Question

\[
\newcommand{j}{\mathbf{\hat{j}}}
\newcommand{k}{\mathbf{\hat{k}}}
\newcommand{jk}[3]{
#1
+
#2
\j
+
#3
\k
}
\]Let \(\j\) and \(\k\) be mathematical entities such that: \[
\j^2=\k\\
\k^2=-\j\\
\j\k =\k\j = -1
\]You can generate linear combinations of \(1,\j,\k\):\[
\jk abc
\]I'm wondering how you would find real valued functions \(f,g,h\) such that: \[
\frac{1}{\jk abc} = \jk{f(a,b,c)}{g(a,b,c)}{h(a,b,c)}
\]I have some ideas of how to do it, but it's late so I won't explore them tonight, so I'm leaving this here for others to explore.

Here is an example of multiplication: \[
(\jk abc)(\jk xyz) \\~~~~~~~= \jk{(ax-bz-cy)}{(ay+bx-cz)}{(az+by+cx)}
\]

Answer 1

I've found a solution, but I want to see what others come up with.

Answer 2

The solution I came up with was: \[
\frac{1}{\jk abc}=\frac{\jk{(a^2+bc)}{(-c^2-ab)}{(b^2-ac)}}{a^2-b^2+c^3+3abc}
\]Sort of similar to how: \[
\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}
\]But I still wanted to explore things with this a bit.

Answer 3

Not exactly a solution, but definitely related is if you want to divide a vector in geometric algebra in a n orthonormal basis then you have: \[\Large V =\frac{1}{a \hat i + b \hat j + c \hat k}\] and then multiply both top and bottom by the same vector, since it's the same as multiplying by 1. \[\Large V =\frac{a \hat i + b \hat j + c \hat k}{(a \hat i + b \hat j + c \hat k)(a \hat i + b \hat j + c \hat k)}\] So let's just look at the bottom and see that it evaluates to a scalar: \[\Large (a \hat i + b \hat j + c \hat k)(a \hat i + b \hat j + c \hat k) \\a^2 \hat i^2 + b^2 \hat j^2 + c^2 \hat k^2 + ab \hat i \hat j + ab \hat j \hat i +... \\ \Large a^2+b^2+c^2\] The key thing is that in clifford algebra a vector squared is its length, so when we get down to the unit vectors like i, j, and k squared, they were all 1. In addition to this, it's anti commutative so ij=-ji and the other terms all cancel each other out.

Why did I write all this out? Cause I think we can probably tweak the rules of clifford algebra to solve your problem.

Another idea I am going to consider working on is making some kind of "extended complex conjugate".

Answer 4

Interesting properties is \[
\j^3=-1\\
\k^3=1
\]In this system, there are no complex numbers, and I'm not sure how many solutions there are to \(z=x^n\), but I think \(z=x^3\) would have two solutions, given how \(\sqrt[3]{1}=1\) and \(\sqrt[3]{1}=\k\).

Answer 5

Another interpretation I came up with is to treat multiplications by \(\j\) and \(\k\) as rotations. It works on this circle at least.

Answer 6

Which would imply these numbers could be converted to complex numbers.

However, I don't think you could go back, because it doesn't seem like a bijective sort of conversion.

Answer 7

Want to know how I came to my solution?

Answer 8

Yeah that would help a lot actually haha.

Answer 9

It brought out a lot of insight.

Answer 10

Okay, so first what I did was multiply by each entity: \[
(\jk abc) \cdot 1 = \jk abc\\
(\jk abc)\cdot \j = \jk{-c}ab\\
(\jk abc)\cdot k=\jk{-b}{-c}a
\]And I converted this into a matrix:\[
\begin{bmatrix}
a&b&c\\
-c&a&b\\
-b&-c&a
\end{bmatrix}
\begin{bmatrix}1\\\j\\\k\end{bmatrix}
\]I did the same for \(1\):\[
1\cdot 1=1\\
1\cdot \j=\j\\
1\cdot k=\k
\]Gives: \[
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\begin{bmatrix}1\\\j\\\k\end{bmatrix}
\]

Answer 11

I interpreted it as a matrix equation: \[
A^{-1}A=I
\]Then I found \(A^{-1}\), and took the top row.

Answer 12

This reminds me of something , maybe it will be useful to us:\[\Large \omega = e^{i \frac{2\pi}{3}}\] Then I made the analogy with how every function is the sum of an even and odd function: \[\Large f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}\] So we can see that the first one is obviously even and the second one is odd: \[\Large g(x)=\frac{f(x)+f(-x)}{2} =\frac{f(-x)+f(x)}{2} = g(-x)\]\[\Large h(-x)=\frac{f(-x)-f(x)}{2} =\frac{-f(x)+f(-x)}{2} = -h(x)\] Sp we can write \[\Large f(x)=g(x)+h(x)\] Similarly, using this omega, we can actually represent every function as the sum of 3 differently symmetric functions, but instead of even/odd it's more like 3n, 3n+1, and 3n+2. \[\Large f(x)=g(x)+h(x)+k(x)\] \[\Large g(x) = \frac{f(x)+ f(\omega x)+ f(\omega^2x)}{3}\]\[\Large h(x) = \frac{f(x)+ \omega f(\omega x)+ \omega^2 f(\omega^2x)}{3}\]\[\Large k(x) = \frac{f(x)+ \omega^2 f(\omega x)+ \omega f(\omega^2x)}{3}\]

It might seem arbitrary, but it really does follow the same pattern, for instance the derivative of an even function is odd, and then the derivative of an odd is even. g'(x) is a h(x) type function, h'(x) is a k type, and k' is a g function. I discovered this on my own, so I don't know if there is better terminology for them unfortunately.

Also the new even/odd sort of analogy relationshis are: \[\Large g(x) = g(\omega x)= g(\omega^2x) \\ \Large h(x) = \omega h(\omega x ) = \omega^2 h(\omega^2 x) \\ \Large k(x) = \omega^2 k(\omega x) = \omega k(\omega^2x)\]

Answer 13

Wow that's pretty awesome! I'll play with what you have and throw my stuff away for now I think lol.

Answer 14

The next thing I did was I looked at how you would express \(1\), \(\j\) and \(\k\) in this matrix form.

Answer 15

Obviously \(1=I\)

Answer 16

And \[
\j = \begin{bmatrix}
0&1&0\\
0&0&1\\
-1&0&0
\end{bmatrix}\\
\k = \begin{bmatrix}
0&0&1\\
-1&0&0\\
0&-1&0
\end{bmatrix}
\]

Answer 17

And when I multiplied these matrices together, they obeyed the rules I had made up.

Answer 18

But there are many things I'm really confused and curious about.

Answer 19

For example, what does the determinate represent here?

Answer 20

Ahh yeah this is related to what I'm talking about after all.

Answer 21

\[
\begin{vmatrix}a&b\\-b&a\end{vmatrix} = a^2+b^2 = zz^* = |z|^2
\]But on the other hand: \[
\begin{vmatrix}a&b&c\\-c&a&b\\-b&-c&a\end{vmatrix}=a^2+3abc-b^2+c^3=?
\]

Answer 22

You can fully represent your algebra with complex numbers: \[\Large \hat j = -\omega^2 \\ \Large \hat k = \omega\]

Answer 23

A fun relationship is that \[\Large \omega ^* = \omega^2\] So similarly this kind of reflection structure should also exist with j and k in your algebra.

Answer 24

Is there a one to one correspondence?

Answer 25

I was thinking about it, because it's a bit strange. I know how to "project" down onto the complex plain, but if I can't go back, then I'm throwing data away, aren't I?

But if I could go back, it would seem to imply that \(\k\) is redundant.

Answer 26

Nope, it's exactly 1 to 1 here, nothing lost in thinking in your way or the other way. They obey the exact same rules.

Answer 27

@wio I think it is redundant in a sense, however it might not necessarily be redundant. I have a hunch that you're on to something, I mean if there are an infinite number of numbers between 0 and 2pi then maybe we can squeeze out 3 times as many numbers rather than 2 times as many like we should? Maybe not.

Answer 28

what did u do exactly http://prntscr.com/5rvgxa

Answer 29

For instance, if you look at the graph of tangent from -pi/2 to pi/2 you contain all the way from - to + infinity. Just scale the domain slightly and you can get two separate trips on that same range squeezed into -pi/2 to pi/2. Here: http://www.wolframalpha.com/input/?i=tan%28x%29+and+tan%28x%2F3%29

Answer 30

I understand that \(|\mathbb R| = |\mathbb C| = |\mathbb R^3|\), it's more of something else.

Answer 31

Okay, well if that is the case, then I would like to know: \[
\jk abc = f(a,b,c)+g(a,b,c)i
\]and even more so: \[
a+bi = \jk{f(a,b)}{g(a,b)}{h(a,b)}
\]

Answer 32

I know we can use\[
\j= \cos(\pi/3)+i\sin(\pi/3)\\
\k=\cos(2\pi/3)+i\sin(2\pi/3)
\]Based on the picture I gave above.

Answer 33

Just plug this in, \[\Large \omega = \frac{-1}{2} + i \frac{\sqrt{3}}{2}\\\Large \hat j = -\omega^2 \\ \Large \hat k = \omega\]

Answer 34

Which implies \[
\jk abc = \bigg( a+b\cos(\pi/3)+c\cos(\pi/3) \bigg)+ \bigg(b\cos(2\pi/3)+c\cos(2\pi/3) \bigg)i
\]I think.

Answer 35

What about the second one?

Answer 36

Wouldn't there be many solutions?

Answer 37

I think it is all contained in just plugging in these to convert around between the three separate representations.

Answer 38

Oh you're right, I see, there would be many solutions but I don't think any of them would be any more meaningful than saying 1/2=2/4=3/6=... etc.

Answer 39

Or a better way to describe it would be like saying 2=3-1=4-2=5-3=... etc it's essentially because you're using 3 vectors when you only need 2 to span the space.

Answer 40

Are you sure?
The reason why I am interested in this is because it defined a type of multiplication for \(\mathbb R^3\) vectors.

Answer 41

Well, one that had the nice property of commutattivity.

Answer 42

Before you were talking about this:
http://en.wikipedia.org/wiki/Quaternion

Would you say they are aren't any different than complex numbers?

Answer 43

Quaternions are totally different because they don't commute, ij=-ji for instance.

Answer 44

Yeah, I was purposefully trying to avoid anti-symmetry, but I don't see why that matters so much.

Answer 45

I wanted to define a modulus/magnitude operation for these numbers. First thing that came to mind was \(\sqrt{a^2+b^2+c^2}\).
I am going to experiment a bit though.

Answer 46

I don't know why it matters so much, I hate "commutative" and "associative" because there's something incredibly fake about them since they seem to be purely artifacts of how we write the math down and no geometric significance at all. I think for some reason it just ends up working, I think I heard that octonions not only aren't commutative, they aren't associative either, which sounds terrible. Maybe we just need a new way to write math so it's not all in a line idk

Answer 47

Maybe anticommutative is geometric, I take that back. It makes sense that doing two things in opposite orders can give you opposite rotation.