# 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*

*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

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:

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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