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.
[Math] the difference between Quaternions and Bicomplex Numbers
complex numbershypercomplex-numbersquaternions
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Here's some information on quaternion analysis. The list of references is probably the most useful part.
http://www.zipcon.net/~swhite/docs/math/quaternions/analysis.html
What most of the algebras that you mentioned have in common is that they are all examples of unital algebras. A unital algebra is an algebra with a multiplicative identity element, and that multiplicative identity element is denoted with the symbol $1,$ though this is a bit of an abuse of notation. We really should denote it as $\vec1.$
I do not know what the double numbers are, and the binarions are just another name for the complex numbers. The dual numbers are an alternative to the complex numbers. Consider the the vector space $\mathbb{R}^2$ with basis $\{e_0,e_1\}.$ To form an algebra, we want a bilinear product $\cdot,$ and so if we can determine the products of the basis vectors, bilinearity is sufficient to determine all the products in the vector space. Bilinearity, to put it simply, is a more technical way of addressing distributivity. We also want the algebra to be unital, i.e, that is, for $e_0\cdot{x}=x\cdot{e_0}=x.$ Hence we know $e_0\cdot{e_0}=e_0$ and $e_0\cdot{e_1}=e_1\cdot{e_0}=e_1.$ The remaining product to determine is $e_1\cdot{e_1}.$ We know that $e_1\cdot{e_1}=e_1^2=ae_0+be_1,$ so $e_1^2-be_1-ae_0=0.$ We can complete the square, so that $e_1^2-be_1+\frac{b^2}4e_0=\left(\frac{b^2}4+a\right)e_0=\left(e_1-\frac{b}2e_0\right)^2.$ Since $\frac{b^2}4+a$ are themselves real numbers, this limits what the possible products $e_1^2$ can be. The three different cases are $\frac{b^2}4+a\lt0,$ $\frac{b^2}4+a=0,$ and $\frac{b^2}4+a\gt0.$ These three cases determine the three possible unital algebras over $\mathbb{R}^2$ up to isomorphism. These cases give the complex numbers, the dual numbers, and the split-complex numbers, respectively.
Of course, nothing is limiting you to make your algebra unital. You can have other number systems, those systems are just not named and not particularly interesting as extensions of the real numbers. But you can study them and they are perfectly valid structures to work with.
Now, let us talk about quaternions and octonions. You should look into the Cayley-Dickson construction. The construction is a construction that systematically extends the real numbers to the complex numbers, and applying the construction again to the complex numbers produces the quaternions, and applying it to the quaternions produces the octonions, and so on. Thus, the reason these algebras have a dimension that is a power of $2$ is because of the way the Cayley-Dickson construction works.
But if you are willing to work with algebras not built from the Cayley-Dickson construction, then there is absolutely nothing wrong with using algebras of other dimensions. On YouTube, there is a very good video about the triplex numbers, a unital associative and commutative algebra over $\mathbb{R}^3.$ The basis is $\{1,i,j\}$ and one has $i\cdot{j}=j\cdot{i}=1,$ and $i^2=j$ and $j^2=i.$ This actually equivalent to making the basis $\{1,i,i^2\}$ and noting the rule that $i^3=1,$ thus uniquely determining the multiplication table.
Now, to answer your last question: why only consider $i^2$? Well, not all algebras focus on only $i^2,$ as shown above. However, if you do uniquely determine $i^2=r$ where $r\in\mathbb{R},$ then you cannot separately control what $i^3$ is, because then $i^3=ri,$ necessarily. You cannot have it any other way. In general, if $i^n=r$ with $r\in\mathbb{R}$ for some $n\in\mathbb{N},$ then $i^{m+n}$ is automatically determined for all $m\in\mathbb{Z},$ so you have no freedom to choose those individually. In the example you gave, you suggested, $i^2=0,$ which makes your algebra equivalent to the dual numbers. Thus, it is already impossible that $i^3=-1,$ because $i^3=i\cdot{i^2}=i0=0.$ You can have some number system where $i^3=-1$ and $i^2=0,$ but that number system will not be an algebra over anything, and the multiplication will not distribute over addition. This opens up a whole can of worms beyond the scope of your question.
If you want more on the subject, you should read on the subject of Clifford algebras, the theory that unifies all of these concepts together.
EDIT 1: I would also like to point out that if you want to get really creative with number systems, you are not limited to algebras over $\mathbb{R}.$ You can form algebras over any field. Why not try making $2$-dimensional algebras over the field with two elements $\mathbb{Z}_2$? What about algebras over $\mathbb{Q}$?
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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.
Sources:
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.