# rotational symmetries of a polyhedron

Given is the following Polyhedron $P$ in the $\mathbb{R}^3$:

I want to define the set $S$ of the rotational symmetries of the Polyhedron in $\mathbb{R}^3$. Hence an element $s \in S$ is a function $s: P \rightarrow P$.

One is obviously the identity function and there are three more. Two like you can find them in equilateral triangles, the rotations $120^\circ$ and $240^\circ$ and the last one $180^\circ$ with which the front becomes the back.

But as you can see my problem is that I am not able to "describe" them adequate enough so it's really clear what I mean. Could you help me with this?

• You have 2 rotations about the axis through the triangular faces, and one rotation about the about the vertical axis equidistant from the triangular faces. You also have the identity rotation. Is that sufficient? – Rocket Man Nov 14 '15 at 4:07
• the whole problem has been worked out for point group symmetries in 2D, 3D, and above. See attached Wikipedia article for a start. en.wikipedia.org/wiki/Point_group – MaxW Nov 14 '15 at 4:44
• The rotations you list indicate that you are assuming the polyhedron to be, specifically, a right prism with equilateral triangle bases. In that case, there are actually three $180^\circ$ rotations "with which the front becomes the back". – Blue Nov 14 '15 at 4:48
• The kind of triangles make a big difference. If they're scalene, I don't believe you get any nontrivial rotational symmetry. – pjs36 Nov 14 '15 at 5:48
• Hello Blue, I am not sure what you are exactly saying since my english is not the best. Did I actually not find 2 of 3 $180^\circ$ symmetries or is my description bad, since there are more than one $180^\circ$ rotations but only one is a symmetry – swizzor Nov 14 '15 at 16:14

One possible form of description for such a kind of symmetry would be using a Coxeter group. Start not with the rotations but with the reflections. You could say that $a$ is the plane through one symmetry axis of each triangle, $b$ likewise so that $a$ and $b$ enclose an angle of $60°$, and $c$ is the plane through the center parallel to both triangles. Then you'd say that $a$ and $b$ together form a $120°$ rotation, so if you repeat that three times you have identity. Formally: $(ab)^3=\text{Id}$. Since $c$ is orthogonal to $a$ and $b$, any two of them form a $180°$ rotation, so you have $(ac)^2=(bc)^2=\text{Id}$. Taking it all together you have the Coxeter group
$$\langle a,b,c\mid a^2=b^2=c^2=(ab)^3=(ac)^2=(bc)^2=\text{Id}\rangle$$
You could draw a Coxeter graph for this, which is just three vertices, two of them connected by an unlabeled line. You could decompose the group into $A_2\times A_1$, i.e. the symmetry of the triangle combined with reflection in a single orthogonal plane. Since $\lvert A_2\rvert=6$ and $\lvert A_1\rvert=2$ you'd know that the full symmetry group has $6\cdot2=12$ elements. You could then say that the group of rotational symmetries is the subgroup of all combinations of an even number of reflections. So half of the elements would be rotations, and you have $6$ in total: the identity, the two rotations which rotate each endcap around itself, and three rotations which exchange the endcaps and are each perpendicular to one of the rectangles.
You could also go to the level of coordinates, and describe all these $6$ rotations as a linear transformation in some vector space. But with a regular triangle in there, you'll usually end up with some nasty square roots in your formulas.