3rd bounded cohomology of Kleinian surface group
게시글 주소: https://orbi.kr/00068516715
Hyperbolic 3-manifold에 대해서 알고 있는 사람도 소수의 사람들 빼고는 bounded cohomology에 대해서는 잘 모른다. (당연하다) 놀랍게도(?) hyperbolizable 3-manifold의 3rd bounded cohomology는 geometrically infinite 인 경우에 대해서는 interplay가 있다. Geometrically finite한 경우에는 3rd bounded cohomology는 아무것도 capture하지 못하는데, 그 기본적인 이유는 convex core가 bounded 되어있기 때문이다.
Theorem A. (Cohomological nearness implies geometric sameness). For a given $i_0>0$, let $(M_0,f_0)$ be any doubly-degenerate element of $AH_+(\Sigma_g)$ with $\mathrm{inj}(M_0)\geq i_0$. Then there exists a constant $\epsilon(g,i_0)>0$ depending only on $g$ and $i_0$ so that any element $(M_1,f_1)$ of $AH_+(\Sigma_g)$ satisfying $\parallel [f_0^*(\omega_{M_0})] - [f^*_1(\omega_{M_1})]\parallel<\epsilon$ in $H^3_b(\Sigma;\Bbb R)$ is equal to $(M_0,f_0)$ in $AH_+(\Sigma_g)$.
Corollary B (Geometric nearness does not imply cohomological nearness). Suppose that $\{M_n,f_n\}_{n=1}^\infty$ is a sequence in $H_+(\Sigma_g)$ such that each $(M_n,f_n)$ is not doubly-degenerate. then, $\{[f_n^*(\omega_{M_n})]_B\}_{n=1}^\infty$ contains no subsequences converging in $HB^3(\Sigma_g;\Bbb R)$ to the (induced) fundamental class of doubly-degenerate elements of $AH_+(\Sigma_g)$.
Theorem C. Suppose $\Gamma$ is a topologically tame Kleinian group such that the volume of $M_\Gamma$ is infinite. Then $[\omega_\Gamma] = 0$ in $H^3_b(M_\Gamma;\Bbb R)$ if and only if $\Gamma$ is either elementary or geometrically finite. If $[\omega_\Gamma]\neq 0$ in $H^3_b(M_\Gamma;\Bbb R)$, then $\parallel \omega_\Gamma\parallel = v_3$, so in particular, $[\omega_\Gamma]_B\neq 0$ in $HB^3(M_\Gamma;\Bbb R)$.
Theorem D. Let $M,M'$ be hyperbolic 3-manifolds with markings $\iota:\Sigma\to M$, $\iota':\Sigma\to M'$ respectively. Suppose that either the $(+)$ or $(-)$-end $\mathcal{E}$ of $M$ with respect to $\iota(\Sigma)$ is totally degenerate. If
$$\parallel\iota^*([\omega_M]) - \iota'^*([\omega_{M'}])\parallel<v_3$$
holds in $H^3_b(\Sigma;\Bbb R)$, then there exists a marking and orientation-preserving homeomorphism $\varphi_0:M\to M'$ and a neighborhood $E$ of $\mathcal{E}$ such that $\varphi_0|_E:E\to E' = \varphi_0(E)$ is bilipschitz. In particular, $\varphi_0$ defines the bijection between the components of $E_{cusp}$ and those of $E'_{cusp}$.
Corollary E. Under the assumptions in theorem A, suppose moreover that all genuine ends of $M$ are simply degenerate. Then $\varphi$ is properly homotopic to an isometry. In particular, $\iota^*([\omega_M]) = \iota'^*([\omega_{M'}])$ in $H^3_b(\Sigma,\Bbb R)$.
Theorem F. Let $M$ be an oriented hyperbolic 3-manifold with a marking of $\Sigma$. Suppose that there exists an orientation-preserving homeomorphism $\varphi$ from $M$ to another hyperbolic $M'$ inducing a bijection between the components of $M_{cusp}$ and those of $M'_{cusp}$. If
$$\parallel [\omega_M] - \varphi^*([\omega_{M'}])\parallel <v_3$$
holds in $H^3_b(\Sigma;\Bbb R)$, then $\varphi$ is properly homotopic to a bilipschitz map.
Theorem G. Let $G$ be a group, and let $\Gamma_1,\ldots,\Gamma_n$ be topologically tame Kleinian groups admitting isomorphisms $\varphi_i:G\to\Gamma_i$. Suppose that, for each $M_{\Gamma_i}$, only one end $E_i$ of $M_{\Gamma_i}$ is geometrically infinite. Let $\Lambda_i$ be a subgroup (uniquely determined up to conjugate) of $\Gamma_i$ corresponding to $E_i$. If, for every integers $i,j$ with $1\leq i\leq j\leq n$, there exists $\lambda_{ij}\in\Lambda_j$ such that $\varphi_i^{-1}(\Lambda_i)\cap C_\infty(\varphi_j^{-1}(\lambda_{ij})) = \emptyset$, then $[\varphi_1^*(\omega_{\Gamma_1})]_B,\ldots,[\varphi^*_n(\omega_{\Gamma_n})]_B$ are linearly independent in $HB^3(G;\Bbb R)$. Here, $C_\infty(g) = \bigcup_{n=1}^\infty C(g^n)$ where $C(g) = \{hgh^{-1}:h\in G\}$.
Theorem 1.1. Suppose $\Gamma$ is a finitely generated group that is isomorphic to a torsion-free Kleinian group without parabolics, and let $\{\rho_\alpha:\Gamma\to\mathrm{PSL}_2\Bbb C:\alpha\in\Lambda\}$ be a collection of discrete and faithful representation without parabolic or elliptic elements such that at least one of the geometrically finite end invariants of $M_{\rho_\alpha}$ is different from the geometrically infinite end invariants of $M_{\rho_\beta}$ for all $\alpha\neq \beta \in \Lambda$. Then $\{[\hat{\omega}_{\rho_\alpha}:\alpha\in\Lambda]\}$ is a linearly independent set in $H^3_b(\Gamma;\Bbb R)$.
Theorem 1.2. Suppose $\Gamma$ is a finitely generated group that is isomorphic to a torsion-free Kleinian group without parabolics. Then there is an $\epsilon = \epsilon(\Gamma)>0$ such that if $\{\rho_i:\Gamma\to\mathrm{PSL}_2\Bbb C:i = 1,\ldots, n\}$ is a collection of discrete and faithful representations without parabolics or elliptic elements such that at least one of the geometrically infinite end invariants of $M_{\rho_i}$ is different from the geometrically infinite end invariants of $M_{\rho_j}$ for all $i\neq j$ then
$$\parallel\sum_{i=1}^n a_i[\hat{\omega}_{\rho_i}]\parallel_\infty >\epsilon\max |a_i|.$$
Theorem 1.3. Let $S$ be an orientable surface with negative Euler characteristic. Then there is a constant $\epsilon = \epsilon(S)$ such that the following holds. Let $\rho:\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ be discrete and faithful, without parabolic elements, and such that $M_\rho$ has at least one geometrically infinite end invariant. If $\rho':\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ is any other representation satisfying
$$\parallel[\hat{\omega}_\rho] - [\hat{\omega}_{\rho'}]\parallel_\infty<\epsilon,$$
then $\rho'$ is faithful.
Corollary 1.4. Let $S$ be a closed orientable surface of genus at least 2. There is a constant $\epsilon = \epsilon(S)$ such that the following holds. Suppose that $\rho:\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ is discrete, faithful, without parabolics, and such that $M_\rho$ has two geometrically infinite ends. Then for any other discrete representation $\rho':\pi_1(S)\to\mathrm{PSL}_2\Bbb C$, if
$$\parallel[\hat{\omega}_\rho] - [\hat{\omega}_{\rho'}]\parallel_\infty<\epsilon,$$
then $\rho$ and $\rho'$ are conjugate.
Corollary 1.5. There is an injective map
$$\psi:\mathcal{EL}(S)\to\overline{H}^3_b(\pi_1(S);\Bbb R)$$
whose image is a linearly independent, discrete set.
Theorem 2.1. If $\rho_1$ and $\rho_2$ are discrete, faithful, without parabolics, and their quotient manifolds $M_{\rho_i}$ are singly degenerate and share one geometrically infinite end invariant, then $\rho_1^*\mathrm{Vol} = \rho_2^*\mathrm{Vol}\in H^3_b(\pi_1(S);\Bbb R)$.
Theorem 2.2. Let $\lambda,\lambda'\in\mathcal{EL}_b(S)$ and $X,Y\in\mathcal{T}(S)$ be arbitrary. We have an equality in bounded cohomology
$$[\hat{\omega}(\lambda',\lambda)] = [\hat{\omega}(\lambda',X)]+[\hat{\omega}(Y,\lambda)]\in H^3_b(S;\Bbb R).$$
Here, $\mathcal{EL}_b(S)\subset\mathcal{EL}(S)$ denotes the lamination that have bounded geometry.
Corollary 2.3. Suppose $\lambda_1,\lambda_2,\lambda_3\in\mathcal{EL}_b(S)$ are distinct. Then we have an equality in bounded cohomology
$$[\hat{\omega}(\lambda_1,\lambda_2)] + [\hat{\omega}(\lambda_2,\lambda_3)] = [\hat{\omega}(\lambda_1,\lambda_3)]\in H^3_b(S;\Bbb R).$$
Theorem 2.4. Fix a closed, orientable surface $S$ of negative Euler characteristic. There is an $\epsilon_S>0$ such that if $\{[\rho_\alpha]\}_{\alpha\in\Lambda}\subset\mathrm{Hom}(\pi_1(S),\mathrm{PSL}_2\Bbb C)/\mathrm{PSL}_2\Bbb C$ are discrete, faithful and without parabolics such that at least one of the geometrically infinite end invaraints of $\rho_\alpha$ is different from the geometrically infinite end invariants of $\rho_\beta$ for all $\beta\neq\alpha\in\Lambda$, then
(1) $\{\rho_\alpha^*\mathrm{Vol}\}_{\alpha\in\Lambda}\subset H^3_b(\pi_1(S);\Bbb R)$ is linearly independent set;
(2) $\parallel\sum_{i=1}^Na_i\rho_{\alpha_i}^*\mathrm{Vol}\parallel_\infty\geq\epsilon_{S}\max |a_i|$.
Corollary 2.5. The bounded fundamental class is a quasi-isometry invariant of discrete and faithful representations of $\pi_1(S)$ without parabolics. In this setting, $\parallel\rho_0^*\mathrm{Vol} - \rho_1^*\mathrm{Vol}\parallel_\infty<\epsilon_S$ if and only if $\rho_0$ is quasi-isometric to $\rho_1$.
Theorem 2.6. Suppose $\rho_0:\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ is discrete, faithful, has no parabolic elements, and at least one geometrically infinite end invariant and let $\rho:\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ be arbitrary. If $\parallel\rho_0^*\mathrm{Vol} - \rho^*\mathrm{Vol}\parallel_\infty<\epsilon_S/2$, then $\rho$ is faithful. If $\rho$ has dense image, then $\parallel\rho_0^*\mathrm{Vol} - \rho^*\mathrm{Vol}\parallel_\infty\geq v_3$ where $v_3$ is the volume of the regular ideal tetrahedron in $\Bbb H^3$.
Corollary 2.7. Suppose $\rho_0:\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ is discrete and faithful with no parabolics and at least one geometrically infinite end invariant and $\rho_1:\pi_1(S)\to\mathrm{PSL}_2\Bbb C$ has no parabolics (but is otherwise arbitrary). If $\parallel\rho_0^*\mathrm{Vol} - \rho_1^*\mathrm{Vol}\parallel_\infty<\epsilon_S/2$, then $\rho_1$ is discrete and faithful, hence quasi-isometric to $\rho_0$.
Theorem 2.8. Suppose $M_0$ and $M_1$ are hyperbolic 3-manifolds without parabolic cusps with representations $\rho_i:\pi_1(M_i)\to\mathrm{PSL}_2\Bbb C$, $M_1$ is totally degenerate, and $h:M_1\to M_2$ is a homotopy equivalence. If $M_1$ is topologically tame and has incompressible boundary, then there is an $\epsilon$ depending only on the topology of $M_1$ such that $h$ is homotopic to an isometry if and only if $\parallel\rho_0^*\mathrm{Vol} - \rho_1^*\mathrm{Vol}\parallel_\infty<\epsilon$.
Definition. Fix a base point $X\in\mathcal{T}(S)$, and define $\iota:\mathcal{EL}(S)\to H^3_b(S;\Bbb R)$ by the rule $\iota(\lambda) = [\hat{\omega}(X,\lambda)]$. Theorem 1.1 guarantees that $\iota$ does not depend on the choice of $X$.
Theorem 2.9. The image of $\iota$ is linearly independent set. Moreover, for all $\lambda,\lambda'\in\mathcal{EL}(S)$, if $\parallel\iota(\lambda) - \iota(\lambda')\parallel_\infty<\epsilon_S$ then $\lambda = \lambda'$. Finally, $\iota$ is mapping class group equivariant, and $\iota(\mathcal{EL}(S))$ is a topological basis for the image of its closure in the reduced space $\overline{H}^3_b(S;\Bbb R)$.
0 XDK (+0)
유익한 글을 읽었다면 작성자에게 XDK를 선물하세요.
-
회원에 의해 삭제된 댓글입니다.좋아요 0
-
이불 안에 파란담요를 넣고 안고 잠 애착이 생긴건지 안고 있으면 마음이 안정됨
-
https://m.site.naver.com/1Abu2
-
직장인 메디컬 수시 교과 궁금한 점 질문 좀 드려도 될까요? 0
안녕하세요. 직장 다니는 친구가 메디컬 진학을 고민중이길래 한번 여쭙고자 글...
-
정신건강빙고 1
.
-
질문 받습니다 3
ㅇㅇ
-
오르비에서 보면 경한 목표가 되게 많던데 왜그런걸까요 2
미적사탐 현실적 목표라 그런건가
-
다들 잘사는구나 0
나만 친구한테 재수하는데 수업료 없어서 대학간 친구한테 돈빌리고 만원짜리...
-
오르비 도시전설인줄 알았더니 에휴다노
-
메가 현우진샘 들을까하는데요 현우진샘 단점이있을가요? 예전 구 신승범샘이 뭐랄까...
-
수능 킬러문제들에서 벽 느끼고 강민철쌤 유기적독해 체화중인 고3 학생입니다.원래는...
-
그래서해봄
-
수학문제 만들기 정도면 취미 아닐까요... 근데 저기 있는 내용 읽다보니 우울해짐 ㅋㅋㅋㅋㅠ
-
물리 고수님들 답변 부탁드려요(열역학 피스톤 질문) 0
이 문제에 대해서 제가 옆부분에 써놓은 (상황이라고 써놓은) 필기가 맞게 설명한건지...
-
시즌2는 언제시작하는지 난이도는 어느정도 되는지 궁금해요
-
서울 연합동아리 부회장까지 해봤습니다
-
근데 솔직히 흙수저까지는 아닌데...
-
10덕 음악 금지
-
상남자 쉽지않네 4
어렵다
-
토익 오픽 토스 0
뭘 해야할까요
-
하층민달성 0
-
누가 디코방 만듦?
-
돌아와
-
이런 거 뭐였지?
-
뱃지 신청 8
님들 대학 뱃지 신청 하려하는데 합격통지서로는 인증 안됨여? 1월부터 계속 넣었는데 안되네..
-
층간소음 하 2
지금 망치박고 있네ㅋㅋ
-
이신혁 시즌2 0
이신혁 시즌2 언제부터임..? 라이브 듣고 시펑
-
ㄱㄱㄱㄱ
-
탈릅했어요?
-
막 나 특정할려고 사진찍는걸까봐 오르비에 내 사진올려서 "오르비꺼라"하면서...
-
02년생 20년도 졸업하고 대학 들어가지 않고 군대 갔다와서 현재 24살입니다....
-
한번이라도 해당한적 있으면 체크함
-
국어 컨텐츠 0
상상온라인 vs시대컨 이유도 설명해주시면 감사하겠습니다..!
-
흑수저 테스트 ㅇㅈ…. 11
가난해서 울었어…ㅜㅜ 빨리 공부해서 아빠 도와야지…
-
작년버전 아직 서점에 남아있어서 풀어볼까하는데... 아니면 그냥 올해버전 나올때...
-
평일 알바할 거면 몇 시대로 잡는 게 좋음??
-
좋아하는 사람 있음? 본곡보다 더 듣기 좋게 편곡이 많긴 하던데
-
정신건강빙고 2
이야 작년에 했으면 거의다 체크해서 빙고 ㅈㄴ 많았을거같은데
-
2시간이면 되는 거 아니었음?
-
ㅈㄱㄴ
-
지구 커리 0
오지훈쌤 커리타서 유자분(작년거)까지 끝냈고 개념 다시 한번 돌리면서...
-
빙고 4
GOAT
-
예상 2
12시 쯤에 취침예상
-
택 1 ㄱㄱ
-
고1때 하는 문법이 언매인지 뭔지는 잘 모르겠는데 고1때 문법은 꽤 열심히(잘)...
-
늙어서 따라가기 힘드네
-
상의는 무슨 단체복 준다던데 캐리어까지 필요없으려나