Is my paper on a number system that allows arithmetic on 3D vectors useful?

I have constructed a number system similar to the quaternions, but with three dimensions, not four, ie vectors of the form $(x, y, z)$. It has fairly well-behaved multiplication and division and every non-zero element has an inverse. My algebra is not commutative, not associative and it only obeys the distributive law on one side.

Where can I publish my paper on this number system?

My corrected paper is here: http://soler7.com/IFAQ/Ternions.doc

Definition of vector multiplication If $v$ and $q$ are two vectors, $v = (a, b, c)$ and $q = (x, y, z)$, then their product, $vq = (a, b, c)(x, y, z)$ is defined to be

$$ (ax - by - cz, b|q| + yw, c|q| + zw ), $$

where $|q| = \sqrt{x^2 + y^2 + z^2}$ and

$$ w = a + \frac{(by + cz)(x - |q|)}{(y^2 + z^2)} $$

provided $y$ or $z$ is non-zero. If $y$ and $z$ are both zero then $vq = (xa, xb, xc)$.

The reciprocal of the vector is given by

$$ (x, y, z)^{-1} = |q|^{-2}(x, -y, -z). $$

Proof that this acts as the reciprocal of $q$ on the right.

$$ qq^{-1} = (x, y, z)|q|^{-2}(x, -y, -z) = |q|^{-2}(x, y, z)(x, -y, -z) $$

using the definition of multiplication, i.e.

$$ (a, b, c)(x, y, z) = (ax - by - cz, b|q| + yw, c|q| + zw), $$

we obtain

\begin{align*} qq^{-1} &= |q|^{-2} \left( \begin{gathered} x^2 + y^2 + z^2, \\ y|q| - y\Bigl(x + \frac{(-y^2 -z^2)(x - |q|)}{y^2 + z^2}\Bigr), \\ z|q| - z\Bigl(x + \frac{(-y^2 -z^2)(x - |q|)}{y^2 + z^2}\Bigr) \end{gathered} \right) \\ &= |q|^{-2} \bigl(x^2 + y^2 + z^2,\ y|q| - yx + y(x - |q|),\ z|q| - zx + z(x - |q|)\bigr) \\ &= |q|^{-2}(x^2 + y^2 + z^2, 0, 0) \\ &= (1, 0, 0) \end{align*}

as required.

Proof that that $|q|^{-2}(x, -y, -z)$ acts as the reciprocal of $q$ on the left.

$$ q^{-1}q = |q|^{-2}(x, -y, -z)(x, y, z). $$

Using the definition of multiplication:

\begin{align*} q^{-1}q &= |q|^{-2}\left( \begin{gathered} x^2 + y^2 + z^2, \\ - y|q| + y\Bigl(x + \frac{(- y^2 - z^2)(x - |q|)}{y^2 + z^2}\Bigr), \\ - z|q| + z\Bigl(x + \frac{(- y^2 - z^2)(x - |q|)}{y^2 + z^2}\Bigr) \end{gathered} \right) \\ &= |q|^{-2} \bigl(x^2 + y^2 + z^2, -y|q| + yx - y(x - |q|), -z|q| + zx - z(x - |q|)\bigr) \\ &= (1, 0, 0) \end{align*}

as required.

So the vector $|q|^{-2}(x, -y, -z)$ behaves correctly as the reciprocal of $(x, y, z)$ on both the left and the right. All non-zero vectors have reciprocals.

If $y = z = 0$ but $x$ is non-zero, then the reciprocal of $(x, 0, 0)$ is simply $(1/x, 0, 0)$.

Although I am saddened to see that my number system is not associative, I am very grateful to Arctic Tern for pointing this out. The beauty and terror of mathematics is that when you are right, no-one can contradict you, and when you are wrong there is no come-back from that either. All I can do is acknowledge the correctness of everything that Arctic Tern has written.

Incidentally, I am very impressed by the level of helpfulness and the positive attitude that is the case here on stackexchange. It is a lovely contrast to the acrimony and pettiness that is so common on other parts of the Net. In particular, I am grateful that people have overlooked my clumsy notation. In the future I promise to use Latex!

Geometry may be basic to mathematics and to mathematical intuition, but unfortunately it is my Achilles heel. I had to laboriously work out the products of (1,0,0), (0,1,0) and (0,0,1) to see that (aa)b is not equal to a(ab), just as Arctic Tern asserted.

I have to withdraw my comment about wikipedia, as my system is not a near-field. Maybe it is a near-near-field?

I have one lingering question, namely where is the error in my faulty proof of associativity? It goes like this:

"The associative law holds for ternion multiplication because, apart from a modulus adjustment, which is clearly associative, it does nothing other than a three-dimensional rotation. Three-dimensional rotations form a group, which means that they are associative."

I agree that someone else has probably discovered the same number system. I have looked, but was unable to find it.

Also, what did Arctic Tern mean by "with only partially retained versions" in the 5th paragraph under "some more comments"?

Some unsolicited advice. A complicated formula makes for an undesirably opaque definition, even if it is useful for performing calculations. Instead, I'd recommend going with a geometric description of the multiplication operation.

Fix a unit vector $e$. Then $vq$ is the rotation of $v$ around the axis perpendicular to $e$ and $q$ by the angle from $e$ to $q$ and also scaled by $q$'s size, unless $q$ is a scalar multiple of $e$ in which case $vq$ is that scalar times $v$.

This makes a number of properties more immediately accessible:

• $e$ is both a left and right identity
• $vq$ is linear in $v$ but shouldn't be linear in $q$
• if a plane contains $e,a,b$ then $(va)b=(vb)a$
• ($\Rightarrow$) the two-sided inverse of $q$ is its reflection across $e$
• $vq$ is discontinuous where $q$ is at negative multiples of $e$
• multiplicativity of vector norm, $\|vq\|=\|v\|\|q\|$
• no zero divisors, $ab=0~\Rightarrow~(a=0\textrm{ or }b=0)$

Indeed, all of the above can be mentally visualized with the geometric definition.

Claiming that Wikipedia is "mistaken" for saying there is no 3D multiplication with inverses is at best uncharitable: Wikipedia of course is talking about multiplication operations that are associative or at least distributive on both sides, unlike a near field. Although your algebraic structure isn't even a near field!

In fact ternion multiplication is not associative. A ternion product $(ab)c$ first rotates $a$ along the $e$-$b$-plane (the oriented plane spanned by $e$ and $b$) and then rotates along the $e$-$c$-plane, whereas $a(bc)$ first rotates $b$ along the $e$-$c$-plane and then rotates $a$ along the $e$-$bc$-plane. If you're looking at a picture of this, you should see no reason for these two procedures to give the same result.

Counterexample. Let $\{e,a,b\}$ be an oriented orthonormal basis. Observe $ab=a$ and $ba=b$ follows from basic geometry. But then

$$\begin{array}{ccccc} (aa)b & = & (-e)b & = & -b \\ a(ab) & = & aa & =& -e \end{array}$$

All is not lost: this algebraic structure is still power-associative. Indeed, if $v$ is any vector not parallel to $e$, then the multiplication operation restricted to $\mathrm{span}\{e,v\}$ makes it an algebraic structure isomorphic to the complex numbers $\mathbb{C}$ (with $v$ being identified with $e^{i\theta}$). This is enough to see that it is power-associative.

Moreover, there is a partial associative property: it is almost right-alternative. Stating this as generally as possible for this number system, that means if $e,a,b$ all lie on the same plane and $ab\ne -e$, then $(va)b=v(ab)$. Note $ab=-e$ occurs when $a,b$ point in the same direction (one is a positive real scaling of the other) and they are perpendicular to $e$ and have reciprocal lengths.

In my counterexample above, it may be tempting to use alternativity to cancel $a$ from both sides of $ab=a$. We can left-multiply by $a^{-1}$ to get $a^{-1}(ab)=e$ and $e,a,a^{-1}$ all lie in a plane after all. But notice we have the almost right alternative property, not a left one.

Some more comments:

I am put off by the swath of tedious calculations and formulas. I only skimmed one or two of them!

One, you can replace many of the calculations with geometric arguments (as I've hopefully demonstrated a bit) to save the reader time and energy.

Two, if a formula is not actually useful to have memorized or written down (maybe because the calculation it involves is obvious to do anyway) then it strikes me as busywork that doesn't contribute anything. So the formulas better actually be useful for something.

It seems you're interested in using this algebraic structure to create some new types of fractals, in particular ones involving recursion and powers. So it may make sense to write down formulas for powers if you code them into some program you write, say.

The multiplication operation is (geometrically) natural enough that I am sure it has been discovered independently many times, although it is unnatural enough (failing associativity, commutativity and distributivity with only partially retained versions, and also discontinuous) to be superseded in utility by matrices and vectors or just quaternions for most applications. (There are fractals generated with quaternions, after all.)

Unfortunately I have little idea of what to search for to find the previous discoveries of this structure. Maybe try looking up past attempts at creating 3D number systems. I recommend trying to find them to the best of your ability, though - ask around (like you're doing here) - and if you still can't find it in any literature then at least say in the paper that you looked really hard.

Since your interest in ternions is for fractal generation, you might as well make some of your paper about actually using them to generate fractals! Then you can find somewhere to publish that is interested in fractals or other recreational mathematics. Other possible types of places to publish include computer graphics and modern algebra, but algebraists would not take interest in a paper about fractals.

Fractals are a kind of recreational mathematics than amateurs may enjoy playing with, but you don't want to be like them and speak with too much "poetic license" lest you set off alarm bells that you're a crank not worth reading. You write well and seem to be within bounds on that mark, but I get the impression that it's a close call at points. (Authorities in a field are granted poetic license by its community, but not outsiders or newcomers.)

Finally, you may want to invest in learning $\LaTeX$. It is the standard for publishing in the sciences, especially in mathematics. Writing a paper using Microsoft Word is not something that mathematicians do, and potential readers will be conscious of that.

Response to edit:

I had to laboriously work out the products of (1,0,0), (0,1,0) and (0,0,1) to see that (aa)b is not equal to a(ab), just as Arctic Tern asserted.

Well, let's try to visualize it. Here is a sphere:

The product $ab$ means $a$ rotated around the $a$-axis... but $a$ is on the $a$-axis, so it doesn't get rotated at all! So $ab=a$. On the other hand, $aa$ represents $a$ rotates around the $b$-axis, which yields $-e$.

See if you can use this picture to visualize the bullet points I gave above.

Also, what did Arctic Tern mean by "with only partially retained versions" in the 5th paragraph under "some more comments"?

The distributive property is partially retained, since $vq$ is linear in $v$ (although not $q$). Meaning $(\lambda v)q=\lambda(vq)$ for scalars $\lambda$ and $(v_1+v_2)q=v_1q+v_2q$ are identities. And the associative property is partially retained with the "almost right alternative" property I mentioned above (which, again, is a good geometry exercise).

I have one lingering question, namely where is the error in my faulty proof of associativity? It goes like this:

"The associative law holds for ternion multiplication because, apart from a modulus adjustment, which is clearly associative, it does nothing other than a three-dimensional rotation. Three-dimensional rotations form a group, which means that they are associative."

The multiplication operation on your number system is not the composition of two rotations though, it is a rotation designated by one vector applied to another vector. Your argument is like saying matrix multiplication is commutative because it uses numbers and numbers satisfy the commutative property.

We can ponder this a little deeper. Consider the unit sphere $S^2$ as an algebraic structure with your multiplication operation. For each unit vector $q$, there is a rotation $R_q$ that rotates around the axis perpendicular to $e$ and $q$ by the angle from $e$ to $q$. Then $vq=R_q(v)$ (except $R_{-e}$ is multiplication by $-1$, i.e. reflection across the origin). Notice this is a single rotation $R_q$ designated by a vector $q$ applied to a vector $v$, it is not a composition of two rotations.

The lack of continuity of $vq$ at $q=-e$ illustrates an interesting mathematical fact.

The group of 3D rotations is called the special orthogonal group $\mathrm{SO}(3)$, and it acts on the sphere. Right away notice the sphere $S^2$ is $2$-dimensional, whereas $\mathrm{SO}(3)$ is $3$-dimensional (it takes three parameters to specify a rotation: an axis picked from a $2$-sphere's worth of directions to choose from, plus also an angle). In fact, $\mathrm{SO}(3)$ happens to be a three-dimensional sphere sitting inside four-dimensional space (if we pretend antipodal points are the same)!

The idea of "orbits" from the theory of group actions suggests we should investigate the way $\mathrm{SO}(3)$ can "wrap" around the sphere $S^2$. Indeed, fix a unit vector $e$, then we can associate any rotation $R$ to the vector $Re$ obtained by applying $R$ to $e$. This gives a continuous map $\mathrm{SO}(3)\to S^2$.

What are the fibers? In other words, which rotations $R$ map $e$ to a given vector $v$? The great circle (equator) which bisects the arc from $e$ to $v$ (or any of the arcs if $v=-e$) is all possible axes we can rotate around to rotate $e$ to $v$.

Your way of associating rotations $R_q$ to vectors $q$ is a partial inverse $S^2\to\mathrm{SO}(3)$ of this wrapping-around map $\mathrm{SO}(3)\to S^2$, called a left-inverse (I'd call it a pre-inverse) or a "section." This is because $R_q(e)=q$, so applying the functions to $q$ in order gives $q\mapsto R_q\mapsto q$.

However, $R$ fails to be continuous at $-e$. Your multiplication operation is an illustration of the failure of $\mathrm{SO}(3)\to S^2$ to have a continuous section: you can put all of $S^2$ "back" inside $\mathrm{SO}(3)$ starting from $e$, except until you get to the last point $-e$ on the other side you can't tie the bow nicely. In order for $R_{-e}$ to follow the pattern of all the other $R_q$'s it'd have to be a $180^\circ$ rotation around an axis perpendicular to $e$, but there's no canonical choice of which axis to use to achieve that effect (unlike for every other $q$ not parallel to $e$).

The image (range) of the section by the way (meaning the set of all rotations of the form $R_q$) will be all rotations around axes contained in the plane perpendicular to $e$ (notice my $a$ and $b$ on the sphere picture above are in the plane perpendicular to $e$). Except of course for $180^\circ$ rotations, which fail to be present.