8
$\begingroup$

Take a triangle in the (Euclidean or hyperbolic) plane, and consider the group of isometries generated by the reflections in the three sides of the triangle. If the angles between adjacent sides are all integral submultiples of $\pi$, then the group generated is discrete and has a presentation as a Coxeter group.

Questions:

  1. What other choices of angles generate a discrete group? For instance, what if the angles of the triangle are just rational multiples of $\pi$?

    As an example, reflections in the Euclidean triangle with angles $\frac{\pi}{6}, \frac{\pi}{6}, \frac{2\pi}{3}$ generate the (discrete) $(2,3,6)$ triangle group.

  2. When the group that is generated is discrete, can one easily find a fundamental domain for the action, e.g. in terms of the angles of the original triangle? Is the fundamental domain necessarily also a triangle?

  3. For discrete groups generated in the hyperbolic plane, which choices of angles yield arithmetic subgroups of $\operatorname{PSL}(2,\mathbb R)$?

    If the discrete groups generated are themselves triangle groups, then Takeuchi classifies which are arithmetic.

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ Coxeter - Discrete groups generated by reflections must be relevant, although I'm not sure if it directly answers your questions 1 and 3. For question 1, Lemma 4.2 is relevant but (as you pointed out on my deleted answer) not decisive. Pp. 595–596 describes the "Fricke–Klein construction for a fundamental region": take a never-fixed point $P$, and consider the region bounded by the perpendicular bisectors of the lines connecting $P$ to its transforms. $\endgroup$
    – LSpice
    Commented Aug 12, 2022 at 1:50
  • 3
    $\begingroup$ If reflections in a triangle generate a discrete group, then it will be tessellated by the fundamental domain. See: en.wikipedia.org/wiki/… $\endgroup$
    – Ian Agol
    Commented Aug 12, 2022 at 10:05
  • 2
    $\begingroup$ Regarding your final comment, I think the resulting discrete group must be a triangle group, since it will have three conjugacy classes of torsion elements and these classes generate. $\endgroup$
    – HJRW
    Commented Aug 12, 2022 at 14:43

1 Answer 1

Reset to default
11
$\begingroup$

The Euclidean case is an easy case-by-case analysis.

The hyperbolic case was resolved by Anna Felikson, encapsulated in this figure from her paper:

enter image description here

A triangle with mirrored sides may be regarded as a cone manifold. If the reflection group generated by reflections in its sides is discrete, then a fundamental domain will be a Coxeter polygon. Hence the triangle will have a “Coxeter decomposition”, which can also be described as a branched cover of the cone manifold over the Coxeter orbifold. The figure shows all the possible Coxeter decompositions of hyperbolic triangles.

The answer to part 2) of your question follows from the classification. The answer to part 3) follows from Takeuchi’s classification as you’ve indicated.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .