Timeline for Triangle groups - uniqueness and trace field
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2013 at 10:51 | comment | added | j0equ1nn | You don't necessarily need something to be unique up to conjugacy for the trace field to be well-defined, just up to commensurability (in the wide sense). But I think the correct answer to Q1 is just: "Always, because a signature defines a group in terms of its presentation" (as explained in the selected answer below). As for Q2, nobody addressed whether there's an algorithm for computing the invariant trace field of this group, but it seems you could start with the information in the answer below, then try to compute it just by its definition. I expect it would help to consider the geometry. | |
| Oct 22, 2010 at 21:03 | answer | added | HJRW | timeline score: 10 | |
| Oct 20, 2010 at 11:54 | comment | added | Dylan Thurston | My immediate reading was: is the triangle group $(p,q,\infty)$ ever isomorphic to $(p',q',\infty)$ (changing $p$ and $q$)? But the second question suggest something more like Pete Clark's answer. | |
| Oct 20, 2010 at 7:24 | comment | added | Pete L. Clark | @Henry: It is part of my guess that the "up to isomorphism" part of the question is a misstatement. Of course I don't know for sure what the OP means, but I have studied triangle groups and this is the one true, nontrivial statement that lives in a small neighborhood of the OP's question, so far as I know. (Note also that you want uniqueness up to conjugacy, not just abstract isomorphism, in order for the trace field to be well-defined.) | |
| Oct 20, 2010 at 4:25 | comment | added | HJRW | Pete - good guess! Although, if so, why 'up to isomorphism'? | |
| Oct 20, 2010 at 0:08 | comment | added | Steve Richards | I do not think the question is clear enough. Should be revised or closed. | |
| Oct 19, 2010 at 23:33 | comment | added | Pete L. Clark | Is it possible that the uniqueness statement in question is (or should be) that the set of Fuchsian groups isomorphic to a given triangle group forms a single $\mathbb{SL}_2(\mathbb{R})$-conjugacy class? | |
| Oct 19, 2010 at 22:34 | comment | added | HJRW | Just to be clear, what I said above is only literally true of the orientation-preserving index-two subgroup of the triangle group (what Wikipedia calls `von Dyck groups'). But the same sort of thing applies for triangle groups. | |
| Oct 19, 2010 at 15:42 | comment | added | HJRW | I don't think I understand your question about 'unique' - of course there's only one triangle group with a give signature, because it's well defined! Or are you worried that the signature isn't unique? It's clear that $\Delta(p,q,\infty)\cong\Delta(p',q',\infty)$ if and only if $\lbrace p,q\rbrace=\lbrace p',q'\rbrace$ because the only torsion in $\Delta(p,q,\infty)$ is $p$-torsion and $q$-torsion. In other words, the signature is unique. | |
| Oct 19, 2010 at 14:55 | history | edited | Ali K | CC BY-SA 2.5 |
added 32 characters in body
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| Oct 19, 2010 at 14:29 | history | asked | Ali K | CC BY-SA 2.5 |