The Peculiar Math That Could Underlie the Laws of Nature

Mike Hewson
Mike Hewson
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The Full Gore Now suppose

The Full Gore Now suppose that p = (a, b, c, d) and q = (e, f, g, h) are arbitrary quaternions. Then

p * q = (a, b, c, d) * (e, f, g, h) = (a + bi + cj + dk) * (e + fi + gj + hk)

= a * (e + fi + gj + hk) + bi * (e + fi + gj + hk) + cj * (e + fi + gj + hk) + dk * (e + fi + gj + hk)

= ae + afi + agj + ahk + bei + bfi2 + bgij + bhik + cej + cfji + cgj2 + chjk + dek + dfki + dgkj + dhk2

= (ae - bf - cg - dh) + (af + be)i + (ag + ce)j + (ah + de)k + (bg - cf)ij + (ch  - dg)jk + (bh - df)ik

Now consider the 3D vectors p' = bi + cj + dk and q' = fi + gj + hk then

p'.q' = bf + cg + dh

aq' = afi + agj + ahk 

ep' = ebi + ecj + edk

p' X q' = (bg - cf)ij + (ch  - dg)jk + (bh - df)ik

giving as the product

p * q  = ae - p'.q' + aq' + ep' + p' X q'

. is the dot product of 3D vectors, X is the cross product of 3D vectors.

This suggest the following strategy. Set ( as a notation ) p = (a, p') and q = (e, q') where we denote a and e as the 'scalar parts', with p' and q' the 'vector parts'. Then the entire product of the two quaternions becomes :

(a, p')*(e, q') = ae - p'.q' + aq' + ep' + p' X q' = (ae - p'.q', aq' + ep' + p' X q')

The scalar part of the product is (ae - p'.q') and its' vector part is (aq' + ep' + p' X q'). The processing/calculation of a quaternion product works out all right in this 'mode'. What space does p', q' and p' X q' span or 'cover' ? Provided neither p' nor q' are zero, or are (anti-)parallel, then the entire 3D subset of 4D quaternion space !

If you overlook the presence of the imaginary units i, j and k then the vector parts of quaternions can represent 3D vectors in ordinary R3. They map one to one and have the same geometric behaviour. We can use that to represent rotations in R3, nominating an angle to rotate through and the vector part an axis to rotate around. Which we will look at next time.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Mike Hewson
Mike Hewson
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Associative Division Ring

Associative Division Ring Over The Reals William Rowan Hamilton must have been great company ! While walking with his wife he had a flash of insight about 3D vectors* and just had to write it down, so he scratched the relevant properties ( that we have discussed ) onto a bridge that they were passing by. Such is genius, it strikes when it can. The problem he was wanting to solve was the division of 3D vectors, in which he failed, but he opened up an area of math that we know as quaternions. He called them 'tensors', which they are not as per modern definitions. He also did much work in the physics of mechanics and that even later penetrated to quantum mechanics where the energy operator is known as the Hamiltonian. 

In any event it turns out there are only three cases of 'associative division ring over the reals'. Associative means that the grouping of terms in an algebraic expression is irrelevant eg.

p(qr) = (pq)r

{ but the order of the variables : p then q then r left to right may be important }

A ring is composed of generic mathematical rules over a set of objects ( the usual arithmetic stuff ). The reals are the real numbers. The three cases are 1,2 and 4 dimensions ie. the reals number line, the complex plane and the 4D space of quaternions. Octonions ( which we will get to one day ) are not associative.

We discovered that generally :

if q = (p0q') then q-1 = (p0, -q')/||q||2

using the scalar & vector representation. Here's how to do a rotation about some 3D axis which we will denote as a for 'axis' ( let that not be the zero vector ). First you need a unit vector in the direction from the tail of the axis vector toward the head, call it u,

u = a/||a||

then decide upon what angle you would like to rotate, call it THETA ( right handed sense of the 'advancing screw' toward the vector's head ). I'll just give you the formula for the quaternion that represents a rotation - the derivation is straight forward but tedious trigonometry :

p = cos(THETA/2) + u * sin(THETA/2)

giving the quaternion conjugate as :

p* = cos(THETA/2) - u * sin(THETA/2)

{ NB cos2(some_angle) + sin2(some_angle) = 1 }

Now given some vertex/point that you want to rotate in 3D space, make a vector to it and call that v', and compose a quaternion from it which is simply :

v = (0, v')

Then ( again avoiding the derivation ) the image of the point in quaternion form is simply p v p* ie. we perform quaternion multiplication. When finished, or finished with a sequence of quaternion multiplications for chained rotations for that matter, we recover the desired result by converting the quaternion back to a 3D vector.

It turns out that a system optimised for quaternion multiplication beats a matrix approach by a significant margin, due to fewer overall operations. It is the favoured approach for alot of software and avoids other problems. There is never an issue with coordinate ambiguity or gimbal lock. Some matrix approaches can suffer badly from rounding errors : a set of orthogonal axes may become not orthogonal after several transformations. Quaternions may become less accurate but the effect on the methodology isn't as drastic.

* Add a dimension !!

Cheers, Mike.

( edit ) I forgot to mention that Euler showed that any displacement transformation of a body where ( at least ) one point on the body remains fixed can be considered as a rotation about some axis ( which goes through that point, in 3D, 'rigid body' etc ).

( edit ) The given transform, p v p*, is from the point of view of the coordinate axes looking at the rotating point(s). If you want the view from the axis of rotation ie. the coordinate axes appear to rotate while the point(s) are fixed, then use p* v p. They are opposites.

( edit ) I will note too that a composition of more than one rotation in sequence can be represented as a single overall rotation.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Mike Hewson
Mike Hewson
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Octonions : I suppose by now

I hadn't forgotten, I've just been busy .....

Octonions : I suppose by now you could guess at the form of an octonion, by extension from the quaternions. Each octonion has eight parts, the first is a purely real number, while remaining seven are like i, j and k for quaternions. A common notation preference is to label the non-real parts i1 though to i7 :

a + b * i1 + c * i2 + d * i3 + e * i4 + f * i5 + g * i6 + h * i7

OR

(a, b * i1, c * i2, d * i3, e * i4, f * i5 , g * i6 , h * i7)

with a, b, c, d, e, f, g & h all real numbers. As you might also guess :

in2 = -1 for n = 1 to 7

ie. each imaginary unit is an independent square root of -1. A key point is the relationship b/w the various in units amongst themselves. After all we have an eight dimensional space, so how does it 'work'? Addition, subtraction and multiplication by a scalar all happen as expected. What about multiplication of one quaternion by another ? It turns out that certain groupings of three units act like the i, j and k of quaternions. There is a diagram that summarises this :

.... which is called the Fano plane or the Fano algebra, and I will begin to unwrap the meaning of it next time.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Mike Hewson
Mike Hewson
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For now : notice that the

For now : notice that the Fano plane/algebra/diagram has seven 'points' and seven 'lines' ( if the interior circle is considered as a line ), each line has three points and each point is at the intersection of three lines. A pleasing symmetry.

I should add that when I say the Fano plane, it is really a Fano plane as there are equivalent representations ( the e1 .... e7 relations don't change ) obtained through various symmetry operations. See if you can find all those operations. ;-)

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Mike Hewson
Mike Hewson
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Sorry for the delay : I've

Sorry for the delay : I've been busy, distracted and then ill in turn. But all good now ! ;-)

The key idea of the Fano plane is to be able to reconstruct an octonion multiplication table by reference to it. In that sense the Fano diagram is a mnemonic. That would be an 8 x 8 table. Any entries in the row or column for 1 is easy, as 1 is the unity element ( multiplying from either side ) :

Octonion Multiplication table
  1 e1 E2 E3 E4 E5 E6 e7
1 1 e1 e2 e3 e4 e5 e6 e7
e1 e1              
e2 e2              
e3 e3              
e4 e4              
e5 e5              
e6 e6              
e7 e7              

Also each ek is an independent square root of -1 :

Octonion Multiplication table
  1 e1 E2 E3 E4 E5 E6 e7
1 1 e1 e2 e3 e4 e5 e6 e7
e1 e1 -1            
e2 e2   -1          
e3 e3     -1        
e4 e4       -1      
e5 e5         -1    
e6 e6           -1  
e7 e7             -1

and that's the end of the easy ones! For the others let's look again at the Fano :

how does this help us with any general entry, say ej times ek with k <> j ? Let's try a specific example : what is e1 x e2 ? First look to find the 'line' that contains both of them : that would be the ( only ) one from the bottom right of the outer triangle ( at e2 ), going thru the centre of the triangle ( at e7 ) and onto the midpoint of the upper left side ( at e1 ). In your mind let the other lines fade away : you are left a single line which you then curve into a circle upon which e2, e7 and e1 are around with the arrow ( don't forget that arrow ! ) giving a sense to 'positive' traversal ( e2 -> e7 -> e1 -> e2 -> e7 -> e1 ... etc ) and for that matter the sense of 'negative' traversal. Clearly e1 x e2 = e7, in fact we could use this cycle to fill out e2 x e1 = - e7 too :

Octonion Multiplication table
  1 e1 E2 E3 E4 E5 E6 e7
1 1 e1 e2 e3 e4 e5 e6 e7
e1 e1 -1 e7          
e2 e2 -e7 -1          
e3 e3     -1        
e4 e4       -1      
e5 e5         -1    
e6 e6           -1  
e7 e7             -1

Keeping that e2 -> e7 -> e1 -> e2 -> e7 -> e1 cycle in mind we can expand to evaluate e1 x e7, e7 x e1, e2 x e7 and lastly e7 x e2 :

Octonion Multiplication table
  1 e1 E2 E3 E4 E5 E6 e7
1 1 e1 e2 e3 e4 e5 e6 e7
e1 e1 -1 e7         - e2
e2 e2 -e7 -1         e1
e3 e3     -1        
e4 e4       -1      
e5 e5         -1    
e6 e6           -1  
e7 e7 e2 -e1         -1

We have now six lines ( -> cycles of three ) remaining to analyse and each produces six distinct multiplication table entries. I'm gunna let you sweat to next time for the full table : you might want to try filling you the remainder yourself. Also you might by now have an inkling of why associativity doesn't work for octonion multiplication ....

Even thus far we already see some structure/pattern and even some symmetries emerging. A question comes to mind : what else in mathematics/physics has 'relationship' patterns with operations upon some set ? The answer is : groups and group theory. Which we will turn to next.

Cheers, Mike.

( edit ) This is a great picture of Gino Fano :

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Mike Hewson
Mike Hewson
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ASIDE : Now here's an

ASIDE : Now here's an absolute gem of a comment I found during my research on octonions, this being made by Paul C. Kainen ( Department of Mathematics, Georgetown University, Washington, DC ) with my emphasis :

"Of course, multiplication in the octaval arithmetic fails to be either commutative or associative, but that could be a blessing in disguise. If multiplication depends on the order of the elements being multiplied together and even on how they are grouped, then at one fell swoop, geometry enters the calculation in an organic way. The Principle of Indeterminacy could then arise in a natural fashion from relativistic considerations, making quantum theory a consequence of an underlying 8-dimensional hidden-variable process, very much in the flavor of the theories of de Broglie and Bohm. Uncertainty of measurement would be a corollary of our inability to absolutely order events or to absolutely control the way in which they are grouped."

.... and if true then cuts the QM Gordian knot to solve, for example, the melding of QM and GR. If Albert Einstein was alive to read this he would probably click his heels together and do a little dance for joy* ! :-)

If so this explains why, for instance, a set of measurements extended in time ( like taking a photograph with a long exposure time ) has apparent indeterminacy per measurement instance because each instance does not reliably have the same onset as another. Not even in principle. A sequence of photon transits is not a sequence of identically repeated experiments : the hidden variables will vary and kick off each photon's flight with sufficient variation to ensure the appearance of indeterminacy. For us the fact of being within the thing ( some high multidimensional spacetime ) we wish to measure will forever prevent us from fixing or sensing the ordering & grouping of interactions. You'd have to be god-like and exist outside the whole construct to do that. Hence the classical and QM viewpoints arise, as they should, merely as scale dependent explanations of the same basic reality.

Also, if true, it would in some gentle way explain QM 'weirdness' like entanglement. Those separated entities with correlated features in entanglement become merely facets of a single extended object and that includes any interacting measurement devices too. So if an interference pattern changes because 'which-way' information is disclosable by some setup, that is simply a different experiment than a similar setup sans 'which-way' data channel(s).

Cheers, Mike.

* In other words probability in QM becomes, like in classical mechanics, just a placeholder for our ignorance of some firm features of reality rather than a truly indeterminate state ( half-alive and half-dead cats for example ).

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Mike Hewson
Mike Hewson
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The Full Horror Octonion

The Full Horror

Octonion Multiplication table
  1 e1 E2 E3 E4 E5 E6 e7
1 1 e1 e2 e3 e4 e5 e6 e7
e1 e1 -1 e7 e5 -e6 -e3 e4 - e2
e2 e2 -e7 -1 -e6 -e5 e4 e3 e1
e3 e3 -e5 e6 -1 e7 e1 -e2 -e4
e4 e4 e6 e5 -e7 -1 -e2 -e1 e3
e5 e5 e3 -e4 e1 e2 -1 e7 -e6
e6 e6 -e4 -e3 e2 e1 -e7 -1 e5
e7 e7 e2 -e1 e4 -e3 e6 -e5 -1

..... which I do hope I have done faithfully. Can you now visualise why it is that multiplication for octonions is not associative ? Consider e1*e2*e3 : what is the result here ? Well, it is an ambiguous product because no grouping of terms is stated or implied, as follows :

firstly try (e1*e2)*e3 = (e7)*e3 = e4

but then e1*(e2* e3) = e1*(-e6) = -e4

.... and you see the trouble. Loss of associativity is not 'a little bit off' as it can be directly opposite !

Also you could/ought perceive some (anti-)symmetries in the pattern of the table entries.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

Jim1348
Jim1348
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Mike Hewson wrote:Can you now

Mike Hewson wrote:
Can you now visualise why it is that multiplication for octonions is not associative ?

I am working on it.  Fortunately, the brain is a wonderful instrument.  Even though the result does not show up in the conscious part, it may sink in to a deeper level by osmosis.

Mike Hewson
Mike Hewson
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Ah yes, osmosis. That ever so

Ah yes, osmosis. That ever so useful learning instrument. ;-)

I visualise/rationalise the associativity problem as changing to new lines/circles/cycles when one re-brackets the factors. Take for instance the product e1 * e2 * e3. These three operands are not on the same line or circle :

Now there are two choices for grouping the factors :

(A) (e1 * e2) * e3

(B) e1 * (e2 * e3)

ie. does e2 group with e1 or e3 ? The general convention is to evaluate the ( innermost ) bracket's contents first, and then evaluate any 'outer' operands. Hence for these instances :

(A) e1 * e2 = e7 leaving the outer multiplication as e7 * e3 which evaluates to e4.

(B) e2 * e= -e6 leaving the outer multiplication as e1 * (-e6) = - (e1 * e6 ) = -e4

So by grouping e2 with a different partner leaves it being part of a product on a different line/circle. Basically the rule is that any product with three ( or more ) consecutive factors un-grouped is ambiguous. For example e1 * e2 * e3 * e4 * e5 * e6 has many possible interpretations when bracketed according to this rule. We are not used to this ( perverse ) behaviour when doing arithmetic on the real ( everyday ) numbers.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

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