\subsection{Diskussion der Parameter f"ur Joseph}


\begin{Bemerkungl}
  Let $G$ be a connected reductive complex algebraic group.
In \cite[1.12]{ABV} we find a definition of what data should be added to
this datum to obtain what is called there an {\bf extended group for $G/\DR$}.
 I prefer to call this
concept an {\bf $\DR$-extension of $G$}, but will still denote such an
$\DR$-extension by $G/\DR$. 
\end{Bemerkungl}



\begin{Bemerkungl}
  Let us agree to call an $\DR$-extension  $(G^\Gamma,\mathcal W)$ 
of $G$ {\bf special} if we have
$\delta^2=1$ for one and equivalently all $(\delta, N,\chi)\in\mathcal W$. 
This implies 
that the conjugation by $\delta$ is a quasisplit involution of $G$. 
\end{Bemerkungl}

\begin{Bemerkungl}
  Special $\DR$-extensions of $G$ can be obtained from the data 
$(\gamma, B,\chi)$ consisting of an antiholomorphic involution
$\gamma$ of $G$, a Borel $B\subset G$ stabilized by it, and a
nondegenerate unitary character $\chi:N^\gamma\ra\DC^\times$ 
on the fixed points of $\gamma$ in the unipotent radical $N\subset B$
of our Borel: We just have
to take as $G^\Gamma$ the semidirect product of $G$ with the two-element
group $\Gamma$ determined by our antiholomorphic involution, and
take the other data in the obvious way.\label{treI}  
We write elements of our semidirect product $(g,\sigma)$ with $g\in G$ and
$\sigma\in\Gamma$ and $(g,\sigma)(h,\tau)=(gh^\sigma,\sigma\tau)$.
It's such a relief that all elements of $\Gamma$ are their own inverses! 
\end{Bemerkungl}



\begin{Bemerkungl}
I claim that 
if our $\DR$-extension  $G/\DR$ of $G$ is special, 
the bijection of \cite[1.18]{ABV} induces  for any finite subgroup
$F\subset {\op{Z}}(G)$ a bijection
$$\Pi(G/\DR)_F\;\sira \;\Xi(G/\DR)_F$$
between subsets defined in the following way:
 For $\Pi(G/\DR)_F$, take in the $\Pi(G/\DR)$ from
\cite[1.14]{ABV} only equivalence classes $(\pi,\delta)$ with
$\delta^2\in F$. For $\Xi(G/\DR)_F$, take in
$\Xi(G/\DR)$ from \cite[1.17]{ABV} only those pairs equivariant under
the finite cover $^\vee G_F$ of the dual group corresponding to $F$. 
This claim is however nothing more than an educated guess, I did not
follow through the bijection constructed in \cite{ABV} to actually check it.  
\end{Bemerkungl}

\begin{Bemerkungl}
 Let us now specialize this claim to $F=1$ and to an $\DR$-extension
$G/\DR$ of $G$ determined by a triple $(\gamma, B,\chi)$ as in \ref{treI}. 
Then we consider only elements $\delta=(g,\gamma)\in G^\Gamma - G$ 
with $\delta^2=1$ alias $gg^\gamma=1$ and thus get a bijection
$\op{Z}^1(\Gamma;G)\sira \{\delta\in G^\Gamma - G\mid \delta^2=1\}$ 
mapping a cycle $z:\Gamma\ra G$ to the pair $(z(\gamma),\gamma)$ for the
nontrivial element $\gamma\in \Gamma$. This map will induce a bijection
$$\op{H}^1(\Gamma;G)\sira 
\{\delta\in G^\Gamma - G\mid \delta^2=1\}/(\op{int}G)$$
and thus \cite[1.15]{ABV} will specialize to a bijection 
$$\coprod_{\delta\in \op{H}^1(\Gamma;G)}\Pi(G(\DR,\delta))\sira
\Pi(G/\DR)_1$$
\end{Bemerkungl}