Post History

There is a [famous result](https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)) about finite-dimensional real associative division algebras. You may find that interesting, but I don't think it's what you want. It is a direct answer to the question of why we "can't" multiply "triplets of numbers" if you require that multiplication to satisfy certain properties. However, this wasn't what made the quaternions "satisfactory" to Hamilton. Instead, he was interested in an "algebra" of rotations.

Interpreting your question instead as "Why do we need a 4D (linear) space to represent 3D rotations?" the answer comes from the fact that rotation happens in planes, and while in 2D there is only one plane, in 3D there are 3 orthogonal planes.

Let's start with 2D. Fix an orthonormal basis set, $\mathbf e_1$ and $\mathbf e_2$. Orthonormal means $\mathbf e_i \cdot \mathbf e_j = \begin{cases}1, &\text{if } i=j \\\\ 0, &\text{if } i\neq j\end{cases}$. We'll implicitly define a bilinear, associative, distributive ***but not closed*** product of vectors. It's behavior is completely defined by its action on basis vectors (and that's all we'll need). We'll have $\mathbf e_i^2 = 1$ and $\mathbf e_i \mathbf e_j = -\mathbf e_j \mathbf e_i$ where $i \neq j$. $\mathbf e_1 \mathbf e_2$ is *not* a vector (hence this product not being closed on vectors), but is what is known as a **bivector** and, in this case, can be identified with the (oriented) plane spanned the vectors $\mathbf e_1$ and $\mathbf e_2$. To make it a closed operation, we'd need the entire (Euclidean) [geometric algebra](https://en.wikipedia.org/wiki/Geometric_algebra) aka Clifford algebra, and what we've just defined is the geometric product.

The interesting this about this for us is that $(\mathbf e_1 \mathbf e_2)^2 = -1$. $$\begin{aligned}
(\mathbf e_1 \mathbf e_2)^2
& = \mathbf e_1 \mathbf e_2 \mathbf e_1 \mathbf e_2 \\\\
& = -\mathbf e_1 \mathbf e_1 \mathbf e_2 \mathbf e_2 \\\\
& = -\mathbf e_1^2 \mathbf e_2^2 \\\\
& = -1
\end{aligned}$$

$\mathbf e_1\mathbf e_2$ behaves exactly like the complex imaginary $i$. Using [Euler's formula](https://en.wikipedia.org/wiki/Euler's_formula), we can write $e^{\alpha \mathbf e_1 \mathbf e_2} = \cos\alpha + \mathbf e_1 \mathbf e_2\sin\alpha$. This represents a rotation by $\alpha$ radians in the $\mathbf e_1$-$\mathbf e_2$ plane. If $\mathbf v$ is a vector, then $e^{\alpha \mathbf e_1 \mathbf e_2}\mathbf v = \mathbf v \cos\alpha + \mathbf e_1\mathbf e_2\mathbf v\sin\alpha$ is $\mathbf v$ rotated by $\alpha$ radians. Notice how the above exponential has a scalar part and a bivector part. This combination of scalar and bivector is called a **rotor** or, more generally, a **spinor** if we don't require it to have unit norm.

When we move to 3D, we have 3 basis vectors, $\mathbf e_1$, $\mathbf e_2$, and $\mathbf e_3$. This gives rise to three orthonormal bivectors representing the 3 orthogonal (oriented) planes, $\mathbf e_1 \mathbf e_2$, $\mathbf e_2 \mathbf e_3$, and $\mathbf e_1 \mathbf e_3$. The exact same Euler formula logic works for each of these planes. However, we see that applying rotors doesn't quite work using the above formula. For example, $e^{\alpha \mathbf e_1 \mathbf e_2}\mathbf v$ *will* perform a rotation as expected ***if*** $\mathbf v$ is in the plane spanned by $\mathbf e_1$ and $\mathbf e_2$, but if it isn't, e.g. $\mathbf v = \mathbf e_3$, then you won't even end up with a vector. The correct formula is $e^{\alpha \mathbf e_1 \mathbf e_2 / 2}\mathbf v e^{-\alpha \mathbf e_1 \mathbf e_2 / 2}$. You can verify that this reduces to the other formula when $\mathbf v$ is in the plane spanned by $\mathbf e_1$ and $\mathbf e_2$. With that adjustment, a rotation is now represented by any unit spinor, i.e. rotor, and the spinors are the 4-dimensional linear space of the 1 scalar component and the 3 bivector components. This is the even subalgebra of the 3D geometric algebra – even meaning containing the parts corresponding to even dimensional objects, i.e. 0-dimensional for scalars and 2-dimensional for bivectors – and it corresponds *exactly* to the quaternion algebra. The significance of the even dimensional subalgebra is that the geometric product (roughly speaking) multiplies dimensions and so the even subalgebra is closed under geometric product. This is just to say that the geometric product of spinors is a spinor, and thus the geometric product of rotors is a rotor.

If we move to 4D, everything works the same except now we'll have 6 planes. However, the even subalgebra of the 4D geometric algebra has 8 components, not 7 = 1 + 6. This is because there is now a single 4-dimensional component as well corresponding to $\mathbf e_1 \mathbf e_2 \mathbf e_3 \mathbf e_4$. We find that our rotor application formula was again a bit too specialized before. The final form (which generalizes all the others) is $R\mathbf v R^{-1}$ where $R$ is our rotor or spinor, e.g. $e^{\alpha \mathbf e_1 \mathbf e_2 / 2}$. This formula means a (non-zero) spinor acts the same as the corresponding rotor produced by normalizing the spinor. One thing worth pointing out is that the 8-dimensional vector space of spinors in 4D does NOT correspond to the [octonions](https://en.wikipedia.org/wiki/Octonion). Multiplication in the octonions is non-associative, while the geometric product is associative by definition.

Another route to rotations is the (special) [orthogonal groups](https://en.wikipedia.org/wiki/Orthogonal_group). In this case, the even subalgebra of 3D geometric algebra is closely related to the [Lie algebra](https://en.wikipedia.org/wiki/Lie_algebra) of the (Lie) group [SO(3)](https://en.wikipedia.org/wiki/SO(3)). The bivectors $\mathbf e_i \mathbf e_j$ for $i\neq j$ are the generators of this Lie algebra for [Spin(3)](https://en.wikipedia.org/wiki/Spin_group) or isomorphically [SU(2)](https://en.wikipedia.org/wiki/Special_unitary_group). The exponential unsurprising corresponds to the [exponential map](https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)) for this Lie algebra. The rotors then correspond to elements of the Lie group. The reason we get Spin(3) or SU(2) instead of SO(3) directly is due to the fact that we have have *two* rotors that correspond to the same rotation, namely $R$ and $-R$. This leads to a "[double cover](https://en.wikipedia.org/wiki/Spin_group#Double_covering)" of SO(3). In general, the Lie algebra of Spin(n) is generated by the bivectors of the geometric algebra on $n$-dimensions, and the Lie group is the resulting rotors. In fact, the Lie algebra of SO(n) and Spin(n) agree, the resulting Lie groups differ because they have different (global) topologies.

So, ultimately, the reason there isn't a corresponding algebra with 3 components is that there isn't a (Euclidean) space with exactly two orthogonal planes. An $n$-dimensional Euclidean space will have it's space of bivectors spanned by $n \choose 2$ bivectors, and this will give rise to an even subalgebra containing $2^{n-1}$ components.