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.
-
더 재밌는 걸 가져와라
-
남은 문제는 3개 예상 소요 시간 90분
-
왜아직도백분위가없노
-
안녕하세요. 2025 혜윰 모의고사 시즌1 정오표를 업로드합니다. 추가 정오 사항이...
-
1시 취침 오전9시 기상 10시까지 산책 씻고 11시쯤 점심 오전...
-
2주남았는데 멀어서 안가요 사물함도 있음
-
엉덩이도 너무 슬퍼서 물이 막 너ㅓㅁ쳐 ㅠㅠ큐ㅠㅠㅠ 설사가 푸드득 푸으읅ㅇ으ㅡㄺ...
-
오늘 두 개 잘려서
-
수학 기출 복습 1
쉬운4점~어려운4점 위주로 기출 복습하고 싶은데 마땅한 교재 있나요?
-
바로 자살… 다른 건 다들 ~에 대한 윤리적 쟁점으로만 써놨는데 자살은 윤리적...
-
설맞이 수2 전문항 다시 보면서 정리하고 국어 마더텅 언매랑 독서 뿌수고 탐구 개념...
-
ㅅㅂ
-
시험지 끝까지 다 못보는 한이 있더라도 선지 5개 다 확인할래 손가락 걸었다가 틀린게 몇개냐...
-
실수가 너무 많고 문학 감 수직하락 이슈로 이감 커하 갱신 실패 문학이랑 독서 실력을 교환한 듯
-
뉴런 질문 3
현 고2인데 뉴런 수1,2를 2달안에 끝낼 수 있을까요.??
-
아무 인간관계없는 재수가 제일 행복해요
-
규동먹고싶다 6
오랜만에 현지의 규동이 먹고싶군아...
-
얼버기 2
-
할복
-
오르비는 왜 하는걸까요?
-
뭔가 이감이 서바보다 답을 도출하기 위한 단서에 대해 친절한 것 같은데 맞나요?
-
처음엔 잠 못자서 수면제받으려고 갔는데 자기 adhd 전문 의사라고 adhd랑 뇌파...
-
번장에서 23,34 실모를 샀는데 답지가 없어서요ㅠ 해설지까지 없어도 되니까...
-
그건 걍 내년 대교협에서 지원금 다 짤려도 할말없는짓임
-
아니면 지문 읽으면서 와리가리함
-
올해 교육청 2
모의고사 푸는게 좋을까요? 수능에 0.1이라도 도움되면 풀거같은데 교육청이랑...
-
그 외 단과에서 ㄱㅇㅇ 선생님이나 ㄱㅁㅊ 쌤이 알려주신 컷 아시는 분 댓글로...
-
운전해보고 싶음 0
근데 무서움 근데 사실 내가 차를 갖고 싶은건지 운전을 하고 싶은건지 헷갈림
-
매실문 2
지금 하면 늦음? 유자분까진 끝냇음
-
더데유데는 오답률이나 1등급 비율 따로 공개안하나요? 시즌 1 알고싶은데 영어 조정식 션티 등급컷
-
고1 과목으로 수능을 치자는 아이디어는 대체 어떤 ㅅㄲ 작품인지 0
곱씹을수록 기도 안 차네 진짜 ㅋㅋㅋ 그럴 거면 차라리 통합 과목을 3년 내내 가르치는 걸로 하든가
-
ㅈㄱㄴ 올해 과탐으로 문과 교차는 예년처럼 이득이 크지 않댔는데 그것도 맞는 얘긴가...
-
88점 독서 2틀 문학 2틀 언매 1틀 사설은 왜 항상 손가락걸기로 찝찝하게 푼...
-
논술로 인하대 의대 입학 메디컬 적성 X 2년 뒤 수능 응시 서울대학교 수리과학부...
-
상당히 자1살 마렵네 내일은 잘봐야지
-
나는 해병대가 아닌디... 후잉...
-
얼리버드기상 6
ㄹㅇ이
-
팝송 잘 아는사람... is it a crime~ 비슷한 느낌 노래ㅠ 2
is it a crime 노래에서 그 가사부분 음정이 조금 더 올라가는 어느...
-
인강안듣구 현돌 실개완 기시감 킬쿼모 6평분석서 n제 이렇게 하고잇는데 혹시 생윤은...
-
어제, 생일날 헤어진 사람인데 아무나 위로 부탁해도 될까? 3
공부도 눈에 안잡히고 정말 그 사람 아니면 안될거 같고 많이 보고싶어. 얘 없으면...
-
수바 18회 0
올해 수바 첫 100인듯 오랜만에 실수 안했네
-
장르 불문하고 너무 유명한 가수는 다 알고있음
-
미미미누가 흑백요리사 패러디로 실모요리사 서바이벌 컨텐츠 열면 재밌을 듯 전국에서...
-
다른 건 모르겠고 그냥 아무 생각 없이 자고 싶음 수능최저 맞추고 그 주에 논술...
-
왜냐하면 지구표면의 한지점에서 구멍을 뚫어서 정확히 지구중심을 지나 정확히 정반대편...
-
제가 봐도 시골 거름냄새라고 생각할 정도로 똥방구가 장난아니게 자주...
-
4000부 판매돌파 지구과학 핵심모음자료를 소개합니다. (현재 오르비전자책 1위)...
-
수능본다고 하니까 대대장님이 그럼 미복귀전역해도 된다고 하셨다 대대대장
첫번째 댓글의 주인공이 되어보세요.