6
$\begingroup$

I was looking back on complex analysis and asked myself: ''Why is there no complex number in 3 dimensions ?''. To place this question let me define with what I mean with 3 dimensions in the following.

When we first explore the complex numbers we start by looking at roots of negative numbers where we define (in order to solve equations with negative roots):$$i=\sqrt{-1}.$$ Using this imaginary unit we can define a complex number as $$z=a+ib,\text{ with } z_1z_2=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1),$$ or written as a 2-tupel:$$z=(a,b),\text{ with } z_1z_2=(a_1a_2-b_1b_2,a_1b_2+a_2b_1).$$The complex number has one real variable and one imaginary variable, so it can be represented as in a 2-dimensional plane.

If we make it more complex (no pun intended), we arrive at the quaternions, which are an extention of the complex numbers, here we have 3 imaginary units $i$, $j$ and $k$ with the property: $$i^2=j^2=k^2=ijk=-1.$$This number can be represented in a 4-dimensional space (so I say it has 4 dimensions).

Now I was wondering if there might exist some kind of complex number with only 2 imaginary units $i$ and $j$ which might be represented as a number in the 3 dimensional plane ? If i look at the way the quaternions are constructed I think the answer is maybe, the reasoning I have for this is that given 2 imaginary numbers $i$ and $j$ we could construct it analogous and state that $$i^2=j^2=-1,$$ but what to do with the product $ij$? The product should square to one so it should be minus one or one:$$ij=-1\quad\text{ or }\quad ij=1.$$If it squared to minus 1 it would be a new complex number, and then we would get that one of the numbers $i$ or $j$ is redundant (unless we itroduce a third one of course), so we continue with the demand that the second relation holds.

Doesn't this give a valid way of constructing 3D-complex numbers ? Now for the quaternions itself you could argue the same and you could also choose that $ijk=+1$, is there a reason that the $-1$ is chosen (maybe this coincides with rotations in 3D ?)?

$\endgroup$

marked as duplicate by Grigory M, user122283, user61527, Mark Bennet, achille hui Apr 12 '14 at 21:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
  • $\begingroup$ See en.wikipedia.org/wiki/…. $\endgroup$ – user122283 Apr 12 '14 at 20:48
  • 1
    $\begingroup$ I disagree with the close-as-duplicate votes. This question poses a specific alternative, "why isn't this a three-dimensional algebra", so I think "because there aren't any" is not a satisfactory answer. $\endgroup$ – Ben Millwood Apr 12 '14 at 21:28
  • 1
    $\begingroup$ @BenMillwood The link answers the OP's question «Why is there no complex number in 3 dimensions?» [and what it means exactly]. $\endgroup$ – Grigory M Apr 13 '14 at 0:34
  • 2
    $\begingroup$ @GrigoryM: it doesn't answer the OP's question "Doesn't this give a valid way of constructing 3D-complex numbers?"... moreover I don't think it's at an appropriate level for someone who hasn't necessarily heard the words "division algebra" or even "field" before. $\endgroup$ – Ben Millwood Apr 13 '14 at 20:41
5
$\begingroup$

If you want $ij = 1$, then multiplying on the left by $-i$ you have $j = -i$, so that's not much use. On the other hand, $ij = -1$ leads to $j = i$, which is equally unexciting.

In fact, no other choices will work either. Consider that whatever system you construct will form a vector space not only over the real numbers, but over the complex numbers as well, and a complex vector space of dimension $n$ is a real vector space of dimension $2n$, so an odd number of dimensions is impossible.

Of course, you can avoid these results by weakening the properties of your system, as with the quaternions whose multiplication is non-commutative, but in that case you've got a lot more work to do to complete your definition and work out what you really want to be going on, and in what ways you want your new numbers to be similar to or different from the existing systems.

$\endgroup$
  • $\begingroup$ is this in some way related with complex manifolds ? $\endgroup$ – Nick Apr 14 '14 at 12:08
  • $\begingroup$ Everything is related to everything else, right? But I can't think of any very enlightening connections here. $\endgroup$ – Ben Millwood Apr 15 '14 at 22:23
  • $\begingroup$ hopefully, that's in my eyes the beauty of math. But no problem, it was worth the shot :). $\endgroup$ – Nick Apr 17 '14 at 14:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.