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

Question

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.

Answer 1

(1) Fix a rotation $R$. It is a product of two reflections :

Proof : WLOG we can assume that $R$ fixes $z$-axis and rotates $\theta$-rotation on $xy$-plane. That is $v_1:=(1,0,0)\rightarrow v_2:=(\cos\ \theta, \sin\ \theta ,0) $

Define $$ w_1=(\cos\ \theta/4, \sin\ \theta/4,0 ),\ w_2:= (\cos\ \theta/2, \sin\ \theta/2,0 ) ,$$ $$ w_3:=(\cos\ 3\theta/4, \sin\ 3\theta/4,0 ),\ w_4 :=(\cos\ 5\theta/4, \sin\ 5\theta/4,0 )$$

If $r_1$ is a reflection fixing $(0,0,1),\ w_1$ and $r_2$ is a reflection fixing $(0,0,1),\ w_3$ so $$ r_2\circ r_1 (v_1 ) = r_2(w_2)=v_2 $$

And $$ r_2\circ r_1 (w_1)=r_2(w_1)=w_4 $$

Hence $$ R= r_2\circ r_1 $$

(2) Given reflection $r $ where $r(v_1)=-v_1$, we have a reflection $r'$ s.t. $r'r$ is $\pi$-rotation :

Proof : Let $r'(0,1,0)=(0,-1,0)$ so that $$ r'r (v_1)=-v_1,\ r'r(0,1,0)=r'(0,1,0)=(0,-1,0)$$

Note : Already $r(v_1)$ is $\pi$-rotation so that $r'(v_1)=v_1$ If $(0,\cos\ t,\sin\ t)$ is fixed by $r'$, then $$ r'r(0,\cos\ t,\sin\ t)=(0,\cos\ t,\sin\ t)$$

That is $r'r$ is $\pi$-rotation. If we summarize this, let $ r(v)=-v,\ |v|$, and fix $w\in v^\perp,\ |w|$ Then we have $r'(v\times w)=-v\times w$ and $r'r(w)=w$

(3) Our strategy is to find a reflection $r$ s.t. $$ R_1:=r\circ r_1,\ R_2:= r_2\circ r $$ are $\pi$-rotations (That is $ R_2\circ R_1=r_2\circ r_1=R$) :

Proof : $r_1((0,0,1)\times w_1)=-(0,0,1)\times w_1$ so that we define $r$ to be a reflection fixing $(0,0,1)\times w_1,\ a(0,0,1) + bw_1 $ where $a^2+b^2=1$

And $$((0,0,1)\times w_1 )\times ( a(0,0,1) + bw_1 )=aw_1 -b(0,0,1) $$ Here $$ r(aw_1 -b(0,0,1) )=-(aw_1 -b(0,0,1)) $$ so that $r_2$ fix $ aw_1 -b(0,0,1) $

Hence $$\bigg(aw_1 -b(0,0,1) \bigg) \perp (0,0,1)\times w_3 $$

Hence $a=0$

(4) Conclusion : $R_1$ fixes $ w_1$ and $R_2$ fixes $w_3$ Then $R=R_2\circ R_1$

Answer 2

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.

Answer 3

Choose an $\alpha\in\>\bigl]0,{\pi\over4}\bigr[\>$, denote by $S_a$ the rotation by $\pi$ around the axis $a:=(\cos\alpha,\sin\alpha,0)$, and by $S_b$ the rotation by $\pi$ around the axis $b:=(cos\alpha,-\sin\alpha,0)$. We now shall analyze the effect of the rotation $T:=S_a\circ S_b$ on the standard basis vectors of ${\mathbb R}^3$.

As $S_a e_3=S_b e_3=-e_3$ it follows that $Te_3=e_3$, which implies that $T$ is a rotation with axis $e_3$. Furthermore geometric considerations show that $S_b e_1=\bigl(\cos(-2\alpha),\sin(-2\alpha),0\bigr)$, and this leads to $$Te_1=S_a\bigl(\cos(-2\alpha),\sin(-2\alpha),0\bigr)=\bigl(\cos(4\alpha),\sin(4\alpha),0\bigr)\ .$$ Altogether this shows that $T$ is a rotation around the $e_3$-axis by the angle $4\alpha\in\>]0,\pi[\>$.

From these computations we can infer that any rotation around any axis can be realized as a product of two suitably chosen rotations with rotation angle $\pi$.

Answer 4

I take that a "rotation of order 2" is a rotation by $\pi$ in some plane?

If so, here's a brief summary of the solution, described in both clifford algebra terms as well as quaternion algebra language:

1. A 180-degree rotation is given by (within a minus sign) bilinear multiplication by a bivector (or, in quaternions, a pure imaginary quaternion). That is, given a unit plane $A$ (unit, pure imaginary quaternion), then $-AaA$ rotates a vector $a$ by 180 degrees.

2. Given a rotor (quaternion) $q$, with bivector (pure imaginary) part $V$, compute $-q BB$, where $B$ is a unit magnitude plane and where the scalar part of $BV$ is zero (in quaternions, $B$ would be a pure imaginary quaternion orthogonal to $V$; in terms of planes, the normal of $B$ is orthogonal to the normal of $v$). Then, $C = -qB$ also corresponds to a plane (a pure imaginary quaternion). Moreover, if $q$ and $B$ are normalized, then so is $C$.

3. Since $-BB = 1$ (compare with quaternions, such as $-ii = 1$), then $-qBB = q$. We then have proven that $q = CB$ for two unit planes (unit, pure imaginary quaternions) $C, B$---which again correspond to 180-degree rotations. QED.