that $P$
is either ramified or inert in $K\left(\sqrt{-t}\right)$. Then there is
$\ell\in\operatorname{Spin}_{f_{s}\restriction U^{\perp}}(K_{P})$ such that
$\ell\varphi\left(T_{t}(K_{P})\right)\ell^{-1}\subseteq\operatorname{Spin}_{f_{s}}(O_{P})$.

###### Proof.

Let $\varpi\in O_{P}$ be a uniformizer and let $n\in\mathbb{N}$ be such that
$\operatorname{val}_{P}(t)=2n$ (if $P$ is inert) or
$\operatorname{val}_{P}(t)=2n+1$ (if $P$ is ramified). The six dimensional
subspace $U+Lv_{1}$ is contained in a twelve dimensional subspace $W$ such
that $v_{2}\in W^{\perp}$ and $W\cap O^{2n}$ is regular over $O$. Since
$\dim(f_{s})-12\geq 8\geq 4$, $f_{s}\restriction_{W^{\perp}}$ is universal
over $O_{p}$ so there exists $u\in O_{P}^{n}\cap W^{\perp}$ such that
$f_{s}(u)=\frac{t}{\varpi^{2n}}$. By Lemma 5.3.1, there is an element
$\ell\in\operatorname{Spin}_{q\restriction W^{\perp}}(K_{P})$ such that and
$\ell(v_{2})=\varpi^{n}u$. Clearly, $\ell(v_{1})=v_{1}$. If $z=x+y\sqrt{-t}\in
T_{t}(K_{P})$, then $\operatorname{val}_{P}(x)=0$ and
$\operatorname{val}_{P}(y)\geq-n$. Therefore,
$\ell\varphi(z)\ell^{-1}=x+y\varpi^{n}v_{1}u\in\operatorname{Spin}_{q}(O_{P})$.
∎

In the next lemma, we use the coroots $h_{1},h_{2}$ from §4.2.

######  Lemma 5.3.3.

Let $t\in O_{+}$, let $z_{1},z_{2}\in T_{t}(K)$, and let $\rho$ be a
trivialization of $T_{t}$ over a finite subset $\mathcal{Q}$ of
$\mathcal{P}_{K}$. There are elements
$g_{1},g_{2}\in\psi\left(\operatorname{Spin}_{f_{a}}(K)\right)$,
$b_{1},b_{2}\in\mathbf{A}^{\times}$, and
$\ell\in\operatorname{Spin}_{f_{s}}(\mathbf{A})$ such that, for every $1\leq
i\leq 2$,

  1. (1)

For every $P\in\mathcal{Q}$, $\left(\ell
g_{i}\ell^{-1}\right)_{P}=h_{i}(b_{i})_{P}$.

  2. (2)

For every $P\in\mathcal{Q}$, $(b_{i})_{P}=\rho_{P}(z_{i})$.

  3. (3)

$\ell g_{i}\ell^{-1}\in\operatorname{Spin}_{f_{s}}(O_{P})$, for every
$P\in\mathcal{P}_{K}$ such that $z_{i}\in T_{t}(O_{P})$.

###### Proof.

Let $b_{i}\in\mathbf{A}^{\times}$ be the element such that

$(b_{i})_{P}=\begin{cases}\rho_{P}(z_{i})&P\in\mathcal{Q}\\\
1&P\notin\mathcal{Q}\end{cases}.$

Clearly, Claim (2) holds.

Since $\dim(f_{a})\geq 20$, by Lagrange’s Four Squares Theorem, there are
$f_{a}$-orthogonal vectors $v_{1},v_{2},v_{3},u_{1},u_{2},u_{3}\in O^{n}$ such
that $f_{a}(v_{1})=f_{a}(v_{2})=f_{a}(v_{3})=1$ and
$f_{a}(u_{1})=f_{a}(u_{2})=f_{a}(u_{3})=t$.

Choose elements $z_{i}^{\prime}=x_{i}^{\prime}+y_{i}^{\prime}\sqrt{-t}\in
T_{t}(\overline{K})$ such that $(z_{i}^{\prime})^{2}=z_{i}$. Define

$g_{1}=\left(x_{1}^{\prime}+y_{1}^{\prime}\psi(v_{1})\psi(u_{1})\right)\cdot\left(x_{1}^{\prime}-y_{1}^{\prime}\psi(v_{2})\psi(u_{2})\right)$

and

$g_{2}=\left(x_{2}^{\prime}+y_{2}^{\prime}\psi(v_{2})\psi(u_{2})\right)\cdot\left(x_{2}^{\prime}-y_{2}^{\prime}\psi(v_{3})\psi(u_{3})\right).$

By the last paragraph of §4.2,
$g_{1},g_{2}\in\psi\left(\operatorname{Spin}_{f_{a}}(K)\right)$.

Let
$\mathcal{R}_{1}=\left\\{P\in\mathcal{P}_{K}\smallsetminus\mathcal{Q}\mid\operatorname{val}_{P}(x_{1}),\operatorname{val}_{P}(y_{1}),\operatorname{val}_{P}(x_{2}),\operatorname{val}_{P}(y_{2})\geq
0\right\\}$. Then $\mathcal{R}$ is cofinite in $\mathcal{P}_{K}$. For each
$P\in\mathcal{R}_{1}$, $g_{1},g_{2}\in\operatorname{Spin}_{f_{s}}(O_{P})$. For
every $P\in\mathcal{R}_{1}$, denote
$\ell_{P}:=1\in\operatorname{Spin}_{f_{s}}(K_{P})$.

Let
$\mathcal{R}_{2}=\left\\{P\in\mathcal{P}\smallsetminus\mathcal{R}_{1}\mid\text{$P$
is either ramified or inert in $K\left(\sqrt{-t}\right)$}\right\\}$. Then
$\mathcal{R}_{2}$ is finite. For every $P\in\mathcal{R}_{2}$, Lemma 5.3.2
implies that there are elements
$\ell_{1},\ell_{2},\ell_{3}\in\operatorname{Spin}_{f_{s}}(K_{P})$ such that

  * •

$\ell_{3}$ fixes $\psi(v_{1}),\psi(u_{1}),\psi(v_{2}),\psi(u_{2})$ and
conjugates $x_{2}-y_{2}\psi(v_{3})\psi(u_{3})$ into
$\operatorname{Spin}_{f_{s}}(O_{P})$.

  * •

$\ell_{2}$ fixes
$\psi(v_{1}),\psi(u_{1}),\ell_{3}\left(\psi(v_{3})\right),\ell_{3}\left(\psi(u_{3})\right)$
and conjugates both $x_{1}-y_{1}\psi(v_{2})\psi(u_{2})$ and
$x_{2}+y_{2}\psi(v_{2})\psi(u_{2})$ into $\operatorname{Spin}_{f_{s}}(O_{P})$.

  * •

$\ell_{1}$ fixes
$\ell_{2}\left(\psi(v_{2})\right),\ell_{2}\left(\psi(u_{2})\right),\ell_{3}\left(\psi(v_{3})\right),\ell_{3}\left(\psi(u_{3})\right)$
and conjugates $x_{1}+y_{1}\psi(v_{1})\psi(u_{1})$ into
$\operatorname{Spin}_{f_{s}}(O_{P})$.

For every $P\in\mathcal{R}_{2}$, denote $\ell_{P}:=\ell_{1}\ell_{2}\ell_{3}$,
we get that
$\ell_{P}g_{1}\ell_{P}^{-1},\ell_{P}g_{2}\ell_{P}^{-1}\in\operatorname{Spin}_{f_{s}}(O_{P})$.

Let
$\mathcal{R}_{3}=\left\\{P\in\mathcal{P}\smallsetminus\mathcal{R}_{1}\mid\text{$P$
splits in $K\left(\sqrt{-t}\right)$}\right\\}$. Then $\mathcal{R}_{3}$ is
finite and contains $\mathcal{Q}$. Extend $\rho$ to $\mathcal{R}_{3}$. By
applying Lemma 5.3.2 several times, we see that for each
$P\in\mathcal{R}_{3}$, there is an element $\ell_{P}$ such that
$\ell_{P}(\psi(v_{1}))=e_{1},\ell_{P}(\psi(v_{2}))=e_{3},\ell_{P}(\psi(v_{3}))=e_{5},\ell_{P}(\psi(u_{1}))=\rho_{P}(\sqrt{-t})e_{2},\ell_{P}(\psi(u_{2}))=\rho_{P}(\sqrt{-t})e_{4},\ell_{P}(\psi(u_{3}))=\rho_{P}(\sqrt{-t})e_{6}$.
It follows that for every $P\in\mathcal{R}_{2}$,
$\ell_{P}g_{i}\ell_{P}^{-1}=h_{i}(\rho_{P}(z_{i}))$.

Finally, define $\ell:=(\ell_{P})_{P\in\mathcal{P}}$. ∎

###  5.4 Approximation in $\operatorname{Spin}_{f_{s}}$

######  Definition 5.4.1.

Let $a\in{\bf O}^{\times}$ and let $\mathcal{Q}\subseteq\mathcal{P}_{K}$ be a
subset. The support of $a$ in $\mathcal{Q}$ is
$\operatorname{supp}_{\mathcal{Q}}(a):=\\{P\in\mathcal{Q}\mid\operatorname{val}_{P}(a)\neq
0\\}$.

######  Definition 5.4.2.

Under Setting 4.4.2, let $a_{1},a_{2},b_{1},b_{2}\in{\bf O}^{\times}$ and
$I_{1},I_{2},J\in\mathcal{I}_{K}$. The pair $(b_{1},b_{2})$ is called an
$(I_{1},I_{2},J)$-approximation of $(a_{1},a_{2})$ if there exist commuting
$g_{1},g_{2}\in\psi(\operatorname{Spin}_{f_{a}}({\bf A}))$ and
$l\in\operatorname{Spin}_{f_{s}}({\bf A})$ such that for every $1\leq i\leq
2$:

  1. (1)

$\mathcal{R}_{1}:=\operatorname{supp}_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(I_{1}I_{2}J)}(a_{1})$
and
$\mathcal{R}_{2}:=\operatorname{supp}_{=\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(I_{1}I_{2}J)}(a_{2})$
are disjoint.

  2. (2)

$(lg_{i}l^{-1})_{\mathcal{P}_{K}(I_{1}I_{2})\cup\mathcal{R}_{1}\cup\mathcal{R}_{2}}=h_{i}(b_{i})_{\mathcal{P}_{K}(I_{1}I_{2})\cup\mathcal{R}_{1}\cup\mathcal{R}_{2}}$.

  3. (3)

$(lg_{1}l^{-1})_{(\mathcal{R}_{1}\cup\mathcal{P}_{K}(I_{1}I_{2}))^{c}}\in\operatorname{Spin}_{f_{s}}({\bf
O})$ and
$(lg_{2}l^{-1})_{(\mathcal{R}_{2}\cup\mathcal{P}_{K}(I_{1}I_{2}))^{c}}\in\operatorname{Spin}_{f_{s}}({\bf
O})[J]$.

  4. (4)

$h_{i}(a_{i}b_{i}^{-1})_{\mathcal{P}_{K}(I_{1}I_{2})}\in\operatorname{Spin}_{f_{s}}({\bf
O})[I_{i}]$.

######  Lemma 5.4.3.

Let $a_{1},a_{2}\in K^{\times}$ and let $I\in\mathcal{I}_{K}$. There exists an
$(I,I,O)$-approximation of $(a_{1},a_{2})$.

###### Proof.

Let $t,z_{1},z_{2},\mathcal{R}_{1},\mathcal{R}_{2},\rho$ be obtained by
applying Lemma 5.2.2 to $a_{1},a_{2},I$. We can assume that for every $1\leq
i\leq 2$,
$\mathcal{R}_{i}:=\operatorname{supp}_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(I)}(b_{i})$.
Let $g_{1},g_{2},b_{1},b_{2},\ell$ be obtained by applying Lemma 5.3.3 to
$t,z_{1},z_{2},\rho,\mathcal{P}(I)\cup\mathcal{R}_{1}\cup\mathcal{R}_{2}$. It
can be readily verified that $(b_{1},b_{2})$ is an $(I,I,O)$-approximation of
$(a_{1},a_{2})$. ∎

######  Lemma 5.4.4.

Let $I,J\in\mathcal{I}_{K}$ such that
$\mathcal{P}_{K}(I)\cap\mathcal{P}_{K}(J)=\emptyset$, $a_{1}\in({\bf
A})^{\times}$ and $a_{2}\in({\bf O})^{\times}$. There exists an $({\bf
O},I,J)$-approximation $(b_{1},b_{2})$ of $(a_{1},a_{2})$, $c\in({\bf
O})^{\times}$ and
$\mathcal{R}\subseteq\mathcal{R}_{1}:=\operatorname{supp}_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(IJ)}(b_{1})$
such that $b_{2}\in({\bf O})^{\times}$ and

$h_{2}(b_{2})_{\mathcal{R}_{1}}=h_{2}(c^{2})_{\mathcal{R}_{1}}h_{2}(-1)_{\mathcal{R}}.$

###### Proof.

Let $a^{\prime}_{2}\in{\bf O}^{\times}$ be such that
$(a_{2})_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(J)}=(a_{2}^{\prime})_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(J)}$
and $(a_{2}^{\prime})_{\mathcal{P}_{K}(J)}=1$. Let $t,\rho,z_{2}$ be obtained
by applying Lemma 5.2.4 to $IJ$ and $a^{\prime}_{2}$(without any $s_{i}$).
Since $t$ is not a perfect square, $\sqrt{-1}\notin K(\sqrt{-t})$.

Denote $\mathcal{Q}=\left\\{P\in\mathcal{P}_{K}\mid
P\notin\mathcal{P}_{K}(I)\wedge t\notin P\wedge|O/P|\equiv 3\text{ (mod
$4$)}\right\\}$. We claim that $\mathcal{Q}$ satisfies the condition of Lemma
5.2.5.

Let $H$ be the Hilbert class field of $K(\sqrt{-t})$. By the assumption, the
set
$X:=\left\\{\sigma\in\operatorname{Gal}\left(H(i)/K\left(\sqrt{-t}\right)\right)\mid\sigma(i)=-i\right\\}$
is the complement of a proper subgroup, so it generates
$\operatorname{Gal}\left(H(i)/K\left(\sqrt{-t}\right)\right)$. If
$Q\in\mathcal{P}_{K\left(\sqrt{-t}\right)}$, $Q\cap
O\notin\mathcal{P}_{K}(I)$, $t\notin Q\cap O$ and
$\left(\frac{H(i)/K\left(\sqrt{-t}\right)}{Q}\right)\in X$, then $Q\cap
O\in\mathcal{Q}$. By Chebotarev’s density theorem, $\mathcal{Q}$ satisfies the
condition of Lemma 5.2.5.

Let $z_{1}$ be obtained by applying Lemma 5.2.5 to
$\mathcal{P}(I),t,\rho,a_{1},\mathcal{Q}$.

For $P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}(I)$, if $z_{1}\notin
T_{t}(O_{P})$, then $P\in\mathcal{Q}$ and, by Proposition 5.1.33, $P$ is split
in $K\left(\sqrt{-t}\right)$. Extend $\rho$ to $P$ arbitrarily. Since
$|O/P|\equiv 3\text{ (mod $4$)}$, either $\rho_{P}(z_{2})$ is a square, or
$-\rho_{P}(z_{2})$ is a square (in $O_{P}^{\times}$). Let

$\mathcal{R}:=\left\\{P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(IJ)\mid
z_{1}\notin T_{t}(O_{P})\text{, $-\rho_{P}(z_{2})$ is a square in
$O_{P}^{\times}$}\right\\},$
$\mathcal{R}^{\prime}:=\left\\{P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(IJ)\mid
z_{1}\notin T_{t}(O_{P})\text{, $\rho_{P}(z_{2})$ is a square in
$O_{P}^{\times}$}\right\\}$

and denote $\mathcal{R}_{1}:=\mathcal{R}\cup\mathcal{R}^{\prime}$. Let
$c\in{\bf O}^{\times}$ be such that $-c_{P}^{2}=\rho_{P}(z_{2})$, for all
$P\in\mathcal{R}$ and $c_{P}^{2}=\rho_{P}(z_{2})$, for all
$P\in\mathcal{R}^{\prime}$.

Let $g_{1},g_{2},b_{1},b_{2},\ell$ be obtained by applying Lemma 5.3.3 to
$t,z_{1},z_{2},\rho,\mathcal{P}(II^{\prime}I^{\prime\prime})$. The elements
$g_{1},g_{2},b_{1},b_{2},\ell$ testify that $(b_{1},b_{2})$ is an
$(O,I,J)$-approximation of $(a_{1},a_{2})$,
$\mathcal{R}_{1}=\operatorname{supp}_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(IJ)}(b_{1})$
and

$h_{2}(b_{2})_{\mathcal{R}_{1}}=h_{2}(c^{2})_{\mathcal{R}_{1}}h_{2}(-1)_{\mathcal{R}}.$

∎

######  Lemma 5.4.5.

Let $a\in{\bf A}^{\times}$ and $I,J\in\mathcal{I}_{K}$ such that
$\mathcal{P}_{K}(I)\cap\mathcal{P}_{K}(J)=\emptyset$. There exists an $({\bf
O},I,J)$-approximation $(b,-1)$ of $(a,-1)$ and $c\in({\bf O})^{\times}$ such
that $h_{2}(-1)_{\mathcal{R}}=h_{2}(c^{2})_{\mathcal{R}}$ where
$\mathcal{R}:=\operatorname{supp}_{\mathcal{P}_{K}\setminus\mathcal{P}_{K}(IJ)}(b)$.

###### Proof.

Choose any $t\in\mathbb{Z}_{+}$ such that $t\equiv-1\text{ (mod $IJ$)}$ and
$t\equiv 3\text{ (mod $4$)}$. By the first requirement on $t$, there is a
trivialization $\rho$ of $T_{t}$ over $\mathcal{P}(IJ)$. Let
$\mathcal{Q}=\left\\{P\in\mathcal{P}_{K}\mid P\equiv 1\text{ (mod
$4$)}\right\\}$. By the second requirement on $t$, $2$ does not ramified in
$K(\sqrt{-t})$. Since the Hilbert class field of a field in an unramified
extension of this field and 2 ramifies in $K(\sqrt{-1})$, $\sqrt{-1}$ does not
belong to the Hilbert class field of $K(\sqrt{-t})$. Thus, by Chebotarev’s
density theorem, ,every co-finite subset of $\mathcal{Q}$ generates
$\operatorname{Cl}_{K\left(\sqrt{-t}\right)}$.

We can assume that $(a)_{\mathcal{P}_{K}(J)}=1_{\mathcal{P}_{K}(J)}$. Let $z$
be the element obtained by applying Lemma 5.2.5 to
$\mathcal{P}_{K}(IJ),t,\rho,a,\mathcal{Q}$. Note that $z\in T_{t}(O_{P})$, for
every $P\in\mathcal{P}(J)$. Let
$\mathcal{R}=\left\\{Q\in\mathcal{P}\smallsetminus\mathcal{P}(I)\mid z\notin
T_{t}(O_{Q})\right\\}$. It follows that:

  * •

$\mathcal{R}$ and $\mathcal{P}_{K}(IJ)$ are disjoint.

  * •

By Item 3 of Lemma 5.1.3, every prime in $\mathcal{R}$ splits in
$K\left(\sqrt{-t}\right)$.

  * •

$\mathcal{R}\subseteq\mathcal{Q}$.

Extend $\rho$ to $\mathcal{P}(IJ)\cup\mathcal{R}$.

Let $z_{2}\in{\bf O}^{\times}$ be such that
$(z_{2})_{\mathcal{P}_{K}(I)}=(-1)_{\mathcal{P}_{K}(I)}$ and
$(z_{2})_{\mathcal{P}_{K}(I)^{c}}=1_{\mathcal{P}_{K}(I)^{c}}$ Let
$g_{1},g_{2},b_{1},b_{2},\ell$ be the elements obtained by applying Lemma
5.3.3 to $t,z_{1}=z,z_{2},\rho,\mathcal{P}(IJ)\cup\mathcal{R}$. For each
$Q\in\mathcal{R}$, $Q\in\mathcal{Q}$, so $-1$ is a square in $O_{Q}$. Choose
an element $c\in{\bf O}^{\times}$ such that $(c^{2})_{Q}=(-1)_{\mathcal{Q}}$,
for all $Q\in\mathcal{R}$.

It follows that $(b,-1)$ is an $(O,I,J)$-approximation of $(a_{1},-1)$,
$\mathcal{R}=\operatorname{supp}_{\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(IJ)}(b)$
and $h_{2}(-1)_{\mathcal{R}}=h_{2}(c^{2})_{\mathcal{R}}$. ∎

###  5.5 Approximation to squares

####  5.5.1 Lenstra’s theorem

We will use a result of Lenstra from [23] under some simplifying assumptions.
In this subsection, we will rephrase the result for the simple case that we
need and explain the differences from the original formulation using the
notations of that paper.

The setup is similar to loc. cit., except that we restrict to the special
where $k=2$ and $W$ contains a primitive element (i.e. an element which is not
a proper power in $K^{\times}$). In the notations of [23]:

######  Setting 5.5.1.

Let $F/E$ be a Galois extension of number fields, let
$\mathcal{C}\subseteq\operatorname{Gal}(F/E)$ be a normal set and let
$W\subseteq E^{\times}$ be a finitely generated group of rank at least 1 which
contains a primitive element (i.e. an element which is not a proper power in
$E^{\times}$). Define $M(E,F,\mathcal{C},W,2)$ as the set of all prime ideals
$\mathfrak{p}$ of $O_{E}$ for which the following conditions hold:

  1. (1)

$\left(\frac{F/E}{\mathfrak{p}}\right)\in\mathcal{C}$.

  2. (2)

$\operatorname{val}_{\mathfrak{p}}(w)=0$, for every $w\in W$.

  3. (3)

The group $W\text{ (mod $\mathfrak{p}$)}$ contains $\\{x^{2}\mid
x\in\left(O_{E}/\mathfrak{p}\right)^{\times}\\}$..

Under Setting 5.5.1, in the notations of [23], the following hold:

  * •

Since $k=2$, for all odd prime $\ell$, $q(\ell)=\ell$ and $q(2)=4$.

  * •

For every prime $\ell$, $E_{l}$ is the field extension of $E$ generated by all
$q(l)$-roots of unity. $L_{l}$ is the field extension of $E$ generated by all
the $q(\ell)$-root of the elements of $W$.

  * •

Since $W$ contains a primitive element of $E^{\times}$, $h=1$.

The following theorem follows from [23, Theorem 3.1 and Theorem 4.1].

######  Theorem 5.5.2 (Lenstra).

Under Setting 5.5.1, the following properties are equivalent:

  1. (1)

There is an element $\sigma\in\mathcal{C}$ such that for all primes $\ell$ for
which $L_{l}\subset F$,
$\sigma\restriction_{L_{\ell}}\neq\operatorname{id}_{\ell}$.

  2. (2)

The analytic density of $M(E,F,\mathcal{C},W,2)$ is positive. In particular,
$M(E,F,\mathcal{C},W,2)$ is infinite.

######  Corollary 5.5.3.

Under Setting 5.5.1, if there is an element $\sigma\in\mathcal{C}$ such that
for all primes $\ell$ for which $E_{l}\subset F$,
$\sigma\restriction_{E_{\ell}}\neq\operatorname{id}_{\ell}$. then the analytic
density of $M(E,F,\mathcal{C},W,2)$ is positive. In particular,
$M(E,F,\mathcal{C},W,2)$ is infinite.

####  5.5.2 Approximation theorem

######  Setting 5.5.4.

Recall that $K=\mathbb{Q}$. Fix $t\in O$ such that $E:=K(\sqrt{-7},\sqrt{-t})$
does not contains neither $\sqrt{3}$ nor $\sqrt{-1}$ nor $\sqrt{-3}$ and
$[E:K]=4$. Let $\gamma:=\frac{3+\sqrt{-7}}{4}\in(O[\sqrt{-7}])^{\times}$. Then
$\gamma$ is a primitive element and $N_{K(\sqrt{-7})/K}(\gamma)=1$. For every
prime $l$, let $q(\ell):=\ell$ if $\ell$ is odd and $q(2):=4$.

######  Lemma 5.5.5.

Let $G_{1},\ldots,G_{s}$ be non-trivial finite cyclic groups and denote
$G=G_{1}\times\cdots\times G_{s}$. Let $L$ be a subgroup of $G$ such that
$G^{2}:=\\{g^{2}\mid g\in G\\}\subseteq L$ and assume that
$A:=\\{(g_{i})_{i}\in L\mid(\forall 1\leq i\leq m)\ g_{i}\neq 1\\}$ is not
empty. Then $G^{4}\subseteq\langle A\rangle$.

###### Proof.

Assume that for every $1\leq i\leq r\leq s$, $|G_{i}|\in\\{2,4\\}$ and for
every $r+1\leq i\leq s$, $|G_{i}|\notin\\{2,4\\}$. For every $i$, denote
$G^{2}_{i}:=\\{g_{i}^{2}\mid g_{i}\in G_{i}\\}$. Note that for every $r+1\leq
i\leq s$, $|G_{i}^{2}|\geq 3$. It is enough to show that $\langle A\rangle$
contains $G_{r+1}^{2}\times\cdots G_{s}^{2}$.

Let $a=(a_{1},\ldots,a_{m})\in A$. For every $r+1\leq i\leq s$ there are at
least $|G_{i}^{2}|-1>\frac{|G_{i}^{2}|}{2}$ elements $g_{i}$ of $G_{i}^{2}$
such that $a_{i}g_{i}\neq 1$. For any such element,
$g_{i}=(ag_{i})a^{-1}\in\langle A\rangle$. Thus, $\langle A\rangle$ contains
$G_{i}^{2}$ and the claim is clear. ∎

######  Remark 5.5.6.

Let $1\leq r^{\prime}\leq r$ be such that for every $r^{\prime}+1\leq i\leq
r$, $|G_{i}|=4$. Since for every $r^{\prime}+1\leq i\leq s$, $L$ contains
$G_{i}^{2}$ and $|G_{i}^{2}|\geq 2$, in order to show that $A$ in non-empty it
is enough to show that there exists $\ell\in L$ such that for every $1\leq
i\leq r^{\prime}$, the projection of $\ell$ to $G_{i}$ is not trivial.

######  Lemma 5.5.7.

Under Setting 5.5.4, suppose that $T\subseteq\\{p\mid p\text{ is an odd
prime}\\}\cup\\{4\\}$ is a finite set. Denote $N=\prod_{m\in T}p$. Then there
exists $\sigma\in\operatorname{Gal}(E(\zeta_{N})/E)$ such that for every
$\ell\in T\cap\\{3,4\\}$, $\sigma(\zeta_{\ell})\neq\zeta_{\ell}$.

###### Proof.

Recall that $K=\mathbb{Q}$ and that $\zeta_{3}=\frac{-1+\sqrt{-3}}{2}$. Since
the field $E$ does not contains neither $\sqrt{3}$ nor $\sqrt{-1}$ nor
$\sqrt{-3}$, there exists $\tau\in\operatorname{Gal}(E(\zeta_{N})/E)$ such
that for every $\ell\in T\cap\\{3,4\\}$, $\tau(\zeta_{\ell})\neq\zeta_{\ell}$.
We can extended $\tau$ to an element
$\sigma\in\operatorname{Gal}(E(\zeta_{N})/E)$. ∎

######  Lemma 5.5.8.

Under Setting 5.5.4, suppose that $T\subseteq\\{p\mid p\text{ is an odd
prime}\\}\cup\\{4\\}$ is a finite set. Denote $N=\prod_{m\in T}p$. Then

  1. (1)

Denote
$A:=\left\\{\sigma\in\operatorname{Gal}\left(E\left(\zeta_{N}\right)/E\right)\mid\left(\forall\ell\in
T\right)\sigma(\zeta_{\ell})\neq\zeta_{\ell}\right\\}$. Then
$\operatorname{Gal}\left(E\left(\zeta_{N}\right)/E\right)^{4}\subseteq\langle
A\rangle$.

  2. (2)

If $P\in\mathcal{P}_{E}$ is unramified in $E\left(\zeta_{N}\right)$ and
$\left(\frac{E\left(\zeta_{N}\right)/E}{P}\right)\in A$ and $\ell\in T$, then
$\ell\not||(O/P)^{\times}|$.

###### Proof.

  1. (1)

Recall that $K=\mathbb{Q}$.
$\operatorname{Gal}\left(E\left(\zeta_{N}\right)/E\right)$ is isomorphic to a
subgroup $L\leq
G:=\operatorname{Gal}\left(K\left(\zeta_{N}\right)/K\right)\cong\left(\mathbb{Z}/N\right)^{\times}\cong\prod_{\ell\in
T}\left(\mathbb{Z}/\ell\right)^{\times}$ and $G^{2}\subseteq L$. The claim
follows from Lemmas 5.5.5 and 5.5.7 and Remark 5.5.6.

  2. (2)

$\left(\frac{E\left(\zeta_{N}\right)/E}{P}\right)(\zeta_{\ell})=\zeta_{\ell}^{|O/P|}$,
and the claim is clear.

∎

######  Lemma 5.5.9.

Under Setting 5.5.4, let $T\subseteq\\{p\mid p\text{ is an odd prime}\\}$ be a
finite set of odd primes. Then there is a set
$Y\subseteq\operatorname{Cl}_{K(\sqrt{-t})}$ such that
$Cl_{K(\sqrt{-t})}^{8}\subseteq\langle Y\rangle$ and, for each $c\in Y$, there
are infinitely many $Q\in\mathcal{P}_{K(\sqrt{-t})}$ such that

  1. (1)

$c=\left(\frac{H/K(\sqrt{-t})}{Q}\right)$ where $H$ is the Hilbert Class Field
of $K(\sqrt{-t})$.

  2. (2)

$P:=Q\cap K$ splits completely in $E$.

  3. (3)

If $Q^{\prime}\in\mathcal{P}_{k(\sqrt{-7})}$ lies above $P$, then
$\frac{|O_{K{\sqrt{-7}}}/Q^{\prime}|-1}{2}$ is relatively prime to every
element of $T\cup\\{2\\}$.

  4. (4)

If $Q^{\prime}\in\mathcal{P}_{k(\sqrt{-7})}$ lies above $P$, then
$\langle\gamma\text{ (mod $Q^{\prime}$)}\rangle$ contains
$\left(O_{K(\sqrt{-7})}/Q^{\prime}\right)^{\times 2}$.

###### Proof.

Let $N=4\cdot\prod_{p\in T}p$, let $F=H(\sqrt{-7},\zeta_{N})$, and let
$K(\zeta_{M})$ be the compositum of all cyclotomic extensions of $K$ that are
contained in $F$. For a prime number $\ell$, $\zeta_{\ell}\in F\iff\ell|M$.

Denote
$A:=\left\\{\sigma\in\operatorname{Gal}\left(E(\zeta_{M}\right)/E)\mid\left(\forall\ell|M\right)\sigma(\zeta_{\ell})\neq\zeta_{\ell}\right\\}$.
By Lemma 5.5.8,
$\operatorname{Gal}\left(E(\zeta_{M})/E\right)^{4}\subseteq\langle A\rangle$.
Denote
$X:=\left\\{\sigma\in\operatorname{Gal}(F/E)\mid\left(\forall\ell|M\right)\sigma(\zeta_{\ell})\neq\zeta_{\ell}\right\\}$.
Since $X$ is the inverse image of $A$ under the restriction epimorphism
$\operatorname{Gal}(F/E)\rightarrow\operatorname{Gal}(E(\zeta_{M})/E)$,
$\operatorname{Gal}(F/E)^{4}\subseteq\langle X\rangle$. Since
$\operatorname{Gal}(F/E)$ is of index two in
$\operatorname{Gal}(F/K(\sqrt{-t}))$,
$\operatorname{Gal}(F/K(\sqrt{-t}))^{8}\subseteq\langle X\rangle$.

Let $Y$ be the image of $X$ under the restriction homomorphism
$\operatorname{Gal}(F/K(\sqrt{-t}))\twoheadrightarrow\operatorname{Gal}(H/K(\sqrt{-t}))\cong\operatorname{Cl}_{K(\sqrt{-t})}$.
Then, $\operatorname{Cl}_{\sqrt{-t}}^{8}\subseteq\langle Y\rangle$.

Let $c\in Y$ and choose some $\sigma\in X$ whose restriction to $H$ is $c$.
Applying Theorem 5.5.2 to the extension $F/E$, the conjugacy class of
$\sigma$, and the cyclic group generated by $\gamma$, we get that the set

$\mathcal{R}:=\left\\{R\in\mathcal{P}_{E}\mid\left(\frac{F/E}{R}\right)=\sigma\text{
and }\langle\gamma\text{ (mod
$R$)}\rangle\supseteq\left(O_{E}/R\right)^{\times 2}\right\\}$

has positive density. Since the set of primes $R\in\mathcal{P}_{E}$ such that
$R\cap K$ does not split completely in $E$ has density zero, the set

$\mathcal{R}^{\prime}:=\\{R\in\mathcal{R}\mid R\cap K\text{ splits completely
in }E\\}$

as positive density. Thus, in order to finish the proof, it is enough to show
that if $R\in\mathcal{R}^{\prime}$, $Q:=R\cap K(\sqrt{-t})$ and $P:=Q\cap K$
then Items (1), (2), (3) and (4) hold. It is clear that Item (2) holds. Item
(2) implies that
$\left(\frac{F/K(\sqrt{-t})}{Q}\right)=\left(\frac{F/E}{R}\right)$ so

$\left(\frac{H/K(\sqrt{-t})}{Q}\right)={\left.\kern-1.2pt\left(\frac{F/K(\sqrt{-t})}{Q}\right)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{H}}=\sigma\restriction_{H}=c$

and Item (1) holds.

We now prove that Item (3) holds. Let
$Q^{\prime}\in\mathcal{P}_{k(\sqrt{-7})}$ be a prime which lies above $P$. By
Item (2), $|O_{K(\sqrt{-7})}/Q^{\prime}|=|O_{E}/R|$. For any
$\ell\in\left\\{4\right\\}\cup T$,

$\zeta_{\ell}^{|O_{E}/R|}=\left(\frac{F/E}{R}\right)(\zeta_{\ell})=\sigma(\zeta_{\ell})\neq\zeta_{\ell}.$

For $\ell=4$, we get $|O_{K{\sqrt{-7}}}/Q|=|O_{E}/R|\equiv\text{ 3 (mod 4)}$.
For odd $\ell\in T$, we get $|O_{K{\sqrt{-7}}}/Q|=|O_{E}/R|\not\equiv\text{1
(mod $\ell$)}$. Thus, Item (3) holds.

Finally, we verify that Item (4) holds. Let
$Q^{\prime}\in\mathcal{P}_{k(\sqrt{-7})}$ be a prime which lies above $P$ and
let $\tau$ be the non-trivial elements of
$\operatorname{Gal}(K(\sqrt{-7})/K))$. Then either $Q^{\prime}=R\cap
K(\sqrt{-7})$ or $Q^{\prime}=\tau(R\cap K(\sqrt{-7}))$. In the later case,
since $N_{K(\sqrt{-7})/K}(\gamma)=1$, $\tau$ induce an isomorphism from
$O_{K(\sqrt{-7})}/Q^{\prime}\rightarrow$ to $O_{K(\sqrt{-7})}/(R\cap
K(\sqrt{-7}))$ which sends $\gamma+Q^{\prime}$ to $\gamma^{-1}+(P\cap
K(\sqrt{-7}))$. Thus, it is enough to prove the claim in the case where
$Q^{\prime}=R\cap K(\sqrt{-7})$. In this case, by Item (2), there is a
isomorphism $O_{K(\sqrt{-7})}/Q^{\prime}\rightarrow O_{E}/R$ which sends
$\gamma+Q^{\prime}$ to $\gamma+R$. The claim follows from the definition of
$\mathcal{R}$. ∎

######  Lemma 5.5.10.

Under Setup 5.5.4, let $J\in\mathcal{I}_{K}$. Then there is a set
$\mathcal{Q}\subseteq\mathcal{P}_{K}$ such that

  1. (1)

For any cofinite subset $\mathcal{Q}_{0}\subseteq\mathcal{Q}$,
$\operatorname{Cl}_{K(\sqrt{-t})}^{8}\subseteq\langle
A_{\mathcal{Q}_{0}}\rangle$ where
$A_{\mathcal{Q}_{0}}:=\left\\{P\in\mathcal{P}_{K\left(\sqrt{-t}\right)}\mid
P\cap O\in\mathcal{Q}_{0}\right\\}$.

  2. (2)

For every distinct $P_{1},\ldots,P_{N}\in\mathcal{Q}$ and every
$Q_{1},\ldots,Q_{N}\in\mathcal{P}_{K\left(\sqrt{-7}\right)}$ such that
$Q_{i}\cap K=P_{i}$, the image of $\langle\gamma^{\phi(|J|}\rangle$ in
$G:=\prod_{i}(O_{K\left(\sqrt{-7}\right)}/Q_{i})^{\times}$ contains $G^{2}$.

###### Proof.

Let $\mathcal{Y}\subseteq 2^{\operatorname{Cl}_{L}}$ be the collection of all
subsets $Y\subseteq\operatorname{Cl}_{L}$ such that the exponent of
$\operatorname{Cl}_{K(\sqrt{-t})}^{8}\subseteq\langle Y\rangle$. For each
$Y\in\mathcal{Y}$, let $c_{Y,0},c_{Y,1},\ldots\in Y$ be a sequence such that
any element of $Y$ occurs infinitely many times among the $c_{Y,i}$s.

Let $T_{0}$ be the union of $\\{2\\}$ and the set of prime divisors of
$\phi(|J|)$ and define by recursion three sequences $T_{n},P_{n},Y_{n}$, where
$T_{n}$ is a finite set of prime numbers, $P_{n}\in\mathcal{I}_{K}$, and
$Y_{n}\in\mathcal{Y}$. Given $T_{n-1}$, let $Y$ be the element obtained by
applying Lemma 5.5.9 to $T_{n-1}$. Denote $a=\\#\left\\{i<n\mid
Y_{i}=Y\right\\}$ and let $P\in\mathcal{P}_{K(\sqrt{-t})}$ be a representative
of $c_{Y,a}$ that satisfies the claims of Lemma 5.5.9. Define $Y_{n}=Y$,
$P_{n}=P$, and $T_{n}=T_{n-1}\cup\left\\{\text{prime divisors of
$|P|-1$}\right\\}$. Finally, let $\mathcal{Q}=\left\\{P_{n}\cap K\mid
n=1,\ldots\right\\}$.

Since $\mathcal{Q}$ is infinite, there is $Z\in\mathcal{Y}$ such that
$Y_{n}=Z$ infinitely often. It follows that $\mathcal{Q}$ contains infinitely
many representatives of each element of $Z$. This proves (1).

Given $P_{i},Q_{i}$ as in (2), by consturction, we have
$O_{K\left(\sqrt{-7}\right)}/Q_{i}\cong O/P_{i}$, so
$\frac{|O_{K\left(\sqrt{-7}\right)}/Q_{i}|-1}{2}$ is relatively prime to
$\phi(|J|)$. Since $\gamma$ generates
$\left(O_{K\left(\sqrt{s}\right)}/Q_{i}\right)^{\times 2}$, the same is true
for $\gamma^{\phi(|J|)}$. By consturction the orders of the groups
$\left(O_{K\left(\sqrt{-7}\right)}/Q_{i}\right)^{\times 2}$ are relatively
prime so $\gamma^{\phi(|J|)}$ generates their product, proving (2). ∎

######  Lemma 5.5.11.

Let $I,J\in\mathcal{I}_{K}$ be coprime, $a_{1}\in{\bf A}^{\times}$ and
$a_{2}\in{\bf O}^{\times}$. There exist an $({\bf O},I,J)$-approximation
$(b_{1},b_{2})$ of $(a_{1}^{8},a_{2})$ such that $b_{2}\in{\bf O}^{\times}$
and for every
$P\in\mathcal{R}:=\operatorname{supp}_{\mathcal{P}_{K^{*}}\smallsetminus\mathcal{P}_{K}^{*}(IJ)}(b_{1})$,
$\operatorname{val}_{P}(b_{1})=-1$, and an $({\bf
O},\prod_{P\in\mathcal{R}}P,J)$-approximation $(c_{1},c_{2})$ of
$(b_{1},b_{2}^{-2})$ such that
$h_{1}(c_{1})_{\mathcal{R}^{c}}\in\operatorname{Spin}_{f_{s}}({\bf O})$ and
$c_{2}\in({\bf O})^{\times}$.

###### Proof.

Let $a_{1}^{\prime}\in{\bf A}^{\times}$ be such that
$(a_{1}^{\prime})_{\mathcal{P}_{K}(I)}=(a_{1})_{{\mathcal{P}_{K}}(I)}$ and
$(a_{1}^{\prime})_{\mathcal{P}_{K}(J)}=1$. Let $a_{2}^{\prime}\in{\bf
O}^{\times}$ be such that
$(a_{2}^{\prime})_{\mathcal{P}_{K}(I)}=(a_{2})_{{\mathcal{P}_{K}}(I)}$ and
$(a_{2}^{\prime})_{\mathcal{P}_{K}(J)}=1$. Apply Lemma 5.2.4 to $IJ$ and
$a_{2}^{\prime}$ to get an element $t\in K$, a trivialization $\rho$ of
$T_{t}$ over $\mathcal{P}(IJ)$, and an element $z_{2}\in T_{t}(O)$ such that

  1. (a)

$\rho_{P}(z_{2})a_{2}^{-1}\in(1+IO_{P})\cap O_{P}^{\times}$ if
$P\in\mathcal{P}(I)$.

  2. (b)

$\rho_{P}(z_{2})\in(1+JO_{P})\cap O_{P}^{\times}$ if $P\in\mathcal{P}(J)$.

  3. (c)

$\sqrt{-3},\sqrt{-1}\notin K\left(\sqrt{-t},\sqrt{-7}\right)$. It follows that
$K\left(\sqrt{-t},\sqrt{-7}\right)$ does not contain any root of unity other
than $\pm 1$. .

Let $\mathcal{Q}$ be the set obtained by applying Lemma 5.5.10 to $J$ and $t$.
Then any cofinite subset $\mathcal{Q}_{0}\subseteq\mathcal{Q}$, the set
$\left\\{P\in\mathcal{P}_{K\left(\sqrt{-t}\right)}\mid P\cap
O\in\mathcal{Q}_{0}\right\\}$ generates a subgroup $L$ of
$\operatorname{Cl}_{K\left(\sqrt{-t}\right)}$ such that
$\operatorname{Cl}_{K\left(\sqrt{-t}\right)}^{8}\subseteq L$. Let $z_{1}$ be
the element obtained by applying Lemma 5.2.6 to
$\mathcal{P}(IJ),t,\rho,a_{1}^{\prime},\mathcal{Q}$. We have,

  1. (d)

For every distinct $P_{1},\ldots,P_{N}\in\mathcal{Q}$ and every
$Q_{1},\ldots,Q_{N}\in\mathcal{P}_{K\left(\sqrt{-7}\right)}$ such that
$Q_{i}\cap K=P_{i}$, the image of $\langle\gamma^{\phi(|J|}\rangle$ in
$G=\prod_{i}(O_{K\left(\sqrt{-7}\right)}/Q_{i})^{\times}$ contains $G^{2}$.

  2. (e)

$\rho_{P}(z_{1})a_{1}^{-8}\in O_{P}^{\times}$, for all $P\in\mathcal{P}(I)$.

  3. (f)

$\rho_{P}(z_{1})\in O_{P}^{\times}$, for all $P\in\mathcal{P}(J)$.

  4. (g)

If $P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(IJ)$ and $z_{1}\notin
T_{t}(O_{P})$, then $P\in\mathcal{Q}$, $\operatorname{val}_{P}(z_{1})=-1$ and
$P\cap K$ splits is $K(\sqrt{-7})$ (by 5.1.33).

Define $I^{\prime}$ as the product of all primes
$P\in\mathcal{P}_{K}\smallsetminus\mathcal{P}_{K}(I)$ such that $z_{1}\notin
T_{t}(O_{P})$. Clearly, $I^{\prime}$ is relatively prime to both $I$ and $J$
and $\mathcal{R}:=\mathcal{P}_{K}(I^{\prime})$ is contained in $\mathcal{Q}$.
By g, $\operatorname{val}_{P}(z_{1})=-1$, for all
$P\in\mathcal{P}(I^{\prime})$. By 5.1.33, every $P\in\mathcal{P}(I^{\prime})$
splits in $K\left(\sqrt{-t}\right)$, so $\rho$ can be extended to a
trivialization of $T_{t}$ over $\mathcal{P}(II^{\prime}J)$.

Let $g_{1},g_{2},b_{1},b_{2},\ell$ be the elements obtained by applying Lemma
5.3.3 to $t,z_{1},z_{2},\rho,\mathcal{P}(II^{\prime}J)$. Then
$g_{1},g_{2},\ell$ are a testment that $(b_{1},b_{2})$ is an
$(O,I,J)$-approximation of $(a_{1}^{8},a_{2})$.

If $P\in\mathcal{P}(I^{\prime})$, then $P$ splits in $K(\sqrt{-7})$ . Choose a
trivialization $\tau$ of $T_{7}$ over $\mathcal{P}(I^{\prime})$. It is known
that $\operatorname{Cl}_{K(\sqrt{-7})}$ is trivial. Let $z_{1}^{\prime}$ be
the element obtained by applying Lemma 5.2.5 to
$\mathcal{P}(I^{\prime}),-7,\tau,b_{1},\emptyset$. Then
$\tau_{P}(z_{1}^{\prime})b_{1}^{-1}\in O_{P}^{\times}$, for all
$P\in\mathcal{P}(I^{\prime})$, and $z_{1}^{\prime}\in T_{7}(O_{P})$ for all
$P\notin\mathcal{P}(I^{\prime})$.

By d, there is a natural number $m$ such that for every
$P\in\mathcal{P}(I^{\prime})$,
$\tau_{P}\left(\gamma^{m\phi(|O/J|)}\right)b_{2}^{2}\in 1+PO_{P}$. Thinking of
$\gamma$ as an element of $T_{7}(O)$, let
$g_{1}^{\prime},g_{2}^{\prime},c_{1},c_{2}^{\prime},\ell^{\prime}$ be the
elements obtained by applying Lemma 5.3.3 to
$7,z_{1}^{\prime},\gamma,\tau,\mathcal{P}_{K}(I^{\prime})$ and denote
$c_{2}:=(c_{2}^{\prime})^{m\phi(|O/J|)}$ We get that
$(c_{1})_{\mathcal{P}_{K}(I^{\prime})^{c}}\in({\bf
O}_{\mathcal{R}})^{\times}$, $c_{2}\in(1+JO_{P})$, for every
$P\in\mathcal{P}_{K}(J)$, and $(c_{1},c_{2})$ is an
$(O,I^{\prime},J)$-approximation of $(b_{1},b_{2}^{-2})$. ∎

##  6 Proof of Theorem 1.1.3

###  6.1 Preliminary lemmas

We will use two lemmas in the proof of Theorem 1.1.3. The first is a version
of Rapinchuk’s lemma (see [35, §3] or [24, Lemma 2.6]):

######  Lemma 6.1.1.

Suppose that $\Gamma$ does not satisfy the Congruence Subgroup Property. Then,
there is a surjection $\widehat{\Gamma}\twoheadrightarrow H$, a finite simple
group $F$, and a short exact sequence

$1\rightarrow F^{\infty}\rightarrow H\rightarrow X\rightarrow 1$

such that $\Gamma$ acts faithfully on $F^{\infty}$. If $F$ is abelian, then
there are $v_{0}\notin S$ and a subgroup $L\subseteq X$ that is commensurable
to $G(O_{v_{0}})$ such that $\left[\rho(X):\rho(L)\right]<\infty$, where
$\rho:X\rightarrow\operatorname{Aut}(F^{\infty})$ is induced by the
conjugation.

###### Proof.

Consider the short exact sequence $1\rightarrow
C\rightarrow\widehat{G(K)}\rightarrow G(\mathbb{A}_{S})\rightarrow 1$ from
§1.3 and note that $C\subseteq\widehat{G(O)}$. By assumption, $C$ is infinite.
Let $C_{0}=\left[\widehat{G(K)},C\right]$. Since

$1\rightarrow C/C_{0}\rightarrow\widehat{G(K)}/C_{0}\rightarrow
G(\mathbb{A}_{S})\rightarrow 1$

is a central metaplectic extension, we get that $C/C_{0}$ is finite.

Since $C$ is infinite, there is a normal subgroup $C_{1}\triangleleft C_{0}$
such that $F:=C_{0}/C_{1}$ is a finite simple group. Let
$L=\bigcap_{g\in\widehat{G(K)}}g^{-1}C_{1}g$. Clearly, $L$ is a normal
subgroup of $\widehat{G(K)}$. If the conjugation action of $\widehat{G(K)}$ on
$C/L$ was trivial, the extension

$1\rightarrow C/L\rightarrow\widehat{G(K)}/L\rightarrow
G(\mathbb{A}_{S})\rightarrow 1$ (22)

would be central, contradicting $L\subsetneq C_{0}$. Hence, the conjugation
action of $\widehat{G(K)}$ on $C/L$ is nontrivial. Since $\widehat{G(K)}$ has
no nontrivial finite quotients, it follows that $C/L$ is infinite. Because
$C_{0}/L$ is both infinite and embeds in $F^{\infty}$, we get that
$C_{0}/L\cong F^{\infty}$.

Consider the short exact sequence

$1\rightarrow C_{0}/L\cong
F^{\infty}\rightarrow\widehat{G(K)}/L\rightarrow\widehat{G(K)}/C_{0}\rightarrow
1.$ (23)

We claim that (23) is not central (this is clear if $F$ is non-abelian).
Indeed, if it were central, then $\widehat{G(K)}$ would act trivially on
$C_{0}/L$ and on $C/C_{0}$. For any $z\in C/L$, the map
$\widehat{G(K)}\rightarrow C_{0}/L$ given by $g\mapsto[z,g]$ would be a
homomorphism. Since $\widehat{G(K)}$ has no finite quotients, we would get
that $C/L$ is central in $\widehat{G(K)}$, which contradicts $L\subsetneq
C_{0}$.

If $F$ is non-abelian, the short exact sequence

$1\rightarrow
F^{\infty}\rightarrow\widehat{G(O)}/L\rightarrow\widehat{G(O)}/C_{0}\rightarrow
1,$

contained in (23), satisfies the requirements of the lemma: if $\Gamma$ does
not act faithfully on $F^{\infty}$, then the action of $G(K)$ on $F^{\infty}$
in (23) is not faithful, hence (by the simplicity of $G(K)$) trivial,
contradicting the claim that (23) is not central.

Now suppose that $F=C_{p}$. The group
$\widetilde{\widetilde{G}}:=\widehat{G(K)}/C_{0}$ is a central extension of
$G(\mathbb{A}_{S})=\widehat{G(K)}/C$ by the finite group $D:=C/C_{0}$. We
denote $U:=\prod_{v\notin S}G(O_{v})$ and, for a subgroup $H\subseteq
G(\mathbb{A}_{S})$, we denote the pre-image of $H$ in
$\widetilde{\widetilde{G}}$ by $\widetilde{\widetilde{H}}$.

Returning to (23), we claim that there is a
$\widetilde{\widetilde{U}}$-invariant subgroup $V\subseteq C_{0}/L$ of finite
index. Indeed, $\widetilde{\widetilde{G}}$ acts continuously on $C_{0}/L$ and,
therefore, acts continuously on the Pontrjagin dual
$\left(C_{0}/L\right)^{\vee}$, which is a discrete space. For any
$\xi\in\left(C_{0}/L\right)^{\vee}$, the orbit
$\widetilde{\widetilde{U}}\cdot\xi$ is compact and hence finite. Taking the
intersection of the kernels of the elements in
$\widetilde{\widetilde{U}}\cdot\xi$, we get the required subgroup $V$.

We now claim that there is a valuation $v_{0}\notin S$ such that the orbit of
$V$ under $\widetilde{G(K_{v_{0}})}$ is infinite. Indeed, since $G(K)$ acts
faithfully on $C_{0}/L$, by a similar argument to the argument we used to
prove that (23) is not central, the set $\left\\{g^{-1}Vg\mid
g\in\widetilde{\widetilde{G}}\right\\}$ is infinite. By continuity, the set-
wise stabilizer $St({\widetilde{\widetilde{G}}},V)$ of $V$ in
$\widetilde{\widetilde{G}}$ is open, so
$St({\widetilde{\widetilde{G(K_{v})}}},V)=St({\widetilde{\widetilde{G}}},V)\cap\widetilde{\widetilde{G(K_{v})}}$
is open. Since $G(K_{v})$ does not have open finite index subgroups, if
$St({\widetilde{\widetilde{G(K_{v})}}},V)$ does not of infinite index in
$\widetilde{\widetilde{G(K_{v})}}$, its index divides $|D|$. Since $G(K)$ is
generated by its $|D|^{th}$ powers and is dense in
$\widetilde{\widetilde{G}}$, it follows that there is $v_{0}\notin S$ such
that $\widetilde{\widetilde{G(K_{v_{0}})}}\cdot V$ is infinite.

Let $W=\bigcap_{g\in\widetilde{\widetilde{G(K_{v_{0}})}}}g^{-1}Vg$. We will
show that the short exact sequence

$1\rightarrow
C_{0}/W\rightarrow\widehat{G(O)}/W\rightarrow\widetilde{\widetilde{U}}\rightarrow
1$ (24)

satisfies the claims of the lemma.

Firstly, $\widetilde{\widetilde{G(K_{v_{0}})}}\cdot V$ is infinite, so
$[C_{0}:W]=\infty$. It follows that $C_{0}/W\cong C_{p}^{\infty}$.

Next, $\Gamma$ acts faithfully in (24). Indeed, (24) extends to the short
exact sequence

$1\rightarrow C_{0}/W\cong C_{p}^{\infty}\rightarrow
Z/W\rightarrow\widetilde{\widetilde{U}}\cdot\widetilde{\widetilde{G(K_{v_{0}})}}\rightarrow
1,$ (25)

where $Z\subseteq\widehat{G(K)}/L$ is the preimage of
$\widetilde{\widetilde{U}}\cdot\widetilde{\widetilde{G(K_{v_{0}})}}$. If
$\gamma\in\Gamma\smallsetminus\left\\{1\right\\}$ fixes $C_{0}/W$ in (24),
then $G(K)$ would fix $C_{0}/W$ in (25), a contradiction.

Finally, let $U^{\prime}=\prod_{v\neq v_{0}}G(O_{v})$. There is an open
subgroup $M$ of $U^{\prime}$ for which

  1. (1)

There is a section $s:M\rightarrow\widetilde{\widetilde{M}}$ of the central
extension $1\rightarrow D\rightarrow\widetilde{\widetilde{M}}\rightarrow
M\rightarrow 1$ (because $D$ is finite).

  2. (2)

$s(M)$ acts trivially on $C_{0}/V$ (because $C_{0}/V$ is finite).

  3. (3)

$s(M)$ commutes with $\widetilde{\widetilde{G(K_{v_{0}})}}$ (because
$\widetilde{\widetilde{G(K_{v_{0}})}}$ is compactly generated and, for every
compact set $X\subseteq\widetilde{\widetilde{G(K_{v_{0}})}}$, $[s(M),X]=1$, if
$M$ is small enough).

It follows that $s(M)$ fixes $C_{0}/W$, as required by the lemma. ∎

######  Lemma 6.1.2.

Let $\mathbf{G}$ be a non-isotropic semisimple group over a non-archimedean
local field $F$, let $L\subseteq\mathbf{G}(F)$ be a compact open subgroup, and
let $p$ be a prime number. There is a constant $D$ such that, for every
homomorphism $\rho:L\rightarrow\operatorname{GL}_{N}(\mathbb{F}_{p})$,
$|\rho(L)|\leq DN^{\dim\mathbf{G}}$.

###### Proof.

Without loss of generality, we can assume that $F=\mathbb{Q}_{\ell}$ and that
$L\subset\mathbf{G}(\mathbb{Z}_{\ell})[\ell]$. For an open subgroup
$H\subseteq\mathbf{G}(\mathbb{Z}_{\ell})$, we denote
$\delta(H)=\min\left\\{k\mid\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{k}]\subseteq
H\right\\}$. For any normal and open subgroup $H\triangleleft L$,

$\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{\delta(H)}]\subseteq
H\subseteq\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{\delta(H)-\delta(L)}].$ (26)

We distinguish between two cases:

  1. Case 1:

$\ell=p$. $L$ is a pro-$p$ group, so, for every $g\in L$, $\rho(g)$ is
nilpotent and, hence, $\rho(g)^{p^{M}}=1$, where
$M=\lfloor\log_{p}(N)\rfloor$. Denote the word $x^{p^{M}}$ by $w$. Since
$\mathbf{G}(\mathbb{Z}_{p})[p^{\delta(L)}]$ is a powerful pro-$p$ group,

$\ker(\rho)\supseteq\left\langle w(L)\right\rangle\supseteq\left\langle
w\left(G(\mathbb{Z}_{p})[p^{\delta(L)}]\right)\right\rangle\supseteq
G(\mathbb{Z}_{p})[p^{\delta(L)+M}],$

so

$|\rho(L)|\leq\left[G(\mathbb{Z}_{p}):G(\mathbb{Z}_{p})[p^{\delta(L)+M}]\right]\leq\left[G(\mathbb{Z}_{p}):G(\mathbb{Z}_{p})[p^{\delta(L)}]\right]\cdot
N^{\dim\mathbf{G}}.$

  2. Case 2:

$\ell\neq p$. Let $\mathbf{T}\cong\mathbb{G}_{m}\subset\mathbf{G}$ be a split
torus in $\mathbf{G}$. There is a unipotent element $u\in L$ that is an
eigenvector of $\mathbf{T}$: for some character
$\lambda:\mathbf{T}\rightarrow\mathbb{G}_{m}$, we have
$t^{-1}ut=u^{\lambda(t)}$, for every $t\in\mathbf{T}(\mathbb{Z}_{\ell})\cap
L$. We can assume that
$u\notin\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{\delta(L)+1}]$. The subgroup
$\lambda(\mathbf{T}(\mathbb{Z}_{\ell})\cap L)$ has finite index in
$\mathbb{Z}_{\ell}^{\times}$. We denote this index by $c$.

Since $\mathbf{G}(\mathbb{Z}_{\ell})[\ell]$ is powerful and since
$u\notin\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{\delta(L)+1}]$, we get that
$u^{\ell^{\delta(\ker\rho)-2\delta(L)-1}}\notin\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{\delta(\ker\rho)-\delta(L)}]$.
By (26), $u^{\ell^{\delta(\ker\rho)-2\delta(L)-1}}\notin\ker(\rho)$.

Since $L$ is a pro-$\ell$ group, the order of $\rho(u)$ is a power of $\ell$
and is at least $\ell^{\delta(\ker\rho)-2\delta(L)}$. Since $\ell\neq p$, this
means that $\rho(u)$ is semisimple and one of its eigenvalues is a root of 1
of order at least $\ell^{\delta(\ker\rho)-2\delta(L)}$. Using conjugation and
by the definition of $c$, $\rho(u)$ has at least
$\frac{1}{c}\ell^{\delta(\ker\rho)-2\delta(L)}$ different eigenvalues, so
$N\geq\frac{1}{c}\ell^{\delta(\ker\rho)-2\delta(L)}$. It follows that

$|\rho(L)|\leq[\mathbf{G}(\mathbb{Z}_{\ell})[\ell]:\mathbf{G}(\mathbb{Z}_{\ell})[\ell^{\delta(\ker\rho)}]]\leq\ell^{\delta(\ker\rho)\cdot\dim\mathbf{G}}\leq\left(c\ell^{2\delta(L)}\right)^{\dim
G}\cdot N^{\dim\mathbf{G}}.$

∎

###  6.2 Proof of Theorem 1.1.3

###### Proof of Theorem 1.1.3.

The implication $CSP\implies FCW$ follows from [2, Lemma 3.20] and [2, Lemma
3.22]. It remains to prove the converse.

Assume that $\Gamma=\mathbf{G}(O)$ does not have CSP. Applying Lemma 6.1.1, we
get a short exact sequence $1\rightarrow F^{\infty}\rightarrow H\rightarrow
X\rightarrow 1$ and, if $F$ is abelian, a place $v_{0}\notin S$ and a subgroup
$L\subseteq X$ satisfying the conditions of Lemma 6.1.1. Denote the
characteristic of the residue field of $K_{v_{0}}$ by $\ell$. We will show
that $H$ has infinite conjugacy width and, therefore, so does
$\widetilde{\Gamma}$.

Suppose first that $F$ is non-abelian. Consider the composition

$\widehat{\Gamma}\rightarrow
H\rightarrow\operatorname{Aut}\left(F^{\infty}\right)\cong\operatorname{Sym}(\infty)\ltimes\operatorname{Aut}(F)^{\infty}\rightarrow\operatorname{Sym}(\infty),$

where $\operatorname{Sym}(\infty)$ denotes the group of all permutations of
$\mathbb{N}$ with the topology of pointwise convergence.

By compactness, the orbits of $\widehat{\Gamma}$ on $\mathbb{N}$ are finite.
Since $\operatorname{Aut}(F)^{\infty}$ has a finite exponent, the sizes of the
orbits of $\widehat{\Gamma}$ are unbounded (otherwise, some non-trivial
elements of $\Gamma$ would act trivially).

Let $O_{1},O_{2},\ldots\subset\mathbb{N}$ be the $\widehat{\Gamma}$-orbits.
Choose representatives $i_{j}\in O_{j}$, choose a non-trivial element $f\in
F$, and let $v\in F^{\infty}$ be the element

$v_{k}=\begin{cases}f&k\in\left\\{i_{1},i_{2},\ldots,\right\\}\\\
1&k\notin\left\\{i_{1},i_{2},\ldots,\right\\}\end{cases}.$

We claim that the conjugacy class of $v$ has infinite width in $H$. Indeed, it
is easy to see that
$\left\langle\operatorname{gcl}_{H}(v)\right\rangle=F^{\infty}$, but, for any
two natural numbers $C,n$, if $g\in\operatorname{gcl}_{H}(v)^{C}$, then
$|\operatorname{supp}(g)\cap O_{n}|\leq C$. Since $|O_{n}|$ are unbounded,
$\operatorname{gcl}_{H}(v)^{C}\neq F^{\infty}$, for any $C$.

Next, suppose that $F$ is abelian. Then $F=C_{p}$, for some prime $p$. We
first claim that there is an element $u\in C_{p}^{\infty}$ such that
$\rho(X)\cdot u$ is infinite. Assume that $\rho(X)\cdot x$ is finite, for
every $x\in C_{p}^{\infty}$. The function $x\mapsto|\rho(X)\cdot x|$ is
unbounded since, otherwise, a non-trivial subgroup of $\Gamma$ would act
trivially. For every $x,y\in C_{p}^{\infty}$, we have

$|\rho(X)(x+y)|\leq|\rho(X)x|\cdot|\rho(X)y|.$ (27)

It follows that the function $x\mapsto|\rho(X)\cdot x|$ is unbounded on every
open subgroup of $C_{p}^{\infty}$. We recursively construct a decreasing
sequence $W_{k}$ of $\rho(X)$-invariant open subgroups of $C_{p}^{\infty}$ and
a sequence $u_{k}$ of elements of $C_{p}^{\infty}$ such that

  1. (1)

$u_{k+1}\in W_{k}$.

  2. (2)

$\bigcap W_{k}=0$.

  3. (3)

The $X$-orbit of the element $(u_{0}+\cdots+u_{k})+W_{k}\in
C_{p}^{\infty}/W_{k}$ has size greater than $k$.

For $k=0$, take $W_{0}=C_{p}^{\infty},u_{k}=0$. Given
$W_{0},\ldots,W_{k},u_{0},\ldots,u_{k}$, choose $u_{k+1}\in W_{k}$ such that
$\left|\rho(X)(u_{1}+\cdots+u_{k})\right|(k+1)<\left|\rho(X)(u_{k+1})\right|$.
Since $u_{k+1}=(u_{1}+\ldots+u_{k+1})-(u_{1}+\ldots+u_{k})$, by (27),
$\left|\rho(X)(u_{1}+\cdots+u_{k+1})\right|>k+1$, so there is an open subgroup
$W_{k+1}$ such that the orbit $\rho(X)(u_{1}+\cdots+u_{k+1}+W_{k+1})$ has size
greater than $k+1$. Shrinking $W_{k+1}$, we can further require that
$W_{k+1}\subseteq W_{k}$, that the first $k$ coordinates of every element in
$W_{k+1}$ are zero (so $\bigcap W_{k}=0$), and that $W_{k+1}$ is
$\rho(X)$-invariant.

It follows from the first two properties that the series $u_{1}+u_{2}+\cdots$
converges. Denote the sum by $u$. Since the projection of $\rho(X)\cdot u$ to
$C_{p}^{\infty}/V_{k}$ has size greater than $k$, for every $k$, it follows
that $\rho(X)\cdot u$ is infinite. We will show that the width of the
conjugacy class of $u$ in $H$ is infinite.

Denote $V=\overline{\langle\rho(X)\cdot u\rangle}$. By compactness, there is a
decreasing sequence $V=V_{0}\supseteq V_{1}\supseteq V_{2}\supseteq\cdots$ of
$\rho(X)$-invariant open subgroups of $V$ such that $\cap V_{i}=0$. Denote the
projection $V\rightarrow V/V_{n}$ by $\pi_{n}$ and denote the representation
of $X$ on $V/V_{n}$ by $\rho_{n}$. We have

$\left|\pi_{n}\left(\langle\rho(X)\cdot
u\rangle\right)\right|=\left|\langle\rho_{n}(X)\cdot\pi_{n}(u)\rangle\right|=\left|V/V_{n}\right|\geq
2^{\dim V/V_{n}},$

since $\pi_{n}(u)$ generates $V/V_{n}$.

On the other hand, by the assumption on $L$, there are finitely many elements
$g_{1},\ldots,g_{m}\in X$ such that $\rho(X)=\cup g_{i}\rho(L)$. Thus, by
Lemma 6.1.2, there exists a constant $D$ such that for every $C$,

$\left|\pi_{n}\left(\left(\rho(X)\cdot
u\right)^{C}\right)\right|=\left|\left(\rho_{n}(X)\cdot\pi_{n}(u)\right)^{C}\right|=\left|\left(\bigcup_{1}^{m}\rho_{n}(g_{i})\rho_{n}(L)\cdot\pi_{n}(u)\right)^{C}\right|\leq$
$\leq
m^{C}\left|\rho_{n}(L)\cdot\pi_{n}(u)\right|^{C}\leq\left(m\left|\rho_{n}(L)\right|\right)^{C}\leq\left(D\cdot\dim
V/V_{n}\right)^{2C}.$

Since $\dim V/V_{n}$ is unbounded, we get

$\langle\operatorname{gcl}_{H}(u)\rangle=\langle\rho(X)\cdot
u\rangle\not\subseteq\left(\rho(X)\cdot
u\right)^{C}=\operatorname{gcl}_{H}(u)^{C},$

so the width of the conjugacy class of $u$ in $H$ is infinite. ∎

######  Proposition 6.2.1.

Suppose that Conjecture 1.1.2 holds for $(K,S,\mathbf{G},\Gamma)$. Then the
pro-finite completion of $\Gamma$ has finite conjugacy width.

###### Proof.

It is enough to show that there is constant $N$ such that, for any finite
index $\Delta\triangleleft\Gamma$ and every $g\in\Gamma$, the width of
$\operatorname{gcl}_{\Gamma/\Delta}(g\Delta)$ in $\Gamma/\Delta$ is bounded by
$CN$. By the assumption, it is enough to show that there is an element
$g^{\prime}\in g\Delta$ such that
$\operatorname{wid}_{\mathbf{G}(K_{v})}\left(\operatorname{gcl}_{\mathbf{G}(K_{v})}(g^{\prime})\right)\leq
N$, for every $v\in S$.

Let $S_{\mathbb{R},cpt}$ be the set of real places for which
$\mathbf{G}(K_{v})$ is compact. For each $v\in S_{\mathbb{R},cpt}$, [2, Lemma
3.17] implies that there is an open set $U_{v}\subseteq\mathbf{G}(K_{v})$ and
a natural number $N_{v}$ such that
$\operatorname{wid}_{G(K_{v})}\left(\operatorname{gcl}_{\mathbf{G}(K_{v})}(x)\right)\leq
N_{v}$, for every $x\in U_{v}$. For $v\in S\smallsetminus S_{\mathbb{R},cpt}$,
one of [2, Lemma 3.18] or [2, Lemma 3.20] implies that there is a natural
number $N_{v}$ such that
$\operatorname{wid}_{G(K_{v})}\left(\operatorname{gcl}_{\mathbf{G}(K_{v})}(x)\right)\leq
N_{v}$ for every $x\in\mathbf{G}(K_{v})$. We show that the claim holds with
$N=\max\left\\{N_{v}\mid v\in S\right\\}$.

Let $g\in\Gamma$. For every $v\in S_{\mathbb{R},cpt}$, $\mathbf{G}(K_{v})$ is
connected. Since $\Delta$ is commensurable to $\mathbf{G}(O_{S})$, $\Delta$ is
dense in $\prod_{v\in S_{\mathbb{R},cpt}}\mathbf{G}(K_{v})$, so there is
$g^{\prime}\in g\Delta$ such that $g^{\prime}\in U_{v}$, for every $v\in
S_{\mathbb{R},cpt}$. The claim follows from the last paragraph. ∎

##  Appendix A Items 2 and 3 of Setting 3.1 hold for
$\operatorname{Spin}_{f}(O)$

In this appendix we prove that Items 2 and 3 of Setting 3.1 hold for many
higher rank spin groups.

###  A.1 Setting

Let $K,S,O$ be as in Setting 2.1. In this appendix, we fix $n\geq 12$ and a
non-degenerate anisotropic quadratic form $q:O^{n}\rightarrow O$. We assume
that there is a place $u\in S$ such that $i_{q}(K_{u}^{n})\geq 2$. We also
assume that $\operatorname{Spin}_{q}(O)$ has the Congruence Subgroup Property
in the strong sense, i.e., that every finite index subgroup of
$\operatorname{Spin}_{q}(O)$ contains a principal congruence subgroup. All
these assumptions hold for
$\operatorname{Spin}_{f_{a}}\left(\mathbb{Z}\left[\frac{1}{2}\right]\right)$.

We will use the following notations:

  * •

$(\cdot,\cdot)_{q}$ (or simply $(\cdot,\cdot)$) denotes the bilinear from
induced by $q$.

  * •

$\theta:\operatorname{SO}_{q}(K)\rightarrow K^{\times}/(K^{\times})^{2}$
denotes the spinor norm.

  * •

Denote
$\Theta_{q}(K)=\\{\alpha\in\operatorname{SO}_{q}(K)\mid\theta(\alpha)=(K^{\times})^{2}\\}$.

  * •

The natural homomorphism
$\operatorname{Spin}_{q}(K)\rightarrow\operatorname{SO}_{q}(K)$ induces an
action of $\operatorname{Spin}_{q}(K)$ on $K^{n}$. For
$\alpha\in\operatorname{Spin}_{q}(K)$ and $v\in K^{n}$, we denote the image of
$v$ under the action of $\alpha$ by $\alpha v$ or $\alpha(v)$.

  * •

We view $\operatorname{Spin}_{q}$ as a subscheme of the Clifford algebra
$\operatorname{Cliff}_{q}$ of $q$. For $a\in K^{n}$ denote the image of $a$ in
$\operatorname{Cliff}_{q}$ by $e_{a}$.

  * •

For every non-isotropic $a\in K^{n}$, denote the reflection in the hyperplane
$a^{\perp}$ by $\tau_{a}\in\mathbb{O}_{q}(K)$.

Note that, for every non-isotropic $a\in K^{n}$ and every $b\in K^{n}$,
$\frac{1}{q(a)}e_{a}e_{b}e^{\prime}_{a}=e_{-\tau_{a}(b)}$ where
${}^{\prime}:\operatorname{Cliff}_{q}\rightarrow\operatorname{Cliff}_{q}$ is
the canonical involution.

######  Remark A.1.1.

The results in this appendix remain true if $q$ is isotropic but the proofs
should be modified.

###  A.2 $\epsilon$-separated elements

######  Definition A.2.1.

Let $S_{real}$ be the set of real places of $K$ and let $S_{def}$ be the set
of places $v\in S_{real}$, for which $\operatorname{Spin}_{q}(K_{v})$ is
compact.

  1. (1)

For every $v\in S_{def}$, let $||\cdot||_{v}$ be the norm on $K_{v}^{n}$
defined by $||a||_{v}:=\sqrt{q(a)}$ for every $a\in(K_{v})^{n}$.

  2. (2)

For every $v\in S_{def}$, let $\operatorname{dist}_{v}(\cdot,\cdot)$ be the
bi-invariant metric on $\operatorname{SO}_{q}(K_{v})$ defined by for all
$\alpha,\beta\in\operatorname{SO}_{q}(K_{v})$,

$\operatorname{dist}_{v}(\alpha,\beta):=\sup\\{||(\alpha-\beta)a||_{v}\mid
a\in K_{v}^{n}\text{ and }||a||_{v}=1\\}.$

  3. (3)

An element $\alpha\in\operatorname{SO}_{q}(K)$ is called $\epsilon$-separated
if for every $v\in S_{def}$,
$\operatorname{dist}_{v}(\alpha,Z(\operatorname{SO}_{q}(K_{v})))\geq\epsilon$.

  4. (4)

An element $\alpha\in\operatorname{Spin}_{q}(K)$ is called
$\epsilon$-separated if its image in $\Theta_{q}(K)$ is $\epsilon$-separated.

######  Definition A.2.2.

Under Setting A.1, let $V\subseteq K^{n}$ be a regular subspace and denote
$M:=V\cap O^{n}$. Define an $O$-subscheme
$\operatorname{Spin}_{q\restriction_{V}}$ of $\operatorname{Spin}_{q}$ by

$\operatorname{Spin}_{q\restriction_{V}}(A):=\\{\alpha\in\operatorname{Spin}_{q}(A)\mid\alpha\text{
pointwise fixes }(M\otimes A)^{\perp}\\},$

for every $O$-algebra $A$.

######  Definition A.2.3.

Under Setting A.1, let $\alpha\in\operatorname{Spin}_{q}(K)$.

  1. (1)

The support of $\alpha$ is the subspace
$\operatorname{supp}(\alpha):=\left\\{a\in
K^{n}\mid\alpha(a)=a\right\\}^{\perp}$.

  2. (2)

We say that $\alpha$ is strongly $\epsilon$-separated if
$\alpha\restriction_{\operatorname{supp}(\gamma)}$ is $\epsilon$-separated in
$\operatorname{Spin}_{q\restriction_{\operatorname{supp}(\gamma)}}(K)$.

######  Lemma A.2.4.

Under Setting A.1, there exists $N$ such that the following holds:

Let $1\leq k\leq n-6$ and let $c_{1},\ldots,c_{k}\in A^{n}$ be non-isotropic
orthogonal vectors such that

$i_{q}(K_{u}c_{0}+\cdots+K_{u}c_{k})\geq 1\text{ and
}i_{q}\left(K_{u}c_{k}+\cdots+K_{u}c_{n}\right)\geq 2.$

For every $1$-separated element $\alpha\in\operatorname{Spin}_{q}(O)$,
$(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha)^{N})(c_{1},\ldots,c_{k})$
contains an open neighborhood of $(c_{1},\ldots,c_{k})$ in
$\operatorname{Spin}_{q}(O)(c_{1},\ldots,c_{k})$ with respect to the
$S$-adelic topology.

###### Proof.

The proof is almost identical to [2, Lemma 4.16] and is omitted. ∎

######  Remark A.2.5.

The differences between [2, Lemma 4.16] and Lemma A.2.4 are:

  * •

[2, Lemma 4.16] is stated for $k=3$. The generalization to $1\leq k\leq n-6$
follows from a simple induction.

  * •

[2, Lemma 4.16] is stated for centerless congruence subgroups of
$\Theta_{q}(K)$. The assumption that the center of the congruence subgroup is
trivial is used in a different part of [2] and is not needed for [2, Lemma
4.16].

######  Lemma A.2.6.

There is $M\in\mathbb{N}$ such that, for every $v\in
S_{real}\cup\left\\{u\right\\}$ and every $g\in\Theta_{q}(K_{v})$,

  1. (1)

If $v\notin S_{def}$, then
$\left(\operatorname{gcl}_{\Theta_{q}(K_{v})}(g)\right)^{M}=\Theta_{q}(K_{v})$.

  2. (2)

If $v\in S_{def}$, then
$\left(\operatorname{gcl}_{\Theta_{q}(K_{v})}(g)\right)^{\frac{M}{\operatorname{dist}_{v}(g,Z(\Theta_{q}(K_{v})))}}=\Theta_{q}(K_{v})$.

###### Proof.

The first claim is a special case of [2, Lemma 3.18].

For the second claim, we first note that, by [2, Lemma 3.17], there is a
constant $M_{1}$ such that, if
$\operatorname{dist}_{v}(g,Z(\Theta_{q}(K_{v})))>\frac{1}{10}$, then
$\left(\operatorname{gcl}_{\Theta_{q}(K_{v})}(g)\right)^{M_{1}}=\Theta_{q}(K_{v})$.
If $\operatorname{dist}_{v}(g,Z(\Theta_{q}(K_{v})))\leq\frac{1}{10}$, then one
of the eigenvalues of $g^{2}$ is not in the ball of radius
$\frac{1}{2}\operatorname{dist}_{v}(g,Z(\Theta_{q}(K_{v})))$ around 1. It
follows that, for some
$N\leq\frac{4}{\operatorname{dist}_{v}(g,Z(\Theta_{q}(K_{v})))}$, $g^{N}$ has
an eigenvalue whose real part is greater than $\frac{1}{5}$, which implies
that $\operatorname{dist}_{v}(g^{N},Z(\Theta_{q}(K_{v})))>\frac{1}{10}$. This
implies that
$\left(\operatorname{gcl}_{\Theta_{q}(K_{v})}(g)\right)^{M_{1}N}=\Theta_{q}(K_{v})$,
proving the claim for $M=4M_{1}$. ∎

######  Lemma A.2.7.

There exists a constant $N$ such that for every $\epsilon>0$, every
$m\geq\frac{N}{\epsilon}$, the following holds.

For every $\epsilon$-separated element $\alpha\in\Theta_{q}(K)$ and every
regular subspace $U$ of $K^{n}$ of dimension at least $6$,
$(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha))^{m}\cap\Theta_{q\restriction_{U}}(K)$
is dense in $\prod_{v\in S_{real}}\Theta_{q\restriction_{U}}(K_{v})$. In
particular, there exists
$\beta\in(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha))^{m}\cap\Theta_{q\restriction_{U}}(K)$
which is strongly $1$-separated.

###### Proof.

Let $M$ be the constant of Lemma A.2.6. Suppose that $g$ is
$\epsilon$-separated. By weak approximation, there is an element
$g_{1}\in\left(\operatorname{gcl}_{\Theta_{q}(K)}(g)\right)^{\frac{M}{\epsilon}}$
and, for every $v\in S_{real}\cup\left\\{u\right\\}$, an element $a_{v}\in
K_{v}^{n}$ such that

  1. (1)

For every $v\in S_{real}$,
$\operatorname{Span}\left\\{a_{v},g_{1}a_{v},g_{1}^{2}a_{v}\right\\}$ is
regular and embeds isometrically into $U\otimes K_{v}$.

  2. (2)

For every $v\in S_{real}\smallsetminus S_{def}$,
$\\{a_{v},g_{1}a_{v},g_{1}^{2}a_{v}{\\}}^{\perp}$ is undefinite.

  3. (3)

For every $v\in S_{def}$, the angles between $a_{v},g_{1}a_{v},g_{1}^{2}a_{v}$
are between $89^{\circ}$ and $91^{\circ}$.

Since $K^{n}$ is dense in $\prod_{v\in
S_{real}\cup\left\\{u\right\\}}K_{v}^{n}$ and the conditions above are open,
by weak approximation, after perturbing $a_{v}$, we can assume that there is a
vector $b\in K^{n}$ such that $b_{v}=a_{v}$, for all $v\in
S_{real}\cup\left\\{u\right\\}$. Since $\dim U\geq 6$, for every non-real
place $v$, there is an isometric embedding
$\operatorname{Span}\left\\{b_{v},g_{1}b_{v},g_{1}^{2}b_{v}\right\\}\rightarrow
U\otimes K_{v}$. By (1) and Hasse–Minkowski, there is an isometric embedding
$\operatorname{Span}\left\\{b,g_{1}b,g_{1}^{2}b\right\\}\rightarrow U$. By (2)
and a similar argument to the one in the proof of Lemma 5.3.1, there is
$k\in\Theta_{q}(K)$ such that
$k\left(\operatorname{Span}\left\\{b,g_{1}b,g_{1}^{2}b\right\\}\right)\subseteq
U$. Let $c=k(b)$ and $g_{2}=kg_{1}k^{-1}$.

Denote $W=\operatorname{Span}\left\\{c,g_{2}c,g_{2}^{2}c\right\\}\subseteq U$.
By weak approximation, there is an element $h\in\Theta_{q\restriction W}$ such
that, for all $v\in S_{def}$, the angle between
$g_{2}h^{-1}g_{2}^{-1}(c),h(c)$ is between $89^{\circ}$ and $91^{\circ}$. It
follows that
$[h,g_{2}]\in\left(\operatorname{gcl}_{\Theta_{q}(K)}(g)\right)^{\frac{2M}{\epsilon}}\cap\Theta_{q\restriction
U}(K)$ is a 1-separated element. The first claim follows from this. ∎

######  Lemma A.2.8.

Let $g\in\operatorname{Spin}_{q}(O)$ be a non-central element. If
$\left(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(g)\right)^{m}$ is dense
in $\prod_{v\in S_{def}}\operatorname{Spin}_{q}(K_{v})$, then $g$ is
$\frac{\sqrt{2}}{m}$-separated.

###### Proof.

If $v\in S_{def}$, the function $D(g,h)=\inf\left\\{\|ga-
ha\|_{v}\mid\|a\|_{v}=1\right\\}$ is a translation invariant pseudometric on
$\operatorname{Spin}_{q}(K_{v})$. For every
$g,h,z\in\operatorname{Spin}_{q}(K_{v})$, $D(gh,z)\leq
D(gh,gz)+D(gz,z)=D(h,z)+D(g,1)$. Thus, for every
$g\in\operatorname{Spin}_{q}(K_{v})$, every $m$, and every
$h\in\left(\operatorname{gcl}_{\operatorname{Spin}_{q}(K_{v})}(g)\right)^{m}$,
$D(h,Z(\operatorname{Spin}_{q}(K_{v})))\leq
mD(g,Z(\operatorname{Spin}_{q}(K_{v})))$. Since there is an element
$g\in\operatorname{Spin}_{q}(K_{v})$ such that
$D(g,Z(\operatorname{Spin}_{q}(K_{v})))=\sqrt{2}$, the claim follows. ∎

###  A.3 Item 2 of Setting 3.1 holds in $\operatorname{Spin}_{q}(O)$

######  Notation A.3.1.

Let $N_{1}$ be the constant from Lemma A.2.4 and $N_{2}$ be the constant from
Lemma A.2.7. Denote $N:=\max(N_{1},N_{2})$

######  Lemma A.3.2.

Let $c_{1},\ldots,c_{n}\in O^{n}$ be non-isotropic orthogonal vectors such
that $i_{q}(K_{u}c_{0}+\cdots+K_{u}c_{6})\geq 1$ and
$i_{q}\left(K_{u}c_{6}+\cdots+K_{u}c_{n}\right)\geq 2$. Denote
$U:=\operatorname{Span}_{K}\\{c_{i}\mid 1\leq i\leq 6\\}$ and let $N$ be as in
Notation A.3.1. There exists a non-zero ideal $J$ of $O$ such that
$\operatorname{Spin}_{q}(O)[J]\subseteq\big{(}\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\operatorname{Spin}_{q\restriction_{U}}(O))\big{)}^{N}$.

###### Proof.

There exists $\alpha\in\operatorname{Spin}_{q\restriction_{U}}(O)$ which is
$1$-separated as an element of $\operatorname{Spin}_{q\restriction_{U}}(O)$
and thus also as an element of $\operatorname{Spin}_{q}(O)$. Lemma A.2.4
implies that there exists a non-zero ideal $J$ such that for every
$\beta\in\operatorname{Spin}_{q}(O)[J]$, there exists
$\gamma\in\big{(}\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha)\big{)}^{N}$
such that $\gamma^{-1}\beta\in\operatorname{Spin}_{q\restriction_{U}}(O)$.
Thus,
$\beta\in\big{(}\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\operatorname{Spin}_{q\restriction_{U}}(O))\big{)}^{N+1}$.
∎

######  Proposition A.3.3.

Let $N$ be as in Notation A.3.1. There exists $J\in\mathcal{I}_{K}$ for which
the following statement holds:

For every $1$-separated element $\alpha\in\operatorname{Spin}_{q}(O)$, there
exists $I\in\mathcal{I}_{K}$ for which

$\\{[\beta_{1},\beta_{2}]\mid\beta_{1}\in\operatorname{Spin}_{q}(O)[I]\text{
and
}\beta_{2}\in\operatorname{Spin}_{q}(O)[J]\\}\subseteq(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{2N^{2}}.$

###### Proof.

Let $\alpha\in\operatorname{Spin}_{q}(O)$ be a 1-separated element. There
exist non-isotropic pairwise orthogonal vectors $v_{1},\ldots,v_{n}\in O^{n}$
such that:

  1. (a)

$i_{q}(K_{u}v_{0}+\cdots+K_{u}v_{6})\geq 1$ and
$i_{q}(K_{u}v_{7}+\cdots+K_{u}v_{n})\geq 1$.

  2. (b)

$i_{q}\left(K_{u}v_{0}+\cdots+K_{u}v_{7}\right)\geq 2$ and
$i_{q}\left(K_{u}v_{6}+\cdots+K_{u}v_{n}\right)\geq 2$.

Denote $U:=K_{u}v_{1}+\cdots+Kv_{6}$ and $V:=Kv_{7}+\cdots+Kv_{n}$ and let $J$
be as in Lemma A.3.2. By Lemma A.2.4 there exists $I\in\mathcal{I}_{k}$ such
that for every $\beta\in\operatorname{Spin}_{q}(O)[I]$, there exists
$\gamma\in(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{N}$ for
which $\beta\gamma\in\operatorname{Spin}_{q\restriction_{V}}(O)$.

Let $\beta\in\operatorname{Spin}_{q}(O)[I]$,
$\delta\in\operatorname{Spin}_{q\restriction_{U}}(O)$ and
$\chi\in\operatorname{Spin}_{q}(O)$. Let
$\gamma\in(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{N}$ such
that $\beta\gamma\in\operatorname{Spin}_{q\restriction_{V}}(O)$. Then
$1=[\beta\gamma,\delta]=\beta[\gamma,\delta]\beta^{-1}[\beta,\delta]$ so, by
the normality and symmetry of
$\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha)$,

$[\beta,\delta]=\beta[\gamma,\delta]^{-1}\beta^{-1}\in(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{2N}.$

Since $\chi^{-1}\beta\chi\in\operatorname{Spin}_{q}(O)[I]$, by the normality
of $\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha)$,

$[\beta,\chi\delta\chi^{-1}]=\chi[\chi^{-1}\beta\chi,\delta]\chi^{-1}\in(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{2N}.$

Similarly,
$[\beta,\chi\delta^{-1}\chi^{-1}]\in(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{2N}$.
By the definition of $N$ and $J$, the identity $[x,yz]=[x,y]y[x,z]y^{-1}$ and
the normality of $\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha)$, we
get that

$\\{[\beta_{1},\beta_{2}]\mid\beta_{1}\in\operatorname{Spin}_{q}(O)[I]\text{
and
}\beta_{2}\in\operatorname{Spin}_{q}(O)[J]\\}\subseteq(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{2N^{2}}.$

∎

######  Lemma A.3.4.

Given $q$ and $w$, there is $N$ such that, for every two ideals $I\subseteq
J$,
$\operatorname{wid}_{w}(\operatorname{Spin}_{q}[J]/\operatorname{Spin}_{q}[I])<N$.

###### Proof.

Denote $G=\operatorname{Spin}_{q}$. By Strong Approximation, it is enough to
find $N$ such that, for every $P\in\mathcal{P}_{K}$ and $a\in\mathbb{N}$,
$\operatorname{wid}_{w}(G(O_{P})[P^{a}])<N$.

For $P\in\mathcal{P}_{K}$ and $g\in G(O_{P})\smallsetminus Z(G(O_{P}))$, let
$v_{P}(g)=\max\left\\{n\mid g\in Z(G(O_{P}))\cdot G(O_{P})[P^{n}]\right\\}$.
Define also

$\mathcal{Q}=\left\\{P\mid\text{the characteristic of $O/P$ is less than
$n^{16}$ or $(O_{P}^{n},q)$ is not regular}\right\\}.$

The primes in $\mathcal{Q}$ are precisely the primes $P$ for which
$G\otimes\operatorname{Spec}(O_{P})$ is not good in the sense of [2,
Definition 3.2]. It is clear that $\mathcal{Q}$ is finite. We use the
following results from [2]:

  1. (1)

[2, Lemma 3.20]: For every $P$ there are integers $N_{P},c_{P}$ such that, for
every $g\in G(O_{P})\smallsetminus Z(G(O_{P}))$,

    1. (1a)

$\left(\operatorname{gcl}_{G(O_{P})}(g)\right)^{N_{P}}=\langle\operatorname{gcl}_{G(O_{P})}(g)\rangle$.

    2. (1b)

$G(O_{P})[P^{v(g)+c_{P}}]\subseteq\langle\operatorname{gcl}_{G(O_{P})}(g)\rangle$.

  2. (2)

[2, Lemma 3.22] If $P\notin\mathcal{Q}$, then, for every $g\in G(O_{P})$,
$\left(\operatorname{gcl}_{G(O_{P})}(g)\right)^{10n^{2}}=\langle\operatorname{gcl}_{G(O_{P})}(g)\rangle$.

We now show that the claim of the lemma holds with

$N:=\max\left\\{10n^{2}+2,N_{P}\cdot|P|^{2c_{P}n^{2}}\mid
P\in\mathcal{Q}\right\\}.$

Let $w$ be a non-trivial word, let $P$ be a prime, and let $a\in\mathbb{N}$.
Since $G(O_{P})[P^{a}]$ contains a free group, $w(G(O_{P})[P^{a}])$ contains
non-central elements. Choose $g\in w(G(O_{P})[P^{a}])$ with minimal $v(g)$. By
definition, $\langle w(G(O_{P})[P^{a}])\rangle\subseteq Z(G(O_{P}))\cdot
G(O_{P})[P^{v(g)}]$. Since $w(G(O_{P})[P^{a}])$ is characteristic in
$G(O_{P})[P^{a}]$, it is invariant under the conjugation action of $G(O_{P})$.
Thus, $\operatorname{gcl}_{G(O_{P})}(g)\subseteq w(G(O_{P})[P^{a}])$.

If $P\in\mathcal{Q}$, then, by (1),

$G(O_{P})[P^{v(g)+c_{P}}]\subseteq\langle\operatorname{gcl}_{G(O_{P})}(g)\rangle=\left(\operatorname{gcl}_{G(O_{P})}(g)\right)^{N_{P}}\subseteq$
$\subseteq\left(w(G(O_{P})[P^{a}])\right)^{N_{P}}\subseteq\langle
w(G(O_{P})[P^{a}])\rangle\subseteq Z(G(O_{P}))\cdot G(O_{P})[P^{v(g)}]$

and since

$\left[Z(G(O_{P}))\cdot
G(O_{P})[P^{v(g)}]:G(O_{P})[P^{v(g)+c_{P}}]\right]\leq|P|^{2c_{P}n^{2}},$

we get that $\left(w(G(O_{P})[P^{a}])\right)^{N}=\langle
w(G(O_{P})[P^{a}])\rangle$.

If $P\notin\mathcal{Q}$, then any normal subgroup of $G(O_{P})$ is sandwiched
between a congruence subgroup and the product of the center and that
congruence subgroup, so $G(O_{P})[P^{v(g)}]\subseteq\langle
w(G(O_{P})[P^{a}])\rangle\subseteq Z(G(O_{P}))\cdot G(O_{P})[P^{v(g)}]$. A
argument similar to the above that uses 2 shows that
$\left(w(G(O_{P})[P^{a}])\right)^{N}=\langle w(G(O_{P})[P^{a}])\rangle$ in
that case as well.

∎

######  Corollary A.3.5.

For every non-trivial word $w$, Item 2 of Setting 3.1 holds.

###### Proof.

  * •

For every $I\in\mathcal{I}_{K}$, $\operatorname{Spin}_{q}(O)[I]$ is dense in
$\prod_{v\in S_{def}}\operatorname{Spin}_{q}(K_{v})$ and thus contains an
$1$-separated element. Proposition A.3.3 implies that there exist a constant
$N_{1}$ and $J_{1}\in\mathcal{I}_{K}$ such that for every
$I\in\mathcal{I}_{K}$, there exists $I^{\prime}\in\mathcal{I}_{K}$ for which

$\\{[\alpha,\beta]\mid\alpha\in\operatorname{Spin}_{q}(O)[J_{1}],\
\beta\in\operatorname{Spin}_{q}(O)[I^{\prime}]\\}\subseteq\big{(}w(\operatorname{Spin}_{q}(O)[I])\big{)}^{N_{1}}.$

  * •

Let $J^{\prime}\in\mathcal{I}_{K}$. Since
$\operatorname{Spin}_{q}(O)[J^{\prime}]$ contains a non-abelian free group,
$(\operatorname{Spin}_{q}(O)[J^{\prime}])_{w}$ is an infinite normal subgroup
of $\operatorname{Spin}_{q}(O)$. By Margulis’ Normal Subgroup Theorem,
$(\operatorname{Spin}_{q}(O)[J^{\prime}])_{w}$ has a finite index in
$\operatorname{Spin}_{q}(O)$. Since we assume that
$\operatorname{Spin}_{q}(O)$ has the CSP in the strong sense, there exists
$J_{2}\in\mathcal{I}_{K}$ such that
$\operatorname{Spin}_{q}(O)[J_{2}]\subseteq(\operatorname{Spin}_{q}(O)[J^{\prime}])_{w}$.

  * •

By Lemma A.3.4, there exists a constant $N_{2}$ such that for every
$I\in\mathcal{I}_{K}$ which is contained in $J^{\prime}$,
$\operatorname{wid}_{w}\big{(}\operatorname{Spin}_{q}(O)[J^{\prime}]/\operatorname{Spin}_{q}(O)[I]\big{)}\leq
N_{2}$.

  *

It follows that Item 2 of Setting 3.1 holds with respect to $J:=J_{1}\cap
J_{2}$ and $N_{w}:=\max(N_{1},N_{2})$.

∎

######  Corollary A.3.6.

Let $N$ be as in Notation A.3.1. For every non-central
$\alpha\in\operatorname{Spin}_{q}(O)$ and every $m\in\mathbb{N}$, if
$\operatorname{wid}_{\operatorname{Spin}_{q}(K_{v})}(\alpha)\leq m$, for every
real place $v$, then there exists $I\in\mathcal{I}_{K}$ such that

$\\{[\beta_{1},\beta_{2}]\mid\beta_{1},\beta_{2}\in\operatorname{Spin}_{q}(O)[I]\\}\subseteq(\operatorname{gcl}_{\operatorname{Spin}_{q}(O)}(\alpha))^{2mN^{3}}$

###### Proof.

Follows from Lemma A.2.8 and Proposition A.3.3.∎

###  A.4 Item 3 of Setting 3.1 holds for $\operatorname{Spin}_{q}(K)$

The proofs in this subsection mimic the ones of [21].

######  Definition A.4.1.

Let $E$ be a field and let $f:E^{n}\rightarrow E$ a regular quadratic from.

  1. (1)

$f$ is called universal if $f(E^{n})=E$.

  2. (2)

A subspace $V$ of $E^{n}$ is called a $f$-universal subspace of $E$ if it is
regular and $f(V)=f(E^{n})$.

######  Lemma A.4.2.

Let $U$ be a $q$-universal subspace of $K^{n}$ such that $\dim U\geq 5$. Then
for every $\alpha\in\operatorname{O}_{q}(K)$, there exist $\beta,\gamma\in
O_{q\restriction_{U}}(K)$ such that $\alpha\beta$ and $\gamma\alpha$ belong to
$\Theta_{q}(K)$.

###### Proof.

Since $(\alpha^{-1}\gamma^{-1})^{-1}=\gamma\alpha$, it is enough to prove that
existence of $\beta$. It is clear that there exists
$\beta_{1}\in\operatorname{O}_{q\restriction_{U}}(K)$ such that
$\alpha\beta_{1}\in\operatorname{SO}_{q}(K)$. Let $a$ be the spinor norm of
$\alpha\beta_{1}$. Let $b,c\in K$ be such that $a=bc$ and for every $v\in
S_{real}$, $b,c\in q(U_{v})$. Since $\dim U\geq 5$, for every non-real $v$,
$q$ is universal over $U_{v}$ so $b,c\in q(U_{v})$. The Hasse principal
implies that $b,c\in q(U)$, say $b=q(b^{\prime})$ and $c=q(c^{\prime})$ for
some $b^{\prime},c^{\prime}\in K^{n}$. Let
$\beta_{2}=\tau_{b^{\prime}}\tau_{c^{\prime}}$. Then
$\alpha\beta_{1}\beta_{2}\in\Theta_{q}(K)$ and $\beta_{1}\beta_{2}\in
O_{q\restriction_{U}}(K)$. ∎

######  Lemma A.4.3.

Let $U$ be a $q$-universal subspace of $K^{n}$ of dimension at least 5 and let
$\alpha\in\Theta_{q\restriction U^{\perp}}(K)$. Then
$\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)=\operatorname{gcl}_{O_{q}(K)}(\alpha)$.

###### Proof.

Let $\beta\in\operatorname{O}_{q}(K)$. We have to show that
$\beta\alpha\beta^{-1}\in\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)$. Lemma
A.4.2 implies that there exists
$\gamma\in\operatorname{O}_{q\restriction_{U}}(K)$ such that
$\beta\gamma\in\Theta_{q}(K)$. The claim follows from the fact that
$\beta\alpha\beta^{-1}=(\beta\gamma)\alpha(\beta\gamma)^{-1}$. ∎

######  Lemma A.4.4.

Let $N$ be the constant from Lemma A.2.7. For every $1$-separated element
$\alpha\in\operatorname{Spin}_{q}(K)$, there exists a $1$-separated element
$\beta\in(\operatorname{gcl}_{\operatorname{Spin}_{q}(\alpha)})^{N}$ such that
$\operatorname{gcl}_{\operatorname{O}_{q}(K)}(\beta)=\operatorname{gcl}_{\Theta_{q}(K)}(\beta)$.

###### Proof.

By the Hasse principal, a subspace $U$ of $K^{n}$ is $q$-universal if for
every place $v$, $U_{v}$ is a $q$-universal subspace $K_{v}^{n}$. For every
non-real place $v$, any regular quadratic space of dimension at least five is
universal. Thus, a regular six dimensional subspace $U$ of $K^{n}$ is
$q$-universal if and only if for every real place $v$, $U_{v}$ is
$q$-universal. A simple approximation argument with respect to the finitely
many real places implies that there exists a regular six dimensional
$q$-universal subspace $U$ of $K^{n}$. By the choice of $N$,
$(\operatorname{gcl}_{\operatorname{Spin}_{q}(K)}(\alpha))^{N}$ contains a
strongly $1$-separated element
$\beta\in\operatorname{Spin}_{K\restriction_{U^{\perp}}}(K)$. Lemma A.4.3
implies that
$\operatorname{gcl}_{O_{q}(K)}(\beta)=\operatorname{gcl}_{\Theta_{q}(K)}(\beta)$.∎

######  Lemma A.4.5.

Let $\gamma\in\Gamma$ and $a,b,c\in K^{n}$ be non-isotropic vectors such that
$q(a)=q(b)=q(c)$ and $(a,\gamma a)=(a,c)=(b,c)$. Then there exists
$\delta\in(\operatorname{gcl}_{O_{q}(K)}(\gamma))^{2}$ such that $b=\delta a$.

###### Proof.

By Witt Lemma, there exist $\sigma,\tau\in O_{q}(K)$ such that
$\sigma(a,\gamma a)=(a,c)$ and $\tau(a,\gamma a)=(b,c)$. Thus $\delta a=b$
where
$\delta:=\tau\gamma^{-1}\tau^{-1}\sigma\gamma\sigma^{-1}\in(\operatorname{gcl}_{O_{q}(K)}(\gamma))^{2}$.
∎

######  Lemma A.4.6.

Let $N$ is the constant from Lemma A.2.7. Let $a,b\in K^{n}$ be non-isotropic
vectors such that, $b\neq\pm a$, $q(a)=q(b)$ and the hyperplane $U:=\\{c\in
K^{n}\mid(a,c)=(b,c)\\}$ is $q$-universal. Then for every $1$-separated
element $\alpha\in\operatorname{Spin}_{q}(K)$, there exists
$\delta\in\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)\right)^{2{N}^{2}}$
such that
$\operatorname{gcl}_{O_{q}(K)}(\delta)\subseteq\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)\right)^{2{N}^{2}}$
and $\delta a=b$.

###### Proof.

By Lemma A.4.4, there exists a $1$-separated element
$\alpha^{\star}\in\big{(}\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)\big{)}^{N}$
such that
$\operatorname{gcl}_{\operatorname{O}_{q}(K)}(\alpha^{\star})=\operatorname{gcl}_{\Theta_{q}(K)}(\alpha^{\star})$.

For every $\beta\in\Theta_{q}(K)$, define a three dimensional quadratic space
$(V_{\beta},f_{\beta})$ in the following way: $V_{\beta}$ is spanned over $K$
by $a^{\prime},b^{\prime},c^{\prime}$,
$f_{\beta}(a^{\prime})=f_{\beta}(b^{\prime})=f_{\beta}(c^{\prime})=q(a)$,
$(a^{\prime},b^{\prime})_{f_{\beta}}=(a,b)_{q}$ and
$(a^{\prime},c^{\prime})_{f_{\beta}}=(b^{\prime},c^{\prime})_{f_{\beta}}=(a,\beta
a)_{q}$. Let $d^{\prime}$ be the orthogonal projection of $c^{\prime}$ on
$(Ka^{\prime}+Kb^{\prime})^{\perp}$. For every $\beta\in\Theta_{q}(K)$ for
which $f_{\beta}$ is regular the following conditions are equivalent:

  1. (1)

There exists an isometric embedding $\iota_{\beta}$ of $V_{\beta}$ in $K^{n}$
under which $a^{\prime}$ and $b^{\prime}$ are mapped to $a$ and $b$.

  2. (2)

$f_{\beta}(d^{\prime})\in q\left((Ka+Kb)^{\perp}\right)$.

Assume for the moment that there exists $\beta\in\Theta_{q}(K)$ for which the
embedding $\iota_{\beta}$ exists (and hence $f_{\beta}$ is regular). By the
definition of $N$, there exists
$\gamma\in\operatorname{gcl}_{\Theta_{q}(K)}(\alpha^{\star})^{N}$ such that
for every real place $v$, $\gamma$ is close enough to $\beta$ so that
$f_{\gamma}(d^{\prime})\in q\left((K_{v}a+K_{v}b)^{\perp}\right)$ and
$f_{\gamma}$ is regular. Since $n\geq 7$, for every non-real $v$,
$(K_{v}a^{\prime}+K_{v}b^{\prime})^{\perp}$ is a universal subspace so
$f_{\gamma}(d^{\prime})\in q\left((K_{v}a+K_{v}b)^{\perp}\right)$. The Hasse
principal implies that $f_{\gamma}(d^{\prime})\in
q\left((Ka+Kb)^{\perp}\right)$ so an isometric embedding
$\iota_{\gamma}:V_{\gamma}\rightarrow K^{n}$ exists. Lemma A.4.5 applied to
$c:=\iota_{\gamma}(c^{\prime})$ implies that there exists

$\delta\in\left(\operatorname{gcl}_{\operatorname{O}_{q}(K)}(\gamma)\right)^{2}\subseteq\left(\operatorname{gcl}_{\operatorname{O}_{q}(K)}\left((\operatorname{gcl}_{\Theta_{q}(K)}(\alpha^{\star}))^{N}\right)\right)^{2}\subseteq\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha^{\star})\right)^{2N}\subseteq\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)\right)^{2{N}^{2}}$

such that $\delta a=b$.

We now prove the existence of $\beta$. Since $U$ is $q$-universal, there
exists $c_{1}\in U$ such that $q(a)=q(c_{1})$. By Witt’s theorem. there exists
$\beta_{1}\in\operatorname{O}_{q}(K)$ such that $\beta_{1}a=c_{1}$. Since $U$
is $q$-universal, Lemma A.4.2 implies that there exists
$\beta_{2}\in\operatorname{O}_{q\restriction_{U}}(K)$ such that
$\beta:=\beta_{2}\beta_{1}\in\Theta_{q}(K)$. Denote $c:=\beta a\in U$. Since
$c\in U$, $(a,c)=(b,c)$ and $(a,\beta a)=(a,c)=(b,c)$. Thus, there exists an
embedding of $V_{\beta}$ in $K^{n}$ which sends $a^{\prime}$, $b^{\prime}$ and
$c^{\prime}$ to $a$, $b$ and $c$. Since $f$ is anisotropic, $V_{\beta}$ is
regular. ∎

######  Lemma A.4.7.

Let $N$ be the constant from Lemma A.2.7. Let $\alpha$ be a $1$-separated
element and $a,b\in K^{n}$ be non-isotropic vectors such that
$\tau_{a}\tau_{b}\in\Theta_{q}(K)$. Then
$\tau_{a}\tau_{b}\in\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)\right)^{4{N}^{2}}$.

###### Proof.

We can assume that $\tau_{a}\tau_{b}\neq 1$ so, since $q$ is anisotropic,
$U:=Ka+Kb$ is a regular subspace. We claim that there exists a non-isotropic
$c\in Ka+Kb$ such that $(Kc)^{\perp}$ is $q$-universal. A simple approximation
argument with respect to the finitely many real places implies that there
exists a non-isotropic $c\in U$ such that for every real place $v$,
$(K_{v}c)^{\perp}$ is $q$-universal. Since $n\geq 6$, for every non-real place
$v$, $(K_{v}c)^{\perp}$ is universal. The Hasse principal implies that
$(Kc)^{\perp}$ is $q$-universal.

Since $\tau_{a}\tau_{b}\tau_{c}\in\operatorname{O}_{q\restriction_{U}}(K)$ and
every element in $\operatorname{O}_{q\restriction_{U}}(K)$ of determinant $-1$
is a reflection, there exists a non-isotropic $d\in(Ka+Kb)$ such that
$\tau_{a}\tau_{b}=\tau_{c}\tau_{d}\in\Theta_{q}(K)$. Since
$\tau_{c}\tau_{d}\in\Theta_{q}(K)$, by multiplying $d$ with a scalar, we can
assume that $q(c)=q(d)$. Since $q$ is anisotropic, $q(e)\neq 0$ and
$d=\tau_{e}(c)$ where $e:=c-d$. Thus,

$\tau_{a}\tau_{b}=\tau_{c}\tau_{d}=\tau_{c}\tau_{\tau_{e}(c)}=[\tau_{c},\tau_{e}]=\tau_{\tau_{c}(e)}\tau_{e}.$

Since $\tau_{a}\tau_{b}\neq 1$, $e\neq\pm\tau_{c}(e)$. Moreover, $\\{x\in
K^{n}\mid(e,x)=(\tau_{c}(e),x)\\}=(Kc)^{\perp}$ is $q$-universal. By Lemma
A.4.6, there exists $\beta\in\Theta_{q}(K)$ such that
$\operatorname{gcl}_{\operatorname{O}_{q}(K)}(\beta)\subseteq\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha)\right)^{2{N}^{2}}$
and $\beta(e)=\tau_{c}(e)$. It follows that

$\tau_{a}\tau_{b}=\tau_{\tau_{c}(e)}\tau_{e}=\tau_{\beta(e)}\tau_{e}=[\beta,\tau_{e}]\in\left(\operatorname{gcl}_{\operatorname{O}_{q}(K)}(\beta)\right)^{2}\subseteq\left(\operatorname{gcl}_{\Theta_{q}(K)}(\Theta[I])\right)^{4{N}^{2}}.$

∎

######  Proposition A.4.8.

Let $N$ be the constant from Lemma A.2.7. Then for every $1$-separated element
$\alpha\in\operatorname{Spin}_{q}(K)$,
$(\operatorname{gcl}_{\operatorname{Spin}_{q}(K)}(\alpha))^{12(n+1){N}^{2}}=\operatorname{Spin}_{q}(K)$.

###### Proof.

Let $\alpha\in\operatorname{Spin}_{q}(K)$ be a $1$-separated element and let
$\alpha^{\star}$ be its image in $\Theta_{q}(K)$. For every non-isotropic
pairwise orthogonal vectors $a,b,c\in K^{n}$ such that $q(a)=q(b)=q(c)$, and
any lifts $\delta_{1}$ and $\delta_{2}$ of $\tau_{a}\tau_{b}$ and
$\tau_{b}\tau_{c}$ to $\operatorname{Spin}_{f_{s}}(K)$,
$[\delta_{1},\delta_{2}]=-1$. Since $\\{-1\\}$ is the kernel of the
epimorphism from $\operatorname{Spin}_{q}(K)$ to $\Theta_{q}(K)$, it is enough
to prove that
$\big{(}\operatorname{gcl}_{\Theta(K)}(\alpha^{\star})\big{)}^{4(n+1){N}^{2}}=\Theta_{q}(K)$.

Let $\gamma\in\Theta_{q}(K)$. Since every element in
$\operatorname{SO}_{q}(K)$ is the product of at most $n$ reflections, there
are an even $2\leq k\leq n$ and non-isotropic $a_{1},\ldots,a_{n}\in K$ such
that $\gamma=\tau_{a_{1}}\cdots\tau_{a_{k}}$. Lemma 101:11 of [34] implies
that there exists a regular 3-dimensional subspace of $K^{n}$ and
$b_{1},\ldots,b_{k}\in U$ such that for every $1\leq i\leq k$,
$q(a_{i})=q(b_{i})$. Denote $\beta:=\tau_{b_{1}}\cdots\tau_{b_{k}}$. Then
$\beta\in\Theta_{q\restriction_{U}}(K)$ and since $\dim U=3$, there exist non-
isotropic $c_{1},c_{2}\in U$ such that $\beta:=\tau_{c_{1}}\tau_{c_{2}}$. For
every $1\leq i\leq k$, denote $d_{i}=\tau_{a_{i+1}}\cdots\tau_{a_{k}}(b_{i})$
(so $d_{k}=b_{k}$). Recall that for every
$\delta\in\operatorname{Spin}_{q}(K)$ and every non-isotropic $x\in K^{n}$,
$\delta\tau_{x}\delta^{-1}=\tau_{\delta(x)}$. By Lemma A.4.7,

$\gamma=\gamma\beta\beta^{-1}=(\tau_{a_{1}}\cdots\tau_{a_{k}})(\tau_{b_{1}}\cdots\tau_{b_{k}})(\tau_{c_{2}}\tau_{c_{1}})=(\tau_{a_{1}}\tau_{d_{1}})\cdots(\tau_{a_{k}}\tau_{d_{k}})(\tau_{c_{2}}\tau_{c_{1}})\in\left(\operatorname{gcl}_{\Theta_{q}(K)}(\alpha^{\star})\right)^{4(k+1)N^{2}}.$

∎

######  Corollary A.4.9.

Under setting A.1, Item 3 of Setting 3.1 holds.

###### Proof.

For every $I\in\mathcal{I}_{K}$, $\operatorname{Spin}_{q}(O)[I]$ contains a
$1$-separated element. Proposition A.4.8 implies that there exists a constant
$N_{1}$ such that for every $I\in\mathcal{I}_{K}$,
$(\operatorname{gcl}_{\operatorname{Spin}_{q}(K)}(\operatorname{Spin}_{q}(O)[I]))^{N_{1}}=\operatorname{Spin}_{q}(K)$.

Since $\operatorname{Spin}_{q}(K)/Z(\operatorname{Spin}_{q}(K))$ is not
abelian, there exists $\alpha,\beta\in\operatorname{Spin}_{q}(K)$ and
$\epsilon>0$ such that $[\alpha,\beta]$ is $\epsilon$-separated. Since the
inverse and every conjugate of a commutator are commutators, Lemma A.2.7 and
Proposition A.4.8 imply that there exists $N_{2}$, such that every element of
$\operatorname{Spin}_{q}(K)$ is the product of at most $N_{2}$ commutators.
The claim holds with respect to $N:=\max(N_{1},N_{2})$. ∎

######  Remark A.4.10.

The proofs of Lemma A.4.7 and of Proposition A.4.8 show that every element of
$\operatorname{Spin}_{q}(K)$ is the product of at most $n+2$ commutators.

## References

  * [1] N. Avni and C. Meiri, Words have bounded width in SL(n, Z), Compos. Math. 155 (2019), no. 7, 1245–1258.
  * [2] N. Avni and C. Meiri, On the model theory of higher rank arithmetic groups. accepted to Duke Math. J., arXiv:2008.01793.
  * [3] N. Avni, A. Lubotzky and C. Meiri, First order rigidity of non-uniform higher rank arithmetic groups. Invent. Math. 217 (2019), no. 1, 219–240.
  * [4] H. Bass, M. Lazard and J.-P. Serre, Sous-groupes d’indice fini dans SL(n, Z). Bull. Amer. Math. Soc. 70 1964 385–392.
  * [5] H. Bass, J. Milnor and J.-P. Serre, Solution of the congruence subgroup problem for $SL_{n}$ ($n\geq 3$) and $Sp_{2n}$ ($n\geq 2$). Inst. Hautes Etudes Sci. Publ. Math. No. 33 (1967), 59–137.
  * [6] O. Becker and M. Chapman Stability of approximate group actions: uniform and probabilistic. arXiv:2005.06652.
  * [7] M. Bestvina, K. Bromberg and K. Fujiwara, The verbal width of acylindrically hyperbolic groups. Algebr. Geom. Topol. 19 (2019), no. 1, 477–489.
  * [8] A. Borel, On free subgroups of semisimple groups. Enseign. Math. (2) 29 (1983), no. 1-2, 151–164.
  * [9] N. Bourbaki, General topology. Chapters 1–4. Translated from the French. Reprint of the 1989 English translation. Elements of Mathematics (Berlin). Springer-Verlag, Berlin, 1998. vii+437 pp.
  * [10] D. Burago, S. Ivanov, and L. Polterovich, Conjugation-invariant norms on groups of geometric origin. Groups of diffeomorphisms, 221–250, Adv. Stud. Pure Math., 52, Math. Soc. Japan, Tokyo, 2008.
  * [11] M. Burger and N. Monod, Bounded cohomology of lattices in higher rank Lie groups, J. Eur. Math. Soc. 1 (1999), no. 2, 199–235.
  * [12] M. Burger, N. Ozawa and A. Thom, On Ulam stability. Israel J. Math. 193 (2013), no. 1, 109–129.
  * [13] D. Carter and G. Keller, Bounded elementary generation of $\operatorname{SL}_{n}(O)$. Amer. J. Math. 105 (1983), no. 3, 673–687.
  * [14] P. Corvaja, A.S. Rapinchuk, J. Ren and U.M. Zannier, Non-virtually abelian anisotropic linear groups are not boundedly generated. Invent. Math. 227 (2022), no. 1, 1–26.
  * [15] R.K. Dennis and L.N. Vaserstein, On a question of M. Newman on the number of commutators. J. Algebra 118 (1988), no. 1, 150–161.
  * [16] I.V. Erovenko and A.S. Rapinchuk, Bounded generation of S-arithmetic subgroups of isotropic orthogonal groups over number fields, J. Number Theory 119 (2006), no. 1, 28–48.
  * [17] K.M. George, Verbal properties of certain groups, Ph.D. thesis, University of Cambridge, (1976).
  * [18] D.R. Heath-Brown, Artin’s conjecture for primitive roots, Quart. J. Math. Oxford Ser. (2), 37 (1986), pp. 27–38.
  * [19] A. Jaikin-Zapirain, On the verbal width of finitely generated pro-p groups. Rev. Mat. Iberoam. 24 (2008), no. 2, 617–630.
  * [20] A. Khelif, Bi-interpretabilite et structures QFA: etude de groupes resolubles et des anneaux commutatifs. C. R. Math. Acad. Sci. Paris 345 (2007), no. 2, 59–61.
  * [21] M. Kneser, Orthogonale Gruppen uber algebraischen Zahlkorpern. J. Reine Angew. Math. 196 (1956), 213–220.
  * [22] M. Larsen and A. Shalev, Word maps and Waring type problems. J. Amer. Math. Soc. 22 (2009), no. 2, 437–466.
  * [23] H. W. Lenstra, On Artin’s conjecture and Euclid’s algorithm in global fields. Invent. Math. 42 (1977), 201–224.
  * [24] A. Lubotzky, Subgroup growth and congruence subgroups, Invent. Math. 119 (1995), 267–295.
  * [25] H. Matsumoto, Sur les sous-groupes arithmetiques des groupes semi-simples deployes. Ann. Sci. Ecole Norm. Sup. (4) 2 (1969), 1–62.
  * [26] J.L. Mennicke, Finite factor groups of the unimodular group. Ann. of Math. (2) 81 (1965) 31–37.
  * [27] J. Milnor, Introduction to algebraic K-theory. Annals of Mathematics Studies, No. 72. Princeton University Press, Princeton, N.J.; University of Tokyo Press, Tokyo, 1971. xiii+184 pp.
  * [28] D.W. Morris, Bounded generation of $SL(n,A)$ (after D. Carter, G. Keller, and E. Paige). New York J. Math. 13 (2007), 383–421.
  * [29] A. Myasnikov and A. Nikolaev, Verbal subgroups of hyperbolic groups have infinite width. J. Lond. Math. Soc. (2) 90 (2014), no. 2, 573–591.
  * [30] W. Narkiewicz, A note on Artin’s conjecture in algebraic number fields, J. Reine Angew. Math. 381 (1987), 110–115.
  * [31] A. Nies, Describing groups. Bull. Symbolic Logic 13 (2007), no. 3, 305–339.
  * [32] N. Nikolov and D. Segal, On finitely generated profinite groups. I. Strong completeness and uniform bounds. Ann. of Math. (2) 165 (2007), no. 1, 171–238.
  * [33] N. Nikolov and D. Segal, On finitely generated profinite groups. II. Products in quasisimple groups. Ann. of Math. (2) 165 (2007), no. 1, 239–273.
  * [34] O. T. O’Meara, Introduction to quadratic forms. Second printing, corrected. Die Grundlehren der mathematischen Wissenschaften, Band 117. Springer-Verlag, New York-Heidelberg, 1971.
  * [35] V.P. Platonov and A.S. Rapinchuk, Abstract properties of S-arithmetic groups and the congruence subgroup problem, Russian Acad. Sci. Izv. Math. 40 (1993), 455–476.
  * [36] G. Prasad and M.S. Raghunathan, On the congruence subgroup problem: Determination of the metaplectic kernel, Invent. Math. 71 (1983), 21–42
  * [37] G. Prasad and A. Rapinchuk, Topological central extensions of semi-simple groups over local fields, Ann. Math. 119 (1984), 143–268.
  * [38] G. Prasad and A. Rapinchuk, Developments on the congruence subgroup problem after the work of Bass, Milnor and Serre. arXiv:0809.1622.
  * [39] M.S. Raghunathan, On the congruence subgroup problem. Inst. Hautes Etudes Sci. Publ. Math. No. 46 (1976), 107–161.
  * [40] M.S. Raghunathan, The congruence subgroup problem. Proc. Indian Acad. Sci. Math. Sci. 114 (2004), no. 4, 299–308.
  * [41] A.S Rapinchuk, The congruence subgroup problem. Algebra, K-theory, groups, and education (New York, 1997), 175–188, Contemp. Math., 243, Amer. Math. Soc., Providence, RI, 1999.
  * [42] A.H. Rhemtulla, A problem of bounded expressibility in free products, Proc. Cambridge Philos. Soc. 64 (1968), 573–584.
  * [43] V.A. Roman’kov, Width of verbal subgroups in solvable groups. Algebra i Logika 21(1982), 60–72 (Russian); Algebra and Logic 21 (1982), 41–49 (English).
  * [44] D. Segal, Words: notes on verbal width in groups. London Mathematical Society Lecture Note Series, 361. Cambridge University Press, Cambridge, 2009. xii+121 pp.
  * [45] Z. Sela, Diophantine geometry over groups. VII. The elementary theory of a hyperbolic group. Proc. Lond. Math. Soc. (3) 99 (2009), no. 1, 217–273.
  * [46] Y. Shalom, Bounded generation and Kazhdan’s property (T), Publ. Math. IHES 90 (2001), 145– 168\.
  * [47] S. Sinchuk and A. Smolensky, Decompositions of congruence subgroups of Chevalley groups. Internat. J. Algebra Comput. 28 (2018), no. 6, 935–958.
  * [48] P.W. Stroud, Topics in the theory of verbal subgroups, Ph.D. Thesis, University of Cambridge, (1966).
  * [49] A.A. Suslin, On the structure of special linear groups over polynomial rings, Math. USSR. Izv., 11 (1977) 221–238.
  * [50] O.I. Tavgen, Bounded generation of Chevalley groups over the rings of S-integers, Izv. Akad. Nauk SSSR, Ser. Mat., 54 (1990), 97–122.
  * [51] S. M. Ulam, A collection of mathematical problems, Interscience Tracts in Pure and Applied Mathematics, no. 8, Interscience Publishers, New York-London, 1960.
  * [52] D. Witte Morris, Bounded generation of SL(n,A) (after D. Carter, G. Keller, and E. Paige). New York J. Math. 13 (2007), 383–421.