Note that $m \in \bG(E)_{y,0} \cap \bG(E)_{x_T,0^+}$, which means that the image of $m$ in the reductive quotient  $\bG(E)_{y,0} /\bG(E)_{y,0^+}$ must be unipotent.   Since $\bG$ is $K$-tame, we can assume $E$ is a tame extension of $K$ and so the image of $m$ in $\bG(E)_{y,0} /\bG(E)_{y,0^+}$ is semisimple.  Thus, the image of $m$ is the trivial element.
Consequently, $m$ fixes an open neighborhood $U$ in $\AA'(\bT,E)$ with $y \in \bar{U}$, hence $m$ must fix all of $\AA'(\bT,E)$ and so $m \in \bT(E)$.  Since $m \in N_{G}(\bT) \cap \bT(E)$, we conclude that $m \in T$.

The proof of Lemma <ref>
shows that, in fact,  $N_{G}(\bT)  \leq N_{\bG(E)_{y,0}}(\bT) \cdot T$.

Tameness may or may not be required for the second statement of Lemma <ref>.   However, it is required for the current  proof.   For example, for $\ff$ of characteristic two we have

1 & 0 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}

in $\SL_2(\ff)$.  That is, without some assumptions on the characteristic of our field, nontrivial elements of the normalizer of a torus can be unipotent!

Thanks to Lemma <ref> when $\bT$ splits over a tame extension, we can interpret $N_{G_{F,0}}(\bT) / (T \cap G_{F,0^+})$ as $\Stab_{\bfG_{F}}(\bT)$ where $\bfG_{F}$ is the connected reductive $\ff$-group corresponding to the quotient $G_{F,0} / G_{F,0^+}$.  By $\Stab_{\bfG_{F}}(\bT)$ we mean the group of $\bar{x} \in \Stab_{\bfG_{F}}(\bT)$ for which there exists a lift $x \in G_{F,0}$  such that $\lsup{x}\bT = \bT$.

§ THE MAIN RESULT AND EXAMPLES

We begin by stating the parameterization theorem of this paper.   We then develop some machinery for calculating the objects that appear in this theorem, and we end  by looking at a number of examples.

§.§ The main result
Before stating the main result, we develop and recall some notation.

Let $(\Wwhistles)^{\Fr \circ N_q}$  denote the set of   $\Fr \circ N_q$-fixed points in the set of  $\sigma$-elliptic tame  $\sigma$-conjugacy classes in $\absW$.  For $c \in (\Wwhistles)^{\Fr \circ N_q}$, choose $w \in c$ and let $n \in \titsW$ be a lift of $w$. We slightly modify the notation of Section <ref>, and denote the order of  $n \sigma \in \titsW \rtimes \EGal$ by $\ell_c$.  Let $\pi$ denote a uniformizer in the tame extension of $K$ of degree $\ell_c$ such that $\pi^{\ell_c} = \varpi$.  Let $\lambda_c $ denote the unique element in $ \X_*(\bA \cap \bG')$  such that

* $n$ is $\sigma$-conjugate to $\lambda_c(\sigma(\pi)/\pi)$ by an element of $\Egroup_{x_0,0}$ (see Lemma <ref>) and

* $x_c := x_0 + \lambda_c/\ell_c$ is an element of  $\bar{C}'$ (see Lemma <ref>).

If $c \in (\Wwhistles)^{\Fr \circ N_q}$, then there exists a tame $K$-minisotropic maximal $k$-torus $\bT^c$ in $\bG$ for which the point associated to $T^c$ is $x_c$  (see Lemma <ref> and Corollaries <ref> and <ref>).     Set $T^c_{x_c} = T^c \cap G_{x_c,0}$.

Recall that $k$ is a nonarchimedean local field.  For each $c \in (\Wwhistles)^{\Fr \circ N_q}$  fix a tame $K$-minisotropic maximal $k$-torus $\bT^c$ in $\bG$ for which the point associated to $T^c$ is $x_c$.

*  The set of $G^{\Fr}$-conjugacy classes of tame $K$-minisotropic maximal $k$-tori in $\bG$ is parameterized by  the set

$$\{ (c, \tilde{c} ) \, | \, \text{ $c \in (\Wwhistles)^{\Fr \circ N_q}$ and $ \tilde{c} \in (\bar{W}_{T^c})_{\sim \Fr}  $}\}.$$

Here $(\bar{W}_{T^c})_{\sim \Fr}$ denotes the set of $\Fr$-conjugacy classes in $\bar{W}_{T^c} := N_{G_{x_c,0}}(\bT^c)/T^c_0$.
More precisely,  for each $\tilde{c} \in (\bar{W}_{T^c})_{\sim \Fr} $ we fix

$k_{\tilde{c}} \in G_{x_c,0}$ such that $k_{\tilde{c}}\inv \Fr(k_{\tilde{c}}) \in N_{G_{x_c,0}}(\bT^c)$ is a lift of a member of  $\tilde{c}$.  Then each $G^{\Fr}$-conjugacy class in $\CTkc$ has exactly one representative in the set

$$  \{ \Ad({k_{\tilde{c}}}) \bT^c \, | \, \tilde{c} \in (\bar{W}_{T^c})_{\sim \Fr} \}.$$

*  The set of $k$-stable classes of tame $K$-minisotropic maximal $k$-tori in $\bG$ is parameterized by  the set

$$\{ (c, \hat{c} ) \, | \, \text{ $c \in (\Wwhistles)^{\Fr \circ N_q}$ and $ \hat{c} \in ({W}_{T^c})_{\sim \Fr}  $}\}.$$

Here $({W}_{T^c})_{\sim \Fr}$ denotes the set of $\Fr$-conjugacy classes in ${W}_{T^c} := N_{G_{x_c,0}}(\bT^c)/T^c_{x_c}$.
More precisely,  for each $\hat{c} \in ({W}_{T^c})_{\sim \Fr} $ we fix
$k_{\hat{c}} \in G_{x_c,0}$ such that $k_{\hat{c}}\inv \Fr(k_{\hat{c}}) \in N_{G_{x_c,0}}(\bT^c)$ is a lift of a member of  $\hat{c}$.  Then each $k$-stable conjugacy class in $\CTkc$ has exactly one representative in the set

$$  \{ \Ad({k_{\hat{c}}}) \bT^c \, | \, \hat{c} \in ({W}_{T^c})_{\sim \Fr} \}.$$

*  Suppose $c \in (\Wwhistles)^{\Fr \circ N_q}$ and $\hat{c} \in ({W}_{T^c})_{\sim \Fr}$.   Let $\bT =  \Ad({k_{\hat{c}}}) \bT^c$ where $k_{\hat{c}}$ is chosen as in (<ref>).  Let $\bar{m}$ denote the image of $k_{\hat{c}}\inv \Fr(k_{\hat{c}})$ in  ${W}_{T^c}$.  The set of $k$-embeddings of $\bT$ into $\bG$ is parameterized by $(\bar{\bfT}^c_{x_c})_{\bar{m}\Fr}$, the group of $\bar{m} \Fr$-coinvariants in
$\bar{\bfT}^c_{x_c} = T^c_{x_c}/T^c_0$.   More precisely,  for each $\bar{c} \in (\bar{\bfT}^c_{x_c})_{\bar{m}\Fr}$ we fix $k_{\bar{c}} \in G_{x_c,0}$ such that $k_{\bar{c}}\inv (\bar{m} \Fr(k_{\bar{c}}) \bar{m}\inv) \in T^c_{x_c}$ is a lift of a member of $\bar{c}$.

$$\{f_{\bar{c}} \, \colon \bT \rightarrow \bG \, | \, \text{ $f_{\bar{c}}(s) = \Ad({k_{\hat{c} }k_{\bar{c}} k_{\hat{c}}\inv}) s$ for $s \in \bT$  and $\bar{c} \in (\bar{\bfT}^c_{x_c})_{\bar{m}\Fr}$}\}$$

is a complete set of representatives for the $G^{\Fr}$-conjugacy classes of $k$-embeddings  of $\bT$ into $\bG$.

If $\hat{c} \in ({W}_{T^c})_{\sim \Fr}$, then the  fiber over $\hat{c}$ of  the natural map

$$(\bar{W}_{T^c})_{\sim \Fr} \rightarrow ({W}_{T^c})_{\sim \Fr}$$

parameterizes the set of $G^{\Fr}$-conjugacy classes of tame $K$-minisotropic maximal $k$-tori that appear in the $k$-stable class corresponding to $\hat{c}$.

From Lemma <ref>  every $G$-conjugacy class of tame  maximal $K$-tori in $\bG$ corresponds to an element of  $\Wtamesimnon$.    Thus the assignment $c \mapsto \lsup{G}{\,}\bT^c$ from   $(\Wwhistles)^{\Fr \circ N_q}$ to the set of $G$-conjugacy classes of tame $K$-minisotropic maximal tori in $\bG$ has image consisting of exactly those   $G$-conjugacy classes of  tame $K$-minisotropic maximal tori  that contain a torus defined over $k$.

Consequently, to establish   statements (<ref>)  and (<ref>) of the theorem it is enough to show that for every $c \in (\Wwhistles)^{\Fr \circ N_q}$ the $G^{\Fr}$-conjugacy classes and $k$-stable classes in $\CTkc$ can be parameterized as claimed.  This follows from Lemmas <ref> and <ref>.

Statement (<ref>) follows from  Corollary <ref> and Lemma <ref>.

§.§ A more concrete realization of bunch of group quotients in the tame setting

Fix $c \in \Wtamesim$, $w \in c$, $n \in \titsW$ with image $w$ in $\absW$.
Choose $h$, $\lambda = \lambda_n$, $g = g_n \in \bG(\nsplits)_{x_n,0}$ as in <ref> and set $\bT = \lsup{g}\Ebtorus$.   Note that the image of $n_\sigma := g \inv \sigma(g)$ in $\absW$ is $w$.  Let $F \subset \BB(G)$ be the facet corresponding to $x_n \in \BB^{\red}(G)$.

Because it will help with the bookkeeping later, we will refer to $w$ as $w_\sigma$.

We have

*  The group homomorphism that sends $m \in N_{G}(\bT)$ to $g\inv m g \in N_{\bG(\nsplits)}(\Ebtorus)$ induces an isomorphism

$$\rho_g \colon N_G(\bT)/T \longrightarrow \absW^{w_\sigma \sigma}.$$

*  The group homomorphism that sends $t \in T$ to $g\inv t g \in \Ebtorus(\nsplits)$ induces an isomorphism

$$\rho_g \colon T/{T_{0^+}} \longrightarrow (\Ebtorus(\nsplits)/\Ebtorus(\nsplits)_{0^+})^{w_\sigma \sigma}.$$

The group homomorphism that sends $t \in T_F$ to $g\inv t g \in \Ebtorus(\nsplits)_0$ induces an injective map

$$\rho_g \colon T_F/{T_{0^+}} \longrightarrow
(\bshA)^{w_\sigma \sigma}.$$

*  If $w_\sigma$ is $\sigma$-elliptic, then for $g$ as in Lemma <ref> the group homomorphism that sends $m \in N_{G_{F,0}}(T)$ to $g\inv m g \in  N_{\bG(\nsplits)}(\Ebtorus)$ induces an injective map

$$\rho_g \colon  N_{G_{F,0}}(\bT)/T_{0^+} \longrightarrow (N_{\Equotient_{x_0}} (\bshA))^{n \lambda(\xi) \sigma}.$$

Moreover, if $\bG$ is simply connected, then $T_0 = T_{0^+}$, $\bar{W}_T = N_{G_{F,0}}(\bT)/T_{0^+}$, and the maps of both (<ref>) and (<ref>) are isomorphisms.

The maps in (<ref>) and (<ref>) both fail to be surjective when $\bG = \PGL_2$.   See also Lemma <ref>.

Recall that ${n_{\sigma}} \in N_{\bG(\nsplits)}(\Ebtorus)$ has image $w_\sigma$ in $\absW$.

We begin by showing (<ref>).
Suppose $m \in N_{G}(\bT)$.

$$ {n_{\sigma}} \sigma(g\inv  m g) {n_{\sigma}}\inv = ( g\inv \sigma(g))( \sigma(g \inv) m \sigma(g)) (\sigma(g\inv) g) =  g\inv  m  g,$$

the map sending $m \in N_{G}(\bT)$ to $g\inv m g \in N_{\bG(\nsplits)}(\Ebtorus)$ induces a homomorphism from $N_{G}(\bT)$ to $\absW^{w_\sigma \sigma}$.  Since the kernel of this homomorphism is $T$, the map descends to an injective map $\rho_g \colon W_\bT \rightarrow \absW^{w_\sigma \sigma}$.

We now show that $\rho_g$ is surjective.  Choose $w' \in \absW^{w\sigma}$.  Fix $n' \in N_{\bG(\nsplits)}(\Ebtorus)$ such that the image of $n'$ in $\absW$ is $w'$.  We have

$$\sigma(\lsup{g}n') = g (g\inv \sigma(g)) \sigma(n') (\sigma(g)\inv g) g\inv = g ( \lsup{n_\sigma} \sigma(n') ) g\inv = \lsup{g} (n'a) = \lsup{g}n' \lsup{g}a$$

for some $a \in \Ebtorus(\nsplits)$.
As $\lsup{g}a \in \bT(\nsplits)$ we conclude that the coset $(\lsup{g}n' )\bT(\nsplits)$ is $\sigma$-invariant.  Since $\coho^1(\sigma, \bT(\nsplits))$ is trivial, we conclude that every $\sigma$-invariant coset in $N_{\bG(\nsplits)} ( \bT) / \bT(\nsplits)$ has a representative in $N_G(\bT)/T$.  Hence, $\rho_g$ is surjective.

We now show that (<ref>) holds.   Note that

\begin{equation*}
\begin{split}
T_{0^+} &= T \cap G_{F,0^+} = \bT(\nsplits)^\sigma \cap ( \bG(\nsplits)_{F,0^+})^\sigma =( \bT(\nsplits) \cap \bG(\nsplits)_{F,0^+})^\sigma \\ &= (\lsup{g}(\Ebtorus(\nsplits)) \cap \bG(\nsplits)_{F,0^+})^\sigma = (\lsup{g}(\Ebtorus(\nsplits) \cap \bG(\nsplits)_{F,0^+}))^\sigma
= (\lsup{g} (\Ebtorus(\nsplits)_{0^+}))^\sigma.
\end{split}
\end{equation*}

Since $\coho^1(\sigma, \bT(\nsplits)_{0^+})$ is trivial, we  conclude that $T/T_{0^+}$ is isomorphic to

$$ (\lsup{g} (\Ebtorus(\nsplits) )/ \lsup{g} (\Ebtorus(\nsplits)_{0^+}))^\sigma =  (\lsup{{g}} (\Ebtorus(\nsplits)/ (\Ebtorus(\nsplits)_{0^+})))^\sigma.$$

Since $w_\sigma$ is the image of $g\inv \sigma(g)$ in $\absW$, we conclude that $(\lsup{{g}} (\Ebtorus(\nsplits)/ (\Ebtorus(\nsplits)_{0^+})))^\sigma$ is isomorphic, via their identification under the map $\Ad(g)$,  to   $(\Ebtorus(\nsplits)/ (\Ebtorus(\nsplits)_{0^+}))^{w_\sigma \sigma}$.

For (<ref>), recall that $T_F = T \cap G_{F,0}$ and $g \in G_{F,0}$, so $g\inv T_F g \leq \Ebtorus(\nsplits)_0$.   Thus, from (<ref>)  we have an injective map from $T_F/{T_{0^+}}$ to $(\Ebtorus(\nsplits)_0/\Ebtorus(\nsplits)_{0^+})^{w_\sigma \sigma}$.    Since $(\Ebtorus(\nsplits)_0/\Ebtorus(\nsplits)_{0^+})$ is isomorphic to $\bshA$, the result follows.

We now show that (<ref>) is valid.   From Lemma <ref> we may assume that $g \inv \sigma(g) = z  \cdot \lsup{\lambda(\pi)}n $ for some $z \in \tilde{Z}_0$.
Suppose $m \in N_{G_{F,0}}(\bT)$.   Since $g \in \bG(\nsplits)_{F,0}$ we have $g\inv m g \in N_{\bG(\nsplits)_{F,0}}(\Ebtorus)$.  Since conjugation by $g$ carries topologically unipotent elements in $G_{F,0}$ to topologically unipotent elements in $\bG(\nsplits)_{F,0}$, we have a well-defined injective group homomorphism from $\bar{W}_T$ to $\Stab_{\bG(\nsplits)_{x_n,0}/ \bG(\nsplits)_{x_n,0^+}} (\Ebtorus)$ with image in $(\Stab_{\bG(\nsplits)_{x_n,0}/ \bG(\nsplits)_{x_n,0^+}} (\Ebtorus))^{{n_{\sigma}} \sigma}$.   Here the stabilizer in the quotient is to be interpreted as in Remark <ref>.
Since $x_n = \lambda(\pi) \cdot x_0$, ${n_{\sigma}} = z \cdot \lsup{\lambda(\pi)}n$, and $\bG(\nsplits)_{x_0,0}/ \bG(\nsplits)_{x_0,0^+}$ is isomorphic to $\Equotient_{x_0}$ the result follows.

Finally, if $\bG$ is simply connected and $x \in \bG(\nsplits)_{x_0,0}$ has image in $(\bG(\nsplits)_{x_0,0}/\bG(\nsplits)_{x_0,0^+})^{n \lambda(\xi) \sigma}$, then  $g \lambda(\pi) x \lambda\inv(\pi) g\inv$ belongs to $\Stab_{\bG(K)}(F) \cdot \bG(\nsplits)_{F,0^+} = \bG(K)_{F,0} \cdot \bG(\nsplits)_{F,0^+}$.  It follows that the maps of both (<ref>) and (<ref>) are surjective, hence isomorphisms.

If $y \in \bG(\nsplits)$ also has the property that $\lsup{y} \Ebtorus = \bT$, then there exists a $\tilde{n} \in N_{\bG(\nsplits)}(\Ebtorus)$ such that $g = y \tilde{n}$.  Thus, $\rho_y(N_G(\bT)/T) = \bar{w} \absW^{w_\sigma \sigma} \bar{w}\inv = \absW^{(\bar{w}w_\sigma \sigma(\bar{w})\inv )\sigma}$ where $\bar{w}$ denotes the image of $\tilde{n}$ in $\absW$.   Similarly,
$\rho_y(T/T_{0^+}) = (\Ebtorus(\nsplits)/ (\Ebtorus(\nsplits)_{0^+}))^{(\bar{w}w_\sigma \sigma(\bar{w})\inv )\sigma}$.

§.§ The action of Fr

Suppose $\lsup{G}\bT$ contains a $k$-torus.  It may happen that $\bT = \lsup{g}\bA$ is not defined over $k$.   However, if $\lsup{G}\bT$ contains a $\Fr$-stable element, then there exists $\ell \in G$ such that $\lsup{\ell g}\bA = \lsup{\ell}\bT$ is defined over $k$.   In fact, if $\bT$ is $K$-minisotropic, then from Lemma <ref> and Lemma <ref> we can also assume $\ell \in G_{x_n,0}$.  Note that $(\ell g)\inv \sigma(\ell g) = g\inv \sigma(g) = n_\sigma$.

Suppose $\bT = \lsup{g}\bA$ is defined over $k$ and $g \in \bG(\nsplits)_{x_n,0}$.
Define $n_{\Fr} \in N_{\bG(K_n)_{x_n,0}}(\Ebtorus)$ by $n_{\Fr} := g\inv \Fr(g)$.
Let $w_{\Fr}$ denote the image of $n_{\Fr}$ in $\absW$.

We have that $n_\Fr \circ \Fr$ stabilizes $N_{\bG(K_n)_{x_n,0}}(\Ebtorus)^{n_\sigma  \sigma}$ while
$w_\Fr \circ \Fr$ stabilizes $\absW^{w_\sigma \sigma}$.

It will be enough to prove the first statement. Suppose $m \in N_{\bG(K_n)_{x_n,0}}(\Ebtorus)$ and $\Ad({n_\sigma}) \sigma(m) = m$.   We have

\begin{equation*}
\begin{split}
\Ad({n_\sigma})  \sigma [\Ad({n_\Fr}) \Fr(m)] &= \Ad [ (g\inv \sigma(g) )\sigma(g\inv \Fr(g)) ] \sigma (\Fr (m))\\
& =  \Ad [ (g\inv \sigma(\Fr(g)) ] \sigma (\Fr (m))\\
& =  \Ad ( n_\Fr ) \Fr [\Ad(g\inv \sigma^q(g))  \sigma^q  (m)]\\
\end{split}
\end{equation*}

$$\Ad(g\inv \sigma^q(g))  \sigma^q  (m) = \Ad(g\inv \sigma^{q-1}(g)) \sigma^{q-1}[ \Ad(g\inv \sigma(g)) \sigma(m)] =  \Ad(g\inv \sigma^{q-1}(g)) \sigma^{q-1}[ m],$$

after $q-1$ iterations we conclude that $\Ad(g\inv \sigma^q(g))  \sigma^q  (m) = m$.  Hence, we have shown

$$\Ad({n_\sigma})  \sigma [\Ad({n_\Fr}) \Fr(m)]  =  \Ad ( n_\Fr ) \Fr (m),$$

as required.

Via $\Ad(g\inv)$ (that is, $\rho_g$), the action of $\Fr$ on $W_T$ and $T/T_{0^+}$ corresponds to the action of $w_{\Fr} \circ \Fr$ on $\absW^{w \sigma}$ and $(\Ebtorus(\nsplits)/\Ebtorus(\nsplits)_{0^+})^{w \sigma}$.

We will verify this for $\absW^{w_\sigma \sigma}$; the proof for  $(\Ebtorus(\nsplits)/\Ebtorus(\nsplits)_{0^+})^{w_\sigma \sigma}$ is similar.    The map from $N_G(T)$ to $N_{\bG(\nsplits)}(\Ebtorus)/\Ebtorus(\nsplits)$ is given by $m \mapsto g\inv m g$.   Thus $\Fr(m)$ maps to

$$g\inv \Fr(m) g = g\inv \Fr(g g\inv) \Fr(m) \Fr(g g\inv) g = (g\inv \Fr(g)) \Fr(g\inv m g) (\Fr(g)\inv g).$$

The result follows from looking at the images of these elements in $\absW^{w_\sigma \sigma}$.

We have

$$\text{$\Fr(N_q(n_\sigma)) = n\inv_\Fr n_\sigma \sigma(n_\Fr)$ \, \, and \, \,   $\Fr(N_q(w_\sigma)) = w\inv_\Fr w_\sigma \sigma(w_\Fr)$.}$$

For the first equality we have:

\begin{equation*}
\begin{split}
\Fr(N_q(n_\sigma)) &= \Fr(n_\sigma \cdot  \sigma(n_\sigma) \cdots \sigma^{q-1}(n_\sigma)) = \Fr(g \inv \sigma(g)  \cdot  \sigma(g \inv \sigma(g) ) \cdots \sigma^{q-1}(g \inv \sigma(g)) )\\
&= \Fr(g\inv \sigma^q(g))  = \Fr( g\inv (\Fr \inv \circ \sigma \circ \Fr) (g)) = \Fr(g\inv) \sigma(\Fr(g)) \\
&= \Fr(g\inv) (g g\inv) \sigma( (g g\inv) \Fr(g)) = (\Fr(g\inv) g) ( g\inv \sigma( g)) \sigma (g\inv \Fr(g))\\
&= n\inv_\Fr n_\sigma \sigma(n_\Fr).
\end{split}
\end{equation*}

The second equality follows from the first.

Recall that $n_\Fr \in N_{\bG(K_n)_{F,0}}(\Ebtorus)$.
The equality

$$\Fr(N_q(n_\sigma)) = n\inv_\Fr n_\sigma \sigma(n_\Fr)$$

of Lemma <ref>
uniquely determines the element $n_\Fr$  up to left multiplication by an element of $(N_{\bG(K_n)_{F,0}}(\Ebtorus))^{n_\sigma \circ \sigma}$. Similarly, the equality

$$\Fr(N_q(w_\sigma)) = w\inv_\Fr w_\sigma \sigma(w_\Fr)$$

uniquely determines the element $w_\Fr$  up to left multiplication by an element of $\absW^{w_\sigma \sigma}$.

It will be enough to establish the first statement. Suppose $y \in N_{\bG(K_n)_{F,0}}(\Ebtorus)$ satisfies

$$\Fr(N_q(n_\sigma)) = y\inv n_\sigma \sigma(y).$$

We then have

$$n_\sigma \sigma (n_{\Fr}y\inv) n_\sigma\inv  = n_\Fr y\inv$$

and the result follows.

Suppose $c$ is $\sigma$-elliptic.  Suppose $\bT' = \lsup{h}\bT \in\CTk$ with $h \in G$.  Set $m_h = h\inv \Fr(h) \in N_G(\bT)$ and let $w_h$ denote the image of $g\inv {m}_h g$ in $\absW$.   Via the map $\Ad(g\inv) \circ \Ad(h\inv)$ ( i.e., the composition $\rho_g \circ \varphi_h$ where $\varphi_h$ is defined in Lemma <ref>) the action of $\Fr$ on $W_T$ and $T/T_{0^+}$ corresponds to the action of $(w_h w_{\Fr}) \circ \Fr$ on $\absW^{w_\sigma \sigma}$ and $(\Ebtorus(\nsplits)/\Ebtorus(\nsplits)_{0^+})^{w_\sigma \sigma}$.

This follows from Lemmas <ref> and <ref>.

If $\bG$ is $K$-split, then from Lemmas <ref> and <ref> we have  $\Fr(w_\sigma^q) = w\inv_\Fr w_\sigma w_\Fr$ and $(W_T)_\simFr$ is in bijection with  the set of $ w_\Fr \circ \Fr$-conjugacy classes in $\absW^{w_\sigma}$.

Note that from Corollary <ref> the element $w_\Fr$ will only be determined (in $\absW$) up to left multiplication by an element of $\absW^{w_\sigma}$, but from Lemma <ref> this is exactly what must happen as different choices of base point for $\lsup{G}\bT$ will result in different realizations of the various actions.

Suppose $\bG$ is semisimple and $c$ is $\sigma$-elliptic.
The group homomorphism that sends $t \in T$ to $g\inv t g \in \Ebtorus(\nsplits)$ induces an isomorphism $\rho_g \colon T/{T_{0^+}} \rightarrow \bshA{}^{w_\sigma \sigma}$.

If $\bG$ is simply connected, then since $\nsplits/K$ is tame we have $\bG(\nsplits)_{F,0}^\sigma = G_{F,0}$.     Thus, in this case,
$\bfT_F  = T/{T_{0^+}}$ and so $\bfT_F$ is isomorphic to $\bshA{}^{w_\sigma \sigma}$ in agreement with Lemma <ref>.

Since $c$ is $\sigma$-elliptic, we have that $T$ is a bounded subgroup of $G$.  It follows that $T$ is bounded in $\bT(\nsplits)$, and so $T \subset \bT(R_{\nsplits})$ where $R_{\nsplits}$ denotes the ring of integers of $\nsplits$.   Since $g \in \bG(\nsplits)_{F,0}$, we have

$$T = \bT(R_{\nsplits})^{\sigma} = (\bT(\nsplits) \cap \bG(\nsplits)_{F,0})^\sigma = (\lsup{g}(\Ebtorus(\nsplits)) \cap \bG(\nsplits)_{F,0})^\sigma = (\lsup{g} (\Ebtorus(R_{\nsplits})))^\sigma.$$

From the proof of Lemma <ref> we have
$T_{0^+} =  (\lsup{g} (\Ebtorus(\nsplits)_{0^+}))^\sigma$.
Identify $\bshA$ with the image of $\Ebtorus(R_{\nsplits})/ \Ebtorus(\nsplits)_{0^+}$ in  $\bG(\nsplits)_{F,0}/\bG(\nsplits)_{F,0^+}$.    Since $\coho^1(\sigma, T_{0^+})$ is trivial, we  conclude that $T/T_{0^+}$ is isomorphic to

$$ (\lsup{g} (\Ebtorus(R_{\nsplits}) / \lsup{g} (\Ebtorus(\nsplits)_{0^+}))^\sigma =  (\lsup{\bar{g}} (\Ebtorus(R_{\nsplits})/ (\Ebtorus(\nsplits)_{0^+})))^\sigma =   (\lsup{\bar{g}} \bshA)^\sigma$$

where $\bar{g}$ denotes the image of $g$ in $ \bG(\nsplits)_{F,0}/\bG(\nsplits)_{F,0^+}$.

Since $w_\sigma$ denotes  the image of  $g \inv \sigma(g) \in N_{\bG(\nsplits)}(\Ebtorus)$ in $\absW$,  we have that $(\lsup{\bar{g}} \bshA)^\sigma$ is isomorphic to  $\bshA{}^{w_\sigma \sigma}$ via their identification under the map $\Ad(\bar{g})$.

Suppose $\bG$ is simply connected and $w_\sigma$ is $\sigma$-elliptic.   The set of $k$-stable classes in $\CTk$ is indexed by $(\absW^{w_\sigma \sigma})_{\sim_{w_\Fr \circ \Fr}}$.  If $\bT'$ is a $k$-torus in the $w_\Fr \circ \Fr$-class corresponding to $w' \in \absW^{w_\sigma \sigma}$,  then, up to $G^{\Fr}$-conjugacy, the set of  $k$-embeddings of $\bT'$ into $\bG$ is indexed by $\bshA{}^{w_\sigma \sigma}_{w' w_\Fr \circ \Fr}$.

This  follows from Lemmas <ref>, <ref>, <ref>, and <ref>.

§.§ The action of Fr when G is k-quasi-split and n is sigma elliptic

In this subsection we refine the results of subsection <ref> under the additional assumptions that $\bG$ is $k$-quasi-split and $n$ is $\sigma$-elliptic.

Since $\bG$ is $k$-quasi-split, we may assume that $x_0$ is $\Fr$-fixed, and so, since $x_n$ is $\Fr$-invariant, we have that $\Fr(\lambda) = \lambda$.   In this section we also assume that $\Fr$ is chosen such that $\Fr(\pi) = \pi$.

Since $n$ is $\sigma$-elliptic, from Lemma <ref> we may assume that $n_\sigma = z \cdot \lsup{\lambda(\pi)}n$.
for some $z \in \tilde{Z}_0$.

Define $n_F \in N_{\bG(\nsplits)_{x_0,0}}(\Ebtorus)$ by $n_F = \lsup{\lambda\inv(\pi)}n_{\Fr}$.

If $\bG$ is $k$-quasi-split and $n$ is $\sigma$-elliptic, then
we have that $n_F \circ \Fr$ stabilizes
$N_{\bG(\nsplits)_{x_0,0}}(\Ebtorus)^{n \lambda(\xi)\sigma}$.

Suppose $x \in N_{\bG(\nsplits)_{x_0,0}}(\Ebtorus)$ such that $\lsup{n \lambda(\xi)}\sigma(x) = x$.  Then

$$\lsup{\lambda(\pi)} x = \lsup{\lambda(\pi) n \lambda(\xi)} \sigma(x) = \lsup{n_\sigma} \sigma (\lsup{\lambda(\pi)} x).$$

Thus, $\lsup{\lambda(\pi)}x \in (N_{\bG(\nsplits)_{x_n,0}}(\Ebtorus))^{n_\sigma \sigma}$.
Since $\lsup{\lambda(\pi)} n_F \circ \Fr = n_{\Fr} \circ \Fr$, the result follows by unwinding definitions and using Lemma <ref>.

If $\bG$ is $k$-quasi-split and $n$ is $\sigma$-elliptic, then we have

$$\text{ $\Fr[N_q(n \lambda(\xi) z)] = n_F\inv ( n \lambda(\xi)z) \sigma(n_F)$.}$$

We have

\begin{equation*}
\begin{split}
\Fr[N_q(n & \lambda(\xi)z )] =
\Fr[N_q(\lsup{\lambda\inv(\pi)} n_\sigma  \lambda(\xi))]
=  \Fr[N_q(\lambda\inv(\pi) n_\sigma  \sigma(\lambda(\pi)))]\\
& = \Fr[( \lambda\inv(\pi) n_\sigma \sigma(\lambda(\pi)) ) \cdot \sigma(   \lambda\inv(\pi) n_\sigma \sigma(\lambda(\pi))) \cdot \sigma^2(  \lambda\inv(\pi) n_\sigma \sigma(\lambda(\pi))) \cdot \cdots \cdot \sigma^{q-1}(  \lambda\inv(\pi) n_\sigma \sigma(\lambda(\pi))]\\
& =\lambda\inv(\pi)  \Fr[ N_q( n_\sigma)  \sigma^q(\lambda(\pi))] = \lambda\inv(\pi)  \Fr[ N_q( n_\sigma) ] \lambda(\pi)  \cdot  \Fr[\lambda\inv(\pi) \sigma^q(\lambda(\pi))] \\
&= \lambda\inv(\pi)  \Fr[ N_q( n_\sigma) ] \lambda(\pi)  \cdot  \lambda(\Fr(\xi^q))
\end{split}
\end{equation*}

From Lemma <ref>, this becomes

\begin{equation*}
\begin{split}
\lambda\inv(\pi) &n_{\Fr}\inv n_\sigma \sigma(n_{\Fr}) \lambda(\pi)  \cdot  \lambda(\Fr(\Fr\inv(\xi)))
= (n_F\inv n z ) \cdot \lambda\inv(\pi) \sigma(n_{\Fr}) \lambda(\pi)  \cdot  \lambda(\xi)\\
&  = (n_F\inv n z ) \cdot \lambda\inv(\pi) \sigma( \lambda (\pi)) \cdot \sigma( \lambda\inv(\pi) n_{\Fr} \lambda(\pi)) \cdot \sigma(\lambda\inv(\pi)) \lambda(\pi)  \cdot  \lambda(\xi)\\
&  = (n_F\inv n z \lambda(\xi) ) \sigma(n_F)  \cdot \lambda\inv(\xi)   \cdot  \lambda(\xi)\\
&=    n_F\inv ( n \lambda(\xi) z) \sigma(n_F).
\end{split}
\end{equation*}

If $\bG$ is $k$-quasi-split and $n$ is $\sigma$-elliptic, then the equality

$$\Fr(N_q(n \lambda(\xi) z)) = n\inv_F n \lambda(\xi) z \sigma(n_F)$$

uniquely determines the element $n_F$  up to left multiplication by an element of $(N_{\bG(K_n)_{x_0,0}}(\Ebtorus))^{n \lambda(\xi) \circ \sigma}$.

The proof is very similar to that of  Corollary <ref>.

Suppose $\bG$ is $k$-quasi-split and simply connected.  Suppose also that  $n$ is $\sigma$-elliptic.     The set of $\bG(k)$-classes in $\CTk$ is indexed by $((N_{\Equotient_{x_0}}(\bshA))^{n \lambda(\xi) \sigma})_{\sim_{n_F \circ \Fr}}$.

This  follows from Lemma <ref>,  Lemma <ref>, and  tracing through how $\Fr$ acts on  $(N_{\Equotient_{x_0}}(\bshA))^{n \lambda(\xi) \sigma}$.

§.§ Example: the Coxeter conjugacy class for K-split groups

Suppose $\bG$ is $K$-split. Since $\bG$ is $K$-split, we have that $\sigma$ acts trivially on $\absW$ and  $\Pi = \Esimple$. Suppose $w_\sigma$ is a Coxeter element $\prod_{\alpha \in  \Esimple} w_\alpha$ in $\absW$; here $w_\alpha$ is the simple reflection in $\absW$ corresponding to $\alpha$.
We have that $\absW^{w_\sigma} = \langle w_\sigma \rangle$.   Let $\ell$ denote the order of $w_\sigma$ and suppose throughout this section that $(p,\ell) = 1$.

Since $(p,\ell) = 1$, we know that $w_\sigma^q$ is $\absW$-conjugate to $w_\sigma$.  Since $\Fr$  preserves $\Eroot$, it must carry $\Esimple$ to another basis for $\Eroot$.   Thus $\Fr(w_\sigma)$ is again a Coxeter element of $\absW$.
Since the set of Coxeter elements form a single $\absW$-conjugacy class, we conclude that the Coxeter conjugacy class in $\absW$ is stable under $\Fr \circ N_q$.  Thus, from Corollary <ref> we may choose a $\Fr$-stable maximal torus $\bT$ of $\bG$ in $\varphi_{\sigma}(\lsup{\absW}w_\sigma)$.

From Lemma <ref> we have that the elements of the set of  $k$-stable conjugacy classes in $\CTk$ are indexed by $(W_{T})_\simFr$.   Since $y = w_\sigma^q$ also generates $\langle w_\sigma \rangle = \absW^{w_\sigma}$, from  Lemma <ref> and the fact that $W_T$ is abelian we have that $(W_{T})_\simFr$ is isomorphic to
$\langle w_\sigma \rangle / \langle y^{-1} (w_\Fr \Fr(y) w_{\Fr}\inv) \rangle$.  From Lemma <ref> we have that
$y^{-1} (w_\Fr \Fr(y)w_{\Fr}\inv)= (w_\sigma\inv)^{q-1}$.  Thus, the elements of the set of $k$-stable conjugacy classes in  $\CTk$ are indexed  by $\langle w_\sigma \rangle / \langle w_\sigma^{q-1}  \rangle \cong \Z/(\ell,q-1)$.

The group $T_F/T_0$ may be identified with $\bfZ_{x_0}$, the center of $\bfG_{x_0}$.  Thus,  from Lemma <ref>, we conclude that, up to $G^{\Fr}$-conjugacy, the group $(\bfZ_{x_0})_{\Fr}$ parameterizes the $k$-embeddings of $\bT$ into $\bG$.

Since $\bfZ_{x_0}$ is central in the reductive quotient $\bfG_{x_0}$, we conclude that $w_\sigma$ acts trivially  on $\bfZ_{x_0}$.    Thus, if $\bT'$ is a $k$-torus in one of the stable classes in $\CTk$, then, up to $G^{\Fr}$-conjugacy, the set of $k$-embeddings of $\bT'$ into $\bG$ is indexed by $(\bfZ_{x_0})_{w_h w_\sigma \circ \Fr} \cong (\bfZ_{x_0})_{\Fr}$.
If we also assume that $\bG$ is $k$-split, then  $(\bfZ_{x_0})_{\Fr} = \bfZ_{x_0} / \{ x^{q-1} \, | \, x \in \bfZ_{x_0} \} $.

§.§ Example: SLn and unramified SUn

Suppose $\bG_+$ is the  $k$-group $\SL_n$ and $\bG_-$ is the $k$-group unramified $\SU_n$.   Suppose $p$ does not divide $n$, and  let $w_\sigma$ be a Coxeter element in $\absW$.  Note that the Coxeter class is the only $\sigma$-elliptic class in $\absW$.

From the discussion in <ref>,  we can choose a $k$-torus $\bT$ in the $G_{\pm}$-orbit $\varphi_\sigma(\lsup{\absW}w_\sigma)$ and the elements of the set of $k$-stable classes in $\CTk$  are indexed by $\langle w_\sigma \rangle / \langle w_\sigma^{q-1}  \rangle$, which is  isomorphic to  $\Z/(n,q - 1)$.

Since $\bG_\pm$ is simply connected, from Remark <ref> we have  $\bfT_F \simeq  \bshA{}^{w_\sigma \sigma}$, which, in this case, may be identified with $\mu_n$, the center of $\bG_{\pm}$.   The action of $w_\Fr$ on $\bshA{}^{w_\sigma \sigma}$ is trivial and the action of $\Fr$ on $\bshA{}^{w_\sigma \sigma}$ is given by $x \mapsto (x^{\pm 1})^q$.   Thus, $(\bshA{}^{w_\sigma \sigma})\sim_{w_\Fr \circ \Fr}$ is isomorphic to  $\Z/(n,q \mp 1)$.
We conclude that there are, up to $G_{\pm}^{\Fr}$-conjugacy, $(n,q \mp 1)$  $k$-embeddings of $\bT$ into $\bG$.

The set $\CTk$ breaks into $(n,q - 1)$ $k$-stable classes.  If  $\bT'$ is a $k$-torus in one of these classes,    then, up to $G_{\pm}^{\Fr}$-conjugacy, it can be embedded into $\bG_\pm$ in  $(n,q \mp 1)$ ways.

§.§ Example: Sp4

Suppose $p > 2$.  We adopt the notation of Example <ref>.

As discussed in Example <ref> there are two $G$-conjugacy classes of  $K$-minisotropic maximal $K$-tori in $\Sp_4$, denoted $\dorbit_{C_2}$ and $\dorbit_{-1}$, corresponding to the two elliptic $\absW$-conjugacy classes  ${C_2}$ and ${-1}$.    Since $\Sp_4$ is $k$-split, from Corollary <ref> each of  $\dorbit_{C_2}$ and $\dorbit_{-1}$ contains tori that are defined over $k$.

A Coxeter element of $\Sp_4$ has order $4$ and the center of $\Sp_4$ is isomorphic to $\mu_2$.  Therefore, $\dorbit^k_{C_2}$ decomposes into $(q-1,4)$ $k$-stable classes.  If $\bT$ is a torus in one of these classes, then, up to $\Sp_4(k)$-conjugacy, it $k$-embeds  into $\Sp_4$ in two ways.  Since $\bar{W}_T$ is isomorphic to $\mu_8$, the number of $\Sp_4(k)$-conjugacy classes in $\dorbit^k_{C_2}$ is $(q-1,8)$.

The set $\dorbit^k_{-1}$ breaks into five $k$-stable orbits, corresponding to the five conjugacy classes in $\absW^{-1} = \absW$.   Note that we can take $w_\Fr$ to be the trivial element in $\absW$.   A  conjugacy class $c$ in $\absW$ will be called $\alpha$-even if for some (hence any) element $w$ in $c$ we have  that $w_\alpha$ appears an even number of times in some (hence any) expression for $w$ in terms of the simple reflections $w_\alpha$ and $w_\beta$.  If $\bT'$ belongs to a $k$-stable class that is indexed by an $\alpha$-even conjugacy class (of which there are three), then, up to $\Sp_4(k)$-conjugacy, $\bT'$ has four $k$-embeddings into $\Sp_4$.  If $\bT'$ belongs to a $k$-stable class that is not indexed by an $\alpha$-even conjugacy class (of which there are two), then, up to $\Sp_4(k)$-conjugacy, $\bT'$ has two $k$-embeddings into $\Sp_4$.

For some choice of $\bT \in \dorbit^k_{-1}$, the group $\tilde{C}_2 := N_{G_{x_T,0}}(\bT)/T_0$ has $32$ elements and is isomorphic to the group, under matrix multiplication, of two-by-two matrices whose elements look like

a & 0 \\
0 & b
\end{bmatrix}
\, \text{ or } \,
\begin{bmatrix}
0 & c \\
d & 0
\end{bmatrix}

with $a,b,c,d \in \{1,i,-1,-i\}$ and $\Fr(i)=\pm i$ depending on whether or not $-1$ is a square in $\ff^\times$.  One calculates that
if $4$ does not divide $q-1$, then  $\dorbit^k_{-1}$ breaks into six $\Sp_4(k)$-conjugacy classes.
If $4$ does      divide $q-1$, then $\dorbit^k_{-1}$ breaks into fourteen $\Sp_4(k)$-conjugacy classes; more specifically, in this case two of the stable classes break into two  $\Sp_4(k)$-orbits each, two break into three $\Sp_4(k)$-orbits each, and one breaks into four $\Sp_4(k)$-orbits.

§.§ Example: PSp4

Suppose $p > 2$.  We adopt the notation of Example <ref>.

There are two $G$-conjugacy classes of  $K$-minisotropic maximal $K$-tori in $\PSp_4$, denoted $\dorbit_{C_2}$ and $\dorbit_{-1}$, corresponding to the two elliptic $\absW$-conjugacy classes  ${C_2}$ and ${-1}$.    Since $\PSp_4$ is $k$-split, from Corollary <ref> each of  $\dorbit_{C_2}$ and $\dorbit_{-1}$ contains tori that are defined over $k$.

A Coxeter element of $\PSp_4$ has order $4$ and the center of $\PSp_4$ is trivial.  Therefore, $\dorbit^k_{C_2}$ decomposes into $(q-1,4)$ $k$-stable classes.  If $\bT$ is a torus in one of these classes, then, up to $\PSp_4(k)$-conjugacy, it $k$-embeds  into $\PSp_4$ in one way; hence, the number of $\PSp_4(k)$-conjugacy classes in $\dorbit^k_{C_2}$ is also  $(q-1,4)$.

As in the case of  $\Sp_4$, the set $\dorbit^k_{-1}$ breaks into five $k$-stable orbits indexed by the $\absW$-conjugacy classes in $\absW$.       If $\bT'$ belongs to a $k$-stable class that is indexed by an $\alpha$-even conjugacy class, then, up to $\PSp_4(k)$-conjugacy, $\bT'$ has two $k$-embeddings into $\PSp_4$.  If $\bT'$ belongs to a $k$-stable class that is not indexed by an $\alpha$-even conjugacy class, then, up to $\PSp_4(k)$-conjugacy, $\bT'$ has one $k$-embedding into $\PSp_4$.

For some choice of $\bT \in \dorbit^k_{-1}$, the group $N_{G_{x_T,0}}(\bT)/T_0$ has $16$ elements and is isomorphic to the $\tilde{C}_2/\mu_2$.  One calculates that
if $4$ does not divide $q-1$, then  $\dorbit^k_{-1}$ breaks into six $\PSp_4(k)$-conjugacy classes.
If $4$ does  divide $q-1$, then $\dorbit^k_{-1}$ breaks into ten $\PSp_4(k)$-conjugacy classes; more specifically, in this case each stable classes breaks into two  $\PSp_4(k)$-orbits each.

§.§ Example: G2

Suppose $p > 3$.  We adopt the notation of Example <ref>.  As discussed in Example <ref> there are three $G$-conjugacy classes of  $K$-minisotropic maximal  $K$-tori in $\Gtwo$,  denoted $\dorbit_{G_2}$, $\dorbit_{A_2}$, and $\dorbit_{-1}$ corresponding to the three elliptic $\absW$-conjugacy classes $G_2$, $A_2$, and $-1$.
Since $\Gtwo$ is $k$-split, from Corollary <ref> each of   $\dorbit_{G_2}$, $\dorbit_{A_2}$, and $\dorbit_{-1}$ contains tori that are defined over $k$.

A Coxeter element of $\Gtwo$ has order $6$ and the center of $\Gtwo$ is trivial.  Therefore, $\dorbit^k_{G_2}$ decomposes into $(q-1,6)$ $k$-stable classes.  Since $\Gtwo$ has trivial center, if $\bT'$ is a torus in one of these classes, then, up to $\Gtwo(k)$-conjugacy, it $k$-embeds  into $\Gtwo$ in one way.   This shows that the number of $\Gtwo(k)$-conjugacy classes in $\dorbit^k_{G_2}$  is $(q-1,6)$; this can also be verified by calculating $\bar{W}_T$.

Suppose $\bT \in \dorbit_{A_2}^k$.  The group $\bar{W}_T$ is isomorphic to $\mu_6$,  and $\bfT_F$ is isomorphic to $\mu_3$.  One calculates that $(\bar{W}_T)_{\Fr} =(\bar{W}_T)_{\simFr}$ is $\mu_6(\ff)$ – that is, there are six $k$-stable classes in $\dorbit_{A_2}^k$ when the cubic roots of unity belong to $\ff$ and two otherwise.  For half of the $k$-stable classes in $\dorbit_{A_2}^k$ a torus $\bT'$ in the class will embed, up to $\Gtwo(k)$-conjugacy, into $\Gtwo$ in three ways.   For the other half of the classes a torus $\bT'$ in the class will embed, up to $\Gtwo(k)$-conjugacy, into $\Gtwo$ in one way.    Finally, there are twelve $G^\Fr$-conjugacy classes of tori in $\dorbit_{A_2}^k$ when $3 = \dabs{\mu_3(\ff)}$ and three otherwise.

Alternatively, if $w_\sigma = w_\alpha w_\beta w_\alpha w_\beta$, then $W_T \cong \absW^{w_\sigma \sigma} = \absW^{w_\sigma}= \langle w_\alpha w_\beta \rangle$.
Up to left multiplication by an element of $\absW^{w_\sigma}$, from Corollary <ref> we have $w_\Fr$ is trivial if $q$ is congruent to $1$ modulo $6$ and $w_\Fr$ is $w_\alpha$ if $q$ is congruent to $-1$ modulo $6$.  (Note that $\mu_3(\ff) = 1$ if and only if $q$ is congruent to $-1$ mod $6$.)  When $w_\Fr = 1$, there are six $k$-stable classes in $\dorbit_{A_2}^k$ and a torus $\bT'$ indexed by  $(w_\alpha w_\beta)^j \in \langle w_\alpha w_\beta \rangle $ will, up to $\Gtwo(k)$-conjugacy, $k$-embed into $\Gtwo$ in three ways when $j$ is even and in one way when $j$ is odd.  When $w_\Fr$ is $w_\alpha$, then there are two $K$-stable classes in $\dorbit_{A_2}^k$ and a torus $\bT'$ indexed by $\bar{w}  \in \langle w_\alpha w_\beta \rangle / \langle ( w_\alpha w_\beta)^2 \rangle  $ will, up to $\Gtwo(k)$-conjugacy, $k$-embed into $\Gtwo$ in three ways if $\bar{w}$ is not trivial and in one way if $\bar{w}$ is trivial.

If $\dabs{\mu_3(\ff)} = 3$, then $\dorbit_{A_2}^k$ breaks into twelve $\Gtwo(k)$-conjugacy classes.  If $\dabs{\mu_3(\ff)} = 1$, then $\dorbit_{A_2}^k$ breaks into three $\Gtwo(k)$-conjugacy classes.

For some choice of $\bT \in \dorbit^k_{-1}$, the group $\bar{W}_T$ is isomorphic to the Weyl group of $\Gtwo$ and $\bfT_F$ is isomorphic to  $\mu_2 \times \mu_2$.    Thus
$\dorbit^k_{-1}$ decomposes into six $k$-stable classes – one for each $w_\Fr$-conjugacy class in $\absW^{-1} = \absW$.
If $\bT'$ belongs to a $k$-stable class that is indexed by a  $w_\Fr$-conjugacy class with one element then, up to $\Gtwo(k)$-conjugacy, $\bT'$ has four $k$-embeddings into $\bG$.
If $\bT'$ belongs to a $k$-stable class that is indexed by a  $w_\Fr$-conjugacy class with three elements then, up to $\Gtwo(k)$-conjugacy, $\bT'$ has two $k$-embeddings into $\bG$.
If $\bT'$ belongs to a $k$-stable class that is indexed by a  $w_\Fr$-conjugacy class with two elements then, up to $\Gtwo(k)$-conjugacy, $\bT'$ has one $k$-embedding into $\bG$.

The group $N_{G_{x_T,0}}(\bT)/T_0$ has $48$ elements and is  isomorphic to a Tit's group of $\Gtwo$ with generators $n_\alpha$ and $n_\beta$ of order four where $\Fr(n_\alpha) = n_\alpha$ and $\Fr(n_\beta) = n_\beta^q$.
Independent of how $\Fr$ acts, $\dorbit_{-1}^l$ breaks into ten $\Gtwo(k)$-conjugacy classes. The $k$-stable classes indexed by a $w_\Fr$-class with an odd number of elements each break into two $\Gtwo(k)$-conjugacy classes.

§.§ Example: ramified SU3
Suppose $p > 3$.
We adopt the notation of Example <ref>.  As discussed in Example <ref> there are two $G$-conjugacy classes of  $K$-minisotropic maximal $K$-tori in ramified $\SU_3$,
denoted $\dorbit_{c_{w_\alpha}}$ and $\dorbit_{c_{w_0}}$, corresponding to the twisted Coxeter element $w_\alpha \sigma$ and the element $-1 = w_0 \sigma$.   Since  $c_{w_\alpha} = (\Fr \circ N_q)(c_{w_\alpha})$ and $c_{w_0} = (\Fr \circ N_q)(c_{w_0})$, from Corollary <ref> we conclude that each of $\dorbit_{c_{w_\alpha}}$ and $\dorbit_{c_{w_0}}$ contains tori that are defined over $k$.

Suppose $\bT \in \dorbit^k_{c_{w_\alpha}}$.  Then ${W}_T$ is isomorphic to $\mu_3$.  Therefore, for the twisted Coxeter class, we have that $\dorbit^k_{c_{w_\alpha}}$ decomposes into $\dabs{(\mu_3)_{\Fr}} = \dabs{\mu_3(\ff)}$ $k$-stable classes. Since $\bshA{}^{w_\alpha  \sigma} \cong \bar{\sfT}_F$ is trivial, for $\bT'$ in a given $k$-stable class there is, up to $\SU_3(k)$-conjugacy, only one  $k$-embedding of $\bT'$ into $\SU_3$.

Suppose $\bT \in \dorbit^k_{c_{w_0}}$.  Then ${W}_T$ is isomorphic to $S_3$, the symmetric group on three letters.   Since $\absW^{-1} = \absW$, we can assume $w_\Fr = 1$.  Thus
$\dorbit^k_{c_{w_\alpha}}$ decomposes into three $k$-stable classes – one for each $\Fr$-conjugacy class in $\absW^{-1} = \absW$.
We have $\bshA{}^{-1}$ is $\mu_2 \times \mu_2$.
If $\bT'$ belongs to a $k$-stable class that is indexed by a  $\Fr$-conjugacy class with one element then, up to $\SU_3(k)$-conjugacy, $\bT'$ has four $k$-embeddings into $\SU_3$.
If $\bT'$ belongs to a $k$-stable class that is indexed by a $\Fr$-conjugacy class with three elements then, up to $\SU_3(k)$-conjugacy, $\bT'$ has two $k$-embeddings into $\SU_3$.
If $\bT'$ belongs to a $k$-stable class that is indexed by a  $\Fr$-conjugacy class with two elements then, up to $\SU_3(k)$-conjugacy, $\bT'$ has one $k$-embedding into $\SU_3$.

§ K-MINISOTROPIC TORI IN ISOGENOUS GROUPS

Motivated by the examples of $\Sp_4$ and $\PSp_4$ discussed in Examples <ref> and <ref> we show that information about $K$-minisotropic maximal tori can, in some cases, be derived from information about the analogous tori in isogenous groups.

§.§ On the surjectivity of isogenies at the level of parahoric groups for tori

Recall that $L$ is the completion of $K$ and $R_L$ denotes its ring of integers.   Let $\bT$ and $\bT'$ be tori over $K$, hence they are tori over $L$. We denote by $\CT$ and $\CT'$ the connected Néron models of $\bT$ and $\bT'$, such that $\bar{T}_0 := \CT(R_L)\subset \bar{T} := \bT(L)$ is the parahoric subgroup of $\bar{T}$ and  $\bar{T}'_0 := \CT'(R_L)\subset \bar{T}' := \bT'(L)$ is the parahoric subgroup of $\bar{T}'$.

Suppose $\rho : \bT \ra \bT'$ is an isogeny.
Since $\CT$ and $\CT'$ are  (lft-)Néron models, $\rho$ extends uniquely to a  morphism $\rho:\CT\ra\CT'$ <cit.>.
From <cit.>, we have
$\bar{T}_{0^+} := \CT(R_L)_{0^+}= \ker(\CT(R_L)\ra\CT(\ffc))$ and $\bar{T}'_{0^+} := \CT'(R_L)_{0^+}= \ker(\CT'(R_L)\ra\CT'(\ffc))$.

If $\rho:\bT\ra \bT'$ is an isogeny whose order is prime-to-$p$, then $\rho[\bar{T}_0]=\bar{T}'_0$, $\rho[\bar{T}_{0^+}] = \bar{T}'_{0^+}$, and  $\rho$ is surjective on special fibers.

We first claim that $C:=\coker(\bar{T}_0 \xra{\rho} \bar{T}_0')$ is finite and prime-to-$p$. Indeed, we have a Snake Lemma diagram

\[
\begin{tikzcd}
&&&\ker(\rho_*)_I \arrow{d}\\
1\arrow{r}& \bar{T}_0\arrow{r}\arrow{d}& \bar{T}\arrow{r}{\kappa_{\bar{T}}}\arrow{d}&\X_*(\bT)_{I}\arrow{d} \arrow{r}&1\\
1\arrow{r}&\bar{T}'_0\arrow{r}\arrow{d}&\bar{T}' \arrow{r}{\kappa_{\bar{T}'}} \arrow{d}&\X_*(\bT')_{I}\arrow{r}&1\\
\end{tikzcd}
\]

where $\ker(\rho_*)$ and $H^1(L,\ker(\rho))$ are both  finite and prime-to-$p$, hence the claim.

Next, we claim that $\rho:\CT\ra\CT'$ is surjective on special fibers. Suppose  on the contrary that it is not.  Since $\ffc$ is infinite, we then have
$\coker(\CT(\ffc)\ra\CT'(\ffc))$ is infinite. Since $\bar{T}_0'=\CT'(R_L)$ surjects onto $\CT'(\ffc)$, we have that $C$ is also infinite, a contradiction.

Looking again at a Snake Lemma diagram

\[
\begin{tikzcd}
&&&\ker(\CT(\ffc)\ra\CT'(\ffc)) \arrow{d}\\
1\arrow{r}&\ker(\CT'(R_L)\ra\CT'(\ffc))\arrow{d}\arrow{r}&\CT'(R_L)\arrow{d}\arrow{r}&\CT'(\ffc)\arrow{r}&1 \\
& \coker(f^{\flat}) \arrow{r} &C&
\end{tikzcd}
\]

we need to show that $\rho^{\flat}$ is surjective.  Since we already know that $\CT(\ffc)$ surjects onto $\CT'(\ffc)$ and that both $C$ and $\ker(\CT(\ffc)\ra\CT'(\ffc))$ are finite and prime-to-$p$, it suffices to show that $\coker(f^{\flat})$ is pro-$p$. This follows from the fact that $\ker(\CT'(R_L)\ra\CT'(\ffc))$ is pro-$p$ <cit.>.

Suppose $\bT$ and $\bT'$ are tori that are defined over $K$. If $\rho:\bT\ra \bT'$ is an isogeny whose order is prime-to-$p$,  then
$\rho[T_0]= T'_0$ and  $\rho[{T}_{0^+}] = {T}'_{0^+}$.

Suppose $\gamma' \in T'_0$.   From Proposition <ref> there exists $\gamma \in \bar{T}_0 = \CT(R_L)$ such that $\rho(\gamma) = \gamma'$.  The fiber of  $\rho$ over $\gamma'$ is a finite scheme over $K$, and so any element of the fiber must be defined over a finite extension, call it $K'$,   of $K$.  Thus, $\gamma \in \bT(K') \cap \bT(L) = \bT(K)$.  We conclude that $\gamma \in T_0$.  A similar argument shows that  $\rho[{T}_{0^+}] = {T}'_{0^+}$.

§.§ On the surjectivity of isogenies at the level of parahoric subgroups

Suppose $\bH$ and $\bL$ are reductive $K$-groups and $\rho \colon \bH \rightarrow \bL$ is a $K$-isogeny.
Note that $\rho$ carries $H$ into $L$ and,  for $x \in \BB(H) = \BB(L)$,
it carries $\Stab_H(x)$ into $\Stab_L(x)$.   Thus, from <cit.> the map $\rho$ carries $H_{x,0}$ into $L_{x,0}$.
We show that, under mild conditions, the latter map is surjective.

A version of the lemma below, but for $k$ rather than $K$, occurs in <cit.>.

If $p$ does not divide the order of $\ker(\rho)$, then the map $\res_{H_{x,0^+}} \rho \colon H_{x,0^+} \rightarrow L_{x,0^+}$ is surjective.  Similarly, the  map
$\res_{H_{x,0}} \rho  \colon  H_{x,0} \rightarrow L_{x,0}$
is surjective.

Let $\bA_H$ denote a maximal $K$-split torus in $\bH$ such that $x$ belongs to the apartment of $A_H$. Let $\bA_L$ denote the corresponding maximal $K$-split torus of $\bL$.
Let $\Ebtorus_H$ denote the centralizer of $\bA_H$ in $\bH$, and let $\Ebtorus_L$ denote the centralizer of $\bA_L$ in $\bL$.   Since $\bH$ and $\bL$ are $K$-quasi-split, both $\Ebtorus_H$ and $\Ebtorus_L$ are maximal $K$-tori.  Moreover, $\rho(\Ebtorus_H) = \Ebtorus_L$.

Since  the restriction of $\rho$ to the group generated by unipotent elements in  $H_{x,0^+}$ is an isomorphism onto the group generated by    unipotent elements in $L_{x,0^+}$, it is enough to show that  the restriction of ${\rho}$ to the pro-unipotent radical of $\EbtorusKrat_H$  surjects onto the pro-unipotent radical of $\EbtorusKrat_L$.   This follows from Corollary <ref>.

Similarly,  since the restriction of $\rho$ to the group generated by unipotent elements in  $H_{x,0}$ is an isomorphism onto the group generated by    unipotent elements in $L_{x,0}$, it is enough to show that  the restriction of ${\rho}$ to the parahoric subgroup of $\EbtorusKrat_H$  surjects onto the parahoric subgroup of $\EbtorusKrat_L$.   This follows from Corollary <ref>.

§.§ k-embeddings of maximal K-anisotropic k-tori in isogenous semisimple groups

Suppose now that $\bH$ and $\bL$ are semisimple $k$-groups and $\rho \colon \bH \rightarrow \bL$ is a $k$-isogeny.

Let $\bT^H$ be a maximal $K$-minisotropic $k$-torus in $\bH$, and let $\bT^L$ be the corresponding maximal $K$-minisotropic $k$-torus in $\bL$.  Let $x_T$ denote the point in $\BB(L) = \BB(H)$ identified by $\bT$ and let $F$ denote the facet to which $x_T$ belongs.  From Remark <ref>
it follows that $(\res_H\rho)\inv[T^L_F]$, the preimage of $T^L_F = T^L \cap L_{x_T,0}$ under $\res_H\rho$, is $T^H_F$ and $(\res_H\rho)\inv[N_{L_{x_T,0}}(T^L)] = N_{H_{x_T,0}}(T^H)$.

Set $Z_F^H = Z^H \cap H_{F,0}$ where $Z^H$ denotes the center of $H$.  Similarly, since $\ker(\rho) \leq \bZ^H$ it makes sense to define $\ker(\rho)_F = \ker(\res_H\rho) \cap H_{F,0}$.

If $T^L_0$ denotes the parahoric subgroup of $T^L$, then $(\res_H\rho)\inv[T^L_0] \leq T^H_F$ and

$$(\res_H\rho)\inv[T^L_0] = \ker(\rho)_F T^H_0.$$

Since $T^L_0 \leq T^L_F$ and $(\res_H\rho)\inv[T^L_F] = T^H_F$, we have $(\res_H\rho)\inv[T^L_0] \leq T^H_F$.

Since $L$ is semisimple, we have that $T^L_0 = T^L_{0^+}.$  Thus, if
$t \in (\res_H\rho)\inv[T^L_0]$, then $\bar{t}$, the image of $t$ in $L$, is topologically unipotent and so $t$ must be topologically unipotent mod $\ker(\res_H \rho)$.   Since $(\res_H\rho)\inv[T^L_0] \leq T^H_F$, we conclude that $t$ must be topologically unipotent mod $\ker(\rho)_F$.

Define $W^H_{T} = N_H(\bT^H)/T^H = N_{H_{x_T,0}}(T^H)/T^H$, $\bar{W}^H_{T} = N_{H_{x_T,0}}(T^H)/T^H_{0}$, and $\bar{\sfT}^{H} = T^H_F/T^H_0$.   Define $W^L_{T}$, $\bar{W}^L_{T}$, and $\bar{\sfT}^{L}$ similarly.

The map $\rho$ induces an isomorphism $W^H_{T} \cong W^L_{T}$.  Moreover,
if $p$ does not divide the order of $\ker(\rho)$, then $\rho$ induces surjections $N_{H_{x_T,0}}(T^H) \twoheadrightarrow N_{L_{x_T,0}}(T^L)$, $T^H_F \twoheadrightarrow T^L_F$,  $\bar{W}^H_{T} \twoheadrightarrow \bar{W}^L_{T}$, and $\bar{\sfT}^{H} \twoheadrightarrow \bar{\sfT}^{L}$.  The kernel of each of these surjections is $\ker(\rho)_F$ or $\ker(\rho)_F T^H_0 /T^H_0$ as appropriate.

Since $\ker(\res_H\rho) \leq T^H$, we have $W^H_{T} \cong W^L_{T}$.
The remainder of the statements follow from  Corollary <ref> and Lemmas <ref> and <ref>.

If $p$ does not divide the order of $\ker(\rho)$, then

we may and do identify $\ker(\rho)_F$ with its image in $\bar{\sfT}^H$.

If $p$ does not divide the order of $\ker(\rho)$,  then

$$(T^L_F)_{\Fr} \cong (T^H_F)_{\Fr}/ (\ker(\rho)_F)_\Fr \text{\, \,  and \, \,}  (\bar{\sfT}^{L})_{\Fr}   \cong   (\bar{\sfT}^{H})_{\Fr} / (\ker(\rho)_F)_\Fr.$$

From Lemma <ref> we know that

$$1 \rightarrow \ker(\rho)_F \longrightarrow T^H_F \longrightarrow T^L_F \longrightarrow 1$$

is exact.
Since taking coinvariants is right exact, we have
$(T^L_F)_{\Fr} \cong (T^H_F)_{\Fr}/ (\ker(\rho)_F)_\Fr$.
Similarly,  from Lemma <ref> we know that

$$1 \rightarrow \ker(\rho)_F T^H_0/T^H_0 \longrightarrow \bar{\sfT}^{H}  \longrightarrow \bar{\sfT}^{L} \longrightarrow 1$$

is exact.   Since taking coinvariants is right exact, from Lemma <ref> we conclude
$(\bar{\sfT}^{L})_{\Fr}   \cong   (\bar{\sfT}^{H})_{\Fr} / (\ker(\rho)_F)_\Fr$.

If $p$ does not divide the order of $\ker(\rho)$,  then

\[\pushQED{\qed}  N_{L_{x_T,0}}(T^L)_{\sim \Fr} \cong (N_{H_{x_T,0}}(T^H)/\ker(\rho)_F)_{\sim \Fr}  \text{\, \,  and \, \,}  (\bar{W}^L_{T})_{\sim \Fr}  \cong (\bar{W}^H_{T}/\ker(\rho)_F)/_{\sim \Fr}.\qedhere  \popQED
\]

If $z \in \ker(\rho)_F$ and $c \in (\bar{W}^H_{T})_{\sim \Fr}$, then, since $z$ is central in $H$, we have $cz \in (\bar{W}^H_{T})_{\sim \Fr}$.
For $c_1$ and $c_2$ in $(\bar{W}^H_{T})_{\sim \Fr}$ we write $c_1 \stackrel{\rho}{\sim} c_2$ provided that there exists $z \in \ker(\rho)_F$ such that $c_1 = c_2z$.  Then  $(\bar{W}^H_{T}/\ker(\rho)_F)/_{\sim \Fr}$ may be identified with the set of $\stackrel{\rho}{\sim}$-equivalence classes in $(\bar{W}^H_{T})_{\sim \Fr}$.   A similar interpretation holds for $(N_{H_{x_T,0}}(T^H)/\ker(\rho)_F)_{\sim \Fr}$.

Suppose $\bG'$ is the derived group of $\bG$.  Recall that  $\bG_{\scon}$ denotes the simply connected cover of $\bG'$ while $\bG_{\ad}$ denotes the adjoint group of $\bG$.     Let $\bT'$ denote a $K$-minisotropic maximal $k$-torus in $\bG'$.

The results of this Section (that is, Section <ref>) show that if $p$ does not divide the order of  the kernel of the isogeny $\bG_{sc} \rightarrow \bG'$, then  the stable classes in the $\bG'$-orbit of $\bT'$,  the rational classes in the $\bG'$-orbit of $\bT'$, and the $k$-embeddings, up to rational conjugacy, of $\bT'$ into $\bG'$ can be calculated in a straightforward manner if we know how  to parameterize these things for the corresponding torus in $\bG_{\scon}$.

§.§ A generalization of Example <ref>

If $\bG$ is $K$-split and $\bT$ is a tame $K$-minisotropic maximal $k$-torus in $\bG$ corresponding to the Coxeter element of $\absW$, then from Example <ref>  the set of $k$-embeddings of $\bT$ into $\bG$ form a single $G_{\ad}^\Fr$-orbit.  This phenomenon can be generalized as follows.

Suppose $\bT$ is a maximal $K$-minisotropic $k$-torus in $\bG$. Let $x_T$ denote the point in $\BB(G)$ identified by $\bT$ and let $F$ denote the facet to which $x_T$ belongs.
If the images of $T \cap G_{F,0}$ and $Z \cap G_{F,0}$ in $\sfG_F$ agree, then $G_{\ad}^{\Fr}$ acts transitively on the set of $k$-embeddings of $\bT$ into $\bG$.   In fact, up to rational conjugacy, the parahoric subgroup $((G_{\ad})_{F,0})^{\Fr}$ acts transitively on the set of $k$-embeddings of $\bT$ into $\bG$.

Let $\bT_{\ad}$ denote the $k$-torus in $\bG_{\ad}$ corresponding to $\bT$.

The $k$-map $\bG \rightarrow \bG_{\ad}$ induces maps $G_{F,0} \rightarrow (G_{\ad})_{F,0}$ and $G_{F,0^+} \rightarrow (G_{\ad})_{F,0^+}$.   Since  the images of $T \cap G_{F,0}$ and $Z \cap G_{F,0}$ in $\sfG_F$ agree, we conclude that the image of $T_{\ad} \cap (G_{\ad})_{F,0}$ in $(\sfG_{\ad})_F$ is trivial.

Suppose $\gamma \in T^{\Fr}$ is strongly regular semisimple and $g \in G$ such that $\lsup{g}\gamma$ is $\Fr$-fixed.  It will be enough to show that there exists $h \in G_{\ad}^\Fr$ such that $\lsup{g}\gamma = \lsup{h} \gamma$.   Let $\bar{g}$ denote the image of $g$ in $G_{\ad}$.
There exists $\ell \in G_{\ad}^{\Fr}$ such that $\ell \bar{g} \cdot x_T  \in \bar{C}$, hence $\ell \bar{g} \cdot x_T = x_T$.  So, without loss of generality, we may assume that $\bar{g} \in \Stab_{G_{\ad}}(x_T)$. From Lemma <ref>  there exist $t \in T_{\ad}$ and $k \in (G_{\ad})_{F,0}$ such that $ \bar{g} = kt$.   Thus, $\lsup{g}\gamma = \lsup{k}\gamma$.   But then $k\inv \Fr(k) \in T_{\ad} \cap (G_{\ad})_{F,0} \leq T_{\ad} \cap (G_{\ad})_{F,0^+} = (T_{\ad})_0$.  Since $\cohom^1(\Fr, (T_{\ad})_{0})$ is trivial, there exists $y \in (T_{\ad})_{0}$ such that $k\inv \Fr(k) = y\inv \Fr(y)$.   But then $h = ky\inv \in ((G_{\ad})_{F,0})^\Fr \leq G_{\ad}^\Fr$ and $\lsup{h}\gamma = \lsup{k} \gamma = \lsup{g}\gamma$.

§ K-MINISOTROPIC COXETER TORI

The  existence of a maximal $K$-minisotropic $k$-torus in $\bG$ was established by Adler  in the Appendix to [12].   We sharpen this result by showing that there exists a maximal $K$-minisotropic $k$-torus  such that $x_T$ is in the interior of $C'$.

Recall that $Z_C = Z \cap G_{C,0}$.  There exists a maximal $K$-minisotropic $k$-torus in $\bG$ such that (a) $x_T \in C'$ and (b)  $T_C = Z_C T_{0^+}$.

Recall that to $z \in \BB(G)$ and $s \in \R_{\geq 0}$ we may define as in  [29, 30] the Moy-Prasad subgroups $G_{z,s}$ and  $G_{z,s^+}$  of $G$.  If $\psi$ is an affine root of $\bA$ with respect to $\bG$, $\bA$, $K$, and our valuation $\nu$, then we  let $U_\psi$ (resp. $U_{\psi^+}$) denote the  subgroup of $U_{\dot{\psi}}$ corresponding to $\psi$.  Recall that for $z \in \AA(\bA)$, we have $U_\psi \leq G_{z,s}$ if and only if $\psi(z) \geq s$ (resp. $U_{\psi^+} \leq G_{z,s^+}$ if and only if $\psi(z) > s$).

Without loss of generality, we suppose that $\bG$ is absolutely almost simple; this implies $\Phi(\bG,k)$ is irreducible and the center of $\bG$ is finite.   Since $\Fr$ preserves $C$, it acts on $\Delta$.
For $\psi \in \Delta$,  choose $u_\psi \in U_{\psi} \setminus U_{\psi^+}$  such that $\Fr (u_\psi) = u_{\Fr (\psi)}$.
Set $u = \prod u_\psi \in G^{\Fr}$.   Since $\Phi(\bG,k)$ is irreducible, there exists a unique point $x$ in $C = C'$ where $\psi(x) = \psi'(x)$ for all $\psi, \psi' \in \Delta$.  If $r = \psi(x)$ for some, hence any, $\psi \in \Delta$, then $u \in G_{y,r} \setminus G_{y,r^+}$ if and only if $x = y$.
In other words, $x$ is the unique point in $\BB(G)$ where the depth of $u$ achieves its minimum; in fact this is true for any element of $u G_{x,r^+}$.  It follows that $C_G(\gamma) \leq \Stab_G(C)$ for all $\gamma \in u G_{x,r^+}$.    Choose strongly regular $\gamma \in uG^{\Fr}_{x,r^+}$. Since $C_G(\gamma)$ is a bounded subgroup of $G$,  we conclude that $\bT := C_\bG(\gamma)$ is a  maximal $K$-anisotropic $k$-torus of $\bG$.  Note that $x_T = x$.

Suppose $t \in T_F = G_{x_T,0} \cap T$, and let $\bar{t}$ denote the image of $t$ in  $\bfG_{x_T}$.  Since $x_T$ is in $C'$, the reductive quotient $\bfG_{x_T}$ is $\bfA$.  Since  $t$ commutes with $\gamma \in u G_{x_T,r^+}$ and $\bar{t} \in \bfA$, we have  $\alpha(\bar{t}) = 1$ for all $\alpha \in \Delta$.  Thus, $T_C = Z_C T_{0^+}$.

The point $x_T$ for $\bT = C_{\bG}(\gamma)$ with strongly regular $\gamma$ as in the proof of Lemma <ref> is,
in the $K$-split tame setting, given by the Kac coordinates of the Coxeter conjugacy class in $\absW$.  For this reason, we make the following definition.

A torus that arises via the construction in the proof of Lemma <ref> is called a $K$-minisotropic Coxeter $k$-torus.   If we don't require that it be defined over $k$ (i.e., we don't require that $\Fr(u) = u$ or $\Fr(\gamma) = \gamma$), then we will call it a $K$-minisotropic Coxeter $K$-torus or just a $K$-minisotropic Coxeter torus.

$K$-minisotropic Coxeter $k$-tori have many desirable properties.  We enumerate some of these properties here.

Suppose $\bT$ is a $K$-minisotropic Coxeter $k$-torus with $x_T \in C'$.

* If a facet $H$ of $\BB(G)$ is contained in the closure of $C$, then the representatives for $\Stab_{G}(H)/G_{H,0}$ can be chosen in $T$.  If $H$ is also $\Fr$-stable, then the representatives for $\Stab_{G^{\Fr}}(H)/G^{\Fr}_{H,0}$ can be chosen in $T^{\Fr}$.

* The group $G_{\ad}^{\Fr}$ acts transitively on the set of $k$-embeddings of $\bT$ into $\bG$.

* Every element of $\Stab_{G_{\ad}}(C)$ is cohomologous to an element of $T_{\ad}$.  Here $\bT_{\ad}$ is the $k$-torus in $\bG_{\ad}$ corresponding to $\bT$.

The first point follows from Corollaries <ref> and <ref>, the second follows from
Lemma <ref>, and  the final point follows from  Lemma <ref> and Remark <ref>

Recall that $\eta \colon \bG_{\scon} \rightarrow \bG$ is the composition $\bG_{\scon} \rightarrow \bG' \rightarrow \bG$.

If $\bT$ is a $K$-minisotropic Coxeter $K$-torus and $x_T \in C'$, then $T_C = \eta[Z_{\scon}] T_0$.

We already know from Corollary <ref> that $T_C = \eta[T_{\scon}] T_0$.  Since $\bT_{\scon}$ is a $K$-minisotropic Coxeter $K$-torus in $\bG_{\scon}$, from Lemma <ref> we have $T_{\scon} = (T_{\scon})_C = Z_{\scon} \cdot (T_{\scon})_{0^+}$ where   $Z_{\scon}$ denotes the center of $G_{\scon}$.   Since $\eta[(T_{\scon})_{0^+}] \subset T_0$, the claim follows.

If $\bT$ is a $K$-minisotropic Coxeter $K$-torus in $\bG$, then $A \cap T = A \cap Z$.

This result is not true for arbitrary  $K$-minisotropic tori.   For example, if $p > 2$ and $\bT$ is a $K$-minisotropic torus in $\Sp_4$ of type  $-1 \in A_1 \times A_1$ as in Example <ref>, then $A \cap T$ has order $4$, not $2$.

Without loss of generality, we may and do assume that $\bG$ is absolutely almost simple.

Choose $g \in A \cap T$.  Let $\bH = C_{\bG}(g)^{\circ}$.   If $\bH = \bG$, then $g \in Z$.

Suppose $\bH \neq \bG$.   Note that $ \Ebtorus \leq \bH$ and $\bT \leq \bH$.

If $\bH$ is a Levi $K$-subgroup, then since $g \in A$ we must have that $\bH$ is the Levi of a parabolic $K$-subgroup.  However, since $\bT$ is elliptic in $\bG$, it cannot then belong to $\bH$.

Thus, $\bH$ must be a generalized Levi $K$-subgroup of $\bG$ that is not the Levi of a parabolic $K$-subgroup.  Consequently, the ranks of the derived groups of $\bH$ and $\bG$ must be the same and so we can identify $\BB^{\red}(H)$ in $\BB^{\red}(G)$.
Since $\bT \leq \bH$, by uniqueness we have $x_T \in \BB^{\red}(H) \subset \BB^{\red}(G)$.  Choose $r > 0$ so that $G_{x_T,0^+} = G_{x_T,r} \neq G_{x_T,r^+}$.  Since $x_T \in \BB(H)$, there exists $h \in H$ such that $hx_T \in \AA'(A)$.   Let $C_1'$ be the $G$-alcove in $\AA'(A)$ that contains $hx_T$, let $x_1$ be an absolutely special vertex in $\bar{C}'_1$, and let $\bB_1$ be the Borel subgroup determined by $\bA$, $\bar{C}_1'$, and $x_1$.
Since $\lsup{h}\bT$ is a $K$-minisotropic Coxeter $K$-torus in $\bG$, it contains an element $u$ that, modulo $G_{hx_T,r^+}$,  looks like $\prod u_\psi$ as $\psi$ runs over ${\Delta}' = \Delta(\bG, \bA, K, \nu, C_1')$ and  $u_\psi \in U_{\psi} \setminus U_{\psi^+}$.
Since  $H_{y,s} = H \cap G_{y,s}$ for all $y \in \AA'(A)$ and $s \in \R_{>0}$, we have that, modulo $H_{hx_T,r^+}$, the element $u \in H$ looks like $\prod u_\psi$  where $\psi$ runs over ${\Delta}'$ and  $u_\psi \in U_{\psi} \setminus U_{\psi^+}$.  We conclude that for each $\psi \in {\Delta}'$ we have $\dot{\psi} \in \Phi(\bH,\bA)$.   In particular, $\Delta(\bG,\bB',\bA)$ is a subset of $\Phi(\bH,\bA)$, hence $\Phi(\bH,\bA) = \Phi(\bG,\bA)$.  This contradicts our assumption that  $\bH \neq \bG$.

The proof of  Corollary <ref> shows that if $\bH$ is a connected reductive $K$-subgroup of $\bG$ that contains both a $K$-minisotropic $K$-torus and the centralizer of a maximal $K$-split torus of $\bG$, then $\bH$ must be $\bG$.

As Lemma <ref> shows, the Lie algebra version of Corollary <ref> holds for any $K$-minisotropic Cartan subalgebra.   Thus, in characteristic zero we can always conclude that if $\bG$ is semisimple and  $\bT$ is any $K$-minisotropic torus $\bG$, then $A \cap T$ is finite.

Suppose $k$ is any field of characteristic zero and $\mathfrak{g}$ is the Lie algebra of a reductive $k$-quasi-split group.   Suppose $\mathfrak{b}$ is a Borel $k$-subalgebra of $\mathfrak{g}$ and $\mathfrak{h}$ is a Cartan $k$-subalgebra  of $\mathfrak{b}$.  If $\mathfrak{t}$ is a $k$-elliptic Cartan subalgebra of $\mathfrak{g}$, then $\mathfrak{t} \cap \mathfrak{h}^k$ is $\mathfrak{z} \cap \mathfrak{h}^k$, where $\mathfrak{z}$ denotes the center of $\mathfrak{g}$.

Here $\mathfrak{h}^k$ denotes the maximal $k$-split toral subalgebra of $\mathfrak{h}$.

Suppose $Y \in \mathfrak{t} \cap \mathfrak{h}^k$.
Then $C_{\mathfrak{g}}(Y)$ is a Levi $k$-subalgebra of a parabolic $k$-subalgebra of $\mathfrak{g}$.  Since $\mathfrak{h} \leq C_{\mathfrak{g}}(Y)$ and $\mathfrak{h}$ is $k$-elliptic, we conclude that $C_{\mathfrak{g}}(Y) = \mathfrak{g}$.  That is, $Y \in \mathfrak{z}$.

In characteristic zero, the proof of Lemma <ref> can be modified to work on the  Lie algebra by choosing, for $\psi \in \Delta$, $X_\psi \in \gg_{\psi} \setminus \gg_{\psi^+}$  such that $\Fr (X_\psi) = X_{\Fr (\psi)}$.    Here the subgroups $\gg_{\psi}$ and $\gg_{\psi^+}$ of the root group $\gg_{\dot{\psi}}$ are the Lie algebra analogues of $U_{\psi}$ and $U_{\psi^+}$.   We then replace $\gamma$ with a regular semisimple $Y \in X+ \gg^{\Fr}_{x,r^+}$ where $X =  \sum X_\psi \in \gg^{\Fr}$.

In the tame setting the $G^{\Fr}$-conjugacy class of $C_{\bG}(\gamma)$ is independent of the choice of strongly regular $\gamma \in uG^{\Fr}_{x,r^+}$.  Similarly, in the tame setting in characteristic zero the $G^{\Fr}$-conjugacy class of the Cartan subalgebra $C_{\bgg}(Y)$ is independent of the choice of regular semisimple $Y \in X + \gg^{\Fr}_{x,r^+}$.   This follows from <cit.>.

§ TWO QUESTIONS ABOUT PARAHORIC SUBGROUPS OF K-MINISOTROPIC TORI

Mitya Boyarchenko, Stephen DeBacker, Anna Spice, Loren Spice, and  Cheng-Chiang Tsai

We continue to use the notation developed in the main body of this paper.  In particular, $\bG$ is a connected reductive $k$-group, $G$ is the group of $K$-points of $\bG$, $C$ is an alcove in the building of $G$, and $G_{C,0}$ is the parahoric subgroup of $G$ corresponding to $C$.   If $\bT$ is a $K$-torus, then  $\BB(T)$ is the alcove of $\BB(T)$, and we denote by $T_0$ the parahoric subgroup of $T$.

We explore two natural questions that arise from results in this paper:

*  Does the sequence

\begin{equation} \label{equ:exactomega}
1 \longrightarrow G_{C,0} \longrightarrow \Stab_{G}(C) \longrightarrow \Stab_G(C)/G_{C,0} \longrightarrow 1
\end{equation}

discussed in Section <ref> always split?

* If $\bT$ is a maximal  $K$-anisotropic torus in the derived group of $\bG$, is the cardinality of  $T/T_0$  independent of the isogeny class of the derived group?

§.§ On the splitting of <ref>

It is known that the sequence <ref> splits whenever $\bG$ is $K$-split and either the derived group of $\bG$ is simply connected or $\bG$ is almost simple [4].
In this appendix we  show that <ref> also
splits when both $\bG$ is adjoint and a $K$-minisotropic Coxeter $k$-torus (see Definition <ref>) in $\bG$ splits over a tame extension of $K$.  We then show that <ref>  does not, in general, split.

Suppose $\bT$ is a $K$-anisotropic maximal torus.  Note that  $T$ is bounded.

Write $T = T_{p'} \times T_p$ where $T_{p'}$ is the subgroup of elements of order prime to $p$ and $T_p$ is the pro-$p$-subgroup of $T$. If $T_p = T_0 = T_{0^+}$, which always happens when $\bT$ splits over a tame extension, then the  sequence

\begin{equation*} %\label{equ:tsplit}
1 \longrightarrow T_0 \longrightarrow T \longrightarrow T/T_0 \longrightarrow 1
\end{equation*}

If $\bG$ is an adjoint group and $\bT$ is a $K$-minisotropic Coxeter $K$-torus in $\bG$ that splits over a tame extension, then sequence <ref> splits.

This follows from the discussion prior to the statement of the lemma together with  Corollary <ref> and Lemmas <ref> and <ref>.

To see that the sequence in <ref> can fail to split in the absence of tameness, even in characteristic zero, consider the following example.  Let $k$ be $\Q_2$, let $i \in \bar{k}$ be a square root of $-1$, and let $E$ be the quadratic ramified extension $K[i]$.  Note that $1+i$ is a uniformizer in $E$.  Let $\bT$ be a $K$-torus such that $\bT(E) = E^\times$ and $T = N_{E/K}^1$, the kernel of the norm map from $E$ to $K$.  From <cit.> we have the following commutative diagram.

E^×= (E) [r, "κ_(E)"] [d, "N"]
_*() [d,  "α" ]

N_E/K^1 = T  [r, "κ_T"]

Here each of the horizontal maps is a surjective Kottwitz map, $N$ is  defined by $N(x) = x \tau({x})\inv$ where $\tau$ is the nontrivial element of $\Gal(E/K)$, and $\alpha$ is the obvious surjective map.  Since $N(1+i) = i$ and $\alpha (\kappa_{\bT(E)} (1 +  i)) \neq 0$ in $\X_*(\bT)_{\Gal(E/K)} \cong \Z/2 \Z$, we have $\kappa_T(i) \neq 0$.  Note that $\kappa_T(-1) = \kappa_T(i^2) = 2 \kappa_T(i) = 0$.  Since the kernel of $\kappa_T$ is $T_0$,  we conclude that $-1 \in T_0$ and $\bar{i}$, the image of $i$ in $T/T_0$, generates the group $T/T_0 \cong \Z/2 \Z$.  Since the only order two element of $T$ is $-1$, we conclude that the sequence in <ref> does not split.

If we replace $\Q_2$ in the above discussion with $\mathbb{F}_2((t))$ where $\mathbb{F}_2$ is the field with two elements and let $E$ be a quadratic extension of  $\mathbb{F}_2((t))$, then $T/T_0$ has order two, but $T$ has no order two elements and so the sequence in <ref> does not split.

§.§ Isogeny and the cardinality of Tquotient for K-anisotropic T

If $\bT$ is a $k$-torus in $\SL_2$ corresponding to the norm one elements of a tamely ramified extension and $\bT'$ is the corresponding $k$-torus in $\PGL_2$, then $T/T_0$ and $T'/T_0'$ have the same cardinality.  This is not a general phenomenon, but it is not rare.

From the results of Haines and Rapoport  and of Kottwitz discussed in Section <ref>  we have $T/T_0 \cong \X_*(\bT)_I$ where $I$ is the inertial group $\Gal(\bar{k}/K)$.  We will study $\X_*(\bT)_I$.

Let \(\Lambda\) be an additive cyclic group of order \(n\), and
\(\tau\) an automorphism of \(\Lambda\).
Let \(m\) be a multiple of \(n\) such that
\(\tau^m\) is trivial,
and put
\(N_{\tau, m}(\lambda) \mathrel{:=} \sum_{i = 0}^{m - 1} \tau^i(\lambda)\)
for all \(\lambda \in \Lambda\).
Then \(N_{\tau, m}\) is identically \(0\) on \(\Lambda\).

By the Chinese Remainder Theorem, we may, and do, assume that \(n\) is a power of some prime \(\ell\), so that \(\Lambda\) is a cyclic module for the \(\ell\)-adic integers \(\Z_\ell\).  Since \(N_{\tau, m}\) factors through \(N_{\tau^{m/n}, n}\), we may, and do, assume that \(m\) is also a power of \(\ell\), as well as a multiple of \(n\) (or even that \(m\) equals \(n\), though we do not need this).

Choose an element \(k \in \Z_\ell\) such that \(\tau\) sends a generator \(\lambda\) of \(\Lambda\) to \(k\lambda\), hence acts as multiplication by \(k\) on all of \(\Lambda\).  Then \(N_{\tau, m}\) acts on \(\Lambda\) as multiplication by the \(\ell\)-adic integer \(\sum_{i = 0}^{m - 1} k^i\).

Since \(\tau^m\), which acts on \(\Lambda\) as multiplication by \(k^m\), is trivial, and the annihilator of \(\Lambda\) in \(\Z_\ell\) is \(n\Z_\ell\), we have that \(k^m\) belongs to \(1 + n\Z_\ell\).  Since \(m\) is a power of \(\ell\) and \(\Z_\ell/(1 + \ell\Z_\ell)\) has order relatively prime to \(\ell\), we conclude that \(k\) belongs to \(1 + \ell \Z_\ell\), and so that the \(\ell\)-adic valuation of \(k^m - 1\) is the sum of the \(\ell\)-adic valuations of \(m\) and \(k - 1\).
That is, \(k^m - 1\) belongs to \(m(k - 1)\Z_\ell\), so \(\sum_{i = 0}^{m - 1} k^i = \frac{k^m - 1}{k - 1}\) belongs to \(m\Z_\ell \subseteq n\Z_\ell\), and hence multiplication  by \(\sum_{i = 0}^{m - 1} k^i \)  annihilates \(\Lambda\); that is, \(N_{\tau, m}\) is identically \(0\)  on \(\Lambda\).

We now present a technical result, Proposition <ref>, on Weyl groups, whose proof seems to require case-by-case analysis.

The quantification in Proposition <ref> seems elaborate, but note that almost every irreducible root system has a cyclic fundamental group, so the result usually says that the order of \(w\) is divisible by the order of the fundamental group, and in the remaining case (of \(D_n\) with \(n\) even) just says that the order of \(w\) is divisible by \(2\).

If \(\Phi\) is an irreducible root system and \(w\) is an elliptic element of \(W(\Phi)\), then the order of \(w\) is divisible by the order of every element of every subquotient of the fundamental group of \(\Phi\).

Put \(W = W(\Phi)\).  We indicate below the possibilities, according to the Borel–de Siebenthal algorithm, for a maximal proper, but non-parabolic, Weyl subgroup \(W'\) of \(W\).

For \(W\) of type \(A_n\), the maximal proper Weyl subgroups of \(W\) are all parabolic.  Since the Weyl groups of types \(B_n\) and \(C_n\) are the same, we do not distinguish them.

We put \(B_1 = C_1 = A_1\) and \(D_2 = A_1 + A_1\), but do not consider \(D_1\), so that any occurrence of a symbol \(D_m\) carries the implicit constraint \(m > 1\).

Type of \(W\)     Type of \(W'\)

\(B_n\) or \(C_n\)     \(C_r + C_{n - r}\), \(D_n\)

\(D_n\)     \(D_r + D_{n - r}\)

\(E_6\)     \(A_1 + A_5\), \(3A_2\)

\(E_7\)     \(A_1 + D_6\), \(A_2 + A_5\), \(A_1 + 2A_3\), \(A_7\)

\(E_8\)     \(A_1 + A_2 + A_5\), \(A_1 + A_7\), \(A_1 + E_7\), \(A_2 + E_6\), \(A_3 + D_5\), \(2A_4\), \(A_8\), \(D_8\)

\(F_4\)     \(A_1 + A_3\), \(A_1 + C_3\), \(2A_2\), \(B_4\)

\(G_2\)     \(A_1 + A_1\), \(A_2\).

Recall that \(W\) is the Weyl group of \(\Phi\).
For each entry in the chart above,
if \(W'\) is the Weyl group of \(\Phi'\) and \(\Phi''\) is an irreducible component of \(\Phi'\), then
in each case except \(D_n\) with \(n\) even, the fundamental group of \(\Phi\) is (cyclic and) isomorphic to a subgroup of the fundamental group of \(\Phi''\).  Even in this exceptional situation, we have that every cyclic subgroup of the fundamental group of \(\Phi\) is isomorphic to a subgroup of the fundamental group of \(\Phi''\).

Therefore, we may, and do, assume that \(w\) belongs to no proper Weyl subgroup of \(W\).  Then, using <cit.> to enumerate the possibilities and <cit.> to compute their orders, we find:

Type of \(W\)     order of \(w\) is divisible by …
\(A_n\)     \(n + 1\)

\(B_n\) or \(C_n\)     \(2n\)

\(D_n\)     \(2\gcd(n - 1, 2)\)

\(E_6\)     \(3\)

\(E_7\)     \(2\)

\(E_8\)     \(1\)

\(F_4\)     \(12\)

\(G_2\)     \(6\).

The result follows.

Suppose $k$ has characteristic zero and
$\bG$ is split and simple over \(K\).  Suppose $\rho \colon \bG \ra \bH$ is an isogeny. Let $\bT$ be a $K$-anisotropic maximal torus in $\bG$ and let $\bT'$ denote the corresponding torus in $\bH$ under $\rho$.   Let $E$ be the splitting field of $\bT$ over $K$, and suppose that $I_E := \Gal(E/K)$ is cyclic.
We have that  $\X_*(\bT)_{I_E}$ and $\X_*(\bT')_{I_E}$ have the same (finite) cardinality.

Since \(k\) has characteristic zero, the isogeny is separable.
Since \(\bG\) is \(K\)-split and \(K\)-simple, its absolute root system is irreducible.

By assumption, \(I_E\) is cyclic, say of order \(m\).  Choose a generator \(\tau\).

We have an exact sequence

\[
\begin{tikzcd}
0 \arrow{r}&\X_*(\bT) \arrow{r}{\rho_*} &\ \X_*(\bT') \arrow{r} &\Lambda \arrow{r}&0,
\end{tikzcd}
\]

and $\Lambda$ is a quotient of the (absolute) fundamental group \(\X_*(\bT')/\Z\Phi^\vee(\bH, \bT')\) of $\bH$, hence a subquotient of the fundamental group of the absolute root system \(\Phi(\bH, \bT')\). The short exact sequence above yields the long exact sequence in homology

\[
\begin{tikzcd}
%\cohom_1(I_E,\X_*(\bT)) \arrow{r}&
\cohom_1(I_E, \X_*(\bT')) \arrow{r} & \cohom_1(I_E, \Lambda) \arrow{r} & \X_*(\bT)_{I_E} \arrow{r}  & \X_*(\bT')_{I_E} \arrow{r} &\Lambda_{I_E} \arrow{r}&0.
\end{tikzcd}
\]

Since $I_E$ is cyclic of order \(m\) with generator \(\tau\), for every $\Z$-module $A$ on which $I_E$ acts we have $\cohom_1(I_E,A) = A^{I_E}/N_{I_E}[A]$  where $N_{I_E}(a) = \sum_{j=0}^{(m-1)} \tau^j(a)$ for $a \in A$.   Since $\bT$, and so $\bT'$, is $K$-anisotropic, we conclude that

$\cohom_1(I_E, \X_*(\bT'))  = 0$.   Thus, the long exact sequence becomes

\[
\begin{tikzcd}
0 \arrow{r} & \Lambda^{I_E}/N_{I_E}[\Lambda] \arrow{r} & \X_*(\bT)_{I_E} \arrow{r}  & \X_*(\bT')_{I_E} \arrow{r} &\Lambda_{I_E} \arrow{r}&0.
\end{tikzcd}
\]

\[
\begin{tikzcd}
0 \arrow{r} & \Lambda^{I_E} \arrow{r} & \Lambda \arrow{r}{1 - \tau} & \Lambda \arrow{r}  &  \Lambda_{I_E} \arrow{r}&0
\end{tikzcd}
\]

is exact, we conclude that $\dabs{\Lambda^{I_E}} = \dabs{\Lambda_{I_E} }$.
Thus, the index of $\X_*(\bT)_{I_E}$ in $ \X_*(\bT')_{I_E}$  is $\dabs{N_{I_E}[\Lambda]}$.

Since $\bG$, hence $\bH$, is $K$-split, it must be the case that there is some elliptic element $w \in \absW$ such that the action of $\Gal(E/K)$ on $\X_*(\bT')$ satisfies $\tau (\lambda) = w(\lambda)$ for all $\lambda \in \X_*(\bT')$.  Then $m$ is the order of $w$, so Proposition <ref> gives that the order of every element of $\Lambda$ divides $m$.

Moreover, the action of \(I_E\) on the fundamental group of the absolute root system \(\Phi(\bH, \bT')\), hence on its subquotient \(\Lambda\), is trivial.
Thus \(N_{I_E}(\Lambda)\) is multiplication by \(m\), which, by assumption, divides the order of every element of \(\Lambda\); so \(N_{I_E}(\Lambda)\) equals \(0\).

If $I_E$ is not cyclic, then  the cardinality of
$\X_*(\bT)_{I_E}$ and  $\X_*(\bT')_{I_E}$ need not be the same, even if \(\Lambda\) is cyclic.   For example, suppose that $k$ has characteristic zero, the residue field of $K$ has characteristic $3$, and  $E$ is a Galois extension of $K$ with Galois group $S_3$.   Let $\bT$ be the maximal $k$-torus in $\mathrm{SL}_3$
determined by the standard action of $S_3$ on

$$\X_*(\bT) = \{\sum a_i e_i \, | \, a_1, a_2, a_3 \in \Z \text{ with } a_1 + a_2 + a_3 = 0\}.$$

Here $(e_i \,|\, 1 \leq i \leq 3)$ is the standard basis of $\Z^3$.
Specifically, \(\bT\) may be taken to be \(\ker(\operatorname{Res}_{E'/k}\operatorname{GL}_1 \to \operatorname{GL}_1)\), where \(E'\) is the fixed field of any one of the order-\(2\) elements of \(I_E\).
Let $\bT'$ denote the corresponding $k$-torus in $\mathrm{PGL}_3$ under the isogeny $\rho \colon \mathrm{SL}_3 \rightarrow \mathrm{PGL}_3$.  Then $\X_*(\bT)_{I_E}$ is trivial while $\X_*(\bT')_{I_E}$ is isomorphic to $\mathbb{Z}/3\mathbb{Z}$.

§ ON -1 INVARIANTS AND TITS GROUPS OF A SPLIT SIMPLY CONNECTED GROUP

Ram Ekstrom

We continue to use the notation of the main body of this paper. Suppose $\bG$ is a simply connected $K$-split group.  This means that $\Esplits = K$ and $\barsigma = 1$.  Since $\bG$ is simply connected, $\X_*(\Ebtorus)$ is spanned by the coroots, ${\Phi}^\vee = \Phi^\vee(\bG,\Ebtorus)$.

Suppose $-1$ belongs to $\absW$ and let $n \in \titsW$ be a representative of $-1$.     In this note we investigate when the group of $n$-fixed points of $N = N_{\Egroup}(\Ebtorus)$ is a Tits group (with respect to some pinning).  We assume $n$ is tame.  Thanks to the next lemma, this  is equivalent to assuming $p$ is not two.

$n^4 = 1$.

Since $(-1)^2 = 1$, we know $n^2 \in \titsW \cap \EbtorusErat$.   Thus, there exists $\mu \in \Z {\Phi}^{\vee}$ such that $n^2 = \mu(-1)$.  The result follows.

Let $\ell$ denote the order of $n$; so $\ell$ is $2$ or $4$.
Let $\xi$ be $-1$ or $i$ depending on whether $\ell$ is  two or four.

§.§ The structure of nt fixed points of N for any t in the rational split torus

Recall that $\bG$ is simply connected and $K$-split.
Let $N  = N_{\Egroup}(\Ebtorus)$.

For all $t \in \EbtorusErat$, the elements $n$ and $n t$ have the same order.  In fact, we have $(n t)^2 = n^2$.

Note that

$$(n t)^2 =\lsup{n} t \cdot \lsup{n^2}t \cdot n^2 =  t\inv \cdot  t \cdot n^2 = n^2.$$

Since $n$ has order 2 or 4, the result follows.

Fix $t \in \EbtorusErat_0$.
There is a pinning $(\bG, \Ebtorus, \bB, \{X^t_a\}_{a \in \Esimple})$ such that if
$n_a^t$ is the unique element of

$$N \, \bigcap \, \bU_{-a}(\Esplits) \exp(X^t_a) \bU_{-a}(\Esplits) = N \, \bigcap \,  \exp(X^t_a) \bU_{-a}(\Esplits) \exp(X^t_a)$$

for $a \in \Esimple$, then

$$N^{n t} = \langle  n^t_a \colon  a \in \Esimple \rangle.$$

That is, $N^{n t}$ is the Tits group for the pinning  $(\bG, \Ebtorus, \bB, \{X^t_a\}_{a \in \Esimple})$.

If $a \in \Esimple$, then

\begin{align*}
\lsup{nt}n_a = \check{a}(a(t))^{-1} \cdot \lsup{n}n_a.
\end{align*}

Recall that $n_a$ is the unique element of $N \cap \bU_{-a}(\Esplits) \exp(X_a) \bU_{-a}(\Esplits)$. Because $\lsup{n} n_a$ and $n_a$ have the same image in $\absW$ and $n$ normalizes $\bG_a$, we have that  $n_a^{-1}\cdot\lsup{n}n_a$ lies in both $\Egroup_a \cap \EbtorusErat$ and $\titsW \cap \EbtorusErat$. So $n_a^{-1}\cdot \lsup{n}n_a$ is in the image of $\check{a}$ and has order either one or two. In the former case $n_a$ is $n$-fixed while in the latter $\lsup{n}n_a = n_a \check{a}(-1)$.
If $\lsup{n}n_a = n_a$, then
define $y_a^t \in R_{K}$ to be a square root of $a(t)$.    If  $\lsup{n}n_a = n_a \check{a}(-1)$, then define $y_a^t$ to be a square root of $-a(t)$.
Observe that these elements exist since the polynomial $X^2 - a(t)$ has roots in $R_{\tilde{K}}^\times = R_K^\times$.  This is because,  modulo $\wp_{\Esplits}$, $X^2 - a(t)$ has simple roots over the algebraicially closed residue field, $p$ is not equal to two, and $\Esplits$ is Henselian. Put $X^t_a = (y_a^t)^{-1} \cdot X_a$ and note that $n^t_a = \lsup{\check{a}(f_a^t)}n_a = n_a \check{a}(y^t_a)$ where $f_a^t$ is a square root of $(y_a^t)^{-1}$. We then calculate that if $\lsup{n}n_a = n_a$, then

\begin{align*}
\lsup{nt}n^t_a = \check{a}(a(t))^{-1}\cdot n_a\cdot \check{a}(y_a^t)^{-1} = n_a \cdot (\check{a}(a(t))\cdot  \check{a}(y_a^t)^{-1}) = n_a\check{a}(y^t_a) = n_a^t.
\end{align*}

While if $\lsup{n}n_a = n_a \check{a}(-1)$, then we similarly find

\begin{align*}
\lsup{nt}n_a^t = n_a \cdot ( \check{a}(a(t)) \cdot \check{a}(-1) \cdot \check{a}(y_a^t)^{-1}) = n_a\check{a}(y_a^t) = n_a^t.
\end{align*}

Consequently the group $\titsW^t =  \langle  n^t_a \colon  a \in \Esimple \rangle$ is a Tits group for the pinning  $(\bG, \Ebtorus, \bB, \{X^t_a\}_{a \in \Esimple})$ and  is $nt$-fixed. Since $N = \titsW^t \EbtorusErat$ and $\bG$ is simply connected, we find

\[
N^{nt} = \titsW^t \cdot (\EbtorusErat)^{nt} = \titsW^t \cdot (\EbtorusErat)^n = \titsW^t. \qedhere \]

Note that  $N^{n} = \titsW^1$, and it can happen that $\titsW^1 \neq \titsW$.

Simply connectedness is necessary in the above lemma. For example for the non-simply connected group $\bG = \PGL_2$ with $\Ebtorus$ the diagonal torus, consider the matrices
$$n =\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}, \, t_0 = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Then $n=n_a$ is a nontrivial representative of the Weyl group $N_{\tilde{G}}(\Ebtorus)/\EbtorusErat$ and one calculates that $(N_{\tilde{G}}(\Ebtorus))^n$ is the four element group generated by $n$ and $t_0$. On the other hand, preserving the notation of the above lemma, one checks that since $n_a$ is $n_a$-fixed we have that $y^1_a$ can be taken to be $1$, and $n_a^1 = n_a \check{a}(y^1_a) = n_a$. Consequently $\langle  n^1_a \colon  a \in \Esimple \rangle = \{1, n\}$ is a Tits group and is a proper subgroup of $(N_{\tilde{G}}(\Ebtorus))^n$.

§.§ Results about n

For all $t \in \EbtorusErat$ we have that
$n^2 = (nt)^2$ is central.

Fix $t \in \EbtorusErat$.   From Lemma <ref>  we know that $n^2 = (nt)^2$.

It will be enough to show that for all $a \in \Esimple$ we have $\lsup{(nt)^2}X^t_a = X^t_a$.

Fix $a \in \Esimple$.
Since $\lsup{nt}n^t_a = n^t_a$ and $\Ad({(nt)^2}) \bU_{-a}(\Esplits) = \bU_{-a}(\Esplits)$, we have

$$N_{\Egroup}(\Ebtorus) \, \bigcap \,  \bU_{-a}(\Esplits) \exp(X^t_a) \bU_{-a}(\Esplits)  =
N_{\Egroup}(\Ebtorus) \, \bigcap \, \bU_{-a}(\Esplits) \exp(\lsup{(nt)^2}X^t_a) \bU_{-a}(\Esplits).$$

A calculation shows this can happen if and only if $\lsup{(nt)^2}X^t_a = X^t_a$.

If $n$ has order four, then for all $t \in \EbtorusErat$ we have $n^2= (n t)^2 = \lambda_n(-1)$.

Recall from Section <ref> that there is an $h \in \Egroup$ such that $h\inv n h = \lambda_n(i)$.   Thus,

$$\lambda_n(-1) = (h\inv n h)(h\inv n h) = h\inv n^2 h.$$

Since $n^2$ is central, the conclusion follows.

If $\bG$ is $k$-quasi-split, $K$-split, then we may assume $\Fr(n) = n$.

If $\bG$ is $k$-quasi-split and $K$-split, then we may assume $\{X_a\}_{a \in \Esimple}$ is a Chevalley-Steinberg system for $\bG$ (see <cit.>).   Note that we then have $\Fr(n_a)= n_{\Fr(a)}$ for all $a \in \Esimple$.

If $w \in \absW$ has a reduced decomposition $w = s_{a_1} \cdots s_{a_r}$ where $a_i \in \Esimple$, then the lift $n_w$ of $w$ to $\titsW$ defined by $n_w = n_{a_1} \cdots n_{a_r}$ is independent of the reduced decomposition of $w$ <cit.>. It thus follows that if $-1 \in \absW$ has a reduced decomposition $-1 = s_{a_1} \cdots s_{a_m}$, then $\Fr(n) = \Fr(n_{a_1} \cdots n_{a_m}) = n_{\Fr(a_1)} \cdots n_{\Fr(a_m)} = n$ since $s_{\Fr(a_1)} \cdots s_{\Fr(a_m)}$ is another reduced decomposition of $-1$.

If $\bG$ is not quasi-split, then it may be the case that $n$ cannot be $\Fr$-fixed. As an example, let $D$ be a central division algebra over $k$ of index $2$ and let $\bG = \mathrm{SL}_{1,D}$. Then our $n$ fixes exactly one vertex of the Bruhat-Tits building $\BB(G)$ while $\Fr$ fixes exactly the barycenter $x$ of an alcove $C$. This latter fact follows from the fact that since $\{x\}$ is the underlying set of $\BB(\bG, k)$, the reductive quotient $\bfG_x$ is defined over $\ff$ and therefore is quasi-split by Lang's theorem. And so a Borel subgroup $\bfB \subset \bfG_x$ defined over $\ff$ is a torus since $\bG$ is anisotropic, whence $\bfG_x$ is an $\ff$-torus. This implies that $x$ is an interior point of $C$ (necessarily the barycenter). As a consequence one concludes that $n$ cannot be $\Fr$-fixed.

§.§ Results about nF

Recall that $n_\Fr$ and $n_F = \lsup{\lambda_n\inv(\pi)}n_\Fr$ are defined in Definitions <ref> and <ref>.
By construction, $n_F \in N_{\bG(K_n)_{x_0,0}}(\Ebtorus)$ normalizes $N^{n \lambda_n(\xi)}$.  In this section we assume $\bG$ is $K$-split, $k$-quasi-split and simply connected.

From Lemma <ref>, we may and do assume $\Fr(n) = n$.

From Lemma <ref> there exists $z \in Z_0 = \tilde{Z}_0$  such that
$\Fr \left[ N_q (n z \lambda_n(\xi)) \right] = n_F\inv (n z \lambda_n(\xi)) n_F$.  Since $-1 \in W$,  the center of $\bG$ is an elementary abelian two group and so
$z^2 = 1$.   Since $Z_0 = Z_{0^+}$ and $p \neq 2$, we conclude that $z = 1$.   Thus  $\Fr \left[ N_q (n  \lambda_n(\xi)) \right] = n_F\inv (n \lambda_n(\xi)) n_F$.

We have  $n_F \in N^{n \lambda_n(-1)}$.

Since $n^2$ is central, we have $\Ad(n) \lambda_n(\xi) \cdot \Ad(n^2) \lambda_n(\xi)$ is trivial.  Thus, $N_q(n \lambda_n(\xi)) = \Ad(n^q)(\lambda_n (\xi)) \cdot n^q$.

Suppose first that $n^2 = 1$.  In this case we have $\xi = -1$, and so
$N_q(n \lambda_n(-1)) = \Ad(n^q)(\lambda_n (-1)) \cdot n^q = \Ad(n) \lambda_n(-1) \cdot n = \lambda_n(-1) n = n \Ad(n\inv) \lambda_n(-1)$.   Thus,
from Corollary <ref> we conclude that

$$ n \lambda_n(-1)   =  n \lambda_n (-1)  = \Fr( N_q( n \lambda_n(-1) ) = (n_F)\inv \cdot (  n \lambda_n(-1) ) \cdot  n_F.$$

Consequently, $n_F \in N^{n \lambda_n(-1)}$.

Suppose now that $n$ has order four. In this case we can take $\xi = i$, and so
$N_q(n \lambda_n(i)) = \Ad(n^q)(\lambda_n (i)) \cdot n^q$.   We have two cases: $q  \equiv 1  \, (4)$ and $q  \equiv 3  \, (4)$

Suppose first that $q  \equiv 1  \, (4)$.
In this case $\Fr(i) = i$ and
$N_q(n \lambda_n(i)) = \Ad(n)(\lambda_n (i)) \cdot n = n \lambda_n (i)$.
Thus, from Corollary <ref> we conclude that

$$ n \lambda_n(i) = n \lambda_n (i)  = \Fr[ N_q(( n  \lambda_n(i)))] = (n_F)\inv \cdot ( n \lambda_n(i) ) \cdot  n_F,$$

and so  $n_F \in N^{n \lambda_n(i)}$.

Suppose now that $q  \equiv 3  \, (4)$.  In this case $\Fr(i) = -i$ and
$N_q(n \lambda_n(i)) = \Ad(n^3)(\lambda_n (i)) \cdot n^3$.  From Lemma <ref> and Corollary <ref>, $n^2 = \lambda_n(-1)$ is central, so

$$N_q(n \lambda_n(i)) = \Ad(n) \lambda_n (i) \cdot n \lambda_n(-1) = \lambda_n(-i) n \lambda_n(-1) = n \cdot \Ad(n\inv) \lambda_n(-i) \cdot \lambda_n(-1) = n \lambda_n(-i).$$

from Corollary <ref> we conclude that

$$n \lambda_n(i)  = \Fr [ n \lambda_n(-i)]  = \Fr[N_q (n \lambda_n(i))]= (n_F)\inv \cdot (  n \lambda_n(i) )\cdot n_F,$$

and so,  $n_F \in N^{n \lambda_n(i)}$.

The following result now follows from the above work and Corollaries <ref> and <ref>.

Suppose $\bG$ is a simply connected, $K$-split, $k$-quasi-split group and  $-1$ belongs to $\absW$.  Let $n \in \titsW$ be a lift of $-1$ such that $\Fr(n) = n$.   The set of $k$-stable classes in $\CTk$ is indexed by the conjugacy classes in $\absW$.
If $\bT'$ is a $k$-torus in the conjugacy-class corresponding to $w' \in \absW$,  then, up to $G^{\Fr}$-conjugacy, the set of  $k$-embeddings of $\bT'$ into $\bG$ is indexed by $\bshA{}^{-1}_{w'  \Fr}$. Finally, the set of $\bG(k)$-classes in $\CTk$ is indexed by $N^{n \lambda_n(\xi)}_{\sim Fr}$.

This Corollary and Lemma <ref>  were used in Example <ref> to compute the number of rational classes in $\CTk$ for $\bT$ corresponding to $-1$ in the Weyl group of $\Gtwo$.  We also used this Corollary to double-check the results about the tori that correspond to $-1$ for $\SL_2$ and $\Sp_4$  in Examples <ref> and  <ref>.

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