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?

  • $\begingroup$ Any element of the form $\;\zeta_n:=e^{\frac{2\pi i}n}\;,\;\;n\in\Bbb N\;$ is a unit in the complex field in the sense that $\;\zeta_n^n=1\;$ ... $\endgroup$ – DonAntonio May 3 '14 at 15:06
  • $\begingroup$ The answer depends a bit on what properties you want your creations to have - whether you want to be able to multiply them as well as add them, for example. You might be interested in this: en.wikipedia.org/wiki/Octonion and mathworld.wolfram.com/DivisionAlgebra.html (Cayley Numbers and Octonions are two names for the same thing) $\endgroup$ – Mark Bennet May 3 '14 at 15:17
  • $\begingroup$ Given what I need, I need to be able to multiply and add them; not much else. The fact of the matter is I need some kind of way to know how to have complex units for finitely dimensional complex space. Edit: That is, space of form $a+bi+ci_2+di_3+...+zi_k$ $\endgroup$ – someuser May 3 '14 at 15:23
  • $\begingroup$ Sum them? Then I think you may want to talk about complex algebras, not merely complex vector spaces... $\endgroup$ – DonAntonio May 3 '14 at 15:25
  • $\begingroup$ Summing them such that the sum can be graphed on a "complex space" $\endgroup$ – someuser May 3 '14 at 15:29

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:


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