Let $Q \in SO_4(\mathbb R)$. Show there exists $u,v \in \mathbb H$ with $|u| = |v| = 1$ such that $Qx = uxv$ for all $x \in \mathbb H$

Question

So the prompt is merely an existence proof--just find a $u$ and $v$ that work. Well, I'm unfortunately a little stuck on getting started.

I know that $Q \in SO_4(\mathbb R) \implies QQ^T = I \text{ and } \det(Q) = 1$.

I tried to solve $Qx = uxv$ for $u,v$ but I was not able to do so successfully. This is because I don't necessarily know if $x$ is has an inverse.

Here's what I managed to deduce successfully:

$$|Qx| = |uxv| = |u||x||v| = 1\cdot |x| \cdot 1 = |x|$$

which means that multiplying $x \in \mathbb H$ by $Q$ doesn't change its length.

Let $Q = [q_1 \,|\, q_2 \,|\, q_3 \,|\, q_4]$, where $q_i$ is the $i^{th}$ column of $Q$, and let $x = a + bi + cj + dk$. Then,

$$\begin{align}Qx & = Q(a + bi + cj + dk)\\& = aQ + bQi + cQj + dQk \\&= aQ + bQ\begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix} + cQ\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix} + d\begin{pmatrix}0 \\ 0 \\ 0 \\ 1\end{pmatrix} \\ & = aQ + bq_2 + cq_3 + dq_4\end{align}$$

And unfortunately I don't see where to go from here. I'm not entirely sure that I did the multiplication of a quaternion by a matrix correctly. If $a \in \mathbb R$, what does $aQ$ mean? Thus, I don't think that's right.

Likely, the solution will boil down to $v = u^{-1}$ or something. But I'm still not quite sure how to arrive there.

Answer 1

Here's a proof that you may find completely unsatisfactory, depending on whether you know about bundles and things like that.

Consider the map $p : SO(4) \to \mathbb S^3 : [q_1, q_2, q_3, q_3] \mapsto q_1$ that selects from a matrix its first column.

This is a fibration, and its fiber ($p^{-1}(q_1)$for $q_1 = \begin{bmatrix}1\\0\\0\\0 \end{bmatrix}$, for example) is just the set of $4 \times 4 $ orthogonal matrices whose first row and column are (1,0,0,0). If you look at the lower right $3 \times 3$ matrix, it must therefore be in $SO(3)$, since all its columns are orthonormal, and a first-row-expansion on the $4 \times 4$ matrix shows that the determinant is $+1$ instead of $-1$.

So now we have $$ SO(3) \to SO(4) \to S^3 $$ as a sequence of maps, where the first is the injection of the fiber and the second is a fibration. That means that $SO(4)$ can be written as a bundle where we look at a trivial $SO(3)$ bundle over the top half of $S^3$ and a trivial $SO(3)$ bundle over the bottom half, and the "gluing map" along the equator is a map from $S^2 \to SO(3)$, i.e., an element of $\pi_2(SO(3)$. Since $SO(3)$ is a Lie group, we know that $\pi_2(SO(3) = 0$.

So the gluing map is trivial, and there's a decomposition of $SO(4)$ as $SO(3) \times S^3$. (I have a feeling that was rather the long way around, but so be it.)

So do this: let $Q$ be your matrix, and let $u$ be the first column of $Q$, and let $L_{u^{-1}}$ be the matrix that represents left quaternion multiplication by $u^{-1}$ (quaternion inverse!), so that $L_{u^{-1}} \cdot e_1 = u^{-1} e_1= u^{-1}$, for instance, and $L_{u^{-1}}u = u^{-1}u = \mathbf {1}$, the quaternion $1$, which corresponds to the vector $e_1$.

What is $L_{u^{-1}}Q$? Well, its first columns is $e_1$, so its $(1,1)$ entry is $1$, so its first row (which must be a unit vector!) is $(1,0,0,0)$. So $$ L_{u^{-1}}Q = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & & & \\ 0 & & A & \\ 0 & & & \end{bmatrix} $$ where $A$ is a $3 \times 3$ matrix that must in fact be in $SO(3)$.

Now every rotation of 3-space, considered as the space of pure-vector quaternions, can be expressed in the form $q \mapsto s^{-1} q s$ for some unit quaternion $s$. So we can write the matrix $A$ in the form $L_{s^-1} R_s$, where these are the matrices of the linear transformations "multiply on the left by $s^{-1}$" and "multiply on the right by $s$", respectively.

So we have \begin{align} L_{u^{-1}} Q &= L_{s^{-1}}R_s \\ Q &= L_u L_{s^{-1}}R_s \\ \end{align} i.e., the matrix $Q$ represents left multiplication by $us^{-1}$ followed by right multiplication by $s$. Picking the $u$ in your solution to be my $us^{-1}$ and $v$ to be my $s$, we're done.

(Apologies for rambling answer...but I got there in the end...)

Answer 2

It is a consequence of the fact that any 4D rotation can be canonically decomposed into a left-isoclinic and a right-isoclinic rotation. As you can see in the reference this means that any $A \in SO_4(\mathbb{R})$ acts, on a vector $\vec x=(x_1,x_2,x_3,x_4) \in \mathbb{R}^4$ as: $$ A \vec x=UV \vec x=U \vec x V^T $$ where $U$ and $V^T$ are matrices of the form: $$ U=\begin{pmatrix} a&-b&-c&-d\\ b&a&-d&c\\ c&d&a&-b\\ d&c&b&a \end{pmatrix} \qquad U^T=\begin{pmatrix} p&q&r&s\\ -q&p&-s&r\\ -r&s&p&-q\\ -s&-r&q&p \end{pmatrix} $$ with:$a^2+b^2+c^2+d^2=p^2+q^2+r^2+s^2=1$ Now we can see that $U$ and $V^T$ are the matrix representation in $M_4(\mathbb{R})$ of two unit quaternions $u,v$ and the vector $\vec x$ can be represented by the quaternion $x=x_1+x_2\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{i}+x_4\mathbf{k}$, so the rotation can be represented as the product $uxv$ .