1. The geometric topology on Kleinian groups we mean giving the discrete subgroup of $\mathrm{PSL}_2\Bbb C$ the Hausdorff topology as closed subsets.

- The sequence of closed subsets $\{Y_i\}$ tends to a closed subset $Z$ in Hausdorff topology of the collection of closed subsets means (1) For every $z\in Z$, there are $y_i\in Y_i$ such that $\lim_{i\to\infty} y_i = z$. (2) For every subsequence $Y_{i_j}$, and elements $y_{i_j}\in Y_{i_j}$, if $y_{i_j}\to z$ then $z\in Z$.

In other words, $\{\Gamma_i\}\to\Gamma$ geometrically if every element $\gamma\in\Gamma$ is the limit of a sequence $\{\gamma_i\in\Gamma_i\}$ and if every accumulation point of every sequence $\{\gamma_i\in\Gamma_i\}$ lies in $\Gamma$.

Rmk. It's known that the set of closed subsets is compact with Hausdorff topology. In particular, passing to a subsequence, one may always assume that a sequence of nonelementary Kleinian groups converges geometrically.

2. Equipping a hyperbolic 3-manifold $M$ with a unit orthonormal frame $\omega$ at a base point $p$ (called a base-frame), $M$ uniquely determines a corresponding Kleinian group without up to conjugacy condition by requiring that the covering projection

$$\pi:(\Bbb H^3,\tilde{\omega})\to(\Bbb H^3,\tilde{\omega})/\Gamma = (M,\omega)$$

sends the standard frame $\tilde{\omega}$ at the origin in $\Bbb H^3$ to $\omega$.

The framed hyperbolic 3-manifolds $(M_n,\omega_n) = (\Bbb H^3,\tilde{\omega})/\Gamma_n$ converge geometrically to a geometric limit $(N,\omega) = (\Bbb H^3,\tilde{\omega})/\Gamma_G$ if $\Gamma_n$ converges to $\Gamma_G$ in the geometric topology stated in 1, i.e,

-For each $\gamma\in\Gamma_G$ there are $\gamma_n\in\Gamma_n$ with $\gamma_n\to\gamma$.

-If elements $\gamma_{n_k}$ in a subsequence $\Gamma_{n_k}$ converges to $\gamma$, then $\gamma$ lies in $\Gamma_G$.

(intrinsic) Manifold formulation

3. $(M_n,\gamma_n)$ converges to $(N,\gamma)$ geometrically if for each smoothly embedded compact submanifold $K\subset N$ containing $\omega$, there are diffeomrophism (or quasi-isometries or biLipschitz) $\phi_n:K\to (M_n,\omega_n)$ so that $\phi_n(\omega) = \omega_n$ and so that $\phi_n$ converges to an isometry on $K$ in the $C^\infty$-topology.

Rmk. Note that one can formulate the above by saying that for $\epsilon>0$, there is a sequence of isometric embeddings $\beta_i: B_{\epsilon}(\phi_i(x))\to\Bbb H^3$ from $\epsilon$-ball around $\phi_i(x)\in M_i$ so that $\beta_i\circ\phi_i$ converges to an isometric embedding of some neighborhood of $x\in N$ into $\Bbb H^3$.

4. A sequence of Kleinain groups $\Gamma_i$ converges geometrically to the Kleinain groups $\Gamma_G$ if there exists a sequence $\{r_i,k_i\}$ and a sequence of maps $\tilde{h}_i:B_{r_i}(0)\subset\Bbb H^3\to\Bbb H^3$ such that the following holds:

(1) $r_i\to\infty$ and $k_i\to 1$ as $i\to\infty$;

(2) the map $\tilde{h}_i$ is a $k_i$-bi-Lipschitz diffeomorphism onto its image, $\tilde{h}_i(0) = 0$, and for every compact set $A\subset\Bbb H^3$, $\tilde{h}_i|_A$ is defined for large $i$ and converges to the identity in the $C^\infty$-topology; and

(3) $\tilde{h}_i$ descends to a map $h_i:Z_i = B_{r_i}(p_G)\to M_i = \Bbb H^3/\Gamma_i$ is a topological submanifold of $M_G$; moreover, $h_i$ is also a $k_i$-bi-Lipschitz diffeomorphism onto its image. Here, $p_G = \pi_G(0)$ where $\pi_G:\Bbb H^3\to M_G$.

Gromov-Hausdroff formulation

5. The sequence of discrete groups $\{G_n\}$ converges polyhedrally to the group $H$ if $H$ is a discrete and for some point $p\in\Bbb H^3$, the sequence of Dirichlet fundamental polyhedra $\{P(G_n)\}$ centered at $p$ converge to $P(H)$ for $H$, also centered at $p$, uniformly on compact subsets of $\Bbb H^3$. More precisely, given $r>0$, set

$$B_r = \{x\in\Bbb H^3:d(p,x)<r\}.$$

Define the truncated polyhedra $P_{n,r} = P(G_n)\cap B_r$ and $P_r = P(H)\cap B_r$. A truncated polyhedron $P_r$ has the property that its faces (i.e. the intersection with $B_r$ of the faces of $P$) are arranged in pairs according to the identification being made to form a relatively compact submanifold, bounded by the projection of $P\cap\partial B_r$. We say that this polyhedral converges if: Given $r$ sufficiently large, there exists $N = N(r)>0$ such that (i) to each face pairing transformation $h$ of $P_r$, there is a corresponds a face pairing transformation $g_n$ of $P_{n,r}$ for all $n\geq N$ such that $\lim_{n\to\infty}g_n = h$, and (ii) if $g_n$ is a face pairing transformation of $P_{n,r}$ then the limit $h$ of any convergent subsequence of $\{g_n\}$ is a face, edge or vertex pairing transformation of $P_r$.

In other words, each pair of faces of $P_r$ is the limit of a pair of faces of $\{P_{n,r}\}$ and each convergence subsequence of a sequence of face pairs of $\{P_{n,r}\}$ converges to a pair of faces, edges, or vertices of $P_r$.

A seuqnece $\{G_n\}$ of Kleinian groups converges geometrically to a nonelementary Kleinian group if and only if it converges polyhedrally to a nonelementary Kleinian group.

Rmk. It's necessary that one needs to assume the limit group nonelementary. It's possible that the geometric limit of nonelementary Kleinian group is an elementary Kleinian group.

6. A sequence $X_k$ of metric spaces converges to a metric space $X$ in a sense of Gromov-Hausdorff if it converges w.r.t. the Gromov-Hausdorff distance. Here, Gromov-Hausdorff means the following:

Let $X$ and $Y$ be metric spaces. A triple $(X',Y',Z)$ consisting of a metric space $Z$ and its two subsets $X'$ and $Y'$, which are isometric respectively to $X$ and $Y$, will be called a realization of the pair $(X,Y)$. We define the Gromov-Hausdorff distance:

$$d_{GH}(X,Y) = \inf\{r\in\Bbb R:\text{ there exists a realization }(X',Y',Z)\text{ of }(X,Y)\text{ such that }d_H(X'.Y')\leq r\}$$

where $d_H$ is a Hausdorff distance.

addendum. A sequence of representations $\varphi_n\in AH(\Gamma)$ converges algebraically to $\varphi\in AH(\Gamma)$ if $\lim_{n\to\infty}\varphi_n(\gamma) = \varphi(\gamma)$ for each $\gamma\in\Gamma$. This is a natural topology once we view $AH(\Gamma) = \mathrm{Hom}(\Gamma,\mathrm{PSL}_2\Bbb C)/\mathrm{PSL}_2\Bbb C\subset \mathrm{Hom}(\Gamma,\mathrm{PSL}_2\Bbb C)//\mathrm{PSL}_2\Bbb C$ as an algebraic variety.

Here, $\mathrm{Hom}$ we implicitly assume it's weakly type preserving but not necessary (strongly) type preserving.

In the manifold term, one can describe the algebraic convergence as follows: Element in $AH(\Gamma)$ can be thought as a homotopy equivalence (called the marking) $h:N\to M$ where $N$ is some fixed hyperbolic 3-manifold with $\pi_1(N) = \Gamma$ such that two elements $(M,h)$ and $(M',h')$ are equivalent if there is an isometry $\psi:M\to M'$ such that $\psi\circ h\simeq h'$. Note that this is equivalent to the discrete faithful representation of $\Gamma$ to $\mathrm{PSL}_2\Bbb C$ by the $K(G,1)$-space property.

Under this view point, a sequence of marked manifolds $(M_i,h_i)$ converges algebraically to $(M,h)$ if there is a smooth homotopy equivalences $H_i: M\to M_i$ compatible with the marking that converges $C^\infty$ to local isometries on compact subsets of $M$.

It's noted that the algebraic convergence of $(M_i,h_i)$ to $(M,h)$ is guaranteed if there is a compact core $K$ of $M$ and a smooth homotopy equivalences $H_i:K\to M_i$ compatible with the markings and which are $L_i$-bilipschitz diffeomorphisms on $K$ with $L_i\to 1$.

Remark/Properties. 1. If $\rho_i:\Gamma\to\mathrm{PSL}_2\Bbb C$ is a sequence of discrete faithful representation that converges algebraically to $\rho$ and geometrically to $\Gamma_G$, then $\rho(\Gamma) = \Gamma_A\subset\Gamma_G$ because by definition, $\Gamma_A$ consists of all convergence sequences $\rho_i(g)$ for fixed $g\in\Gamma$ whereas $\Gamma_G$ contains all convergence sequences of the form $\rho_i(g_i)$ for $g_i\in\Gamma$.

2. Although after passing to a subsequence, algebraically convergence sequence implies geometric convergence, geometric convergence itself does not imply algebraic convergence.

3. Suppose a sequence of discrete faithful representations $\rho_i:\Gamma\to\mathrm{PSL}_2\Bbb C$ converge algebraically to $\rho$ and geometrically to $\Gamma_G$. Then there is not $\gamma\in\Gamma_G - \rho(\Gamma)$ with $\gamma^k\in\rho(\Gamma)$ for some $k\geq 2$. In particular, if the image $\rho(\Gamma)$ of the algebraic limit has finite index in the geometric limit $\Gamma_G$, then $\rho(\Gamma) = \Gamma_G$.

$(\because)$ Suppose there is $g\in\Gamma_G - \rho(\Gamma)$ with $g^k = \rho(\eta)$ for some $\eta\in\Gamma$ for $k\geq 2$. Since $g\in\Gamma_G$, there is a sequence $\gamma_i\in\Gamma_i$ that $\rho_i(\gamma_i)\to g$. Taking power $k$ gives

$$\lim_{i\to\infty}\rho_i(\gamma_i^k) = g^k = \rho(\eta) = \lim_{i\to\infty}\rho_i(\eta).$$

It can be shown (via nontrivial argument) that $\rho_i(\gamma_i^k) = \rho_i(\gamma)$ using the fact that $\rho_i$ converges algebraically to $\rho$. Since the representation is faithful, this implies $\gamma_i^k = \gamma$ for large $i$. It can be shown also that the set of roots $\gamma = \gamma_i^k$ is finite in general. Hence, after passing to a subsequence, $\gamma_i = \gamma_j$ for all $i,j$ so that $g\in\rho(\Gamma)$ which is a contradiction. $\square$