Elements of $SO_3(\mathbb{R})$ are product of two rotations

Question

I am trying to prove the simplicity of $SO_3 \mathbb{R}$. The first step in my exercise book is to prove that every element of $SO_3 \mathbb{R}$ can be written as the composite of two rotations of angle $\pi$.

My geometric intuition in space is very rusty, I have been trying to visualize what happens for hours, but in vain. Can anyone explain why this is true with an image for example?

I know that all elements in $SO_3 \mathbb{R}$ are rotations with axis $D$ and an angle $\theta$. The rotation acts on a point $M\in \mathbb R^3$ as a planar rotation of angle $\pi$ in the affine plan $D^\perp+M$.

Answer 1

I am not entirely sure if you want a proof that every proper isometry is a product of two rotations of angle $\pi$, or if you already have a proof but you want a geometric intuition of this result. I am not good at LaTeXing pictures, so I just provide you a proof. I tried to translate the proof in a geometric way at this end of each step. Maybe it will help you to find your own geometric intuition. Apologies if it is not the kind of answer you're looking for.

Let $u\in SO_3(\mathbb{R})$. Since $u$ is an isometry, its only possible real eigenvalues are $\pm 1$. Since its characteristic polynomial has degree $3$, it has a root, so $u$ has at least one eigenvalue $\varepsilon.$ Let $e_1$ be a unit eigenvector, $F=\mathbb{R}e_1$.

Since $u$ is an isometry and $F$ is stable by $u$, the same holds for $F^\perp$. Let $\mathcal{B'}=(e_2,e_3)$ be an orthonormal basis of $F^\perp$, that we may want to choose more carefully in the sequel.

Set $\mathcal{B}=(e_1,e_2,e_3)$. This is an orthonormal basis of $\mathbb{R}^3$.

If $M=\mathrm{Mat}(u; \mathcal{B})$ and $M'=\mathrm{Mat}(u_{F^\perp}; \mathcal{B}')$, we have $M=\begin{pmatrix} \varepsilon & 0 \cr 0 & M'\end{pmatrix}.$

In particular, $\det(u_{F^\perp})=\det(M')=\varepsilon$.

• Assume that $\varepsilon=-1$. Then $u_{F^\perp}$ is an improper isometry of the euclidean plane $F^\perp$, so it is an orthogonal symmetry. In this cas, we may choose $\mathbb{B}'$ in such a way that $M'=\begin{pmatrix} -1 & 0 \cr \phantom{-}0 & 1\end{pmatrix}$, and so $M=\begin{pmatrix}-1 & & \cr & -1 & \cr & & 1\end{pmatrix}.$

Note for later that this is the matrix of a rotation of angle $\pi$. However, we also have $$M=\begin{pmatrix}-1 & & \cr & 1 & \cr & & -1\end{pmatrix}\begin{pmatrix}1 & & \cr & -1 & \cr & & -1\end{pmatrix}.$$ It follows that $u$ is the product of two rotations of angle $\pi$ (because the basis $\mathcal{B}$ is orthonormal).

Geometric translation. Assuming that $u(e_1)=-e_1$ for some non zero vector, the proof shows that in fact there is a plane on which $u$ restricts to $-Id$. Let $P$ be this plane, and let $D$ be normal to the plane $P$. Take an orthonormal basis $(e_1,e_2)$ of $P$ and a unit vector $e_3$ directing $D$. Then $u$ is a rotation of plabe wrt to axis $D$ and angle $\pi$.

This is also the product of two rotations of angle $\pi$: one in the plane generated by $e_1,e_3$, one in the plan generated by $e_2,e_3$.

• Assume that $\varepsilon=1$. In this case, write $M=\begin{pmatrix}-1 & & \cr & -1 & \cr & & 1\end{pmatrix}N$, where $N=\begin{pmatrix}-1 & & \cr & -1 & \cr & & 1\end{pmatrix}M$.

Since $\mathcal{B}$ is orthonormal, $N$ represents an isometry, and trivial computations show that $\det(N)=1$ and has $-1$ has eigenvalue (just look at the upper left corner). By the previous case, $N$ is a rotation of angle $\pi$, and we are done again.

Geometric translation. If $u(e_1)=e_1$ ($e_1$ is a unit vector) , let $D$ be the line directed by $e_1$, and $P$ be the plane with normal $D$. Take an orthonormal basis $(e_2,e_3)$ of $P$. Then let $v$ be the rotation of angle $\pi$ in the plane generated by $(e_1,e_2)$. Then $vu$ is a rotation of angle $\pi$ and $u=v (vu)$.