Is there infinitely many "complex units"

Question

As we know, $i$ = $\sqrt{-1}$, a simple complex unit. In complex space of two dimensions, you graph an axis of $a+bi$ where $i$ is your second dimension axis.

Now, you also know, in three and four dimensional space, you use the quaternions, such that $i^2 = j^2 = k^2 = ijk = -1$. Then, in three dimensional complex space, you can have it in form $a+bi+cj$.

Yet, are there "hexternions" such that $i^2 = j^2 = k^2 = l^2 = m^2 = ijklm=-1$

And also, are there k-ternions such that $t_1^2 = t_2^2 = ... = t_k^2 = t_1 t_2 ... t_k = -1$? And you can have any finite amount of dimensions in complex space?

Answer 1

Usually they come as R $^{2^k}$, where, for $k=0$ we have the real numbers R, for $k=1$ we have the complex numbers C, for $k=2$ we have the quaternions H, for $k=3$ we have the octonions O, for $k=4$ we have the sedenions S, etc. See also this question, as well as the articles on:

• hypercomplex numbers, bicomplex numbers, multicomplex numbers,

• biquaternions, split-quaternions, split-biquaternions, dual quaternions, dual numbers,

• Musean hypernumbers $\big($conic quaternions/octonions/sedenions$\big)$, hyperbolic quaternions,

• Clifford algebras, Cayley-Dickson constructions, Pauli matrices, etc.

Answer 2

You can build tessarines with any number of dimensions of the form $2^n$, and they still will be commutative and associative.