Geometric convergence
게시글 주소: https://iu.orbi.kr/00068642663
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.
-
하... 사관학교 공남문 208인데 합격 가능할까요? 가능하다면 추합일까요..?
-
넌 아니라니까 ?
-
어느새 나도 치킨 한 마리 혼자 다 먹는 어엿한 으른이라구
-
슬슬 가야겠구나 2
이번주도 화이팅입니다 며칠뒤 평일 밤이나 아님 주말쯤 다시 올듯...
-
계신가요… 이런거 어케 해소하죠? 잇올에 있다보면 약간 더 그러는듯
-
딱히 주제는 없고 현재 저의 심정...?하고 싶은말?임요.. ㅅㄴㅇ ㅇㅁ ㄴㅈ...
-
거기까진 잘 모르겠는데 최소한 20대 초반은 걸어버렸음…
-
못생긴 오뿌이들만 왔네
-
12413으로 중경외시를 가고 싶구나... 사실 스나이핑이 성공해서 서성한을 가면...
-
[칼럼] 수능 1교시, 여러분이 반드시 해야 할 행동 13
국어 칼럼은 또 오랜만이네요. 바로 시작해 보겠습니다 뻔한 이야기는 하지 않겠습니다...
-
재수생이고 매일 30분씩 하는데 수특 수완은 다풀었어요(얼마전에 끝남) 이제 뭐...
-
글 들어가보면 댓글이 없긴함.. 조용하군
-
오르비를 정상화 2
신 창 섭
-
엉덩이 종기 7
엉덩이 종기가 분기마다 한번씩 제발하는 것 같은데 현재 고2인데 고3 되기 전에...
-
말벌 먹방 전후 2
-
7모 4등급, 기하 응시해왔으며 3점도 겨우 한두 개 (풀어서)맞추길래...
-
전문성있게 정리된 정보가 많아 영어.. 으악
-
1. 3특은 그래도 서울대는 ㅂㄱㄴ하고 종합대학 최고로 갈 수 있는데는 연고대고...
-
요즘은 어떠려나 뭐 소문에 의하면 개빡센 교수 들어간 이후로 더 이상 헤븐까지는 아니더라는데
-
손해가 훨씬 많아진 것 같다고 생각함 음모론을 믿지는 않지만 만약 실제 세계를...
-
죽이되든 밥이되든 일단 던진다
-
에반가 하고나면 기진맥진 할거같은데
-
커피한잔 독서인가 뭔지 그분을 그만 상대해 주기로 했다. 내일 경산여자고등학교 교장 좀 보고와야겠다. 1
헌법을 무시하는 그분을 그만 상대해 주고 싶습니다. 상식이 통하지 않는 분이네요....
-
오늘의 느낀점 1
시험이 끝났다고 채점을 매지말고 남은 시간 열심히 검토하자!!
-
현재 기출 2회독/ 박선-코어특강 완강 6월 모의고사 2등급이였음 오지훈 실전문제,...
-
쳐다보지도말자
-
https://orbi.kr/00068628771 대성학력개발연수오에서 평가원 양식...
-
강기분 다 했는데 연계를 하나도 안함 새기분 안하고 강e분 우기분 실모만 ㄱㄱ??
-
사1과1 수능보고 원서 쓸 때 사2보다 불리한 점이 있나요? 다들 사1과1은 애매한...
-
고삼 남자고 서울 학군지에서 재학중입니다 갓반고까지는 잘 모르겠네요 그냥 평반고.....
-
1. 올해 2월 이감 시즌 5 6 구매하기로 예약하고 입금함 2. 6월에 제 단순...
-
노무현이 대전지방판사 반년도 못하고 부산 내려가서 변호사 한 이유를 알거 같다....
-
앱스키마 드랍 2
이제 TIM끝나는데 지금 앱스키마 시작하기에는 늦은 걸까요..? 앱스키마 볼륨이...
-
이감 현강생인데 하라는거 다 할수가 있긴 함 일주일안에? 간쓸개 주간복습북(지문 꽤...
-
이재명 끝났다! 이낙연 끝났다! 김두관 끝났다! 민주당 끝났다!
-
어떰?
-
대성마이맥 과탐 강모K+ 언제쯤 나오는지 아시는 분 계신가요?? 0
언제쯤 나올까요...?
-
전형태 풀커리 탐
-
시대 라이브 복영 재생화면에서 안나가면 기간 초과되고도 볼수잇나용...? 오늘까지...
-
어그로 ㅈㅅ하구여 혹시 사설 수학실모 풀만한거 추천해주실 수 있나요? 본인실력은...
-
호감 오르비언 특 20
스카이 서성한 뱃지에 예쁜 여자 프사에 말도 이쁘게 하면 내 레이더망에 포착 설령...
-
국어 93 영어 86 미적분 80인데 수능으로 치면 어느정도일까요....대학다니다가...
-
옯붕이 사고쳤다 23
실모보관함 들고 빙글빙글 돌아가다가 실모 봉인함 ㅠㅠ 이거 어떻게 열지
-
15분 걸리는데 너무 오래걸리나..
-
첫 정답자 1000덕 드리겠습니다!
-
매달 돈은 빠져나가는데 사용을 안하니 원,..
-
전북대?
첫번째 댓글의 주인공이 되어보세요.