Let $$M$$ be a closed $$m_0$$-manifold, and let $$Rep(M)=Hom(\pi_1(M),SL^{\pm}(m_0+1,R))$$, $$\chi(M)=\{tr\circ\rho\colon \pi_1(M)\to R \mid \rho\in Rep(M)\}$$, $$Rep_c(M)=\{\rho\in Rep(M)\mid \rho\text{ is holonomy of properly convex projective structure on }M\}$$, and $$\chi_c(M)=\{tr\circ \rho\mid \rho\in Rep_c(M)\}$$

Theorem 1: $$\chi_c(M)$$ is closed in $$\chi(M)$$.

Chuckrow’s theorem: $$G$$ a finitely generated group that does not contain an infinite nilpotent normal subgroup (eg $$G=\pi_1(M)$$, $$M$$ a closed hyperbolic manifold of dimension at least 2). Then the subset of $$Hom(G,GL(k,R))$$ consisting of discrete, faithful homomorphisms is closed.

Idea: if $$A,B\in GL(n,R)$$ are $$\epsilon$$-close to the identity, their commutator is $$\epsilon^2$$-close.

Overview of theorem 1: Sequence of projective structures by taking $$\Gamma_n=\rho_n(\pi_1(M))$$, and consider $$\Omega_n/\Gamma_n$$.

Open bounded convex set $$K\subseteq R^n$$. Define $$\widetilde{q}_K\colon R^n\to R$$ with $$\widetilde{q}_K=\int_K\|x-y\|^2dV$$. This attains the minimum at a unique point $$\mu(K)$$, called the center of mass. Define the moment of inertia (tensor) to be $$q_K(y)=\widetilde{q}_K(y)-\widetilde{q}_K(\mu(K))$$, which is a positive-definite quadratic form (when considering $$\mu(K)$$ as the origin). There is a unique ellipsoid $$\Omega$$ with $$q_\Omega=q_K$$, called the ellipsoid of inertia.

Let $$\Omega\subseteq S^n$$ is open and properly convex so that $$\overline{\Omega}$$ is disjoint from some great sphere. A point $$p\in\Omega$$ is called a center of $$\Omega$$ if, letting $$\overline{\Omega}\subseteq U_p=\{x\in S^n\mid \langle x,p\rangle >0\}$$, and $$\pi_p\colon U_p\to R^n$$ by $$\pi_p(x)=\frac{x}{\langle x,p\rangle}$$, we have $$\pi_p(p)=\mu(\pi_p\Omega)$$.

Prop: If $$\Omega\subseteq S^n$$ is properly convex and open, then $$\Omega$$ has a unique center, which is in $$\Omega$$.

Proof: Prove when $$\Omega$$ is round, meaning strictly convex and $$\partial\Omega$$ is $$C^1$$. Then approximate arbitrary $$\Omega$$ by round ones and use continuity.

Let $$\Omega^*$$ be the dual in $$S^n$$, using the inner product to identify $$R^n$$ with its dual. Namely, $$\Omega^*=\{y\in S^n\mid d_{S^n}(x,y)<\pi/2\forall x\in\Omega\}$$.

Define $$m\colon \Omega^*\to \Omega$$ by $$m(p)=\pi_p^{-1}(\mu(\pi_p(\Omega)))$$. Then $$p$$ is the center if and only if $$m(p)=p$$.

Define $$\psi\colon \partial \overline{\Omega}^*\to\partial \overline{\Omega}$$ by $$\psi(x)=x^\perp\cap\overline{\Omega}$$. Strict convexity implies $\psi$ is one-to-one, and roundness implies $$\psi$$ is continuous, and $$\psi$$ is a continuous extension of $$m$$. Extend $$m$$ by $$m\vert_{\partial\Omega}=\psi$$. If $$m$$ has no fixed point, let $$\ell$$ be the line through $$p$$ and $$m(p)$$ and $$r(p)$$ the point on $$\ell\cap\partial\overline{\Omega}^*$$ closer to $$p$$. This gives $$r\colon \overline{\Omega}^*\to \partial \overline{\Omega}^*$$ and $$r\vert_{\partial \overline{\Omega}^*}$$ the identity, a contradiction.

A box is a product of intervals in $$R^n$$; given a box $$B$$ and $$k>0$$, $$kB=\{kx\mid x\in B\}$$, which is another box. The unit box is $$B_1=\prod_{i=1}^n[-1,1]$$.

Fact 1: For any dimension $$n$$, there is a $$k$$ so that for any $$\Omega\subseteq R^n$$ which is open, bounded, and convex with inertia tensor $$x_1^+\dots +x_n^2$$ and the origin the center of mass, we have $$k^{-1}B_1\subseteq \Omega\subseteq kB_1$$.

Corollary: If $$\Omega\subseteq S^n$$ is open and properly convex, there is an $$A\in SO(n+1)$$ and a box $$B\subseteq R^n$$ with $$B\subseteq \pi_{e_1}(A\Omega)\subseteq k^2B$$.

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