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Apr 13, 2017 at 12:58 history edited CommunityBot
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May 6, 2012 at 23:38 vote accept Lee Mosher
May 6, 2012 at 21:32 answer added Autumn Kent timeline score: 2
May 5, 2012 at 4:30 comment added Misha @Lee: Of course, you are right, I just somehow misread your question. Now I am not sure what the answer is...
May 3, 2012 at 19:57 comment added Lee Mosher @Misha: I suspect that for "random" choices of $B$, all loops in $\Gamma_B$ project to nonsimple loops in $S$, in which case your argument shows that the image of $T(\Gamma) \to T(\Gamma_B)$ entirely misses the thin part of $T(\Gamma_B)$. So your reformulation in such cases is equivalent to my formulation because the restricted projection map from the thick part of $T(\Gamma_B)$ to $M(\Gamma_B)$ is a quasi-isometry.
May 3, 2012 at 18:28 comment added Misha @Lee: If $B$ contains a simple non-peripheral loop $\beta$ which projects to a non-simple loop in $S$ then the restriction map $f$ will always miss the thin part of $T(\Gamma_B)$ corresponding to $\beta$. Thus, the map is never coarsely surjective unless $B$ is a subsurface or a pair of pants. Also, for dimension reasons, the image is "too small" if $-\chi(B)> -\chi(S)$. An interesting question then would be if $f$ is (sometimes/always) coarsely onto thick part of $T(\Gamma_B)$ when $-\chi(B)\le -\chi(S)$ and $B$ is not embedded. "Sometimes" is probably true, "always" is probably false.
May 3, 2012 at 18:22 history edited Lee Mosher CC BY-SA 3.0
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May 3, 2012 at 18:21 comment added Lee Mosher added: also, as the boundary length goes off to infinity, all lengths of all closed curves would have a lower bound that goes off to infinity.
May 3, 2012 at 18:20 comment added Lee Mosher @Misha: just because the lengths of all elements of $B$ are uniformly long does not mean that the image is unbounded in the marking complex $M(\Gamma_B)$. For example, $\Gamma_B$ could be a one-holed torus with three very short properly embedded arcs hitting the boundary at right angles, cutting the torus into two right-angled hexagons. On this $\Gamma_B$, one could force the boundary length off to infinity without changing which three arcs are short, and the marking would be unchanged.
May 3, 2012 at 18:15 history edited Lee Mosher CC BY-SA 3.0
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May 3, 2012 at 18:11 comment added Lee Mosher I edited to try to clarify what "rigid" means, although it is still a somewhat vague notion.
May 3, 2012 at 18:10 history edited Lee Mosher CC BY-SA 3.0
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May 3, 2012 at 18:05 comment added Misha @Lee: It will never be bounded (except for the pair of pants example): Just apply powers your favorite pseudo-Anosov to a non-peripheral element of $B$. Could you clarify what do you mean by rigidity in Question 1? The setup sounds somewhat similar to McMullen's work on theta-operator (or, even more remotely, skinning map). The restriction map $T(\Gamma)\to T(\Gamma_B)$ is holomorphic, so it contracts Teichmuller metric. It might follow from McMullen's work that the map is a strict contraction.
May 3, 2012 at 15:45 history asked Lee Mosher CC BY-SA 3.0