Let's take a modified version of the axioms above. We want a coordinate system for 3D rotations with the following features:

1.  The coordinates form a set equal to $\mathbb R^n$ for some $n$.
2.  All non-zero coordinates represent some rotation.
3.  Every rotation is represented by some coordinate.
4.  If $v$ represents a 3D rotation, then any $\lambda v$ where $\lambda$ is a non-zero scalar, represents the same rotation. The motivation for this axiom is that we want to find a *projective* coordinate system for 3D rotations.
5.  The composition of rotations corresponds to a *bilinear* product between coordinates. In particular, such a product consists of only the operations $+, \times, -$.

Let $\mathbf 1$ be the coordinate for a null rotation, $i$ be the coordinate for a half-turn around the $\vec i$ vector, and $j$ and $k$ be likewise for the $\vec j$ and $\vec k$ vectors. This all follows from axiom 3.

Let's consider the point $i$. Immediately, it's clear that $i^2 = \lambda \mathbf 1$ for some scalar $\lambda$. This scalar could be anything at this point except $0$.

By axiom (1) and (2), every non-zero linear combination of $\mathbf 1$ and $i$ represents some rotation. It is *weird* to take linear combinations of projective coordinate, but it is certainly *possible*, and it allows us to narrow down the space of possibilities. We see that these linear combinations generate a 2D algebra. It turns out there are only three 2D algebras up to isomorphism: The complex numbers, the dual numbers, and the split-complex numbers. We see that up to isomorphism, the algebra we've generated *must* be the complex numbers as the other ones contain zero divisors, and all rotations are invertible. Alternatively, we may argue using the *orders* of the elements: Rotations contain elements of any order.

The subalgebra generated by $\{\mathbf 1, i\}$ must correspond to some continuous subgroup of the 3D rotations. The only such subgroups that exist are rotations about some axes. Since $i$ represents a rotation about the $\vec i$ axis by a half turn, the group we've generated must be the set of all rotations about $\vec i$.

A similar argument applies to $j$ and $k$.

Finally, what is $ij$? Well, for geometric reasons it must be $\lambda_{ij} k$ for some scalar $\lambda_{ij}$. In fact, up to scalar multiples, the group generated by $\{1,i,j,k\}$ must be the Klein 4-group. By classification of small groups, the smallest candidate must be the Quaternion group of order 8. The products can't actually be the Klein 4-group because rotations aren't commutative. The quaternion group works, so we are done.

# What's the motivation?

Essentially, I posit that a transformation group can often be endowed with a coordinate system or representation. This coordinate system is either *cartesian* or *projective*.

A Cartesian coordinate system for a transformation group is one where (except for a negligible subset) all elements of $\mathbb R^n$ for some $n$ are in one-to-one correspondence with that transformation group. We insist that the product be bilinear. The rotation-dilation group can be represented by the Cartesian coordinate system $\mathbb C$. The complex number $i$ now represents a quarter-turn.

In fact, the classic examples of Cartesian systems for transformation groups are rings of square matrices.

Now consider as an alternative *projective* coordinate systems for transformation groups. Here, a scalar multiple of a coordinate represents the same transformation. Again, we insist that almost all elements of $\mathbb R^n$ stand for some element of the group, and that the product be bilinear.

The classic example is the matrix representation of the group of *affine transformations*. Here, the matrices are specified *up to a ratio*. We are generalising this because we may want our coordinate system to be more specific for computational efficiency and accuracy reasons.

The complex numbers $\mathbb C$ are also a *projective* coordinate system for rotations (not dilations). But then $i$ represents a half-turn, not a quarter turn.

# Subjective criticism of Clifford algebra

Feel free to tell me in the comments whether I should remove what I'm about to write. It may perhaps be opinion and not mathematics.

The following is entirely subjective. My issue with Clifford algebra is that a general element of a Clifford algebra lacks meaning. Normally, you restrict your attention to the even subalgebra or the odd subalgebra. The even subalgebra represents rotation-like transformations, whereas the odd subalgebra represents reflection-like transformation (general roto-reflections). A linear combination of these two "subalgebras" is often meaningless, unless you go to higher dimensions, where they are revealed to be rotations in one dimension higher; but then the mystery simply repeats itself at a higher dimension. It's also not clear why an element and its negation represents the same transformation, unless you approach it as I have: as a form of homogeneous or projective representation.

Actually, my discomfort is with the idea of formally adding a scalar to a vector to a bivector. But I'll stop here. The approach I'm taking has the advantage that *every* element has some meaning as a transformation, instead of *some* elements being vectors, and *some* being bivectors, and *some* apparently not meaning anything. But it's also possible that understanding all elements as transformations causes one to miss out on exterior algebra.