We note that if $a=\operatorname{Diag}\left(a_{1},\ldots,a_{n}\right)\in
A_{\mathbb{F}}$, then also

$\sqrt[n+m]{a}:=\operatorname{Diag}(\sqrt[n+m]{a_{1}},\ldots,\sqrt[n+m]{a_{n}})\in
A_{\mathbb{F}},$

since this is a first order statement that is true over the reals. Then

$g.\sqrt[n+m]{a^{n}a^{\prime m}}.o=\sqrt[n+m]{b^{n}b^{\prime
m}}.o\in\mathbb{A}$

completes the proof. ∎

We now introduce a notion of regularity. Elements $a.o\in f_{0}(\mathbb{A})$
with $a\in A_{\mathbb{F}}$ are represented by diagonal matrices with entries
$a_{1},\ldots,a_{n}\in\mathbb{F}$. Consider the amount of distinct entries
$(-v)(a_{1}),\ldots,(-v)(a_{n})$ as elements in
$\Lambda=(-v)(\mathbb{F}_{>0})$. Let $\boldsymbol{a}\in B$ be a maximal
element with respect to the amount of distinct entries. The intuition is that
elements with $a_{i}=a_{j}$ for $i\neq j$ lie on a wall. The more walls they
lie on, the less regular they are. Other elements in $B$ may have as many
distinct entries as $\boldsymbol{a}$ but the next Lemma states that those are
the same ones as the ones for $\boldsymbol{a}$.

######  Lemma A.3.

Let $b\in B$. If $(-v)(b_{i})\neq(-v)(b_{j})$ for some $i,j$, then
$(-v)(\boldsymbol{a}_{i})\neq(-v)(\boldsymbol{a}_{j})$. Equivalently if
$(-v)(\boldsymbol{a}_{i})=(-v)(\boldsymbol{a}_{j})$, then
$(-v)(b_{i})=(-v)(b_{j})$.

###### Proof.

For any element $b\in B$, we define
$I_{b}=\\{(i,j)\colon(-v)(b_{i})\neq(-v)(b_{j})\\}$. The maximality of
$\boldsymbol{a}$ means that $|I_{b}|\leq|I_{\boldsymbol{a}}|$ for all $b\in
B$. We want to prove that $I_{b}\subseteq I_{\boldsymbol{a}}$ for all $b\in
B$. We note that there are certainly less than $n^{2}$ elements in
$I_{\boldsymbol{a}}$, and define for $k\in\\{0,1,\ldots,n^{2}\\}$ the element

$c(k)=\sqrt[n^{2}]{\boldsymbol{a}^{k}b^{n^{2}-k}}\in B$

which is in $B$ by Lemma A.2. Let $(i,j)\in I_{\boldsymbol{a}}\cup I_{b}$.
Thus $(-v)(\boldsymbol{a_{i}}/\boldsymbol{a_{j}})\neq 0$ and
$(-v)(b_{i}/b_{j})\neq 0$. We define the map

$k\mapsto(-v)(c(k)_{i}/c(k)_{j})\in\Lambda$

which is either constant, but not $=0$, or strictly monotonous ascending or
descending. Either way, there is at most one value of $k$ for which the map is
$0$. Let $k_{ij}$ be this value, if it exists.

Now pick a $k$ which does not appear as $k_{ij}$ for any $(i,j)\in
I_{\boldsymbol{a}}\cup I_{b}$. This means that $c(k)_{i}\neq c(k)_{j}$ for all
$(i,j)\in I_{\boldsymbol{a}}\cup I_{b}$. We conclude that
$I_{\boldsymbol{a}}\cup I_{b}\subseteq I_{c(k)}$. But since $\boldsymbol{a}$
is maximal, we have $I_{\boldsymbol{a}}=I_{c(k)}$ and $I_{b}\subseteq
I_{\boldsymbol{a}}$.

∎

Recall that $O\subsetneqq\mathbb{F}$ is a strict subring of a non-Archimedean
real closed field $\mathbb{F}$. Denote the units of $O$ by $O^{\times}$. The
following is a standard fact about valuation rings.

######  Lemma A.4.

In any valuation ring $O$, the set of nonunits $O\setminus O^{\times}$ is an
ideal.

###### Proof.

Let $x,y\in O\setminus O^{\times},r\in O$. Since $x$ is not a unit,
$x^{-1}=r\cdot(rx)^{-1}\not\in O$. Therefore $(rx)^{-1}\not\in O$, so $rx\in
O\setminus O^{\times}$. If $x$ and $y$ are nonzero, then $x/y\in O$ or $y/x\in
O$, since $O$ is a valuation ring. Writing $x+y=y\cdot(1+x/y)=x\cdot(1+y/x)$,
we see that also $x+y\in O\setminus O^{\times}$. ∎

The determinant of a matrix is a polynomial of its entries. By the previous
Lemma, if all entries of a matrix are in $O\setminus O^{\times}$, then so is
its determinant. We conclude that if the determinant of a matrix with entries
in $O$ is $1$, then at least one entry has to be a unit. Actually, even more
holds.

######  Lemma A.5.

Let $g$ be a matrix with entries in $O$ and $\operatorname{det}(g)=1$. Then in
every row and every column there exists at least one entry that is a unit. In
fact, there is a permutation $\sigma$, such that $g_{i\sigma(i)}\in
O^{\times}$.

###### Proof.

Consider the $i$’th row $(g_{i1},g_{i2},\ldots,g_{in})$ of $g$. Using
Laplace’s formula for the determinant

$\det(g)=\sum_{j=1}^{n}(-1)^{i+j}g_{ij}\cdot\det M_{ij},$

where $M_{ij}$ are the minor matrices after deleting row $i$ and column $j$,
we see that since $\det(g)=1\in O^{\times}$, at least one of the $g_{ij}$ for
$j=1,\ldots,n$ has to be a unit. Equivalently there has to be a unit in every
column. We can extract a permutation by alternating the row and column
argument. ∎

We will use Lemma A.3 and Lemma A.5 to prove a certain rigidity of the action
of $G_{\mathbb{F}}$ on $f_{0}(\mathbb{A})$, namely when an isometry $g\in
G_{\mathbb{F}}$ fixes the identity, then its action on $f_{0}(\mathbb{A})$ can
be described by an element in the spherical Weyl group $W_{s}$. The apartment
$\mathbb{A}$ itself could have many other symmetries, but the isometry can
only be extended to all of $\mathcal{B}$, when it is one of the finitely many
elements in $W_{s}$. Consider the fundamental sector

$s_{0}=\\{a.o\in\mathbb{A}\colon a_{1}\geq\ldots\geq a_{n}\\}$

based at $o$. The sectors in $f_{0}(\mathbb{A})$ based at $o$ are fundamental
domains of the action of $W_{s}$, see for instance [Hum72, Chapter 10.3]. So
two points that lie inside a common sector based at $o$ can only be related by
an element in $W_{s}$ if they are the same. This implies a first restricted
version of axiom (A2).

######  Proposition A.6.

Let as before $\boldsymbol{a}\in\mathcal{B}$ be maximal with respect to the
amount of distinct entries and $g\in G_{\mathbb{F}}$ such that $g.o=o$. If
both $\boldsymbol{a}.o$ and $g.\boldsymbol{a}.o$ lie in the same sector based
at $o$, then $\boldsymbol{a}.o=g.\boldsymbol{a}.o$ and moreover the action of
$g$ is the identity on the set $\Omega:=f_{0}^{-1}\left(f_{0}(\mathbb{A})\cap
g.f_{0}(\mathbb{A})\right)$, i.e. $f_{0}|_{\Omega}=g.f_{0}|_{\Omega}$.

###### Proof.

Let
$\boldsymbol{a}=\operatorname{Diag}\left(\boldsymbol{a}_{1},\ldots,\boldsymbol{a}_{n}\right)$
and
$\boldsymbol{b}=\operatorname{Diag}\left(\boldsymbol{b}_{1},\ldots,\boldsymbol{b}_{n}\right)\in
A_{\mathbb{F}}$ such that $g.\boldsymbol{a}.o=\boldsymbol{b}.o$. Since
$\boldsymbol{a}.o$ and $\boldsymbol{b}.o$ lie in the same sector based at $o$,
their entries satisfy the same ordering. Formally we can find a permutation
$\rho$, such that

$\displaystyle(-v)(\boldsymbol{a}_{\rho(1)})$
$\displaystyle\leq(-v)(\boldsymbol{a}_{\rho(2)})\leq\ldots\leq(-v)(\boldsymbol{a}_{\rho(n)})\quad\text{and}$
(A.1) $\displaystyle(-v)(\boldsymbol{b}_{\rho(1)})$
$\displaystyle\leq(-v)(\boldsymbol{b}_{\rho(2)})\leq\ldots\leq(-v)(\boldsymbol{b}_{\rho(n)}).$

From $g.o=o$ follows that $g\in G_{\mathbb{F}}(O)$ by Proposition 7.11. By the
previous Lemma A.5 we get a permutation $\sigma$, such that $g_{i\sigma(i)}\in
O^{\times}$. In coefficients, $\boldsymbol{b}^{-1}g\boldsymbol{a}\in
G_{\mathbb{F}}(O)$ implies
$\boldsymbol{b}_{i}^{-1}g_{i\sigma(i)}\boldsymbol{a}_{\sigma(i)}\in O$ and
therefore also $\boldsymbol{b}_{i}^{-1}\boldsymbol{a}_{\sigma(i)}\in O$. This
means that
$\operatorname{Diag}(\boldsymbol{a}_{\sigma(1)},\ldots,\boldsymbol{a}_{\sigma(n)}).o=\operatorname{Diag}\left(\boldsymbol{b}_{1},\ldots,\boldsymbol{b}_{n}\right).o$.
So although the elements $\boldsymbol{a}_{\sigma(i)}$ and $\boldsymbol{b}_{i}$
may not be exactly the same, they satisfy
$(-v)(\boldsymbol{a}_{\sigma(i)})=(-v)(\boldsymbol{b}_{i})$. So we can order
the entries as

$\displaystyle(-v)(\boldsymbol{b}_{\rho(1)})$
$\displaystyle\leq(-v)(\boldsymbol{b}_{\rho(2)})$
$\displaystyle\leq\ldots\leq(-v)(\boldsymbol{b}_{\rho(n)})\quad\text{and
thus}$ (A.2) $\displaystyle(-v)(\boldsymbol{a}_{\rho(\sigma(1))})$
$\displaystyle\leq(-v)(\boldsymbol{a}_{\rho(\sigma(2))})$
$\displaystyle\leq\ldots\leq(-v)(\boldsymbol{a}_{\rho(\sigma(n))}).$

But we already have a decreasing ordering of the diagonal entries of
$\boldsymbol{a}$ in equation (A.1), so they have to be the same, i.e. for all
$i$, $(-v)(\boldsymbol{a}_{\rho(i)})=(-v)(\boldsymbol{a}_{\rho(\sigma(i))})$
and thus
$(-v)(\boldsymbol{a}_{i})=(-v)(\boldsymbol{a}_{\sigma(i)})=(-v)(\boldsymbol{b}_{i})\in\Lambda$.
We now know that $\boldsymbol{a}.o=\boldsymbol{b}.o$, but we still need to
show that $g$ is the identity on all of $\Omega$.

If all the inequalities in (A.2) were strict, then $\sigma$ would necessarily
be the identity. For indices $i,j$ where there is an equality, $\sigma$ could
change these entries. However by Lemma A.3 we know that whenever
$(-v)(\boldsymbol{a}_{i})=(-v)(\boldsymbol{a}_{j})$, then also
$(-v)(a_{i})=(-v)(a_{j})$ for any
$a=\operatorname{Diag}\left(a_{1},\ldots,a_{n}\right)\in B$. So for all $i$,
$(-v)(a_{\sigma(i)})=(-v)(a_{i})$. In fact we see

$g_{i\sigma(i)}\frac{a_{\sigma(i)}}{a_{i}}=g_{i\sigma(i)}\in O$

and thus $g.a.o=a.o$ for all $a\in B$, so $f_{0}|_{\Omega}=g.f_{0}|_{\Omega}$.
∎

We will now allow slightly more general points $\boldsymbol{a}.o$ and
$\boldsymbol{b}.o$.

######  Proposition A.7.

Let $g\in G_{\mathbb{F}}$ with $g.o=o\in\mathcal{B}$, then there exists an
element $w\in W_{s}$ such that $f_{0}|_{\Omega}=g.f_{0}\circ w|_{\Omega}$,
where $\Omega:=f_{0}^{-1}\left(f_{0}(\mathbb{A})\cap
g.f_{0}(\mathbb{A})\right)$.

###### Proof.

Use Lemma A.3 to get $\boldsymbol{a}\in B$ maximal with respect to the amount
of distinct entries. Let $\boldsymbol{b}\in A_{\mathbb{F}}$ such that
$\boldsymbol{b}.o=g.\boldsymbol{a}.o\in f_{0}(\Omega)$. Since the sectors in
$f_{0}(\mathbb{A})$ are fundamental domains for $W_{s}$, there is a $w\in
W_{s}$ such that $w.\boldsymbol{b}.o$ and $\boldsymbol{a}.o$ are in the same
sector. Equivalently $\boldsymbol{b}.o$ and $w^{-1}.\boldsymbol{a}.o$ are in
the same sector. Note that $gw.o=o$,
$\boldsymbol{b}.o=(gw).(w^{-1}\boldsymbol{a}).o$,
$f_{0}(\Omega)=f_{0}(\mathbb{A})\cap
g.f_{0}(\mathbb{A})=f_{0}(\mathbb{A})\cap(gw).f_{0}(\mathbb{A})$ (since
$w.\mathbb{A}=\mathbb{A}$) and $w^{-1}\boldsymbol{a}$ is maximal with respect
to the amount of distinct entries, so we can apply Proposition A.6 to get
$f_{0}|_{\Omega}=(gw).f_{0}|_{\Omega}=g.f_{0}\circ w|_{\Omega}$. ∎

To show that axiom (A2) holds, there is one more obstacle, namely $g$ may not
preserve $o$ in general. In fact, $\Omega$ may not even contain $o$, so we
have to first translate $\Omega$ to be able to use the previous propositions.

######  Proposition A.8.

Let $g\in G_{\mathbb{F}}$ and define
$\Omega:=f_{0}^{-1}\left(f_{0}(\mathbb{A})\cap g.f_{0}(\mathbb{A})\right)$.
Then there exists an element $\overline{w}\in W_{a}$ such that
$f_{0}|_{\Omega}=g.f_{0}\circ\overline{w}|_{\Omega}$.

###### Proof.

If $\Omega=\emptyset$, then any element of $W_{a}$ will suffice. Otherwise
choose and fix $\boldsymbol{a}\in B$ (not necessarily maximal) and
$\boldsymbol{b}\in A_{\mathbb{F}}$ with
$\boldsymbol{b}.o=g.\boldsymbol{a}.o\in f_{0}(\Omega)$. Now consider any
$a,b\in A_{\mathbb{F}}$ with $b.o=g.a.o\in f_{0}(\Omega)$. We translate the
problem by $\boldsymbol{b}^{-1}$ and get

$\boldsymbol{b}^{-1}.b.o=(\boldsymbol{b}^{-1}.g.\boldsymbol{a}).\boldsymbol{a}^{-1}.a.o$

where $(\boldsymbol{b}^{-1}.g.\boldsymbol{a}).o=o$. We are almost in the
situation of Proposition A.7, but we have a different $\Omega$. In fact since
$\boldsymbol{b}^{-1}.f_{0}(\mathbb{A})=f_{0}(\mathbb{A})=\boldsymbol{a}.f_{0}(\mathbb{A})$,

$\displaystyle
f_{0}(\mathbb{A})\cap(\boldsymbol{b}^{-1}.g.\boldsymbol{a}).f_{0}(\mathbb{A})$
$\displaystyle=\boldsymbol{b}^{-1}.f_{0}(\mathbb{A})\cap\boldsymbol{b}^{-1}.g.f_{0}(\mathbb{A})$
$\displaystyle=\boldsymbol{b}^{-1}.f_{0}(\Omega)=f_{0}(\boldsymbol{b}^{-1}.\Omega),$

we can apply Proposition A.7 to get a $w\in W_{s}$ such that

$f_{0}|_{\boldsymbol{b}^{-1}.\Omega}=(\boldsymbol{b}^{-1}.g.\boldsymbol{a}).f_{0}\circ
w|_{\boldsymbol{b}^{-1}.\Omega}\ \ ,$

which can be rewritten as

$\boldsymbol{b}^{-1}.f_{0}|_{\Omega}=\boldsymbol{b}^{-1}.g.f_{0}\circ(\boldsymbol{a}w\boldsymbol{b}^{-1})|_{\Omega}.$

Renaming $\overline{w}=\boldsymbol{a}w\boldsymbol{b}^{-1}\in W_{a}$ and
translating back results in the required formula

$f_{0}|_{\Omega}=g.f_{0}\circ\overline{w}|_{\Omega}.$

∎

We can now conclude the proof of axiom (A2).

######  Proposition A.9.

The axiom

  1. (A2)

For all $f_{1},f_{2}\in\mathscr{F}$, if $f_{1}(\mathbb{A})\cap
f_{2}(\mathbb{A})\neq\emptyset$, then the set $\Omega:=f_{2}^{-1}\circ
f_{1}(\mathbb{A})$ is a $W_{a}$-convex set and there exists $w\in W_{a}$ such
that $f_{2}|_{\Omega}=f_{1}\circ w|_{\Omega}$.

holds for $(\mathcal{B},\mathscr{F})$.

###### Proof.

Let $g,h\in G_{\mathbb{F}}$ with $f_{1}=g.f_{0}$ and $f_{2}=h.f_{0}$ such that
$g.f_{0}(\mathbb{A})\cap h.f_{0}(\mathbb{A})\neq\emptyset$. We define
$\Omega:=(h.f_{0})^{-1}(h.f_{0}(\mathbb{A})\cap g.f_{0}(\mathbb{A}))$. We will
first show the second part of (A2). To be able to apply Proposition A.8, we
act with $h^{-1}$ on

$h.f_{0}(\Omega)=h.f_{0}(\mathbb{A})\cap g.f_{0}(\mathbb{A}),$

to get

$f_{0}(\Omega)=f_{0}(\mathbb{A})\cap(h^{-1}g).f_{0}(\mathbb{A}).$

We can apply Proposition A.8 with $h^{-1}g$ to get an element $w\in W_{a}$
such that $f_{0}|_{\Omega}=(h^{-1}g).f_{0}\circ w|_{\Omega},$ and thus
$f_{2}|_{\Omega}=h.f_{0}|_{\Omega}=g.f_{0}\circ w|_{\Omega}=f_{1}\circ
w|_{\Omega}$.

It remains to show that $\Omega$ is $W_{a}$-convex. Elements $a.o\in
f_{0}(\Omega)$ are exactly those elements which have the special property that
$h.a.o=g.w.a.o,$ which is equivalent to $a^{-1}h^{-1}gwa\in G_{\mathbb{F}}(O)$
by Theorem 7.11. If $a=\operatorname{Diag}\left(a_{1},\ldots,a_{n}\right)\in
A_{\mathbb{F}}$, then

$(h^{-1}gw)_{ij}\frac{a_{j}}{a_{i}}\in O$

in matrix entries. Taking the valuation we obtain

$(-v)(\chi_{\alpha_{ij}}(a))=(-v)(a_{i}/a_{j})\geq(-v)((h^{-1}gw)_{ij})\in\Lambda$

and these are exactly the inequalities that define affine half-apartments

$H_{\alpha_{ij},k}^{+}:=\left\\{a.o\in\mathbb{A}\colon(-v)(\chi_{\alpha_{ij}}(a))\geq
k\right\\}$

for $k=(-v)((h^{-1}gw)_{ij})$. Since there are only finitely many pairs
$(i,j)$, $\Omega$ is a finite intersection of half-apartments, i.e.
$W_{a}$-convex. ∎

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