# Constraints on the factorization of a unit quaternion rotation operator

In Jack B. Kuipers' Quaternions and Rotation Sequences page 194-195, section 8.7.1 explains how to decompose a tracking sequence quaternion $q = q_0 + \vec{i}q_1+\vec{j}q_2+\vec{k}q_3$ into two other quaternions $a^3 = a_0 + \vec{k}a_3$ and $b^2 = b_0 + \vec{j}b_2$ (the 3 and 2 denote axes of rotation, not quaternion exponentiation). Equation 8.10 states that

$$q = q_0 + \vec{i}q_1+\vec{j}q_2+\vec{k}q_3 = a^3b^2$$ $$= a_0b_0 - \vec{i}a_3b_2 + \vec{j}a_0b_2+\vec{k}a_3b_0$$

which makes sense in the context of quaternion multiplication. However, he then states

In Equation 8.10, it is easily verified, that $$q_0q_1 + q_2q_3 = 0 \tag{8.11}$$

Where did this constraint come from?

Edit: This text continues by tabulating this factorization for all combinations of $a^ib^j$ for $i,j \in [1,2,3]$. Associated with each factorization is an associated constraint of the form $q_lq_m \pm q_oq_p=0$. Where do these come from?

It turns out that the constraint is nothing more complicated than seeing that $$q_0q_1 = -a_0b_0a_3b_2$$ while $$q_2q_3 = a_0b_0a_3b_2$$