Timeline for Quotient of the hyperbolic plane with respect to commutator group of $\pi_1(\Sigma_g)$
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2014 at 4:19 | vote | accept | Mohammad Farajzadeh-Tehrani | ||
| Dec 21, 2014 at 16:03 | answer | added | Misha | timeline score: 5 | |
| Dec 21, 2014 at 1:00 | history | edited | Sam Nead |
edited tags.
|
|
| Dec 21, 2014 at 0:23 | answer | added | Sam Nead | timeline score: 7 | |
| Nov 20, 2014 at 17:10 | answer | added | Holonomia | timeline score: 2 | |
| Oct 20, 2014 at 10:23 | comment | added | BS. | Unfortunately I didn't found any recent textbook on this subject. By the way, the classification theorem proves that the semi-infinite Jacob ladder is homeomorphic to the "Loch Ness monster"! en.wikipedia.org/wiki/Loch_Ness_monster_surface | |
| Oct 20, 2014 at 0:45 | comment | added | Mohammad Farajzadeh-Tehrani | For fun: Finite Jacob ladder can be seen here cambridgeblog.org/2013/01/into-the-intro-games-and-mathematic. | |
| Oct 20, 2014 at 0:26 | comment | added | Mohammad Farajzadeh-Tehrani | @ BS: interesting, I had never heard of such terminologies. Is there any somewhat recent book discussing this sort of things! | |
| Oct 19, 2014 at 15:06 | comment | added | BS. | It seems I was wrong. Kerekjarto says (if I correctly understood, p. 183 of "Vorlesungen uber Topologie) that this "homology covering" of a genus $g\geq 2$ surface has infinte genus and only one end (so $E=F$=point). So it is homeomorphic to the half-infinite "Jacob ladder". | |
| Oct 19, 2014 at 8:21 | comment | added | BS. | The classification of noncompact boundaryless surfaces is more complicated than that. If orientable, they are classified genus (maybe infinite) and (homeo class of) a pair of compact totally disconnected metrisable spaces of ends $(E\supset F)$, with $F$ the subspace of nonplanar ends. This is due to Kerekjarto (1923). In your case, I would guess that $E$ is a Cantor and $F$ is empty, but this is only a guess. | |
| Oct 18, 2014 at 14:04 | comment | added | Mohammad Farajzadeh-Tehrani | I think the following describes the generators:mathoverflow.net/questions/38413/… | |
| Oct 18, 2014 at 11:37 | comment | added | Mohammad Farajzadeh-Tehrani | Of course it is infinite, but it should be still describe-able in terms on number of ends and genus; e.g. a pair of pants is genus zero with 3 ends. In this case, I feel it should be a genus 0 curve with infinity many ends dictated by the type of free graph. | |
| Oct 18, 2014 at 11:22 | comment | added | Mohammad Farajzadeh-Tehrani | Thanks Seirios: But what is rank of it? Can we say that? at least for g=2? | |
| Oct 18, 2014 at 5:36 | comment | added | Ryan Budney | Is there any particular dictionary for surfaces you would like to use, Mohammad? The surface has fairly direct descriptions in most ways I can think of thinking of surfaces. | |
| Oct 18, 2014 at 5:10 | comment | added | Qiaochu Yuan | I would run that argument the other way: $\mathbb{H}/[G, G]$ is an infinite covering space of $\Sigma_g$, hence it is noncompact. As a noncompact surface, it's homotopy equivalent to a $1$-dimensional CW complex, hence its fundamental group is free. | |
| Oct 18, 2014 at 3:50 | comment | added | Seirios | The commutator subgroup of a surface group is free. See corollary 4 at chiasme.wordpress.com/2014/08/27/on-subgroups-of-surface-groups In particular, it can be deduced that $\mathbb{H} / [G,G]$ is not compact from the abelianization of $G$. | |
| Oct 18, 2014 at 0:55 | history | edited | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |
added 45 characters in body
|
| Oct 18, 2014 at 0:46 | history | asked | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |