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% specific math abbreviations
\newcommand\R{\mathbb{R}}
\newcommand\C{\mathbb{C}}
\newcommand\Z{\mathbb{Z}}
\newcommand\Q{\mathbb{Q}}
\newcommand\HH{\mathcal{H}}
\newcommand\EE{\mathcal{E}}
\newcommand\arrangement{\mathcal{A}}
\newcommand\smoothforms{\mathcal{C}_0}
\newcommand\smoothmoduli{M_0}
\newcommand\orb{{\rm orb}} % for orbifold fundamental group
\DeclareMathOperator{\Out}{Out}
\begin{document}
\title{Orthogonal complex hyperbolic arrangements}
%
\author{Daniel Allcock}
\address{Department of Mathematics\\Harvard
University\\Cambridge, MA 02138}
\email{allcock@math.harvard.edu}
\urladdr{http://www.math.harvard.edu/\textasciitilde allcock}
%
\author{James A. Carlson}
\address{Department of Mathematics\\University of Utah\\Salt
Lake City, UT 84112}
\email{carlson@math.utah.edu}
\urladdr{http://www.math.utah.edu/\textasciitilde carlson}
%
\author{Domingo Toledo}
\address{Department of Mathematics\\University of Utah\\Salt
Lake City, UT 84112}
\email{toledo@math.utah.edu}
\urladdr{http://www.math.utah.edu/\textasciitilde toledo}
%
\date{September 2, 2001}
\thanks{First author partly supported by NSF grant DMS~0070930.
Second and third authors partly supported by NSF grant DMS~9900543.}
\dedicatory{To Herb Clemens on his 60th birthday}
%
\maketitle
\section{Introduction}
\label{intro}
The purpose of this note is to study the geometry of certain remarkable
infinite arrangements of hyperplanes in complex hyperbolic space which we
call \emph{orthogonal arrangements}: whenever two hyperplanes meet, they meet
at right angles.
A natural example of such an arrangement appears in
\cite{ACT}; see also
\cite{ACTCR}. The concrete theorem that we prove here is that
the fundamental group of the complement of an orthogonal
arrangement has a presentation of a certain sort. As an
application of this theorem we prove that the fundamental group of the
quotient of the complement of an orthogonal arrangement by a
lattice in $PU(n,1)$ is
not a lattice in any Lie group with finitely many connected components. One
special case of this result is that the fundamental group of the moduli
space of smooth cubic surfaces is not a lattice in any Lie group with
finitely many components. This last result was the motivation for the
present note, but we think that the geometry of orthogonal arrangements
is of independent interest.
To state our results, let $B^n$ denote complex hyperbolic $n$-space, which
can be described concretely as either the unit ball in $\C^n$ with its
Bergmann metric, or as the set of lines in $\C^{n+1}$ on which the
hermitian form
$ h(z) = -|z_0|^2 + |z_1|^2 + \dots + |z_n|^2 $
is negative definite, with its unique (up to constant scaling factor)
$PU(n,1)$-invariant metric. Let $\arrangement = \{H_1, H_2, H_3, \dots \}$ be a
non-empty locally finite collection of totally geodesic complex hyperplanes
in
$B^n$. We call
$\arrangement$ a complex hyperbolic arrangement and write $\HH$ for $H_1\cup H_2\cup H_3
\cup \dotsb$. We are interested in $\pi_1(B^n-\HH)$, the
fundamental group of the complement. It is clear that if $\arrangement$ is infinite
(the case of interest here), this group is not finitely generated. (For
instance, its abelianization $H_1(B^n-\HH,\Z)$ is the free abelian
group on the set $\arrangement$.) If $n =1$, $\HH$ is a discrete subset of $B^1$ and
$\pi_1(B_1-\HH)$ is a free group, and we have nothing further to
say in this case. We thus assume throughout the paper that $n\ge 2$. We
say that $\arrangement$ is an \emph{orthogonal arrangement} if any two
distinct $H_i$'s are either orthogonal or disjoint.
The main purpose of this note is to prove the following
theorem:
\begin{theorem}
\label{presentation}
Let $\arrangement$ be an orthogonal complex hyperbolic
arrangement. Then the group
$\pi_1(B^n-\HH)$ has a presentation
$\langle\gamma_1,\gamma_2,\dots|\,r_1,r_2,\dots\rangle$
where each relator $r_k$ has the form $r_k = [\gamma_i,
l_{ij} \gamma_j l_{ij}^{-1}]$, where $l_{ij}$ is a word in
$\gamma_1,\dots,\allowbreak\gamma_{\mathop{\rm max}\nolimits\{i,j\}-1}$.
\end{theorem}
At this point the conclusion of the theorem may seem completely
technical. We hope that a look at the proof will convince the
reader that the conclusion reflects some beautiful complex
hyperbolic geometry. We will use Theorem~\ref{presentation}
to
prove the following theorem.
\begin{theorem}
\label{notLattice}
Suppose an orthogonal hyperplane arrangement
in $B^n$ is preserved by a lattice $\Gamma\subseteq PU(n,1)$, $n>1$.
Then the orbifold fundamental group
$\pi_1^\orb(\Gamma\backslash\allowbreak (B^n- \HH))$ is not a
lattice in any Lie group with finitely many connected
components.
\end{theorem}
This allows us to solve the problem which motivated this work,
concerning the moduli space of cubic surfaces in $\C P^3$.
Following \cite{ACT}, let $\smoothforms$ denote the space of
smooth cubic forms in $4$ variables, let $P\smoothforms$ denote
its image in the projective space of all cubic forms in $4$
variables, and let $\smoothmoduli = PGL(4,\C)\backslash
P\smoothforms$ denote the moduli space of smooth cubic surfaces.
\begin{corollary}
\label{cubicModuli}
Neither $\pi_1(P\smoothforms)$ nor
$\pi_1^\orb(\smoothmoduli)$ is a lattice in any Lie group with finitely many
connected components.
\end{corollary}
The corollary follows from Theorem~\ref{notLattice} because the
main result of \cite{ACT} is that there is an orbifold
isomorphism $\smoothmoduli\cong \Gamma\backslash(B^4-\HH)$ where
$\Gamma$ is a certain lattice in $U(4,1)$ and $\HH$ is a
certain $\Gamma$-invariant orthogonal arrangement in $B^4$.
Namely, let $\EE$ denote the ring of Eisenstein integers (the
integers $\Z[{\root 3 \of 1}]$ in $\Q(\sqrt{-3})$), let $h$ be the
above Hermitian form in $n+1$ variables, let $\arrangement$ be the
arrangement $ \{v^\perp : v\in
\EE^{n+1}, h(v) = 1\}$, and let $\HH$ be the union of the
hyperplanes. It is easy to see (see Lemma 7.29 of
\cite{ACT}) that $\arrangement$ is an orthogonal arrangement. Let
$\Gamma$ denote the lattice $PU(h,\EE)$ in $PU(n,1)$, which
obviously preserves $\HH$. Then $\Gamma\backslash B^n$ is a
quasi-projective variety and a complex analytic orbifold, and in
the case $n=4$ we have the orbifold isomorphism
$\smoothmoduli\cong
\Gamma\backslash (B^4-\HH)$.
\section{Proof of Theorem~\ref{presentation}}
\label{proof}
Let $\arrangement$ be an orthogonal arrangement, let $\HH$ denote the union of the
hyperplanes in $\arrangement$, and choose the basepoint
$p_0\in B^n-
\HH$. We will now show that $p_0$ can be chosen to satisfy two
genericity conditions:
(G1) for any two distinct nonempty sub-balls $X$ and $Y$ of $B^n$,
each of which is an intersection of members of
$\arrangement$, $d(p_0,X)\neq d(p_0,Y)$, and
(G2) for any $i\neq j$, the minimal geodesic from $p_0$ to
$H_i$ does not meet $H_j$. (We write $d$ for the complex
hyperbolic distance.)
It is clear that condition (G1) holds on the complement of a
countable collection of equidistant hypersurfaces, which are
closed real analytic subvarieties of real codimension one. The
same holds for (G2), namely, let $V_{i,j}$ denote the union of
all geodesic rays from $H_i$ which are perpendicular to $H_i$
and which meet $H_j$. Then $V_{i,j}$ is a locally closed real
analytic subvariety of real codimension one, whose closure is
the nowhere dense semi-analytic set of geodesics perpendicular
to $H_i$ and which meet the closure of $H_j$ in the closed unit
ball $\bar B^n$. Thus the complement of $\bigcup_{i,j}V_{i,j}$
is not empty. We denote $d(p_0,H_i)$ by $d_i$, and label the hyperplanes $H_1, H_2, \dots
\in\arrangement$ according to increasing distance from $p_0$, so that
$d_1 point at 0 0
% THE GRID
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\putrule from -1 0 to 6 0
\putrule from 0 -.3 to 0 3.7
\linethickness=.7pt
%\putrule from 0 1 to 5 1
%\putrule from 0 2 to 5 2
\putrule from 0 3 to 5 3
\putrule from 1 0 to 1 3
\putrule from 2 0 to 2 3
\putrule from 3 0 to 3 3
\putrule from 4 0 to 4 3
\putrule from 5 0 to 5 3
% THE LITTLE `PERPENDICULAR' MARKS
\linethickness=.3pt
% the horizontal bits
\putrule from 0 .2 to .2 .2
\putrule from .8 .2 to 1 .2
\putrule from 1.8 .2 to 2 .2
\putrule from 2.8 .2 to 3 .2
\putrule from 3.8 .2 to 4 .2
\putrule from 4.8 .2 to 5 .2
%\putrule from 0 .8 to .2 .8
%\putrule from 0 1.8 to .2 1.8
\putrule from 0 2.8 to .2 2.8
% the vertical bits
\putrule from .2 0 to .2 .2
\putrule from .8 0 to .8 .2
\putrule from 1.8 0 to 1.8 .2
\putrule from 2.8 0 to 2.8 .2
\putrule from 3.8 0 to 3.8 .2
\putrule from 4.8 0 to 4.8 .2
%\putrule from .2 .8 to .2 1
%\putrule from .2 1.8 to .2 2
\putrule from .2 2.8 to .2 3
% THE MARKED POINTS
\put {$\bullet$} at 0 0
\put {$\bullet$} at 5 0
\put {$\bullet$} at 5 3
\put {$\bullet$} at 0 3
% TEXT
\put {$p_{ij}$} [br] <-2pt,2pt> at 0 0
\put {$p_j$} [r] <-2pt,0pt> at 0 3
\put {$p_i$} [t] <0pt,-3pt> at 5 0 \put {$p_0$} [bl] <2pt,2pt> at 5 3
%\put {$Q$} at 2.5 1.5
\put {$\sigma_i$} [l] <1pt,0pt> at 5 1.5
\put {$\sigma_j$} [b] <0pt,1pt> at 2.5 3
\put {$H_i$} [b] <0pt,1pt> at 6 0
\put {$H_j$} [l] <1pt,0pt> at 0 3.7
\endpicture
}%
\caption{A picture of $Q$.}
\label{vertical foliation}
\end{figure}
\begin{lemma}
Suppose that $k\ne i,j$ and that the hyperplane $H_k$
meets either $H_i$ or $H_j$. Then $H_k\cap Q =\emptyset$.
\label{ThirdHyperplaneDisjoint}
\end{lemma}
\begin{proof}
Suppose, say, that $H_k\cap H_i\ne \emptyset$. We will first show
that $H_k\cap Q$ is a vertical segment. Note that $H_k$ and
$H_i$ intersect at right angles, and that $H_k$ is foliated by
totally geodesic discs (complex lines) perpendicular to $H_i$ at
the points of $H_i\cap H_k$. Call these discs \emph{vertical
discs}. Suppose that $H_k\cap Q\ne\emptyset$, and let $q\in
H_k\cap Q$. Then $q$ is in a unique vertical segment and a
unique vertical disc, and it is easy to see that the segment is
contained in the disc. Thus $H_k\cap Q$ is a vertical segment.
Since $Q$ is foliated by these vertical segments, the segment
$H_k\cap Q$ must intersect the side $\sigma_j$ of $Q$ opposite
to $H_i\cap Q$; see Figure~\ref{vertical foliation}. Thus $H_k\cap
\lambda_j\ne\emptyset$, contradicting assumption (G2) of the choice of
$p_0$. Thus we must have
$H_k\cap Q = \emptyset$, and the lemma is proved.
\end{proof}
\begin{lemma}
\label{TransversIntersection}
Suppose that $k\ne i,j$ and $H_k\cap Q\ne\emptyset$. Then
$H_k\cap Q$ consists of a single point in the interior of $Q$, and hence the
intersection is transverse.
\end{lemma}
\begin{proof}
It is clear that $H_k$ cannot intersect the boundary of $Q$: by
the choice (G2) of $p_0$ it cannot intersect
the edges $\sigma_i$ or $\sigma_j$ of $Q$, and by the previous lemma
it cannot intersect the edges $H_i\cap Q$,
$H_j\cap Q$. If $H_k\cap Q\ne \emptyset$, then since the
intersection is totally geodesic in $Q$, it must be either a
point or the intersection of $Q$ with a full geodesic in the
real 2-ball containing $Q$.
Since $H_k$ cannot intersect the boundary of $Q$, the intersection must be an
interior point, and the proof is complete.
\end{proof}
\begin{lemma}
\label{ThirdHyperplaneCloser}
Suppose that $k\ne i, j$ and $H_k\cap Q\ne\emptyset$.
Then $d(p_0, H_k\cap Q) < \mathop{\rm max}\nolimits\{d_i,d_j\}$.
\end{lemma}
\begin{proof}
Observe that the quadrilateral $Q$ lies in a unique totally geodesic
complex hyperbolic plane $B^2\subset B^n$, so by intersecting all these
hyperplanes with this $B^2$ we may reduce to the case of lines in $B^2$.
Consider this $B^2$ concretely as the unit ball $\{|z|^2 + |w|^2 < 1\}
\subset \C^2$, with $p_{ij}$ at the origin, $H_i$ and $H_j$ as the
intersection of $B^2$ with the $z$ and $w$-axes respectively, and with the
quadrilateral $Q$ lying in the quadrant
$$
\{(x,y): x,y\in\R,\,
x^2 + y^2< 1,\ x\geq0,\ y\geq 0\}
$$
of the real subspace $B_{\R}^2$.
The complex hyperbolic geometry of $B^2$
restricts to the Klein model of real hyperbolic geometry of $B_{\R}^2$, so
geodesics in $B^2_\R$
are real line segments and convex sets are the Euclidean convex
sets.
We suppose without loss of generality that $i point at 0 0
% THE RECTANGLE
\linethickness=1.6pt
\putrule from -1 0 to 6 0
\putrule from 0 -.3 to 0 3.7
\linethickness=.7pt
\putrule from 0 3 to 5 3
\putrule from 5 0 to 5 3
% THE CURVE AND LINES
\circulararc 37 degrees from 0 3 center at 5 3
\setlinear
\plot 0 3 1 0 /
\setdashes
\plot 5.8 .9 3 3.7 /
% THE MARKED POINTS
\put {$\bullet$} at 0 0
\put {$\bullet$} at 5 0
\put {$\bullet$} at 5 3
\put {$\bullet$} at 0 3
\put {$\bullet$} at 4.8 2.5
% TEXT
\put {$0$} [tr] <-2pt,-2pt> at 0 0
\put {$p_j$} [r] <-2pt,0pt> at 0 3
\put {$H_j$} [l] <1pt,0pt> at 0 3.7
\put {$p_i$} [t] <0pt,-3pt> at 5 0
\put {$H_i$} [t] <0pt,-2pt> at 6 0
\put {$\sigma_i$} [r] <-1pt,0pt> at 5 .9
\put {$\sigma_j$} [b] <0pt,1pt> at 2 3
\put {$x+y=1$} [l] <3pt,0pt> at 3 3.7
\put {$p_0$} [bl] <2pt,2pt> at 5 3
\put {$(x_0,y_0)$} [b] <-9pt,4pt> at 4.8 2.5
% these are for placing the letters C and L in a scale-invt way;
% found by trial and error
%\put {$\bullet$} at .5 .8
%\put {$\bullet$} at .5 1.5
\put {$C$} [tr] <0pt,0pt> at .5 .8
\put {$L$} [bl] <1pt,0pt> at .5 1.5
\endpicture
}%
\caption{}
\label{intersection figure}
\end{figure}
Now let $ H_k\cap Q=\{(x_0,y_0)\}$. We must show that
$d(p_0,(x_0,y_0))1$. Therefore
$(x_0,y_0)$ lies in the connected component of $Q- C$
containing $p_0$, so $(x_0, y_0)$ is interior to the $d_j$-ball
about $p_0$, and the lemma is proven.
\end{proof}
We can now finish the proof of Theorem~\ref{presentation}; we
continue to assume $i point at 0 0
% THE AXES; USED FOR AID IN COMPOSITION; DRAWN FOR REAL BELOW
% \linethickness=1.6pt
% \putrule from -1 0 to 6 0
% \putrule from 0 -.3 to 0 3.7
% THE 4 MARKED POINTS AND LINES JOINING THEM
\put {$\bullet$} at 5 3
\put {$\bullet$} at 5 .3
\put {$\bullet$} at .3 3
\put {$\bullet$} at .3 .3
\linethickness=.7pt
\putrule from 5 3 to .3 3
\putrule from 5 3 to 5 .3
\putrule from .3 3 to .3 .3
\putrule from 5 .3 to .3 .3
% THE LOOPS AT THE MARKED POINTS
\ellipticalarc axes ratio 1:4 -250 degrees from 5 .3 center at 5 0
\ellipticalarc axes ratio 1:4 70 degrees from 5 .3 center at 5 0
%
\ellipticalarc axes ratio 1:4 -250 degrees from .3 .3 center at .3 0
\ellipticalarc axes ratio 1:4 70 degrees from .3 .3 center at .3 0
%
\ellipticalarc axes ratio 4:1 -250 degrees from .3 .3 center at 0 .3
\ellipticalarc axes ratio 4:1 70 degrees from .3 .3 center at 0 .3
%
\ellipticalarc axes ratio 4:1 -250 degrees from .3 3 center at 0 3
\ellipticalarc axes ratio 4:1 70 degrees from .3 3 center at 0 3
% THE AXES (FOR REAL THIS TIME)
\linethickness=1.6pt
\putrule from -1 0 to .33 0
\putrule from .42 0 to 5.03 0
\putrule from 5.12 0 to 6 0
%
\putrule from 0 3.7 to 0 2.97
\putrule from 0 2.88 to 0 .27
\putrule from 0 .18 to 0 -.3
% MARKED POINTS INSIDE THE SQUARE, AND ASSOCIATED DECORATIONS
% circles have radius .15; points on them to which lines are
% drawn were found by paper computations
\linethickness=.7pt
\setlinear
\put{$\bullet$} at 3 2.4
\circulararc 360 degrees from 3 2.55 center at 3 2.4
\plot 5 3 3.143 2.443 /
%
\put{$\bullet$} at 3.6 1.9
\circulararc 360 degrees from 3.6 2.05 center at 3.6 1.9
\plot 5 3 3.717 1.992 /
%
\put{$\bullet$} at 4 1.7
\circulararc 360 degrees from 4 1.85 center at 4 1.7
\plot 5 3 4.092 1.819 /
% TEXT
\put {$H_j$} [r] <-1pt,0pt> at 0 3.7
\put {$c_j$} [r] <-1pt,0pt> at -.3 3
\put {$e_j$} [r] <-1pt,0pt> at -.3 .3
\put {$q_j$} [bl] <2pt,2pt> at .3 3
\put {$\lambda_{ji}$} [l] <1pt,0pt> at .3 1.5
\put {$q_{ij}$} [bl] <2pt,2pt> at .3 .3
\put {$e_i$} [t] <0pt,-2pt> at .3 -.3
\put {$\lambda_j$} [b] <0pt,1pt> at 2.5 3
\put {$\lambda_{ij}$} [b] <0pt,1pt> at 2.5 .3
\put {$p_0$} [bl] <2pt,2pt> at 5 3
\put {$\lambda_i$} [l] <1pt,0pt> at 5 1.5
\put {$q_i$} [bl] <2pt,2pt> at 5 .3
\put {$c_i$} [t] <0pt,-2pt> at 5 -.3
\put {$H_i$} [t] <0pt,-2pt> at 6 0
\put {$l_1$} [r] <-1pt,0pt> at 2.85 2.4
\put {$l_{r-1}$} [tr] <8pt,0pt> at 3.49 1.79
\put {$l_r$} [t] <0pt,-2pt> at 4 1.55
% ELLIPSIS BETWEEN l_1 and l_{r-1}
\setdots
\plot 3.15 2.275 3.45 2.025 /
\endpicture
}%
\caption{}
\label{conjugation figure}
\end{figure}
The only consequence of Theorem~\ref{presentation} we will use
is the following corollary. For each $m = 1, 2,
\dots$, we define
$\HH_m = H_1\cup\dots\cup H_m$, and observe that the inclusion
$B^n- \HH\subset B^n-\HH_m$ induces a surjection
$\pi_1(B^n-\HH)\to \pi_1(B^n-\HH_m)$.
\begin{corollary}
\label{HasNontrivialImage}
Let $\gamma\in\pi_1(B^n-\HH)$, $\gamma\ne
1$. Then there exists an $m$ such that $\gamma$ has non-trivial image in
$\pi_1(B^n-\HH_m)$.
\end{corollary}
\begin{proof}
Theorem~\ref{presentation} clearly implies that for each $m$ the
map that sends each generator $\gamma_1,\dots,\gamma_m$ of
$\pi_1(B^n-\HH_m)$ to the same element of
$\pi_1(B^n-\HH)$ defines a group homorphism
$\pi_1(B^n-\HH_m)\to \pi_1(B^n-\HH)$ which
splits the surjection $\pi_1(B^n-\HH)\to
\pi_1(B^n-\HH_m)$ and has image
$\langle\gamma_1,\dots,\gamma_m\rangle$, the subgroup of
$\pi_1(B^n-\HH)$ generated by
$\gamma_1,\dots,\gamma_m$. Thus the subgroup
$\langle\gamma_1,\dots,\gamma_m\rangle$ maps isomorphically
onto $\pi_1(B^n-\HH_m)$. Given $\gamma\ne 1$ as in the
statement of the corollary, choose an $m$ so that $\gamma\in
\langle\gamma_1,\dots,\gamma_m\rangle$, and the corollary follows.
\end{proof}
\begin{remark}
It is easy to see that the relations
$[\gamma_i,l_{ij}\gamma_jl_{ij}^{-1}]$ of
Theorem~\ref{presentation} in general cannot be simplified to
$[\gamma_i,\gamma_j]$ by choosing the basepoint $p_0$ so that
all the loops $l_{ij}$ are homotopically trivial. We show this
for the arrangement $\arrangement$ associated to the Eisenstein
integers that we defined in the introduction. To see this for
$n=2$, let us follow the conventions of the proof of
Lemma~\ref{ThirdHyperplaneCloser}, so that $H_i$ and $H_j$ are the
$z$ and $w$ axes respectively. Choose a third hyperplane $H_k$
to have equation $z+w =1$, and note that this is also in the
collection $\HH$. Choose a point $(z_1,w_1)\in \partial B^2$
so that $1$ is in the interior of parallelogram in $\C$ spanned
by $z_1$ and $w_1$. Observe that this is equivalent to saying that
the ideal Lambert quadrilateral $\{sz_1 + tw_1: 0\le s,t\le 1\}$
meets the hyperplane $H_k = \{z+w = 1\}$ in the interior of the
quadrilateral. By the density of $\Q(\sqrt{-3})$-rational
points in $\partial B^2$, we may assume that $(z_1,w_1)$ has
coordinates in $\Q(\sqrt{-3})$, for instance we may take
$(z_1,w_1) = ((9 + 3\sqrt{-3})/16, (11 - 3\sqrt{-3})/16).$ Then
there exists a neighborhood $N$ of $(z_1,w_1)$ so that for all
$p_0 = (z_0,w_0)\in N\cap B^2$, $1$ is in the interior of the
parallelogram spanned by $z_0$ and $w_0$. Thus for all $p_0\in N\cap B^2$,
the Lambert quadrilateral $Q = \{sz_0+tw_0:0\le s,t\le 1\}$ with
acute angle at $p_0$ meets the hyperplane $H_k$. Since every
neighborhood $N\cap B^2$ of a cusp point in $\partial B^2$
contains a fundamental domain for $\Gamma_2$, it follows that
for any choice of basepoint $p_0$ there are hyperplanes $H_i,
H_j$ so that the loop $l_{ij}$ is not homotopically trivial.
Suitable modifications of this argument show the necessity of
the $l_{ij}$ for any $n>2$.
\end{remark}
\section{Proof of Theorem~\ref{notLattice}}
In this section we write $K$ for
$\pi_1(B^n-\HH)$ and $\Phi$ for $\pi_1^\orb(\Gamma\backslash
(B^n-\HH).$ Observe that these groups fit into the exact
sequence
$$
1\to K\to \Phi \to \Gamma
\to 1,
$$
which gives rise to a natural homomorphism $\Gamma\to \Out(K)$, where
$\Out$ denotes the group of outer automorphisms. Theorem~\ref{notLattice}
follows from well known theorems on lattices in semi-simple
groups, together with the
following lemmas.
\begin{lemma}
\label{bigCentralizer}
For $n>1$, the centralizer in $\Phi$ of each generator
$\gamma_i\in K$ contains a non-abelian free group.
\end{lemma}
\begin{proof}
Let $M_i\subset B^n- H_i$ be the boundary of a tubular
neighborhood of the hyperplane $H_i$, and suppose that $M_i$ contains
the point
$q_i$ of \S\ref{proof}.
Let $\Gamma_i$ denote the subgroup of $\Gamma$ that preserves
$H_i$. Then $\gamma_i$ is freely homotopic to the loop $c_i$ of \S 2 based
at
$q_i$, which is clearly central in the orbifold fundamental group
$\pi_1^\orb(\Gamma_i\backslash (M_i-\HH))$. Since this latter
group surjects (to $\pi_1^\orb(\Gamma_i\backslash M_i)$ and hence) to
$\Gamma_i$, which is a lattice in $PU(n-1,1)$, it contains a
non-abelian free group if $n>1$. It then follows that the centralizer of
$\gamma_i$ in $\Phi$ contains a non-abelian free group if $n>1$.
\end{proof}
\begin{lemma}
\label{normalSubgroupsContainFree}
Let $H\ne\{1\}$ be a subgroup of $K$ which is
normal in
$\Phi$. Then
\newcounter{bean}
\begin{enumerate}
%
\item $H$ contains a
non-abelian free group.
%
\item The image of the natural map $\Gamma\to \Out(H)$
contains elements of infinite order.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\mu\in H$ and $\mu\ne id$. We will produce $\nu\in H$ so that
$\mu,\nu$ generate a non-abelian free group. By
Corollary~\ref{HasNontrivialImage}, there exists an $m$ so that
$\mu$ has non-trivial image in $\pi_1(B^n-\HH_m)$.
Choose a hyperbolic element of $\Gamma$ neither of whose limit
points in $\partial B^n$ lie in any of the boundaries $\partial
H_i\subset\partial B^n$ for $i=1,\dots,m$. This can be done
because $\Gamma$ is a lattice. Sufficiently high powers of this
transformation map $\HH_m$ far away
from itself; choose one such power and denote it by $\phi$. Then
$B^n$ can be written as the union of two
topological $n$-sub-balls intersecting along a topological
$(n-1)$-sub-ball, where one of the balls contains $\HH_m$ and
the other contains $\phi(\HH_m)$. It follows that $\pi_1(B^n
- (\HH_m\cup \phi(\HH_m))$ is the free product of the
groups $K_m = \pi_1(B^n-\HH_m)$ and $\tilde K_m =
\pi_1(B^n-\phi(\HH_m))$. Since $\phi\in\Gamma$,
$\phi(\HH_m)\subset\HH$. Thus there is a natural surjection
$K\to K_m * \tilde K_m$. Since $H$ is normal in
$\pi_1^\orb(\Gamma\backslash (B^n-\HH))$, the group
$\Gamma$ acts by outer automorphisms on $H$, and there must be
an element $\nu\in K$ which maps to the image of $\phi(\mu)$ in $K_m*\tilde
K_m$. Thus $H$ contains the free product of the cyclic groups
generated by $\mu$ and $\nu$, which is a free non-abelian group
since $\mu$ and $\nu$ must be of infinite order, say by the
result of
\cite{Allcock} that $B^n-\HH$ is aspherical. (One may
avoid using \cite{Allcock} by first applying the above
construction to show that $H$
has an element of infinite order, and then applying the construction
to this element.) This proves the first part of the lemma. Iterating this
reasoning, it is clear that $\phi$ has infinite order in $\Out(H)$, so the
second part of the lemma is proved.
\end{proof}
It is now easy to prove Theorem~\ref{notLattice}. First we show that the
first part of Lemma~\ref{normalSubgroupsContainFree} precludes $\Phi$ from being a lattice in
a connected Lie group $G$ with non-trivial solvable radical $R$.
Otherwise, in the exact sequence of Lie groups
$$
1\to R \to G\to G/R \to 1
$$
there would be two possibilities: (1)~The image of $\Phi$ in
$G/R$ is discrete. In this case, by Theorem 1.13 of
\cite{Raghunathan}, $R\cap\Phi$ would be a lattice in $R$, thus
$R\cap\Phi$ would be a non-trivial normal solvable subgroup of
$\Phi$, which is precluded by the first part of
Lemma~\ref{normalSubgroupsContainFree}. (2)~The image of
$\Phi$ in $G/R$ is not discrete. In this case, by a theorem of
Auslander, Theorem 8.24 of \cite{Raghunathan}, the identity
component of the closure of its image is a solvable group. The
intersection of $\Phi$ with the pre-image in $G$ of this group
is then a non-trivial normal solvable subgroup of $\Phi$, which
we have already excluded. We can then exclude $\Phi$ from being
a lattice in a Lie group $G$ with finitely many components and
with identity component having non-trivial radical by observing
that Lemma~\ref{normalSubgroupsContainFree} also holds for
subgroups of finite index in $\Phi$.
It remains only to exclude the possibility that $\Phi$ is a lattice in a
semisimple group $G$, again easily reduced to a connected $G$. First, from
the the fact that $\Gamma$ is not virtually a product and from the second
part of Lemma~\ref{normalSubgroupsContainFree} (which also holds for subgroups of finite index of
$\Phi$), it is clear that $\Phi$ is not virtually a product. Thus if it
were a lattice in $G$, it would be an irreducible one. If $G$ had real rank
at least $2$, by a deep theorem of Margulis \cite{Margulis} any normal
subgroup would have to be either of finite index or central (note that this
holds even if
$G$ is not linear), which is precluded by the infinite-index non-central
subgroup
$K$. If $G$ were a rank $1$ group, linear or not, the centralizer
of any non-central element would be virtually nilpotent. Since the $\gamma_i$
are clearly not central, this contradicts Lemma~\ref{bigCentralizer} (Note this is the only
point where the hypothesis
$n>1$ is used, and that Theorem~\ref{notLattice} does not hold for
$n=1$). This completes the proof of Theorem~\ref{notLattice}.
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\end{thebibliography}
\end{document}