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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