What's the intuition for extending $\mathbb{C}$ to $\mathbb{H}$?

Question

It seems to me that there is a clear, intuitive reason for extending the real number system to the complex number system. Namely, some polynomial equations that have no solutions in $\mathbb{R}$ become soluble in $\mathbb{C}$. When we do this, we lose almost none of the nice algebraic properties of the reals, and pick up some nice new ones along the way (e.g. fundamental theorem of algebra).

However, I cannot see such an intuitive reason for similarly extending $\mathbb{C}$ to $\mathbb{H}$, other than "because we can." In so doing, we actually lose the important property of commutative multiplication. So precisely what problems do we solve by moving from $\mathbb{C}$ to $\mathbb{H}$?

I am of course aware of the quaternions' numerous applications to 3D geometry; what I am really interested in here are the analytic properties they provide. Does quaternionic analysis offer results comparable to those of complex analysis?

Answer 1

I assume that you're referring solely to solving equations (rather than applications of quaternions, like modelling 3D vectors), in which case there is not really much point in extending $\mathbb{C}$ to $\mathbb{H}$; in terms of solving polynomial equations, it's actually superfluous to solve in $\mathbb{H}$. e.g. $w^2=-1$ has two solutions in $\mathbb{C}$, but infinitely many in $\mathbb{H}$. Also, some (non-polynomial) equations cannot be solved in $\mathbb{C},$ but can be solved in $\mathbb{H}$- take, for example, $(xi-ix)^2=-1.$ In $\mathbb{C}$, $xi=ix$, so you get $0=-1$, which clearly has no solution, but in $\mathbb{H},$ using the fact that $xi \neq ix,$ we can, indeed, solve this. So, whilst you imply a lack of commutativity is a bad thing, we see that, in this case, it's necessary in order to solve this equation.

Another example, regarding associativity this time is: $(xi)y-x(iy)=1$. Note that, in $\mathbb{H}$ (and any subsets therof), there are no solutions to this equation; we end up with $0=1,$ which has no solutions. Now, in the octonions, $\mathbb{O}$, this equation does have a solution, since $(xi)y \neq x(iy)$ (due to lack of associativity in the octonions).

So, to conclude, purely with regard to equation solving, extending $\mathbb{C}$ to higher dimensions ($\mathbb{H, O, S, }$ etc.) allows us to solve previously-unsolvable equations (for example, by exploiting a lack of commutativity and/or lack of associativity). It also lets us find even more equations to polynomials (although why anyone would want so many solutions is beyond me).

Answer 2

The main utility and the main reason Hamilton sought something like the quaternions, as far as I know, was to find a $3$-d analogue of the complex field's use on the plane.

Analytically, things are less nice. It's well known that some of the key equivalences of complex analysis no longer hold for the quaternions, and they need to be modified.

As far as solutions to equations are concerned, things are radically different from the complex numbers. For one thing, $x^2=-1$ has uncountably many roots! It remains true that any polynomial of the form $\sum q_ix^i$ with $q_i\in \Bbb H$ has a root in $\Bbb H$, but a factor theorem is not possible owing to the lack of commutativity in the quaternions.

If you really intend to talk about polynomials with the intent of evaluating the indeterminate at quaternion values, then you can't allow the coefficients to commute with $x$ anymore, and the polynomials as described above aren't even closed under multiplication.

There is an analogue to the fundamental theorem of algebra for $\Bbb H$ that allows us to conclude a large subset of these generalized polynomials has a root in $\Bbb H$, but there are in fact generalized polynomials over $\Bbb H$ which lack roots in $\Bbb H$.

A simple example is that $qx-xq$ always has real part $0$, so it's impossible to make $qx-xq+1=0$.

Answer 3

To provide a slightly different perspective, $\Bbb{H}$ is important because taking $\Bbb{R}$, $\Bbb{C}$, and $\Bbb{H}$ together is such a fruitful idea. In studying algebras with a quadratic form over the reals and its extensions, one is lead to consider the Clifford algebra, the "most-free" (unital, associative) algebra on the underlying vector space $V$, satisfying the relation $$ v^2 = Q(v) \cdot 1 \qquad \text{for all } v \in V. $$ Using the polarization identity, this is equivalent to the relation using the symmetric, bilinear form: $$ vw + wv = 2 \langle v, w \rangle \cdot 1 \qquad \text{for all } v,w \in V. $$

The classification of these algebras requires the three real, normed, (associative) division algebras: $\Bbb{R}$, $\Bbb{C}$, and $\Bbb{H}$, and matrix rings over them, and has a nice period $8$ symmetry. This is one of the starting points of the study of the homotopy type of the real orthogonal groups $O_n(\Bbb{R})$ and Bott's fascinating Periodicity Theorem.

Answer 4

Sammy Black has offered an answer using clifford algebra. I will as well because I think he's missed the big point: that complex numbers and quaternions both arise from considering rotation operations in a real clifford algebra.

Consider the real clifford algebra $\mathbb G^2$, formed on top of $\mathbb R^2$. The basis elements of the algebra are $1, e_1, e_2, e_1 e_2$. Using the geometric product, the basis element $e_1 e_2$ squares to $-1$:

$$e_1 e_2 e_1 e_2 = -e_1 e_2 e_2 e_1 = -1$$

It's natural to consider $e_1 e_2$ as a geometrically significant quantity that behaves similarly to the imaginary unit in complex numbers. The subalgebra of linear combinations of $1$ and $e_1 e_2$ is identical to the algebra of complex numbers.

But most critically, such objects arise naturally in rotation operations. A rotation can be derived from a composition of reflections. A reflection can be written in clifford algebra as follows: a reflection across a line with normal vector $n$ maps a vector $a \mapsto -nan$. So two reflections can be written as $(mn)a(nm)$ for another vector $m$. The product $mn$ is also a linear combination of $1, e_1 e_2$, so $mn$ is "complex number".

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The rotation argument works the same way for 3d, for the clifford algebra $\mathbb G^3$ that is built on top of $\mathbb R^3$. In particular, the geometric product of two vectors $mn$ is in general a linear combination of four basis elements: $1, e_1 e_2, e_2 e_3, e_3 e_1$. As before, each of these last three basis elements squares to $-1$, so they all correspond to imaginary units that are used in quaternions.