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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?

someuser
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  • 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 $;\zeta_n^n=1;$ ... – DonAntonio May 03 '14 at 15:06
  • 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: http://en.wikipedia.org/wiki/Octonion and http://mathworld.wolfram.com/DivisionAlgebra.html (Cayley Numbers and Octonions are two names for the same thing) – Mark Bennet May 03 '14 at 15:17
  • 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 $a+bi+ci_2+di_3+...+zi_k$

     – someuser May 03 '14 at 15:23
  • Sum them? Then I think you may want to talk about complex algebras, not merely complex vector spaces... – DonAntonio May 03 '14 at 15:25
  • Summing them such that the sum can be graphed on a "complex space" – someuser May 03 '14 at 15:29
  • I've no the slightest idea what you can possibly mean by that, @someuser ... – DonAntonio May 03 '14 at 15:41
  • So, you know how you can graph numbers of form $a+bi$ right... What I need is a way to graph numbers of form $a+bi+ci_2+di_3+...$ as somewhat of an "expansion" of $a+bi$. (where $i_k$ for any natural $k$ is complex) – someuser May 03 '14 at 15:45
  • Is this even possible? @DonAntonio – someuser May 03 '14 at 16:08
  • Have you looked up the references I suggested? There is also a paper here: http://arxiv.org/abs/math/0105155 which is a bit less basic, but has some good discussion in between some of the more advanced mathematics. – Mark Bennet May 03 '14 at 16:14
  • @someuser, perhaps you're thinking of Hamilton's Quaternions, a $;4$-dimensional real algebra? – DonAntonio May 03 '14 at 16:16
  • @DonAntonio Yes. – someuser May 03 '14 at 16:25
  • Also, how they can be expanded to any $2^k$-dimensional real algebra consisting of $2^k$ complex units. – someuser May 03 '14 at 16:28
  • Look up the Cayley-Dickson construction. – Mark Bennet May 03 '14 at 16:37
  • Not if you want to keep some characteristics, @someuser. For example, the only division real algebras (i.e., whithout non-zero zero divisors) are $;\Bbb R;,;;\Bbb C;,;;\Bbb H;$ . If you're willing to give up associativity (hello!) you also have the octonions $;\mathcal O;$ , etc. – DonAntonio May 03 '14 at 16:39

2 Answers2

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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:

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

Anixx
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