Hyperbolic $3$-manifold groups that embed in compact Lie groups Is there a closed hyperbolic $3$-manifold whose fundamental group is isomorphic to a subgroup of some compact Lie group?
It is known that every surface group can be embedded into any semisimple connected Lie group.
I would be interested in similar results for fundamental groups of closed nonpositively curved manifolds of any dimension, but this is probably much harder. All I know is how to embed the fundamental groups of tori, closed surfaces, and their products.
 A: All closed hyperbolic 3-manifold groups embed into a compact Lie group. 
To prove this, note first of all that given a hyperbolic 3-manifold $M$, it suffices to show that a finite-index subgroup $G\leq \pi_1(M)$ of index $m$ embeds into a compact Lie group. Then the representation $\rho: G\hookrightarrow O(n)$ will induce a represenation $Ind_G^{\pi_1(M)} \rho : \pi_1(M) \hookrightarrow O(nm)$. 
Now, by the proof of the virtual Haken conjecture, $\pi_1(M)$ has a finite-index subgroup $G$ which is the fundamental group of a special cube complex, which implies that $G$ embeds into a right-angled artin group $A$, and hence into a right-angled Coxeter group $\Gamma$. 
Finally, right-angled Coxeter groups have faithful embeddings into $O(n)$. This follows from a result of Vinberg, which gives a faithful linear action by reflections on $\mathbb{R}^n$. We'll review this construction following section 7 of this paper. 
Fix a right-angled Coxeter group
$$\Gamma = \langle \gamma_1,\dots,\gamma_k ~|~ (\gamma_i \gamma_j)^{m_{i,j}} = 1\quad \forall i,j\rangle,$$
where $m_{i,i}=1$ and $m_{i,j}\in\{ 2,\infty\}$ for all $i\neq j$.
For $t \in \mathbb{R}$, the matrix $M_t=(M_t(i,j))_{1\leq i,j\leq k}$ where
$$M_t(i,j) = \left\{ \begin{array}{cl}
1 & \text{if }m_{i,j}=1, \text{ i.e.  $i=j$},\\
0 & \text{if }m_{i,j}=2,\\
-t & \text{if }m_{i,j}=\infty
\end{array}\right.$$
 defines a symmetric bilinear form $\langle\cdot,\cdot\rangle_t$ on $\mathbb{R}^k$.
Note that $\mathrm{det}(M_t)$ is a nonzero polynomial in $t$ (take $t=0$), hence it is nonzero outside of some finite set $F$ of exceptional values of $t$.
For any $t\in \mathbb{R}-F$, the form $\langle\cdot,\cdot\rangle_t$ is nondegenerate.
Define the representation $\rho_t : \Gamma\to\mathrm{Aut}(\langle\cdot,\cdot\rangle_t)\leq \rm{GL}(k,\mathbb{R})$ by
$$\rho_t(\gamma_i) : v \mapsto v - 2\langle v, e_i \rangle_t \, e_i $$
for all $i$. Each generator is a reflection in a hyperplane perpendicular to $e_i$ with respect to the metric $\langle\cdot,\cdot\rangle_t$.
For $t>1$, the convex cone
$$\widetilde{\Delta}_t = \{ v\in\mathbb{R}^k ~|~ \langle v, e_i\rangle_t \leq 0 \ \,\forall i\}$$
descends to a convex polytope $\Delta_t$ in an affine chart of $\mathbb{P}(\mathbb{R}^{k})$.
By Theorem 2 of Vinberg, the representation $\rho_t$ is discrete, faithful, the set $\rho_t(\Gamma)\cdot\Delta_t$ is convex in $\mathbb{P}(\mathbb{R}^k)$, and the action of $\Gamma$ on the open set
$$\mathcal{U}_t:=\mathrm{Int} \left ( \rho_t(\Gamma)\cdot\Delta_t \right )$$ is properly discontinuous.
Now, let $t$ be close to $0$ and transcendental, so that $M_t$ is positive definite. Then $\rho_t : \Gamma\to \mathrm{Aut}(\langle\cdot,\cdot\rangle_t) \cong O(k)$ is faithful (it is Galois conjugate to a representation for some transcendental $t>1$). 
Remark: I think the first step of taking an induced representation can be eliminated. If we assume $G\lhd \pi_1(M)$, and $K=\pi_1(M)/G$, then one may embed $G$ into a right-angled Coxeter group which admits an action of $K$ by permuting its generators, and so that the embedding of $G$ is $K$-equivariant. Then $\pi_1(M)$ should embed in this Coxeter group extended by $K$, which clearly still also has a faithful representation.
Remark 2: Regarding your second question, as should be clear from the discussion, this holds for any cubulated hyperbolic group. This includes uniform arithmetic hyperbolic lattices of simple type, and the examples of Gromov-Piatetskii-Shapiro.   
A: Yes, there exists such closed hyperbolic (= constant curvature $-1$) manifolds with this property, in arbitrary dimension.
For $d\ge 1$, let $q_t$ be a quadratic form of rank $d$ with coefficients in $\mathbf{Q}[t]$. Suppose that $q=q_{\sqrt{2}}$ has signature $(d,1)$ while $q'=q_{-\sqrt{2}}$ has signature $(d+1,0)$, for instance, $$q(x)=-\sqrt{2}x_0^2+x_1^2+\dots+x_d^2.$$ Then, by the Borel-Harish-Chandra theorem, the ring embedding $$\mathbf{Z}[\sqrt{2}]\to\mathbf{R}\times\mathbf{R},\quad (a+b\sqrt{2})\mapsto (a+b\sqrt{2},a-b\sqrt{2}),$$
induces an inclusion of
$\Gamma=\mathrm{SO}(q)(\mathbf{Z}[\sqrt{2}]$ as a cocompact lattice into $$\mathrm{SO}(q)(\mathbf{R})\times \mathrm{SO}(q')(\mathbf{R})\simeq \mathrm{SO}(d,1)\times\mathrm{SO}(d+1).$$
The left projection realizes $\Gamma$ as a lattice in $\mathrm{SO}(d,1)$, while the right projection realizes $\Gamma$ as a subgroup of the compact Lie group $\mathrm{SO}(d+1)$.
Many cocompact arithmetic lattices in semisimple Lie groups are embeddable into some compact Lie group in some similar way.
I don't know about non-arithmetic lattices (except in $\mathrm{SL}_2(\mathbf{R})$), and don't know beyond the locally symmetric setting.

Edit: for $d\ge 3$, 

if $\Gamma$ is a lattice in $\mathrm{SL}_d(\mathbf{R})$ with traces in $\mathbf{Q}$, then $\Gamma$ has no embedding into any compact Lie group. 

There are indeed cocompact lattices with this property, for instance those corresponding to $\mathrm{SL}_1(D)$ when $D$ is a division algebra of degree $n$ over $\mathbf{Q}$ that splits over $\mathbf{R}$.
The idea is to use Margulis' superrigidity in its stronger form. If we had such an embedding, up to changing $\Gamma$ to a finite index subgroup we would have an injective homomorphism $f$ with dense image from $\Gamma$ to some compact connected simple Lie group $H(\mathbf{R})$. Then the image is contained in $H(K)$ for some finitely generated subfield $L$ of $\mathbf{R}$, and in turn, we can (Kronecker-Tits) embed $L$ into some nondiscrete locally compact field $\mathbf{K}$ so that the image of $\Gamma$ becomes unbounded. Then super-rigidity will imply that $\mathbf{K}$ is Archimedean, so we can suppose that it equals $\mathbf{C}$ and hence superrigidity again says that $H$ is $\mathbf{C}$-isomorphic to $\mathrm{SL}_d$ and hence $f$ is virtually given by a conjugation (possibly composed by inverse-transposition). In particular, $f$ (or $f\circ (\mathrm{inverse})$ preserves traces. Since $\Gamma$ is closed under inversion and has bounded traces, it follows that $f(\Gamma)$ has bounded traces, hence $f(\Gamma)$ is bounded, contradiction.
