You took a shortcut weeks ago in the program you're writing and now it's biting you, and you know you have to go back and rewrite stuff, but to put that off you start reading Conway and Smith's "On Quarternions and Octonions", and then you wake up in the middle of the night thinking about bi-quarternions. What is the Clifford algebra way of thinking about them?
In Clifford algebra, rotations are given by elements of the even sub-algebra (acting on the vectors to be rotated by the sandwich product). It forms a sub-algebra because the clifford product sums the dimensions so even goes to even. Multiples of the same even element result in the same rotation with the sandwich product, so the degrees of freedom of rotation is one less than the dimension of the even subalgebra.
The grades of the n-dimensional Clifford algebra follow Pascal's triangle and give a total dimension of 2ⁿ. The 0-grade is scalars. There is only one part of the highest grade, so it is called a pseudo-scalar. The 1-grade is the vectors of the base vector space. Orthogonal to each vector is an (n-1)-grade element that is a pseudo-vector which behaves a lot like a vector.
So let's start with 2-d. 1-d is left as an exercise. We have a 2-d vector space. The Clifford algebra consists of: scalars (1-d); vectors (2-d); oriented area elements (1-d, the pseudo scalar). 1+2+1=4. The square by clifford multiplication of the unit area element is the scalar -1. That's suggestive! The even sub-algebra is the scalars plus the area elements. Yes it is the Complex numbers. Because the complex numbers and the vectors are both 2-d it is easy to get them confused.
In 3-d the Clifford algebra consists of: scalars (1-d); vectors (3-d); bivectors (pseudo-vectors, so also 3-d); oriented volume element (1-d pseudo-scalar). 1+3+3+1=8. The even sub-algebra is the scalars and the bivectors. Yes it is the quarternions. Once again there is potential confusion, this time because the vectors and the bivectors have the same dimensions.
The product of n vectors is called a versor. Up to 3-d there are no Clifford algebra elements that aren't versors. The square of a versor is a scalar. So it makes sense to take the sum of squares of the components of an even subalgebra element, then take the square root to get a norm. With a bit of other calculation we find that they form normed division algebras. This breaks down in 4-d where there are bivectors which are not versors.
In 4-d we have: scalars (1-d); vectors (4-d); bivectors (6-d); trivectors (pseudo-vectors 4-d); oriented hypervolume (1-d pseudo-scalar). 1+4+6+4+1=16. The even subalgebra is the scalars, the bivectors and the pseudo-scalar. But we are told that rotations can also be represented by a pair of quarternions. Here's a way to see two quarternions in the even subalgebra:
We like Clifford algebras because so much can be done without picking distinguished directions to be basis vectors. But somehow we keep finding it convenient to specify a basis, as we will here. Let e1, e2, e3 and e4 be a basis of our 4-d vector space. Now consider just the 3-d subspace formed by e1, e2 and e3. The even subalgebra is the quarternions consisting of the scalar and the bivectors generated by e1e2, e2e3 and e3e1.
This leaves from the bivector basis: e1e4, e4e2 and e3e4. Plus the pseudoscalar e1e2e3e4. Now if we define a new multiplication as the Clifford product times (or divided by) e1e2e3e4 then we get a new model of the quarternions: this time with the pseudoscalar as the scalar. So I've divided the even subalgebra into 2 quarternions.
Well I should check this out more carefully, but I'd better get back to my program that needs fixing.