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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