MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Thank you very much for your answer Ali

share|cite|improve this question
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

One page 159 of The Arithmetic of Hyperbolic Manifolds by Maclachlan and Reid:-

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).

More generally, Maclachlan and Reid should have the answers to all your questions about trace fields.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.