Where are the 3D Complex Numbers?

September 23, 2023

It's a natural question. We learn about the number line in elementary school, and later we learn about the 2D complex plane. So what's next? $a + bi + cj$?

What Are We Asking?

If you stop to think, it might not be that obvious what we mean by "3D complex numbers". What would they look like? How can we define 2D complex numbers in a way easily extensible to 3D?

Our intuitive understanding of $\mathbb C$ as a 2D plane lends itself to its description as a 2D vector space, like $\mathbb R^2$. You can't multiply vectors though, so we also give the complex numbers a product:

which you're probably familiar with -- if not, multiply out $(a + bi)(c + di)$ and see what you get.

So really, the complex numbers are just a vector space spanned by $\{1, i\}$ and an additional multiplication operator (we call such an object an algebra). Would the 3D complex numbers then be a vector space spanned by $\{1, i, j\}$ with a special multiplication rule?

Let's work out what such a multiplication rule would look like. We know how to multiply $1 \cdot i$ and $1 \cdot j$, but what about $ij$? There are a few options, none of which work:

• $ij = i \mathrm{,\ so\ } j = 1$
• $ij = j \mathrm{,\ so\ } i = 1$
• $ij = -1 \mathrm{,\ so\ } j = i$
• $ij = 1 \mathrm{,\ so\ } j = -i$

In every case, we can't come up with a reasonable value for $ij$. What we really have to do is create a fourth symbol, $k$, and set $ij = k$ -- if you've heard of the quaternions, that's what we're doing here.

In fact, every time we want to add a new symbol, we have to define how it interacts with every other symbol before it. For instance, if we wanted to add $\ell$ to $\{1, i, j, k\}$, we'd have set values for $1\ell$, $i\ell$, $j\ell$, and $k\ell$. Generally, the number of symbols we need will double every time, so the next algebra after the quaternions is 8-dimensional, then 16, then 32, etc. So the real answer to "why aren't there 3D complex numbers?" is "3 isn't a power of 2".

What About the Cross Product?

You might be thinking "ok, but the cross product exists. Why doesn't that count?" And although the computer graphics part of me loves the cross product, I must admit that it fails on almost all fronts at being well-behaved. There are a few things we expect a product to do:

• Commutativity: $a \times b = b \times a$
• Associativity: $(a \times b) \times c = a \times (b \times c)$
• Zero-product property: if $a \times b = 0$, then $a = 0$ or $b = 0$.
• Bilinearity: $k(a \times b) = ka \times b = a \times kb$

By this measure, the cross product is an abomination; it's not commutative, it's not associative, and it definitely doesn't have the zero-product property (in fact, anything squared is 0 under the cross product!). The only reason we call it a product at all is because it follows the bilinearity condition.^[1]

But how do we know there's really nothing there? Sure, we tried a couple options and they didn't work, but we didn't check every possible value for $ij$. Could there be something that works?

Search all you want, but we actually know that there isn't. In 1877, Ferdinand Georg Frobenius proved that the only algebras following the four rules listed earlier are $\mathbb R$ and $\mathbb C$. If we drop the commutativity requirement, we also get the quaternions $\mathbb H$, but that's it. Any other algebra will break one of those four rules.

1. Most things that feel "product-y" are actually just bilinear. If you stop and think, it's not that clear what the cross product, dot product, tensor product, and matrix product have in common; indeed, it's just that they're bilinear. ↩