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I've tought using split complex and complex numbers toghether for building a 3 dimensional space (related to my previous question). I then found out using both together, we can have trouble on the product $ij$. So by adding another dimension, I've defined $$k=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$ with the property $k^2=1$. So numbers of the form $a+bi+cj+dk$ where ${{a,b,c,d}} \in \Bbb R^4$, $i$ is the imaginary unit, $j$ is the elementry unit of split complex numbers and k the number defined above, could be represented on a 4 dimensinal space. I know that these numbers look like the Quaternions. They are not! So far, I came out with the multiplication table below : $$\begin{array}{|l |l l l|}\hline & i&j&k \\ \hline i&-1&k&j \\ j& -k&1&i \\ k& -j&-i&1 \\ \hline \end{array}$$

We can note that commutativity no longer exists with these numbers like the Quaternions. When I showed this work to my math teacher he said basicaly these :

  1. It's not coherent using numbers with different properties as basic element, since $i^2=-1$ whereas $j^2=k^2=1$
  2. 2x2 matrices doesn't represent anything on a 4 dimensional space

Can somebody explains these 2 things to me. What's incoherent here?

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What about considering $\{\alpha I+\beta k\,|\,\alpha,\beta\in\Bbb C\}$ where $I=\pmatrix{1&0\\0&1}$? I think, it's different from your original question, but makes some sense for sure. Over $\Bbb C$ the extension is $2$ dimensional, so over $\Bbb R$ it's four. –  Berci Apr 26 '13 at 10:20
    
Is your multiplication table right? By it, both the products $ij$ and $jk$ are anti-symmetric, however the product $ik$ is not. For example, $ik = i$, but $ki=-j$. Is this correct? –  Tpofofn Apr 26 '13 at 10:31
    
Sorry, mistyped a character on the table. $ik=-j$, $ki=j$ –  moray95 Apr 26 '13 at 11:01

3 Answers 3

up vote 9 down vote accepted

Congratulations: the multiplication table for basis elements that you have laid out indicate that you have independently discovered the Clifford algebra of a two dimensional vector space with metric signature $(1,-1)$, also denoted as $C\ell_{1,1}(\Bbb R)$!

This algebra is isomorphic to the full ring of $2\times 2 $ real matrices $M_2(\Bbb R)$ as an algebra. So, it is completely coherent.

The quatnerions, split complex numbers, and this structure you are describing are united by the Clifford algebra perspective:

$$ \begin{bmatrix}C\ell_{0,0}(\Bbb R)&&|&&\Bbb R\\ C\ell_{0,1}(\Bbb R)&&|&&\Bbb C\\ C\ell_{0,2}(\Bbb R)&&|&&\Bbb H\\ C\ell_{1,0}(\Bbb R)&&|&& \text{split complex numbers}\cong \Bbb R\times\Bbb R\\ C\ell_{1,1}(\Bbb R)&&|&& \text{your algebra}\cong M_2(\Bbb R)\end{bmatrix} $$

If you find all this incomprehensible at the moment, then I totally understand. I only started learning about Clifford algebras about a year ago. I don't even know if you have any abstract algebra training, either.

I just want to reassure you that what you described here is perfectly sensible thing in ring theory. It looks like your teacher dismissed it, but that may be understandable: teachers often see a lot of ideas by students that do fall flat!

At any rate, the two objections you included in the OP are quite vague.


To find an explicit isomorphism with $M_2(\Bbb R)$, you can use this mapping: $$ 1\mapsto \begin{bmatrix}1&0\\0&1\end{bmatrix}\ \ i\mapsto \begin{bmatrix}0&-1\\1&0\end{bmatrix}\\ j\mapsto\begin{bmatrix}0&1\\1&0 \end{bmatrix}\ \ k=ij\mapsto \begin{bmatrix}-1&0\\0&1\end{bmatrix}\ \ $$

These four matrices clearly are a basis of $M_2(\Bbb R)$ and fit your table.

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Thank you for your remarkable response! Accepted and upvoted! I'd like to have a little bit more information about this algebra. What does the metric signature means exactly? I can't understand anything in the wikipedia article. I'm in high school so I don't know much about abstract algebra. I know complex numbers and a little bit about split-complex and quaternions just from curiosity. –  moray95 Apr 26 '13 at 13:19
    
@moray95 I'm sorry that I can't reduce the level of the answer veyr much! It is not something (even ring theorists) all learn about, so lots of people will be in the same position. I'm glad to see you chase your curiosity, and I'll try to help you with whatever questions you have as you chase this topic (since I am interested in this topic too!). My advice for a first step toward understanding this would be to study linear algebra and try to develop your geometric sense while doing so. This will make you well-prepared for starting abstract algebra. –  rschwieb Apr 26 '13 at 13:40
    
@moray95 To be honest, I think it'll take a little time for you to build up to this solution. If you are interested in investing in a book, Basic Algebra I by Nathan Jacobson is a solid, cheap book with a good route for you to take. –  rschwieb Apr 26 '13 at 13:53
    
@moray95 And Willie Wong's recommendation of MacDonald's linear algebra text is doubly a good suggestion since you can learn both linear algebra and this the connections with complex numbers :) –  rschwieb Apr 26 '13 at 13:58
    
I've just found that these numbers are Biocmplex numbers! (or at least with a slide difference on k). –  moray95 May 3 '13 at 16:00

rschwieb already gave you the high powered answer. Here let me give you the low-powered version of what he wrote.

Consider the collection of $2\times 2$ matrices with real entries. We can write each matrix as $$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} $$ and if we re-organize the presentation, it can be identified with an element of $\mathbb{R}^4$ $$ \begin{pmatrix} A \\ B \\ C \\ D\end{pmatrix} $$ By writing it as a matrix, you allow yourself to do "multiplication" by matrix multiplication.

Now, we can write $$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} = A \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} + B \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} + C \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} + D \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} $$ which, if you know a bit of linear algebra, is just expressing a $2\times 2$ matrix in a basis.

As it turns out, what you've done is basically just choosing a different basis for the $2\times 2$ matrices. You chose

$$ \mathbf{1} = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \quad \mathbf{i} = \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix} $$ and $$ \mathbf{j} = \begin{pmatrix} 0 & -1 \\ -1 & 0\end{pmatrix} \quad \mathbf{k} = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix} $$

We can solve for the "standard" basis $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ etc. in terms of this new basis. Plugging it back in to the expression then we have

$$ \begin{pmatrix} A & B \\ C & D\end{pmatrix} = \frac{A}{2} (\mathbf{1} + \mathbf{k}) + \frac{B}{2} (-\mathbf{i} - \mathbf{j}) + \frac{C}{2} (\mathbf{i} - \mathbf{j}) + \frac{D}{2} (\mathbf{1} - \mathbf{k}) $$

This identification can be reversed (exercise for you!). But in any case your identification of $a\mathbf{1} + b\mathbf{i} + c\mathbf{j} + d\mathbf{k}$ with the $\mathbb{R}^4$ vector $(a,b,c,d)$ corresponds then, to identifying the matrix $\begin{pmatrix} A & B \\ C & D\end{pmatrix}$ with the element $$\begin{pmatrix} \frac12 (A + D) \\ \frac12 (C - B) \\ -\frac12 (B+C) \\ \frac12 (A-D) \end{pmatrix}$$ which can be realized as the linear transformation of $\mathbb{R}^4$ that can be realized by a matrix multiplication $$ \begin{pmatrix} A \\ B \\ C \\ D\end{pmatrix} \mapsto \begin{pmatrix} \tfrac12 & 0 & 0 &\tfrac12 \\ 0 & -\tfrac12 & \tfrac12 & 0 \\ 0 & -\tfrac12 & \tfrac12 & 0 \\ \tfrac12 & 0 & 0 & -\tfrac12 \end{pmatrix}\begin{pmatrix} A \\ B \\ C \\ D\end{pmatrix}$$


What is the lesson behind all this? Given any four real numbers, you can of course identify them with an element of $\mathbb{R}^4$. The real question starts when you ask "how is this identification meaningful"? The first thing you can do is to try a little bit of linear algebra like I outlined above. But things get real exciting when you start connecting the algebra to geometry, and that's where the power of the Clifford Algebra that rschwieb mentioned really shines.

For the time being, if you cannot completely absorb the abstract nonsense in the definitions of Clifford algebras, it may be worthwhile to set your goal a tiny bit lower and think only about geometric algebra. (Unfortunately the Wikipedia link is not the best way to learn about this, read this first, and if you are interested, perhaps follow a textbook such as this.)

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That's a good net link, and there are lots of other pages like it, since geometric algebra is kind of a grassroots movement. That is also exactly the textbook I was going to recommend, since I know Macdonald does a good job explaining :) It also kills two birds with one stone by teaching linear algebra with it! –  rschwieb Apr 26 '13 at 13:58
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@rschwieb: that's precisely why I recommended that book. (I only did so after I saw the OP's comments on your question; one cannot learn Clifford algebra without some not-so-basic linear algebra too!) –  Willie Wong Apr 26 '13 at 14:27
    
Yes: there is a long journey ahead of this student, but hopefully interest will lend him/her speed :) It's unfortunate that modern mathematics curricula and tradition combine to bury this fun stuff under a lot of prerequisites! –  rschwieb Apr 26 '13 at 14:37

You can build numbers generated by such $\mathbf{i},\mathbf{j},\mathbf{k}$, I see no incoherency. They will form a subalgebra of complex $2\times2$ matrices, in which the role of $\mathbf{i},\mathbf{j},\mathbf{k}$ will be played by $$ \mathbf{i}=\left(\begin{array}{cc} 0 & i \\ i & 0\end{array}\right),\qquad \mathbf{j}=\left(\begin{array}{cc} 0 & -i \\ i & 0\end{array}\right), \qquad \mathbf{k}=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right).$$ This is a four-dimensional subspace of the eight-dimensional (over $\mathbb{R}$) space of all $2\times2$ complex matrices.

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Why did you wrote $i=\left(\begin{array}{cc} 0 & i \\ i & 0\end{array}\right), \mathbf{j}=\left(\begin{array}{cc} 0 & -i \\ i & 0\end{array}\right)$? The matricial representation of i is $\left(\begin{array}{cc} 0 & -1 \\ 1 & 0\end{array}\right)$ and the one of j is $\left(\begin{array}{cc} 0 & -1 \\ -1 & 0\end{array}\right)$ –  moray95 Apr 26 '13 at 11:39
    
it doesn't really matter as far as this gives an equivalent algebra (you can for example exchange $j$ and $k$). And it seems to me that you had $ij=-k$ in the first version of the multiplication table. –  O.L. Apr 26 '13 at 13:43

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