Teaching Clifford Algebra Instead of Imaginary/Complex Numbers

Question

For those unaware, Clifford Algebra (also known as Geometric Algebra) is able to generalize vectors and rotations in n-dimensional space, and simplifies a great many formulas. However, I was curious if anyone had ever thought to skip complex numbers and move straight to Clifford Algebra, or create a Clifford Algebra textbook which assumes only a high-school level knowledge?

Answer 1

One of the pinnacle goals in high school and basic college algebra is to be able to express the fundamental theorem of algebra, which is everywhere expressed in terms of complex numbers.

Complex numbers are the first and simplest algebraically closed field, so everyone mathematically literate should be familiar with them.

Thus, it's exceedingly unlikely that "skipping" complex numbers would be fruitful or justified in almost any conceivable curriculum.

Answer 2

There is a reason why mathematics is taught from more concrete to more abstract. Students need to figure out what the role of proofs, axioms, etc. is before they can start abstraction first. Some very few students have figured this out before university, while some go through their master thesis without really getting it. Or at least to the master-level courses. They will be difficult.

So, in general, there is a good reason to increase the abstraction step by step - first real numbers, than complex numbers, then into more abstraction. With the complex numbers the students have already learned some tools and intuitions about missing tools that they can take further, as well as routine for calculations and just plain general mathematical maturity due to the course happening later.

That said, in principle, there is no problem having a specialist course, or maybe an elite university could indeed hop over some stages in the step-by-step-progression. I was not at an elite university, but we had a first year course on abstract metric spaces as an alternative to a course on Euclidian spaces. Of the seven or so people taking that course, about five have doctor's degree.

Answer 3

Yes. David Hestenes has been pushing for this for many years, and wrote the textbook 'New Foundations for Classical Mechanics' with the idea of starting that process off. It's probably more advanced than high school level - I think it's intended as a first year university course - but the first few chapters probably wouldn't pose any great difficulty to a bright high school student.

One of the major selling points of geometric algebra is that it already includes the complex numbers as a natural part of the number system, and gives them a clear, intuitive, geometrical interpretation. Geometric algebra encompasses and unifies the reals, complex numbers, quaternions, spinors, vectors, pseudovectors, directed areas, directed volumes, reflections, rotations, scaling, points, lines, planes, circles, spheres, and more. You only have to learn one number system, and all the others are special cases or simple extensions of it. It also fixes a lot of the problems with vector algebra, which is valuable in itself.

It does not require a great deal of abstraction (at least to start with), being only a little more complicated than vectors. We start with the vectors, and we define a new sort of multiplication on them, that combines the dot product and something that works a lot like the cross product together (except that the cross product only works in 3 dimensions, while the geometric product generalises to any number of dimensions). This can produce scalars (like the dot product), but also produces new higher dimensional objects called bivectors, trivectors, etc. that correspond to directed areas, directed volumes, etc. in the same way that vectors represent directed lengths. If you start with the 2D vectors, then the combinations of scalars and bivectors they produce turn out to be identical to the complex numbers. If you start with the 3D vectors, then the combinations of scalars and bivectors produced turn out to be the quaternions.

So, you would start with vectors, introduce multiplication, look at the new sorts of geometrical objects you can get, and then point to this particular subset of them as a straightforward way to introduce the complex numbers. They're easily visualised geometrical objects, that you can apply your inbuilt geometric intuition to.

The principal advantage of geometric algebra is the deep intuitive insights the geometric interpretation provides in a lot of otherwise quite opaque and abstract topics. It is often criticised by mathematicians as not offering anything new, that was not already well known from the theory of Clifford algebras, and they're probably right about that. (Although I've not seen the generalisations of Cauchy's theorem for complex numbers to higher dimensions elsewhere.) Physicists, on the other hand, usually love it, because it provides simple pictures of things. And one of the particular things they like most is that it means you can do physics without needing complex numbers!

There are lots of introductions to geometric algebra around, although in my view most of them leap ahead too rapidly to the more advanced, abstract topics. Here is one by Lasenby, that specifically deals with complex numbers. The first third of the document is quite basic, introducing multiplication of vectors and showing how complex numbers arise, but then he suddenly accelerates and by the end is covering Maxwell's equations and advanced electrodynamics. The start demonstrates that there is no reason why you couldn't write a basic textbook for highschool students. It's just that most of the books on it are aimed at professionals who already know the standard vector methods, trying to persuade them to pick up a new and unfamiliar tool.

(For anyone interested in what else it is capable of, this is a more comprehensive introduction.)

Answer 4

If we want to solve the quadratic equation $az^2+bz+c=0$ where $a,b,c \in \mathcal{A}$ and $a^{-1}$ exists where $\mathcal{A}$ is a semi-simple associative unital algebra then since $\mathcal{A}$ is isomorphic to $\mathbb{R}^n \times \mathbb{C}^m$ the quadratic equation given translates into $n$-real quadratic equations paired with $m$-imaginary quadratic equations. Therefore, understanding the real and complex case is, well, basic. If the Clifford algebra viewpoint makes it easier, then I wait to see it.

For example, hyperbolic numbers $\mathcal{H} = \{ x+jy \ | \ x,y \in \mathbb{R}, j^2=1 \}$ are isomorphic to $\mathbb{R} \times \mathbb{R}$ and as such a quadratic equation in $\mathcal{H}$ amounts to a pair of real quadratic equations. It follows there is either no solution, one solution, two solutions, or four solutions to the quadratic equation in the hyperbolic number system. These correspond respectively to:

• both underlying real quadratic equations have only complex solutions (no solution)
• one or both of the underlying real quadratic equations have repeated root solution (one solution)
• both of the underlying real quadratic equations have distinct real solutions (four solutions)

For the Clifford Algebra approach to be interesting to me, it would have to also allow for the natural demarcation into real and complex aspects of a semi-simple real associative algebra.

For what it's worth, I have similar gripes with algebraists who insist on always working over $\mathbb{C}$ since $\mathbb{R}$ are "easy" to derive from the $\mathbb{C}$ theory... this is little consolation when you're looking for a breakdown into what is possible over $\mathbb{R}$.