Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?

Question

I'm reading a paper about Hamilton's discovery of quaternions and it explains why he failed in his 'theory of triplets' where he tried to make a vector with $3$-dimensions, as an analogy to the complex field, where we can see a number as a $2$-dimensional vector. In this paper, he explains why it is impossible to create a field with $3$ components, that is an extension of the complex field (in other words, it respects addition, and multiplication in the same way...).

Here it is

As you can see, it goes through all the possibilities and proves that it is impossible. The paper, however, does not explain why $j^2=-1$. It could be anything! Why $-1$?** The article itself is pretty intuitive, but this aspect kills me. Later, in the article, it says that we should instead consider a 4th component called $k$, such that $k^2=-1$ (also, $i$ and $j$ too). Here is the paper.

EDIT: This paper by Rupert Shuttleworth turned out to be extremely helpful (mirrored here on archive.org)

Answer 1

It seems to me that we can achieve a contradiction more quickly without assuming this anyway: Since $(1, i, j)$ is a basis for the field, $ij = a + bi + cj$ for some unique $a, b, c \in \mathbb{R}$. Then, on the one hand $i^2 j = -j$, and on the other it is

$i(ij) = i(a + bi + cj) = -b + ai + c(ij) = -b + ai + c(a + bi + cj) = (-b + ac) + (a + bc)i + c^2 j .$

Comparing coefficients of the $j$ term gives that $c^2 = -1$, which is not true for any $c \in \mathbb{R}$.

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Anyway, the question is certainly not an idle one: Indeed, it leads to something interesting when carrying out this sort of analysis for four-dimensional algebras over $\mathbb{R}$, which famously yields the quaternion algebra $\mathbb{H}$ (NB that since $\mathbb{H}$ is not commutative, it is not a field but a division algebra).

Interestingly, one can also try to construct such an algebra taking $i^2 = -1$ but $j^2 = k^2 = 1$ and find there's a coherent and interesting way to define an associative product, giving an algebra $\widetilde{\mathbb{H}}$ sometimes called the split quaternions. This is perhaps nonobviously isomorphic (as an $\mathbb{R}$-algebra) to the ring $M(2, \mathbb{R})$ of $2 \times 2$ real matrices. Unlike $\mathbb{H}$ this is not a division ring (there are nonzero matrices that square to the zero matrix), but like $\mathbb{H}$ it has a nondegenerate quadratic form $Q$ that satisfies $Q(xy) = Q(x) Q(y)$, namely the determinant.

Answer 2

I don't know if this is what Hamilton had in mind, but: suppose you've decided that what you want to do is to write down a division algebra $D$ over $\mathbb{R}$ bigger than the complex numbers $\mathbb{C}$. Any element $d \in D$ of $D$ generates a subalgebra $\mathbb{R}[d]$ of $D$ which is a commutative division algebra (in other words, a field) over $\mathbb{R}$ and hence (because $\mathbb{C}$ is algebraically closed) must be isomorphic to either $\mathbb{R}$ or $\mathbb{C}$.

If $d \not \in \mathbb{R}$, then this subalgebra $\mathbb{R}[d]$ must be isomorphic to $\mathbb{C}$. Any such isomorphism realizes $d$ as a complex number $a + bi, b \neq 0$, and hence up to a real change of coordinates we might as well have chosen $d$ to correspond to $i$ under this isomorphism; that is, to satisfy $d^2 = -1$. With a bit more work this argument shows that $D$ is generated by elements that square to $-1$.