[Math] Show that every rotation in $\mathbb{R^3}$ can be written as the product of two rotations of order 2.

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Show that every rotation in $\mathbb{R^3}$ can be written as the
product of two rotations of order 2.

Here's my attempt at a solution:

We know that any rotation in $\mathbb{R^3}$ can be represented as the product of two reflections. So we write our rotation as $R_1R_2$ where the $R_j$ are reflections in $\mathbb{R^3}$.

We would like to show that $R_1R_2$=$(R_aR_b)(R_cR_d)$=$R_aR_bR_cR_d$ where $R_aR_b$ and $R_cR_d$ are rotations of order 2 in $\mathbb{R^3}$.
I think I have shown that a rotation $R_1R_2$, which is the product of reflections in the planes $\Pi_1$ and $\Pi_2$ respectively, has order two if and only if $\Pi_1$ and $\Pi_2$ are perpendicular. Next I thought it sufficient to show that $R_1$=$R_aR_b$ (and similarly for $R_2$ with $R_c$ and $R_d$) and as $R_1$ and $R_2$ are just any reflections, I now attempt to show that any reflection can be written as the product of two reflections in perpendicular planes.

The form of a reflection in the plane $x.n=d$ in $\mathbb{R^3}$ is given as $R(x)=x+2(d-x.n)n$.

Suppose that we have $R_a$ and $R_b$ as reflections in the perpendicular planes (that is, the planes have perpendicular normals), $x.n_a=d_a$ and $x.n_b=d_b$ respectively. As $n_a.n_b=0$, we can show that $R_aR_b(x)= x+2(d_1+d_2)-(x.(n_1+n_2))(n_1+n_2)$ which is of the required form. So that for any reflection $R$ in $x.n=d$, we can set $n_1,n_2,d_1,d_2$ such that $d_1+d_2=d$ and $n_1+n_2=n$.

First off, is this correct, have I shown what I was required to show? I was also wondering if there may be a nicer, perhaps geometric way of going about the question.

Apologies if my attempt at a solution is difficult to follow, I have next to zero experience in writing formal solutions.

Best Answer

A rotation $R$ in $\Bbb R^ 3$ about an axis $a$ by an angle $\alpha$ can be written as product $R_1R_2$ of two reflections at planes $\Pi_1,\Pi_2$, where these planes intersect in $a$ at an angle of $\frac\alpha2$. Let $\Pi_3$ be perpendicular to $a$ and $R_3$ the reflection at $\Pi_3$. Then $$R=R_1R_2=R_1R_3R_3R_2 $$ Now $R_1R_3$ and $R_3R_2$ are again rotations (i.e., orientation preserving) about the intersection of the two reflecting planes involved and by an angle twice the angle between the two plaens, hence these are rotations by $180^\circ$, in other words: of order two.