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\begin{document}
Let $\Omega$ be a domain in ${\Bbb R}P^n$. Suppose that there
exist a neighborhood $U$ of $x\in\partial\Omega$ and a real-valued
function $\rho$ defined on $U\cap\Omega$ satisfying\par

\begin{itemize}
\item $\rho$ is twice differentiable on $U\cap\Omega$.
\item  $U\cap\Omega$ is defined by $\rho(x)<0$.
\item The gradinet $\nabla\rho={\d\left(\frac{\partial\rho}{\partial
x_1},\cdots,\frac{\partial\rho}{\partial x_n}\right)}$ is never zero at any point of
$U\cap\partial\Omega$.
\end{itemize}
In such a case, $\rho$ is called a defining function of $\Omega$
at $x$. Notice that by the Implicit Function Theorem, the boundary
$U\cap\partial\Omega$ is $C^2$ smooth.\par\pagebreak Let $Q$ be a
quadratic form

\begin{eqnarray*}
\lefteqn{Q(a_2\cdots,a_n)=}\\ &&\{[x_0:\cdots:x_n]\mid
x_0x_1+a_2x_2^2+\cdots+a_nx_n^2<0\} \end{eqnarray*} where
$a_i\in{\Bbb R}$. We denote the Hessian of $\rho(x)$ at $p$ by
$Hess_{\rho}(p)$. First, we consider a projective domains with
$\det Hess_\rho(p)\ne 0$ at a boundary accumulation
point.\par\pagebreak For $x\in{\Bbb R}P^n$, we define {\it
$n-$simplex at $x$} to be a convex $n-$simplex with a vertex $x$
and we call {\it the end point of the $n-$simplex at $x$} to be
the other vertex of the $n-$simplex at $x$ and {\it the axis of
the $n-$simplex at $x$} to be the line segment connecting $x$ and
the other vertex of the $n-$simplex at $x$. We call the simpex
with vertexes $(1,0,\cdots,0)$, $\cdots$, $(0,\cdots,0,1)$ by the
standard unit simplex at $0$.\pagebreak
\begin{thm}
Let $\Omega$ be a projective domain not necessarily hyperbolic
with non-compact automorphism group $Aut(\Omega)$. Assume that
there is a sequence $\{\varphi_\nu\}\subset Aut(\Omega)$ such that
for a subsequential limit $\varphi$,
$R(\varphi)=p\in\partial\Omega$. If $\partial\Omega$ is $C^2$ in a
neighborhood of $p$ and the Hessian at $p$ is non-singular, then
$\Omega$ is projectively equivalent to one of $Q(a_2,\cdots,a_n)$
with $a_\ell\ne 0$ for all $\ell\in\{2,\cdots,n\}$.
\end{thm}\par\pagebreak
{\em PROOF}\quad We'll proove the Theorem for $n=2$. One can
assume that $a_2>0$. Then $p$ is a strictly convex point. One can
see that $q\not\in K(\varphi)$ for a subsequential limit $\varphi$
of $\{\varphi_\nu\}$. If one takes an open neighborhood
$N\subset\Omega$ of $q$ such that $N\cap K(\varphi)=\emptyset$,
the limit $\lim\limits_{\nu\ra\infty} \varphi_\nu(N)$ is open in
$R(\varphi)$ containing $p$. Clearly, $R(f)$ is tangential to
$\partial\Omega$ at $p$. It follows that $\dim
R(\varphi)\cap\partial\Omega=\dim R(\varphi)$. By strictly
convexity, one obtains $R(\varphi)=\{p\}$.\par\pagebreak Take a
simplex $E_0$ at $p_0\in\partial\Omega$ such that
$\overline{E}_0\cap K(\varphi)=\emptyset$ and one axis is tangent
to $\partial\Omega$ at $p_0$. Then the other axis is transversal
to $T_{p_0}\partial\Omega$.\par Take an affine coordinate
$\phi:U_0\longrightarrow{\Bbb R}^2$ with $\phi(p)=0$ and $p_0\in
U_0$. Denote $\varphi_\nu(p_0)=(c_1^\nu,c_2^\nu)=c^\nu$ and let
$\{e_1^\circ,e_2^\circ\}$ be the end points of the axes of $E_0$,
where the axis having end point $e_1^\circ$, is not tangential to
$\partial\Omega$. Let
$d^\nu=(d_1^\nu,d_2^\nu)=\varphi_\nu(e_1^\circ)$ be the end point
of the axis of $\varphi_\nu(E_0)$ which is transversal to
$T_{\varphi_\nu(p_0)}\partial\Omega$. We know that
$\varphi_\nu(\overline{E_0})\rightarrow\varphi(\overline{E_0})=p$
uniformly. Hence $\varphi_\nu(p_0)\rightarrow p=0$ and
$d^\nu\rightarrow 0$ uniformly.\par From the Talyor expansion of
second order, there exists a neighborhood $U$ of $p$ such that
$U\cap\Omega$ is defined by

$$ \rho(x)=x_1+\delta_2x_2^2+o(|x_2|^2)<0 $$ where $\rho$ is the
defining function and $\delta_2\in{\Bbb R}$. Note that one has
$\delta_2\ne0$ since $\det Hess_{\rho}(p)\ne0$.
\par\pagebreak\noindent (I) For each $\nu$,

$$ \Phi_\nu=\left\{\begin{array}{lr} y_1=
x_1-c_1^\nu-a_2^\nu(x_2-c_\ell^\nu)&\\
y_2=x_2-c_2^\nu\end{array}\right. $$ where $a_2^\nu$ are
determined so that, in the coordinate $(y_1,y_2)$, we have
\begin{enumerate}
\item $\Phi_\nu(c^\nu)$ is the point $(0,0)\in\partial\Omega$
\item the tangent space to $\partial\Omega$ at $(0,0)$ is given
by $\{y\mid y_1=0\}$.
\end{enumerate}
If $\Phi_\nu(d^\nu)=(\beta_1^\nu,\beta_2^\nu)$, $\beta_1^\nu=
d_1^\nu-c_1^\nu-a_2^\nu(d_2^\nu-c_2^\nu)$. Moreover, one can take
the simplex $E_0$ at $p_0$ so that $\beta_1^\nu>0$ for all $\nu$.
For all sufficiently large $\nu$, ${\d\frac{\partial\rho}{\partial
x_1}(c^\nu)}\ne 0$ since $c^\nu\rightarrow 0$ as
$\nu\rightarrow\infty$.\par\pagebreak Determine $a_2^\nu$ so that

$$\begin{array}{lll}0=\frac{\partial\rho_\nu}{\partial y_2}(0)&=
&\sum_{k=1}^2\frac{\partial\rho}{\partial x_k}(c^\nu)
\frac{\partial\Phi_k^{-1}}{\partial y_2}(0)\\ &&\\
&=&a_2^\nu\frac{\partial\rho}{\partial x_1}(c^\nu)+
\frac{\partial\rho}{\partial x_2}(c^\nu)\end{array}$$ It follows
that $$ 0\neq\frac{\partial\rho_\nu}{\partial y_1}(0)=
\sum_{k=1}^2\frac{\partial\rho}{\partial x_k}(c^\nu)
\frac{\partial\Phi_k^{-1}}{\partial y_1}(0)=
\frac{\partial\rho}{\partial x_1}(c^\nu)\\ $$ Since all
$c_\ell^\nu$'s, $d_\ell^\nu$'s, $a_\ell^\nu$'s approach $0$ as
$\nu\rightarrow\infty$, we have $\beta_1^\nu\rightarrow 0$ as
$\nu\rightarrow\infty$.\par The defining function $\rho_\nu(y)$
is, in this coordinate,

$$ \rho_\nu(y)=y_1+A_2^\nu y_2^2+\delta_2y_2^2+ o(|y_2|^2)<0 $$
where $A_2^\nu$ is a function of $c_2^\nu$ satisfying

$$
\begin{array}{lr}
A_2^\nu\ra 0&\mbox{ as }\nu\ra\infty.
\end{array}
$$ \par\pagebreak One axis of
$E_\nu=(\Phi_\nu\circ\varphi_\nu)(E_0)$ lie in the tangent space
$T_{(\Phi_\nu\circ\varphi_\nu)(p_0)}\partial\Omega$ and the other
axis is transversal to this tangent space by definition of
$\Phi_\nu$ and $\varphi_\nu\in Aut(\Omega)$.  So, $\beta_1^\nu$
cannot be $0$ for all $\nu$. Define $\epsilon_\nu>0$ to be
$\beta_1^\nu$. Note that $\epsilon_\nu\rightarrow 0$ as
$\nu\rightarrow\infty$ since $\varphi_\nu(E_0)\rightarrow p$ by
$R(\varphi)=p$.\par \noindent (II) For each $\nu$, define

$$ \Psi_\nu=\left\{\begin{array}{lr} u_1=y_1&\\
u_2={\d-\frac{\beta_2^\nu}{\beta_1^\nu}y_1+y_2}\\
\end{array}\right.
$$ This linear transformation send the axis of the simplex $E_\nu$
to the normal direction at $0$ i.e. on the $x_1$-axis and the
other axis are invariant i.e. belong to the tangent space of the
boundary $\partial\Phi_\nu(\Omega)$ at $0$.\par\pagebreak The
defining function $\rho_\nu(u)$ is, in $u-$coordinate,

$$
\begin{array}{rr}{\d\rho_\nu(u)=u_1+A_2^\nu(u_2
+\frac{\beta_2^\nu}{\beta_1^\nu}u_1)+\delta_2(u_2
+\frac{\beta_2^\nu}{\beta_1^\nu}u_1)^2}&\\ +{\d
o\left(|(u_2+\frac{\beta_2^\nu}{\beta_1^\nu}
u_1)|^2\right)<0}&\end{array} $$ \noindent (III) By two affine
coordinate changes above, one can see that
$(\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(e_2^\circ)$ has the first
coordinate $0$ and we also have
$(\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(e_1^\circ)=(\beta_1^\nu,0)$
with $\beta_1^\nu>0$. Denote
$B_\ell^\nu=|(\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)
(e_\ell^\circ)|$. Let $(0,r_{2\ell}^\nu)=
(\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(e_2^\circ)$.\par Define
$R_\nu$ to be the affine transformation defined by the matrix
${\d\left(\frac{r_{ab}^\nu}{B_b^\nu}\right)^{-1}}$ on
$T_0(\partial D_\nu)$, where
$D_\nu=(\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(\Omega)$, so that

$$ R_\nu((\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(e_\ell^\circ)\in
u_j{\mbox -axis} $$ for some $j$ which depends on
$\ell$.\par\pagebreak In this $w-$coordinate,

$$\begin{array}{ll}
\rho_\nu(w)=w_1&\\
&+A_2^\nu{\d\left(\frac{\beta_2^\nu}{\beta_1^\nu}w_1
+\frac{r_{22}^\nu}{B_2^\nu}w_2\right)^2}\\
&{\d +\delta_2\left(\frac{\beta_2^\nu}{\beta_1^\nu}w_1
+\frac{r_{22}^\nu}{B_2^\nu}w_2\right)^2}\\
&+{\d o\left(|(\frac{\beta_2^\nu}{\beta_1^\nu}w_1+
\frac{r_{22}^\nu}{B_2^\nu}w_2)|^2\right)}<0\end{array}
$$
\noindent (IV) Define an affine coordinate change $\Gamma_\nu$ as follows.

$$
\Gamma_\nu=\left\{\begin{array}{lr}\tilde{x}_1=w_1&\\
\tilde{x}_2={\d\frac{\sqrt{\epsilon_\nu}}{B_2^\nu}w_2}
\end{array}\right.
$$ Now, all axes of the simplex $(\Gamma_\nu\circ
R_\nu\circ\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(E_0)$ which are
tangential to $(\Gamma_\nu\circ
R_\nu\circ\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu)(\Omega)$ at $0$
are of length $\sqrt{\epsilon_\nu}$.\par\pagebreak In this
$\tilde{x}-$ coordinate,

$$
\begin{array}{ll}
\rho_\nu(\tilde{x})=\tilde{x}_1&\\
&+{\d A_2^\nu\left(
\frac{\beta_2^\nu}{\beta_1^\nu}\tilde{x}_1
+\frac{r_{22}^\nu}{\sqrt{\epsilon_\nu}}\tilde{x}_2\right)^2}\\
&{\d +\delta_2\left(\frac{\beta_2^\nu}{\beta_1^\nu}\tilde{x}_1
+\frac{r_{22}^\nu}{\sqrt{\epsilon_\nu}}\tilde{x}_2\right)^2}\\
&{\d+o\left(|({\d\frac{\beta_2^\nu}{\beta_1^\nu}\tilde{x}_1+
r_{22}^\nu\frac{\tilde{x}_2}{\sqrt{\epsilon_\nu}})|^2}\right)<0}
\end{array}
$$ Let $D_\nu$ be the intersection of a small neighborhood $U$ of
$p=0$ and the domain defined by $\rho_\nu(\tilde{x})<0$. $$
L_\nu=\left\{\begin{array}{lr
}\hat{x}_1=\tilde{x}_1/\epsilon_\nu&\\
\hat{x}_2=\tilde{x}_2/\sqrt{\epsilon_\nu}\end{array}\right. $$
Recall that $\epsilon_\nu=\beta_1^\nu$. This sequence preserves
the tangent space and acts as a homothety on the tangent space at
$0$ and each transformation sends the last simlex to the standard
unit simplex at $0$.\par\pagebreak Then the defining function
$\rho_\nu(\hat{x})$ of the domain $L_\nu(D_\nu)$ can be written as

$$
\begin{array}{ll}
\rho_\nu(\hat{x})=\epsilon_\nu\hat{x}_1&\\
&{\d+ A_2^\nu
\left(\beta_2^\nu\hat{x}_1+r_{22}^\nu\hat{x}_2\right)^2}\\
&{\d+\delta_2\left(\beta_2^\nu\hat{x}_1
+r_{22}^\nu\hat{x}_2\right)^2}\\
&{\d+o\left(|(\beta_2^\nu\hat{x}_1
+r_{22}^\nu\hat{x}_2)|^2\right)<0}
\end{array}
$$
By our construction, we see that all $\sqrt{\epsilon_\nu}$, $\beta_\ell^\nu$'s
and $r_{ab}^\nu$'s
approach $0$ as $\nu\rightarrow\infty$. Compare the decreasing rates of these values
and divide the above equation by $\{|\alpha_\nu|^2\}$, where $\{\alpha_\nu\}\in
\{\{\sqrt{\epsilon_\nu}\},\{\beta_2^\nu\},\{r_{2}^\nu\}\}$
has the slowest decreasing rate by extracting a subsequence if necessary.

$$
\begin{array}{ll}
\rho_\nu(\hat{x})={\d\frac{\epsilon_\nu}{|\alpha_\nu|^2}\hat{x}_1}\\
&{\d+A_2^\nu
\left(\frac{\beta_\sigma^\nu}{|\alpha_\nu|}\hat{x}_1
+\frac{r_{2}^\nu}{|\alpha_\nu|}\hat{x}_k\right)^2}\\
&{\d+\delta_\ell\left(\frac{\beta_2^\nu}{|\alpha_\nu|}\hat{x}_1+
\frac{r_{22}^\nu}{|\alpha_\nu|}\hat{x}_2\right)^2}\\
&{\d+\frac1{|\alpha_\nu|^2}o\left(|(\beta_2^\nu\hat{x}_1
+r_{22}^\nu\hat{x}_2)|^2\right)<0}\end{array}
$$
Then all limits satisfy

$$
\begin{array}{lll}
\lim\limits_{\nu\rightarrow\infty}
{\d\left|\frac{\sqrt{\epsilon_\nu}}{\alpha_\nu}\right|\leq 1}&
\lim\limits_{\nu\rightarrow\infty}{\d\left|\frac{\beta_\ell^\nu}{\alpha_\nu}\right|\leq 1}&
\lim\limits_{\nu\rightarrow\infty}{\d\left|\frac{r_{ab}^\nu}{\alpha_\nu}\right|\leq 1}
\end{array}
$$
for $\ell,a,b=2$. Also, observe that

$$
\begin{array}{ccc}
&c_\ell^\nu\rightarrow 0&\\
&A_{22}^\nu\rightarrow 0&\end{array}
$$
$$
{\d\frac1{|\alpha_\nu|^2}o\left(|(\beta_2^\nu\hat{x}_1+
r_{22}^\nu\hat{x}_2)|^2\right)}\rightarrow 0
$$
as $\nu\rightarrow\infty$.
Hence the limiting defining function becomes a quadratic since $\delta_2\ne0$
and by a linear change of coordinate if necessary,
it must be

$$
\begin{array}{lr}
\hat{\rho}(x)=x_1+{\d\delta_2 x_2^2}<0&\mbox{ with }
\delta_2\ne 0
\end{array}\leqno{(*)}
$$
The domain $L_\nu(D_\nu)$ converges to the domain $\hat{\Omega}$ defined by $(*)$.
Note that $\hat{\Omega}=Q(a_2)$. Combining all the coordinate changes above,
we have a sequence of projective transformation

$$
G_\nu:=L_\nu\circ\Gamma_\nu\circ R_\nu\circ\Psi_\nu\circ\Phi_\nu\circ\varphi_\nu
$$
A subsequential limit yields a mapping

$$ G:\Omega\longrightarrow\hat{\Omega}\in P{\frak g}{\frak
l}(3,{\Bbb R}) $$ \par\pagebreak Here is the crucial step but it
is a very short proof.
\begin{lem}
$G:\Omega\longrightarrow\hat{\Omega}$ is a projective transformation i.e. $\hat{\Omega}$
is projectively equivalent to $\Omega$.
\end{lem}
{\em PROOF}\quad By our construction of $\Phi_\nu$, $\Psi_\nu$,
$R_\nu$, $\Gamma_\nu$, and $L_\nu$, $G_\nu(E_0)$ is the standard
unit simplex at $0$ for all $\nu$. So, $\overline{E}_0\cap
K(G)=\emptyset$ if $G$ were singular. Otherwise, there must exist
an affine full line contained in
$\lim\limits_{\nu\rightarrow\infty}
G_\nu(\overline{E}_0)$\qquad=the closed standard unit simplex at
$0$. Since $G_\nu\mid_{\overline{E}_0}\rightarrow
G\mid_{\overline{E}_0}$ uniformly, $G$ cannot be singular.
\rule{2mm}{2mm}\par


\end{document}
