3D representation of A4 that preserves the unit ball

Question

The subgroup of $\operatorname{GL}_3\left(\mathbb{R}\right)$ generated by the matrices $$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ -1 & -1 & -1 \\ \end{array} \right)\,,\hspace{10pt}\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & -1 & -1 \\ \end{array} \right)\,, \hspace{5pt}\text{and}\hspace{5pt}\left( \begin{array}{ccc} 0 & 0 & 1 \\ -1 & -1 & -1 \\ 1 & 0 & 0 \\ \end{array} \right)$$ is isomorphic to $A_4$.

When acting on $\mathbb{R}^3$, this representation does not preserve the unit $3$-ball (i.e. map the unit $3$-ball into itself). Given a $\mathbf{v}=(x,y,z)$ such that $\left\lvert\left\lvert \mathbf{v} \right \rvert\right \rvert<1$, we either have $\left\lvert\left\lvert A\mathbf{v} \right \rvert\right \rvert=\left\lvert\left\lvert \mathbf{v} \right \rvert\right \rvert$ or, up to permutation of the coordinates, $$\left\lvert\left\lvert A\mathbf{v} \right \rvert\right \rvert=\sqrt{x^2+y^2+(x+y+z)^2}$$ which achieves a maximum of $\sqrt{2+\sqrt{3}}$ on the unit $3$-ball.

Is there another representation of $A_4$ in $\operatorname{GL}_3\left(\mathbb{R}\right)$ that does preserve the unit $3$-ball?

Answer 1

Lord Shark the Unknown gave a fine answer. I want to supplement it with the following generalization.

Assume that you have a finite group $G$ acting on $V=\Bbb{R}^n$ by linear transformations. Let $(\ ,\ )$ be the usual inner product on $\Bbb{R}^n$. Let us define another bilinear form $\langle\ ,\ \rangle$ on $V$ by taking the average: $$ \langle x,y\rangle=\frac1{|G|}\sum_{g\in G}(gx,gy). $$ The following are easy to see:

• $\langle\ ,\ \rangle$ is bilinear and symmetric.
• for all $x\in V$ we have $\langle x,x\rangle\ge0$ with equality only when $x=0_V$.
• $\langle gx, gy\rangle=\langle x,y\rangle$ for all $g\in G$ and all $x,y\in V$.

Therefore the new metric given to $V$ by this averaged is invariant under the action of the group $G$. Call the resulting inner product space $V'$. All this immediately implies that balls in $V'$ are stable under $G$.

But, the space $V'$ is really the space $V$, we only changed the metric. You can easily find an orthonormal basis of $V'$. Rewriting the matrices of $G$ w.r.t. such a basis results in orthogonal matrices, i.e. $G$ acts by distance and angle preserving transformations.

You can turn any finite subgroup of $GL(n,\Bbb{R})$ into a distance preserving group (one that thus also preserves balls) by redefining "distance" in a suitable way.

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Closing remarks:

• This can be generalized to complex matrices and Hermitian inner products with the obvious modifications.
• If a subspace $U$ of $V'$ is stable under $G$, then so is the orthogonal complement $U^\perp$ leading to Maschke's theorem.
• Averaging arguments FTW.

Answer 2

Take a regular tetrahedron inscribed in the unit ball. Work out the coordinates of its vertices $P_1,\ldots,P_4$. Work out the matrix $R_i$ of a rotation with angle $2\pi/3$ about the axis $OP_i$ with $O$ the origin. Then the $R_i$ generate a copy of $A_4$ (you only need two of them) preserving the unit ball.