Thin right-angled Coxeter groups in some uniform arithmetic lattices

The proof of the main theorem in [BdlH04] applies here, so we only sketch the argument provided there. Let $d\geq D$ and let $G_{d}$ be the Zariski-closure of $W_{d}$ in $\mathrm{O}(M_{d};\mathbb{R})$. Denote by $\mathfrak{g}$ and $\mathfrak{h}$ the Lie algebras of $\mathrm{O}(M_{d};\mathbb{R})$ and $G_{d}$, respectively. It is enough to show that $\mathfrak{g}=\mathfrak{h}$, since the Zariski-closure of $\mathrm{SO}(M_{d};\mathbb{R})^{\circ}$ is $\mathrm{SO}(M_{d};\mathbb{R})$ and since $W_{d}\not\subset\mathrm{SO}(M_{d};\mathbb{R})$.

For each distinct pair $i,j\in\{1,\ldots,n\}$, let $E_{i,j}$ be the orthogonal complement of $\langle e_{i},e_{j}\rangle$ in $\mathbb{R}^{n}$ with respect to $M_{d}$. The subgroup of $\mathrm{O}(M_{d};\mathbb{R})$ consisting of all elements that fix each vector in $E_{i,j}$ is a $1$-dimensional closed subgroup of $\mathrm{O}(M_{d};\mathbb{R})$ whose identity component $T_{i,j}$ corresponds to a subspace $\langle X_{i,j}\rangle$ of $\mathfrak{g}$ for some ${X_{i,j}\in\mathfrak{g}}$. Since $M_{d}$ is nondegenerate, the elements $X_{i,j}$ form a basis for $\mathfrak{g}$ as a vector space [BdlH04, Lemme 7]. Thus, to show $\mathfrak{g}=\mathfrak{h}$, it suffices to show that $X_{i,j}\in\mathfrak{h}$ for each distinct pair $i,j\in\{1,\ldots,n\}$.

To that end, let $i,j\in\{1,\ldots,n\}$, $i\neq j$, and suppose first that $v_{i}$ and $v_{j}$ are adjacent in $\Sigma_{1}$. Then $\sigma_{d}(\gamma_{i}\gamma_{j})$ generates an infinite cyclic subgroup of $T_{i,j}$, so that $T_{i,j}\subset G_{d}$. It follows that $X_{i,j}\in\mathfrak{h}$ in this case. One now verifies that, since $\Sigma_{1}$ is connected, any Lie subalgebra of $\mathfrak{g}$ that contains $X_{i,j}$ for all $i,j$ such that $v_{i},v_{j}$ are adjacent in fact contains $X_{i,j}$ for each distinct pair $i,j\in\{1,\ldots,n\}$ [BdlH04, Preuve du Théorème, second cas]. ∎

We show that $W_{\alpha}$ is a thin subgroup of $\Gamma\subset\mathrm{O}(M_{\alpha};\mathbb{R})$. By Lemma 2, it suffices to show that $W_{\alpha}$ is of infinite index in $\Gamma$. Indeed, suppose otherwise. Then $W_{\alpha}$ is a uniform lattice in $\mathrm{O}(M_{\alpha};\mathbb{R})$. If $n=3$, then immediately we obtain a contradiction, since in this case $W_{\alpha}$ is virtually a closed hyperbolic surface group, whereas $W$ is virtually free. Now suppose $n>3$. There is some $\beta>\alpha$ and a path $[\alpha,\beta]\rightarrow\mathrm{GL}_{n}(\mathbb{R}),d\mapsto h_{d}$ such that $h_{d}^{T}M_{d}h_{d}=M_{\alpha}$ for all $d\in[\alpha,\beta]$ (this follows, for example, from the fact that $\mathrm{GL}_{n}(\mathbb{R})$ acts continuously and transitively on the set $\Omega\subset M_{n}(\mathbb{R})$ of symmetric matrices with the same signature as $M_{\alpha}$, and so the orbit map $\mathrm{GL}_{n}(\mathbb{R})\rightarrow\Omega,g\mapsto g^{T}M_{\alpha}g$ is a fiber bundle). Setting $g_{d}=h_{d}h_{\alpha}^{-1}$ for $d\in[\alpha,\beta]$, we have that $g_{\alpha}=I_{n}$ and $g_{d}^{T}M_{d}g_{d}=M_{\alpha}$ for $d\in[\alpha,\beta]$. For $d\in[\alpha,\beta]$, let $\rho_{d}=g_{d}^{-1}\sigma_{d}g_{d}$, and note $\rho_{d}(W)\subset g_{d}^{-1}\mathrm{O}(M_{d};\mathbb{R})g_{d}=\mathrm{O}(g_{d}^{T}M_{d}g_{d};\mathbb{R})=\mathrm{O}(M_{\alpha};\mathbb{R})$.

Let $\rho_{d}^{+}=\rho_{d}\bigr{|}_{W^{+}}$ and $\sigma_{d}^{+}=\sigma_{d}\bigr{|}_{W^{+}}$ for $d\in[\alpha,\beta]$. Then $\rho_{\alpha}^{+}(W^{+})$ is a uniform lattice in the connected non-compact simple Lie group $\mathrm{SO}(M_{\alpha};\mathbb{R})^{\circ}$, and the latter is not locally isomorphic to $\mathrm{SO}(2,1)^{\circ}$ by our assumption that $n>3$. Thus, by Weil local rigidity [Wei60, Wei62], up to choosing $\beta$ closer to $\alpha$, we may assume that for each $d\in[\alpha,\beta]$ there is some $a_{d}\in\mathrm{SO}(M_{\alpha};\mathbb{R})^{\circ}$ such that

But $\rho_{d}^{+}=g_{d}^{-1}\sigma_{d}^{+}g_{d}$, so we obtain from (1) that the trace $\mathrm{tr}(\sigma_{d}(\gamma_{i}\gamma_{j}))$ remains constant as $d$ varies within $[\alpha,\beta]$, where $i,j\in\{1,\ldots,n\}$ are chosen so that the vertices $v_{i},v_{j}$ are adjacent in $\Sigma_{1}$.

We claim, however, that $\mathrm{tr}(\sigma_{d}(\gamma_{i}\gamma_{j}))=4d^{2}-4+n$ for $d\geq D$. Indeed, let $d\geq D$. Then $M_{d}$ is nondegenerate, so that $\mathbb{R}^{d}$ splits as a direct sum of the $2$-dimensional subspace $\langle e_{i},e_{j}\rangle\subset\mathbb{R}^{n}$ and its orthogonal complement $E_{i,j}$ with respect to $M_{d}$. Each of $\gamma_{i}$ and $\gamma_{j}$ acts as the identity on $E_{i,j}$, so our claim is equivalent to the assertion that $\mathrm{tr}\left(\sigma_{d}(\gamma_{i}\gamma_{j})\bigr{|}_{\langle e_{i},e_{j}\rangle}\right)=4d^{2}-2$, and the latter follows from the fact that, with respect to the basis $(e_{i},e_{j})$ of ${\langle e_{i},e_{j}\rangle}$, the matrices representing $\sigma_{d}(\gamma_{i}),\sigma_{d}(\gamma_{j})$ are $\begin{pmatrix}-1&2d\\ 0&1\end{pmatrix},\begin{pmatrix}1&0\\ 2d&-1\end{pmatrix},$ respectively. ∎