Geometric convergence
Here, we state the equivalent formulations of the Geometric convergence
Group theoretic formulation (Hausdorff/Chabauty topology)
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. $\square$
0 XDK (+0)
유익한 글을 읽었다면 작성자에게 XDK를 선물하세요.
-
진짜 수능이었으면 지금 텔그정도로 가는건가요?
-
4회에서 처음으로 45점 맞음 ㅜㅜ 기쁘다
-
덕코를 모으면 12
덕코가 증가합니다 그게 덕코니까
-
혼자 개노가다 뺑뺑이 수학 N제 실모 공부하다가 해설강의 들으니까 개좋다......
-
생각 외로 실질적인 교차지원 비율은 줄지 않을듯. 교차지원 할 애들은 이미 사탐런 했거든.
-
수능날 시간부족할까봐 무서운데 왜 난이도가 점점 하향하는거지....
-
이제 5등급만 찍으면 되는거냐구 캬캬
-
왜냐하면 의대 안썼으니까..
-
RF 1
Residually finite: For any nontrivial element...
-
ㅋㅋ 딱 저느낌임 오만하게 살다가 멀리 못간다 ㅋㅋ
-
오공완 8
오늘 독감 주사 맞아서 힘들어요
-
D-41 오공완 4
이것도 힘들어 죽겠는데 정오에 7시간 찍히는 분들은 대체 어떤 분들이실까요…. 대단….
-
마지막게시물로 서울대 성적표 올려놓고 끝내고싶다 엉엉
-
님들은대학왜감 10
원하는 학과?명성?행복?
-
저만 열품타 5
멈추면 다시키는거 까먹나요,,, 시간 측정 매일 실패중이에요
-
병훈이햄도 25분동안 푸네.. 어떤 문젠거노
-
작년에 월화 풀실모에 추가 공부 더하니까 목요일에 컨디션이 별로였어서 저는 토요일...
-
9모 언매 4 확통 5 영어 4 정법 4 사문 4 농어촌 버프로 현역 서울대...
-
버스땜에 시간을 1시간 정도 날렸네요… 내일부터는 더 열심히 하겠습니다!
-
잇올에서 강K 샀던거 오늘 왔는데 화학 K+는 답지 같이 왔는데 K는 왜 답지...
-
오늘의공부 1
국어:비문핟 3지문 문학 3지문 언매한세트(다맞음ㅎㅎ) 영어 10지뮨 수학...
-
오르비가 야해지고 있다 -> 제 댓글 중 너무 야해요의 비중이 급증했어요 너무 야해요
-
한병훈쌤 왤케 훈훈하심 큐앤에이도 하트남발이고 큩보이야완전~..프로필사진올블랙어쩜ㅜㅜ 나미친듯
-
바이퍼존나멋있다 0
박쌤이랑사귀고싶다
-
뭐 외우면 잘풀리는게 있는건가요? 진짜 준킬러가 너무 안풀려요 ㅜㅜ 공부법이라도...
-
카톡을 차단시켜버려서 그렇게 차단하고 못 푼 사람만 근 20명임 ㅋㅋㅋㅋㅋ 다시...
-
개념 특강도 어려움? 상위권 위주인가
-
일단 기만은 아니고 이 성적에 이게 말이 되나요..? 걍 애초에 신빙성이 떨어지긴...
-
9모 전까진 강k 수학 조져도 80점 언저리고 그랬는데 9모 조진 이후로 60점대...
-
오르비언이랑 만나서 술마시기
-
교토식 화법이라기보단 일본식 화법 ㅇㅇ 함 무리한 부탁 해봐바 댓글로
-
갈까 말까
-
김범준쌤 질문점 6
지금 김범준쌤 강k현강 들을까 고민중인데 수업 이해할 수 있나요?? 지금은 걍 굳이...
-
이거지 사문의 신 폼 되찾았노 20번 좃같이 어려워서 유기함 ㅅㅃ
-
ㅈㄴ 떨리네 ㅈ박을듯
-
전 수능 수험생은 아니지만 오늘 공부 할당량 다 채우느라 이제 귀가했네요 수험생들 ㅍㅇㅌ
-
옆에 있으면 되게 서운한 일이 많은거 같아요 예민한것 같아서 나름 눈치 많이...
-
평가원 수능 언매만큼의 위압감을 따라오지 못하는거같음... 특히 24수능 지문형...
-
오랜만에 오운완 2
시원하니좋네요
-
인생 쓰네요...
-
하루에 실모하나 하프모하나 정도 할생각인데 ㄱㅊ나요?
-
거유 영화배우 9
-
이3끼들은 왜 컨텐츠 설명을 똑바로 안적어두는지
-
역겨워서 토나옴 2
어떻게 생각이 그리 어린지
-
오늘한거 8
14시간실패!
-
미적, 과탐런 문과 현역입니다. 국어 어쩌죠? 고대 하위과라도 갈 수 있을까요.?
-
우와 은테 달았다 32
오르비 가입한지 거의 한달만에 은테찍었네요 기쁘다 뉴비인데 이렇게 해주셔서 고마울따름
-
날 묶어줘 14
보채고 혼내줘너의 강아지처럼 길들여줘어ㅓ어ㅓㅓㅓ어어ㅓ네 침대에 네 품에...
-
D-66에 최선을 다하겠다 다짐을 한지 이 주가 넘게 지났네요. 그래서 최선을...
첫번째 댓글의 주인공이 되어보세요.