Timeline for Fixed point set of an isometric group action on an hyperbolic manifold
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| Apr 27, 2013 at 9:59 | comment | added | Misha | at least $n-3$, so each round sphere containing the limit set has dimension $\ge n-3$. The proof of the result about Hausdorff dimension is in my paper: math.ucdavis.edu/~kapovich/EPR/cd.pdf You may also want to read my survey math.ucdavis.edu/~kapovich/EPR/klein.pdf for further background on Kleinian groups in higher dimensions. | |
| Apr 27, 2013 at 9:55 | comment | added | Misha | Selim: Yes, this is a correct reasoning: each qi embedding induces continuous injection from the limit set to the sphere at infinity. This fact is one of the standard tools in geometric group theory, due to Efremovich and Tikhomirova (1964). There is an alternative argument to this, which does not need a qi embedding. Namely, Let $\Gamma$ be a Kleinian group without parabolic elements, whose homological dimension (over ${\mathbb R})$ equals $k$. Then Hausdorff dimension of the limit set of $\Gamma$ is at least $k-1$. In our case, homological dimension is $n-2$, so H.d. of limit set is | |
| Apr 27, 2013 at 8:15 | comment | added | Selim G | hidden or if for some algebraic reason, the limit set of the stabilizer of a group isomorphic to the fundamental group of a closed hyperbolic $n-2$-manifolds has to be of dimension $n-3$ ? Anyway thank you very much to take some time to answer my questions, this is very helpfull to me | |
| Apr 27, 2013 at 8:10 | comment | added | Selim G | Ok, it's this last argument which is mysterious to me. I know that since $M$ can be endowed with a negatively curved metric in which $V$ is totally geodesic, the inclusion $i : \tilde V_j \longrightarrow \mathbb{H}^n $ is a quasi-isometry(for the repsective hyperbolic structures), and using Thurston's arguments, we see that we get a continuous injection $\partial_{\infty} \tilde{V_j} \longrightarrow \partial_{\infty}$ so the limit set of $\tilde{V_j}$ must be of topological dimension $n-3$. What is not clear to me in what you say is if some kind of analytic argument as above is ... | |
| Apr 25, 2013 at 15:31 | comment | added | Misha | The fixed point set of a Moebius transformation is always a round sphere of some dimension. In your case, this set contains the limit set of the stabilizer of $\tilde{V}_j$, which is a fundamental group of a closed hyperbolic $n-2$-manifold. Such limit set has to be topological sphere of dimension $n-3$. | |
| Apr 25, 2013 at 14:44 | comment | added | Selim G | ok those last points are clear. Can I bother you with one last question ? Why is the fixed point set of $\tilde\beta$ a round sphere ? It is the accumulation set of $\tilde{V_j}$ in $\tilde{\mathbb{H}^n}$ but why can't it be degenerate ? | |
| Apr 25, 2013 at 12:39 | comment | added | Misha | But then $\pi_1(V_j)\subset \pi_1(M')$ also admits a finite index proper extension. This contradicts the fact that $V_j'$ is embedded in $M'$. The same arguments work for $\tilde\alpha$. | |
| Apr 25, 2013 at 12:36 | comment | added | Misha | Yes, $\pi_1(V_j)$ is exactly the set of elements commuting with $\tilde\beta$. One inclusion is clear. This shows that $\tilde\beta$ fixes pointwise a codimension 2 round sphere $S$ in the sphere at infinity. If there are extra elements which commute with $\tilde\beta$ there are two possibilities: (1) Their fixed points are not in $S$. Then $\tilde\beta$ is either identity or reverses orientation. This is not the case. (2) Their fixed points are in $S$. Then you take such element $g$ and consider a group $H$ it generates together with $F=\pi_1(V_j)$. Then $H$ is a finite extension of $F$... | |
| Apr 25, 2013 at 12:21 | comment | added | Selim G | I have been looking at the detail of your answer and two details remain not clear to me, maybe because I don't know enough hyperbolic geometry : _ You claim that $\Pi_1(V_j)$ consists of elements commuting with $\tilde{\beta}$, but not of THE elements commuting with $\tilde{\beta}$ right ? In that case, where does the isomorphism $\Pi_1(V_j) \simeq \Pi_1(F_j)$ comes from ? _ Why do $F_j$ and $V_j$ have the same dimension ? Or maybe you don't use that fact, in that case if $dim(V_j) = 2$ and $dim(F_j) = 3$ how do you conclude ? | |
| Apr 24, 2013 at 9:09 | vote | accept | Selim G | ||
| Apr 23, 2013 at 13:11 | history | edited | Misha | CC BY-SA 3.0 |
added 7 characters in body
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| Apr 23, 2013 at 12:27 | history | answered | Misha | CC BY-SA 3.0 |