\section{Conjugacy classes of standard representatives}\label{sec:conjugacy}


In this section we consider $\eW$-conjugacy classes of standard representatives for $\sigma$-conjugacy classes, motivated by the conjugates appearing in Theorem~\ref{thm:ADLVChimneys}. Figure~\ref{fig:conjugates}, repeated from the introduction here with more detailed annotation, illustrates an example of the definitions and results of this section in type $\tilde{A}_2$. In Section~\ref{sec:fillOut}, we give a precise description of the $\eW$-conjugacy classes of standard representatives for non-integral Newton points with a rank 1 parabolic subgroup. This description is based on the notion of transverse subspaces introduced in Section~\ref{sec:transverse}. 


\subsection{Transverse subspaces and conjugacy classes} \label{sec:transverse}

For each Newton point $\nu$, we now introduce a collection of associated affine subspaces of the standard apartment $\mathfrak{a}$.  We then relate these subspaces to the $\eW$-conjugacy class of the standard representative $b_\nu$.  

\begin{definition}[Transverse subspaces]\label{defn:transverse}  Let $[b] \in B(G)_P$, and denote by $\nu$ the corresponding Newton point and by $b_\nu = t^\eta v$ the standard representative.  The affine subspace of $\mathfrak{a}$  given by 
\[
\cT_\nu := \nu + \Ker A_v
\]
is called \emph{the subspace transverse to $\nu$}.  For any $u \in \sW$, the affine subspace $u\cT_\nu$ is called \emph{a transverse subspace of $\nu$}.  
\end{definition}

In the special cases we treat in this paper, the next two results provide a concrete description of these transverse subspaces. See Figure~\ref{fig:conjugates} for an example of transverse subspaces in parabolic rank $1$, illustrating Corollary \ref{cor:nonintegralTransverse} below.

\begin{corollary}\label{cor:integralTransverse}  If either $[b_\nu] \in B(G)_P$ with $P=B$, or $[b_\nu] \in B(G)_P$ with $P = P_i$ and $\nu$ integral, then the subspace transverse to $\nu$ is the single point $\cT_\nu = \{\nu\}$.
\end{corollary}

\begin{proof}
By Example~\ref{ex:Pnu}(1) and Proposition~\ref{prop:stdReps}(1), respectively, in these two cases $b_\nu = t^\nu$.  Thus $A_v = A_\id = \Id$, and so $\Ker A_v$ is trivial.
\end{proof}

\begin{corollary}\label{cor:nonintegralTransverse}
Suppose $[b_\nu] \in B(G)_P$ with $P = P_i$ and $\nu$ non-integral, and write  $b_\nu = t^\eta s_i$ as in Proposition~\ref{prop:stdReps}(2).  Then 
\[
\cT_\nu = \nu + \{ r \alpha_i^\vee \mid r \in \R \}.
\]
\end{corollary}

\begin{proof}
The result follows from Definition \ref{defn:transverse} and the fact that $\Ker A_{s_i} = \{ r \alpha_i^\vee \mid r \in \R \}$, since by Lemma~\ref{lem:AwProperties}(4), the operator $A_{s_i}$ is the orthogonal projection onto $H_{\alpha_i}$.
\end{proof}

For general Newton points, we observe the following. 


\begin{corollary}\label{cor:stdRepTransverse}
Let $[b] \in B(G)_P$, and denote by $\nu$ the Newton point and by $\omega$ the Kottwitz point.  Let $b_\nu = t^\eta v$ be the standard representative for $[b]$.  Then $\eta \in \cT_\nu \cap (\omega+R^\vee)$.
\end{corollary}

\begin{proof}
First recall by property (1) of the standard representative from Lemma \ref{lem:stdRepList} that $\kappa_G(b_\nu) = \kappa_G(\eta) = \omega$, and so $\eta \in \omega+R^\vee$.  By definition, $\eta \in \cT_\nu$ if and only if $\eta - \nu \in \Ker A_v$.  Thus we want to show that $A_v \eta = A_v \nu$.  We have $A_v \eta = \nu$ by Lemma~\ref{lem:stdRepAvDom}, so it suffices to prove that $A_v$ fixes $\nu$.  Let $P = MN$ be the parabolic subgroup associated to $\nu$.  Then $\nu \in \cap_{\alpha_i \in \Delta_M} H_{\alpha_i}$.  Now $v \in W_M$, and it follows that $v$ fixes $\nu$. From the definition of $A_v$, it is now easy to see that $A_v$ fixes $\nu$, as desired.
\end{proof}


We now explain the connection between transverse subspaces and $\eW$-conjugacy classes of standard representatives.


\begin{lemma}\label{lem:containedTransverse} Let $[b] \in B(G)_P$, and denote by $\nu$ the Newton point and by $\omega$ the Kottwitz point.  Let $b_\nu = t^\eta v$ be the standard representative for $[b]$. Then for any $y = t^\mu u \in \eW$ with $\mu \in Q^\vee$ and $u \in \sW,$ the conjugate $b_\nu^y$ is of the form $t^\xi u v u^{-1},$ where $\xi \in (u\cT_\nu) \cap (\omega + R^\vee)$. 
\end{lemma}

\begin{proof}  First compute that $b_\nu^y = t^\xi u v u^{-1}$
where
\[
\xi = \mu + u \eta - u v u^{-1} \mu.
\]
Recall from Corollary \ref{cor:stdRepTransverse} that $\eta \in \omega + R^\vee$.  Since the action of $\sW$ on $\mathfrak{a}$ preserves $R^\vee$, then $\xi \in \omega + R^\vee$ as well, independent of $\mu \in Q^\vee$.

It remains to show that $\mu + u \eta - u v u^{-1} \mu$ lies in $u\cT_\nu$.  Multiplying on the left by $u^{-1}$ and replacing $u^{-1}\mu$ by $\mu'$, it is then enough to show that for all $\mu' \in R^\vee$, we have $\mu' + \eta - v \mu' \in \cT_\nu$.
Now by Corollary~\ref{cor:stdRepTransverse}, we have $\eta \in \cT_\nu$.  It follows that $\nu + \Ker A_{v} = \eta + \Ker A_{v}$.  It now suffices to show that $\mu' - v\mu' \in \Ker A_{v}$.  But $\mu' - v\mu' = (\Id - v)\mu' \in \Range(\Id -v) \subseteq \Ker A_{v}$ by Lemma~\ref{lem:AwProperties}(5), so we are done. 
\end{proof}

We rephrase the previous lemma in the following immediate corollary.


\begin{corollary}\label{cor:containedTransverse}  With all notation as in Lemma~\ref{lem:containedTransverse}, the $\eW$-conjugacy class of $b_\nu$ is contained in the set
\[
\bigcup_{u \in \sW} \left\{ t^\xi u v u^{-1} \mid \xi \in (u\cT_\nu) \cap (\omega+R^\vee) \right\}.
\]
\end{corollary}


\begin{figure}[ht]
\centering
\begin{overpic}[width=0.75\textwidth]{conjugates}
\put(41,46){\footnotesize{\color{red}$\alpha_2^\vee$}}
\put(54,46){\footnotesize{\color{red}$\alpha_1^\vee$}}
\put(49,52){\footnotesize{\color{black}$\check{\rho}$}}
\put(42,58){\footnotesize{\color{black}$b_\nu$}}
\put(45,53.5){\footnotesize{\color{blue}$\nu$}}
\put(49,57){\footnotesize{\color{blue}$\eta$}}
\put(18,35){\footnotesize{\color{blue}$\cT_\nu = s_1 T_\nu$}}
\put(39,30){\footnotesize{\color{DarkGreen}$s_1 s_2\cT_\nu = w_0 T_\nu$}}
\put(59,52){\footnotesize{\color{Fuchsia}$s_2 \cT_\nu = s_2 s_1 T_\nu$}}
\end{overpic}
\caption{The $\eW$-conjugacy class of $b_\nu = t^{\check{\rho}} s_1$, with sectors in the same color representing the corresponding chimneys.}
\label{fig:conjugates}
\end{figure}


We note that in general the union in the statement of Corollary~\ref{cor:containedTransverse} will not be disjoint.  For example, in Figure~\ref{fig:conjugates}, if $\xi \in \cT_\nu \cap R^\vee$ and $u = s_1$ then $\xi \in u \cT_\nu = \cT_\nu$ and $t^\xi s_1 = t^\xi u s_1 u^{-1}$.  In this figure, the colored alcoves, which lie along the transverse subspaces,  are exactly those which are (labeled by) the elements of $\eW$ given in the statement of  Corollary~\ref{cor:containedTransverse}. 


Motivated by Corollary~\ref{cor:containedTransverse}, we now make the following definition.


\begin{definition}\label{defn:fillsOut}  Let $[b] \in B(G)_P$, and denote by $\nu$ the Newton point and by $\omega$ the Kottwitz point.  Let $b_\nu = t^\eta v$ be the standard representative for $[b]$. We say that the $\eW$-conjugacy class of $b_\nu$ \emph{fills out the transverse subspaces} if it is equal to
\[
\bigcup_{u \in \sW} \left\{ t^\xi u v u^{-1} \mid \xi \in (u\cT_\nu) \cap (\omega + R^\vee) \right\}.
\]
\end{definition}

The next lemma gives a convenient condition equivalent to verifying the set equality in Definition \ref{defn:fillsOut}. Recall that by Corollary~\ref{cor:AwKerRange}, for any $v \in \sW$ we have $\Ker A_v = \Range(\Id - v)$. However, we will see later in Examples \ref{ex:C2} and \ref{ex:s1s3} that Condition~\eqref{eq:kerRangeZ} below is a stronger requirement, and by no means automatic from this equality.


\begin{lemma}\label{lem:fillsOut}   
Let $[b] \in B(G)_P$, and denote by $\nu$ the corresponding Newton point. Let $b_\nu = t^\eta v$ be the standard representative for $[b]$. The $\eW$-conjugacy class of $b_\nu$ fills out the transverse subspaces if and only if the following condition holds: 
\begin{equation}\label{eq:kerRangeZ}
\Ker A_{v} \cap R^\vee \subseteq (\Id - v)R^\vee.
\end{equation}
\end{lemma}

\begin{proof} Throughout the proof, denote the Kottwitz point of $[b]$ by $\omega$. First suppose that the $\eW$-conjugacy class of $b_\nu$ fills out the transverse subspaces.  Then by definition, for all $u \in \sW$ and  $\xi \in (u\cT_\nu) \cap (\omega + R^\vee)$, there exists a $\mu \in Q^\vee$ such that for $y = t^\mu u\in \eW$, we have $b_\nu^y = t^\xi u v u^{-1}$.  As in the proof of Lemma~\ref{lem:containedTransverse}, we have
\[
\xi = \mu + u \eta - u v u^{-1} \mu.
\]
Multiply on the left by $u^{-1}$ and replace $u^{-1}\xi$ by $\xi'$ and $u^{-1}\mu$ by $\mu'$.  Then for all $\xi' \in \cT_\nu \cap (\omega+R^\vee)$, there exists $\mu' \in R^\vee$ such that 
\[
\xi' -\eta = \mu' - v \mu'.
\]
In other words, for all $\xi' \in \cT_\nu \cap (\omega+R^\vee)$, we have $\xi' - \eta \in (\Id - v)R^\vee$.


Now consider an arbitrary element $\sum_{j = 1}^n c_j \alpha_j^\vee \in \Ker A_{v} \cap R^\vee$, where $c_j \in \Z$. Define $\zeta:= \eta + \sum_{j = 1}^n c_j \alpha_j^\vee$.  Since $\eta \in \cT_\nu \cap (\omega+R^\vee)$ by Corollary~\ref{cor:stdRepTransverse}, it follows that $
\zeta \in \cT_\nu \cap (\omega+R^\vee)$.  
So by the previous paragraph, $\sum_{j = 1}^n c_j \alpha_j^\vee = \zeta - \eta \in (\Id - v)R^\vee.$
Therefore Condition~\eqref{eq:kerRangeZ} holds.


For the reverse containment, suppose Condition~\eqref{eq:kerRangeZ} holds, and let $u \in \sW$ and $\xi \in (u\cT_\nu) \cap (\omega+R^\vee)$.  We will construct an element $\mu \in Q^\vee$ such that for $y = t^\mu u \in \eW$, we have $b_\nu^y = t^\xi u v u^{-1}$.
Write $\xi' = u^{-1}\xi$, so that $\xi' \in \cT_\nu \cap (\omega+R^\vee)$.   Since $\eta \in \cT_\nu \cap (\omega+R^\vee)$ by Corollary \ref{cor:stdRepTransverse}, it follows that 
\[ \xi' - \eta \in \Ker A_{v} \cap R^\vee.\]  
Then by Condition~\eqref{eq:kerRangeZ}, there exists $\mu' \in R^\vee$ such that
\[
\xi' - \eta = (\Id - v)\mu'.
\]

Finally, let $\mu = u\mu'$, and recall that $\xi = u\xi'$.  Then 
\[
\xi = \mu + u\eta - u v u^{-1}\mu.
\]
Thus if $y = t^\mu u$, we obtain $b_\nu^y = t^\xi u v u^{-1}$, as in the proof of Lemma~\ref{lem:containedTransverse}.  So the $\eW$-conjugacy class of $b_\nu$ fills out the transverse subspaces, as required.
\end{proof}


 Condition~\eqref{eq:kerRangeZ} from the statement of Lemma~\ref{lem:fillsOut} is quite subtle, as indicated by the following examples.  These examples use the indexing from the plates in Bourbaki~\cite{Bourbaki4-6}.  


\begin{example}\label{ex:C2}  Suppose that $G$ is of type $\tilde{C}_2$, and consider $v=s_2$.  By definition, we then have $A_{s_2}= \frac{1}{2}\left( I + s_2\right)$.  Compute that 
\[ A_{s_2}(\alpha_2^\vee) = \frac{1}{2}\left(I+s_2\right)\alpha_2^\vee = \frac{1}{2}\left(\alpha_2^\vee + s_2(\alpha_2^\vee)\right) = 0,\]
and so $\alpha_2^\vee \in \Ker(A_{s_2}) \cap R^\vee.$ (Alternatively, directly apply Lemma \ref{lem:AwProperties}(4) or Equation \eqref{eq:nuProj} from Lemma \ref{lem:NPimageAsi}.)

On the other hand, for any $c_1\alpha_1^\vee + c_2\alpha_2^\vee \in R^\vee$, we have
\begin{equation}\label{eq:s2ActC2} s_2(c_1\alpha_1^\vee + c_2\alpha_2^\vee) =  c_1s_2(\alpha_1^\vee) + c_2s_2(\alpha_2^\vee) =   c_1(\alpha_1^\vee+2\alpha_2^\vee) -c_2\alpha_2^\vee.
\end{equation}
Therefore, 
\[(I-s_2)(c_1\alpha_1^\vee + c_2\alpha_2^\vee) = (c_1\alpha_1^\vee + c_2\alpha_2^\vee)-(c_1(\alpha_1^\vee+2\alpha_2^\vee) -c_2\alpha_2^\vee)=2(c_2-c_1)\alpha_2^\vee. \]
Since $c_1,c_2 \in \Z$, the image $(I-s_2)R^\vee$ consists only of even multiples of $\alpha_2^\vee$, and in particular $\alpha_2^\vee \notin (I-s_2)R^\vee$. Since $\alpha_2^\vee  \in \Ker(A_{s_2}) \cap R^\vee$ but $\alpha_2^\vee \notin (I-s_2)R^\vee$,  Condition~\eqref{eq:kerRangeZ} does not hold for $v=s_2$ in type $\tilde{C}_2$. (Dually, Condition~\eqref{eq:kerRangeZ} does not hold for $v=s_1$ in type $\tilde{B}_2$.) 
\end{example}


As the next lemma suggests, the fact that Condition~\eqref{eq:kerRangeZ} fails for $v=s_2$ in type $\tilde{C}_2$ is not problematic for our purposes.  In type $\tilde{C}_2$, for all $[b] \in B(G)_P$ with $P=P_2$, Lemma \ref{lem:BCintegral} below says that the corresponding Newton point $\nu$ is integral, and so in fact $b_\nu = t^\nu$ and we are instead in the case of Corollary \ref{cor:integralTransverse}, rendering Lemma \ref{lem:fillsOut} for $v=s_2$ (and thus Example \ref{ex:C2}) irrelevant for our purposes. This feature is illustrated in Figure \ref{fig:C2StdReps}, where we see that all of the pink Newton points  $\nu \in H_{\alpha_2}$ are represented by the pink translation alcoves based at $\nu$. 
We remark that one can prove a version of Lemma \ref{lem:BCintegral} in greater generality, but we include here only those cases which arise directly in the proof of Proposition \ref{prop:conjRank1}. 


\begin{lemma}\label{lem:BCintegral}
Suppose $[b] \in B(G)_P$ with $P=P_i$, and denote by $\nu$ the Newton point.  If the Kottwitz point $\kappa_G(b) = 0$, and either of the following conditions hold:
\begin{enumerate}
    \item $G$ is of type $\tilde{B}_2$, and $i=1$; or
    \item $G$ is of type $\tilde{C}_2$, and $i=2$; 
\end{enumerate}
then $\nu$ is integral.
\end{lemma}

\begin{proof}
We include the proof for $G$ of type $\tilde{C}_2$ and $i=2$, since the argument for type $\tilde{B}_2$ with $i=1$ follows by duality.  Since $\kappa_G(b)=0$, the Newton point $\nu$ is the image of an element of $R^\vee$.  It thus suffices to prove that $A_{s_2}(R^\vee) \subseteq R^\vee$.  To this end, consider any $c_1\alpha_1^\vee+c_2\alpha_2^\vee \in R^\vee$, and compute using \eqref{eq:s2ActC2} that 
\[ (I+s_2)(c_1\alpha_1^\vee+c_2\alpha_2^\vee) = (c_1\alpha_1^\vee+c_2\alpha_2^\vee)+(c_1(\alpha_1^\vee+2\alpha_2^\vee) -c_2\alpha_2^\vee))=2c_1(\alpha_1^\vee+\alpha_2^\vee).\]
Therefore, $\nu = A_{s_2}(c_1\alpha_1^\vee+c_2\alpha_2^\vee) = \frac{1}{2}(I+s_2)(c_1\alpha_1^\vee+c_2\alpha_2^\vee) = c_1(\alpha_1^\vee+\alpha_2^\vee) \in R^\vee$. Since $\nu \in R^\vee$ and $\kappa_G(b)=0$, then $\nu$ is integral.
\end{proof}


The lucky phenomenon demonstrated by Lemma \ref{lem:BCintegral}, in which Condition~\eqref{eq:kerRangeZ} fails precisely when no non-integral Newton points actually exist, does not always persist when the rank of the parabolic subgroup is greater than 1.  In fact, Example \ref{ex:s1s3} below illustrates the primary obstruction to providing a straightforward generalization of our methods to treat parabolic subgroups of higher rank.  More specifically, above parabolic rank 1, Example \ref{ex:s1s3} shows that the $\eW$-conjugacy class of a standard representative does not always fill out the transverse subspaces.

\begin{example}\label{ex:s1s3}  Suppose that $G = SL_4$, and consider the element $b = t^{(1,0,0,-1)}s_1s_3 \in \eW$. Compute using \eqref{eq:avgFormula} that $\nu := \nu_G(b) = \left(\frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)$.  By \eqref{eq:PnuRoots}, we have $P_\nu = MN$ where the Levi subgroup $M$ is isomorphic the product $GL_2 \times GL_2$ inside $SL_4$.  To distinguish these two copies of $GL_2$, we write $M_1 \times M_3$, where $\sW_{M_i} = \langle s_i \rangle$ are each of type $A_1$.

We claim that $b$ is the standard representative of $[b]$, which we prove using Lemma \ref{lem:stdRepList}. First note that $\kappa_G(b) = 0$ since $(1,0,0,-1) \in R^\vee$ and $G = SL_4$, verifying property (1).  Next, note that $b = (t^{\varpi_1}s_1) \cdot (t^{\varpi_3}s_3)^{-1} \in \Omega_M$, verifying property (2).  Therefore, $\nu_M(b) = \nu_G(b) = \nu$, satisfying property (3).   By Lemma \ref{lem:stdRepList}, we now have that $b = t^{(1,0,0,-1)}s_1s_3$ is the standard representative of $[b]$; compare \cite[Ex.~7.2.5]{GHKRadlvs}.


For $v=s_1s_3$, we have shown thus far that $v$ is the spherical direction of the standard representative $b_\nu =t^{(1,0,0,-1)}s_1s_3$.  Since $\nu \in H_{\alpha_1} \cap H_{\alpha_3}$, note that the associated parabolic subgroup is not of rank~$1$.  We proceed to show that Condition~\eqref{eq:kerRangeZ} does not hold for $v = s_1 s_3$. With respect to the basis $\{ \alpha_1^\vee, \alpha_2^\vee, \alpha_3^\vee \}$, the matrix for $v$ is
\[
v = \begin{pmatrix} 
-1 & 1 & 0 \\
0 & 1 & 0  \\
0 & 1 & -1 
\end{pmatrix}.
\]
Since $v$ has order $2$, 
\[
A_v =  \frac{1}{2}\left(\Id + v\right) =  \frac{1}{2}\begin{pmatrix} 
0 & 1 & 0 \\
0 & 2 & 0  \\
0 & 1 & 0 
\end{pmatrix}
\quad \mbox{and} \quad 
\Id - v = \begin{pmatrix} 
2 & -1 & 0 \\
0 & 0 & 0 \\
0 & -1 & 2 
\end{pmatrix}.
\]  
Thus,
\[
\Ker A_v \cap R^\vee = \{ c_1 \alpha_1^\vee + c_3 \alpha_3^\vee \mid c_1, c_3 \in \Z \}.
\]  
In \eqref{eq:kerRangeZ}, we are thus considering whether for all $c_1, c_3 \in \Z$ the equation
\[
c_1 \alpha_1^\vee + c_3 \alpha_3^\vee = (\Id - v)\left( \sum_{i=1}^3 d_i \alpha_i^\vee \right)
\]
has a solution with $d_1, d_2, d_3 \in \Z$.  
This is equivalent to the system
\begin{eqnarray*}
2 d_1 - 2d_3 & = & c_1 - c_3 \\
d_2 - 2d_3 & = & -c_3,
\end{eqnarray*}
and this system has a solution $d_1, d_2, d_3 \in \Z$ if and only if the integer $c_1 - c_3$ is even.   Thus Condition~\eqref{eq:kerRangeZ} does not hold for $v = s_1 s_3$ in $\sW$ in type $\tilde{A}_3$, showing that $\eW$-conjugacy classes of standard representatives do not necessarily fill out the transverse subspaces if the rank of the parabolic subgroup is larger than 1.
\end{example}


\subsection{Conjugacy classes of standard representatives in parabolic rank 1}\label{sec:fillOut}

We now consider $\eW$-conjugacy classes of standard representatives $b_\nu$ where $\nu$ is non-integral and the associated parabolic subgroup has rank $1$.  The main results are   Corollary~\ref{cor:conjRank1}, which gives a precise description of these $\eW$-conjugacy classes in terms of the transverse subspaces introduced in Section~\ref{sec:transverse}, and Proposition~\ref{prop:inverseRank1}, which partly solves the ``inverse problem" for an element of the $\eW$-conjugacy class of $b_\nu$.


In contrast to Example~\ref{ex:s1s3}, when the parabolic subgroup is rank 1, we have the following.

\begin{prop}\label{prop:conjRank1}  Suppose $[b_\nu] \in B(G)_P$ with $P = P_i$ and $\nu$ non-integral.  Then 
\[
\Ker A_{s_i} \cap R^\vee \subseteq (\Id - s_i)R^\vee;
\]
that is, Condition~\eqref{eq:kerRangeZ} holds. 
\end{prop}

\begin{proof} Since $\Ker A_{s_i} \cap R^\vee = \{ c_i \alpha_i^\vee \mid c_i \in \Z \}$, it suffices to show that there are integers $d_j \in \Z$ such that 
\begin{equation}\label{eq:rank1}
\alpha_i^\vee = (\Id - s_i) \left( \sum_{j = 1}^n d_j\alpha_j^\vee\right).
\end{equation}
First note that 
\[
(\Id - s_i)\alpha_i^\vee = \alpha_i^\vee - s_i(\alpha_i^\vee) = 2\alpha_i^\vee.
\]
Recall more generally that $s_i(\alpha_j^\vee) = \alpha_j^\vee - \langle \alpha_i, \alpha_j^\vee \rangle \alpha_i^\vee$ by definition.  Using the conventions in \cite{Bourbaki4-6}, the value of the pairing $\langle \alpha_i, \alpha_j^\vee \rangle = C_{ij}$ equals the $ij$-entry of the Cartan matrix. 
Now, if there exists an $\alpha_j \in \Delta$ such that $\langle \alpha_i, \alpha_j^\vee \rangle = -1$, then
\[
(\Id - s_i)\alpha_j^\vee= \alpha_j^\vee - s_i(\alpha_j^\vee) =\alpha_j^\vee - \left( \alpha_j^\vee - \langle \alpha_i, \alpha_j^\vee \rangle \alpha_i^\vee\right) =  \alpha_j^\vee - (\alpha_j^\vee + \alpha_i^\vee) = -\alpha_i^\vee,
\]
and so Equation~\eqref{eq:rank1} holds.  

Using the plates from Bourbaki~\cite{Bourbaki4-6}, the only cases in which $C_{ij}\neq -1$ for any $j \in [n]$ are $G$ of type $\tilde{B}_2$ with $i=1$; type $\tilde{C}_2$ with $i=2$; and type $\tilde{G}_2$ with $i=2$. In type $\tilde{B}_2$ with $i=1$, Lemma \ref{lem:BCintegral} says that $\nu$ is always integral, so there is nothing to prove, since $\nu$ is non-integral by hypothesis.  Similarly, in type $\tilde{C}_2$ with $i=2$, there is nothing to check.

Finally, suppose that $G$ of type $\tilde{G}_2$ and $i=2$.  We then have
\[(\Id - s_2)\alpha_1^\vee = \alpha_1^\vee - s_2 (\alpha_1^\vee) = \alpha_1^\vee - (\alpha_1^\vee - \langle \alpha_2, \alpha_1^\vee \rangle\alpha_2^\vee) = \alpha_1^\vee - (\alpha_1^\vee +3\alpha_2^\vee)=-3\alpha_2^\vee. \]
Recalling from above that $(I-s_2)\alpha_2^\vee = 2\alpha_2^\vee$, we have
 $\{ (\Id - s_2)\alpha_j^\vee \mid \alpha_j \in \Delta \} = \{ -3\alpha_2^\vee, 2 \alpha_2^\vee \}$ in this case.  Taking for example $d_1 = d_2 = -1$, we see that Equation~\eqref{eq:rank1} holds, completing the proof.
\end{proof}


We immediately obtain the following precise description of the $\eW$-conjugacy classes of the standard representatives in parabolic rank $1$. For example, in Figure~\ref{fig:conjugates}, the (labels of) the colored alcoves are exactly the elements of the $\eW$-conjugacy class of $b_\nu$.

\begin{corollary}\label{cor:conjRank1}  Suppose $[b_\nu] \in B(G)_P$ with $P = P_i$ and $\nu$ non-integral, and let $\omega$ denote the Kottwitz point.  Then the $\eW$-conjugacy class of $b_\nu$ is equal to
\[
\bigcup_{u \in \sW} \{ t^\xi u s_i u^{-1} \mid \xi \in (u\cT_\nu) \cap (\omega+R^\vee) \}.
\]
That is, the $\eW$-conjugacy class of $b_\nu$ fills out the transverse subspaces.
\end{corollary}

\begin{proof}
Combine Lemma~\ref{lem:fillsOut} and Proposition~\ref{prop:conjRank1}.
\end{proof}


Given an element $z$ of the $\eW$-conjugacy class of $b_\nu$, a natural problem is to find all $y\in \eW$ such that $z = b_\nu^y$.  The next result provides one explicit solution $y \in \eW$ to this ``inverse problem" for every such $z$. We do not require a complete solution to the inverse problem for our current purposes, and so do not pursue this further.  We simply remark that, given one solution, finding all solutions would be equivalent to understanding the centralizer of $b_\nu$ in $\eW$. As far as we are aware, this centralizer is not known (even in the parabolic rank $1$ case).

\begin{prop}\label{prop:inverseRank1}
Suppose $[b_\nu] \in B(G)_P$ with $P = P_i$ and $\nu$ non-integral, and let $\omega$ denote the Kottwitz point.  For any $u \in \sW$ and $\zeta \in (u\cT_\nu) \cap (\omega+R^\vee)$, note by Corollary~\ref{cor:conjRank1} that $z = t^\zeta us_i u^{-1}$ is in the $\eW$-conjugacy class of $b_\nu$.  Then:
\begin{enumerate}
    \item there exists an integer $d_i \in \Z$ such that $u^{-1}\zeta - \eta = d_i \alpha_i^\vee$; and
    \item if $y = t^\mu u \in \eW$ with $\mu = d_i u \eta\in Q^\vee$ for $d_i \in \Z$ as in (1), then $z = b_\nu^y$.
\end{enumerate}
\end{prop}

\begin{proof}
We have $u^{-1}\zeta \in \cT_\nu \cap (\omega+R^\vee)$ by hypothesis, and $\eta \in \cT_\nu$ by Corollary~\ref{cor:stdRepTransverse}.  Since $\eta \in  \omega + R^\vee$ by Corollary \ref{cor:stdRepTransverse}, using Corollary~\ref{cor:nonintegralTransverse}  we see that $u^{-1}\zeta - \eta = d_i \alpha_i^\vee$ for some $d_i \in \Z$, and so (1) holds.

For (2), let $y = t^\mu u$, where $\mu = d_i u \eta$ and $d_i\in \Z$ is as in part (1).  Then the translation part of $b_\nu^y$ is equal to 
\[
\mu + u \eta - u s_i u^{-1}\mu = d_i u \eta + u \eta - d_i u s_i \eta = d_i u (\eta - s_i \eta) + u \eta.
\]
Since $\eta \in H_{\alpha_i,1}$ by Proposition~\ref{prop:stdReps}, then $s_i\eta = \eta -\alpha_i^\vee$, and so $\eta - s_i \eta = \alpha_i^\vee$.  Therefore the translation part of $b_\nu^y$ equals
\[
u( d_i \alpha_i^\vee + \eta) = u(u^{-1}\zeta) = \zeta
\]
by part (1). Since the spherical direction of $b_\nu^y$ is equal to $us_i u^{-1}$, we have that $b_\nu^y= z$, as required.
\end{proof}

In Figure~\ref{fig:conjugates}, for each alcove $\z$ corresponding to an element of the $\eW$-conjugacy class of $b_\nu = t^{\check{\rho}}s_1$, the shaded region of the same color is a sector representing the $(P_1,y)$-chimney, for $y$ as in the statement of Proposition~\ref{prop:inverseRank1}.  (We do not indicate the alcoves $\y$, as this would make the picture too cluttered.)
{"config": "arxiv", "file": "2006.16288/conjugacy.tex"}