\begin{document}

\title{The space of strictly-convex real-projective structures on a closed manifold}
\date{\today}

\noindent\address{DC: Department of Mathematics, University of California, Santa Barbara, CA 93106, UlSA}\\
\address{ST: School of Mathematics and Statistics, The University of Sydney, NSW 2006, Australia}
\address{}\\
\email{cooper@math.ucsb.edu}\\
\email{stephan.tillmann@sydney.edu.au}

\author{Daryl Cooper  and Stephan Tillmann}
\begin{abstract} We give an expository proof of the fact that, if $M$ is a compact $n$--manifold with no boundary, then the set
of holonomies of strictly-convex real-projective structures on $M$ is a subset
of  $\operatorname{Hom}(\pi_1M,\operatorname{PGL}(n+1,\mathbb{R}))$ that is both open and closed.
\end{abstract}
\maketitle

If $M$ is a compact $n$--manifold, then $\operatorname{Rep}(M)=\operatorname{Hom}(\pi_1M,\operatorname{GL}(n+1,{\mathbb R}))$ is a real algebraic variety. Let
$\operatorname{Rep}_{P}(M)$ and $\operatorname{Rep}_{S}(M)$ be, respectively, the subsets of $\operatorname{Rep}(M)$ of holonomies of properly-convex, and of strictly-convex structures  on $M$.
Then $\operatorname{Rep}_S(M)\subset\operatorname{Rep}_P(M)$. Throughout this paper we use the Euclidean topology everywhere, and not the {\em Zariski topology}.
In the following, $M$ is closed and $n\ge 2$.

\begin{open}\label{open} 
$\operatorname{Rep}_{P}(M)$
 is open in $\operatorname{Rep}(M)$.
  \end{open}
  
  \begin{closed}\label{closed} 
$\operatorname{Rep}_{S}(M)$
 is closed in $\operatorname{Rep}(M)$.  
 \end{closed}

\begin{clopen}
$\operatorname{Rep}_S(M)$ is a union of connected components
of $\operatorname{Rep}(M)$.
\end{clopen}

It follows from (\ref{noinvt}) that the holonomy of a strictly-convex structure determines a projective manifold up to projective isomorphism.
The Open Theorem is due to Koszul~\cite{Kos1, Kos2}. Our proof is distilled from his, and proceeds by showing that  $M$ is properly-convex if and only if
there is a projective $(n+1)$--manifold $N\cong M\times I$ with $M\times 0$  flat, and $M\times 1$
  convex, and triangulated so that adjacent $n$-simplices  
  are never coplanar. This type of convexity  is easily shown to be preserved by small deformations.

The Closed Theorem is due to Choi and Goldman~\cite{CW2} when $n=2$, to Kim~\cite{Kim} when $n=3$, and to Benoist~\cite{Ben5} in general.
Our proof is new, and based on a geometric argument called {\em the box estimate} (\ref{boxestimate}). This might be viewed
as related to Benzecri's compactness theorem \cite{Benz} for a properly-convex domain $\operatorname{O}mega$, but pertaining to $\operatorname{Aut}(\operatorname{O}mega)$. 
It also uses an elementary geometric fact (\ref{centerexists}) about an analogue of centroids for subsets of the sphere.

The Clopen Theorem follows from  the Open Theorem and the fact,  due to Benoist,  
 that a properly-convex manifold that is homeomorphic to a strictly
convex manifold is also strictly-convex (\ref{stricthomeo}). 
We have made an effort to make the paper  self-contained by including  proofs
of all the foundational results needed in \Cref{background,sec:cones}. 
We have endeavoured to simplify these proofs as much as possible.
See \cite{MR2464391} for an excellent survey.

There are extensions of the Open and Closed Theorems
when $M$ is the interior of a compact manifold with boundary, see \cite{CLT2, CT2, CT1}. 
Indeed, the techniques in this paper were developed to handle this more general situation for which the pre-existing
methods do not suffice.
But it seems useful to present some of the main ideas
in the simplest setting. We have liberally borrowed from \cite{CT2}.

 The first author thanks the University of Sydney Mathematical Research
Institute (SMRI) for partial support and hospitality while writing this paper. He also thanks the audience at SMRI that participated in a  presentation
of this material.
  Research of the second author is supported in part under the Australian Research Council's ARC Future Fellowship FT170100316.
 

\section{Properly and strictly-convex}\label{background}

This section and the next review some well known results concerning properly and strictly-convex projective manifolds that are needed to prove the theorems. 
The results needed subsequently are   (\ref{stricthomeo}), (\ref{noinvt}), (\ref{unique}), (\ref{nilpotent}), and  
  (\ref{C1}), (\ref{dualcpct}).
The only mild innovation
is that, to avoid appealing to results about word-hyperbolic groups, a more direct approach was taken to the proof of (\ref{stricthomeo}).
 The early sections of \cite{CLT1}  greatly expand on the background in this section.

 If $M$ is a manifold, the universal cover is $\pi_{M}:\widetilde M \to M$, and if $g\in\pi_1M$ then $\tau_g:\widetilde M\to\widetilde M$
 is the covering transformation  corresponding to $g$. A {\em geometry} is a pair $(G,X)$, where $G$ is a group that acts analytically and transitively 
 on a manifold $X$.
 A $(G,X)$-structure on a manifold $M$ is determined by a {\em development pair} $(\operatorname{dev},\rho)$ that consists of the {\em holonomy} $\rho\in \operatorname{Hom}(\pi_1M,G)$ and 
 the {\em developing map} $\operatorname{dev}:\widetilde M\to X$ which  is a local homeomorphism. The pair
 satisifies for all $x\in\widetilde M$ 
and $g\in\pi_1M$ that $\operatorname{dev}(\tau_{g}x)=(\rho g)\operatorname{dev} x$.

 
In what follows $V={\mathbb R}^{n+1}$, and $V^*=\operatorname{Hom}(V,{\mathbb R})$ is the dual vector space, and ${\mathbb R}^{n+1}_0=V_0=V\setminus \{0\}$.
{\em Projective space} is $\operatorname{{\mathbb P}} V=V_0/{\mathbb R}_0$, and $[A]\in\operatorname{Aut}(\operatorname{{\mathbb P}} V)=\operatorname{PGL}(V)$ acts on $\operatorname{{\mathbb P}} V$ by $[A][x]=[Ax]$.
{\em Projective geometry}  
 is $(\operatorname{Aut}(\operatorname{{\mathbb P}} V),\operatorname{{\mathbb P}} V)$ and is also written $(\operatorname{Aut}({\mathbb R}P^n),{\mathbb R}P^n)$.

We use the notation ${\mathbb R}^+=(0,\infty)$. {\em Positive projective space} is ${\mathbb R}P^n_+=\operatorname{{\mathbb P}}_+V=V_0/{\mathbb R}^+$ and 
$[x]_+=\{\lambda x:\lambda>0\}$ for $x\in V_0$. Sometimes we identify $\operatorname{{\mathbb P}}_+V$ with the unit sphere $S^n\subset{\mathbb R}^{n+1}$
via $[x]_+\equiv x/\|x\|$.
The group $\operatorname{Aut}(\operatorname{{\mathbb P}}_+V):=\operatorname{SL}(V)\subset\operatorname{GL}(V)$ is the subgroup  with $\det=\pm1$. 
{\em Positive projective geometry} $(\operatorname{Aut}(\operatorname{{\mathbb P}}_+V),\operatorname{{\mathbb P}}_+V)$ is the double cover of projective geometry. 
We will pass back and forth between projective geometry and positive projective geometry without mention, often omitting the term {\em positive}. 

If $U$ is a vector subspace of $V$ then $\operatorname{{\mathbb P}} U$ is a {\em projective subspace} of $\operatorname{{\mathbb P}} V$, and
is a {\em (projective) line} if $\dim U=2$ and a {\em (projective) hyperplane} if $\dim U=\dim V-1$.
The {\em dual} of $U$ is the projective subspace $\operatorname{{\mathbb P}} U^0\subset\operatorname{{\mathbb P}} V^*$ where  $U^0=\{\phi\in V^* :\phi(U)=0\}$.
We use the same terminology in positive projective geometry.
By lifting developing maps one obtains:

\begin{proposition}\label{holonomylifts} Every projective structure on $M$ lifts to a
positive projective structure. 
\end{proposition}
 
The {\em frontier} of a subset $X\subset Y$ is $\operatorname{Fr} X=\operatorname{cl}(X)\setminus\operatorname{int}(X)$, and the {\em boundary} is $\partial X=X\cap\operatorname{Fr} X$. 
A {\em segment} is a  connected, proper subset of a projective  line that contains more than one point.
In what follows $\operatorname{O}mega\subset{\mathbb R}P^n$. If $H$ is a hyperplane and $x\in H\cap\operatorname{Fr}\operatorname{O}mega$ and $H\cap\operatorname{int}\operatorname{O}mega=\emptyset$, then $H$ is a called a {\em supporting
hyperplane (to $\operatorname{O}mega$) at $x$}. 
The set $\operatorname{O}mega$
\begin{itemize}
\item is {\em convex} if every pair of points in $\operatorname{O}mega$ is contained in a segment in $\operatorname{O}mega$.
\item is {\em properly-convex} if it is convex,  and $\operatorname{cl}\operatorname{O}mega$ does not contain a projective line.
\item  is {\em strictly-convex} it it is properly-convex and $\operatorname{Fr}\operatorname{O}mega$
does not contain a segment. 
\item is {\em flat} if it is convex and $\dim\operatorname{O}mega<n$.
\item is a {\em convex domain} if it is a properly convex open set.
\item is a {\em convex body} if it is the closure of a convex domain.
\item is $C^1$ if  for each $x\in\operatorname{Fr}\operatorname{O}mega$
there is a unique supporting hyperplane  at $x$.
\end{itemize}

If $V=U\oplus W$,  then {\em projection (along $\operatorname{{\mathbb P}} W$) onto $\operatorname{{\mathbb P}} U$} is $\pi:\operatorname{{\mathbb P}} V\setminus\operatorname{{\mathbb P}} W\longrightarrow\operatorname{{\mathbb P}} U$
given by $\pi[u+w]=[u]$. {\em Duality} is the map that sends each point $\theta=[\phi]\in\operatorname{{\mathbb P}} V^*$ to the hyperplane
$H_{\theta}=\operatorname{{\mathbb P}}\ker\phi\subset\operatorname{{\mathbb P}} V$. If $L$ is a line in $\operatorname{{\mathbb P}} V^*$ the hyperplanes $H_{\theta}$ dual to the points 
$\theta\in L$
are called a {\em pencil of hyperplanes}. Then $Q=\cap H_{\theta}$ is the projective dual of $L$ and is called the {\em core of the pencil}.

\begin{lemma} A convex subset $\operatorname{O}mega$ is properly-convex  if and only if $\operatorname{cl}\operatorname{O}mega$ is disjoint from some hyperplane.
\end{lemma}
\begin{proof} Without loss of generality, suppose $\operatorname{O}mega$ is closed. The result is obvious if $n=1$. Since $\operatorname{O}mega$ is convex and contains no projective line, it is simply connected, and so lifts to $\operatorname{O}mega'\subset\mathbb R\operatorname{P}_+^n$.
Let $H\subset \mathbb R\operatorname{P}_+^n$ be a hyperplane. Then $H\cap \operatorname{O}mega'$ is empty or properly-convex. By induction
on dimension, $H$ contains a projective subspace $Q$  with $\dim Q=n-2$ that is disjoint from $H\cap \operatorname{O}mega'$. There is a pencil of hyperplanes $H_{\theta}$ with core $Q$. Now $\operatorname{O}mega'\cap H_{\theta}$ is contained in one of the two components of $H_{\theta}\setminus Q$. As $\theta$
moves half way round $\mathbb R\operatorname{P}_+^1$ the component must change. Thus for some $\theta$ the intersection is empty.
\end{proof}

In what follows $\operatorname{O}mega\subset {\mathbb R}P^n$ is a properly-convex domain, and $\operatorname{Aut}(\operatorname{O}mega)\subset\operatorname{Aut}({\mathbb R}P^n)$ is the subgroup that preserves $\operatorname{O}mega$.
  The {\em Hilbert metric $d_{\operatorname{O}mega}$} on $\operatorname{O}mega$ is  defined as follows.
If $\ell\subset{\mathbb R}P^n$ is a line and $\alpha=\operatorname{O}mega\cap\ell\ne\emptyset$ then $\alpha$ is a 
{\em proper segment in $\operatorname{O}mega$} and $\alpha=(a_-,a_+)$
with $a_{\pm}\in\operatorname{Fr}\operatorname{O}mega$. There is a projective isomorphism 
$f:\alpha\rightarrow{\mathbb R}^+$
and 
\[
d_{\operatorname{O}mega}(x,y)=\frac{1}{2}\left|\;\log \frac{f(x)}{f(y)}\;\right|
\] for $x,y\in\alpha$. A {\em geodesic} in $\operatorname{O}mega$ is a  curve whose length is the distance between its endpoints.
It follows that line segments are geodesics.
It is immediate that $\operatorname{Aut}(\operatorname{O}mega)$ acts by isometries of $d_{\operatorname{O}mega}$. 

 Let $H_{\pm}$ be supporting hyperplanes at $a_{\pm}$ and 
 $P=H_+\cap H_-$.  {\em Projection (along $P$) onto $\alpha$}  is the map $\pi:\operatorname{O}mega\to\alpha$ given by $\pi[u+v]=[v]$ 
 where $[u]\in P$ and $[v]\in\ell$.
The choice of $H_{\pm}$ is unique only if $a_{\pm}$ are $C^1$ points.  {\em In general $\pi x$ is not the closet point on $\alpha$
to $x$.} 

\begin{lemma}\label{projlemma} (i) $d_{\operatorname{O}mega}(x,y)\ge d_{\operatorname{O}mega}(\pi x,\pi y)$. \\
(ii) If $\operatorname{O}mega$ is strictly
convex then geodesics are segments of lines.
\end{lemma}
\begin{proof} \Cref{projection} represent  two {\em or more} dimensions.
Let $\delta=(b_-,b_+)$ be the proper segment in $\operatorname{O}mega$ containing $x$ and $y$, and $\pi:\operatorname{O}mega\rightarrow\alpha$ as above. Then $d_{\operatorname{O}mega}(x,y)=d_{\delta}(x,y)$. Since $(\pi|\delta):\delta\to\alpha$ is a projective embedding
$d_{\delta}(x,y)=d_{\pi\delta}(\pi x,\pi y)$. Also $d_{\pi\delta}(\pi x,\pi y)\ge d_{\alpha}(\pi x,\pi y)$ because 
$\pi\delta\subset\alpha$. Finally $d_{\alpha}(\pi x,\pi y)=d_{\operatorname{O}mega}(\pi x,\pi y)$, which proves (i).
\begin{figure}
\caption{Projecting onto a line}
\label{projection}
\end{figure}

Equality implies $\alpha=\pi\delta$. Thus, after relabelling if needed,  $b_{\pm}\in H_{\pm}$. Thus $[a_{\pm},b_{\pm}]\subset H_{\pm}\cap\operatorname{Fr}\operatorname{O}mega$.
If $\operatorname{O}mega$ is strictly-convex it follows that $b_{\pm}=a_{\pm}$. This gives (ii).
\end{proof}

In a strictly convex domain $\operatorname{O}mega$ will use the term {\em geodesic} to mean  a proper segment in $\operatorname{O}mega$.
A subset $\mathcal C\subset V_0$ is a {\em cone} if $t\cdot\mathcal C=\mathcal C$ for all $t>0$.
  The {\em dual cone} $\mathcal C^*=\operatorname{int}\left(\{\phi\in V^*:\ \phi(\mathcal C)\ge 0\}\right)$ is convex.  
  If $\operatorname{O}mega\subset\operatorname{{\mathbb P}}_+ V$ then $\mathcal C\operatorname{O}mega$ is the cone $\{v\in V_0:\ [v]_+\in\operatorname{O}mega\}$.
  If $\operatorname{O}mega$ is open and properly-convex, then the {\em dual domain} 
  $\operatorname{O}mega^*=\operatorname{{\mathbb P}}(\mathcal C\operatorname{O}mega^*)$ is open, and properly-convex. Using the natural identification $V\cong V^{**}$
  it is immediate that $(\operatorname{O}mega^*)^*=\operatorname{O}mega$.
 \begin{corollary}\label{C1domain} If $\operatorname{O}mega$ is open and properly-convex,  
 then $\operatorname{O}mega$ is $C^1$ (resp. strictly convex)
 if and only if $\operatorname{O}mega^*$ is strictly-convex (resp. $C^1$).
 \end{corollary}
 \begin{proof} We have $[\phi]\in\operatorname{Fr}\operatorname{O}mega^*$ if and only if the dual hyperplane $H=[\ker\phi]$ 
 supports $\operatorname{O}mega$.
If $p\in\operatorname{Fr}\operatorname{O}mega$, then there are
two distinct supporting hyperplanes $H_0=[\ker\phi_0]$ and $H_1=[\ker\phi_1]$ for $\operatorname{O}mega$ at $p$ 
if and only if 
all the hyperplanes $H_t=[\ker \phi_t]$ contain $p$ and support $\operatorname{O}mega$, where $\phi_t=(1-t)\phi_0+t\phi_1$ and $0\le t\le 1.$
This happens if and only if the segment
$\sigma=\{[\phi_t]:\ 0\le t\le 1\}$ is in $\operatorname{Fr}\operatorname{O}mega^*$. Hence $\operatorname{O}mega$ is $C^1$ if and only if $\operatorname{O}mega^*$ is strictly
convex. Replacing $\operatorname{O}mega$ by $\operatorname{O}mega^*$ and using $\operatorname{O}mega^{**}=\operatorname{O}mega$ gives the other result.
\end{proof}
 

 If $\Gamma\subset\operatorname{Aut}(\operatorname{O}mega)$
is  a discrete and torsion-free subgroup, then
 $M=\operatorname{O}mega/\Gamma$   is a {\em properly (resp. strictly) convex} manifold if $\operatorname{O}mega$ is properly (resp. strictly) convex.
Then $\widetilde M=\operatorname{O}mega$
and there is a development pair $(\operatorname{dev},\rho)$, where $\operatorname{dev}$ is the inclusion map, and the holonomy $\rho:\pi_1M\to\Gamma$ is an isomorphism.

 \begin{lemma}\label{htpyequiv} Suppose $M$ and $M'$ are properly-convex manifolds and
 $\pi_1M\cong\pi_1M'$. If $M$ is
closed, then $M'$ is closed.
\end{lemma}
\begin{proof} Let $\pi:\operatorname{O}mega\to M$ be the projection. This is the universal covering space of $M$, so $M$ is
a $K(\pi_1M,1)$. The same holds for $M'$. Hence $M$ and $M'$ are homotopy equivalent. If $n=\dim M$, then $M$ is closed if and only if 
$H_n(M;{\mathbb Z}_2)\cong{\mathbb Z}_2$. Since homology is an invariant of homotopy type the result follows.\end{proof}

A compactness argument shows that if $\operatorname{O}mega$ is the universal cover of a closed strictly-convex manifold, then projection 
(along a projective subspace) onto a geodesic in $\operatorname{O}mega$ is
uniformly distance decreasing in the following sense.
 
 \begin{lemma}\label{strictprojection} Suppose $M=\operatorname{O}mega/\Gamma$ is a strictly-convex closed manifold. Given  $b>1$ there is  $R=R(b)>0$ so the following holds.
Suppose $\pi:\operatorname{O}mega\to\alpha$ is a projection  onto a geodesic $\alpha$.  Suppose that $\eta$ is a rectifiable arc in
 $\operatorname{O}mega\setminus N_R(\alpha)$
of length $\ell$ with endpoints $x$ and $y$. Then $d_{\operatorname{O}mega}(\pi x,\pi y)< 1+(\ell/b)$ \end{lemma}
 \begin{proof} By decomposing $\eta$ into  finitely many subarcs, with one of length at most $b$, and $\lfloor \ell/b \rfloor$ of length $b$,
 it suffices to prove if $length(\eta)\le b$ then $d_{\operatorname{O}mega}(\pi x,\pi y)\le 1$.
  
 Write $d=d_{\operatorname{O}mega}$. If no such $R$ exists  then there are sequences $x_k,y_k\in\operatorname{O}mega$ and projections $\pi_{_k}:\operatorname{O}mega\to\alpha_k$
 with $d(x_k,y_k)\le b$ and $d(x,\alpha_k)\ge k$ and $1\le d(\pi_{_k} x_k,\pi_{_k} y_k)\le b$. Then $\beta_k=[\pi x_k,x_k]$ and $\gamma_k=[\pi_{_k} y_k,y_k]$
 are geodesic segments and $\pi_{_k}\beta_k=\pi_{_k} x_k $ and $\pi_{_k}\gamma_k=\pi_{_k}y_k$. 
 
   There is a compact $W\subset\operatorname{O}mega$ such that $\Gamma\cdot W=\operatorname{O}mega$. By applying an element of $\Gamma$ we may assume that $\pi_{_k}x_k\in W$.
  After subsequencing we may assume $\beta=\lim \beta_k$ and $\gamma=\lim \gamma_k$  and $\alpha=\lim\alpha_k$ and $\pi=\lim \pi_{_k}$ all exist and $\pi:\operatorname{O}mega\to\alpha$ is
   projection. Refer to \Cref{projection}. Thus $\beta$ are $\gamma$ are geodesic rays in $\operatorname{O}mega$ that start on $\alpha$, and end on $\operatorname{Fr}\operatorname{O}mega$, with
  $\pi(\beta)=\alpha\cap\beta\ne \alpha\cap\gamma=\pi(\gamma)$.
  

 Since $\operatorname{O}mega$ is strictly-convex, and $d(x_k,y_k)\le b$, it follows that $x_k$ and $y_k$ 
  limit on the same point  $p\in \operatorname{Fr}\operatorname{O}mega$. Hence $p$ is the endpoint of both $\beta$ and $\gamma$ on $\operatorname{Fr}\operatorname{O}mega$. 
  Also $p\in P=H_-\cap H_+$, where $H_{\pm}$ are the supporting hyperplanes to $\operatorname{O}mega$ at the endpoints of $\alpha$.
   Since $\operatorname{O}mega$ is strictly-convex, $p\notin P$. It follows that $\pi(\beta)=\pi(p)=\pi(\gamma)$ which is a contradiction.
   \end{proof}
 
 If $X$ is a subset of a metric space $Y$, the $r$-neighborhood of $X$ is $N_r(X)=N(X,r)=\{y\in Y: d(y,x)\le r\}$.
 

  
 The following is due to Benoist \cite{MR2094116}.
    A {\em triangle}  is a disc $\Delta$ in a projective plane bounded by
 three segments.   If $\Delta\subset\operatorname{cl}\operatorname{O}mega$ and $\Delta\cap\operatorname{Fr}\operatorname{O}mega=\partial\Delta$, then $\Delta$ is
  called a {\em properly embedded triangle} or {\em PET}.
  A properly-convex set $\operatorname{O}mega$ has  {\em thin triangles} if there is $\delta>0$ such that for every triangle
 $T$ in $\operatorname{O}mega$, each side of $T$ is contained in a $\delta$-neighborhood of the union of the other two sides with respect to the Hilbert metric.
  
  \begin{proposition}\label{PETstrict} If $M=\operatorname{O}mega/\Gamma$ is a properly-convex closed manifold, then the following are equivalent:
  \begin{enumerate}
\item  $M$ is strictly-convex,
\item $\operatorname{O}mega$ does not contain a PET, 
\item $\operatorname{O}mega$ has thin triangles.
\end{enumerate}
 \end{proposition}
 \begin{proof}  If $\operatorname{O}mega$ is not strictly-convex, there is a maximal segment $\ell\subset\operatorname{Fr}\operatorname{O}mega$. Choose $x\in\operatorname{O}mega$ and let $P\subset\operatorname{O}mega$ 
be the interior of the triangle that is the convex hull of $x$ and $\ell$. Choose a sequence $x_n$ in $P$ that limits on the midpoint of $\ell$.
Then $d_{\operatorname{O}mega}(x_n,\operatorname{Fr} P)\to\infty$ because $\ell$ is maximal. Since $M$ is compact there is compact set $W\subset\operatorname{O}mega$
such that $\Gamma\cdot W=\operatorname{O}mega$. Thus there is $\gamma_n\in\Gamma$ with $\gamma_nx_n\in W$. After
choosing a subsequence  $\gamma_nx_n\to y\in W$, and $\gamma_n P$ converges to the interior of a PET
 $\Delta\subset\operatorname{O}mega$ that contains $y$. 
 Since $\Delta$ is flat $d_{\operatorname{O}mega}|\Delta=d_{\Delta}.$ Now $\operatorname{PGL}(\Delta)$ contains a subgroup $G\cong{\mathbb R}^2$ that acts transitively on $\Delta$.
Therefore $(\Delta,d_{\Delta})$ is isometric to a normed vector-space, thus does not
have thin triangles.

Conversely if $\operatorname{O}mega$ does not have thin triangles, then there is a sequence  of triangles $T_k$ in $\operatorname{O}mega$
and points $x_k$ in $T_k$
with $d_{\operatorname{O}mega}(x_k,\partial T_k)>k$. As above, after applying elements of $\Gamma$, we may assume $T_k$
converges to a PET, so $\operatorname{O}mega$ is not strictly-convex.
 \end{proof}

   A {\em $(K,L)$--quasi-isometric embedding}  is a map $f:X\to Y$ between metric spaces $(X,d_X)$ and $(Y,d_Y)$ such that
  $$K^{-1}\;d_X(x,x')-L\;\le\; d_Y(fx,fx')\;\le\; K\;d_X(x,x')+L $$
for all $x,x'\in X.$
     If $X=[a,b]\subset {\mathbb R}$, then $f$ is called {\em quasi-geodesic}.    
  The map $f$ is a {\em quasi-isometry} or {\em QI} if $Y\subset N(fX,L)$. 
 
  If $g:Y\to Z$ is a $(K',L')$--QI, then $g\circ f$ is a $(KK',K'L+2L')$--QI.   
           In particular, if $g$ is a QI 
   and $f$ is a quasi-geodesic, then $g\circ f$ is a quasi-geodesic
   with QI constants that only depend on
those of $f$ and $g$.
  

  If $G$ is a finitely generated group then  a choice of finite generating set gives a {\em word metric} on $G$.
   A different choice of generating set gives a bi-Lipschitz equivalent metric. The \v{S}varc-Milnor lemma in this setting is

  \begin{proposition}\label{QIhomeo} If $M=\operatorname{O}mega/\Gamma$ is a closed and properly-convex manifold,
 then $(\operatorname{O}mega,d_{\operatorname{O}mega})$ is QI to $\pi_1M$. \end{proposition}
 \begin{proof} Fix $x\in\widetilde M=\operatorname{O}mega$. Then $f:\pi_1M\rightarrow\widetilde M$ given by $f(g)=\tau_g(x)$ is a QI.
  \end{proof}
 
   A metric space $(Y,d_Y)$ is {\em ML} if it satisfies the  {\em Morse Lemma} that for all $(K,L)$ there is $S=S(K,L)>0$, called a {\em tracking constant}, such that if $\alpha$ and $\beta$ are $(K,L)$-quasi-geodesics in $Y$
  with the same endpoints,  then  $\alpha\subset N_{S}(\beta)$ and $\beta\subset N_{S}(\alpha)$. 
  Since quasi-geodesics are sent to quasi-geodesics by quasi-sometries, it follows that the property of ML is preserved by quasi-isometry.

  Clearly ${\mathbb R}^2$ is not ML, and any norm on ${\mathbb R}^2$ is QI to the standard norm, so ${\mathbb R}^2$ with any norm is not ML.
  A geodesic in the Cayley graph picks out a sequence in $G$ that is a quasi-geodesic.
 It follows from (\ref{QIhomeo}) that whether or not a closed properly convex manifold is ML only depends on $\pi_1M$. 
  
 
 \begin{proposition}\label{stricthyperbolic} If $M=\operatorname{O}mega/\Gamma$ is a properly-convex closed manifold, then $M$ is strictly-convex if and only if $(\operatorname{O}mega,d_{\operatorname{O}mega})$ is {\em ML}.
 \end{proposition}
 \begin{proof} If $\operatorname{O}mega$ is not strictly-convex, then by (\ref{PETstrict}) it contains a PET $\Delta$ 
 which is isometric to a norm on ${\mathbb R}^2$,  and therefore is not {\em ML}.

\begin{figure}
\caption{Quasi-geodesics}
\label{quasigeodesic}
\end{figure}

 Now suppose $\operatorname{O}mega$ is strictly-convex. Given $(K,L)$  let $R=R(2K)$ be given by (\ref{strictprojection}). Let $S'=R+4RK+K+L$.
 We show that $S=2KS'+2L$ is a tracking constant.
Refer to \Cref{quasigeodesic}. Let $\alpha'$ be a $(K,L)$--quasi-geodesic in $\operatorname{O}mega$ with endpoints $p$ and $q$, and $\beta'=[p,q]$.
For simplicity we assume $\alpha'$ is continuous.  Let $\beta^+$ be the geodesic that contains $\beta'$ and $\pi:\operatorname{O}mega\rightarrow\beta^+$ be the projection.\\
 Claim 1) $\alpha'\subset N(\beta^+,S')$.\\
Claim 2) $\alpha'\subset N(\beta',S/2)$ and $\beta'\subset N(\alpha',S/2)$.

 Assuming this, if $\gamma'$ is another $(K,L)$-quasi-geodesic with endpoints $p$ and $q$ then
 $\beta'\subset N(\gamma',S/2)$ so $$\alpha'\subset N(\beta',S/2)=N(N(\gamma',S/2),S/2)=N(\gamma',S)$$
 which proves that $S$ is a tracking constant.

 
For Claim 1 suppose that $\alpha$  is the closure of a component of $\alpha'\setminus N(\beta^+,R)$. The endpoints
 $x$ and $y$ of $\alpha$ satisfy $d(x,\beta^+)=d(y,\beta^+)=R$. Let $x',y'\in\beta^+$ be chosen so that $d(x,x')=R=d(y,y')$ and 
 let $\beta=[\pi x,\pi y]\subset\beta^+$.  Then $$\operatorname{length}(\beta)\le  1+\operatorname{length}(\alpha)/2K$$ by definition of $R$. 
Since $\pi$ is distance non-increasing $d(x',\pi x)\le d(x', x)=R$, and similarly $d(y',\pi y)\le R$. Thus
 $$d(x,y)\le d(x,x')+d(x',\pi x)+d(\pi x,\pi y)+d(\pi y,y')+d(y',y)\le 4R+\operatorname{length}(\beta)$$
  Since $\alpha$
 is a $(K,L)$-quasi-geodesic $$\begin{array}{rcl}
 \operatorname{length}(\alpha) &\le & K d(x,y)+L \\
 & \le & K(4R +\operatorname{length}(\beta))+L\\
 & \le & K(4R+1+\operatorname{length}(\alpha)/2K)+L\\
 & = &4RK+K+L+\operatorname{length}(\alpha)/2
 \end{array}$$
 Hence $\operatorname{length}(\alpha)\le 2(4RK+K+L)$. It follows that the distance  of every point on $\alpha'$ from $\beta^+$ is at most
 $$R+(1/2)\operatorname{length}(\alpha)=R+4RK+K+L=S'$$
 This proves Claim 1.
  
For Claim 2, there is a maximal subarc $\delta\subset\alpha'$ that starts at $p$ and ends
 at a point $w\in\alpha'$ such that $\pi w=p$. Possibly the arc is trivial with $w=p$.  By the above 
 $d(w,\beta^+)\le S'$. The arc $\delta$ is a $(K,L)$--quasi-geodesic because it is a subarc of $\alpha$, so $d(w,\beta^+)\ge K^{-1}(\operatorname{length}(\delta)-L)$.
 Thus $\operatorname{length}(\delta)\le K S'+L=S/2$. Using $\pi$ is distance non-increasing, gives $\operatorname{length}(\pi\delta)\le S/2$.
  Now $\pi\alpha'$ extends beyond $\beta'$ by $\pi\delta$ so $\pi\alpha'\subset N(\beta',S/2)$.
  Also $\beta'\subset\pi\alpha'$ so $\beta'\subset N(\alpha',S')$ and $K\ge 1$, so $\beta'\subset N(\alpha',K S'+L)$.
  This proves the claim. \end{proof}
 
 
 
\begin{corollary}\label{stricthomeo} Suppose $M$ and $N$ are closed and properly-convex, and that $\pi_1M\cong\pi_1N$. If $M$ is strictly-convex, then $N$ is strictly-convex.
\end{corollary}
\begin{proof} Since ML is preserved by quasi-isometry, it follows from (\ref{stricthyperbolic}) and (\ref{QIhomeo}) that  $M$ is strictly convex if and only if  $\pi_1M$ is ML, and this is determined by $\pi_1M$.
\end{proof}

\begin{lemma}\label{noinvt} If $M=\operatorname{O}mega/\Gamma$ is a closed properly-convex manifold and  
$\operatorname{O}mega'\subset\operatorname{O}mega$ is a non-empty properly-convex subset
that is preserved by $\Gamma$, then $\operatorname{O}mega'=\operatorname{O}mega$.
\end{lemma}
\begin{proof} Otherwise the function $F:\operatorname{O}mega\to {\mathbb R}$ given by $F(x)=d_{\operatorname{O}mega}(x,\operatorname{O}mega')$ is continuous, unbounded, and $\Gamma$--invariant.
Thus it covers a continuous unbounded function $f:M\to{\mathbb R}$, contradicting the compactness of $M$.
\end{proof}

    The {\em displacement distance of $\gamma\in\operatorname{Aut}(\operatorname{O}mega)$} is
$t(\gamma)=\inf\{d_{\operatorname{O}mega}(x,\gamma x): x\in\operatorname{O}mega\}$. The element $\gamma$ is {\em hyperbolic} if $t(\gamma)>0$. 

\begin{lemma}[Hyperbolics]\label{hyperbolic} Suppose $\operatorname{O}mega\subset \operatorname{{\mathbb P}} V$ and $M=\operatorname{O}mega/\Gamma$ 
is a strictly-convex closed manifold and $1\ne \gamma\in\Gamma$. 

Then
 $\gamma$ is hyperbolic and there are
$a_{\pm}\in\operatorname{Fr}(\operatorname{O}mega)$ such that for all $x\in\operatorname{{\mathbb P}} V\setminus (H_+\cup H_-)$ we have
\[\lim_{n\to\pm\infty}\gamma^nx=a_{\pm}\]
where $H_{\pm}$ is the supporting hyperplane to $\operatorname{O}mega$ at $a_{\pm}$.
\end{lemma}
\begin{proof} Since $M$ is compact, the Arzela-Ascoli Theorem implies  there is a closed
 geodesic $C$ in $M$ that is conjugate to $\gamma$ in $\pi_1M$
and $t(\gamma)=\operatorname{length}(C)>0$. Hence $\gamma$ is hyperbolic. By (\ref{projlemma}) $C$ is covered by a proper segment $\alpha=(a_-,a_+)$ in $\operatorname{O}mega$
that is preserved by $\gamma$.
By (\ref{dualcpct}) $M$ is $C^1$ so there are   unique supporting hyperplanes $H_{\pm}$ 
to $\operatorname{cl}\operatorname{O}mega$ that contain $a_{\pm}$ respectively. 
 Then $Q=H_+\cap H_-$
is a codimension--2 subspace, and it is disjoint from $\operatorname{O}mega$ by {\em strict} convexity. Moroever $Q$ is
 preserved by $\gamma$. 

 The pencil of hyperplanes $\{H_t:t\in L\}$ in $\operatorname{{\mathbb P}} V$ that contain $Q$ is dual to a line $L\subset\operatorname{{\mathbb P}}(V^*)$.
Since the dual of $\gamma$ acts on  $L\cong{\mathbb R}P^1$ projectively and non-trivially, it
 only fixes  the two points $[H_{\pm}]\in L$. The other hyperplanes $[H_t]$ are moved by $\gamma$ away from $[H_-]$ and towards $[H_+]$. Since $\operatorname{O}mega$ is strictly-convex, $\gamma$ moves all points in $\operatorname{O}mega$  towards $a_+$. 
Suppose $\ell$ is a projective line that contains $a_-$. If $\ell$ is not contained
in $H_-$ then, since $H_-$ is the unique supporting hyperplane at $a_-$, it follows that $\operatorname{O}mega\cap \ell\ne\emptyset$.
Thus $\gamma^k\ell\to\ell'$ where $\ell'$ is the projective line containing $\alpha$. Hence $\gamma^k(\ell\setminus a_-)\to a_+$
as $k\to\infty$.
  This reasoning applied to $\gamma^{-1}$ gives the corresponding statements for $a_-$, and gives the second conclusion.\end{proof}

The point $a_+$ is called the {\em attracting}, and $a_-$ is the {\em repelling}, fixed point of $\gamma$,
and $(a_-,a_+)\subset \operatorname{O}mega$ is called the {\em axis} of $\gamma$.  This axis is the only proper segment in $\operatorname{O}mega$ preserved by $\gamma$. The attracting fixed point of $\gamma^{-1}$ is the repelling fixed point of $\gamma$.

    
  \begin{proposition}[Unique domain]\label{unique} Let $n\ge 2.$ Suppose $\operatorname{O}mega\subset{\mathbb R}P^n$ and $M=\operatorname{O}mega/\Gamma$ is a strictly-convex, closed $n$--manifold.
If $\operatorname{O}mega'$ is open and properly-convex and preserved by $\Gamma$, then $\operatorname{O}mega'=\operatorname{O}mega$.\end{proposition}
\begin{proof} 
 Let $X$ be union, over all $1\ne\gamma\in\Gamma$,  of attracting  fixed points of $\gamma$.
 Then $X\subset\operatorname{Fr}\operatorname{O}mega$ by (\ref{hyperbolic}).
Let $W$ be the  convex hull of $X$ in $\operatorname{cl}\operatorname{O}mega$.
Then $U=W\cap\operatorname{O}mega$ contains the axis of each hyperbolic in $\Gamma$, so $U$ is non-empty, convex, 
$\Gamma$-invariant. Since $U\subset\operatorname{O}mega$, we have $\operatorname{O}mega=U$ by (\ref{noinvt}).
Since $\partial\operatorname{O}mega$ is strictly-convex it follows $\operatorname{Fr}\operatorname{O}mega=\operatorname{cl} X$.

Since $\operatorname{O}mega'$ is preserved by $\Gamma$ it follows that $\operatorname{cl}\operatorname{O}mega'\supset \operatorname{cl} X$. Now $\operatorname{Fr}\operatorname{O}mega$ is a
convex hypersurface of dimensions $n-1>0$. 
Only one side of $\operatorname{Fr}\operatorname{O}mega$ is locally convex.
Hence $\operatorname{O}mega'$ contains points on the same side of $\operatorname{Fr}\operatorname{O}mega$ as $\operatorname{O}mega$. Thus $U=\operatorname{O}mega'\cap\operatorname{O}mega\ne\emptyset$ is
properly convex and
 is preserved by $\Gamma$.
 By (\ref{noinvt}) $\operatorname{O}mega=U=\operatorname{O}mega'$.\end{proof}

\begin{corollary}[nilpotent subgroups]\label{nilpotent}  If $M=\operatorname{O}mega/\Gamma$ is a closed, strictly-convex, projective manifold,  then 
every nilpotent subgroup of $\Gamma$ is cyclic.\end{corollary}
\begin{proof}  By (\ref{hyperbolic}) every element of $\Gamma$ is hyperbolic, and the axis is the only segment preserved by a non-trivial hyperbolic in $\Gamma$.
Suppose $1\ne \alpha,\beta\in\Gamma$ and $[[\alpha,\beta],\beta]=1$. Let $\ell$ be the axis of $\beta$. Since $[\alpha,\beta]$ 
commutes with $\beta$, it preserves $\ell$.
Now $\beta$ preserves $\ell$ and $[\alpha,\beta]=(\alpha\beta\alpha^{-1})\beta^{-1}$ so $\alpha\beta\alpha^{-1}$ also preserves $\ell$.  
But $\alpha\beta\alpha^{-1}$ is a hyperbolic that preserves $\alpha\ell$, thus  $\alpha\ell=\ell$.
If follows by induction on the length of the upper central series
 that if $\Gamma'\subset\Gamma$ is nilpotent then $\Gamma'$ preserves an axis.

The action of $\Gamma$ on $\operatorname{O}mega$
is free, so $\Gamma'$ acts freely on $\ell$. A discrete group acting freely by homeomorphisms on ${\mathbb R}$ is cyclic.
\end{proof}

\section{Convex Cones}\label{sec:cones}

This section is based on work of Vinberg, as simplified by Goldman.
Write $V={\mathbb R}^{n+1}$ and
fix an inner product $\langle\;\cdot\;,\;\cdot\;\rangle$ on $V$. This determines a norm on $V$, and  induces a Riemannian metric
and associated volume form
on every smooth submanifold of $V$.

Given $\phi\in\mathcal C\operatorname{O}mega^*$, the centroid of $\mathcal C_{\phi}=\phi^{-1}(1)\cap\mathcal C\operatorname{O}mega$ is a point $\mu(\mathcal C_{\phi})\in\mathcal C\operatorname{O}mega$.
Define $\Theta(\phi)=\mu(\mathcal C_{\phi})$. We show this is a  homeomorphism $\Theta:\mathcal C\operatorname{O}mega^*\rightarrow\mathcal C\operatorname{O}mega$. 

\begin{figure}
\caption{$\Theta:\mathcal C\operatorname{O}
\label{Thetapicnew}
\end{figure}

 Let $S^n=\{x\in V:\ \|x\|^2=1\}$.
Suppose $\operatorname{O}mega$ is an open, properly-convex subset of $S^n$ and $\mathcal C=\mathcal C\operatorname{O}mega$ is the corresponding cone in $V$.
 The {\em dual cone} is 
$$\mathcal C^*=\operatorname{int}\left(\{\phi\in V^*:\ \phi(\mathcal C)\ge0\ \}\right)$$
 The {\em centroid} of a bounded convex set
$K$ in $V$ is the point $\mu(K)$ in $K$ given by
 $$\mu(K)=\left.\int_K x\operatorname{dvol}_K\right/\int_K \operatorname{dvol}_K$$
Here $\dim K\le \dim V$ and $\operatorname{dvol}_K$ is the induced volume form on $K$. The centroid is independent of the inner product.

Given $\phi\in\mathcal C^*$, the set $\phi^{-1}(1)$ is an affine hyperplane in $V$, and
$\mathcal C_{\phi}=\{x\in\mathcal C: \phi(x)=1\}$ is a bounded subset of this hyperplane that separates $\mathcal C$.
The subset of $\mathcal C$ below this hyperplane, $\mathcal C\cap\phi^{-1}(0,1]$, has finite volume and  boundary $\mathcal C_{\phi}$.
 The {\em volume function} $\mathcal V:\mathcal C^*\to{\mathbb R}$  is defined by
 \begin{align}\label{Vdefa}\mathcal V(\phi)=\operatorname{vol}(\mathcal C\cap\phi^{-1}(0,1])=\int_{\mathcal C\cap\phi^{-1}(0,1]}\operatorname{dvol}\end{align}Given $\phi\in\mathcal C^*$, 
 there is $v\in V$ such that $\phi(x)=\langle v,x\rangle$ for all $x\in V$. 
Let $\operatorname{dB}$ be the  volume form on $S^n$. We  compute this integral in
polar coordinates, so
 $\operatorname{dvol}=r^{n} \operatorname{dr}\wedge\operatorname{dB}$.
Given $x\in \mathcal C_{\phi}$, let $y=x/\|x\|$ be the corresponding point in $S^n$.
Using $\phi(x)=1$ gives  \begin{align}\label{reqtn}
r(x)=\|x\|=\phi(x)/\phi(y)=\langle y,v\rangle^{-1}\quad{\rm and}\qquad x=\langle y,v\rangle^{-1}y\end{align}
Using polar coordinates
\begin{align}\label{Vformula}\mathcal V(\phi)=\int_{C_{\phi}}r^n\operatorname{dr}\wedge\operatorname{dB}=
\int_{\operatorname{O}mega}\left(\int_0^{\langle y,v\rangle^{-1}} r^n \operatorname{dr}\right)\operatorname{dB}_y
=(n+1)^{-1}\int_{\operatorname{O}mega}\langle y,v\rangle^{-n-1}\; \operatorname{dB}_y\end{align}
For $q\in\mathcal C$, the set $\mathcal C^*_q=\{\phi\in\mathcal C^*:\ \phi(q)=1\ \}$ 
is the intersection of a hyperplane in $V^*$ with $\mathcal C^*$, and has compact closure. Property (iv) below is very useful later.

\begin{proposition}\label{Vprop}The volume function has the following properties:
\begin{itemize}
\item[(i)]  $\mathcal V$ is smooth and strictly-convex.\\

\item[(ii)] $\mathcal V(\phi)\to\infty$ as $\phi\to\operatorname{Fr}\mathcal C^*$.\\

\item[(iii)] $\mathcal V(t\cdot\phi)=t^{-n-1}\mathcal V(\phi)$ for all $t>0$.\\

\item[(iv)]
There is a unique $\phi\in \mathcal C^*_q$ at which $\mathcal V|\mathcal C^*_q$ attains a minimum, and   $q=\mu(\mathcal C_{\phi})$.
\end{itemize}
\end{proposition}
\begin{proof} We use the inner product to identify $V$ with $V^*$
so that $\mathcal C^*$ is a subset of $V$, and $\mathcal C_q^*=\{x\in \mathcal C^*:\langle x,q\rangle=1\}$.
 We also
 regard $\mathcal V(\phi)$ as the function $\mathcal W(v)$  given by \Cref{Vformula} and prove  corresponding statements
for $\mathcal W$. 

For fixed $y$ the integrand $f(v)=\langle y,v\rangle^{-n-1}$ in \Cref{Vformula} is a smooth and convex function of $v$.
It follows that $\mathcal W(v)$ is a smooth, {\em strictly} convex function of  $v$. This proves (i).

If $0\ne\psi\in\operatorname{Fr}\mathcal C^*$, then there is $0\ne x\in\operatorname{Fr}\mathcal C$ with $\psi(x)=0$.
Thus ${\mathbb R}^+\cdot x\subset\operatorname{cl}\mathcal C_{\psi}$, so $\mathcal C_{\psi}$ is not compact. Moreover $\mathcal C_{\psi}$ is convex
and has non-empty interior, so $\operatorname{vol}(\mathcal C_{\psi})=\infty$. It easily follows that $\mathcal V(\phi)\to\infty$
as $\phi\to\psi$. This proves (ii), and (iii) follows from \Cref{Vdefa}. It follows from convexity and (ii)
that $\mathcal V|\mathcal C^*_q$ has a unique critical
point, and it is a minimum.

   The gradient of $\mathcal W(v)$ is
\begin{align}\label{gradV} \nabla \mathcal W =-\int_{\operatorname{O}mega} \langle y,v\rangle^{-n-2} y\;\operatorname{dB}_y
\end{align}
 From the definition of $\mathcal C^*_q$ it follows that the condition for a critical point  is that $\nabla \mathcal W\in{\mathbb R}\cdot q$.
 
 \begin{figure}
\caption{Volume forms on $\operatorname{O}
\label{dAdBfig}
\end{figure}
 Refer to \Cref{dAdBfig}. Let $v\in V$ be dual to $\phi$, so $\phi(x)=\langle v,x\rangle$. Thus $\phi(v\|v\|^{-2})=1$ and
$v\|v\|^{-2}$ is the point on $\phi^{-1}(1)$ closest to $0$. Thus
the distance of $\phi^{-1}(1)$ from $0$ is $\|v\|^{-1}$.  Let 
$\pi:\mathcal C_{\phi}\to \operatorname{O}mega=S^n\cap\mathcal C$ be the radial projection $\pi(x)=x/\|x\|$. Given $x\in \mathcal C_{\phi}$ set $y=\pi(x)$ so $\|y\|=1$,
and define $\cos\theta=\langle y,v\rangle/\|v\|$. Since $\langle x,v\rangle=1$ it follows that  $x=y/\langle y,v\rangle$ and
 so $\|x\|=1/\langle y,v\rangle$.

  Let $S_x\subset{\mathbb R}^{n+1}$ denote the sphere with center $0$ and radius $\|x\|$. The volume form on $S_x$ is $d_{S_x}=\|x\|^{n}\pi_1^*dB$, where 
  $\pi_1:S_x\rightarrow S^n$ is radial projection. 
 Let 
 $d A_x$ be  the  volume element on $\mathcal C_{\phi}$. Then $dA_x=(\cos\theta)^{-1}\pi_2^*d_{S_x}$,
 where $\pi_2:\mathcal C_{\phi}\rightarrow S_x$ is radial projection.  Then 
  $\pi=\pi_1\circ\pi_2:\mathcal C_{\phi}\rightarrow S^n$ is radial projection, and
  combining gives
\begin{equation}\label{dAdB}
d A_x=(\cos\theta)^{-1}\|x\|^{n}\; \pi^*d B_y=  \|v\|\langle y,v\rangle^{-n-1}\; \pi^*d B_y
\end{equation}
It follows from \Cref{gradV,dAdB} that $$-\nabla \mathcal W =\|v\|^{-1}\int_{\mathcal C_{\phi}}x \; \operatorname{dA}_x=\|v\|^{-1}\mu(\mathcal C_{\phi})$$
Thus  $\mathcal W$ has a minimum when
$$\|v\|^{-1}\mu(\mathcal C_{\phi})\in{\mathbb R}\cdot q$$
Since the centroid $\mu(\mathcal C_{\phi})$ is in $\mathcal C_{\phi}$, and $q\in \mathcal C_{\phi}$, it follows that at the minimum we have $\mu(\mathcal C_{\phi})=q.$
\end{proof}

The volume function gives a natural, strictly-convex hypersurface in $\mathcal C$:
\begin{definition}\label{SPdef} 
Suppose $\mathcal C$ is a  properly-convex cone.  Let $\mathcal D\subset\mathcal C$ be the intersection of all affine halfspaces with the property that the volume of the subset
of $\mathcal C$ outside the halfspace equals one.
\end{definition} 
 In particular $\partial\mathcal D$ is strictly-convex
and meets every ray in $\mathcal C$ once. Clearly $\partial\mathcal D$ is preserved by $\operatorname{SL}(\mathcal C)$.
If $q\in \partial\mathcal D$, then with $\phi$ as in (\ref{Vprop})(iv) the tangent hyperplane to $\partial\mathcal D$ at $q$
is $\phi^{-1}(1)$, and $q=\mu(\mathcal C_{\phi})$. 
In other words, the part inside $\mathcal C$ of the tangent plane to $\mathcal D$ at a point $q\in\partial\mathcal D$
  has centroid $q$.

 
\begin{corollary}\label{Thetaprop} If $\operatorname{O}mega$ is properly-convex, then $\Theta:\mathcal C\operatorname{O}mega^*\rightarrow\mathcal C\operatorname{O}mega$ is a homeomorphism that maps  rays to rays, and $[\Theta]:\operatorname{O}mega^*\rightarrow\operatorname{O}mega$ is a homeomorphism.\end{corollary}
\begin{proof} 
Given $p\in\operatorname{O}mega$ there is a unique $x\in\partial\mathcal D$ 
with $p=[x]$. By (\ref{Vprop})(iv) there is a unique $\phi\in\mathcal C^*$ with $\phi(x)=1$
and $\mathcal V(\phi)=1$. Moroever $\mu(\mathcal C_{\phi})=x$ and so $[\Theta][\phi]=p$.
Hence $[\Theta]$ is a bijection.
 Clearly $\Theta$ is continuous. By (\ref{Vprop})(ii)   $[\Theta]$ is proper, and thus a homeomorphism.
  Since $\Theta(t x)=t^{-n-1}\Theta(x)$, rays are mapped to rays. It follows that $\Theta$ is also a homeomorphism.
 \end{proof}
The dual action of $A\in\operatorname{SL} V$ on $\operatorname{{\mathbb P}} V^*$
 is given by $A[\phi]=[\phi\circ A^{-1}]$. If $\Gamma\subset\operatorname{SL} V$ and preserves $\operatorname{O}mega$, then the dual action 
 of $\Gamma$ preserves $\operatorname{O}mega^*$. It is clear that $\Theta$ is equivariant with respect to these actions.
 It directly follows from (\ref{C1domain}) and (\ref{Thetaprop}) that:

  \begin{corollary}[Vinberg]\label{dualmfd} If $M=\operatorname{O}mega/\Gamma$ is a closed, properly-convex manifold, then $M^*=\operatorname{O}mega^*/\Gamma$
 is a properly-convex manifold that is homeomorphic to $M$. Thus $\pi_1M\cong\pi_1M^*$. 
  \end{corollary}

If $M=\operatorname{O}mega/\Gamma$, then $M$ is called {\em $C^1$} if  $\operatorname{O}mega$ is $C^1$. It follows from
 (\ref{C1domain}) that:
\begin{corollary}\label{C1}  $M$ is strictly-convex if and only if $M^*$ is $C^1$.\end{corollary}

  \begin{corollary}\label{dualcpct} If $M=\operatorname{O}mega/\Gamma$ is a closed, strictly-convex manifold, then $M^*=\operatorname{O}mega^*/\Gamma$
 is a closed, strictly-convex manifold. We have $\pi_1M\cong\pi_1M^*$ and call  $M^*$ the {\em dual } of $M$
 Moreover, $M$ is $C^1$.
  \end{corollary}
   \begin{proof}   Since $M$ is strictly convex, it is properly convex, so $M^*$ is a properly-convex manifold,
   and $\pi_1M\cong\pi_1M^*$. By (\ref{stricthomeo}) $M^*$ is closed, and 
 strictly-convex. Thus $M=(M^*)^*$ is $C^1$ by (\ref{C1domain}).
    \end{proof}
    

The centroid of a bounded open convex set $\operatorname{O}mega\subset {\mathbb R}^n$ is a distinguished point in $\operatorname{O}mega$.
We wish to define something similar for  certain subsets of the sphere $S^n=\{x\in{\mathbb R}^{n+1}:\|x\|=1\}$. Imagine a room that contains a transparent globe close to one wall, with a light source at the center of the globe. The shadow of Belgium appears on the wall.
You can  rotate the globe so that the centroid of this shadow is  the point $p$ on the wall closest to the center of the globe. 
The point on the globe that projects to $p$ is called the
{\em (spherical) center} of Belgium.

The open hemisphere that is
the $\pi/2$ neighborhood of $y\in S^n$ is $U_y=\{x\in S^n: \langle x,y\rangle>0\}$.
Radial projection $\pi_y:U_y\longrightarrow {\operatorname{T_y}}S^n$ 
from the origin onto the tangent space
to $\mathbb R\operatorname{P}_+^n$ at $y$ is given by $$\pi_y(x)=\frac{x-\langle x,y\rangle y}{\langle x,y\rangle}$$
Identify $S^n$ with $\mathbb R\operatorname{P}_+^n$ using $\pi_{\xi}:S^n\to\mathbb R\operatorname{P}_+^n$.

\begin{definition} If $\operatorname{O}mega\subset \mathbb R\operatorname{P}_+^n$ is open, then $y\in \operatorname{O}mega^*$ is called
a {\em (spherical) center of $\operatorname{O}mega$}  if 
 $\hat{\mu}(\pi_y(\operatorname{O}mega))=\pi_y(y)$.
 \end{definition}

If $A\in\operatorname{O}(n+1),$ then $A$ is an isometry of the inner product and so $(A\operatorname{O}mega)^*=A(\operatorname{O}mega^*)$, and  $Ay$ is a center of $A\operatorname{O}mega$
if  $y$ is a center of $\operatorname{O}mega$. The following property of $\partial\mathcal D$ is important for the proof of the Closed Theorem.
\begin{theorem}\label{centerexists} If $\operatorname{O}mega\subset S^n$ is properly-convex, then  it has a unique spherical center $[x]$. 
Moreover
 $x$ is the point on $\partial\mathcal D$ that minimizes $\|x\|$.
\end{theorem}
\begin{proof} Let $\phi\in V^*$ be given by $\phi(y)=\langle y,x\rangle$.
The tangent plane to $\partial\mathcal D$ at $x$ is $\phi^{-1}(1)$ and is orthogonal to $x$. 
Thus $\pi_x(\operatorname{O}mega)=\mathcal C_{\phi}$ and $\mu(\mathcal C_{\phi})=x$.
\end{proof}
 
 \begin{corollary} If $\operatorname{O}mega\subset{\mathbb R}P^n$ is properly-convex and $p\in \operatorname{O}mega$, then
 there is an affine patch ${\mathbb R}^n\subset{\mathbb R}P^n$ such that $p$ is the centroid of $\operatorname{O}mega$ in ${\mathbb R}^n$.
\end{corollary}
\begin{proof} There is a unique $x\in\partial\mathcal D$ with $p=[x]$.
Choose an inner product on $V$ so that $x$ is the closest point to $0$ on $\partial\mathcal D$.
The required affine patch is ${\mathbb R}P^n\setminus\operatorname{{\mathbb P}}(x^{\perp})$.
\end{proof}

 \section{Open}\label{opensec}

A set is {\em locally convex} if every point
has a convex neighborhood. There is a basic local-to-global principle for convexity. 
\begin{proposition}\label{convexhyper} If $K$ is a closed, connected, locally convex subset of ${\mathbb R}^n$, then $K$ is convex.
\end{proposition}
 If  $K$ has non-empty interior, then local convexity only needs to be checked at points in $\partial K$. Suppose that $K$ is contained in the upper halfspace $x_1\ge0$. Then $\operatorname{O}mega=\operatorname{int}(K\cap(0\times{\mathbb R}^{n}))$ 
is a convex subset of ${\mathbb R}^n$ if the subset  $S\subset \partial K$ 
where $x_1>0$ is a locally convex hypersurface in ${\mathbb R}^{n+1}$. Informally: the base of a convex mountain is convex.
We now explain how to use this to show that a manifold is properly-convex.

 A smooth hypersurface $S\subset{\mathbb R}^n$ is {\em Hessian-convex} if 
the surface is locally the zero-set of a smooth, real-valued function with positive-definite Hessian. Suppose $\operatorname{O}mega\subset{\mathbb R}Pn$ is properly-convex and $\mathcal C\operatorname{O}mega=\{v\in {\mathbb R}^{n+1}_0:[v]\in\operatorname{O}mega\}$ is the corresponding convex cone. 
Suppose $M=\operatorname{O}mega/\Gamma$ is a compact, and properly-convex, $n$--manifold.
Then $\widetilde W=\mathcal C\operatorname{O}mega/\Gamma\cong M\times (0,\infty)$ 
is a properly-convex affine $(n+1)$--manifold. We may divide out by a homothety to obtain a compact affine $(n+1)$--manifold $W\cong M\times S^1$.

Suppose there is a  hypersurface $S\subset \mathcal C\operatorname{O}mega$ that is $\Gamma$--invariant and Hessian-convex away from $0$.
Then $Q=S/\Gamma$ is a compact, Hessian-convex, codimension--1 submanifold of $W$. 
Let $U$ be the component of $\mathcal C\operatorname{O}mega\setminus S$ whose closure does not contain $0$. 
Let $K\subset{\mathbb R}P^{n+1}$ be
the closure of $U$ and regard ${\mathbb R}P^{n+1}={\mathbb R}^{n+1}\sqcup{\mathbb R}P^n_{\infty}$. The interior of $K\cap{\mathbb R}P^n_{\infty}$ can be identified with $\operatorname{O}mega$.
The existence of such $Q$ {\em implies} the existence of $S\subset{\mathbb R}^{n+1}$, which {\em implies} $\operatorname{O}mega$ is properly-convex,  by the reasoning above.

For general reasons (the Ehresmann-Thurston principle) deforming $\Gamma$ a small amount to $\Gamma'$
 gives a new projective $n$--manifold $M'\cong M$, and a new affine $(n+1)$--manifold $W'\cong M\times S^1$. 
 Now $W'$ contains  a  submanifold $Q'$ that is Hessian-convex, provided
the deformation of the developing map is small enough in $C^2$. It then follows
that  $M'$ is properly-convex. This is the approach taken in \cite{CLT2} for non-compact manifolds.

Here, we work with a piecewise linear submanifold $Q$ in place of a smooth one.  Hessian convexity 
 is replaced by the condition
 that at each vertex of the triangulation the determinants, of certain matrices formed by the relative positions of vertices, are strictly positive.
This ensures local convexity near the vertex. This version of convexity is preserved by small $C^0$--deformations
of the developing map.
We start by reviewing these ideas for hyperbolic manifolds.

The hyperboloid model of the hyperbolic plane  is the action of $\operatorname{SO}(2,1)$ on
  the surface $S\subset{\mathbb R}^3$  given by $x^2+y^2-z^2=-1$. If we identify ${\mathbb R}^3$ with ${\mathbb R}P^3\setminus{\mathbb R}P^2_{\infty}$
 then we can regard this as an affine action on 
  ${\mathbb R}P^3$ by
using the embedding $\operatorname{SO}(2,1)\oplus(1)\subset\operatorname{SL}(4,{\mathbb R})$.
The open disc  $$D=\{[x:y:z:0]:\ x^2+y^2<z^2\}\subset{\mathbb R}P^2_{\infty}$$ has  the same frontier as $S$
   and $(\operatorname{SO}(2,1),D)$ is the projective (Klein) model  of the hyperbolic plane.
 
\begin{figure}
\caption{The Hyperboloid and Klein models of ${\mathbb H}
\label{SandOmega}
\end{figure}

 

 
 The surface $\operatorname{cl}(S\cup D)\subset{\mathbb R}P^3$ bounds a closed $3$-ball $B\subset{\mathbb R}P^3$. 
 Let  $\operatorname{O}mega=B\setminus\operatorname{Fr} D\cong D\times I$.
 The fact that $S$ is
 a strictly-convex surface {\em implies} $\operatorname{cl}(S\cup D)$
 is a convex surface in some affine patch which {\em implies} $\operatorname{O}mega$ is properly-convex, and this {\em implies} $D$ is properly-convex.

  The action of $\operatorname{SO}(2,1)\oplus(1)$ preserves $B$.
    Suppose $\Gamma\subset\operatorname{SO}(2,1)$ and $\Sigma=D/\Gamma$ is a properly-convex projective (and hyperbolic) surface. Let $\Gamma'=\Gamma\oplus(1)$ then $N=B/\Gamma'\cong\Sigma\times[0,1]$
is a properly-convex manifold with one flat boundary component $\Sigma=D/\Gamma$ and one
strictly-convex boundary component $M=S/\Gamma$. The fact that $M$ is a strictly-convex surface {\em implies} $\Sigma$ is properly-convex.
We will generalize this construction to arbitrary properly-convex manifolds in place of $\Sigma$. But first we 
divide out by a homothety.

Consider the cone $\mathcal C=\{\lambda\cdot x:x\in S,\ \ \lambda>0\}$. There is a product structure $\widetilde \phi:S\times{\mathbb R}^+ \to\mathcal C$ given by $\widetilde\phi(x,\lambda)=\lambda^{-1}\cdot x$
 on $\mathcal C$ that is preserved by $\Gamma'$. Let $H\subset\operatorname{GL}(4,{\mathbb R})$ be the cyclic
group generated by $\exp (\operatorname{Diag}(0,0,0,1))$. Then $\Gamma'$ centralizes $H$ and the group $\Gamma^+\subset\operatorname{GL}(4,{\mathbb R})$ generated by $\Gamma'$ and $H$
preserves $\mathcal C$ and $W:=\mathcal C/\Gamma^+$ is a $3$--manifold. 

In what follows ${\mathbb R}^+=\{x\in{\mathbb R}:x>0\}$, and $S^1={\mathbb R}^+/\exp({\mathbb Z})$, has universal cover ${\mathbb R}^+$.
There is a product structure $\phi:\Sigma\times S^1\to W$
covered by $\widetilde \phi$, and the surfaces $\phi(\Sigma,\theta)$ are convex.
The reader might contemplate all this in the case of one dimension lower, where $\Gamma\subset\operatorname{SO}(1,1)$
is generated by a hyperbolic and $\mathcal C/\Gamma^+$ is an affine structure on $S^1\times S^1$. 

We now return to the general setting. Regard ${\mathbb R}^+$ as a subgroup of $\operatorname{GL}(1,{\mathbb R})$, and write an element  as either $x$ or as $(x)$.
Let $\operatorname{FG}(n)=\operatorname{SL}(n,{\mathbb R})\oplus{\mathbb R}^+\subset\operatorname{GL}(n+1,{\mathbb R})$. We also use the notation  ${\mathbb R}^+$ to denote the subgroup
$1\oplus {\mathbb R}^+\subset \operatorname{FG}(n)$. {\em Flow geometry} is the subgeometry 
$(\operatorname{FG}(n),{\mathbb R}^n_0)\subset(\operatorname{GL}(n+1,{\mathbb R}),{\mathbb R}P^n)$ of projective geometry where ${\mathbb R}^n_0\equiv\{[x:1]:   x\in{\mathbb R}^n_0\}\subset{\mathbb R}P^n$.  

The
subgroup ${\mathbb R}^+\subset \operatorname{FG}(n)$ is called the {\em homothety flow} and  the action of $1\oplus(t)$ on ${\mathbb R}^n_0$ is given
by $(1\oplus(t))[x:1]=[x:t]$. In affine coordinates, this action is $(1\oplus(t))x=t^{-1}\cdot x$. Thus points in ${\mathbb R}^n_0$ move {\em towards} $0$ as $t$ increases. 
The projection  $\pi_{hor}:\operatorname{FG}(n)\to\operatorname{SL}(n,{\mathbb R})$ is called the {\em horizontal holonomy} and $\pi_{rad}:\operatorname{FG}(n)\to{\mathbb R}^+$ is called the {\em radial holonomy}.

\begin{definition} If $M$ is a closed $n$--manifold, then a {\em flow product structure} on $M\times S^1$
is a flow geometry structure $(\operatorname{dev},\rho)$ such that:
\begin{itemize}
\item $\operatorname{dev}(x,t)=t^{-1}\cdot \operatorname{dev}(x,1)$ where $(x,t)\in\widetilde M\times{\mathbb R}^+$
\item $\pi_{rad}(\rho\;\pi_1M)=1$ regarding $\pi_1M\subset\pi_1(M\times S^1)$
\end{itemize}
\end{definition}

It follows that $\pi_{rad}(\rho\;\pi_1M\times S^1)=\exp{\mathbb Z}$.
The  action of ${\mathbb R}^+$ on the right factor of $\widetilde W=\widetilde M\times {\mathbb R}^+$  is conjugate, by the developing map, to the homothety action on ${\mathbb R}^{n+1}_0$.
The action of $S^1$ on the right factor of $W=M\times S^1$  is covered by this action on $\widetilde W$. We call call all of these actions {\em homothety}.
Flow geometry is also a subgeometry of affine geometry, so we may use affine notions.

Suppose $W$ is an affine $n$--manifold, and $S$ is a hypersurface in $W$. Then
$S$
 is a {\em convex hypersurface},  and it is {\em locally convex}, if
 there is a submanifold $V\cong S\times [0,1)\subset W$ with $\partial V=S$, and
every point  $p\in S$ has a small neighborhood $U\subset V$, such that $\operatorname{dev}(U)\subset{\mathbb R}^n$ is convex.

  
If $S$ is locally convex and, in addition, every maximal flat subset of the universal cover $\widetilde S$ is compact,
then $S$ is called {\em strongly locally convex}. This condition automatically holds if $S$ contains no segment.

A  hypersurface, $S\subset W$, is  {\em simplicial}  if it has a triangulation
by flat simplices.
Strongly locally convex simplicial surfaces are a substitute for {\em strictly convex} in this context.

If $W=M\times S^1$ has a flow product structure, then $S$ is   {\em outwards} locally-convex
if, in addition, $t\cdot \partial V\subset V$ for some $t>1$. In other words $\operatorname{dev}\widetilde S$ is locally-convex {\em away} from $0$ in ${\mathbb R}^{n+1}$.

\begin{definition} A flow product structure $(\operatorname{dev},\rho)$ on $M\times S^1$ is {\em flow convex}
if $M\times\theta$ is a strongly outwards convex hypersurface in $M\times S^1$ for some (and hence all) $\theta\in S^1$.
\end{definition}

If $\sigma,\tau\subset W$ are flat $(n-1)$-simplices
and $\sigma\cap\tau\ne\emptyset$ then they are {\em coplanar} if they have lifts $\widetilde\sigma,\widetilde\tau\subset\widetilde W$
with $\widetilde\sigma\cap\widetilde\tau\ne\emptyset$ and $\operatorname{dev}(\widetilde\sigma\cup\widetilde\tau)$ is contained in a hyperplane.

\begin{definition}  A simplicial hypersurface in an affine manifold  is {\em generic-convex}  if it is locally convex, and whenever two $(n-1)$-simplices   intersect then they 
 are not coplanar.
 \end{definition}

It is immediate that generic-convex implies strongly locally convex. Moreover a small movement of the vertices of a generic-convex hypersurface produces a new generic-convex hypersurface. We state these properties in the following:

\begin{lemma}\label{convexlemma} If $S\subset M$ is a generic-convex simplicial hypersurface, then $S$ is strongly locally convex. 
Moreover, for each vertex $v$ of $S$, there is a neighborhood $U(v)\subset M$,
such that the hypersurface $S'$ obtained  
 by moving each vertex $v$ inside $U(v)$ is generic-convex. \end{lemma}
 \begin{proof}  
  The first conclusion is immediate.  
 The generic-convex condition is equivalent to the positivity of the determinants
 of certain $n\times n$ matrices formed by vectors of the form $u-v$ for certain vertices, $u$, that are adjacent to $v$ in $S$.
 If the vertices of $S$ are moved a small enough distance, these determinants remain positive. Thus $S'$
is generic-convex. Now we explain the determinant condition.

 Let $K$ be the union of the simplices in $S$ that contain $v$. Then $\dim K=n-1$.
 Suppose $\sigma$
 is a simplex in $K$ of dimension $(n-1)$.  Then $\sigma$ is contained in a hyperplane $H$.
 If $K$ is  generic convex
then the closure, $W$, of one component of the complement of $H$
satisfies $W\cap K=\sigma$.

 Conversely, $S$ is generic convex at $v$ if this condition holds
 for each such $\sigma$. Let $\{v_0,\cdots,v_{n-1}\}$ be the vertices of $\sigma$.
 Let $f(u)=\det(v_0-u:v_1-u:\cdots:v_{n-1}-u)$. The condition for generic convexity is
 that the sign of $f(v)$ is constant as $u$ ranges over all vertices in $K$ that are connected by an edge to $\sigma$
and that are not in $\sigma$. This ensures the simplices adjacent to $\sigma$ in $K$ all lie on the same side
of the hyperplane that contains $\sigma$.
 \end{proof}

The {\em convex hull, $\operatorname{CH}(Y)$,} of $Y\subset {\mathbb R}^n$ is  the intersection of all the {\em closed} convex sets that contain $Y$.
Let $\pi_{\xi}:V_0\to\operatorname{{\mathbb P}}_+(V)$ be the projection $\pi_{\xi}(x)=[x]_+$.
The  next result can be proven by using the existence of  a {\em characteristic hypersurface}  
~\cite{Koecher1957, Vinberg1963homog}, or of an {\em affine sphere} \cite{MR437805, MR2743442}.  
\begin{theorem}\label{chulllemma} Suppose $\mathcal C\subset{\mathbb R}^{n+1}$ is a convex cone, $\operatorname{O}mega=\pi_{\xi}(C)\subset{\mathbb R}P_+^n$ and $M=\operatorname{O}mega/\Gamma$ is a properly-convex closed manifold. 

Then there 
 is a convex proper submanifold $P\subset \mathcal C$  with $t\cdot P\subset P$ for all $t\ge 1$,
  and $(\pi_{\xi}|\partial P):\partial P\to \operatorname{O}mega$ is a homeomorphism.
 Moreover, $P$ can be chosen such that $\partial P/\Gamma\subset\mathcal C/\Gamma$ is a compact, generic-convex, simplicial hypersurface.
 \end{theorem}
\begin{proof} Let $\mathcal D\subset\mathcal C$  be the closed convex subset given by (\ref{SPdef}).
Then $N=\mathcal D/\Gamma$ is a convex submanifold of $\mathcal C/\Gamma$ with boundary
 a  strictly-convex hypersurface. Let $H\subset {\mathbb R}^{n+1}$ be an affine hyperplane
that separates $\mathcal D$ into two components, such that the bounded component, $Y$,  projects to a subset $\pi(Y)\subset N$ that is contained 
in a small
ball in $N$.
Replace $N$ by $N\setminus \pi(Y)$. This produces a flat part in the boundary. 
This can be done finitely many times, to produce a
 submanifold $Q\subset N$
with simplicial boundary. The preimage of $Q$ in $\mathcal C$ is the required  $P$.  \end{proof}
    

\begin{corollary}[\cite{MR3500374}] A closed properly-convex manifold $\operatorname{O}mega/\Gamma$ has a convex polyhedral fundamental domain
in $\operatorname{O}mega$.
  \end{corollary}
  \begin{proof} Use the notation in the proof of (\ref{chulllemma}).
  Let $H$ be a closed halfspace that contains $\mathcal D$ and such that $(\partial\mathcal D)\cap\partial H$ is a single point. 
  Let $Y=\cap(\gamma\cdot H)$ where the intersection is over all $\gamma\in\Gamma$. Then
  $\mathcal D\subset Y\subset\mathcal C\operatorname{O}mega$, and $Q=Y\cap\partial H$ is a convex polytope.
  Moreover $\partial Y=\Gamma\cdot Q$ is a locally finite union of images of $Q$.
 It follows that $\pi_{\xi}(Q)$
  is a convex polyhedral fundamental domain.
  \end{proof}

  
 

Suppose that $M=\operatorname{O}mega/\Gamma$ is properly-convex with $\operatorname{O}mega\subset{\mathbb R}P_+^n$ and $\Gamma\subset\operatorname{SL}(n+1,{\mathbb R})$. 
Then $\mathcal C=\{x\in{\mathbb R}^{n+1}: [x]_+\in\operatorname{O}mega\}$ is a convex cone. 
Let $\Gamma^+=\Gamma\oplus\exp{\mathbb Z}\subset \operatorname{FG}(n+1)$.
Then $W=\mathcal C/\Gamma^+$ is a flow product structure on $M\times S^1$ by (\ref{chulllemma}). Conversely if $(\operatorname{dev},\rho)$ is a flow product structure
 on $M\times S^1$ then $(\pi_{\xi}\circ\operatorname{dev},\pi_{hor}\circ\rho)$ is a properly-convex structure on $M\times\theta$ because of:

 
 \begin{theorem}\label{sufficient} If $M\times S^1$ is compact and flow-convex, then $M\times S^1$ is properly-convex.
 \end{theorem}
 \begin{proof} 
Let $N'=M\times S^1$ and $\pi_{_{N'}}:\widetilde N'=\widetilde M\times{\mathbb R}^+\rightarrow N'$ be the universal cover,
and $\operatorname{dev}:\widetilde N'\rightarrow{\mathbb R}^{n+1}$ the developing map for $N'$.
Let $R'=\widetilde M\times 1$ and choose a basepoint $p'\in R'$.
Let $V\subset{\mathbb R}^{n+1}$ be a $2$-dimensional vector subspace that contains $p=\operatorname{dev}(p')$. 
 Then $\operatorname{dev}^{-1}V = X\times {\mathbb R}^+$ for some $1$-submanifold
$X\subset \widetilde M$.
Let $C'$
be the component of $R'\cap \operatorname{dev}^{-1}V$ that contains $p'$. Then $C'$ is a 
connected curve in $\operatorname{dev}^{-1}V$ without endpoints
  that is transverse to
the homothety flow and convex outwards. 

The curve $C=\operatorname{dev}(C')$ is immersed in $V_0$, and is everywhere transverse to the radial direction, and convex outwards: radially away from $0$.
Let $\pi:V_0\to S^1$ be radial projection $\pi(x)=x/\|x\|$. 

{\bf Claim 1:}  $\pi:C\to S^1$ is injective. Since $C'$ is transverse to the radial direction in $N'$, it follows that
 $\theta=\pi\circ \operatorname{dev}:C'\rightarrow S^1$ is an immersion. Let $\ell\subset V$ be the tangent line to $C$ at $p$. Suppose
 $q\in C\cap\ell$ is distinct from $p$. Then at some point $r$ in $C$ between $p$ and $q$, the distance 
 of $r$ from $\ell$ is a maximum. This contradicts that $C$ is convex outwards at $r$.
  Hence $\pi(\ell)$ is an open semi-circle in $S^1$ that contains $\pi (C)$. Thus $\theta$ is an immersion of $C'$ into another arc,  therefore $\theta$ is injective, and it follows that $\pi$ is injective. This proves Claim~1.

Define $R\subset{\mathbb R}^{n+1}$ to be  the union, as $V$ varies,
of all the curves $C$ above. Since the developing map is a local homeomorphism,  each curve 
$C$ is an open arc, and  $R$ is the developing image of a connected open subset of $R'$. Thus $R$ is
a locally convex hypersurface without boundary.
 

 
{\bf Claim 2:} $R$ is a closed subset of ${\mathbb R}^{n+1}$. Otherwise there is a sequence $r_k\in R$ that converges to a point
$r\in {\mathbb R}^{n+1}\setminus R$. Define some two dimensional vector spaces of ${\mathbb R}^{n+1}$ by $V_k=\langle p,r_k\rangle$, and $V=\langle p,r\rangle$. Then $V_k$ converges to  $V$.
The segment $\gamma_k=[p,r_k]$ in $V_k$ is the developing image of a segment $\gamma'_k=[p',r'_k]$ in $\widetilde N'$. 
This is because $\operatorname{dev}({\mathbb R}^+\cdot R')$ contains ${\mathbb R}^+\cdot (R\cap V_k)$, and $\gamma_k$ is in the latter.
\begin{figure}
\caption{$\kappa$ approaching $e^tC$ from above or below}
\label{kappa}
\end{figure}

Let $C= V\cap R$ and let $C'\subset R'$ be the curve with $\operatorname{dev}(C')=C$. Let $C^+\subset V$
be the Hausdorff limit of the curves $C_k=R\cap V_k$ in ${\mathbb R}^{n+1}$. Then $r\in C^+\setminus C$.
Let $w$ be the limit point of $C$ in $C^+$ closest along $C^+$ to $r$. Let $\kappa:[0,1)\rightarrow V$ be an affine homeomorphism
with image
$[p,w)$.
There is an affine ray $\widetilde\kappa':[0,1)\rightarrow \widetilde N'$ with $\kappa=\operatorname{dev}\circ\widetilde\kappa'$
that starts at $\widetilde\kappa'(0)=p'$  and is the limit of segments $[p',c'_n]$ in $\widetilde N'$
 where $c'_n\in C'$
and $\lim_{n\to\infty} \operatorname{dev}(c_n)= w$.  The points $c_n'$ in $\widetilde N'$ leave every compact set.
In $V$ the point $w$ is on $e^tC$ for some $t$, and $\kappa$ limits on $w$, either from above or from below. See \Cref{kappa}.

  It follows that $\kappa'=\pi_{_{N'}}\circ\widetilde\kappa':[0,1)\rightarrow N'$ is an affine ray that
 spirals in towards, and accumulates on, 
  some subset $F$ of $M\times t\subset N'$. In other words $F$ is the forward limit set of $\kappa'$.
  Now $F$ is flat because $\kappa'$
 is affine. Some component of $\pi_{_{N'}}^{-1}(F)$ in $\widetilde N'$ is not compact, otherwise $\kappa'$
converges to a limit point in $F$ that maps to $w$. This
  contradicts that $M\times t$ is strongly locally convex, and  thus contradicts that $M\times S^1$ is flow-convex. 
 This proves Claim 2.
 
 

 Hence $R$ is a hypersurface that is properly embedded in ${\mathbb R}^{n+1}$ and that is transverse to the radial direction
 and convex outwards. Then $X=\cup_{t\ge 1} t\cdot R$ is  the closure of the component of ${\mathbb R}^n\setminus R$ that does not contain $0$.
 Thus $X$ is closed and has boundary $R$. Since $R=\partial X$ is a convex hypersurface, it follows that $X$ is convex, and the image of 
 $X$ in ${\mathbb R}P^n$ is properly convex. Now ${\mathbb R}^+\cdot R'$ is a clopen subset of $\widetilde N'$, so these sets are equal.
 Hence $M\times S^1$ is also properly convex.\end{proof}
 

 
Let $\operatorname{FC}(M\times S^1)\subset \operatorname{Hom}(\pi_1M,\operatorname{SL}(n+1,{\mathbb R}))$ consist of all $\sigma$ such that there is a flow convex structure $(\operatorname{dev},\rho)$ on $M\times S^1$
with $\sigma=\pi_{hor}\circ\rho$.

\begin{theorem}\label{developing} If $M\times S^1$ is compact, then $\operatorname{FC}(M\times S^1)$ is an open subset of
$\operatorname{Hom}(\pi_1M,\operatorname{SL}(n+1,{\mathbb R}))$ where $n=\dim M$.
\end{theorem}
\begin{proof}  Suppose $(\operatorname{dev}_0,\rho_0)$ is a flow convex structure on $W=M\times S^1$. By (\ref{sufficient}) $W$ is properly-convex. Thus $M$ is properly
convex. Then by (\ref{chulllemma}) there is a triangulation, $\mathcal T$, of $W$ such that $M\times 0$ is a generic-convex subcomplex
 of  $\mathcal T$ that is outwards  locally convex. 
 
 Let $\widetilde\mathcal T$ be the lifted triangulation on the universal cover $\widetilde W$.
Let $\mathcal C\subset{\mathbb R}^{n+1}_0$ be the image of $\operatorname{dev}_0$ and $\Gamma=\rho_0(\pi_1W)$. Then $\mathcal C$ is triangulated by $\operatorname{dev}\widetilde\mathcal T$.
Suppose  $K\subset\mathcal C$ is a compact set.
Let $D$ be a finite subcomplex that contains $K$ such that $\Gamma\cdot D=\mathcal C$.
 Let $\mathcal P$ be the (finite) set of vertices of $D$ and choose
a subset $\mathcal P^-\subset\mathcal P$ that consists of one point in each $\Gamma$-orbit.

 For each $v\in\mathcal P$ there is a unique $g_v\in\pi_1M$
such that $\rho_0(g_v)v\in\mathcal P^-$. 
 Given  $v\in\mathcal P$ define $f_v:\operatorname{Rep}(M)\to{\mathbb R}^{n+1}_0$ by \begin{equation}\label{eq9}f_v(\rho)=\rho(g_v^{-1})\rho_0(g_v)v\end{equation} for $\rho\in \operatorname{Rep}(M)$.
This  is  continuous and $f_v(\rho_0)=v$. Define $\mathcal P(\rho)=\{f_v(\rho):v\in\mathcal P\}$,
 then $\mathcal P(\rho_0)=\mathcal P$. If $\rho$ is close enough to $\rho_0$, there is a simplicial complex $D(\rho)\subset{\mathbb R}^{n+1}$ with vertex set $\mathcal P(\rho)$,
 and a simplicial homeomorphism $F:D\to D(\rho)$ defined as follows.
  
Given $X\subset D$ a map $F:X\to {\mathbb R}^{n+1}$ is called {\em equivariant}
if whenever $a,b\in X$ and $g\in\pi_1M$ and $\rho_0(g)a=b$ then $\rho(g)F(a)=F(b)$. Since
the action of $\Gamma$ is free there is at most one such $g$.

Define $F:\mathcal P\to\mathcal P(\rho)$ by $F(v)=f_v(\rho)$.
Two vertices $u,v\in\mathcal P$ have the same image in $M$ if and only if $\rho_0(g_v)v=\rho_0(g_{u})u\in\mathcal P^-$.  
Since $\rho_0(g_v)v=\rho(g_v)f_v(\rho)$ it follows from \Cref{eq9} that $\rho(g_v)f_v(\rho)=\rho(g_{u})f_u(\rho)$,
thus $\rho(g_v)F(v)=\rho(g_{u})F(u)$. Hence $F$ is equivariant. 
Extend $F$ over each simplex $\sigma\in D$ using the affine map determined by the images of the vertices. This
extension is equivariant. If $\rho$ is close enough to $\rho_0$ then $F$ is a homeomorphism.

  
  Whenever $g\in\pi_1M$, and $\rho_0(g)$ identifies two simplices in $\partial D$, we use
  $\rho(g)$ to identify the corresponding simplices of $D(\rho)$. Set $M=D/\sim$ and $N=D(\rho)/\sim$. Then $F$ covers a simplicial homeomorphism $f:M\to N$.
Using $M=\mathcal C/\Gamma$, then $\widetilde M=\mathcal C$, and $D\subset\widetilde M$.
  It follows that $N$ is an affine structure on $M$ for some developing pair $(\operatorname{dev},\rho)$ with $\operatorname{dev}|D=F.$
   Since $K\subset D$,  by choosing $\rho$ close
  enough to $\rho_0$, then $\operatorname{dev}|K$ is close to $\operatorname{dev}_0|K$.
   By choosing $K$ large enough, it follows from (\ref{convexlemma}) that the subcomplex $W\times 0$ is generic-convex, and thus locally convex. Thus $(\operatorname{dev},\rho)$ is a flow convex structure on $M$.
   \end{proof}

\subsection{Proof of (\ref{open}. Open)} Suppose $\rho_0\in\operatorname{Rep}_P(M)$ and $(\operatorname{dev},\rho_0)$ is a development pair for $M$. 
 By (\ref{chulllemma})   there is a flow convex structure on $M\times S^1$ with horizontal holonomy $\rho_0$. By (\ref{developing})
 if $\rho$ is close to $\rho_0$ there is a flow convex structure on $M\times S^1$ with horizontal holonomy $\rho$.
 By (\ref{sufficient}) this is a properly-convex structure. Thus $M$ has a properly-convex structure with holonomy $\rho$.
 \qed

\section{Closed}

We first give an outline of the proof of the Closed Theorem.
Suppose $\rho_k$ is a sequence of holonomies of properly-convex real projective structures on $M$,
so that $M\cong \operatorname{O}mega_k/\rho_k(\pi_1M)$ with $\operatorname{O}mega_k$ properly-convex.
Suppose the holonomies converge pointwise to
 $\lim\rho_k=\rho_{\infty}$. If $M$ is strictly-convex, a special case of Chuckrow's theorem (\ref{discfaith}) implies $\rho_{\infty}$
 is discrete and faithful; in general $\rho_{\infty}$ is neither.
 After taking a subsequence we may assume $\operatorname{O}mega_{\infty}=\lim\operatorname{O}mega_k\subset{\mathbb R}Pn$ exists.
 If $\operatorname{O}mega_{\infty}$ is properly-convex then $\operatorname{O}mega_{\infty}/\rho_{\infty}(\pi_1M)$ is a properly-convex structure on $M$.
But $\operatorname{O}mega_{\infty}$
might have smaller dimension, or it might not be properly-convex.  We describe this by saying {\em the domain
has degenerated}.  

The {\em box estimate} (\ref{boxestimate}) implies that one may replace the original sequence $\rho_k$
by conjugates $\rho'_k$ which preserve domains $\operatorname{O}mega'_k$ and  $\lim\rho_k'=\sigma$,
and 
$\lim \operatorname{O}mega'_k=\operatorname{O}mega$ is properly-convex.
  Then $N=\operatorname{O}mega/\sigma(\pi_1M)$ is a properly-convex manifold
 homotopy equivalent to $M$. Hence $N$ is closed  and $\pi_1M\cong\pi_1N$. It then follows from
 Gromov-Hausdorff convergence that $M$ is
 homeomorphic to $N$.
Also $N$ is strictly-convex
 by (\ref{stricthomeo}). Finally, $\sigma$ is irreducible by (\ref{irreducible}), which implies that, in fact,
 the original domains $\operatorname{O}mega_k$ did not degenerate. Thus $\rho_{\infty}$ is the holonomy
 of a strictly-convex structure on $M$.  
 
 

 
 We supply the missing details, starting with the algebraic preliminaries.
  
\begin{theorem}[Irreducible]\label{irreducible} Suppose $M=\operatorname{O}mega/\Gamma$ is a strictly-convex closed manifold and $\dim M\ge 2$.
Then  $\Gamma$ does not preserve any proper projective subspace. 
 \end{theorem} 
\begin{proof}  Suppose ${\mathbb R}^{n+1}=U\oplus V$ with $U\ne 0\ne V$ and $\Gamma$ preserves $U$. Then
$\Gamma$ preserves $Y=\operatorname{cl}\operatorname{O}mega\cap\operatorname{{\mathbb P}}_+ U$. 
Now  $Y\cap \operatorname{O}mega= \emptyset$ by (\ref{noinvt}).
Suppose $Y=\emptyset$. By the hyperplane separation theorem, there is $\phi\in ({\mathbb R}^{n+1})^*$ such that $\ker \phi$ contains
$U$, and $\phi(\operatorname{cl}\operatorname{O}mega)>0$. 
Thus $[\phi]\in \operatorname{{\mathbb P}}_+ U^0\cap\operatorname{O}mega^*$.
By (\ref{dualcpct}) the dual manifold $M^*=\operatorname{O}mega^*/\Gamma$ is compact and strictly-convex, and the dual action of $\Gamma$ preserves $U^0$, 
which  contradicts (\ref{noinvt}). Thus $Y\ne\emptyset$, so
 $Y\subset\operatorname{Fr}\operatorname{O}mega$.
 Since $M$ is strictly-convex, $Y=c$ is a single point that is fixed by $\Gamma$. 
 
 Thus every non-trivial element of $\Gamma$ has an axis
 with one endpoint at $Y$. 
  If $\ell$ and $\ell'$ are the axes of $\gamma,\gamma'\in\Gamma$ then they limit on $Y$. Now $\operatorname{O}mega$ is $C^1$ at $Y$ by (\ref{C1})
 so $d_{\operatorname{O}mega}(p,\ell')\to0$ as the point $p$ on $\ell$ approaches $c$. Let $\pi:\operatorname{O}mega\to M$
 be the projection.  Then $C=\pi(\ell)$ and $C'=\pi(\ell')$ are closed geodesics that become arbitrarily close to each other. It follows that $C=C'$, so $\ell=\ell'$. Thus $\Gamma$
 preserves $\ell$ which contradicts (\ref{noinvt}) unless $\operatorname{O}mega=\ell$, but then $\dim M=1$.\end{proof}
 

The following is a special case of Chuckrow's theorem (see \cite{CHU} or \cite{KAP}(8.4)).
\begin{lemma}\label{discfaith} If $M$ is a closed, strictly-convex, projective manifold
and $\dim M\ge 2$, then  the closure of $\operatorname{Rep}_S(M)$ in $\operatorname{Rep}(M)$ consists of discrete faithful
representations.
\end{lemma}
\begin{proof} Let $d$ be the metric on $G=\operatorname{GL}(m+1,{\mathbb R})$  given by $d(g,h)=\max|(g-h)_{ij}|$ and set $\|g\|=d(I,g)$.
Let $\mathcal W\subset\pi_1M$ be a finite generating set. 
Suppose the sequence $\rho_n\in \operatorname{Rep}_S(M)$ converges to $\rho_{\infty}\in \operatorname{Rep}(M)$.
Then there is a compact set $K\subset G$ such that $\rho_n(\mathcal W)\subset K$
for all $n$. 

The  map $\theta:G\times G\to G$ given by $\theta(g,h)=[[g,h],h]$  has zero derivative on $G\times\operatorname{I}$.
Thus there is a neighborhood $U\subset G$ of the identity such that if $k\in K$ and $u\in U$,  then 
$\|\; [[k,u],u]\;\|\le \|\;u\;\|/2$.

Since $\rho=\rho_n$ is discrete, there is $1\ne\alpha\in\pi_1M$ that minimizes $\|\rho(\alpha)\|$.
Suppose that $\rho(\alpha)\in U$. If $\beta\in \mathcal W$, then $\rho(\beta)\in K$, so $\|\;\rho[[\beta,\alpha],\alpha]\;\|\le \|\;\rho \alpha\;\|/2$. By minimality,
$\rho[[\beta,\alpha],\alpha]=\operatorname{I}$, and
since $\rho$ is injective,
$[[\beta,\alpha],\alpha]=\operatorname{I}$.  
 By (\ref{nilpotent}) $\alpha$ and $\beta$ commute.
Since $\mathcal W$ is a generating set, $\alpha$ is central in $\pi_1M$. Thus the entire group preserves the axis of $\alpha$.
This contradicts (\ref{noinvt}) because $\dim M \ge 2$.
Thus $\rho(\alpha)\notin U.$ Since $U$ is an open neighbourhood of the identity, it contains an open metric ball $U'$, and since $\rho(\alpha)$ is of minimal norm, we have $\rho(\gamma)\notin U'$ for all $1\neq \gamma \in \pi_1M.$ Since $U$ and $U'$ are independent of $n,$ we have $\rho_{\infty}(\alpha)\notin U$ for all $1\neq \gamma \in \pi_1M.$ 
This implies that $\rho_{\infty}$ is discrete and faithful.
\end{proof}

  

Let $\mathfrak B=\prod_{i=1}^n[-1,1]\subset{\mathbb R}^n\subset{\mathbb R}P^n$. For each $K>0$, the set $K\cdot \mathfrak B=\prod_{i=1}^n[-K,K]$ is called a {\em box}.

\begin{lemma}[Box estimate]\label{boxestimate} If $A=(A_{ij})\in \operatorname{GL}(n+1,{\mathbb R})$
 and $K\ge1$ and $[A](\mathfrak B)\subset K.\mathfrak B$ then $$|A_{ij}|\le 2K\cdot|A_{n+1,n+1}|$$
\end{lemma}
\begin{proof} Set $\alpha=A_{n+1,n+1}$. Using the standard basis we have
$$[x_1e_1+\cdots +x_ne_n+e_{n+1}]=[x_1:x_2:\cdots:x_n:1]=(x_1,\cdots,x_n)\in\mathfrak B\quad \Leftrightarrow\quad  \max_i |x_i|\le 1$$ 
First consider the entries $A_{i,n+1}$ in the last column of $A$. Since $[e_{n+1}]=0\in \mathfrak B$ we have 
$$[Ae_{n+1}]=[A_{1,n+1}e_1+A_{2,n+1}e_2+\cdots A_{n,n+1}e_n+\alpha e_{n+1}]\in K\cdot\mathfrak B$$ 
It follows that $|A_{i,n+1}/\alpha|\le K$. This establishes the bound
when $j=n+1$ and $i\le n$.

Next consider the entries $A_{n+1,j}$ in the bottom row with $j\le n$. Observe that 
$$p=[t e_j+e_{n+1}]\in\mathfrak B\quad\Leftrightarrow\quad |\;t\;|\le 1$$
Then $[A]p=[A(te_j+e_{n+1})]\in K\cdot \mathfrak B$. This is in ${\mathbb R}^n$ so the $e_{n+1}$ component
is not zero. Hence $t A_{n+1,j}+\alpha\ne 0$ whenever $|\;t\;|\le 1$ and it follows that $|A_{n+1,j}| <|\alpha|$.
Since $K\ge 1$ the required bound follows when $i=n+1$ and $j\le n$.

The remaining entries are $1\le i,j\le n$. Since  $p=[p_1:\cdots:p_{n+1}]=[A(te_j+e_{n+1})]\in K\cdot \mathfrak B$ it follows that
$$|t|\le1\quad{\mathbb R}ightarrow\qquad \left|\frac{p_i}{p_{n+1}}\right|=\left|\frac{A_{i,n+1}+t A_{i,j}}{\alpha + t A_{n+1,j}}\right| \le K$$
 For all $|t|\le1$ the denominator is not zero hence $|A_{n+1,j}|<|\alpha|$. It follows that
$$|\alpha + t A_{n+1,j}|\le 2|\alpha|$$
Thus 
$$|t|\le1\quad{\mathbb R}ightarrow\qquad\left|A_{i,n+1}+t A_{i,j}\right| \le 2K\cdot|\alpha|$$
We may choose the sign of $t=\pm1$ so that $A_{i,n+1}$ and $t A_{i,j}$ have the same sign. Then
$$ \left| A_{i,j}\right|\le \left|A_{i,n+1}+t A_{i,j}\right| $$
This gives the result
$ \left| A_{i,j}\right|\le  2|\alpha|\cdot K$ in this remaining case.
\end{proof}

If $\operatorname{O}mega\subset{\mathbb R}^n$ has finite positive Lebesgue measure and the {\em centroid} of $\operatorname{O}mega$ is
 $\hat{\mu}(\operatorname{O}mega)=0$, then
$$Q_{_{\operatorname{O}mega}}(y)=\int_K\left(\|x\|^2\|y\|^2-\langle x,y\rangle^2\right) \operatorname{dvol}_x$$  
 is a positive definite quadratic form on  ${\mathbb R}^n$ called the {\em inertia tensor}.
 
 
\begin{lemma}[Uniform estimate]\label{Klemma} For each dimension $n$ there is $K=K(n)>1$ such if $\operatorname{O}mega\subset{\mathbb R}^n$ is an open bounded convex
set with inertia tensor $Q_{_{\operatorname{O}mega}}=x_1^2+\cdots+x_n^2$ and centroid at the origin, then $K^{-1}\mathfrak B\subset\operatorname{O}mega\subset K\cdot\mathfrak B$. Moreover, if $A\in\operatorname{GL}(\operatorname{O}mega)$, then $|A_{ij}|\le K\cdot|A_{n+1,n+1}|$.
\end{lemma}
\begin{proof} The first conclusion follows from the theorem of Fritz John~\cite{Fritz}, see also  \cite{ball}.
Let $D\in\operatorname{GL}(n+1,{\mathbb R})$ be the diagonal matrix $\operatorname{Diag}(K,\cdots,K,1)$ then $\mathfrak B\subset D(\operatorname{O}mega)\subset K^2\mathfrak B$.
Set $A'=D\cdot A\cdot D^{-1}$ then $A'\in\operatorname{GL}(D(\operatorname{O}mega))$, thus $|A'_{ij}|\le 2K^2\cdot|A'_{n+1,n+1}|$ by  (\ref{boxestimate}).
Now $|A'_{n+1,n+1}|=|A_{n+1,n+1}|$ and $|A_{i,j}|\le K^2 |A'_{i,j}|$ thus
$|A_{ij}|\le 2K^4\cdot|A_{n+1,n+1}|$. The result now holds using the constant $2K^4$.
  \end{proof}

\subsection{Proof of (\ref{closed}. Closed)} Suppose $\rho\in\operatorname{Rep}(M)$ is the limit of the sequence $\rho_k\in\operatorname{Rep}_s(M)$.
Let $\operatorname{O}mega_k\subset{\mathbb R}P^n$ be the properly-convex open set preserved by $\Gamma_k=\rho_k(\pi_1M)$, then $M\cong M_k=\operatorname{O}mega_k/\Gamma_k$.
Choose
an affine patch ${\mathbb R}^n\subset{\mathbb R}P^n$. Then by (\ref{centerexists}) there is $\alpha_k\in \operatorname{PO}(n+1)$
such that $\alpha_k(\operatorname{O}mega_k)\subset{\mathbb R}^n$ has center $0\in {\mathbb R}^n$. We may choose $\alpha_k$ so that the
interia tensor $Q_k=Q(\alpha_k(\operatorname{O}mega_k))$ is diagonal in the standard coordinates on ${\mathbb R}^n$, and the entries on the main
diagonal of $Q_k$ are non-increasing going down the diagonal.
Since $\operatorname{PO}(n+1)$ is compact, after subsequencing
we may assume the conjugates of $\rho_k$ by $\alpha_k$ converge. 
We now replace the original sequence of representations
and domains by this new sequence.

Let $K=K(n)$ be given by (\ref{Klemma}). There is a unique positive diagonal matrix $D_k$
such that $Q_k=D_k^{-2}$. Set $\operatorname{O}mega'_k=D_k\operatorname{O}mega_k$, then $Q_{\operatorname{O}mega'_k}=x_1^2+\cdots+x_n^2$.
By (\ref{Klemma}), there is $K>1$ depending only on $n$, such that $$K^{-1}\mathfrak B_1\subset \operatorname{O}mega'_k\subset K\cdot\mathfrak B_1$$
Given $g\in\pi_1M$, then $A=A(k,g)=\rho_k(g)\in \operatorname{SL}(n+1,{\mathbb R})$ preserves $\operatorname{O}mega_k$.
 The matrix $B=B(k,g)=D_k A(k,g) D_k^{-1}$ preserves $\operatorname{O}mega'_k$. By (\ref{Klemma})  $$\forall i,j\quad |B_{i,j}|\le K\cdot |B_{n+1,n+1}|$$ Since $D_k$ is diagonal it follows that $$B_{n+1,n+1}=A_{n+1,n+1}$$
Now $A_{n+1,n+1}=A(k,g)_{n+1,n+1}$ converges as $k\to\infty$ for each $g$. Hence the entries of $B(k,g)$ are uniformly bounded for fixed
$g$ as $k\to\infty$.
 Thus we may pass to a subsequence where $B(k,g)=D_k\rho_k(g)D_k^{-1}$ converges
for every $g\in\pi_1M$, and this gives a limiting representation $\sigma=\lim D_k\rho_kD_k^{-1}$. 

The space consisting of properly-convex open sets $\operatorname{O}mega$
with $K^{-1}\cdot \mathcal B_1\subset\operatorname{O}mega\subset K\cdot\mathcal B_1$ is compact. Therefore there is a subsequence
 so that
$\operatorname{O}mega=\lim\operatorname{O}mega_k'$ exists. 
 Then $K^{-1}\mathfrak B_1\subset \operatorname{O}mega\subset K\cdot\mathfrak B_1$, hence
$\operatorname{O}mega$ is open and properly-convex. Then $\sigma$ is discrete and faithful by (\ref{discfaith}), and $\Gamma=\sigma(\pi_1M)$
preserves $\operatorname{O}mega$,
so $N=\operatorname{O}mega/\Gamma$ is a properly-convex manifold. Since $M$ is closed and $\pi_1M\cong \pi_1N$, then by (\ref{htpyequiv})
 $N$ is also closed. Since $M$ is strictly-convex, (\ref{stricthomeo}) implies
 $N$ is
 strictly-convex. 
 

Since $\operatorname{Rep}_P(N)$ is open, for $D_k\rho_k(g)D_k^{-1}$ close enough to $\sigma$, there is a properly-convex open set $\operatorname{O}mega'_k\subset{\mathbb R}P^m$
that is preserved by $\Gamma_k$ and $N_k=\operatorname{O}mega'_k/\Gamma_k$
 is homeomorphic to $N$. By (\ref{stricthomeo}) $N_k$ is strictly-convex. By (\ref{unique}) $\operatorname{O}mega'_k=\operatorname{O}mega_k$ so $N\cong N_k=M_k\cong M$. Thus $\sigma\in\operatorname{Rep}_S(M)$. 
 
 If $D_k$ does not remain bounded, since the entries are non-increasing going down the diagonal,  and $\rho_k(g)$ remains bounded,
 $\sigma(g)$ is block upper triangular  and therefore $\sigma$ is reducible. Since $N$ is strictly-convex,  (\ref{irreducible}) implies $\sigma$ is irreducible,
therefore $D_k$ stays bounded. Hence we may subsequence the $D_k$ so they converge. Then $\sigma=\rho$, so $\rho\in\operatorname{Rep}_S(M)$. 
 \qed

\small

 

\end{document}