Lecture 20: Closedness Part II

Lemma (Box estimate). Let $B = \prod^n_{i = 1} [-\lambda_i, \lambda_i] \subset \mathbb R^n$ and $\lambda_{n + 1} = 1$ , where $\lambda_i > 0$ for each $i$ and $\mathbb R^n$ is identified with an affine patch in $\mathbb R\mathrm P^n$ . Suppose $A = (A_{ij}) \in \mathrm{SL}_\pm(n + 1; \mathbb R)$ and $\kappa > 1$ are given such that $[A](B) \subset \kappa \cdot B$ , and let $\alpha = A_{n + 1, n + 1}$ . Then

$|A_{ij}| \leq 2 |\alpha| \kappa \frac{\lambda_i}{\lambda_j} \quad \text{for all }1 \leq i, j \leq n + 1.$

Proof. Let $\{e_1, \dots, e_{n + 1}\}$ be the standard basis for $\mathbb R^{n + 1}$ so that $\{e_1, \dots, e_n\}$ is the standard basis for $\mathbb R^n \subset \mathbb R^{n + 1}$ . If

$p = (x_1, \dots, x_n) = x_1 e_1 + \dots + x_n e_n = [x_1 e_1 + \dotsb + x_n e_n + e_{n + 1}] \in \mathbb R^n,$

then

$p \in B \Longleftrightarrow |x_i| \leq \lambda_i \quad \text{for each } 1 \leq i \leq n.$

Step 1: We prove for the last column of $A$ .

Observe that $[e_{n + 1}] = 0 \in B$ . The last column of $A$ is $A e_{n + 1}$ , and

$[A e_{n + 1}] = [A_{1, n + 1} e_1 + \dotsb + A_{n, n + 1} e_n + \alpha e_{n + 1}] \in \kappa \cdot B,$

so $\alpha \neq 0$ . Scaling by $\alpha$ gives

$\left| \frac{A_{i, n + 1}}{\alpha} \right| \leq \kappa \lambda_i,$

so that

$|A_{i, n + 1}| \leq |\alpha| \kappa \lambda_i \leq 2 |\alpha| \kappa \frac{\lambda_i}{\lambda_{n + 1}}$

for each $1 \leq i \leq n + 1$ , since $\lambda_{n + 1} = 1$ .

Step 2: We prove for the last row of $A$ .

Let $1 \leq j \leq n$ , and let $p$ be a point in $B$ on the line through $[e_{n + 1}]$ in the direction of $e_j$ . Then

$p = [t e_j + e_{n + 1}] \quad \text{for some } |t| \leq \lambda_j.$

We have

$[A] p = [A(t e_j + e_{n + 1})] \in \kappa \cdot B,$

so the coefficient of $e_{n + 1}$ in $[A] p$ is nonzero. It follows that

$t A_{n + 1, j} + \alpha \neq 0 \quad \text{for all } |t| \leq \lambda_j,$

whence

$|A_{n + 1, j}| < \frac{|\alpha|}{\lambda_j} < 2|\alpha| \kappa \frac{\lambda_{n + 1}}{\lambda_j}$

for each $1 \leq j \leq n + 1$ , since $\lambda_{n + 1} = 1$ .

Step 3: We prove for the remaining entries of $A$ .

Let $1 \leq i, j \leq n$ , and let ${\mid}t{\mid} \leq \lambda_i$ . Observe that $A (t e_j + e_{n + 1})$ is $t$ times the $j$ th column of $A$ plus the $(n + 1)$ st column of $A$ . Write

$[p_1 : \dots : p_{n + 1}] = [A(t e_j + e_{n + 1})] \in \kappa \cdot B.$

Then

$\left| \frac{p_j}{p_{n + 1}} \right| = \left| \frac{t A_{ji} + A_{j, n + 1}}{t A_{n + 1, i} + \alpha} \right| \leq \kappa \lambda_j.$

Note that the denominator above is never zero, so that ${\mid}t A_{n + 1, j}{\mid} < {\mid}\alpha{\mid}$ , and hence ${\mid}A_{n + 1, j}{\mid} \leq {\mid}\alpha{\mid} / \lambda_j$ . Therefore,

$|t A_{n + 1, i} + \alpha| \leq 2|\alpha|,$

so that

$|t A_{ji} + A_{j, n + 1}| \leq 2|\alpha| \kappa \lambda_j.$

Now choose the sign of $t \in [-\lambda_i, \lambda_i]$ so that $t A_{ji}$ and $A_{j, n + 1}$ have the same sign. Then

$\lambda_i |A_{ji}| \leq 2|\alpha|\kappa \lambda_j.$

Dividing by $\lambda_i$ and switching $i$ and $j$ completes the proof. $\square$

Theorem. Let $M$ be a closed properly convex real projective $k$ -dimensional manifold. Suppose that $\pi_1 M$ has no infinite normal nilpotent subgroup. Then $\chi_c(M)$ is closed in $\chi(M)$ .

Remark 1. If representations $\rho, \rho' : G \to \mathrm{GL}(k; \mathbb R)$ have the same character and $\rho$ is irreducible, then $\rho$ and $\rho'$ are conjugate.

Remark 2. If $\pi_1 M$ is word-hyperbolic (i.e., it’s finitely generated and its Cayley graph has $\delta$ -thin triangles for some $\delta > 0$ ) and $\sigma \in \mathrm{Rep}_c(M)$ , then $\sigma$ is irreducible.

Proof of theorem. Let $\rho_n \in \mathrm{Rep}_c(M)$ be a sequence of holonomies of properly convex projective structures on $M_n = \Omega_n / \Gamma_n$ , where $\Gamma_n = \rho_n(\pi_1 M)$ , and let $\rho_n \to \rho_\infty$ . We claim there is a $\sigma \in \mathrm{Rep}_c(M)$ with the same character as $\rho_\infty$ .

Suppose that such a $\sigma$ exists. We show there exists a sequence $A_n \in \mathrm{SO}(k + 1)$ such that the sequence $\Omega'_n = A_n(\Omega_n)$ converges in the Hausdorff topology to a properly convex domain $\Omega'_\infty$ , and the sequence $\rho'_n = A_n \rho_n A^{-1}_n$ subconverges to $\sigma$ . The $\rho'_n$ ’s are holonomies of properly convex projective structures and hence discrete and faithful. Since $\pi_1 M$ has no infinite normal nilpotent subgroup, $\sigma$ is discrete and faithful by Chuckrow’s theorem.

Now $M'_\infty = \Omega'_\infty / \sigma(\pi_1 M)$ is a properly convex projective manifold, and $M'_\infty$ is diffeomorphic to $M$ because $M_n$ Gromov–Hausdorff converges to $M'_\infty$ , so that $M_n$ is diffeomorphic to $M'_\infty$ for large $n$ . $\square$

Proof of claim. By performing a rotation if necessary, we may assume without loss of generality that for each $n$ , there exists a domain $\Omega'_n = A_n(\Omega_n)$ , with $A_n \in \mathrm{SO}(k + 1)$ , and a center for which there is a new domain $\Omega''_n \subset \mathbb R^k \subset \mathbb R\mathrm P^k$ and a box $B_n$ such that we have the following diagram:

$\begin{matrix} B_1 & \subseteq & \Omega'_n & \subseteq & \kappa \cdot B_1 \\ \downarrow & & \downarrow & & \downarrow \\ B_n & \subseteq & \Omega''_n & \subseteq & \kappa \cdot B_n \end{matrix}$

Here $B_1$ is the unit box, $\kappa$ depends only on the dimension $k$ ,

$B_n = \prod^k_{i = 1} [-\lambda_{n i}, \lambda_{n i}]$

for some $\lambda_{n i} > 0$ ,

$\Omega'_n = D^{-1}_n(\Omega''_n), \qquad \rho'_n = D^{-1}_n \rho''_n D_n, \qquad \rho'_n \curvearrowright \Omega'_n, \qquad \rho''_n \curvearrowright \Omega''_n$

and

$D_n = \mathrm{diag}(\lambda_{n1}, \dots, \lambda_{nk}, 1)$

is the vertical map in the diagram above. Since $\mathrm{SO}(k + 1)$ is compact, the sequence $\rho''_n$ subconverges, so the sequence $\Omega'_n$ subconverges to $\Omega'_\infty$ .

By the box estimate lemma, the entries of each matrix $A_n$ are bounded by a constant multiple of ${\mid}(A_n)_{k + 1, k + 1}{\mid}$ . It is possible that ${\mid}(A_n)_{k + 1, k + 1}{\mid} \to \infty$ ; however, since we are considering projective maps, we can rescale if necessary so that $(A_n)_{k + 1, k + 1} = 1$ for all $n$ . Then the sequence $\rho'_n$ is bounded, so it subconverges. $\square$

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