Lecture 19: Closedness Part I

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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