I read the Wikipedia article on rotations in 4-dimensional space, where it says that 4D rotations can be represented as the product of two unit quaternions, one multiplying on the left and the other on the right. After choosing an orthonormal basis, this action takes a vector $v\in {\bf R}^4$ to $AvB$, where $$A = \pmatrix{ a_1 & -a_2 & -a_3 & -a_4 \cr a_2 & a_1 & -a_4 & a_3 \cr a_3 & a_4 & a_1 & -a_2 \cr a_4 & -a_3 & a_2 & a_1 \cr }\quad\text{and}\quad B = \pmatrix{ b_1 & -b_2 & -b_3 & -b_4 \cr b_2 & b_1 & b_4 & -b_3 \cr b_3 & -b_4 & b_1 & b_2 \cr b_4 & b_3 & -b_2 & b_1 \cr }, $$ where ${a_1}^2 + {a_2}^2 + {a_3}^2 + {a_4}^2 = {b_1}^2 + {b_2}^2 + {b_3}^2 + {b_4}^2 = 1$. (Edit. I think I wrote this wrong. The multiplication should be $ABv$, but this represents an action $gvh^{-1}$ where $g$ and $h$ are unit quaternions.)
The rotation matrix, an element of ${\rm SO}(4)$, is allegedly the product of these two matrices. I can see that the determinant should be $1$, but I'm not sure how this product $C = AB$ satisfies $C\cdot C^{\rm T} = I$, which it's supposed to if it is a member of the orthogonal group. I used Maple to expand out the product and it is something gross and certainly not the identity. Would anyone be able to point out where my misunderstanding is?
Also, how would one show that this map from $v = (a_1, a_2, a_3, b_1, b_2, b_3)\in {\bf R}^6$ is surjective on ${\rm SO}(4)$? (The $a_4$ and $b_4$ components are excluded because they can be retrieved, up to signs, once we pick the other coordinates.) I was hoping that I could get a formula for exactly the matrix defined by the vector $v$, and then I could eyeball it and conclude that every matrix in ${\rm SO}(4)$ is expressible, but I'm not sure if that's a good way to go about it. Thanks in advance!