Exercise 1.2.9 (Group of rigid motions of a regular tetrahedron)

Question

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of the regular tetrahedron. Show that $|G| = 12$.

Richard Ganaye · Accepted Answer

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Note: A rigorous proof of Exercises 9, 10, 11,12 requires many geometric preliminaries. To keep in the spirit of these exercises, I only give the outline of such proofs. For the tetrahedron, I give a little more geometric precisions.

\begin{proof} 
\begin{itemize}
\item
Let $G$ be the group of rotations of the tetrahedron ${\cal T} = \{A,B,C,D\}$. $G$ acts on $T$ by $g\cdot M = g(M)$ ($g\in G, M \in T$).

This action is transitive (there is only one orbit). Indeed, if $O$ is the center of $T$ (the barycenter of the vertices with same weights), the rotations $ho, ho^{-1}$ of axe $OA$ and angles $\pm2\pi/3$ send $B$ on $C$ and $D$. Therefore $|{\cal O}_A| = |T| = 4$.

The fundamental theorem of group actions gives
$$(G : G_A) = |O_A|,$$
where $G_A$ is the stabilizer of $A$. Since $G_A = \{e,ho, ho^{-1}\}$, we obtain
$$|G| =|G_A|\cdot  |O_A| = 3 \cdot 4 = 12.$$

\item Alternatively, we can use the action of $G$ on the set $\mathscr{F}$ of faces of the tetrahedron (each face is a set $F = \{M,N,P\}$ of three summits, and there are $4$ faces). Since this action is transitive, and the stabilizer of $F_0 = \{A,B,C\}$ is a group $G_{F_0}$ of three rotations, we obtain
$$|G| = |G_{F_0}| \cdot |O_{F_0}| = 3 \cdot  4.$$

\end{itemize}
\end{proof}

and a little more ...

\bigskip
{\large GROUP OF ISOMETRIES OF THE REGULAR TETRAHEDRON.}

\bigskip

Prerequisites: knowledge of Euclidean affine spaces (in dimension 3), affine mappings, orthogonal endomorphisms, and isometries.

We call ``displacements'' the direct affine isometries (rigid motions), and ``antidisplacements'' the indirect isometries.

In particular, we will use the proposition:
\begin{prop} Let $E$ be an affine space of dimension 3, and $A_1,A_2,A_3,A_4,A'_1,A'_2,A'_3,A'_4$ be points of $E$.

If $A_1,A_2,A_3,A_4$ are not coplanar, there exists one and only one affine mapping $f$ such that $f(A_i) = A'_i,\ i=1,2,3,4$. \end{prop}

\begin{proof}
(Brief reminder of the proof)

Since $A_1,A_2,A_3,A_4$ are not coplanar,
$\left(\overrightarrow{A_1A_2},\overrightarrow{A_1A_3},\overrightarrow{A_1A_4}ight)$ is a basis of the underlying vector space $\vec{E}$.

If $f$ is the affine map such that $f(A_1) = A'_1$, and whose associated linear map $	ilde{f}$ is determined by
$$	ilde{ f} (\overrightarrow{A_1A_2})= \overrightarrow{A'_1A'_2},\qquad 	ilde{ f} (\overrightarrow{A_1A_3}) = \overrightarrow{A'_1A'_3}, \qquad 	ilde{ f} (\overrightarrow{A_1A_4}) = \overrightarrow{A'_1A'_4},$$
then
$$f(A_i) = A'_i,\qquad i=1,2,3,4.$$
\end{proof}

\begin{dfn}{\it
Let ${\cal T} = \{A,B,C,D\}$ a set of 4 points of $E$.

${\cal T}$ is a regular tetrahedron if and only if
$$AB = AC =BC = AD = BD = CD 
e 0.$$}
\end{dfn}
We will call the value $a = AB \in \R_{+}^*$ the side of the tetrahedron.

Regular tetrahedra with side $a$ exist! (see "The Ants" by Bernard Werber).

Let us start with an equilateral triangle $(A,B,C)$ in a plane $P$ such that $AB = BC = CA = a$, and $\Delta$ the axis orthogonal to $P$ passing through the center of gravity $H$ of $(A,B,C)$. We know that $AH = \frac{2}{3} \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{3} a $. A point $D$ on this axis such that
$$HD = \sqrt{\frac{2}{3}} a$$
satisfies $AD^2 = DH^2+AH^2 = \frac{2}{3} a^2 + = \frac{1}{3} a^2 = a^2 $, or $AD = a$, and similarly $BD = CD = a$, so that $\{A,B,C,D\}$ is a regular tetrahedron.

Moreover, the four points of a regular tetrahedron ${\cal T} = \{A,B,C,D\}$ with side $a>0$ cannot be coplanar, otherwise $D$ verifying $DA=DB=DC$ would be equal to the center of gravity of the equilateral triangle $(A,B,C)$, and then $AD = \frac{\sqrt{3}}{3} a 
e a$.

\begin{prop} If ${\cal T} = \{A,B,C,D\}$ and ${\cal T'} = \{A',B',C',D'\}$ are two regular tetrahedra with side $a$, then they are congruent. More precisely, there exists a unique isometry $f$ of $E$ such that
$$f(A) = A', \qquad f(B) = B', \qquad f(C) = C', \qquad f(D) = D'.$$
\end{prop}

\begin{proof}
Since $A, B, C, D$ are not coplanar, there exists a unique affine map $f$ such that $f(A) = A', f(B) = B', f(C) = C', f(D) = D'$. Let us show that $f$ is an isometry, which is equivalent to the associated linear map $	ilde{f}$ being an orthogonal endomorphism, in other words, that $||	ilde{f}(\vec{u}) || = ||\vec{u} ||$ for all $\vec{u} \in \vec{E}$.

Let $\vec{u}$ be any vector in $\vec{E}$, and $M \in E$ such that $\overrightarrow{AM} = \vec{u}$. Since $\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}$ is a basis of $\vec{E}$, there exists $x,y,z \in \R$ such that $\overrightarrow{AM} = x \overrightarrow{AB}+ y \overrightarrow{AC} + z \overrightarrow{AD}$.

So
$$||\vec{u}||^2 = ||\overrightarrow{AM}||^2 = x^2 a^2 + y^2 a^2 + z^2 a^2 + 2xy \overrightarrow{AB}\cdot \overrightarrow{AC} + 2yz\overrightarrow{AC}\cdot \overrightarrow{AD} + 2xz\overrightarrow{AB}\cdot \overrightarrow{AD}.$$
Since $	ilde{f}(\vec{u}) = 	ilde{f}(\overrightarrow{AM}) = \overrightarrow{A'M'} = x \overrightarrow{A'B'} + y \overrightarrow{A'C'} + \overrightarrow{A'D'}$,
$$||	ilde{f}(\vec{u})||^2 = ||\overrightarrow{A'M'}||^2 = x^2 a^2 + y^2 a^2 + z^2 a^2 + 2xy \overrightarrow{A'B'}\cdot \overrightarrow{A'C'} + 2yz\overrightarrow{A'C'}\cdot \overrightarrow{A'D'} + 2xz\overrightarrow{A'B'}\cdot \overrightarrow{A'D'}.$$
To prove $||	ilde{f}(\vec{u}) || = ||\vec{u} ||$, it therefore remains to prove that
$$\overrightarrow{AB}\cdot \overrightarrow{AC} = \overrightarrow{A'B'}\cdot \overrightarrow{A'C'},\qquad \overrightarrow{AC}\cdot \overrightarrow{AD} =\overrightarrow{A'C'}\cdot \overrightarrow{A'D'} ,\qquad \overrightarrow{AB}\cdot \overrightarrow{AD} = \overrightarrow{A'B'}\cdot \overrightarrow{A'D'}.$$

$A,B,C$ is an equilateral triangle: $AB =AC = BC$. Like $BC^2 = (\overrightarrow{AC} - \overrightarrow{AB})^2 = AC^2 + AB^2 - 2 \overrightarrow{AB} \overrightarrow{AC}$,
$$\overrightarrow{AB}\cdot \overrightarrow{AC} = \frac{1}{2} (AB^2+AC^2 - BC^2) = \frac{a^2}{2},$$
And
$$\overrightarrow{A'B'}\cdot \overrightarrow{A'C'} = \frac{1}{2} (A'B'^2+A'C'^2 - B'C'^2) = \frac{a^2}{2}.$$
Therefore, $\overrightarrow{AB}\cdot \overrightarrow{AC} = \overrightarrow{A'B'}\cdot \overrightarrow{A'C'}$, and similarly $\overrightarrow{AC}\cdot \overrightarrow{AD} = \overrightarrow{A'C'}\cdot \overrightarrow{A'D'} ,\ \overrightarrow{AB}\cdot \overrightarrow{AD} = \overrightarrow{A'B'}\cdot \overrightarrow{A'D'}$.

In conclusion, for all $\vec{u} \in \vec{E}$, $||	ilde{f}(\vec{u}) || = ||\vec{u} ||$, so $	ilde{f}$ is an orthogonal endomorphism, and therefore $f$ is an isometry.
The uniqueness of $f$ comes from Proposition 1.
\end{proof}

Let us move on to the study of the group of isometries of a regular tetrahedron ${\cal T} = (A_1,A_2,A_3,A_4)$ with side $a$. Let $\mathrm{Is}(E)$ denote the group of isometries of $E$, and $\mathrm{Is}_{\cal T}$ the subgroup of $\mathrm{Is}(E)$ consisting of the isometries $f$ of $E$ preserving the tetrahedron, i.e. such that $f({\cal T}) = {\cal T}$. Let us also denote $S({\cal T})$ the set of bijections from $\cal T$ into $\cal T$, in other words the set of permutations of the vertices of $\cal T$.

\begin{prop}
Consider the map
$$\varphi:
\left\{
\begin{array}{ccc}
\mathrm{Is}_{\cal T} & 	o & S({\cal T}) \ 
f & \mapsto & f |_{\cal T} =
\left(
\begin{array}{cccc} 
A_1 & A_2 & A_3 & A_4 \ 
f(A_1)& f(A_2) & f(A_3) & f(A_4)
\end{array}
ight) .
\end{array}
ight.
$$
Then $\varphi$ is an isomorphism of groups.
\end{prop}
\begin{proof}
\be
\item[$\bullet$] The application is well-defined: if $f \in \mathrm{Is}_{\cal T}$, then $f(\{A_1, A_2,A_3,A_4\}) = \{A_1, A_2,A_3,A_4\}$, and the restriction of $f$ to $\cal T$ permutes the 4 points of $\cal T$. \item[$\bullet$] $\varphi$ is a homomorphism of groups: if $f,g \in \mathrm{Is}_{\cal T} $, $$(\varphi( f ) \circ \varphi(g))(A_i) = (f \circ g)(A_i) = \varphi(f \circ g)(A_i),\ i=1,2,3,4,$$
therefore $\varphi( f ) \circ \varphi(g) = \varphi(f \circ g)$.
\item[$\bullet$] $\varphi$ is bijective: if $\sigma = \left (
\begin{array}{cccc}
A_1 & A_2 & A_3 & A_4 \
A'_1& A'_2 & A'_3 & A'_4
\end{array} ight) \in S({\cal T}) $, where $ \{A'_1,A'_2,A'_3,A'_4\} = {\cal T}$, then by Proposition 2 applied to ${\cal T} = {\cal T}'$, there exists one and only one $f \in \mathrm{Is}(E)$ such that $f(A_i)=A'_i,\ i=1,2,3,4$, and therefore $f |_{\cal T} = \sigma$ and $\varphi(f) = \sigma$.
\ee
\end{proof}
Now $S({\cal T}) \simeq S_4$, where $S_4$ denotes the group of permutations of the integers $1,2,3,4$. If $u : \{1,2,3,4\} 	o {\cal T}$ is the bijection defined by $i \mapsto A_i$, corresponding to the choice of vertex numbering, then the isomorphism $\psi : S({\cal T})	o S_4$ is defined by $\psi(\sigma) = u^{-1} \circ \sigma \circ u = 	au$, where $	au$ is such that $	au(i) = j \iff \sigma(A_i) = A_j$. We therefore obtain the following corollary.

\medskip
{\bf Corollary.} {\it The isometry group of a regular tetrahedron is isomorphic to $S_4$.}

\medskip
What is the isometry $f$ corresponding to the transposition $	au = (1\, 2)$? $f$ exchanges $A_1$ with $A_2$ and leaves the points $A_3,A_4$ fixed. It is therefore the reflection with respect to the medial plane of $(A_1,A_2)$, which is an antidisplacement. Similarly, any transposition corresponds in the isomorphism $S({\cal T}) \simeq S_4$ to a plane reflection, and therefore to an antidisplacement. Consequently, even permutations, the product of an even number of transpositions, correspond to displacements, and odd permutations to antidisplacements. We have proven the following proposition:

\begin{prop}
The isometry $f$ defined by $A_i \mapsto A_{	au(i)}$, where $	au \in S_4$ is an antidisplacement if $	au \in A_4$, and an antidisplacement if $	au \in S_4 \setminus A_4$.
\end{prop}

Note that the isobarycenter $O$ of ${\cal T} = \{A_1,A_2,A_3,A_4\}$, characterized by $\overrightarrow{OA_1}+ \overrightarrow{OA_2}+ \overrightarrow{OA_3} + \overrightarrow{OA_4} = \vec{0}$, is invariant under the action of the isometries of $\mathrm{Is}_{\cal T}$. Indeed, if $f \in \mathrm{Is}_{\cal T}$, $f$ is an affine map, therefore $O' = f(O)$ and $A'_i = f(A_i)$ satisfy
$$\vec{0} = 	ilde{f} (\overrightarrow{OA_1}+ \overrightarrow{OA_2}+ \overrightarrow{OA_3} + \overrightarrow{OA_4}) =
\overrightarrow{O'A'_1}+ \overrightarrow{O'A'_2}+ \overrightarrow{OA'_3} + \overrightarrow{OA'_4} = \overrightarrow{O'A_1}+ \overrightarrow{O'A_2}+ \overrightarrow{O'A_3} + \overrightarrow{O'A_4},$$
since $\{A_1,A_2,A_3,A_4\} = \{A'_1,A'_2,A'_3,A'_4\}$. Therefore $O = O' = f(O)$ for all $f \in \mathrm{Is}_{\cal T}$.

We can therefore identify the group of isometries leaving $\cal T$ invariant with the group of orthogonal endomorphisms leaving the set of vectors $\vec{u}_i = \overrightarrow{OA_i}$ invariant.

Thus, the classification theorem for orthogonal endomorphisms of $\vec{E}$ shows that the displacements of $\mathrm{Is}_{\cal T}$ are axial rotations with an axis passing through $O$, or the identity, and the antidisplacements of $\mathrm{Is}_{\cal T}$ are either reflections or commutative composites of a rotation with an axis $\Delta$ and a reflection with a plane $P$ orthogonal to $\Delta$ passing through $O$.

Let $\mathrm{Is}^+_{\cal T}$ be the subgroup of displacements leaving $\cal T$ invariant, and let $f_0$ be a fixed antidisplacement in $\mathrm{Is}_{\cal T}$, for example, the reflection with respect to the medial plane of $A_1A_2$. Then $\mathrm{Is}_{\cal T}$ is the disjoint union
$$\mathrm{Is}_{\cal T} = \mathrm{Is}^+_{\cal T} \cup f_0 \,\mathrm{Is}^+_{\cal T}.$$
Furthermore, the map $ \mathrm{Is}^+_{\cal T} 	o f_0 \,\mathrm{Is}^+_{\cal T}$ defined by $f \mapsto f_0 \circ f$ is a bijection, so $|\mathrm{Is}^+_{\cal T} |= |S_4|/2 = 12$ . In other words, there are 12 moves that preserve the tetrahedron, and 12 antimoves.

The following geometric properties of the regular tetrahedron, left as an exercise for the patient reader, allow us to explain these displacements:
\be
\item[$\bullet$] The line $A_1A_2$ is orthogonal to the line $A_3 A_4$

\item[$\bullet$] The axis joining the midpoint $I$ of $A_1A_2$ to the midpoint $J$ of $A_3 A_4$ is orthogonal to the two lines $A_1A_2,A_3 A_4$.

\item[$\bullet$] The axis joining $A_1$ to the center of gravity $H$ of $(A_2,A_3,A_4)$ is orthogonal to the plane $A_2A_3A_4$.
\ee
We thus obtain the list of displacements that leave the tetrahedron invariant.
\be
\item[$\bullet$] The identity $e$,
\item[$\bullet$] The 3 half-turns (rotations of angle $\pi$) around the axes joining pairs of opposite edges (corresponding to the products of two disjoint transpositions $(i\, j)(k\,l)$.
\item[$\bullet$] The 8 rotations of angle $\pm 2\pi/3$ around the axes connecting a vertex to the center of gravity of the opposite face (corresponding to the 3-cycles $(i\,j\,k) = (i,j)(j,k)$).
\ee

The antidisplacements are the composites $f_0 \circ f$, where $f_0$ is the reflection with respect to the medial plane of $A_1A_2$, and $f$ describes the set of displacements.
In particular, we obtain
\be
\item[$\bullet$] The 6 reflections with respect to the medial planes of the edges (corresponding to the $\binom{4}{2} = 6$ transpositions $(i\, j)$.
\item[$\bullet$] 6 reflection-rotation compounds (corresponding to the 4-cycles $(i\,j\,k\,l)= (i\,j)(j\,k)(k\,l)$).
\ee

As an example, let us decompose the isometry $ho$ corresponding to the 3-cycle $(1\,2\,3\,4)$, characterized by
$$ho(A_1) = A_2, \qquad ho(A_2) = A_3,\qquad ho(A_3) = A_4,\qquad ho(A_4) = A_1.$$
Let's choose the orientation so that the base $(\overrightarrow{A_1 A_2}, \overrightarrow{A_1 A_3},\overrightarrow{A_1 A_4})$ is a direct base, which allows us to specify the angle measurements.

Since $(1\, 2\,3\,4) = (1\,2)(2\,3)(3\,4) = (1\,2\,3)(3\,4)$, $ho$ is the composite $u \circ v$ of the rotation of angle $2\pi/3$, of axis $HA_4$ oriented by $\overrightarrow{HA_4}$, $H$ center of symmetry of triangle $A_1 ??A_2 A_3$, and $v$ the reflection with respect to the medial plane of $A_3 A_4$. Unfortunately, the line and the plane are not orthogonal, so $u \circ v 
e v \circ u$, and this is therefore not a good description of $ho$.

%\begin{figure}[htbp]
%\begin{center}
%\includegraphics[width=11cm,height=8cm] {tetracube}
%\end{center}
%\end{figure}

The correct decomposition is found if we inscribe the tetrahedron in a cube. Consider the cube whose 8 vertex coordinates are $(\pm 1, \pm 1, \pm 1)$ in the direct orthonormal coordinate system $(0,\vec{e}_1,\vec{e_2}, \vec{e_3})$ (see figure). It contains a regular tetrahedron $$A_1(-1,-1,-1),\qquad A_2(1,1,-1),\qquad A_3(-1,1,1),\qquad A_4(1,-1,1).$$ The remaining points form the dual tetrahedron $$B_1(1,-1,-1),\qquad B_2(-1,1,-1),\qquad B_3(1,1,1),\qquad B_4(-1,-1,1)$$ (The figure composed of these two tetrahedra is called Kepler's Stella Octangula). If we want the tetrahedron to be inscribed in a sphere of radius $1$, we divide all the coordinates by $\sqrt{3}$.

Let $P$ be the medial plane of $A_1 B_1$, with equation $x=0$, and $s$ be the reflection with respect to this plane. Then $s(A_i) = B_i,\ i=1,2,3,4$. Let $\Delta$ be the axis passing through 0, orthogonal to this plane, therefore with direction vector $\vec{e}_1$, and $r$ be the rotation of angle $\pi/2$ around this axis oriented by $e_1$. Then
$$r(B_1) = A_2, \qquad r(B_2) = A_3, \qquad r(B_3) = A_4, \qquad r(B_4) =A_1,$$ so that $ho = s \circ r$ verifies $$ho(A_1) = A_2,\qquad ho(A_2) = A_3,\qquad ho(A_3) = A_4,\qquad ho(A_4) = A_1,$$ and therefore $ho$ corresponds to the cycle $(1\,2\,3\,4)$.
$ho(O) = O$, and in the chosen basis $\cal B$ the matrix of $	ilde{ho}$ is
$$A = {\cal M}_{\cal B} (	ilde{ho}) =
\begin{pmatrix}
-1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0\
0 & 0 & -1\
0 & 1 & 0
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 & 0\
0 & 0 & -1\
0 & 1 & 0
\end{pmatrix}
.$$

\bigskip
To conclude, the group of rigid motions of a regular tetrahedron has $12$ elements.

Exercise 1.2.9 (Group of rigid motions of a regular tetrahedron)

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Exercise 1.2.9 (Group of rigid motions of a regular tetrahedron)

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