14
$\begingroup$

The set of Complex numbers can be used in lots of domains like geometry, vectorial calculations, solving equation with no real solution etc. But what are the uses of split-complex number that can't be done with complex numbers? I think you could do the same works in geometry or vectorial calculation in a "split-complex" plane but what advantages gives us to know that j is a solution of the equation $x^2=1$?

What I've thought so far is that using complex numbers and split-complex numbers together, we can have numbers of the form $a+bi+cj$ so everything that can be done in the complex-plan could be extended to 3 dimensional space by adding a split-complex part.

$\endgroup$
10
  • 4
    $\begingroup$ Sadly, your idea doesn't work; there are no three-dimensional numbers analogous to the two-dimensional complex numbers. You'll get into trouble when you try to decide what $i\cdot j$ is. William Hamilton famously struggled for many years to try to find a way to make it work before discovering the (four-dimensional) quaternions. $\endgroup$
    – MJD
    Apr 25, 2013 at 18:58
  • $\begingroup$ If j is a solution to x²=1, then j is real, so what does cj really add as an extra dimension? $\endgroup$
    – imranfat
    Apr 25, 2013 at 19:00
  • 1
    $\begingroup$ @imranfat No, here j stands for the elementary unit of split-complex numbers, j≠1 and j≠-1 so j is not real. $\endgroup$
    – moray95
    Apr 25, 2013 at 19:02
  • $\begingroup$ @MJD If you think that multiplying by i is a rotation of π/2, then $ji$ should be i or -1. Using matrix representations of i and j, ij= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. Which is of course hard to define as a number. $\endgroup$
    – moray95
    Apr 25, 2013 at 19:06
  • $\begingroup$ It seems like you want $i=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and $j=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. But if you do that, then $i+j=\begin{pmatrix}0&2\\0&0\end{pmatrix}$ and then you get $i+j\ne 0$ but $(i+j)^2 = 0$, which isn't very much like numbers any more. $\endgroup$
    – MJD
    Apr 25, 2013 at 20:18

2 Answers 2

Reset to default
7
$\begingroup$

I think the known uses of split-complex numbers are probably going to be addressed by the Wiki page which MJD linked in the comments above, and other "fan pages" on the internet. So, I wanted to address this question in the post:

But what are the uses of split-complex number that can't be done with complex numbers?

In Clifford algebra (or geometric algebra, as called by a small segment of the population that uses them) these two algebras are used to encode the geometry of $\Bbb R$ under two different geometries.

The long story short is that a bilinear form gives rise to a geometry on a vector space. The "signature" of a real bilinear form determines its basic character, and since there are lots of forms with different signatures, you get different geometries.

The complex numbers study $\Bbb R$ with a bilinear form $B(x,y)=-xy$.

For the split-complex numbers, the bilinear form on $\Bbb R$ is just $B(x,y)=xy$.

The quaternions study $\Bbb R\oplus \Bbb R$ with the bilinear form $B((x_1,x_2),(y_1,y_2))=x_1y_1-x_2y_2$.

$\endgroup$
1
  • $\begingroup$ I think the quaternions' bilinear form should be negative-definite, $-x_1y_1-x_2y_2$. $\endgroup$
    – mr_e_man
    Feb 4, 2020 at 22:59
1
$\begingroup$

What I've thought so far is that using complex numbers and split-complex numbers together

You will get exactly tessarines, a commutative associative algebra. But it is 4-dimensional: $ij$ is a separate unit, similar to complex unit. $i^2=-1$, $j^2=1$, $(ij)^2=-1$.

So, your number will look $a+bi+cj+dij$ or $(a+bi)+j(c+di)$ or $(a+bj)+i(c+dj)$.

$\endgroup$
2
  • $\begingroup$ So basciazlly tessarines are a union of complex numbers and split complex numbers. As quaternions are another superset of complex numbers, complex numbers are the intersection of tessarines and quaternions. Can tessarines and quaternions be added and subtracted? $\endgroup$
    – Michael Ejercito
    Mar 8 at 8:48
  • 1
    $\begingroup$ @MichaelEjercito you can add a split-complex unity to quaternions and get a 8-dimensional algebra, for sure. It will not be commutative because quaternions are not commutative. It will include tessarines as a subset. $\endgroup$
    – Anixx
    Mar 8 at 9:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .