So, I know Quaternions are basically 4 dimensional Complex numbers, and the dimensions can double forever to Octonions, Sedinions, etc. I recently heard about bicomplex numbers, which are also sort of 4 dimensional complex numbers. I looked into it a bit, and I found that bicomplex numbers are a commutative version of quaternions. But I wanted to know how exactly that works. How is the multiplication defined? Also can that be extended to 8-Dimensional biquaternions, 16-D bioctonions, etc? And how are those defined exactly? I know with regular hypercomplex numbers, you lose properties as you go up in dimensions, so does the same thing happen with the bi- ones? If so does it just lose the property that it would normally lose one dimension lower? Or do they not lose properties at all? Any answers would be greatly appreciated.

  • $\begingroup$ Quaternions are basically 4 dimensional real numbers? $\endgroup$
    – Curiosity
    Commented Mar 12, 2015 at 18:38
  • $\begingroup$ Well, however you want to say I guess. I was more just saying that I knew what Quaternions were, not trying to give an in-depth description. Really they are their own thing, 4-Dimensional Hypercomplex Numbers. $\endgroup$
    – RothX
    Commented Mar 12, 2015 at 18:45

2 Answers 2


I'd never heard of a bicomplex number, but I like algebra, so let's go for it. From what I understand, reading Wikipedia and playing around:

We assume that, in addition to $i$, our favorite square root of $-1$, we have another, $h$, so that $h^2 = -1 = i^2$. Further, we force $i$ and $h$ to commute, so that $ih = hi$.

You can see for yourself that $(ih)^2 = 1$, since $h$ and $i$ commute. So, $ih$ is a square root of $1$. We know of some other square roots of $1$, let's see what happens when we multiply something by $ih$. At this point, we'll assume our numbers are associative (just like we assumed $h$ and $i$ commute).

$(ih)h = ih^2 = i(-1) = -i$. Thus, multiplication by $ih$ sent $h$ to $-i$. This means $ih$ can't be $1$ or $-1$, the square roots of $1$ we already know and love, since $i$ and $h$ are distinct (that is, $h \neq -i$).

It turns out the set $\{1, i, h, ih\}$ forms a basis for these numbers. Thus, a generic "bicomplex number" looks like $a + bi + ch + dih$.

But, these bicomplex numbers are certainly don't act like quaternions. For one, they form a commutative algebra; the order of multiplication never matters! But more seriously, we have zero-divisors:

$$(1 + ih)(-1 + ih) = -1 + ih - ih +(ih)^2 = -1 +ih - ih - 1 = 0.$$

This is a pretty serious departure from the quaternions.

It turns out that the biquaternions are isomorphic to the direct sum $\mathbb{C} \oplus \mathbb{C}$ of the field of complex numbers with itself (here, addition and multiplication are performed component-wise, like vector addition). This is not obvious to me, but it's believable.

Direct sums are fairly uninteresting: Just slap two copies of something together, and there you go! It explains why we have zero-divisors, and commutativity. We just do everything we normally would, but do it twice; business as usual.

The quaternions on the other hand - now that's an algebra! Associative, but not commutative; no zero divisors; we can define a norm. Generally, life with the quaternions is great.

To build the complex numbers, quaternions, or octonians from the reals, the situation is a lot more subtle; the construction is an example of a "twisted group algebra." The complex numbers, quaternions, and octonions are all examples of twisted group algebras over the real numbers. In particular, we get the quaternions by introducing a "twisted" way of multiplying elements of the group algebra $\mathbb{R}[\mathbb{Z}_2^2]$.

Unfortunately, I'm still processing all the details myself, so it's up to you to figure out exactly what "twisting" something entails. Essentially, it's like a direct sum, but with the terms interacting multiplicatively more than just "component-wise," we need some extra magic thrown in.


John Baez' write up on octonions, where I first saw the term "twisted group algebra." Lacking on details of the construction, lots of attention to the big picture.

John Bales' write up on "Properly Twisted Group Algebras", which is way more detail than I'd hoped for, but could fill in some of the gaps.

  • 1
    $\begingroup$ Ok, thanks, that explains it to an extant. They are more different than I expected. Also, I'm glad my question introduced you to something new as well! $\endgroup$
    – RothX
    Commented Mar 14, 2015 at 0:53
  • 1
    $\begingroup$ Also, you said $−1+ih−ih+(ih)^2 = −1+ih−ih−1$, when it should be $−1+ih−ih+(ih)^2 = −1+ih−ih+1$ $\endgroup$
    – RothX
    Commented May 18, 2017 at 13:46
  • $\begingroup$ @RothX Correct you are, and thank you! I'll hold off on changing anything at the moment -- while it is certainly an error, maybe in a day or two I'll be able to improve the post in a more significant way (although probably not :)). $\endgroup$
    – pjs36
    Commented May 18, 2017 at 14:09

Bicomplex numbers, or tessarines is a commutative algebra as other people said.

It has zero divisors. And I would say, being commutative is a lot more important than disadvantages from zero divisors!

Moreover, zero divisors are not a disadvantage at all, because it allows bicomplex numbers to generalize not only the concept of complex numbers, but also hyperbolic (split-complex) numbers. They are included in bicomplex numbers and have zero divisors.

There is nothing essentially bad regarding zero divisors: this means that we have some elements by which we cannot divide, but we anyway had one such real number: zero.

They can be represented as real matrices of certain structure with usual matrix operations, so one can experiment with them in any computer algebra system that allows to apply analytic functions to matrices.

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Images source:https://citeseerx.ist.psu.edu/viewdoc/download?doi=

  • $\begingroup$ Despite the zero divisors, what you say about dual numbers is incorrect. There are no nonzero solutions of $z^2=0$ in the bicomplex numbers. $\endgroup$
    – J.G.
    Commented Feb 14, 2021 at 10:28
  • $\begingroup$ @J.G. but if one adds dual unity over it, the algebra remains commutative and associative. $\endgroup$
    – Anixx
    Commented Feb 14, 2021 at 10:34
  • $\begingroup$ @J.G. This paper describes tessarines and dual hyperbolic: jcscm.net/fp/126.pdf $\endgroup$
    – Anixx
    Commented Feb 14, 2021 at 10:35

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