For me Newton, De Moivre and Cotes are the most significant triple in modern mathematical science.
N(x,y,z) = xñ + yč + z$
Where n is the unity in honour of Newton
Č is the Cotes root of unity and $ is ths De Moivre root of unity
Č$ = -ñ, č*-ñ= -č and $*-ñ =-$, č2=$ , $2=-č
N2(x,y,z)= x2ñ – z2č +y2$ + 2*( -yzñ + xyč+ xz$)
It is clear that these are not quaternions, but they demonstrate that quaternions are based on the roots of unity modulo 8, these of course being based on roots of unity modulo 6.
However I awake this morning and realise the quaternion 8 group is not commutative. In Normans introduction to algebraic topology he overviews commutative groups, and implies the direct sum of 2 cyclic groups is commutative under addition. This means that within a cyclic group the logarithmic addition commutes with the underlying factorisation in tha case of the roots of unity, but, this is not the case in a direct sum of cyclic groups based on the roots of unity. The logarithmic commutativity does not cross the divide between the two groups. Thus when we identify isomorphic and homeomorphic groups, the distinction jumps out. The direct sum of cyclic 2 and cyclic 4 roots is not isomorphic to the cyclic 8 group of the roots of unity. It may be homeomorphic.
The implication is that we have several cyclic 8 groups which can be bases for the quaternion space, and we may have to explore them all to see which models our reality the best. Then again, some believe the encompassing Clifford algebras are a better model. I do not know enough yet about them to possibly comment, but I suspect finite or even infinite algebras which are improperly distinguished by such small details such as factorisation structure and logarithmic structure are going to be equally misleading.
A model is a model in the end, and we need to enjoy our rich life experiences unencumbered by any ideological constraint, where we can. The scientist/ empiricist cannot entirely exclude the possibility of gods, nether can the believer entirely exclude the possibility of resident systems of motion acting automatically. That both sides frequently do just that is the litmus test of dogmatic belief systems which are homeomorphic to the same degree! There is always a middle way.
The division operation in triples is involved, but my attempt goes as follows:
Originally posted by author:
For me Newton, De Moivre and Cotes are the most significant triple in modern mathematical science.
N(x,y,z) = xñ + yč + z$
Where n is the unity in honour of Newton
Č is the Cotes root of unity and $ is the De Moivre root of unity
Č$ = -n, č*-n= -č and $*-n =-$, č2=$ , $2=-čN2(x,y,z)= x2ñ – z2č +y2$ + 2*( -yzñ + xyč+ xz$)
It is clear that these are not quaternions, but they demonstrate that quaternions are based on the roots of unity modulo 8, these being based on roots of unity modulo 6.
As my computer is down at the moment I wonder if anyone would be so kind as to code this up and render a mandelbulb for me. Just replace ñ,ç,$ by 1,i,j using the given transform.
The formula for higher powers I will discuss later.
For the theoretical background and a bit of historical blather see my blog
Modulo 6 for the roots of unity mean the circle is divided into 6 sectors. Thus i can enscribe in one sector an equilateral triangle and that enables me to read off the root of unity.
Actually i have a desktop image that gives me them anyway.
č=(1/2,√3/2)=1/2+i√3/2
conj(č)=(1/2,-√3/2)=1/2-i√3/2=-$
The division in analogy with the dual complex notation is going to be based on the factorisation of x3+y3+z3
The corresponding factorisation is going to be more involved, more polynomial.
let r=(y3+z3)1/3
x3+y3+z3=(x+r)(x–čr)(x–conj(č)r)
Compare this pattern with the pattern for x2+y2
x2+y2=(x+iy)(x+conj(i)y)
Thus we can expect the same pattern when we factorise r
r=(y3+z3)1/3={(y+z)(y–čz)(y–conj(č)z)}1/3
finally, in terms of the axes labels or vectors or even Strecken
r={(y+ñz)(y–čz)(y+$z)}1/3
x3+y3+z3=(x+ñr)(x–čr)(x+$r)
All coefficients are labeled, but i have chosen not to label the leading coefficient in each bracket with ñ to provide a distinctive mnemonic
Applying these factors appropriately should estabish the denominator of a division as a real value, i just need to check what state it leaves the numerator in.
