# How many “super imaginary” numbers are there?

How many "super imaginary" numbers are there? Numbers like $i$? I always wanted to come up with a number like $i$ but it seemed like it was impossible, until I thought about the relation of $i$ and rotation, but what about hyperbolic rotation? Like we have a complex number $$z = a + bi$$ can describe a matrix $$\begin{bmatrix} a & -b \\ b & a\end{bmatrix}$$ You can "discover" $i$ by doing (which is used for another discovery) $$\begin{pmatrix} a \\ b \end{pmatrix} \cdot \begin{bmatrix} c & -d \\ d & c\end{bmatrix} = \begin{pmatrix} ac - bd \\ ad + bc \end{pmatrix}$$ $$(a + bi) \cdot (c + di) = ac + adi + bci + bdi^2$$ From here on you can infer that $i^2 = -1$.

So what if we do the same thing, but a different matrix? $$z = a + bh$$ can describe a matrix $$\begin{bmatrix} a & b \\ b & a\end{bmatrix}$$ and we can discover it the same way $$\begin{pmatrix} a \\ b \end{pmatrix} \cdot \begin{bmatrix} c & d \\ d & c\end{bmatrix} = \begin{pmatrix} ac + bd \\ ad + bc \end{pmatrix}$$ $$(a + bh) \cdot (c + dh) = ac + adh + bch + bdh^2$$ From here we infer that $h^2 = 1$.

Also $$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots$$ \begin{align} e^{xh} & = 1 + \frac{xh}{1!} + \frac{(xh)^2}{2!} + \frac{(xh)^3}{3!} + \frac{(xh)^4}{4!} + \frac{(xh)^5}{5!} + \cdots \\ & = 1 + \frac{xh}{1!} + \frac{x^2}{2!} + \frac{x^3h}{3!} + \frac{x^4}{4!} + \frac{x^5h}{5!} + \cdots \\ & = \cosh{x} + h \cdot \sinh{x} \end{align}

How many more numbers like this are there? And does that mean that for each set of trigonometric functions there exists a number which can turn multiplication into a rotation using those trigonometric functions?

(Sorry if I got some things wrong)

• One need not even restrict themselves to using $2\times 2$ matrices. There are for example the quaternions, like the complex numbers, but with three distinct imaginary units $i,j,k$ such that $i^2=j^2=k^2=ijk=-1$ (note: they aren't commutative). The question of "how many" different ways can we extend the real numbers to include numbers with similar properties to $i$, of course there are infinitely many ways. Many of them won't be useful or ever studied, but there are some extensions like the quaternions which have received some attention. – JMoravitz Dec 31 '17 at 20:30
• Apparently edited onto the current Wikepedia Page. – Paul LeVan Dec 31 '17 at 21:10
• Not essential, but do you write matrix multiplication in the reverse order, like $\begin{pmatrix} a \\ b \end{pmatrix} \cdot \begin{bmatrix} c & d \\ d & c\end{bmatrix}$? – Jeppe Stig Nielsen Jan 1 '18 at 23:20
• Jeppe Stig Neilsen: I see computer graphics versus mathematics going opposite ways on handedness, row vs column matrix, left vs right, etc. When reading computer graphics papers, you have to be careful about interpreting what you read. – Rob Jan 3 '18 at 1:39
• And because the quats are not multiplicatively commutative, you cant have a function like logarithm that maps multiplication into addition (which IS commutative). – richard1941 Jan 3 '18 at 4:20

Your $h$-based number system is called split-complex numbers, and what you called $h$ is usually called $j$. A related system introduces an $\epsilon$ satisfying $\epsilon^2=0$, and this gives dual numbers. Linear transformations guarantee these two systems and complex numbers are the only ways to extend $\mathbb{R}$ to a $2$-dimensional commutative associative number system satisfying certain properties. However:

• The Cayley-Dickson construction allows you to go from real numbers to complex numbers and thereafter double the dimension as often as you like by adding new square roots of $-1$, taking you to quaternions, octonions, sedenions etc.;
• Variants exist in which some new numbers square to $0$ or $1$ instead, e.g. you can have split quaternions and other confusingly named number systems;
• If you really like, you can take any degree-$d$ polynomial $p\in\mathbb{R}[X]$ with $d\ge 2$ and create a number system of the degree-$<d$ polynomial functions of a non-real root of $p$ you've dreamed up, e.g. $\mathbb{C}$ arises from $p=X^2+1$.
• That is so interesting! I never thought this existed. Thank you for your response. – EEVV Dec 31 '17 at 21:22
• It's called the "hyperbolic numbers" too; and $h$ does get used for the "hyperbolic unit". – user14972 Jan 1 '18 at 0:07
• Hm, I've heard of multicomplex numbers, but not split-complex numbers... – user541686 Jan 1 '18 at 7:12
• Hurkyl, the funny thing is that I called it the hyperbolic number too! That's why it is $h$ -- for hyperbolic. – EEVV Jan 1 '18 at 11:09
• Wikipedia Split-complex number § Synonyms lists so many other names (many of which are not that old). – Jeppe Stig Nielsen Jan 1 '18 at 23:22

There are stockpiles of algebraic constructions out there... Direct sums, direct products, quotients, sub-structures, free algebras, polynomial rings, localisations, algebraic closures, completions, to name a few. This gives us so many different ways to create new algebraic structures, i.e. produce new and beautiful objects that you can calculate with. Some of those can be called "numbers", if you wish, but this is just a question how to label them: the real push is to investigate the constructed structures, see what they are useful for, and apply them to solving various problems.

Your construction is a commutative subring of $M_2(\mathbb R)$. Another construction gives you the same structure: quotient $\mathbb R[x]/(x^2-1)$, i.e. residues of polynomials in one variable under division by the polynomial $x^2-1$. What you've got there is a ring with zero divisors, for example, $0=h^2-1=(h-1)(h+1)$ but $h-1\ne 0$ and $h+1\ne 0$. This makes it harder to solve equations with those numbers. Thus, this structure, still being interesting, is harder to work with (and produces fewer results) than e.g. complex numbers.

My big point was: with the machinery mathematics has these days, it is not that hard to invent new numbers, but it is as hard as ever to invent new useful numbers. A whole other challenge is to invent new constructions, which would produce new structures in ways never seen before.

Geometric Algebra (GA) allows for infinite-dimensional analogues of complex numbers. It subsumes from scalars, through vectors, to normals (ie: bivectors), through quaternions, tensors, etc. The whole basis of it is that you perform algebra in a coordinate-free way, and yet constructs such as imaginary numbers and quaternions just show up as special cases. The fact that it is coordinate-free makes it easy to work with high dimensional cases. GA is a mathematics language designed to align with geometric intuition. The key to all of it is joining together the dot product and the cross-product in a way that generalizes to all dimensions.

$$u v = (u \cdot v) + (u \wedge v)$$

The geometric product has a commutative part, and an anti-commutative part. So the geometric product does not commute in general. But scalars commute with everything. If $e_1$ is perpendicular to $e_2$, and $e_3$ is perpendicular to them both, then they form a basis for doing 3D geometry. The basis vectors are akin to $x$, $y$, and $z$ axis. Multiplication of these basis vectors anti-commute and self-annihilate like this:

$$e_1 e_2 = -e_2 e_1$$

The same basis vector times itself cancels out. $$e_1 e_1 = 1$$

Which causes determinants to just fall out of the definition for instance. Multiply two 2D vectors, where we have scalar coefficients:

$$(a_1 e_1 + a_2 e_2) (b_1 e_1 + b_2 e_2)$$

Just distribute across them as typical, but don't commute anything yet: $$a_1 e_1 b_1 e_1 + a_1 e_1 b_2 e_2 + a_2 e_2 b_1 e_1 + a_2 e_2 b_2 e_2$$

Collect scalars together $$a_1 b_1 e_1 e_1 + a_1 b_2 e_1 e_2 + a_2 b_1 e_2 e_1 + a_2 b_2 e_2 e_2$$

Anti-commute and cancel vectors to simplify $$a_1 b_1 + a_1 b_2 e_1 e_2 + -a_2 b_1 e_1 e_2 + a_2 b_2$$ $$(a_1 b_1 + a_2 b_2) + (a_1 b_2 - a_2 b_1) e_1 e_2$$ Note that we multiplied a pair of 1D objects (vectors), and got back a sum that is a 0D object (scalar) plus a 2D object (bivector). A bivector represents a plane for rotation. The bivector is the dual of a cross-product vector. But with GA, you can have the functionality of a cross-product in all dimensions - not just 3D.

In 2D, $e_1 e_2$ functions as $I$, one of the imaginary planes.

• This is very nice. I never really understood the non-commutative property. Thank you. – EEVV Jan 1 '18 at 11:04
• If it's not apparent, the dot product is related to $cos$ $$(a_1 b_1 + a_2 b_2) = \| \mathbf{a} \| \| \mathbf{b} \| cos(\theta)$$ $$(a_1 b_2 - a_2 b_1) e_1 e_2 = \| \mathbf{a} \| \| \mathbf{b} \| sin(\theta) I$$ And if you divide $\mathbf{a} \mathbf{b}$ by those lengths, you just get a familiar formula relating angles to vectors. – Rob Jan 2 '18 at 2:39
• The justification for calling two perpendicular unit vectors an I (of which there are many in multidimensional space), is that they square to -1: $$(e_1 e_2) (e_1 e_2) = e_1 e_2 e_1 e_2 = -e_1 e_2 e_2 e_1 = -1$$ – Rob Jan 2 '18 at 2:53