Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Looking at http://en.wikipedia.org/wiki/Triangle_group I begin to wonder why the definition explicitly excludes the tessellation of the Euclidean plane by 30-30-120 triangles? In terms of the Wallpaper groups, I am thinking of the group p6 ( http://en.wikipedia.org/wiki/Wallpaper_group#Group_p6 ).

Is it just an exceptional case? The definition of the triangle group asks that the "order" at each vertex to be even, which is natural, as for odd orders only isosceles triangles can "close" under reflections. So if one of the angles of a tessellating triangle is $2\pi / k$ for $k$ odd, it is necessary that there exists some integer $l$ such that $$\frac{1}{k} + \frac{2}{l} = \frac12$$ and the only solution is $k = 3$ and $l = 12$.

But for less rigid geometries (say hyperbolic), this seems to introduce a large number of additional tessellations. (Though not really more groups, I think, since geometrically replacing the obtuse triangle by two congruent acute triangles leads to one of the already defined cases.)

Motivation: I am trying to describe an arts/craft project for demonstrating hyperbolic geometry. As such it is more natural to start from the tessellation picture, rather than the reflection group picture. Therefore it was a bit surprising to me that despite the introductory paragraph on Wikipedia, the two points of view are not exactly the same.

My apologies for the somewhat muddy wording of the question.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

The answer is already contained in your question. You do not describe more reflection groups, since replacing the obtuse triangle by two congruent acute triangles leads to one of the already defined symmetry groups.

Not directly related to your question: here are two cool java applet having to do with hyperbolic tesselations.
Applet 1.
Applet 2.

share|improve this answer
Oh... hum, so does this meant that the reflection group generated by these odd elements are necessarily somehow bigger than the symmetry group of the tiling? (Thinking of $p6 \subset p6m$.) Also, thanks for the applets, though I am looking more at tilings by non regular polygons. –  Willie Wong Jun 25 '10 at 16:22
Ah... after thinking about it some more, I think I understand. Thanks. –  Willie Wong Jun 25 '10 at 17:14

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.