Timeline for uniform properness of lifts of uniform proper maps
Current License: CC BY-SA 3.0
11 events
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| May 21, 2012 at 19:21 | comment | added | Damiano Lupi | Maybe I understood what you said: for every sequence $(x_i,y_i)$ of pairs of points in $\tilde{S}$ such that $d_{\tilde{N}}(x_i, y_i)\le A$ we project to the quotient. If $p\colon\tilde{N}\to N$ is the projection map, denote $x'_i=p(x_i)$ and $y'_i=p(y_i)$. Clearly the inequality $d_{N}(x'_i,y'_i)\le A$ still holds, therefore we may use uniform properness to get $d_{S}(x'_i,y'_i)\le B$ for some positive constant $B$. Now we use the cocompactness of the action of $\pi_1(N)$ on the set $\left\{(x,y)\in \tilde{S}\times\tilde{S}\mid d_S(p(x),p(y))\le B \right\}$ and we're done. Is it correct? | |
| May 21, 2012 at 17:48 | comment | added | Damiano Lupi | Shouldn't we prove that if $d_{\tilde{N}}(x,y) \le A$ then $d_{tilde{S}}(x,y)\le B$? The metric on $\tilde{S}$ is induced by the path metric on $\tilde{N}$ so that $d_{\tilde{S}}(x,y)\ge d_{\tilde{N}}(x,y)$ as a path in $\tilde{S}$ is also a path in $\tilde{N}$ so it's enough to take $A=B$ to get that $d_{\tilde{S}}(x,y)\le A$ implies $d_{\tilde{N}}(x,y)\le B=A$. Am I wrong with that? | |
| May 21, 2012 at 16:19 | comment | added | Lee Mosher | That was not a proof by contradiction. That was a direct proof of the statement that for each $A$ there exists $B$ so that if $d_{\widetilde S}(x,y) \le A$ then $d_{\tilde N}(x,y) \le B$. | |
| May 21, 2012 at 16:14 | comment | added | Damiano Lupi | distance in $\tilde{N}$ is bounded but their distance in $\tilde{S}$ goes to infinity. But this generates a contradiction since the quotient is compact, so they must be at bounded distance (why? isn't the universal cover of a compact surface isometric to $\mathbb{H}^2$?). It doesn't sound very sound to me, though. I guess I misunderstood. Could you please clarify this? | |
| May 21, 2012 at 16:10 | comment | added | Damiano Lupi | I understand why the hypothesis is trivially true in this case, but thanks for pointing it out. I'm still confused about the second part: why do we have to consider the set of pairs with $d_S(x,y)< A$? I mean: the inequality holds when we look at the distance in the bigger space, so in $\tilde{N}$, and we are assuming that it doesn't hold in $\tilde{S}$, by contradiction. Perhaps, using your hint, we may say the following: if $\tilde{S}\hookrightarrow \tilde{N}$ were not uniformly proper, than we would have this sequence of pairs of points $(x_i,y_i)$ in $\tilde{S}$ such that their... | |
| May 21, 2012 at 15:27 | comment | added | Lee Mosher | First, the hypothesis of "the case" that you consider is trivially true: every map from a compact metric space to another metric space is uniformly proper. Second, the set of pairs $(x,y) \in \tilde S$ with $d_S(x,y) \le A$ is cocompact under the diagonal action of $\pi_1(S)$. So $max d_{H^2}(x,y)$ on this set is bounded. | |
| May 21, 2012 at 15:04 | comment | added | Damiano Lupi | OK... done... sorry! | |
| May 21, 2012 at 15:01 | comment | added | Damiano Lupi | That's what I'm trying to do but for some strange reason in the preview everything is shown correctly... I'll try again! Sorry! | |
| May 21, 2012 at 15:00 | history | edited | Damiano Lupi | CC BY-SA 3.0 |
added 1 characters in body; deleted 4 characters in body; deleted 5 characters in body
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| May 21, 2012 at 14:58 | comment | added | Lee Mosher | Eek! Fix the TeX errors! | |
| May 21, 2012 at 14:54 | history | asked | Damiano Lupi | CC BY-SA 3.0 |