%% Title: Almost-minimal nonuniform lattices of higher rank
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%% Authors: Vladimir Chernousov, Lucy Lifschitz, and Dave Witte Morris
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%% Date of last revision: 3 September 2007
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%% to appear in Michigan Mathematical Journal
%% submitted 28 May 2007




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\begin{document}

\title[Almost-minimal lattices of higher rank]
{Almost-minimal nonuniform lattices of higher rank}

\author[V.\,Chernousov]{Vladimir Chernousov}
\address{Department  of Mathematical and Statistical Sciences,
University of Alberta,
Edmonton,  Alberta, T6G~2G1, Canada}
\email{chernous@math.ualberta.ca}

\author[L.\,Lifschitz]{Lucy Lifschitz}
\address{Department of Mathematics,
University of Oklahoma,
Norman, Oklahoma, 73019, USA}
\email{LLifschitz@math.ou.edu,
http://www.math.ou.edu/$\sim$llifschitz/}

\author[D.\,W.\,Morris]{Dave Witte Morris}
\address{Department of Mathematics and Computer Science,
University of Lethbridge,
Lethbridge, Alberta, T1K~3M4, Canada}
\email{Dave.Morris@uleth.ca,
http://people.uleth.ca/$\sim$dave.morris/}
\thanks{V.\,Chernousov was partially supported by the Canada Research
Chairs Program and an NSERC research grant.
D.\,W.\,Morris was partially supported by research grants from NSERC  
and the NSF}

\date{September 3, 2007} %%???

\maketitle

\section{Introduction}

We find the minimal elements in three different (but essentially  
equivalent) partially ordered categories of
mathematical objects:
\begin{enumerate} \renewcommand{\theenumi}{\Alph{enumi}}
\item \label{category-geom}
finite-volume, noncompact, complete, locally symmetric spaces of  
higher rank,
\item \label{category-Lie}
nonuniform, irreducible lattices in semisimple Lie groups of higher  
real rank,
and
\item \label{category-Qgrp}
isotropic, simple algebraic $\rational$-groups of higher real rank.
\end{enumerate}
The main interest is in categories \pref{category-geom} and~\pref 
{category-Lie}, but the proof is carried out using the machinery of~ 
\pref{category-Qgrp}.
(For completeness, we also provide a generalization that applies to  
algebraic groups over any number field, not only~$\rational$.)
Justification of the examples and facts stated in the introduction  
can be found in \S\ref{justify}.

\subsection{Locally symmetric spaces} \label{SymmspaceSubsec}

It is well known that if $G$ is a connected, semisimple Lie group,  
and $\Rrank G \ge 2$, then $G$ contains a closed subgroup that is  
locally isomorphic to either $\SL_3(\real)$ or $\SL_2(\real) \times  
\SL_2(\real)$. Passing from semisimple Lie groups to the  
corresponding symmetric spaces yields the following geometric  
translation of this observation.

\begin{fact} \label{GlobalFact}
Let $\widetilde{X}$ be a symmetric space of noncompact type, with no  
Euclidean factors, such that $\rank \widetilde{X} \ge 2$. Then $ 
\widetilde{X}$ contains a totally geodesic submanifold~$\widetilde 
{X}'$, such that $\widetilde{X}'$ is the symmetric space associated  
to either $\SL_3(\real)$ or $\SL_2(\real) \times \SL_2(\real)$.
In other words, $\widetilde{X}'$ is isometric to either
\begin{enumerate} \renewcommand{\itemsep}{\medskipamount}
\item \label{GlobalFact-SL3}
$\SL_3(\real)/\SO(3) \iso
\left\{ \begin{matrix}
\mbox{$3 \times 3$ positive-definite symmetric} \\
\mbox{real matrices of determinant~$1$}
\end{matrix}
\right\}$,
or
\item the product $\mathbb{H}^2 \times \mathbb{H}^2$ of $2$  
hyperbolic planes.
\end{enumerate}
\end{fact}

In short, among all the symmetric spaces of noncompact type with rank  
$\ge 2$, there are only two manifolds that are minimal with respect  
to the partial order defined by totally geodesic embeddings. Our main  
theorem provides an analogue of this result for noncompact finite- 
volume spaces that are locally symmetric, rather than globally  
symmetric, but, in this setting, the partial order has infinitely  
many minimal elements.

\begin{thm} \label{LocSymmThm}
Let $X$ be a finite-volume, noncompact, irreducible, complete,  
locally symmetric space of noncompact type, with no Euclidean factors  
{\rm(}locally\/{\rm)}, such that $\rank X \ge 2$. Then there is a  
finite-volume, noncompact, irreducible, complete, locally symmetric  
space~$X'$, such that $X'$ admits a totally geodesic, proper  
immersion into~$X$, and the universal cover of~$X'$ is the symmetric  
space associated to either\/
$\SL_3(\real)$, $\SL_3(\complex)$, or a direct product\/ $\SL_2(\real) 
^m \times \SL_2(\complex)^n$, with $m + n \ge 2$.
\end{thm}

\begin{rem}
The symmetric space associated to $\SL_3(\real)$ is given in~\fullref 
{GlobalFact}{SL3}. The others are:
\begin{enumerate} \renewcommand{\itemsep}{\medskipamount}
\item $\SL_3(\complex)/\SU(3) \iso
\left\{ \begin{matrix}
\mbox{$3 \times 3$ positive-definite Hermitian} \\
\mbox{matrices of determinant~$1$}
\end{matrix}
\right\}$,
and
\item the product $(\mathbb{H}^2)^m \times (\mathbb{H}^3)^n$ of $m 
$~hyperbolic planes and $n$~hyperbolic $3$-spaces.
\end{enumerate}
\end{rem}

\begin{rem} \
\begin{enumerate}
\item Our main result actually provides a precise description of~$X'$  
(modulo finite covers), not only its universal cover. It does this by  
specifying the fundamental group $\pi_1(X')$; the possible  
fundamental groups appear in \S\ref{LatticeSubsec}.
\item Our proof of the theorem is constructive: for a given locally  
symmetric space~$X$, our methods produce an explicit locally  
symmetric space~$X'$ that embeds in~$X$.
\item Our theorem assumes $X$ is \emph{not} compact. It would be  
interesting to obtain an analogous result that assumes $X$ is compact  
(and $X'$ is also compact).
\end{enumerate}
\end{rem}

The Mostow Rigidity Theorem tells us that any locally symmetric space~ 
$X$ as discussed above is determined by its fundamental group. This  
means that the above geometric result can be reformulated in group- 
theoretic terms. This restatement of the result is our next topic.



\subsection{Lattices in semisimple Lie groups}
\label{LatticeSubsec}

\begin{defn}
Let us say that an abstract group~$\Gamma$ is a \emph{nonuniform lattice
of higher rank} if
there exists a connected, semisimple, linear (real) Lie group~$G$,
such that
\begin{itemize}
\item $\Gamma$ is isomorphic to an irreducible, \emph{nonuniform}
lattice in~$G$,
and
\item $\Rrank G \ge 2$.
\end{itemize}
(Recall that a discrete subgroup~$\Gamma'$ of~$G$ is a \emph{nonuniform
lattice} if $G/\Gamma'$ has finite volume, but is \emph{not} compact.  
The
lattice~$\Gamma'$ is \emph{irreducible} if no finite-index subgroup
of~$\Gamma'$ is isomorphic to a direct product $\Gamma_1' \times
\Gamma_2'$ with both $\Gamma_1'$ and~$\Gamma_2'$ infinite.)
\end{defn}

\begin{rem}
The nonuniform lattices of higher rank have made many appearances in  
the literature.
For example, the Margulis Arithmeticity Theorem \cite{[M73]} was  
first proved for this class of groups, and M.\,S.\,Raghunathan \cite 
{[R76],[R86]} proved the Congruence Subgroup Property for these groups.
\end{rem}

It is obvious that the collection of all nonuniform lattices of  
higher rank is
closed under passage to finite-index subgroups, so
it has no elements that are minimal under inclusion. Thus, it is
natural to consider a slightly weaker notion of minimality that
ignores finite-index subgroups.

\begin{defn}
A nonuniform lattice~$\Gamma$ of higher rank is \emph{almost minimal}  
if no
subgroup of infinite index in~$\Gamma$ is a nonuniform lattice of  
higher rank.
\end{defn}
Our main result describes all the almost-minimal nonuniform lattices  
of higher rank.
The significance of this result lies in the fact that every  
nonuniform lattice of
higher rank must contain an almost-minimal one, so, for example, they  
can be the base
cases in a proof by induction.

\begin{thm}
Every almost-minimal nonuniform lattice of higher rank is isomorphic  
to a nonuniform, irreducible lattice in either\/ $\SL_3(\real)$, $ 
\SL_3(\complex)$, or a direct product\/ $\SL_2(\real)^m \times \SL_2 
(\complex)^n$, with $m + n \ge 2$.
\end{thm}

We now describe the almost-minimal lattices more explicitly.

\begin{eg} \label{Qrank2eg}
$\SL_3(\integer)$ is an almost-minimal nonuniform lattice of higher  
rank.
\end{eg}


\begin{rem}
$\SL_3(\integer)$ is an arithmetic group whose $\Qrank$
is~$2$. It is well known that any irreducible lattice~$\Gamma$ with $ 
\Qrank \Gamma
\ge 2$ must contain a finite-index subgroup of either $\SL_3(\integer) 
$ or
$\Sp_4(\integer)$, and one can show that $\Sp_4(\integer)$ is \emph{not}
almost minimal. Therefore, up to finite-index, $\SL_3(\integer)$ is  
the only
almost-minimal lattice of higher rank whose $\Qrank$ is $\ge 2$.
\end{rem}

Although (up to finite index) there is only one almost-minimal  
nonuniform lattice whose
$\Qrank$ is~$2$, there are infinitely many whose $\Qrank$ is~$1$.

\begin{eg} \label{SL2Egs} \
\begin{enumerate} \itemsep=\smallskipamount
\item \label{SL2Egs-quad}
If $r$ is any square-free integer $\ge 2$, then $\SL_2 \bigl(
\integer \bigl[\sqrt{r} \bigr] \bigr)$ is an almost-minimal  
nonuniform lattice of higher rank.
\item \label{SL2Egs-any}
More generally, let $\ints_K$ be the ring of integers of an algebraic
number field~$K$, and assume $K$ is neither $\rational$ nor an imaginary
quadratic extension of~$\rational$. Then $\Gamma = \SL_2(\ints_K)$ is a
nonuniform lattice of higher rank. (We remark that if $K$ is a  
totally real extension
of~$\rational$, as is the case in~\pref{SL2Egs-quad}, then
$\Gamma$ is called a \emph{Hilbert modular group}.) This nonuniform  
lattice is almost minimal
if and only if each proper subfield of~$K$ is either~$\rational$ or an
imaginary quadratic extension of~$\rational$.
\end{enumerate}
\end{eg}

\begin{eg} \label{SU2Egs} \
Let
\begin{itemize}
\item $F$ be either the field~$\rational$ or an imaginary quadratic
extension of~$\rational$,
\item $F_v =
\begin{cases}
\real & \text{if $F = \rational$} , \\
\complex & \text{if $F \not\subset \real$} ,
\end{cases}$
\item $L$ be any quadratic extension of~$F$, such that $L \subset
F_v$,
\item $\tau$ be the nontrivial Galois automorphism of~$L$ over~$F$,
\item $f$ be the $\tau$-Hermitian form on~$L^3$ defined by
$$ f(x,y) = \tau(x_1) \, y_1 - \tau(x_2) \, y_2 - \tau(x_3) \, y_3 ,$$
and
\item $\ints_L$ be the ring of integers of~$L$.
\end{itemize}
Then
$$ \SU_3(\ints_L, f, \tau) = \{\, A \in \SL_3(\ints_L) \mid f(Ax,Ay) =
f(x,y), \ \forall x,y \in L^3 \,\} $$
is a nonuniform lattice in $\SL_3(F_v)$, so it is a nonuniform  
lattice of higher rank.
   It is almost minimal if and only if either $F = \rational$ or $L \cap
\real = \rational$.
\end{eg}

The preceding examples are well known (and are of classical type).  
Our main result shows there are no others:

\begin{thm} \label{LattThm}
Every nonuniform lattice of higher rank contains a subgroup that is  
isomorphic to a finite-index
subgroup of a lattice described in Example\/ \ref{Qrank2eg}, \fullref 
{SL2Egs}{any}, or\/~\ref{SU2Egs}.
\end{thm}

\begin{rem}
Theorem~\ref{LattThm} is a fundamental ingredient in the proof \cite 
{LifschitzMorris-RO} that if all nonuniform lattices of higher rank  
are boundedly generated by unipotent elements, then no nonuniform  
lattice of higher rank can be right ordered.
\end{rem}

The Margulis Arithmeticity Theorem tells us that (modulo finite  
groups) any nonuniform lattice of higher rank can be realized as the  
integral points of a simple algebraic $\rational$-group. Also, the  
Margulis Superrigidity Theorem tells us that any embedding $\Gamma'  
\hookrightarrow \Gamma$ extends to an embedding of the corresponding  
algebraic $\rational$-groups (modulo finite groups). This means that  
the classification of almost-minimal nonuniform lattices of higher  
rank is logically equivalent to a result on simple algebraic $ 
\rational$-groups.



\subsection{Simple algebraic $\rational$-groups} \label{QgroupSubsec}

Let $\aG$ be a connected algebraic group over~$\rational$ that is  
almost simple. (Recall that, by definition, this means every proper,  
normal $\rational$-subgroup of~$\aG$ is finite.) It is well known  
that if $\Qrank \aG \ge 2$, then $\aG$ contains a $\rational$-split  
almost simple subgroup~$\aH$, such that $\Qrank \aH = 2$. (Indeed,  
one may choose $\aH$ to be isogenous to either $\aSL_3$ or $\aSp_4$.)  
If we replace the assumption that $\aG$ has large $\Qrank$ with the  
weaker assumption that $\aG$ has large $\Rrank$, then one cannot  
expect to find a subgroup of large $\Rrank$ that is split over~$ 
\rational$. (In any $\rational$-split subgroup, the $\rational$-rank  
and $\Rrank$ are equal.) However, our main result states that if we  
add the obvious necessary condition that $\aG$ is $\rational$- 
isotropic, then there is always a subgroup of large $\Rrank$ that is  
\emph{quasi}split over~$\rational$:

\begin{thm} \label{MainQuasisplit}
Suppose\/ $\aG$ is an isotropic, almost simple algebraic group over\/~ 
$\rational$, such that\/ $\Rrank \aG \ge 2$. Then\/ $\aG$ has a  
connected, isotropic, almost simple\/
$\rational$-subgroup\/~$\aH$, such that\/ $\aH$ is quasisplit over\/~$ 
\rational$, and\/ $\Rrank \aH \ge 2$.
\end{thm}
It was mentioned above that $\aH$ can be chosen to be isogenous to  
either $\aSL_3$ or $\aSp_4$ if $\Qrank \aG \ge 2$. So the theorem is  
only interesting when $\Qrank \aG = 1$. Because there are very few  
quasisplit groups of $\Qrank$~$1$ (and it is not difficult to find  
quasisplit proper subgroups of $\aSp_4$, as will be seen in the proof  
of Lemma~\ref{FrankMust1}), we can restate the result in the  
following more precise form.
\begin{defn}
Suppose $\aG$ is an isotropic, almost simple algebraic group over\/~$ 
\rational$, such that\/
$\Rrank \aG \ge 2$.
For convenience, let us say that $\aG$ is \emph{minimal} if no  
proper, isotropic, almost simple\/ $\rational$-subgroup of~$\aG$ has  
real rank $\ge 2$.
\end{defn}

\begin{notation}
$\res{K/F}$ denotes the Weil restriction of scalars functor from~$K$  
to~$F$.
\end{notation}

\begin{thm} \label{QgrpExplicitThm}
Suppose\/ $\aG$ is an isotropic, almost simple algebraic group over\/~ 
$\rational$, such that\/
$\Rrank \aG \ge 2$. If\/ $\aG$ is minimal, then\/ $\aG$ is isogenous  
to either:
\begin{enumerate} \renewcommand{\theenumi}{\roman{enumi}}
\item \label{QgrpExplicitThm-SL3}
$\aSL_3$,
or
\item \label{QgrpExplicitThm-SUreal}
$\aSU_3(L,f,\tau)$, where $L$ is a real quadratic extension of\/~$ 
\rational$, $\tau$~is the Galois automorphism of~$L$ over\/~$\rational 
$, and
	\begin{equation} \label{QgrpExplicitThm-f}
	f(x_1,x_2,x_3) = \tau(x_1) \, x_1 - \tau(x_2) \, x_2 - \tau(x_3) \, x_3
	, \end{equation}
or
\item \label{QgrpExplicitThm-SUcplx}
$\res{K/\rational} \aSU_3(L,f,\tau)$, where $K$ is an imaginary  
quadratic extension of\/~$\rational$, $L$~is a quadratic extension of~ 
$K$, $\tau$~is the Galois automorphism of~$L$ over~$K$, and $f$~is  
given by \pref{QgrpExplicitThm-f},
or
\item \label{QgrpExplicitThm-SL2}
$\res{K/\rational} \aSL_2$, for some finite extension~$K$ of\/~$ 
\rational$, such that $K$ is neither\/~$\rational$, nor an imaginary  
quadratic extension of\/~$\rational$.
\end{enumerate}
\end{thm}
\begin{rem} \label{QgrpsAreMin}
Conversely:
\begin{enumerate}
\renewcommand{\theenumi}{\roman{enumi}}
\renewcommand{\itemsep}{\smallskipamount}
\item $\aSL_3$ is minimal.
\item The groups described in \fullref{QgrpExplicitThm}{SUreal} are  
minimal.
\item A group as described in \fullref{QgrpExplicitThm}{SUcplx} fails  
to be minimal if and only if $L$ contains a real quadratic extension  
of~$\rational$.
\item A group as described in \fullref{QgrpExplicitThm}{SL2} fails to  
be minimal if and only if $K$ contains a proper subfield that is  
neither~$\rational$ nor an imaginary quadratic extension of~$\rational$.
\end{enumerate}
\end{rem}

\begin{rem}
Under the additional assumption that some minimal parabolic $\real$- 
subgroup of~$\aG$ is defined over~$\rational$, Theorem~\ref 
{QgrpExplicitThm} was proved long ago by G.\,A.\,Margulis \cite 
[Lem.~2.4.2]{[M75]} and M.\,S.\,Raghunathan \cite[Lem.~3.2(ii)] 
{[R86]} (independently).
\end{rem}

\begin{rem}
Theorem~\ref{MainForF} provides a generalization of Theorem~\ref 
{QgrpExplicitThm} that applies to algebraic groups over any algebraic  
number field.
\end{rem}



\subsection*{Outline of the paper}
Section~\ref{justify} justifies statements made in the above  
introduction.
The remaining sections of the paper state and prove our main result  
(Theorem~\ref{MainForF}).
Section~\ref{PrelimSect} covers some preliminaries, ands deals with  
groups that either have global rank $\ge 2$ or are of type $E_7$, $E_8 
$, or~$G_2$.
We treat
groups of classical type in \S\ref{ClassicalSect},
groups of type~$F_4$ in \S\ref{F4Sect},
groups of type  $\out{3,6}D4$ in~\S\ref{TrialitySect},
and
groups of type  $\out{1,2}E6$ in~\S\ref{E6Sect}.







\section{Justification of the introduction} \label{justify}

In this section, we provide brief justifications for the assertions  
made in the introduction. The order of the topics there was chosen  
for purposes of exposition; they will now be treated in reverse order  
(\S\ref{QgroupSubsec}, \S\ref{LatticeSubsec}, \S\ref{SymmspaceSubsec}).

\subsection*{Justification of \S\ref{QgroupSubsec}}
The observation that $\rational$-simple groups of higher $\rational$- 
rank contain subgroups isogenous to either $\aSL_3$ or~$\aSp_4$  
appears in \cite[Prop.~I.1.6.2, p.~46]{MargulisBook}.

The following sections will present a proof of (a generalization of)  
Theorem~\ref{QgrpExplicitThm}. Because all of the groups in the  
conclusion of \pref{QgrpExplicitThm} are quasisplit, Theorem~\ref 
{MainQuasisplit} is an immediate consequence.

To verify the observations in Remark~\ref{QgrpsAreMin}, note that:
\begin{itemize} \renewcommand{\itemsep}{\smallskipamount}

\item The groups described in \fullref{QgrpExplicitThm}{SL3} and~ 
\fullref{QgrpExplicitThm}{SUreal} are isomorphic to $\aSL_3$ over the  
algebraic closure~$\overline{\rational}$. Since $\aSL_3$ has no  
semisimple, proper subgroups of absolute rank $\ge 2$, it is  
immediate that these groups are minimal.

\item Let $\aG$ be one of the groups described in  \fullref 
{QgrpExplicitThm}{SUcplx}.
If $L$ contains a real quadratic extension~$F$ of~$\rational$, then $ 
\aG$ contains $\aSU_3(F,f,\tau|_F)$, so $\aG$ is not minimal.

Conversely, if $\aG$ is not minimal, then there is an isotropic,  
almost simple, proper $\rational$-subgroup~$\aH$ of~$\aG$, such that $ 
\Rrank \aH \ge 2$. Since $\aG$ is isomorphic to $\aSL_3 \times \aSL_3 
$ over~$\overline{\rational}$, we know that $\aH$ must be isogenous  
to either $\aSL_3$ or $\aSL_2 \times \aSL_2$ over~$\overline 
{\rational}$. In either case, because $\Rrank \aH \ge 2$, there is a  
real quadratic extension~$F$ of~$\rational$, such that $\fieldrank{F}  
\aH = 2$. Therefore $\fieldrank{F} \aG \ge 2$.
If $F \not\subset L$, then $\tau$ extends to an automorphism of $L  
\cdot F$ that is trivial on $K \cdot F$, and $\aG$ is $F$-isomorphic to
	$$\res{K \cdot F/F} \aSU_3(L \cdot F, f, \tau) .$$
So $\fieldrank{F} \aG = 1$. This is a contradiction, so we conclude  
that $L$ does contain the real quadratic extension~$F$.

\item Let $\aG$ be one of the groups described in  \fullref 
{QgrpExplicitThm}{SL2}. Any $\rational$-subgroup of~$\aG$ that is  
almost simple is isogenous to $\res{L/\rational} \aSL_2$, for some  
subfield~$L$ of~$K$. Thus, $\aG$ fails to be minimal if and only if $ 
\Rrank \bigl(  \res{L/\rational} \aSL_2 \bigr) > 1$ for some proper  
subfield~$L$ of~$K$.
\end{itemize}

\subsection*{Justification of \S\ref{LatticeSubsec}}
The Margulis Arithmeticity Theorem \cite[Thm.~IX.1.16, p.~299, and  
Rem.~IX.1.6(iii), p.~294]{MargulisBook} states that (up to  
isomorphism of finite-index subgroups) the collection of nonuniform  
lattices of higher rank is the same as
	$$ \{\, \aG(\integer) \mid \text{$\aG$ is an isotropic almost simple  
$\rational$-group with $\Rrank \aG \ge 2$} \,\} .$$
For isotropic almost simple $\rational$-groups $\aG$ and~$\aG_1$ with  
$\Rrank \aG_1 \ge 2$,
the Margulis Superrigidity Theorem \cite[Thm.~IX.5.12(ii), p.~327,  
and Rem.~IX.1.6(iv), p.~295]{MargulisBook} implies there is a finite- 
index subgroup of $\aG_1(\integer)$ that is isomorphic to a subgroup  
of $\aG(\integer)$ if and only if $\aG_1$ is isogenous to a subgroup  
of $\aG$. Hence, $\aG(\integer)$ is almost minimal (as a nonuniform  
lattice of higher rank) if and only if $\aG$ is minimal (as an  
algebraic $\rational$-group). Therefore, all the assertions of \S\ref 
{LatticeSubsec} are simply translations of results in \S\ref 
{QgroupSubsec}. For example, because $\aSL_3$ is minimal, it is  
immediate that $\SL_3(\integer)$ is almost minimal.

\subsection*{Justification of \S\ref{SymmspaceSubsec}}
Let $X$ be as in Theorem~\ref{LocSymmThm}. It is well known (cf.\  
\cite[\S2.2, pp.~70--71]{EberleinBook} and \cite[Thm.~5.6, p.~222] 
{HelgasonBook}) that, up to isometry, we have $X = \Gamma \backslash  
G/K$, where
\begin{itemize}
\item $G$ is a connected, semisimple, adjoint Lie group with no  
compact factors,
\item $K$ is a maximal compact subgroup of~$G$,
and
\item $\Gamma$ is a (torsion-free) nonuniform, irreducible lattice in~ 
$G$.
\end{itemize}
We have $\Rrank G = \rank X$ (cf.\ \cite[\S2.7, pp.~76--77] 
{EberleinBook}), so, since $\rank X \ge 2$, we see that $\Gamma$ is a  
nonuniform lattice of higher rank. Hence, Theorem~\ref{LattThm}  
implies that $\Gamma$ contains a subgroup~$\Gamma'$ that is  
isomorphic to a nonuniform, irreducible lattice in a connected,  
semisimple, adjoint Lie group~$H$, and $H$ is locally isomorphic to  
either $\SL_3(\real)$, $\SL_3(\complex)$, or a product $\SL_2(\real) 
^m \times \SL_2(\complex)$, with $m + n \ge 2$. The Margulis  
Superrigidity Theorem \cite[Thm.~IX.5.12, p.~327]{MargulisBook}  
implies that (after passing to a finite-index subgroup of~$\Gamma'$),  
the inclusion $\Gamma' \hookrightarrow \Gamma$ extends to an  
embedding $H \hookrightarrow G$, so we may assume $H \subset G$ and $ 
\Gamma' = H \cap \Gamma$. We may choose a Cartan involution~$\sigma$  
of~$G$, such that $\sigma(H) = H$ \cite[Thm.~7.3]{MostowSelfAdjoint}.  
Let
	\begin{itemize}
	\item $g_0 \in G$, such that $g_0 K g_0^{-1}$ is the maximal compact  
subgroup of~$G$ on which $\sigma$ is trivial \cite[Thm.~2.2(i),  
p.~256]{HelgasonBook},
	\item $K' = (g_0 K g_0^{-1}) \cap H$, so $K'$ is a maximal compact  
subgroup of~$H$,
	and
	\item $X' = \Gamma' \backslash H / K'$.
	\end{itemize}
Then $X'$ is a a finite-volume, noncompact, irreducible locally  
symmetric space whose universal cover is $H/K'$. The immersion
	$$ X' \to X \colon \Gamma' h K' \mapsto \Gamma h g_0 K $$
is proper \cite[Thm.~1.13, p.~23]{RaghunathanBook} and has totally  
geodesic image
\cite[Prop~2.6.2, p.~74]{EberleinBook}.





\section{Preliminaries} \label{PrelimSect}

Throughout the remainder of this paper, $\aG$ is a connected,  
isotropic, almost simple algebraic group over an algebraic number  
field~$F$.

\begin{rem}
Our notation and terminology for discussing algebraic groups  
generally follows \cite{[PR]}.
However, we use boldface letters ($\aG$, $\aH$, $\aT$, etc.) to  
denote algebraic groups. Also, if $A$ is a central simple algebra,  
and $f$ is a Hermitian (or skew-Hermitian) form on~$A^m$, with  
respect to an involution~$\tau$, we use $\aSU_m(A,f,\tau)$ to denote  
the corresponding special unitary group, whereas \cite{[PR]} writes  
merely $\aSU_m(f)$.
\end{rem}

\begin{notation}
Let $\SG$ be the set of all archimedean places~$v$ of~$F$, such that $ 
\locrank \aG \ge 2$.
\end{notation}

\begin{defn}
We say $\aG$ is \emph{minimal} if $\SG \neq \emptyset$, and there  
does not exist a proper, isotropic, almost simple $F$-subgroup~$\aH$  
of~$\aG$, such that $\locrank \aH \ge 2$ for every $v \in \SG$.
\end{defn}

The following is a natural generalization of Theorem~\ref 
{QgrpExplicitThm}.

\begin{thm} \label{MainForF}
Suppose\/ $\aG$ is an isotropic, almost simple algebraic group over  
an algebraic number field~$F$, such that $\SG \neq \emptyset$. If\/ $ 
\aG$ is minimal, then\/ $\aG$ is isogenous to either:
\begin{enumerate} \renewcommand{\theenumi}{\roman{enumi}}
\item \label{MainForF-SL3}
$\aSL_3$,
or
\item \label{MainForF-SU}
$\aSU_3(L,f,\tau)$, where
\begin{itemize}
\item $L$ is a quadratic extension of~$F$, such that $L \subset F_v$,  
for some archimedean place~$v$ of~$F$,
\item $\tau$~is the Galois automorphism of~$L$ over~$F$,
and
\item $f(x_1,x_2,x_3) = \tau(x_1) \, x_1 - \tau(x_2) \, x_2 - \tau 
(x_3) \, x_3 $,
\end{itemize}
or
\item \label{MainForF-SUres}
$\res{K/F} \aSU_3(L,f,\tau)$, where
\begin{itemize}
\item $K$ is a quadratic extension of~$F$, such that $K \not\subset  
F_v$, for some archimedean place~$v$ of~$F$,
\item $L$~is a quadratic extension of~$K$,
\item $\tau$~is the Galois automorphism of~$L$ over~$K$,
and
\item $f(x_1,x_2,x_3) = \tau(x_1) \, x_1 - \tau(x_2) \, x_2 - \tau 
(x_3) \, x_3 $,
\end{itemize}
or
\item \label{MainForF-SL2}
$\res{K/F} \aSL_2$, for some nontrivial finite extension~$K$ of~$F$,  
such that either $|K:F| > 2$, or $K \subset F_v$, for some  
archimedean place~$v$ of~$F$.
\end{enumerate}
\end{thm}

\begin{cor}
Suppose\/ $\aG$ is an isotropic, almost simple algebraic group over  
an algebraic number field~$F$, such that $\SG \neq \emptyset$. Then $ 
\aG$ contains an isotropic, almost simple $F$-subgroup~$\aH$, such  
that $\locrank \aH \ge 2$ for every $v \in \SG$, and $\aH$ is  
isogenous to a subgroup described in \pref{MainForF-SL3}, \pref 
{MainForF-SU}, \pref{MainForF-SUres}, or~\pref{MainForF-SL2} of  
Theorem~\ref{MainForF}.
\end{cor}

The remainder of this paper provides a proof of Theorem~\ref{MainForF}.

\begin{notation}
For algebraic groups~$\aG_1$ and~$\aG_2$ over a field~$K$, we write $ 
\aG_1 \approx \aG_2$ if they have the same simply connected covering.
\end{notation}

Let us record an observation that will be used repeatedly.

\begin{lem} \label{SO=SL_2(F[a])}
If $a \in F^*$ and $\sqrt{a} \notin F$, then
	$$\aSO_4(x_1^2 - x_2^2 - x_3^2 + a x_4^2) \approx \res{F[\sqrt{a}]/ 
F} \aSL_2 .$$
\end{lem}

\begin{proof}
$\aSO_4$ is of type $D_2 = A_1\times A_1$. Since the
discriminant of the quadratic form under consideration is not a
square, we know that the associated orthogonal group is an outer
form. Thus, it is isogenous to $\res{F[\sqrt{a}]/F} \aSL_1(A)$, where
$A$ is a quaternion
algebra over $F[\sqrt{a}]$. Since the group is isotropic over $F$,
the algebra $A$ must be split, so $\aSL_1(A) \iso \aSL_2$.
\end{proof}


Recall that a connected algebraic $F$-group is \emph{absolutely  
almost simple} if it remains simple over an algebraic closure~$ 
\overline{F}$ of~$F$. The following basic observations allow us to  
assume that $\Frank \aG = 1$, and that $\aG$ is absolutely almost  
simple.

\begin{lem} \label{FrankMust1}
If\/ $\aG$ is minimal, then either:
\begin{enumerate}
\item \label{FrankMust1-Frank1}
$\Frank \aG = 1$,
or
\item \label{FrankMust1-SL3}
$\aG$ is isogenous to $\aSL_3$ {\rm(}so \fullref{MainForF}{SL3} holds 
\/{\rm)}.
\end{enumerate}
\end{lem}

\begin{proof}
Assume $\Frank \aG \ge 2$. It is well known that $\aG$ contains an $F 
$-subgroup that is isogenous to either $\aSL_3$ or $\aSp_4$ \cite 
[Prop.~I.1.6.2, p.~46]{MargulisBook}. By minimality, $\aG$ itself  
must be isogenous to either $\aSL_3$ or $\aSp_4$.

Suppose $\aG$ is isogenous to $\aSp_4$. Then $\aG$ is a split group  
of type $C_2 = B_2$, so it is also isogenous to
	$$ \aSO_5( x_1^2 - x_2^2 - x_3^2 + x_4^2 + a x_5^2) ,$$
for any $a \in F$.
It therefore contains a subgroup isogenous to
	$$ \aSO_4( x_1^2 - x_2^2 - x_3^2 + a x_5^2) .$$
By Weak Approximation, we may choose $a$ so that $a$ is a square in~ 
$F_v$, for every $v \in \SG$, but $a$ is not a square in~$F$. Then $ 
\aH$ is isogenous to $\res{F[\sqrt{a}]/F} \aSL_2$ \see{SO=SL_2(F 
[a])}, so it is isotropic and $\locrank \aH  = 2$ for every $v \in \SG 
$. This contradicts the minimality of~$\aG$.
\end{proof}


\begin{lem} \label{AbsSimple}
If\/ $\aG$ is minimal, then either:
\begin{enumerate}
\item \label{AbsSimple-SL}
$\aG$ is isogenous to $\res{K/F} \aSL_2$, with $K$ as described in  
Theorem~\fullref{MainForF}{SL2},
or
\item \label{AbsSimple-Simple}
$\aG$ is absolutely almost simple,
or
\item \label{AbsSimple-Imag}
$\aG$ is isogenous to $\res{K/F}\aG_0$, where $\aG_0$ is an  
absolutely almost simple group over a quadratic extension~$K$ of~$F$,  
such that $K \not\subset F_v$, for some $v \in \SG$.
\end{enumerate}
\end{lem}

\begin{proof}
Assume \pref{AbsSimple-Simple} does not hold. Then there is an  
algebraic number field $K \supset F$, and an absolutely almost simple  
group~$\aG_0$ over~$K$, such that $\aG$ is isogenous to $\res{K/F}  
\aG_0$ \cite[Thm.~26.8, p.~365]{BookInvols}. Since $\aG$ is isotropic  
over~$F$, we know $\aG_0$ is isotropic over~$K$, so $\aG_0$ contains  
a subgroup that is isogenous to $\aSL_2$. Therefore, $\aG$ contains a  
subgroup~$\aH$ that is isogenous to $\res{K/F} \aSL_2$.

If $\locrank \aH \ge 2$, for every $v \in \SG$, then the minimality  
of~$\aG$ implies $\aG = \aH$, so \pref{AbsSimple-SL} holds. On the  
other hand, if $\locrank \aH = 1$, for some $v \in \SG$, then $K$ is  
a quadratic extension of~$F$, and $K \not\subset F_v$, so \pref 
{AbsSimple-Imag} holds.
\end{proof}

\begin{lem} \label{CanAbsSimple}
If Theorem~\ref{MainForF} holds\/ {\rm(}for all algebraic number  
fields\/{\rm)} under the additional assumption that\/ $\aG$ is  
absolutely almost simple, then it holds in general.
\end{lem}

\begin{proof}
Suppose $\aG$ is minimal, but is not absolutely almost simple. From  
Lemma~\ref{FrankMust1}, we see that $\Frank \aG = 1$. We may assume  
\fullref{AbsSimple}{Imag} holds (for otherwise \fullref{MainForF} 
{SL2} holds). Since $\aG$ is minimal (as an $F$-group), it is clear  
that $\aG_0$ is minimal (as a $K$-group). The only absolutely almost  
simple group of global rank~$1$ in the conclusion of Theorem~\ref 
{MainForF} is in \fullref{MainForF}{SU}. Thus, we conclude that $\aG_0 
$ is as described in \fullref{MainForF}{SU}, but with $F$ replaced by~ 
$K$. Then $\aG = \res{K/F} \aG_0$ is as described in \fullref 
{MainForF}{SUres}.
\end{proof}

Lemma~\ref{FrankMust1} immediately rules out some types of  
exceptional groups:

\begin{cor} \label{MostExceptCor}
If\/ $\aG$ is minimal, then\/ $\aG$ is not of type $E_7$, $E_8$, or~ 
$G_2$.
\end{cor}

\begin{proof}
The Tits Classification \cite[pp.~59--61]{[T66]} shows there are no  
rank~1 forms of any of these types over a number field.
\end{proof}

The following useful observation is well known, and easy to prove.

\begin{lem} \label{NormalizeHerm}
Let
\begin{itemize}
\item $D$ be a quaternion algebra over a field~$L$,
\item $\tau$ be an involution of~$D$ {\rm(}of either the first or  
second kind{\rm)},
\item $f(x,y) = \tau(x_1) \, a_1 \, y_1 + \tau(x_2) \, a_2 \, y_2 +  
\cdots + \tau(x_n) \, a_n \, y_n$ be a nondegenerate $\tau$-Hermitian  
form on~$D^n$, for some~$n$,
\item $d \in D$, such that $\tau(d) = d$,
\item $\tau' = \int(d) \circ \tau$, where $\int(d)$ is the inner  
conjugation in~$D$ by~$d$,
and
\item $f'(x,y) = d \, f(x,y) = \tau'(x_1) \, d a_1 \, y_1 +  
\tau'(x_2) \, d a_2 \, y_2 + \cdots + \tau'(x_n) \, d a_n \, y_n$.
\end{itemize}
Then:
\begin{enumerate}
\item $\tau'$ is an involution {\rm(}of the same kind as~$\tau${\rm)},
\item $f'$ is $\tau'$-Hermitian,
and
\item $\aSU_n(D, f', \tau') = \aSU_n(D, f, \tau)$.
\end{enumerate}
\end{lem}

\begin{defn}[{\cite[\S2.2, p.~69]{[T66]}}]
Recall that if $\aS$ is a maximal $F$-split torus in~$\aG$,
then the semisimple part of the centralizer $\aC_\aG(\aS)$ is called the
\emph{semisimple $F$-anisotropic kernel} of~$\aG$. It is unique up
to $F$-isomorphism.
\end{defn}

\begin{defn}
A connected, semisimple subgroup~$\aH_0$ of~$\aG$ is
\emph{standard} if $\aH_0$ is normalized by a maximal torus~$\aT$
of~$\aG$. (We remark that neither~$\aH_0$ nor~$\aT$ is assumed
to be defined over~$F$.) Equivalently, there exist
roots $\beta_1,\ldots,\beta_r$ of~$\aG$ (with respect to~$\aT$),
such that $\aH_0$ is generated by the root subgroups
$U_{\pm \beta_1}, \ldots, U_{\pm \beta_r}$.
For short, we may say that $\aH_0$ is \emph{generated by the
roots $\pm\beta_1,\ldots,\pm\beta_r$}.
\end{defn}

The following useful observation is well known (cf.\ \cite[pp.~353] 
{[PR]}).

\begin{prop} \label{Malpha}
Let
\begin{itemize}
\item $\aM$ be an anisotropic, semisimple group over~$F$, such that  
$-1$ is in the Weyl group of\/~$\aM$,
\item $L$ be a quadratic extension of~$F$, such that\/ $\aM$ is  
quasisplit over~$L$,
and
\item $\alpha$ be a simple root of\/~$\aM$ that is fixed in the $*$- 
action shown in the Tits index of\/~$\aM$.
\end{itemize}
Then there is a maximal $F$-torus\/ $\aT$ of\/~$\aM$, such that the  
standard subgroup\/~$\aM_{\alpha}$ generated by the roots $\pm \alpha 
$ is defined over~$F$.

Furthermore, if\/ $\aM$ is split over~$L$, then\/ $\aT$ may be chosen  
to be split over~$L$.
\end{prop}

\begin{proof}
Letting $\sigma$ be
the Galois automorphism of~$L$ over~$F$, there is a Borel
$L$-subgroup~$\aB$ of~$\aM$, such that $\aT = \aB \cap \sigma(\aB)$ is a
maximal torus of~$\aM$ \cite[Lem.~6.17, p.~329]{[PR]}.
The Borel subgroup~$\aB$ determines an ordering of the roots of~$\aM$
(with respect to~$\aT$). Note that the negative roots are precisely  
those
that appear in $\sigma(\aB)$.
Let $K$ be a Galois splitting field of~$\aM$ that contains~$L$.
Since $\aT$ is defined over~$F$, the Galois group $\Gal(K/F)$
permutes the root spaces of~$\aM$. Furthermore, for any $\tau \in \Gal 
(K/F)$,
either $\tau$ sends every positive root to a positive root (if $\tau 
(\aB) = \aB$),
or $\tau$ sends every positive root to a negative root (if $\tau(\aB)  
= \sigma(\aB)$).
Since $\alpha$ is fixed in the
$*$-action shown in the Tits index, and $-1$~belongs to the
Weyl group, %\cite[VI.4.8(X), p.~221]{[B68]}
this implies that $\tau(\alpha) = \pm \alpha$.
Therefore, $\aM_{\alpha}$  is stable
under $\Gal(K/F)$; thus, it is defined over~$F$.
\end{proof}




\section{Groups of classical type} \label{ClassicalSect}

\begin{assump}
We assume in this section that $\aG$ is a group of classical type,  
and that $\aG$ is minimal. Furthermore, with Lemmas~\ref{FrankMust1}  
and~\ref{CanAbsSimple} in mind, we assume that $\Frank \aG = 1$ and  
that $\aG$ is absolutely almost simple.
\end{assump}

We know $\aG \neq \aSp_n$ (because $\aSp_n$ is $F$-split, but $\Frank  
\aG = 1 < \locrank \aG$ for any $v \in \SG$).
Thus, $\aG$ is either a special linear group, an orthogonal group, or  
a unitary group of either the first or second kind \cite[\S2.3.4]{[PR]}.
We consider each of these possibilities separately.



\subsection{Special Linear Groups}
\label{SLSect}

\begin{assumps} \
\begin{itemize}
\item $D$ is a central division algebra over~$F$,
and
\item $\aG = \aSL_2(D)$.
\end{itemize}
\end{assumps}

Let $K$ be a maximal subfield of~$D$. For $v \in \SG$, we have
$$\locrank \aG > 1 = \locrank \aSL_2 .$$
Therefore $D \neq F$, so $K$ is a proper extension of~$F$.  Because $ 
\res{K/F} \aSL_2 \subseteq \aG$, the minimality of~$\aG$ implies  
there exists $w \in \SG$, such that $\fieldrank{F_w} ( \res{K/F}  
\aSL_2 ) = 1$. Therefore $|K:F| = 2$, so $D$ is a quaternion algebra  
over~$F$.

Write $D = (a,b)_F$.
By Weak Approximation, there exist $c_1,c_2,c_3 \in F$, such that
	\begin{equation} \label{SLSect-allpos}
	 \text{for every $v \in \SG$, $a c_1^2 + b c_2^2 -ab c_3^2$ is a  
nonzero square in~$F_v$.}
	 \end{equation}
Let $c = a c_1^2 + b c_2^2 - ab c_3^2 \in F$, so $c$ has a square  
root in~$D$.
Thus, letting $\aH = \res{F[ \sqrt{c}]/F} \aSL_2$, we have
$\aH \subseteq \aG$.
Also, for every $v \in \SG$, we know $c$ is a square in~$F_v$ \see 
{SLSect-allpos}, so $\locrank \aH \ge 2$.
This contradicts the minimality of~$\aG$.


\subsection{Orthogonal groups}
\label{orthogonal}

\begin{assumps} \
\begin{itemize}
\item $f$ is a nondegenerate quadratic form on~$F^n$, for some~$n \ge  
5$,
\item $\aG = \aSO_n(f)$,
and
\item the maximal totally isotropic $F$-subspace of~$F^n$ is $1$- 
dimensional
(in other words, $\Frank \aG = 1$).
\end{itemize}
\end{assumps}

After a change of basis, to diagonalize the form, we may write
$$ f(x) = x_1^2 - x_2^2 + a_3 x_3^2 + a_4 x_4^2
   + \cdots + a_n x_n^2  .$$
(We may assume the form begins with $x_1^2 - x_2^2$, because it is  
isotropic.)
By normalizing the form, we may assume $a_3 = -1$.
By Weak Approximation, there exist $b_4,b_5,\ldots,b_n \in F$, such that
	\begin{equation} \label{orthogonal-allpos}
	 \text{for every $v \in \SG$, $a_4 b_4^2 + \cdots + a_n b_n^2$ is a  
nonzero square in~$F_v$.}
	 \end{equation}
Let $a = a_4 b_4^2 + \cdots + a_n b_n^2$, so,
after a change of basis, we may assume $a_4 = a$. Then
	$$ \aH =  \aSO_4(x_1^2 - x_2^2 - x_3^2 + a x_4^2) \subset \aG. $$
For any $v \in \SG$, we know
	$\locrank \aH = 2$
(since $a$ is a square in~$F_v$). Hence, the minimality of~$\aG$ implies
	$$\aG = \aH \approx \res{F[\sqrt{a}]/F} \aSL_2 . $$
So \fullref{MainForF}{SL2} holds.




\subsection{Unitary groups of the second kind}
\label{2ndKindSect}

\begin{assumps} \
\begin{itemize}
\item $L$ is a quadratic extension of~$F$,
\item $D$ is a central division algebra over~$L$,
\item $\tau$ is an involution of~$D$ that fixes every element of~$F$,  
but fixes no other elements of~$L$,
\item $f$ is a $\tau$-Hermitian form on $D^n$, for some~$n$,
\item $\aG = \aSU_n(D,f,\tau)$,
and
\item the maximal totally isotropic $D$-subspace of~$D^n$
is $1$-dimensional (in other words, $\Frank \aG = 1$).
\end{itemize}
\end{assumps}

After a change of basis, to diagonalize the form, we may write
$$ f(x,y) = x_1^\tau y_1 - x_2^\tau y_2 + x_3^\tau a_3 y_3 + x_4^\tau
a_4 y_4 + \cdots + x_n^\tau a_n y_n  ,$$
where $a_j^\tau = a_j$ for each~$j$.
(We may assume the form begins with $x_1^\tau y_1 - x_2^\tau y_2$,  
because it is isotropic.)

\setcounter{case}{0}

\begin{case}
Assume $D = L$.
\end{case}
By normalizing the form $f(x,y)$, we may assume $a_3 = -1$.
We may also assume $n \ge 4$; otherwise \fullref{MainForF}{SU} holds.

For each $v \in \SG$:
\begin{itemize}
\item let $L_v = F_v \otimes_F L$,
\item identify $F_v$ with $F_v \otimes_F F \subset L_v$,
and
\item let $\tau_v$ be the extension of~$\tau$ to an involution of~$L_v 
$ with fixed field~$F_v$.
\end{itemize}
\begin{claim}
For every $v \in \SG$, there exist
$b_{v,4},b_{v,5},\ldots,b_{v,n} \in L_v$, such that
$$ \text{$a_4 \, b_{v,4}^{\tau_v} \, b_{v,4} + a_5 \, b_{v,5}^ 
{\tau_v} \, b_{v,5} + \cdots + a_n \, b_{v,n}^{\tau_v} \, b_{v,n}$ is  
a nonzero square in~$F_v$.} $$
\end{claim}
We consider two possibilities:
\begin{itemize}
\item If $L \not\subset F_v$, then $L_v$ is a field extension of~$F_v 
$, and $\aG$ is isomorphic over~$F_v$ to
$$\aSU_n(L_v, x_1^{\tau_v} y_1 - x_2^{\tau_v} y_2 - x_3^{\tau_v} y_3
+ a_4 x_4^{\tau_v} y_4 + a_5 x_5^{\tau_v} y_5 + \cdots + a_n x_n^ 
{\tau_v} y_n, \tau_v) .$$
The desired conclusion follows from the fact that $\locrank \aG \ge 2$.
\item If $L \subset F_v$, then there is an isomorphism $\varphi_v  
\colon (L_v,\tau_v) \to (F_v \oplus F_v, \overline{\tau})$, where $ 
\overline{\tau}(x_1,x_2) = (x_2,x_1)$. Since
	$ (x,1)^{\overline{\tau}} \, (x,1) = (x,x)$ is an arbitrary element  
of~$\varphi_v(F_v)$,
the desired conclusion is obvious.
\end{itemize}
This completes the proof of the claim.

\medskip

Combining the above claim with Weak Approximation yields
$b_4,b_5,\ldots,b_n \in L$, such that, for every $v \in \SG$,
$$ \text{$a_4 \, b_4^{\tau_v} \, b_4 + a_5 \, b_5^{\tau_v} \, b_5 +  
\cdots + a_n \, b_n^{\tau_v} \, b_n$ is a nonzero square in~$F_v$.} $$
Let $a = a_4 \, b_4^{\tau_v} \, b_4 + a_5 \, b_5^{\tau_v} \, b_5 +  
\cdots + a_n \, b_n^{\tau_v} \, b_n$, so, after a change of basis, we  
may assume $a_4 = a$. Then
	$$ \aH = \aSO_4(x_1^1 - x_2^2 - x_3^2 + a x_4^2) \subset \aG .$$
 From the choice of~$a$, we know $\locrank \aH = 2$ for every $v \in  
\SG$.
(Also, since $\Frank \aG = 1$, we know that $a = a_4$ is not a square  
in~$F$, so
	$ \aH \approx \res{F[\sqrt{a}]/F} \aSL_2 $
is almost simple.) This contradicts the minimality of~$\aG$.


\begin{case} \label{2ndKindSect-DnotL}
Assume $D \neq L$.
\end{case}
Choose a maximal subfield~$K$ of~$D$, such that $K$ is invariant
under~$\tau$, and let $K_0$ be the fixed field of~$\tau$ in~$K$. Then
	\begin{equation} \label{2ndKindSect-DnotL-GhasSL2}
\aG
\supset \res{K_0/F} \aSU_2(K,x_1^\tau y_1 - x_2^\tau y_2, \tau|_K)
% \iso \res{K_0/F} \aSU_2(K,x_1^\tau y_1 - x_2^\tau y_2, \tau)
\approx \res{K_0/F} \aSL_2
	 . \end{equation}
Thus, we may assume $K_0$ is a quadratic extension of~$F$,
for otherwise minimality implies \fullref{MainForF}{SL2} holds.
Then, since $|L:F| = 2 = |K:K_0|$, we have
	$$ 2 = |K_0:F|
	= \frac{ |K:L| \cdot |L : F|}{|K:K_0|}
	= |K:L| ,$$
so $D$ is a quaternion algebra.

There is a quaternion algebra $D'$ over~$F$, such that $D = D'  
\otimes_F L$, and $\tau|_{D'}$ is the canonical involution \cite 
[Thm.~11.2(ii), p.~314]{Scharlau}.

\begin{subcase}
Assume $n = 2$.
\end{subcase}
For every $v \in \SG$, we know $F_v$ splits~$D$ (because $n = 2$ and $ 
\locrank \aG \ge 2$).
Therefore, by Weak Approximation and the Hasse Principle, there is a  
quadratic extension~$E$ of~$F$, such that $E$ splits~$D'$ and, for  
each $v \in \SG$,
	\begin{equation} \label{2ndKindSect-DnotL-EinFv<>LinFv}
	E \subset F_v \quad \iff \quad L \subset F_v
	. \end{equation}
Then $D$ splits over $E \cdot L$, so we may assume $K = E \cdot L$.
Since $\tau$ is nontrivial on both $E$ and~$L$, we see from \pref 
{2ndKindSect-DnotL-EinFv<>LinFv} that the fixed field
$K_0$ of~$\tau$ is contained in~$F_v$, for every $v \in \SG$. So the  
minimality of~$\aG$ (together with \pref{2ndKindSect-DnotL-GhasSL2})  
implies \fullref{MainForF}{SL2} holds.

\begin{subcase}
Assume $n \ge 3$.
\end{subcase}
By replacing $\tau$ with $\int(a_3^{-1}) \circ \tau$, we may assume  
$a_3 = 1$ \cf{NormalizeHerm}.
By Weak Approximation and the Hasse Principle, there is a quadratic  
extension~$E$ of~$F$, such that $E$ splits~$D'$ and, for each $v \in  
\SG$,
	\begin{equation} \label{2ndKindSect-DnotL-LnotinFv->EnotinFv}
	L \not\subset F_v \quad\implies\quad E \not\subset F_v
	. \end{equation}
Then $D$ splits over $E \cdot L$, so we may assume $K = E \cdot L$.

Let $\aH_0 = \aSU_3(K, x_1^\tau y_1 - x_2^\tau y_2 + x_3^\tau y_3 ,  
\tau|_K)$ and $\aH = \res{K_0/F} \aH_0 \subset \aG$. For any $v \in  
\SG$:
\begin{itemize}
\item If $K_0 \subset F_v$, then $K_0 \otimes_F F_v \iso F_v \oplus  
F_v$. Therefore, it is clear that $\locrank \aH \ge 2$ (since $\aH_0$  
is isotropic).
\item If $K_0 \not\subset F_v$, then, from \pref{2ndKindSect-DnotL- 
LnotinFv->EnotinFv} and the fact that $\tau$ is nontrivial on both $E 
$ and~$L$, we see that $L \subset F_v$. Therefore $\aH_0$ is inner  
(hence, split) over the field $K_0 \otimes_F F_v$, so $\locrank \aH  
\ge 2$.
\end{itemize}
This contradicts the minimality of~$\aG$.



\subsection{Unitary groups of the first kind}

\begin{assumps} \
\begin{itemize}
\item $D$ is a quaternion algebra over~$F$,
\item $\tau$ is the canonical involution of~$D$,
\item $f$ is a $\tau$-Hermitian or $\tau$-skew Hermitian form on $D^n$,
for some~$n$,
\item $\aG = \aSU_n(D, f, \tau)$,
and
\item the maximal totally isotropic $D$-subspace of~$D^n$
is $1$-dimensional (in other words, $\Frank \aG = 1$).
\end{itemize}
\end{assumps}

After a change of basis, to diagonalize the form, we may write
$$ f(x,y) =
\begin{cases}
x_1^\tau y_1 - x_2^\tau y_2 + x_3^\tau a_3 y_3 + x_4^\tau
a_4 y_4 + \cdots + x_n^\tau a_n y_n
& \text{if $f$ is Hermitian}, \\
x_1^\tau y_2 - x_2^\tau y_1 + x_3^\tau a_3 y_3 + x_4^\tau
a_4 y_4 + \cdots + x_n^\tau a_n y_n
& \text{if $f$ is skew-Hermitian}
. \end{cases} $$
(We may assume the form begins with $x_1^\tau y_1 - x_2^\tau y_2$ or
$x_1^\tau y_2 - x_2^\tau y_1$ respectively, because  $\Frank \aG = 1  
\neq
0$.)
Note that $a_3,\ldots,a_n$ are
\begin{itemize}
\item elements of~$F$ if $f$ is Hermitian,
and
\item purely imaginary if $f$ is skew-Hermitian.
\end{itemize}

\setcounter{case}{0}

\begin{case}
Assume $n \le 3$.
\end{case}
The quaternion algebra~$D$ must split over~$F_v$, for each $v \in \SG 
$ (because $\locrank \aG \ge 2$).

\begin{subcase}
Assume $f$ is Hermitian.
\end{subcase}
Let
$$ \aG' =  \aSU_2(D, x_1^\tau y_1 - x_2^\tau y_2, \tau) \subset \aG .$$
Then $\aG'$ is of type~$\aC_2$ (see \cite[Prop.~2.15(2), p.~86] 
{[PR]}), so it is also of type~$B_2$. Therefore, it has a realization  
to which \S\ref{orthogonal} applies.


\begin{subcase}
Assume $f$ is skew-Hermitian.
\end{subcase}
Because $\aG$ is absolutely almost simple, we know it is not of type  
$D_2 = A_1 \times A_1$. Therefore $n = 3$. Then $\aG$ is of type~$D_3 
$, so it is also of type~$A_3$. Therefore, it
has a realization to which either \S\ref{SLSect} or~\S\ref{2ndKindSect}
applies.


\begin{case}
Assume $n \ge 4$.
\end{case}

\begin{subcase}
Assume $f$ is Hermitian.
\end{subcase}
By Weak Approximation and the Hasse Principle, there is a quadratic  
extension~$E$ of~$F$, such that $E$ splits~$D$ and, for $v \in \SG$,
	\begin{equation} \label{1stKindSect-n>4-Herm-E}
	\text{$F_v$ splits $D$} \quad\implies\quad E\subset F_v
	. \end{equation}

By normalizing, we may assume $a_3 = -1$.
By Weak Approximation, there exist $b_4,b_5,\ldots,b_n \in F$, with  
the property that, for every $v \in \SG$, such that $F_v$ does \emph 
{not} split~$D$, we have
	\begin{equation} \label{1stKindSect-n>4-Herm-a}
	\text{$a_4 b_4^2 + a_5 b_5^2 + \cdots + a_n b_n^2 > 0$ in~$F_v$.}
	\end{equation}
Let $a = a_4 b_4^2 + a_5 b_5^2 + \cdots + a_n b_n^2$, so, after a  
change of basis, we may assume $a_4 = a$.

Let
	$$ \aH = \aSU_4( E, x_1^\tau y_1 - x_2 y_2^\tau - x_3 y_3^\tau + a  
x_4^\tau y_4, \tau|_E) \subset \aG .$$
For any $v \in \SG$:
\begin{itemize}
\item If $E \subset F_v$, then $\aH$ is split over~$F_v$, so $ 
\locrank \aH = 3$.
\item If $E \not\subset F_v$, then $F_v$ does not split~$D$ \see 
{1stKindSect-n>4-Herm-E}, so $a > 0$ in~$F_v$ \see{1stKindSect-n>4- 
Herm-a}. Hence $\locrank \aH = 2$.
\end{itemize}
This contradicts the minimality of~$\aG$.



\begin{subcase}
Assume $f$ is skew-Hermitian.
\end{subcase}
Because $f$ is skew-Hermitian, we know that $a_3$ and~$a_4$ are
purely imaginary elements of~$D$, so there exists a nonzero, purely
imaginary $\alpha \in D$, such that $a_3$~and~$a_4$ both negate~$ 
\alpha$;
that is, $a_3 \alpha = - \alpha a_3$ and $a_4 \alpha = - \alpha a_4$.  
(To
see this, note that $a_3$ and~$a_4$ each negate a $2$-dimensional  
space of
imaginary elements of~$D$. Since the imaginary elements form only a
$3$-dimensional space, there must be nonzero intersection.) Hence, $a_3$
and~$a_4$ act by conjugation on $F[\alpha]$.

Let
\begin{itemize}
\item $F' = F[\alpha] \subset D$,
\item $\{e_1, e_2, e_3, e_4\}$ be an orthogonal basis of~$D^4$, such  
that
$$ \text{$f(e_1,e_1) = -f(e_2,e_2) = f(e_3,e_3) = a_3$ and $f(e_4,e_4) =
a_4$} $$
(namely, $e_1 = \frac{1}{2}(a_3,1,0,0)$, $e_2 =
\frac{1}{2}(-a_3,1,0,0)$, $e_3 = (0,0,1,0)$, and $e_4 = (0,0,0,1)$),
\item $V'$ be the $F'$ span of $\{e_1, e_2, e_3, e_4\}$,
and
\item $f'$ be the restriction of $a_3^{-1} f$ to~$V'$.
\end{itemize}
Note that $a_3^{-1} a_4$ centralizes~$\alpha$, so it must belong
to~$F'$.

Then
\begin{itemize}
\item $f'(V' \times V') \subseteq F'$,
\item $f'$ is a nondegenerate, symmetric $F'$-bilinear form on~$V'$,
   \item $f'$ is isometric to the form $f'' = x_1 y_1 - x_2 y_2 + x_3  
y_3 +
(a_3^{-1} a_4) x_4 y_4$ on~$(F')^4$,
and
\item $\aG
=\aSU_n(D,f,\tau)
\supset \res{F'/F} \aSO(f')
\approx \res{K/F} \aSL_2$,
where $K = F'\left[ \sqrt{-a_3^{-1} a_4} \, \right]$.
\end{itemize}
Since $\Frank \aG = 1$, we know that $f'$ has no $2$-dimensional totally
isotropic subspace, so $-a_3^{-1} a_4$ is \emph{not} a square in~$F'$.
Hence, we have
$$|K : F|
= |K : F'| \cdot |F' : F|
= 2 \cdot 2
 > 2 .$$
This contradicts the minimality of~$\aG$.









\section{Groups of type $F_4$} \label{F4Sect}

\begin{prop} \label{F4NotMin}
If\/ $\aG$ is of type~$F_4$, then $\aG$ is not minimal.
\end{prop}

\begin{proof}
We may assume $\Frank \aG = 1$, and that $\aG$ is absolutely almost  
simple \seeand{FrankMust1}{CanAbsSimple}. The Tits Classification  
\cite[p.~60]{[T66]} tells us that the Tits index of~$\aG$ is
$$
\begin{picture}(85,10)(-5,-5)
\put(00,00){\line(1,0){20}} \put(20,1.1){\line(1,0){30}}
\put(20,-1.2){\line(1,0){30}} \put(30,-3){$<$}
\put(50,00){\line(1,0){20}}
\put(00,0){\circle*{5}}
\put(20,0){\circle*{5}}
\put(50,0){\circle*{5}}
\put(70,0){\circle*{5}}
\put(0,0){\circle{10}}
\put(-5,10){$\alpha_4$} \put(15,10){$\alpha_3$}
\put(45,10){$\alpha_2$} \put(65,10){$\alpha_1$}
\end{picture}
$$
(We number the simple roots as in \cite[p.~223]{[B68]}.)
Let $\aS$ be an $F$-split $1$-dimensional torus in
$\aG$ and  let $\aM$ be the corresponding semisimple $F$-anisotropic
kernel.
%Now $\aM$ is of type $B_3$, so it is isogenous to $\aSO_7(f)$, where  
$f$ is a
%quadratic form on~$F^7$.

For each $v \in \SG$, we have $\locrank \aG > 1$, so
the Tits Classification \cite[p.~60]{[T66]} implies $\aG$ is
split over~$F_v$. Hence, by Weak Approximation and the Hasse  
Principle, there is a quadratic extension~$L$ of~$F$, such that $L$  
splits~$\aG$ and
	\begin{equation} \label{F4NotMin-LinFv}
	\text{$L \subset F_v$, for every $v \in \SG$.}
	\end{equation}
Since $L$ splits~$\aG$ (and hence splits~$\aM$), there is an $L$- 
split maximal $F$-torus~$\aT$ of~$\aM$, such that the standard  
semisimple subgroup~$\aG_{\alpha_1}$ generated by the roots $\pm 
\alpha_1$ is defined over~$F$ \see{Malpha}.
Let $\aR = \aT \cap \aG_{\alpha_1} \subset \aM$, so $\aR$ is a $1$- 
dimensional, $L$-split, anisotropic $F$-torus.

Let $\aH$ be the semisimple part of the identity component of $\aC_ 
{\aG}(\aR)$.
We know that $\aH$ is defined over~$F$ (since $\aR$ is defined over~$F 
$).
It is easy to see that $\aH$ is the standard semisimple subgroup  
generated by the roots
	$$ \pm \alpha_3, \pm \alpha_4, \pm (\alpha_1 + 2 \alpha_2 + 2  
\alpha_3). $$
Thus, $\aH$ is of type $C_3$, so it is (absolutely) almost simple  
over~$F$.
Furthermore, since $\aH$ has absolute rank~$3$, we have $\aC_{\aG} 
(\aR) = \aH \aR$.
\begin{itemize}
\item Since $\aR \subset \aM$, we know $\aS \subset \aC_{\aG}(\aR) =  
\aH \aR$.
Since $\aS$ is isotropic over~$F$, and $\aR$ is anisotropic, this  
implies $\aH$ is isotropic over~$F$.
\item By construction, $L$ splits both~$\aG$ and~$\aR$; therefore, $ 
\aC_{\aG}(\aR)$ contains an $L$-split maximal torus of~$\aG$. Since $ 
\aC_{\aG}(\aR) = \aH \aR$, we conclude that $\aH$ splits over~$L$.  
 From \pref{F4NotMin-LinFv}, we conclude that $\aH$ splits over~$F_v 
$, for every $v \in \SG$, so $\locrank \aH > 1$.
\end{itemize}
Therefore $\aG$ is not minimal.
\end{proof}








\section{Groups of type $\out{3,6}D4$} \label{TrialitySect}

The following theorem may be of independent interest. The proof makes  
no use of our standing assumption that $F$ is an algebraic number  
field --- it suffices to assume only that $\mathop{\rm char} F \neq 2$.

\begin{thm} \label{TrialityHasSL2}
Let\/ $\aG$ be an absolutely almost simple $F$-group of type $\out3D4 
$ or~$\out6D4$, such that $\Frank \aG = 1$.
Then there exists an extension field~$K$ of~$F$, such that\/ $\res{K/ 
F} \aSL_2$ is isogenous to an $F$-subgroup of\/~$\aG$, and\/ $|K:F| =  
4$.
\end{thm}

\begin{proof}
We start with notation:
\begin{itemize}
\item Let $\aS$ be a maximal $F$-split torus of~$\aG$.
\item Let $\aM = [\aC_{\aG}(\aS),\aC_{\aG}(\aS)]$ be the semisimple $F 
$-anisotropic kernel of~$\aG$.
\item It is well known \cite[Thm.~43.8 and Prop.~43.9, p.~555] 
{BookInvols} that there exist
	\begin{itemize}
	\item a cubic extension~$L$ of~$F$,
	and
	\item a quaternion algebra $D = (a,b_1)_L$ over~$L$,
	\end{itemize}
	such that
	\begin{itemize}
	\item $\aM$ is isogenous to $\res{L/F} \aSL_1(D)$,
	\item $a \in F$,
	\item $b_1 \in L$,
	and
	\item $N_{L/F} (b_1) = 1$.
	\end{itemize}
Because $\res{L/F} \aSL_1(D)$ is anisotropic, we know that $D$ is a  
division algebra.
\item Let $P = L \bigl[ \sqrt{b_1} \bigr]$, so $P$ is isomorphic to a  
maximal subfield of~$D$.
\item Let $\widetilde{P}$ be the Galois closure of $P$ over~$F$.
\item There is a maximal $F$-torus $\aT$ of~$\aM$ that is isogenous  
to $\res{L/F} \left( \res{P/L}^{(1)} \Gmult \right)$.
\item Let
	$$\Phi_H^+ = \{ \alpha_2, \alpha_2 + \alpha_1 + \alpha_3, \alpha_2 +  
\alpha_1 + \alpha_4, \alpha_2 + \alpha_3 + \alpha_4 \}$$
and
	$$ \Phi_H = \{\, \pm \alpha \mid \alpha \in \Phi_H^+ \,\} ,$$
where $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ is a base of the  
roots of~$\aG$ with respect to the maximal torus $\aS \aT$, numbered  
as in Figure~\ref{TrialityTitsFig}.
\item Let $\aH$ be the standard subgroup of~$\aG$ generated by the  
roots in~$\Phi_H$.
Since the roots in $\Phi_H^+$ are pairwise orthogonal, it is obvious  
that $\aH$ is a semisimple group that is of type $A_1 \times A_1  
\times A_1 \times A_1$ over the algebraic closure~$\overline{F}$.
\end{itemize}

\begin{figure}[ht]
\begin{picture}(110,40)(-25,-15)
%\put(-45,-2.75){$\times$} \dashline{5}(-40,0)(0,0)
%\put(-45,6){$\mu$}
\put(00,00){\line(1,0){40}} \put(40,0){\oval(80,30)[l]}
\put(0,0){\circle*{5}} \put(40,0){\circle*{5}}
\put(40,15){\circle*{5}} \put(40,-15){\circle*{5}}
\put(00,0){\circle{12}}  %\put(40,0){\oval(15,40)}
\put(-17,7){$\alpha_2$} \put(45,-2){$\alpha_3$}
\put(45,13){$\alpha_1$} \put(45,-17){$\alpha_4$}
\end{picture}
\caption{The Tits index of the trialitarian group~$\aG$.}
\label{TrialityTitsFig}
\end{figure}

Since $\aM$ and $\aT$ are defined over~$F$, the Galois group $\Gal 
(\widetilde{P}/F)$ acts on the set
	$$\Phi_M = \{ {\pm \alpha_1}, \pm \alpha_3, \pm \alpha_4 \} $$
of roots of~$\aM$. Letting $b_3$ and~$b_4$ be the Galois conjugates  
of~$b_1$ (over~$F$), it is clear that $\Gal(\widetilde{P}/F)$ also  
acts on
	$$ B = \left\{ {\pm \sqrt{b_1}}, \pm \sqrt{b_3}, \pm \sqrt{b_4} \, 
\right\} .$$
It is easy to see that these two actions are isomorphic (because both  
are transitive and have $\Gal(\widetilde{P}/P)$ as the stabilizer of  
a point). Therefore, after renumbering and choosing the signs of the  
square roots appropriately, we know, for any $\varphi \in \Gal 
(\widetilde{P}/F)$, that there exist $\varepsilon_1, \varepsilon_3,  
\varepsilon_4 \in \{0,1\}$ and a permutation~$\sigma$ of $\{1,3,4\}$,  
such that
	$$ \text{$\varphi(\alpha_i) = (-1)^{\varepsilon_i} \alpha_{\sigma(i)} 
$ and $\varphi \bigl( \sqrt{b_i} \bigr) = (-1)^{\varepsilon_i} \sqrt 
{b_{\sigma(i)}}$ for $i = 1,3,4$.} $$
Since
	$$\sqrt{b_1} \sqrt{b_3} \sqrt{b_4} = \pm \sqrt{b_1 b_3 b_4} = \pm  
\sqrt{N_{L/F}(b_1)} = \pm \sqrt{1} \in F ,$$
we know that $\sqrt{b_1} \sqrt{b_3} \sqrt{b_4}$ is fixed by~$\varphi 
$; therefore $\varepsilon_1 + \varepsilon_3 + \varepsilon_4$ is even.  
Hence,
	\begin{equation} \label{Epsilon0or2}
	\text{$\#\{\, i \mid \varepsilon_i \neq 0  \,\}$ is either $0$ or $2$.}
	\end{equation}
Let $\mu  = 2 \alpha_2 + \alpha_1 + \alpha_3 + \alpha_4$ be the  
maximal root of~$\aG$. The restriction of~$\mu$ to~$\aS$ is different  
from the restriction of any other root, so $\mu$ must be fixed by  
every element of $\Gal(\widetilde{P}/F)$. Therefore
	\begin{align*}
	2 \alpha_2 + \sum_{i \in \{1,3,4\}} \alpha_i
	&= \varphi \left( 2 \alpha_2 + \sum_{i \in \{1,3,4\}} \alpha_i  
\right) \\
	&= 2 \varphi(\alpha_2) + \sum_{i \in \{1,3,4\}} (-1)^{\varepsilon_i}  
\alpha_{\sigma(i)}
	, \end{align*}
so
	$$\varphi(\alpha_2) = \alpha_2 + \sum_{i \in \{1,3,4\}} \varepsilon_ 
{\sigma^{-1}(i)} \alpha_i .$$
 From \pref{Epsilon0or2}, we conclude that $\varphi(\alpha_2) \in  
\Phi_H^+$.

Since $\varphi$ is an arbitrary element of $\Gal(\widetilde{P}/F)$,  
the conclusion of the preceding paragraph implies that $\Phi_H^+$  
contains the entire orbit of~$\alpha_2$ under $\Gal(\widetilde{P}/F) 
$. In fact, it is easy to see that this orbit must be all of~$\Phi_H^+ 
$. Since $\aT$, and hence~$\aG$, is obviously split over~$\widetilde 
{P}$, this implies that $\aH$ is defined over~$F$ and is almost $F$- 
simple. Also, since $\aS \subset \aH$, it is obvious that $\aH$ is  
isotropic over~$F$. Because $\aH$ is of type $A_1 \times A_1 \times  
A_1 \times A_1$ over the algebraic closure, it  is now clear that $\aH 
$ is isogenous to $\res{K/F} \aSL_2$, where $K$ is an extension of  
degree~$4$ over~$F$.
\end{proof}

\begin{rem}
The specific choice of the maximal subfield~$P$ of~$D$ is crucial in  
the above proof; it is important that $|\widetilde{P} : \widetilde 
{L}| = 4$, where $\widetilde{P}$ and $\widetilde{L}$ are the Galois  
closures of~$P$ and~$L$ over~$F$. If $P$ is chosen differently, then  
the action of the Galois group on the roots of~$\aG$ is different,  
and the standard subgroup generated by the roots in $\Phi_H$ is not  
defined over~$F$.
\end{rem}


\begin{cor}
If\/ $\aG$ is of type $\out3D4$ or $\out6D4$, then\/ $\aG$ is not  
minimal.
\end{cor}






\section{Groups of type $\out{1,2}E6$} \label{E6Sect}

We assume, in this section, that $\aG$ is of type~$E_6$. By Lemma~\ref 
{FrankMust1}, we may also assume $\Frank \aG = 1$. Then there are  
only two possibilities for~$\aG$ in the
Tits Classification \cite[pp.~58--59]{[T66]}:
$$
\begin{array}{ccc}
\out2E{6,1}^{29}& &\out2E{6,1}^{35} \\
\begin{picture}(70,50)(00,-20)
\put(05,02){\line(1,0){20}} \put(65,02){\oval(80,14)[l]}
\put(05,02){\circle*{5}} \put(25,02){\circle*{5}}
\put(65,-5){\circle*{5}} \put(65,9){\circle*{5}}
\put(45,9){\circle*{5}} \put(45,-5){\circle*{5}}
\put(65,02){\oval(10,25)} \put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$} \put(40,14){$\alpha_3$}
\put(62,20){$\alpha_1$} \put(62,-20){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}
& \qquad &
\begin{picture}(65,50)(00,-20)
\put(05,02){\line(1,0){20}} \put(65,02){\oval(80,14)[l]}
\put(05,02){\circle*{5}} \put(25,02){\circle*{5}}
\put(65,-5){\circle*{5}} \put(65,9){\circle*{5}}
\put(45,9){\circle*{5}} \put(45,-5){\circle*{5}}
\put(05,02){\circle{10}} \put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$} \put(40,14){$\alpha_3$}
\put(62,14){$\alpha_1$} \put(62,-16){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}
\end{array}
$$
(We number the simple roots as in \cite[p.~230]{[B68]}.)

The two possible forms will be considered individually
(in Theorems~\ref{E6(D4kernel)} and~\ref{E6(A5kernel)}).
The proofs assume somewhat more background than those in previous  
sections.
\begin{thm} \label{E6(D4kernel)}
If\/ $\aG$ is an absolutely almost simple $F$-group of type $\out2E 
{6,1}^{29}$, then\/ $\aG$ contains an absolutely almost simple $F$- 
subgroup\/~$\aH$ of type $\out2A5$, such that $\locrank \aH \ge 2$,  
for every archimedean place~$v$ of~$F$.
\end{thm}

Before proving this theorem, we recall the following result (and, for  
completeness, we provide a self-contained proof based on Galois  
cohomology). It does not require our standing assumption that $F$ is  
an algebraic number field.

\begin{lem}[{\cite[Rem.~2.10]{GaribaldiPetersson}}] \label{anis(E6) 
=spin}
If\/ $\aG$ is an absolutely almost simple $F$-group of type $\out2E 
{6,1}^{29}$, then the semisimple anisotropic kernel of\/~$\aG$ is  
isomorphic to\/ $\Spin_8(f)$, for some quadratic form~$f$ on~$F^8$  
with nontrivial discriminant.
\end{lem}

\begin{proof}
There is no harm in assuming $\aG$ is adjoint.
Let
\begin{itemize}
	\item $K$ be the (unique) quadratic extension of~$F$ over which
$\aG$ becomes inner,
	\item $\aG^q$ be a quasisplit, absolutely almost
simple, adjoint $F$-group of type $\out2E6$ that splits over~$K$
(so the Tits index of~$\aG^q$ is the diagram on the right in \pref 
{TitsIndex2E6overImag} below),
	\item $\aP$ be a minimal parabolic $F$-subgroup of~$\aG$,
	\item $\aP^q$ be a parabolic $F$-subgroup of~$\aG^q$ that is of the  
same type as~$\aP$ (so the semisimple part of~$\aP^q$ is the standard  
subgroup generated by the roots $\pm\alpha_2,\pm\alpha_3,\pm\alpha_4, 
\pm\alpha_5$),
	\item $\aR^q\aM^q$ be the the Levi subgroup of~$\aP^q$, where $\aR^q 
$ is its central torus and $\aM^q$ is its semisimple part,
	and
	\item $\xi\in Z^1(F,\aG^q)$, such that $\aG$ is (isomorphic to) the  
twisted group ${^{\xi}}\aG^q$.
\end{itemize}
\setcounter{step}{0}

\begin{step} \label{xiInM0}
The class of $\xi$ is in the image of the map $H^1(F,\aM^q)\to
H^1(F,\aG^q)$.
\end{step}
It is well known that the image of the map
$H^1(F,\aP^q)\to H^1(F,\aG^q)$ consists of the classes of the
$1$-cocycles $\eta$ with the property that the twisted group $^{\eta} 
\aG^q$ has a
parabolic $F$-subgroup  of the same type as~$\aP^q$.
Thus, we may assume that $\xi\in Z^1(F,\aP^q)$.
Then, since the unipotent radical of $\aP^q$ has trivial
cohomology in dimension~$1$,
we may assume
	$$ \xi\in Z^1(F,\aR^q \aM^q) .$$

Since the center of the universal cover of~$\aM^q$ has order~$4$,
and the center of the universal cover of~$\aG^q$ has order~$3$, which is
relatively prime, we know that $\aM^q$ is simply connected.
It is easy to check that $\aR^q$ is of the form
$\res{K/F}(\Gmult)$, and that $\aR^q\cap \aM^q$ is the entire center  
of~$\aM^q$,
which is isomorphic to $(\integer/2\integer)\times (\integer/2 
\integer)$,
so
$\aR^q \cap \aM^q$ is precisely the $2$-torsion part ${_2\aR^q}$ of $ 
\aR^q$; hence
	$$
	\frac{\aR^q \aM^q}{\aM^q}
	\iso \frac{\aR^q}{\aR^q\cap \aM^q}
	= \frac{\aR^q}{{_2\aR^q}}
	\iso \aR^q
	\iso \res{K/F}(\Gmult)
	. $$
Therefore
	$$ H^1 \left( F, \frac{\aR^q \aM^q}{\aM^q} \right)
	\iso H^1\bigl( F,\res{K/F}(\Gmult) \bigr)
	\iso H^1(K,\Gmult)
	= 0
	.$$
Since  $\xi\in Z^1(F,\aR^q \aM^q)$, the desired conclusion now  
follows from the exact sequence
	$$ H^1(F,\aM^q)
	\to H^1(F,\aR^q \aM^q)
	\to H^1 \left( F, \frac{\aR^q \aM^q}{\aM^q} \right)
	.$$

\begin{step}
Completion of the proof.
\end{step}
Recall that~$\aM^q$ is the standard subgroup of~$\aG^q$ generated by  
the roots
$\pm\alpha_2,\ldots,\pm\alpha_5$. Thus, $\aM^q$ is a simply connected,
quasisplit group of type
$\out2D4$, so it is $F$-isomorphic to $\Spin_8(f_0)$,
where $f_0$ is a quasisplit quadratic form on~$F^8$. From Step~\ref 
{xiInM0},
we may assume that
	$$\xi\in Z^1(F,\aM^q) = Z^1 \bigl( F, \Spin_8(f_0) \bigr)  .$$
Then the semisimple
$F$-anisotropic kernel~$\aM$ of~$\aG$ is the
twisted group
	$${^\xi}\aM^q = {^\xi}{\Spin_8(f_0)} \iso \Spin_8(f) ,$$
for some quadratic form~$f$ on~$F^8$.
\end{proof}

\begin{proof}[{}{\bf Proof of Theorem~\ref{E6(D4kernel)}}]
We may assume $\aG$ is simply connected. Let
\begin{itemize}
\item $\aS$ be a $1$-dimensional $F$-split torus in~$\aG$,
\item $\aM$ be the corresponding semisimple $F$-anisotropic kernel,  
so we may identify $\aM$ with $\Spin(f)$, for some quadratic form~$f$  
on~$F^8$ with nontrivial discriminant \see{anis(E6)=spin},
\item $K$ be the (unique) quadratic extension of~$F$ over which $\aG$  
becomes a group of inner type,
\item $L$ be a totally imaginary quadratic extension of~$F$ (such  
that $L \neq K$),
\item $a \in F$, such that $L = F[\sqrt{a}]$,
and
\item $\aR$ be the central torus in the reductive group $\aC_\aG(\aS) 
$, so $\aC_\aG(\aS) = \aR\aM$ is an almost-direct product, and $\aR$  
is isogenous to $\res{K/F} \Gmult$.
\end{itemize}

Since $L \neq K$, we know that $\aG$ remains outer over~$L$.
It is well known \cite[p.~385]{[PR]} that there are only two  
possibilities for the Tits index of a group of type $\out2E6$ over a  
totally imaginary number field:
\begin{equation} \label{TitsIndex2E6overImag}
\begin{array}{ccc}
\begin{picture}(65,50)(00,-20)
\put(05,02){\line(1,0){20}} \put(65,02){\oval(80,14)[l]}
\put(05,02){\circle*{5}} \put(25,02){\circle*{5}}
\put(65,-5){\circle*{5}} \put(65,9){\circle*{5}}
\put(45,9){\circle*{5}} \put(45,-5){\circle*{5}}
\put(05,02){\circle{10}} \put(25,02){\circle{10}}
\put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$} \put(40,14){$\alpha_3$}
\put(62,14){$\alpha_1$} \put(62,-16){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}
& \qquad &
\begin{picture}(70,50)(00,-20)
\put(05,02){\line(1,0){20}} \put(65,02){\oval(80,14)[l]}
\put(05,02){\circle*{5}} \put(25,02){\circle*{5}}
\put(65,-5){\circle*{5}} \put(65,9){\circle*{5}}
\put(45,9){\circle*{5}} \put(45,-5){\circle*{5}}
\put(05,02){\circle{10}} \put(25,02){\circle{10}}
\put(45,02){\oval(10,25)} \put(65,02){\oval(10,25)}
\put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$} \put(40,20){$\alpha_3$}
\put(62,20){$\alpha_1$} \put(62,-20){$\alpha_6$}
\put(40,-20){$\alpha_5$}
\end{picture}
\end{array}
\end{equation}
Furthermore, since the roots $\alpha_1$ and~$\alpha_6$ are circled in  
the Tits index
over~$F$, they must also be circled in the Tits index over~$L$.
Therefore, $\aG$ is quasisplit over~$L$; this means that $\aM$ is  
quasisplit over~$L$.
Hence, after a change of basis to diagonalize the form appropriately,  
we may write
	$$ f = \langle a_1, -a_1 a, a_2, -a_2 a, a_3, -a_3 a, b_1, b_2  
\rangle .$$
Let $f' = \langle a_1, -a_1 a, a_2, -a_2 a \rangle$ be the  
restriction of~$f$ to the first $4$ coordinates.
By normalizing, we may assume $a_1 = 1$. Then $f'$ is the norm form  
of the quaternion algebra $D = (a,-a_2)_F$. (In other words, $f'$ is  
the 2-fold Pfister form $\langle 1,-a\rangle \otimes \langle 1,a_2  
\rangle$.) Hence,
	\begin{equation} \label{SO(f')=SL1DxSL1D}
	\aSO_4(f') \approx \aSL_1(D) \times \aSL_1(D)
	. \end{equation}
Let $\aM_1$ and $\aM_2$ be the two simple factors of~$\aSO_4(f')$.

Writing $f = f' \oplus f''$, we see that $\aM_1$ is normalized by $ 
\aSO_4(f') \times \aSO_4(f'')$, which contains a maximal torus of~$\aM 
$. So $\aM_1$ is a standard subgroup. Since all roots of~$\aM$ are  
conjugate under the Weyl group, we may assume $\aM_1 = \aG_{\alpha_2} 
$ is the standard subgroup generated by the roots $\pm\alpha_2$.

Let $\aH$ be the identity component  of $\aC_\aG(\aM_1)$.
Because $\aM_1$ is defined
over~$F$, we know that $\aH$ is defined over~$F$.
Furthermore, since $\aM_1 = \aG_{\alpha_2}$, it is easy to see that $ 
\aH$ is the standard subgroup
of~$\aG$ generated by the roots
	$$ \pm\alpha_3 , \pm\alpha_1,
	\pm(\alpha_2 + \alpha_3 + 2\alpha_4 + \alpha_5),
	\pm\alpha_6, \pm\alpha_5 .$$
Thus, $\aH$ is semisimple, and is of type $\out2A5$. Also, since $\aH$
contains the $F$-split torus $\aS$, we know that $\aH$ is $F$-isotropic.

All that remains is to show, for every archimedean place~$v$ of~$F$,  
that $\locrank \aH \ge 2$.

\setcounter{case}{0}

\begin{case}
Assume $\aG$ is inner over~$F_v$.
\end{case}
This assumption implies $K \subset F_v$. Since $\aH$ contains the $2$- 
dimensional torus~$\aR$, which splits over~$K$ (because it is  
isogenous to~$\res{K/F} \Gmult$), we have $\locrank \aH \ge 2$.

\begin{case}
Assume $f'$ is isotropic over~$F_v$.
\end{case}
Since $\aSO_4(f')$ is isotropic, and its two simple factors $\aM_1$  
and $\aM_2$ are isogenous \see{SO(f')=SL1DxSL1D}, we see that $\aM_2$  
is isotropic. Since $\aM_2$ centralizes $\aM_1$, and is contained in~$ 
\aM$, we see that $\aM_2$ is contained in the $F$-anisotropic kernel  
of~$\aH$. Thus, the $F$-anisotropic kernel of~$\aH$ is isotropic over~ 
$F_v$; so $\locrank \aH \ge 2$.

\begin{case}
The remaining case.
\end{case}
Recall that $f = f' \oplus f''$. Since $f'$ is anisotropic over~$F_v 
$, we must have $F_v \iso \real$, and we may assume all of the  
coefficients of $f' = \langle a_1, -a_1 a, a_2, -a_2 a \rangle$ are  
positive in~$F_v$.

Since $\aG$ is isotropic over~$F_v$ (indeed, it has been assumed to  
be isotropic over~$F$), we see, from the Tits Classification \cite 
[pp.~58--59]{[T66]} of real forms of~$E_6$, that $\locrank \aG \ge 2 
$. Hence, $\aM$ is isotropic over~$F_v$, so $f$ is isotropic over~$F_v 
$. Thus, some coefficient of $f''$ must be negative. On the other  
hand, because $\aG$ is outer over~$F_v$, we know that the  
discriminant of~$f$ is not~$1$, so the coefficients of~$f''$ cannot  
all be negative. Thus, $f''$ has both positive and negative  
coefficients, so $\Spin_4(f'')$ is isotropic over~$F_v$. Since $ 
\Spin_4(f'')$ obviously centralizes $\Spin_4(f') \supset \aM_1$, and  
is contained in~$\aM$, we see that $\Spin_4(f'')$ is contained in the  
$F$-anisotropic kernel of~$\aH$. Thus, the $F$-anisotropic kernel of~$ 
\aH$ is isotropic over~$F_v$; so $\locrank \aH \ge 2$.
\end{proof}




\begin{thm} \label{E6(A5kernel)}
If\/ $\aG$ is an isotropic, absolutely almost simple $F$-group of  
type $\out2E{6,1}^{35}$, then\/ $\aG$ contains an isotropic,  
absolutely almost simple $F$-subgroup~$\aH$ of type $\out3D4$ or $ 
\out6D4$.
\end{thm}

\begin{proof}
We may assume $\aG$ is simply connected. We fix:
\begin{itemize}
\item a maximal $F$-split torus~$\aS$ of~$\aG$,
\item a maximal $F$-torus~$\aT$ that contains~$\aS$,
and
\item an ordering of the roots of~$\aG$ (with respect to the maximal  
torus~$\aT$).
\end{itemize}
Let
	$$ \mu = -(\alpha_1 + 2 \alpha_2 + 2 \alpha_3 + 3 \alpha_4 + 2  
\alpha_5 + \alpha_6) , $$
be the minimal root of $\aG$, so $\aG$ has the following extended  
Tits index.
$$
\begin{picture}(105,44)(-35,-17)
\put(-30,-2){$\times$}
\dashline{5,1}(-23,1)(0,1)
\put(00,01){\line(1,0){20}}
\put(40,01){\oval(40,14)[l]}
\put(40,8){\line(1,0){20}}
\put(40,-6){\line(1,0){20}}
\put(00,01){\circle*{5}}
\put(20,01){\circle*{5}}
\put(40,-6){\circle*{5}}
\put(40,8){\circle*{5}}
\put(60,8){\circle*{5}}
\put(60,-6){\circle*{5}}
%\put(65,02){\oval(10,20)}
\put(0,01){\circle{12}}
\put(-30,14){$\mu$}
\put(-4,14){$\alpha_2$}
\put(15,14){$\alpha_4$}
\put(35,14){$\alpha_3$}
\put(55,14){$\alpha_1$}
\put(55,-16){$\alpha_6$}
\put(35,-16){$\alpha_5$}
\end{picture}
$$
The standard subgroup $\aG_\mu$ of $\aG$
generated by the roots~$\pm\mu$ is isomorphic to
$\aSL_2$ over $F$, so $\aS \subset \aG_\mu$.

Let
\begin{itemize}
\item $K$ be the (unique) quadratic extension of~$F$ over
which $\aG$ becomes a group of inner type,
and
\item $\aM=[\aC_\aG(\aS),\aC_\aG(\aS)]$ the semisimple $F$- 
anisotropic kernel of~$\aG$, so~$\aM$ is generated by the roots
$\{\pm\alpha_1,\pm\alpha_3,\pm\alpha_4,\pm\alpha_5,\pm\alpha_6\}$.
\end{itemize}
Therefore $\aM$ is of type $\out2A5$ and becomes inner over~$K$, so,  
as is well known \cite[Prop.~2.18, p.~88]{[PR]}, we have
	$$ \mbox{$\aM$ is $F$-isomorphic to $\aSU_m(D,f,\tau)$,}$$
where $D$ is a central division algebra of index $d = 6/m$ over~$K$,  
with involution~$\tau$ of the second kind, such that $F$ is the fixed  
field of the restriction of~$\tau$ to~$K$, and $f$ is a nondegenerate  
Hermitian form on~$D^m$.

\begin{claim}
$D$ is a cubic division algebra over~$K$ {\rm(}and $m = 2${\rm)}.
\end{claim}
(This is known, but we provide a proof for completeness.)
We know that $\aG$ is a twisted form $\aG={^{\xi}\aG^q}$
of a quasisplit, almost simple, simply connected $F$-group $\aG^q$
of type~$\out2E6$ splitting over~$K$, where $\xi$ is a $1$-cocycle with
coefficients in the adjoint group~$\overline{\aG}^q$.
In fact, there is a $1$-dimensional $F$-split torus~$\aS^q$ of~$\aG^q 
$, such that we may take $\xi$ to have its values in $\aC_{\overline 
{\aG}^q}(\overline{\aS}^q)$ (cf.\ \cite[Prop.~6.19, p.~339]{[PR]}).  
Write $\aC_{\aG^q}(\aS^q) = \aS^q \aM^q$, where $\aM^q$ is  
semisimple. Now $H^1 \bigl(F, \aC_{\overline{\aG}^q}(\overline{\aS} 
^q)/\overline{\aM}^q \bigr) = 0$ (because the coefficient group is an  
$F$-split torus), so we may take $\xi$ to have its values in~$ 
\overline{\aM}^q$. Therefore $\aM =  {^{\xi}\aM^q}$.
Let $\aZ$ be the center of $\aG^q$ (note that $\aZ$ is contained in $ 
\aM^q$), and let $\partial \colon H^1(F,\overline{\aM}^q)\to H^2(F, 
\aZ)$ be the connecting morphism. There is a cubic extension~$E$ of~$F 
$, such that the image of~$\partial \xi$ in $H^2(E,\aZ)$ is trivial  
\cite[Prop.~6.14, p.~334]{[PR]}. This means that the image of~$\xi$  
in $H^1(E,\overline{\aM}^q)$ lifts to an element of $H^1(E,\aM^q)$,  
so $\aM$ is isomorphic over~$E$ to $\aSU_6(K \cdot E, f', \tau')$,  
where $\tau'$ is the Galois automorphism of $K \cdot E$ over~$E$, and  
$f'$~is a Hermitian form on $(K \cdot E)^6$. Therefore, $D \otimes_K  
(K \cdot E)$ is a matrix algebra. So $D$ is either~$K$ or a cubic  
division algebra over~$K$.

To complete the proof of the claim, we need only show $D \neq K$.  
Assume the contrary. Then $\tau$ is the Galois automorphism of~$K$  
over~$F$, $f$~is a Hermitian form on~$K^6$, and $\aM \iso \aSU_6(K,f, 
\tau)$. For any archimedean place~$v$ of~$F$, the Tits Classification  
\cite[pp.~58--59]{[T66]} implies
	$$\locrank \aG > 1 = \Frank \aG ,$$
so $\locrank \aM \ge 1$; thus, $f$ is $F_v$-isotropic. Then, since  
any Hermitian form in $\ge 3$ variables is isotropic at every  
nonarchimedean place, the Hasse Principle tells us that $f$~is $F$- 
isotropic. This contradicts the fact that $\aM$ is the $F$- 
anisotropic kernel. This completes the proof of the claim.

\bigbreak

Choose a basis $\{e_1,e_2\}$ of~$D^2$ that is orthogonal with respect  
to~$f$. By making a change of coordinates, we may assume $e_1 = (1,0) 
$ and $e_2 = (0,1)$. Then, letting
	$$ \text{$d_1=f(e_1,e_1)$ and $d_2=f(e_2,e_2)$,} $$
we have
	$$f(x_1,x_2) = \tau(x_1) \, d_1 \, x_1 + \tau(x_2) \, d_2 \, x_2 .$$
Let $d= d_1^{-1} d_2\in D$. Then Lemma~\ref{NormalizeHerm} implies  
that we may assume $d_1 = 1$ and $d_2 = d$ (by replacing $\tau$ with $ 
\int(d_1^{-1}) \circ \tau$). That is,
	$$ f = \langle 1, d \rangle .$$
For convenience, let us identify $\aM$ with $\aSU_2(D,f,\tau)$.

It is not difficult to see that there exists $x \in D$, such that $ 
\tau(x) \, x \notin F$. Thus,
for a generic choice of the orthogonal basis $\{e_1,e_2\}$ (or merely  
multiplying $e_2$ by a generic element of~$D$), we have $d \notin F$.  
Since $\tau(d) = d$, this implies $d \notin K$. Therefore
\begin{itemize}
\item $E=K[d]$ is a maximal subfield in~$D$ (so it is cubic over~$K$),
\item $E$ is stable under~$\tau$,
and
\item $L= F[d]$ is a subfield of~$D$ that is cubic over~$F$.
\end{itemize}
Consider the subgroup $\aM'=\aSU_2(E,f,\tau|_E)$ of~$\aM$.
Writing $K = F[\sqrt{a}]$, for some $a \in F$, and
letting $T$ be the quaternion algebra $T=(a,-d)_F$ over~$F$, we have
	$$\aM'\iso \res{L/F} \bigl( \aSL_1(T) \bigr) .$$

Let $\overline {K}$ be an algebraic closure of~$K$. Then $D \otimes_K  
\overline{K} \iso \Mat3(\overline{K})$, and the isomorphism may be  
taken so that $E \otimes_K \overline{K}$ maps to the diagonal  
matrices. Then the algebra $\Mat2(E)$, viewed as a subalgebra of $ 
\Mat2(D) \otimes_K \overline{K} \iso \Mat6(\overline{K})$,
consists of matrices of the form
$$
\begin{bmatrix}
\ast & 0 & 0 & \ast & 0 & 0\\
0 &\ast & 0 & 0 &\ast &0\\
0 & 0 & \ast & 0 & 0 & \ast\\
\ast & 0 & 0 & \ast & 0 & 0\\
0 &\ast & 0 & 0 &\ast & 0\\
0 & 0 & \ast & 0 & 0 & \ast
\end{bmatrix}
.$$
Hence, $\aM'$ is the standard subgroup generated by the roots
$$
\pm(\alpha_1+\alpha_3+\alpha_4),\ \pm(\alpha_3+\alpha_4+\alpha_5),
\ \pm(\alpha_4+\alpha_5+\alpha_6).
$$
Let $\aH$ be the subgroup of $\aG$ generated by $\aM'$ and
$\aG_{\alpha_2}$. One easily checks that $\aH$ has type $D_4$,
contains $\aG_\mu$, and is defined over~$F$.
Since $\aG_\mu\simeq {\rm SL_2}$, we know $\aH$
is $F$-isotropic. Also, since $\aH$ contains $\aM'$, it is a
trialitarian group.
\end{proof}

\begin{cor} \label{E6(A5kernel)notmin}
If\/ $\aG$ is of type $E_6$, then\/ $\aG$ is not minimal.
\end{cor}

\begin{proof}
The conclusion is immediate from Theorem~\ref{E6(D4kernel)} if $\aG$  
is of type $\out2E{6,1}^{29}$.

When $\aG$ is of type $\out2E{6,1}^{35}$, it suffices to observe that  
the subgroup $\aH$ provided
by Theorem~\ref{E6(A5kernel)} satisfies $\locrank \aH \ge 2$, for  
every archimedean
place~$v$ of~$F$. This follows from Theorem~\ref{TrialityHasSL2}, but  
it is also
easy to give a short direct proof. Note that, because $\aH$ is a  
trialitarian group of rank~$1$, its
Tits index is as shown in Figure~\ref{TrialityTitsFig}; thus, the  
root $\alpha_2$ is circled.
So $\alpha_2$ is also circled in the Tits index of~$\aH$
over~$F_v$. From the Tits Classification \cite[pp.~56--58]{[T66]}
of groups of type $D_4$ over~$\real$,
we see that this implies at least two roots are circled,
so $\locrank \aH \ge 2$.
\end{proof}


\begin{acks}
We thank an anonymous referee for detailed comments that clarified  
and shortened some of the proofs, and strengthened Theorem~\ref{E6 
(A5kernel)}.
We also thank Gopal Prasad for suggesting that our main theorem  
should be proved over general number fields. D.W.M.\ would like to  
thank Skip Garibaldi for helpful discussions.
\end{acks}






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