Is it possible to "get" quaternions without specifically postulating them?

Question

I know that quaternions were first invented to handle description of 3D and 4D rotations, just as 2D rotations can be described by complex numbers. On the other hand, non-natural numbers can be "deduced" in a sense.

For example, starting with natural numbers (including 0), so that subtraction can be done for all numbers, we add negative numbers to get integers. To make division work the same way (well, except for division by 0), we introduce rational numbers. To "fill in the remainders", we enter real numbers. Finally, in order for any polynomials to have a solution, we add an imaginary unit and get complex numbers.

However, complex numbers are algebraically closed, which leads us to the fact that we can no longer obtain other numbers in the same way. But perhaps there are other branches of mathematics where quaternions "jump out of nowhere" at some point, just like the imaginary unit when solving $x^2 = -1$?

Answer 1

Let $S_2$ denote the set of all integers that are representable as a sum of two squares. Then $S_2$ is closed under multiplication, because of the identity $$(a^2 + b^2)(\alpha^2 + \beta^2) = (a\alpha - b\beta)^2 + (a\beta + b\alpha)^2,$$ which comes from the fact that the norm of a product of two complex numbers is the product of their norms.

Let $S_4$ denote the set of all integers that are representable as a sum of four squares. Analogously, you might hope that there is an identity which proves that $S_4$ is closed under multiplication. With some luck and/or ingenuity, you might find Euler's four-square identity: $$(a^2 + b^2 + c^2 + d^2)(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) = (a\alpha - b\beta - c\gamma - d\delta)^2 + (a\beta + b\alpha + c\delta - d\gamma)^2 + (a\gamma - b\delta + c\alpha + d\beta)^2 + (a\delta + b\gamma - c\beta + d\alpha)^2.$$ You might then wonder if there is a number system analogous to the complex numbers that "explains" this identity, and thereby might be led to discover the quaternions.

By the way, this is not an isolated fact, because there are deep connections between the theory of quaternion algebras and the theory of quadratic forms.

Answer 2

Quaternions—in the sense of objects that obey the relations of the quaternion group—arise automatically in Lie Theory. The most elementary noncommutative continuous group is $SO(3)$, and if you want to understand rotations in three dimension, you obviously have to understand the representations of this group—how it can act on scalar, vector, and higher tensor objects, for instance. The most natural way to do this is to study the Lie algebra; even if you are not going to fully generalize to abstract Lie algebras, you are going to be led to a study of the generators $T_{i}$, which obey a commutation relation $[T_{i},T_{j}]=\epsilon_{ijk}T_{k}$.

However, if you explore the representations of this Lie algebra, $\mathfrak{so}(3)$, you find that there are a whole bunch of group representations missing—just like there are missing roots to real polynomials! There is one finite-dimensional irreducible representation of the algebra $\mathfrak{so}(3)$ for dimension $n$ for each integer $n\geq1$. Yet the group $SO(3)$ only has representations with odd dimension! We know the reason for this now is that there is a topological obstruction; $SO(3)$ is not simply connected. However, even without understanding the general topological issue involved, it straightforward to locate another group—the universal cover $SU(2)$ of $SO(3)$—which has the same Lie algebra $\mathfrak{su}(2)\approx\mathfrak{so}(3)$ but which does have all the representations, including a two-dimensional fundamental representation. The generators of this two-dimensional representation (from which, via tensor products, all the representations can be built up) $$i\sigma_1=\left[\begin{array}{cc} 0 & i \\ i & 0 \end{array}\right] \\ i\sigma_2=\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] \\ i\sigma_3=\left[\begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right]$$ then have precisely the multiplication table of the quaternions $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$.

Of course, the connection between the quaternion algebra and the theory of continuous groups goes on to be much deeper. However, this is the first place that it is likely to be evident that the introduction of objects that obey the defining relations for quaternions is necessary.

Answer 3

The group algebra over $\mathbb{R}$ of every finite group is a direct sum of simple algebras. If you decide to compute this decomposition for some small groups, then you will discover that for the 8-element quaternion group $Q_8$, you get a direct sum of four copies of $\mathbb{R}$ and one copy of the quaternion algebra. (Over $\mathbb{C}$, you get four copies of $\mathbb{C}$ and one of the $2\times 2$ matrix algebra $M_2(\mathbb{C}))$.

Answer 4

They can be deduced via Dirac Belt Trick, see Understanding Quaternions and the Dirac Belt Trick by Mark Staley. This Trick is also known as Plate Trick.

Answer 5

Given a normed vector space, with the basic operators of addition, scalar multiplication, and squared norm, you can generalize to an algebra by taking the free algebra over the vectors modulo $\mathbf u+\mathbf v=\mathbf w$, $a\mathbf v=\mathbf w$, and $a = \mathbf v^2$ for all such identities that hold in the usual sense.* The result is called the Clifford algebra of the vector space.

Within the algebra there is a subalgebra called the even algebra, which is generated by the products of even numbers of vectors.

The even Clifford algebra of $\mathbb R^3$ is isomorphic to the quaternions. If you take every nonzero vector in $\mathbb R^3$ to represent a reflection in the plane normal to itself, then the product of an even number of them is equal to a quaternion that represents that rotation in the usual way.

* There is a bit of awkwardness with the zero vector, which is equal to the zero scalar in the Clifford algebra. I'm not sure how to deal with that properly, but perhaps all that's necessary is to add the identity $0=\mathbf 0$.