# Triangle groups - uniqueness and trace field

Dear all,

again I need your help for the following two questions: Suppose we have a triangle group of signature (p,q,\infty).

1) When is such a group unique (up to isomorphism)? 2) Do you have a method how to calculate the trace field of such a group? Is the trace field unique?

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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. – HJRW Oct 19 '10 at 15:42
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. – HJRW Oct 19 '10 at 22:34
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? – Pete L. Clark Oct 19 '10 at 23:33
I do not think the question is clear enough. Should be revised or closed. – Steve Richards Oct 20 '10 at 0:08
@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.) – Pete L. Clark Oct 20 '10 at 7:24

In more detail, suppose that $\Gamma$ is a $(l,m,n)$-triangle group where $1/l+1/m+1/n<1$ so that $\Gamma$ has the presentation
$\langle x,y\mid x^l=y^m=(xy)^n=1 \rangle$.
Then $\mathbb{Q}(\mathrm{tr}\Gamma)=\mathbb{Q}(\cos\pi/l,\cos\pi/m,\cos\pi/n)$ (see (3.25)), and the invariant trace field is a subfield of this totally real number field (see Exercise 4.9, No. 1).
Although HJRW's answer uses the notation for a co-compact triangle group, non co-compact triangle groups of the form $(p,q, \infty)$ (and $(p, \infty, \infty)$ also apply here. The necessary adjustments are removing the relation $(xy)^n$ (and $y^m$) and $\mathbb{Q}(tr \Gamma) = \mathbb{Q}(cos \pi/l, cos \pi/m)$ (and $\mathbb{Q}(tr \Gamma) = \mathbb{Q}(cos \pi/l$)). – Neil Hoffman Jul 5 at 21:42